Notes on Abstract Algebra

Daniel C. Bastosdbastos@math.utoledo.edu

Jan 10, 2006

Abstract

These are my notes on abstract algebra; if you see somethingwrong, let me know, please. If you found this document with thehelp of search engines and you’re studying about algebra, I suggestyou look for a book on the subject.

1 Groups

Definition 1.1 (binary operation). A binary operation * on a set Sis a function of S×S → S. That is, for each pair (a, b) ∈ S×S, thereis a single element s ∈ S such that a ∗ b = s. This implies S closedunder *.

Definition 1.2 (group). A group G is a nonempty set with a binaryoperation * satisfying: (i) * is associative, (ii) G has an identity e, (iii)for each g ∈ G, there is a g−1 such that gg−1 = g−1g = e.

Exercise 1.1. Are the even positive integers closed under addition?Yes. Proof. Because 2n + 2n = 4n is even and positive. Are theyclosed under multiplication? Yes. Proof. Because 2n ∗ 2n = 4n2 iseven and positive.

Exercise 1.2. For rectangles, there corresponds to each pair (lengthmeasurement, width measurement), an area measurement. Still, com-puting the areas is not, strictly speaking, a binary operation. Why?Because area is something different than length.

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Exercise 1.3. Are the odd positive integers closed under addition?Yes. Proof. Because 2n+1+2n+1 = 4n+1 is odd and positive. Undermultiplication? Yes. Proof. Because (2n + 1)(2n + 1) = 4n2 + 4n + 1is odd and positive.

Theorem 1.1. Let r be a nonzero integer. Let D be the set of alldivisors of r. The system of all fractions, in lowest terms, with theproperty that their denominators is di ∈ D, is closed under addition.

Proof. Let 0 6= r ∈ Z. Let D be the set of divisors of r. Let d1 =ar, d2 = br. Then,

p

d1+

q

d2=

ap + bq

lcd(d1, d2),

where lcd(d1, d2) ∈ D and p, q ∈ Z.

Exercise 1.4. Show that the system of fractions having denominatorsof 1, 2, or 4 when written in lowest terms is closed under addition.Proof. By Theorem 1.1, it is closed. Here, r = 4 and D = {1, 2, 4}which are all divisors of 4.

Theorem 1.2. The identity of a group is unique.

Proof. Let e and e′ be identities. So, ee′ = e′ because e is an identity.Also, ee′ = e because e′ is an identity as well. Since the binaryoperation of a group is a function, and assigns the operation to asingle element in the group, there can only be one identity and so,e = e′.

Exercise 1.5. Consider Z with the binary operation of subtraction.Is Z a group? If so, what is the identity? If not, what axioms do nothold? At first I thought that 0 could be the identity, because z−0 = z,but for 0 to be an identity, it must be true that z − 0 = 0 − z = z.That is not true because z − 0 = z 6= 0 − z = −z. Also, notice thatsubtraction is not associative: (a− b)− c 6= a− (b− c).

Theorem 1.3. Let G be finite with order 2n, n ∈ N. Then, there’sg ∈ G such that g2 = e.

Proof. Let S = {g1, ..., g2n}. Step one: remove from S all pairs(g−1

i , gi) such that gig−1i = g−1

i gi = e with gi 6= g−1i . This removes an

even number of elements. We’re left with an even number of elementsin S. After step one, e ∈ S because e−1 = e. Step two: remove e

2

from S. We’re left with an odd number of elements in S. So at leastone of these left will have to be its own inverse — since they come inpairs.

Theorem 1.4. Let G = {1, 2, ..., p − 1}, p prime. Let θ be multipli-cation mod p on G. Then (G, θ) is a group.

Proof. Unfortunately, I haven’t proved this entirely; but here’s whatI can say: because p is prime, we have no g1g2 ∈ G such that g1g2

divides p, so it never happens that giθgj = 0 mod p. Multiplicationis a binary operation, but we would need to show closure with modp. The identity is always present by hypothesis. Multiplication isassociative, but we would need an argument for inverses.

Theorem 1.5. The multiplication table of a group is a latin square;that is, each group element occurs once and only once in each row ofthe body of the table and in each column of the body of the table.

Proof. Consider the elements of the ith row. They are gig1, ..., gigj .The theorem claims that each element occurs once; so, suppose onedoes not show up once; that is, two elements of the row must beequal. In other words, suppose gig1 = gig2, but notice that g1 6= g2.Then, since we’re dealing with a group, g−1

i is present; so, g−1i (gig1) =

g−1i (gig2) =⇒ (g−1

i gi)g1 = (g−1i gi)g2 =⇒ eg1 = eg2 =⇒ g1 = g2.

So, each element occurs once.

If the group is finite, we’re done. If the group is not finite, we must stillshow that each element still occurs on a row. To prove for columns,we can repeat the proof replacing the appropriate subscripts.

Definition 1.3 (permutation). A permutation P of a set S is a one-to-one mapping of S into itself. In other words, P implies the existenceof a function f : S → S such that f is one-to-one and onto.

Exercise 1.6. Write the 4! = 24 possible arrangements of all fourletters A, E, P, T. Check which permutations are English words. Whatis the probability of making an English word from these four letters bychance arrangement? I’m not going to write them all here, but I willtell the algorithm I’d use. Fix A and permute EPT; then fix E andpermute APT, then fix P and permute AET; then fix T and permuteAPT. This will generate 3! + 3! + 3! + 3! = 24 permutations, and it’seasy to do it this way. To permute EPT, the same can be done: fixE, permute PT, fix P permute ET, fix T permute EP.

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It turns out, we find the English words PATE, PETA, PEAT, TAPE,TEPA. They mean, respectively: the human head, 1015, carbonizedvegetable matter, tape, a chemical compound: C6H12N2OP . So, wehave 5 chances out of 24 of making a word by chance arrangement;that’s a probability of 5/24 = 0.208 which is aproximatelly 20% ofchances of making a word out of random permutation; a pretty goodchance, if you will.

Exercise 1.7. Let Mr. 1, Mrs 2, Mr. 3 and Mrs 4 be seated on a table.How many seating arrangements are possible? 4! = 24. Write down allseating arrangements in which men and women are alternated. Thereare 8 possibilities because once you fix 1, there are two possibilitiesto follow: 2 and 4. Fix 2, there are two possiblities: 1 and 3. Fix 3,there are two possibilities; fix 4, there are two possibilities. What’sthe probabilitity that a random permutation of the four people willalternate men and women? That’d be 8/24 = 1/3 which means 33%.

Theorem 1.6. The permutations of a set S form a group under com-position.

Proof. Let P and Q be two permutations; then, for any s ∈ S, thereis a unique Q(s) ∈ S because for any s1, s2 ∈ S, if Q(s1) = Q(s2),then s1 = s2. The fact that Q(s) ∈ S comes from the fact that Q is afunction from S into S. Also, because P is one-to-one and because Qis onto, for all s ∈ S, there is a unique P (Q(s)) ∈ S.

Last paragraph implies P (Q(s)) is one-to-one. Now, to show compo-sition is a binary operation, we need to show that the composition ofP and Q is onto; because that would be showing that it is closed inS.

For all s ∈ S, there is q ∈ S such that q = Q(s). For all q ∈ S, thereis p ∈ S such that p = P (q). Then, for all s ∈ S, there is a q ∈ S suchthat q = P (Q(s)) = (P ◦Q)(s).

To show it forms a group, we state that composition is associative bysome theorem — which we haven’t proved yet. The identity wouldbe the permutation which maps each s ∈ S into itself. The inversesof each permutation is the same permutation with domain and rangeswaped.

Definition 1.4. If S is finite with n elements, then the set of allpermutations on S is called the symmetric group Sn on n elements.Sn has n! members.

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Exercise 1.8. Let

P =(

1 2 3 4 52 3 1 4 5

)

What is the image of 3 under the permutation? It’s 1. Which elementsare fixed under the permutation? 4 and 5. Which are moved? 1, 2, 3.Write the inverse permutation with respect to composition.

P−1 =(

2 3 1 4 51 2 3 4 5

).

Exercise 1.9. Let

Q =(

1 2 3 4 51 4 3 5 2

).

Find P ◦Q and Q ◦ P . Find Q−1.

P ◦Q =(

1 2 3 4 52 4 1 5 3

)

Q ◦ P =(

1 2 3 4 54 3 1 5 3

)

Q−1 =(

1 4 3 5 21 2 3 4 5

).

Exercise 1.10. Find Q−1 ◦ P−1. Find (P ◦Q) ◦ (Q−1 ◦ P−1).

Q−1 ◦ P−1 =(

2 3 1 4 51 5 3 2 4

)

(P ◦Q) ◦ (Q−1 ◦ P−1) =(

2 3 1 4 52 3 1 4 5

)

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Permutations can be written in cycle notation, which is easier to write.For the permutation P above, we would write (123)(4)(5) meaning 1goes to 2, 2 goes to 3, 3 goes to one and we close. We close because ithas cycled. The same happens to 4 and 5 because they didn’t moveat all; we may as well omit the ones which don’t move.

In general, start the first cycle with the first element of the permuta-tion; go on writing which follows which until the first element appears;then we close the cycle. If the first cycle has not exhausted all ele-ments, start a new cycle with an element that has been used. Repeatuntil all elements have been used.

If we omit the 1-cycles, then someone who only sees the permutation incycle notation would not know possible elements which were omited.But usually this is not a problem.

Cycle multiplication is cycle composition. You apply the right sidefirst, then the left. For example,

(1234)(2465) =(

2 4 6 5 3 11 6 5 3 4 2

),

because 2 goes to 4 which goes to 1, 4 goes to 6 which goes to 6 (sinceit does not appear), 6 goes to 5 which goes to 5 (since it does notappear), 5 goes to 2 which goes to 3, 1 goes to 1 which does to 2, 3goes to 3 which goes to 4.

Exercise 1.11. Q used previously in cycle notation is (245).

Exercise 1.12. Write the permutations of S3 as cycles. S3 is the sym-metric group of all permutations on n = 3 elements. Then, S3 has 3! =6 elements; they are: e, (12), (13), (23), (123), (132). It’s a bit toughto be sure that this is correct; but I built the table and it does work.For example, (12)−1 = (12), (13)−1 = (13), (23)−1, (23), (123)−1 =(132), (132)−1 = (123). It’s not commutative, though: (12)(13) =(132) 6= (123) = (13)(12).

Exercise 1.13. Write the following cycles in permutation form.

(134) =(

1 3 4 23 4 1 2

),

(12)(34) =(

3 4 1 24 3 2 1

),

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(1423) =(

1 4 2 34 2 3 1

),

(13)(12) =(

1 2 3 42 3 1 4

),

(14)(13)(12) =(

1 2 3 42 3 4 1

),

Definition 1.5. A transposition is a 2-cycle.

Theorem 1.7. Let G be a group in which every element g ∈ G hasthe property that g2 = e. Then G is abelian.

Proof. Let a, b ∈ G. Then: (ab)2 = e =⇒ a2b(ab) = a =⇒ b(ab) =a =⇒ b2(ab) = ba =⇒ ab = ba.

Definition 1.6 (group isomorphism). Let (G, !) and (H, ∗) be groups.We say G is isomorphic to H if and only if there exists a bijectivefunction f : G→ H such that for all u, v ∈ G, it’s true that f(u!v) =f(u) ∗ f(v). We write H ∼= G.

Theorem 1.8. If m ≤ n, there is a subgroup of Sn isomorphic to Sm.

Proof. Since m ≤ n, the set of all permutations of Sm is a subsetof Sn, and Sm is closed because it is a group, so it is a subgroup ofSn; then, one isomorphism exists between Sm and Sm: the identitymap.

Theorem 1.9. The identity of a subgroup H of G is necessarily theidentity of the group G.

Proof. Let H be a subgroup G; for all h ∈ H, eHh = heH = h. Sinceevery element of H is also in G, we have eGh = heG = h, for all h ∈ H.So, eGh = eHh =⇒ eG = eH by cancellation law.

Theorem 1.10. Inverses are unique.

Proof. Let h ∈ G. Now h−11 h = e and h−1

2 h = e =⇒ h−11 = h−1

2 .

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Exercise 1.14. Find a collection of eight permutations in S4 that isa subgroup isomorphic to the dihedral group D4 = {e, r, r2, r3, s, sr,sr2, sr3}. Give an explicit isomorphism from D4 to the subgroup ofS4.

Let P4 = {e, (1234), (13)(24), (1432), (12)(34), (14)(23), (13), (24)}.Map eP with e, (1234) with r, (13)(24) with r2, (1432) with r3,(12)(34) with s, (13) with sr, (14)(23) with sr2, (24) with sr3.

Theorem 1.11. GL(n,R) denotes the group of invertible n×n matri-ces with real entries. The collection H of matrices with determinants{±1} forms a subgroup of GLn(R).

Proof. We only need to show that H is closed; that is, det(AB) =±1 = det(A)det(B), for any A,B ∈ H. Let E be an elementarymatrix; det(E) = −1 whenever E comes from I by interchanging tworows; det(E) = 1 whenever E comes from I by adding a row to amultiple of another row of I; every invertible matrix can be expressedas a product of elementary matrices.

Let P ∈ H; express P as Ek...E1. For any Ei, with 1 ≤ i ≤ k,det(Ei) = ±1, then det(P ) = det(Ek...E1) = det(Ek)...det(E1) =±1. So, for two matrices P, Q ∈ H, det(PQ) = det(Ek...E1Q) =det(Ek)...det(E1)det(Q) = det(P )det(Q) = ±1. Then H is a subgroupof GL(n,R).

Okay, this proof is not correct because we must also show that thegroup is closed under inverses; the group is infinite. Also, det(AB) =det(A) det(B) for any square matrices A and B, so we need not useelementary matrices here.

On this next example, there’s a new notation there. When we writeσ(1) = 2, we mean that it’s a cycle in which 1 always goes to 2. Sothe set described includes all permutations of S6 in which 1 goes to 2.Examples: (12), (123), (1234), (12345), and so on.

Exercise 1.15. Let C = {σ ∈ S6 : σ(1) = 2}. Is C ≤ S6? No,because (1256)(1256) = (15...) 6= σ(1) = 2.

Definition 1.7. Let H ≤ G. Define as left cosets of H in G the setsgH = {gh1, gh2, ..., ghk, ...} where g ∈ G and hi ∈ H.

As you pick different elements g ∈ G, you create |G|/|H| cosets. Thisis called the index of H in G.

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Definition 1.8. Let H ≤ G. Generate all left cosets of H in G.Count the number of left cosets; call this number the index of H inG.

These cosets are mutually exclusive and exhaustive, like equivalenceclasses; in fact, each coset is an equivalence class. Cosets, however,can have many names; but since they’re equivalence classes, one wayto know if two names are actually the same coset is to check if theirg’s are left equivalent modulo H. See the next example.

Exercise 1.16. Let V = {e, (12)(34),(13)(24), (14)(23)}. Left cosets:(12)V = {(12), (34), (1324), (1423)}, (13)V = {(13), (1234), (24),(1432)}, (14)V = {(14), (1243), (1342), (23)}, (123)V = {(123), (134),(243), (142)}, (124)V = {(124), (143), (132), (234)}, (12)(34)V ={(12)(34), e, (14)(23), (13)(24)}. What are the right cosets?.

Let W = {e, (12), (34), (12)(34)}. Left cosets: (12)W = {(12), e,(34)(12), (34)}, (13)W = {(13), (123), (134), (1234)}, (14)W = {(14),(124), (143), (1243)}, (23)W = {(23), (132), (1342), (234)}, (24)W ={(24), (142), (1432), (243)}, (1324)W = {(1324), (14)(23), (13)(24),(1423)}. What are the right cosets?

Take a look at the first coset of V — (12)V . Now, notice that thecoset (1423)V is the same coset, though with a different name, as(12)V . Why? Because (1423) is left congruent to (12) modulo H.Why? Recal the definition of the left congruence. Two elements, iand j, of a group P are left congruent (mod Q ≤ P ) to each other ifthere’s q ∈ Q such that i = jq. So notice that the multiplication by qhappens on the right. In the example above, the subgroup is V , and(12) is left congruent to (1423) because (12) = (1423)[(14)(23)] and(14)(23) ∈ V .

Definition 1.9. Let H ≤ G. Define gi ≡L gj (mod H) if and onlyif there is a hk ∈ H such that gi = gjhk. Call this relation “leftcongruence.”

Theorem 1.12. Left congruence is an equivalence relation.

Definition 1.10. Let H ≤ G. Define gi ≡R gj (mod H) if and onlyif there is a hk ∈ H such that gi = hkgj . Call this relation “rightcongruence.”

Theorem 1.13. Right congruence is an equivalence relation.

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Proof. Let H be a subgroup of G. Suppose that for some g1, g2 ∈ G,there is h ∈ H such that g1 = hg2; that is, g1 ≡ g2 (mod H). Theng ≡ g because g = eHg (reflexive). Suppose g1 = hg2; then g2 =h−1g1 (symmetric). Suppose g1 = h1g2 and g2 = h2g3. Then g1 =(h1h2)g3 implying g1 ≡ g3 (mod H) (transitive). It is an equivalencerelation.

Theorem 1.14. In a group G, g, g−1 ∈ G have the same order.

Proof. Let gt = e. Then (g−1)tgt = (g−1)te which implies e = (g−1)t.Now we must show that t is the smallest such number in which(g−1)t = e. Suppose not; then, there is n < t such that (g−1)n

= e. Then, write (g−1)t as g−1 multiplied t times: (g−1)1...(g−1)t.Since n < t, we have that (g−1)1...(g−1)n = e. So, (g−1)1...(g−1)t =e(g−1)n+1...(g−1)t. Now what? Where is the contradiction?

Definition 1.11 (cyclic). Call a “cyclic group” the group generatedby some element g, and write 〈g〉. If 〈g〉 is finite, then call the “orderof g” the least positive element f for which gf = e, where e is theidentity of the group. Call an “infinite cyclic group” the cyclic groupin which gs = gt if and only if s = t.

Notice that the distinct elements of 〈g〉 are g, g2, ..., gf = e. Also,notice that the powers are not always positive. For example, 1 is agenerator for the integers under addition because 10 = 0, 11 = 1, 12

= 2, 13 = 3, and so on. But where are the inverses? Well, we do 1−1

= −1, 1−2 = −2, and so on.

Theorem 1.15. Let G be finite and cyclic. Then G is abelian.

Proof. Because G is finite, it has a finite order; call it n. Any elementof G is expressed gk, where 1 ≤ k ≤ n. Let gi, gj ∈ 〈g〉. Now gjgi =gj+i = gi+j = gigj .

Exercise 1.17. Write, in cycle form, the members of the cyclic groupof order 8 and compare with the octic group. Are both groups abelian?No; a cyclic group of order 8 is abelian, by the last theorem, butD4 is non-abelian. The cycles of the cyclic group of order 8 are e,(12345678), (1357)(2468), (14725836), (15)(26)(37)(48), (16385274),(1753)(2864), (18765432).

Theorem 1.16. Let G be finite with g ∈ G. Order of g divides |G|.

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Proof. The “order of g” is the natural number n such that gn = e;since G is finite, |G| and n are natural numbers.

Suppose f is a natural number such that gf = e; since n is the smallestsuch number, f ≥ n. By the division algorithm, f = nq + r, where q,r ∈ Z with r < n. So gf = gnq+r = (gn)qgr = eqgr = egr = gr = e.But r < n by the division algorithm, and since n is the smallest naturalsuch that gn = e, r must be zero, so n divides f .

The distinct powers of g form a subgroup of G, because they’re closedunder the binary operation. Lagrange’s theorem states that, here inparticular, |G| = |〈g〉| · |G : 〈g〉|, and |〈g〉| = n, so n divides |G| with|G : 〈g〉| as quotient.

Exercise 1.18. Write down all elements in D4, S4 and S5 which haveorder 2. In D4, with order 2, we have: (13), (24), (13)(24), (14)(23),(12)(34). In S4 we have (12), (13), (14), (23), (24), (34), (12)(34),(13)(24), (14)(23). In S5 we have (12), (13), (14), (15), (23), (24),(25), (34), (35), (45), (12)(34), (12)(35), (12)(45), (13)(24), (13)(25),(13)(45), (14)(23), (14)(25), (14)(35), (15)(23), (15)(24), (15)(34),(23)(45), (24)(35), (25)(34).

Exercise 1.19. We know that Sn can be generated by transpositions;that is, every σ ∈ Sn can be written as a product of transpositions.It is actually the case that Sn can be generated by adjacent trans-positions; that is, those of the form (i, i + 1). Prove this for n = 5by expressing each (a,b) ∈ S5 as a product of adjacent transpositions.Proof. (12) = (12), (13) = (12)(23)(12), (14) = (12)(23)(34)(23)(12),(15) = (12)(23)(34)(45)(34)(23)(12), (23) = (23), (24) = (23)(34)(23),(25) = (23)(34)(45)(34)(23), (34) = (34), (35) = (34)(45)(34), (45) =(45).

Exercise 1.20. Let H be a subgroup of a finite group G. If H isabelian, then H is normal in G. Show this is not true with an example.Consider the dihedral group with its subgroup V = {1, (12)(34)}. LetW = {e, (12), (34), (12)(34)} ≤ S4, which is abelian. Now the rightcoset W (1234) = {(1234), (234), ...} 6= (1234) W = (1234), (134),(123), (13...).

Exercise 1.21. Let K = {1, (12)(34), (14)(23), (13)(24)} ⊂ D4 = G.Show that V = {1, (12)(34)} is normal in K and that K is normal inG, but V is not normal in G. We list the cosets; the left cosets areeV = V and (14)(23)V = {(14)(23), (13)(24)} and the right cosets

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are V e = V and V = {(14)(23), (13)(24)}, so V normal in K. Nowthe left cosets of K in G are eK = K and (1234)K = {(1234), (13),(24), (1432)}; the right cosets of K in G are Ke = K and K(1234)= (1234)K because there are four elements left to be taken, so K isnormal in G. Now V is not normal in G because (1234)V = {(1234),(13)} 6= V (1234) = {(1234), (24)}.

Theorem 1.17. The intersection of two subgroups is a group.

Proof. Let H1,H2 ≤ G, then e ∈ H1 ∩H2 because e1,2 ∈ H1,2 is thee of G, so it’s the same identity. Let h ∈ H1 ∩ H2; because H1,2 isa group, h−1 ∈ H1,2. Associativity comes from the fact that G isassociative. Closure: let h, g ∈ H1 ∩ H2. Since h, g ∈ H1 and H1 isa group, it is closed, then hg ∈ H1. Similarly, hg ∈ H2. Therefore,hg ∈ H1 ∩H2.

Exercise 1.22. Let N be a normal subgroup of a group G. Prove thatif n is an element of N , then g−1ng is an element of N for each elementg ∈ G. We start by letting n ∈ N . Since N is normal subgroup ofG, we have G/N as the group of cosets of N in G. So, gN ∈ G/H;since n ∈ N , nN ∈ G/H and g−1N ∈ G/H. Since G/H is a group,g−1N ·nN ·gN = g−1ngN . Since n ∈ N , it follows that nN = eN = N .Then g−1NnNgN = g−1NeNgN = g−1egN = g−1gN = eN = N .Therefore, g−1ngN = N which implies g−1ng ∈ N .

Exercise 1.23. Show that the converse of Lagrange’s Theorem isfalse. For example, show that A4 is a group with 12 elements that con-tains no subgroup with 6 elements. We start by listing the twelve per-mutations of A4: e, (12)(34), (13)(24), (14)(23), (123), (132), (124),(142), (134), (143), (234), (243). There are 3 2-cycles, and 1 identitywhich amounts to 4 elements, so to get to 6, we will need 3-cycles.

Theorem 1.18. A4 has no subgroup with 6 elements.

Proof. Let H ≤ A4 such that |H| = 6. We want to show a contra-diction. We know that H P A4 because the index is 2. H containsa 3-cycle, for sure. Now, ghg−1 ∈ H for all g ∈ A4, for some h ∈ H.Let’s say that (123) ∈ H. Then (124)(123)(142) ∈ H which implies(243) ∈ H. Then (243) = g (123) g−1. Let’s say (234) ∈ H, then(234)(123)(243) = (134). Then (134) = g′(123)g′−1. If (123), (134),(243) ∈ H, then we can square each of them because H is a subgroup.We square the elements until we actually see that we don’t get only 6elements, which we assumed we had.

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Theorem 1.19. The intersection of two normal subgroups is normal.

Proof. Let subgroups H1,H2 P G. Then H1 ∩ H2 is a subgroup byTheorem 1.17; so must only show normality. Let g ∈ G; we must showg(H1∩H2) = (H1∩H2)g, which is true if ghg−1 ∈ (H1∩H2) for someh ∈ (H1 ∩H2) — see some example above. So, we have that gh = h′gfor some h′ ∈ (H1 ∩H2). We multiply g−1 on the right to get ghg−1

= h′e = h′ for some h′ ∈ (H1 ∩H2). So, ghg−1 ∈ (H1 ∩H2) and we’redone.

Definition 1.12. The center Z(G) of a group is the set of all elementsof G which commute with every element of G. That is, Z(G) = {z ∈ Gsuch that zg = gz for all g ∈ G}.

Definition 1.13. A subgroup H of G is a normal subgroup of G ifgH = Hg for all g ∈ G. We write H P G.

Theorem 1.20. Z(G) P G.

Proof. Associativity comes from the fact that G is a group. The iden-tity is in Z(G) because eg = ge = g for all g ∈ G. Let z ∈ Z(G). Now,since e ∈ Z(G) and e = zz−1, we have (zz−1)g = (z−1z)g =⇒ z(z−1g)= (gz)z−1. Since z ∈ Z(G), it commutes, so z(z−1g) = (zg)z−1. Bycancellation law and associativity, z−1g = gz−1, so z−1 ∈ Z(G). Now,normality comes from the fact that gz = zg for all g ∈ G, then gZ(G)= Z(G)g.

To understand normality, we must understand cosets well. Let’s thinkof cosets for a while. Consider a subgroup H ≤ G. What can we sayabout the cosets of H? We can say that |gH| = |H|; that is, thegroup formed by the cosets of H has the same order as H. In fact,such group has a name. Is it really a group? It is, as we prove on thenext theorem, and it also has a name.

Definition 1.14. The cosets of H in G is the quotient group G/H.

Theorem 1.21. The set of left cosets of H in G form a group underthe operation defined as (g1H)(g2H) = g1g2H whenever gH = Hg forall g ∈ G.

Proof. This theorem defines the binary operation in which the groupis formed upon. This operation can only be used whenever somecondition about the left and right cosets holds; that is, that the left

13

cosets match the right cosets exactly. In other words, H must benormal in G.

The definition of the binary operation gives us the fact that (g1g2H) isa coset. Do we have an identity? Yes, because g1HeH = g1eH = g1H.So, H is the identity. Is it associative? Since (g1g2)g3H = g1(g2g3)H,then (g1Hg2H)g3H = g1H(g2Hg3H). So yes, it is associative. Dowe have inverses? Yes, because (gH)−1 = g−1H; the proof of that is(gH)(gH)−1 = gg−1H = eH = H. So we have inverses, and we havea group.

Because cosets have many names, it would be nice to have theoremsthat would tell us when they’re equal. For instance, if I show youg1H and g2H, such that g1 6= g2, does that mean that the cosets aredifferent? Not necessarily. So, how can we tell?

One way, of course, is to take a look at the contents. Since cosetsare sets, if what they have as elements are all the same, in any order,they’re equal.

Theorem 1.22. Let H ≤ G; now g1H = g2H if and only if g−12 g1 ∈ H.

Proof. Because cosets are equivalence classes, g1H = g2H if g1 ≡L g2

(mod H) — we’re dealing with left cosets here. If this holds, then it’sbecause there is an h ∈ H such that g1 = g2h. Now, because G isa group, we have the inverse of g2 available; multiply it on the left:g−12 g1 = eh = h. Then, if g1H = g2H, then g−1

2 g1 ∈ H.

Now we must show that once g−12 g1 ∈ H, it follows that g1H = g2H.

Well, because g−12 g1 ∈ H, we can write g−1

2 g1 = h, for some h ∈ H.Then, multiply g2 on the right, so eg1 = g2h =⇒ g1 = g2h. Now, thismeans that there’s an h ∈ H, such that g1 = g2h. Therefore, g1 ≡L g2

(mod H). That is, g1H = g2H.

Theorem 1.23. Let H ≤ G; now g1H = g2H if and only if g−11 g2 ∈ H.

Proof. Repeat last proof changing subscripts appropriately.

So, these two theorems give us a nice way to check if cosets are equal.Next we prove a theorem about normal subgroups. It says that a nor-mal subgroup of a group is normal only when an element of the grouptimes an element of the subgroup times the inverse of the element ofthe group is in the subgroup. Verbal mathematics is very hard tounderstand, but it’s a good exercise.

Theorem 1.24. A subgroup H of G is normal in G if and only ifghg−1 is in H, for all g in G.

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Proof. Suppose H is normal in G. Let h in H and g in G be arbitraryelements. Then, gh = h′g for some h′ in H. Now multiply by g−1 onthe right to get ghg−1 = h′ which is in H.

By hypothesis, ghg−1 is in H, so let ghg−1 = h′ be in H. Multiply byg on the right, so gh = h′g, which means that gh is in Hg, and sinceh is an arbitrary element of H, gH is a subset of Hg.

Now since g−1 and g are inverses of each other, we can write thehypothesis as g−1hg = h′. Now multiply by g on the left to get hg =gh′ which means that hg is in gH and since h is an arbitrary elementof H, Hg is a subset of gH. So, gH = Hg, for all g in G.

Now, if you read this theorem carefully, you may notice that it givesyou a nice message. The message is, if G is abelian, then any subgroupH of G is normal. How does it say that? Well, suppose G is abelianand that H ≤ G. Then for any g ∈ G, gg−1 = e ∈ H, right? Now,multiply h on the right to get gg−1h = eh. But, since every elementof G commutes with every element of G, including those of H becauseH ≤ G, we can alter their order; so gg−1h = eh =⇒ ghg−1 = eh =h ∈ H and whenever this happens, H P G.

Now, how do we actually build a quotient group? Let’s look at anexample. Let G = (Z, +), H = nZ. Because addition is the binaryoperation for this group, mathematicians like to write the cosets inthe form g + H to make explicit that the operation is addition. If itwere any other operation, they’d probably write gH as we have beenso far.

Let’s try to build the quotient group Z/nZ, for n = 3, so we’ll buildZ/3Z. The elements of Z/3Z are {3Z, 1+3Z, 2+3Z}. If you’re askingwhy, consider the integers mod 3. By the division algorithm, z ÷ 3 =nq + r, where r < 3. So r is either 0, 1 or 2. That explains why themembers are as they are. Now, we set up the Cayley table.

θ 3Z 1 + 3Z 2 + 3Z3Z 3Z 1 + 3Z 2 + 3Z

1 + 3Z 1 + 3Z 2 + 3Z 3Z2 + 3Z 2 + 3Z 3Z 1 + 3Z

You may quickly notice that this group is abelian. How would onequickly notice such thing? Think of the group as a matrix A. Now,write AT and notice that they are equal. The fact that this group isabelian may get us a new theorem. Once a group G is abelian, is ittrue that G/H is also abelian? Yes; if G is abelian, every subgroup isnormal, and so G/H is also abelian.

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Theorem 1.25. Let H P G, with G abelian; then G/H is abelian.

Proof. Let g1H, g2H ∈ G/H. Now (g1H)(g2H) = (g1g2)H, right?That’s the definition for the multiplication (the binary operation) ofa quotient group. But, because G is abelian, every g commutes withevery element of G, so (g1g2)H = g2g1H = g2Hg1H and this showsthat G/H is abelian.

Once I say H P G, does it automatically follow that G is abelian? Itcan’t be, because let H = {e} which is normal in G and is a subgroupof any G non-abelian. But what if H is a proper normal subgroup ofG? Does that make G abelian automatically? If so, can you prove? Ifnot, can you show an example?

But now, once G is abelian, every subgroup of G is normal, and everysubgroup of G is abelian; and once a subgroup is normal, a quotientgroup with respect to the subgroup is also abelian. So, abelianism isa pretty powerful and nice property that a group may have.

Since we’re talking about abelianism, notice that when G is cyclic,then there’s no way that a subgroup won’t be abelian. Think aboutit; there’s just no way because all elements are expressed as a factorof the generator, so it doesn’t matter which order you multiply any ofthem.

Exercise 1.24. Let H P G. What is the order of gH ∈ G/H? Recallthat the order of gH is the smallest k such that (gH)k = H = gkHwhich happens when gk ∈ H. Then, the order of gH is the order ofH; that is, |gH| = |H|.

Theorem 1.26. Let [G : H] = 2. Then H P G.

Proof. What do we have to show here? We have to show that oncethe index of H is 2, the right cosets of H in G are the same as theleft cosets of H in G. Since the index of H in G is the number of leftcosets of H in G, there are 2 left cosets. Then, each gh is either inH or in G−H. Then we have two left cosets: g1H = H and g2H =G−H. Now, consider... I don’t clearly see this one right now.

Let’s talk about ways to produce groups quickly. Which groups dowe know? We know the number systems: (R, +), (R 6= 0, ·), (Z, +),(Q 6= 0, ·), (Q, +).

We know how to generate some groups, like the dihedral group D4 ={r,s : r4 = s2 = e, rs = sr3}. And we can always write the Cayley

16

table for any group, but it’s very hard to check associativity for crazygroups in which we may build a table for.

Associativity, then, makes is a hard enough thing to check so that wemay wish to build groups out of things which we already know willbe associtive. For example, any group under composition is associa-tive, so we like composition as a binary operation. So, a group canbe formed out of invertible functions from a set S into itself undercomposition; what kind of groups are those? These are the symmetricgroups Sn; they’re built out of the symmetry of objects.

For example, think of a rectangle. What can we do with a rectangle?We can flip it vertically and horizontally: we create two elements oforder 2. The identity is any of these two elements squared. We canrotate it 180 degrees; which if we square this element, we also getthe identity. This is the symmetries of a rectangle in which we onlyhave elements which square to the identity; so each element is its owninverse. This is the Klein 4-group.

Other ways to get away from checking associativity is to form groupsout of other groups. For example, let H ⊆ G. Then, by using the sameoperation of G, we may form a group (H, op(G)) in which associativitywill be given by G, at no charge — a pretty good deal. Not to mentionthat to check that H ≤ G we only need to check that for all h, h′ ∈ H,hh′ ∈ H and h−1 ∈ H.

We can use generators. Pick some g ∈ G, and take a look at allpowers of g. This algorithm generates the group 〈g〉 which is cyclicand, therefore, abelian, which makes every possible subgroup normal.

For example, let G = SL2(R). This is the special linear group ofmatrices 2× 2, under matrix multiplication as binary operation, withreal entries such that the determinant of each of these matrices is ±1.Let

g =(

1 10 1

),

and notice that 〈g〉 ∼= (Z, +).

Another interesting example is the center of every group. Every groupG has a center as a subgroup. Moreover, every center subgroup isabelian. Let Z(G) = {z ∈ G : zg = gz for all g ∈ G}, then Z(G) is anabelian subgroup of G. Next, we show an interesting theorem aboutcenters.

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Theorem 1.27. Suppose G/Z(G) is cyclic. Then G is abelian and soZ(G) = G.

Proof. Since G/Z(G) is cyclic, we can let gZ(G) generate G/Z(G).So G/Z(G) = {... g−1Z(G), Z(G), gZ(G), g2Z(G), ...}. This meansthat every element of G is in a coset gaZ(G). Now, pick g1, g2 ∈ G;these elements look like gaz and gbz′ in which each one is in some cosetof Z(G) in G, so z, z′ ∈ Z(G). Because they are in the center of Z,they both commute with every other element of G; therefore, gazgbz′

= gazz′gb = gaz′zgb = gaz′gbz which shows that every element ofG commutes with every other element of G, therefore G is abelian.Now Z(G) = G because once G is abelian, every element satisfy thecondition to belong in the center. In other words, abelianism impliesglobal centralization.

Before we move on to rings and other algebras, let’s take a look atthe quaternion group. The quaternion group is a non-abelian groupof order 8. It is usually denoted by Q8 = {1, −1, i, −i, j, −j, k, −k}.The quaternions are actually a number system, similarly constructedto the complex numbers, but the quaternions don’t commute, whichis unsual for number systems.

An interesting thing about the quaternion group is that once we saythat i2 = j2 = k2 = ijk = −1, you can get the whole group. There’sanother way to name the elements of the group, in which I happen toprefer right now.

Let Q8 = {1, a, a2, a3, b, ab, a2b, a3b}. Let a4 = e, b2 = a2, and sob4 = e. Then a−1 = a3 and it follows that ba = a−1b. From this, wecan draw the whole multiplication table.

θ 1 a a2 a3 b ab a2b a3b

1 1 a a2 a3 b ab a2b a3b

a a a2 a3 1 ab a2b a3b b

a2 a2 a3 1 a a2b a3b b ab

a3 a3 1 a a2 a3b b ab a2b

b b a3b a2b ab a2 a 1 a3

ab ab a2b a3b b a3 a2 a 1a2b a2b a3b b ab 1 a3 a2 a

a3b a3b b ab a2b a 1 a3 a2

They don’t commute because, for example, ab 6= ba = a3b. However,notice that all of its proper subgroups are abelian, because they’re allcyclic. They are: {1}, 〈a〉, 〈a2〉, 〈b〉. So non-abelianism doesn’t implynon-abelian subgroups.

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With a quick glance at the table and at the nature of the elements ofthe quaternion, we could think that they have something to do withthe dihedral D4 group, but they are not that similar. The quaterniongroup has two elements which square to the identity, while D4 has atleast five, so D4 is not isomorphic to Q8.

Exercise 1.25. Let G abelian, let H = {g ∈ G : g2 = e}. ProveH ≤ G. Give an example to show G abelian is necessary for suchH. Proof. We have associativity from G; e2 = e, so we have identity.We have closure on H because: since G is abelian, and H is subset ofG, all elements of H also commute, then e2 = h2h′2 = (hh′)(hh′) =(hh′)2 ∈ H. We also have inverses, because for each h ∈ H, h−1 =h ∈ H. Then H ≤ G. For the example, we consider S4 which is notabelian. Both (12) and (14) square to the identity. But (12)(14) =(142), but (142)2 = (124) 6= e, so we don’t have closure. So abelianismis necessary.

Exercise 1.26. Let G be a group and let a be one fixed elementof G. Show that Ha = {x ∈ G : xa = ax} is a subgroup of G.Proof. Associativity is given by G. The identity is present becauseit commutes with every element of G, so it will commute with a justfine. For inverses now, let x ∈ H, so ax = xa, then x−1ax = x−1xa=⇒ x−1axx−1 = x−1xax−1 =⇒ x−1ae = eax−1 =⇒ x−1a = ax−1,so x−1 ∈ H. Now we need to show closure; let x, y ∈ Ha. We mustshow xy ∈ Ha; xa = ax =⇒ (yx)a = yax =⇒ yax = (yx)a =⇒a(yx) = (yx)a.

Exercise 1.27. Show that Sn is non-abelian for any n ≥ 3. Proof.Any Sn, where n ≥ 3, has (12) and (23) as elements; but since (12)(23)= (123) 6= (132) = (23)(12), then no Sn, with n ≥ 3, is abelian.

Exercise 1.28. Give an example of a group G which has no elementsof finite order greater than 1, but has a factor group G/H, all of whoseelements have finite order. Let (Z,+) be the group which none of itselements have finite order; that’s possible because there’s no powert > 1 such that zt = 1. However, Z/2Z is such that all of its cosetshave order not greater than 2.

Exercise 1.29. Let H ≤ G and fix g ∈ G. Define gHg−1 = {ghg−1

: h ∈ H}. Prove (a) that gHg−1 ≤ G. It is called a conjugate of H.Suppose G is finite; then prove (b) that |gHg−1| = |H|. Now suppose

19

G is a group and H is the unique subgroup of size |H| in G; then,prove (c) that H P G.

For (a), we show first the identity is present. Since e ∈ H, then geg−1

= gg−1 = e ∈ gHg−1. Associativity comes from G because gHg−1

is composed only of elements of G. We construct the inverses thisway: e = gg−1 = geg−1 = ghh−1g−1 = gheh−1g−1 = ghg−1gh−1g−1

= (ghg−1)(gh−1g−1) = e. And e = gg−1 = geg−1 = gh−1hg−1 =gh−1ehg−1 = gh−1g−1ghg−1 = (gh−1g−1)(ghg−1). So we have in-verses; therefore, gHg−1 ≤ G.

For (b), let H = {h1, h2, ..., hk} with H ≤ G finite. Then

gHg−1 = {gh1g1 , ..., ghkg−1}.

Now, if I can show that all elements in gHg−1 are distinct, then I’veshown that there’s one and only one ghg−1 for each h ∈ H. Therefore,they have the same number of elements, and therefore, the same order.So, suppose gh′g−1 = gh′′g−1. By cancellation law, h′ = h′′ and thenthey’re all distinct, so they have the same order.

For (c), we have |gHg−1| = |H| ≤ G, by (b) and (a) respectively.The right coset (gHg−1)g = {gh1g

−1g, ..., ghkg−1g} = {gh1, ..., ghk}

= gH. Now, since H is the unique subgroup with |H| elements, thengHg−1 = H, and so gH = Hg — because Hg = (gHg−1)g — implyingH P G. However, (b) assumes G is finite, but I think that gHg−1 willalways have the same elements even when G is not finite; so I thinkthat G-finite in (b) is not a requirement. There’s a not so hard way toshow the same cardinality here, so that we may drop the requirementthat G-finite.

2 Homomorphisms

Definition 2.1. A group homomorphism from (G, ×) to (H, ·) is amapping φ that maps every g ∈ G to some h ∈ H so that the image ofan ×-product is the corresponding (·)-product in H; that is φ(g1×g2)= φ(g1) · φ(g2).

Definition 2.2. Let φ : G→ H be a homomorphism. The set φ(G) ⊆H of all images h ∈ H of elements g ∈ G is called the homomorphicimage of G. The various elements g sharing the common image φ(g)= h form the pre-image of h. Also, we say φ is a homomorphism ontoH if the mapping is onto; that is, if φ(G) = H, so that every element

20

of H appears as the image of at least one g ∈ G. If φ is onto andone-to-one, then φ is an isomorphism.

Definition 2.3. Let φ : G → G′ be a homomorphism. Let H ≤ G.Then we call the set φ[H] = {φ(h) : h ∈ H} the homomorphic imageof H on G′.

Definition 2.4. Let φ : G → G′ be a homomorphism. The kernel ofφ is ker(φ) = {g ∈ G : f(g) = e} = f−1[e].

Theorem 2.1. φ[H] ≤ G′.

Definition 2.5. Let φ : G → G′ be a homomorphism. Let H ′ ≤ G′.Then we call the set φ−1[H ′] = {g ∈ G : φ(g) ∈ H ′} the pre-image ofH ′.

Definition 2.6. Let φ be a homomorphism from a group G to agroup H. The group K of pre-images of eH is called the kernel of thehomomorphism.

Exercise 2.1. The integers can be mapped onto the integers Zn ={0, 1, 2, ..., n − 1}, and this mapping is a homomorphism for theoperation +. Is it also a homomorphism for ×? Let f(i + rn) = i(mod n), and show that f((i + rn)(j + sn)) = ij (mod n).

Proof. We let φ(p = i + rn) = i (mod n), φ(q = i + sn) = j (mod n).Is φ(pq) = ij (mod n)? φ((i+ rn)(j + sn)) = φ(ij + isn+ jrn+ rsn2)= φ(ij + (is + jr + rsn)n) = ij (mod n). So it is a homormorphismfor × as well.

Exercise 2.2. Let φ : G → H be a group homomorphism. Provethat φ[G] is abelian if and only if for all x, y ∈ G we have xyx−1y−1 ∈ker(φ).

Proof. Suppose xyx−1y−1 ∈ ker(φ) for all x, y ∈ G. Let a, b in φ[G],so we have x, y ∈ G such that a = φ(x), b = φ(y). We use the φ’sproperties, such as φ(x−1) = φ(x)−1 to write

φ(xy)φ(x−1y−1) = e

φ(x)φ(y)φ(x−1)φ(y−1) = e

φ(x)φ(y)φ(x)−1φ(y)−1 = e

φ(x)φ(y)φ(x)−1 = φ(y)φ(x)φ(y) = φ(y)φ(x)

ab = ba

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Therefore, φ[G] is abelian. Now suppose φ[G] is abelian. Let x, y ∈ G.Must show φ(xyx−1y−1) = e. So

φ(x)φ(y) = φ(y)φ(x)

φ(x)φ(y)φ(x)−1 = φ(y)

φ(x)φ(y)φ(x)−1φ(y)−1 = e

φ(x)φ(y)φ(x−1)φ(y−1) = e

φ(xyx−1y−1) = e

Therefore xyx−1y−1 ∈ ker(φ) for all x, y ∈ G.

Exercise 2.3. Let f : G→ H be a group homomorphism. Prove thatf is one-to-one if and only if ker(f) = {e}.Proof. Assume f is one-to-one. Let x ∈ ker(f). Then f(x) = e =f(e), and since f is one-to-one, we have that x = e, so ker(f) = {e}.Now assume ker(f) = {e}. Let x, y ∈ G. Suppose f(x) = f(y). Wenow use the fact that f is a homomorphism, to write

f(x) = f(y)

f(x)f(y)−1 = e

f(x)f(y−1) = e

f(xy−1) = e

So xy−1 ∈ ker(f). Therefore, x = y.

Exercise 2.4. Let G be a group and fix g ∈ G. Define a map φ :G→ G by

φ(x) = gxg−1.

Show that φ is (a) one-to-one and (b) onto. Prove φ is a (c) homo-morphism.

For (a), let x, y ∈ G. Suppose φ(x) = φ(y). So φ(x) = gxg−1 = φ(y)= gyg−1. Apply cancellation law on the left and on the right; then x= y. So φ is one-to-one.

For (b) let y ∈ G. We must show that there is an x ∈ G such thaty = φ(x) = gxg−1. Multiply by g on the right, and by g−1 on the left,so g−1yg = x; so, φ is onto.

For (c), let x, y ∈ G. Then φ(x) = gxg−1 and φ(y) = gyg−1. So

φ(x)φ(y) = gxg−1gyg−1

= gxyg−1

= φ(xy)

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So φ is a homomorphism.

Exercise 2.5. Let φ : G→ H be a homomorphism and K = ker(φ),so K P G. Let g ∈ G. Prove gK = φ−1[φ(g)]. That is, we are provingthat the elements of the coset gK are exactly the elements of G thatmap to φ(g) ∈ {φ(g)}.Proof. Let φ : G→ H be a (group) homomorphism, K = ker(φ). LetK P G. Let g ∈ G. Let gk ∈ gK for some k ∈ K. Must show gk ∈φ−1[φ(g)]. Since G is a group, gk ∈ G; since φ is a homomorphism,

φ(gk) = φ(g)φ(k) = φ(g) ∈ φ[G].

Since φ(gk) ∈ φ[G], gk ∈ φ−1[φ(g)].

Now, since f is a homomorphism,

f(g−1u) = f(g−1)f(u)

= f(g)−1f(u)

But f(u) = f(g) because u ∈ f−1[{f(g)}]. So

f(g)−1f(u) = f(g)−1f(g)= e

So g−1u is mapped to e, which means g−1u ∈ K. So g−1u = k forsome k ∈ K. Multiply g on the right, so u = gk; so u ∈ gK.

Exercise 2.6. Let H P G. Define φ : G → G/H by φ(g) = gH.Show that φ is onto and the kernel of φ is H.

Proof. As defined, φ takes each g ∈ G into a coset gH ∈ G/H; sinceH is normal, G/H’s operation is defined as (g1H)(g2H) = (g1g2)H.

By definition, φ(g) = gH, so

φ(g1g2) = g1g2H

= g1Hg2H

= φ(g1)φ(g2)

Therefore, φ is a homomorphism.

The kernel of φ will be all elements which land in the identity of G/H;that is, they land in H. Which elements g will satisfy φ(g) = H? Bydefinition, φ(g) = gH which is equal to H if and only if (g−1e) ∈ H.

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Let h ∈ H. So φ(h) = hH = H because h−1 ∈ H. Therefore,H = ker(φ).

To show φ is onto, let gH ∈ G/H. I must find g′ ∈ G such thatφ(g′) = gH; because g ∈ G, we certainly have one; let g = g′, so φ isonto.

Definition 2.7. A group G 6= {e} is simple if and only if for everyN P G, we have N = G or N = {e}.

Exercise 2.7. Prove that a finite simple group G which is abelianmust be a cyclic group of prime order.

Proof. By definition, G 6= {e}, so pick g 6= e; we know 〈g〉 ≤ G.Suppose 〈g〉 6= G; since G is abelian, 〈g〉 P G, so 〈g〉 is normal andproper in G, but G is simple, so that’s not possible. So, 〈g〉 = G.

By Theorem 5-3, if some prime p divides |G|, then G has a subgroupH of order p. Since G is abelian, H must be normal. Since G issimple, it’s not true that H is proper, so |G| = p.

Theorem 2.2. Suppose H is a subgroup of G and N is normal in G.Define HN a subset of G by HN = {hn : h ∈ H, n ∈ N}. Then HNis a subgroup of G.

Proof. We know e is in both H and N , so e2 = e ∈ HN ; hence HNis not empty.

For closure, let a, b be in HN . Then there are h, h′ ∈ H and n, n′ ∈ Nsuch that a = hn and b = h′n′, so ab = hnh′n′. Now, since Nis normal, hN = Nh for all h ∈ G, so nh′ ∈ Nh′ = h′N ; hence,nh′ = h′n∗ for some n∗ ∈ N , so ab = hh′n∗n′. Since hh′ ∈ H andn∗n′ ∈ N , ab ∈ HN .

For inverses, let a be in HN , so a = hn for some h ∈ H and n ∈N . We know a−1 = n−1h−1, so we multiply h on the left to getha−1 = hn−1h−1 which is in N because N is normal. Since ha−1 ∈ N ,h−1(ha−1) ∈ HN , so a−1 ∈ HN ; hence, HN is a subgroup of G.

Corollary 2.3. If H P G, then HN P G.

Proof. Let x in HN be arbitrary; then x = uv for some u in H and vin N . Let g in G be arbitrary; we want to show that gxg−1 is in HN .Now gxg−1 = guvg−1 = gug−1gvg−1. But H and N are normal in G,so gug−1 is in H and gvg−1 is in N , so gxg−1 is in HN , as desired.

24

Exercise 2.8. Let G be a group. Define ∼ on G by x ∼ y if thereexists g ∈ G such that

gxg−1 = y.

Prove (a) ∼ is an equivalence relation,(b) describe the conjugacyclasses of G if G is abelian, (c) write down the conjugacy classes ofS3.

Proof. Let x ∈ G. Then x ∼ x because xxx−1 = x. Suppose x ∼ y.So exists g such that

gxg−1 = y

xg−1 = g−1y

x = g−1y(g−1)−1.

Suppose x ∼ y, y ∼ z. So there exists g, h such that

gxg−1 = y

hyh−1 = z

So h(gxg−1)h−1 = z. Let g′ = hg, and notice g′−1 = g−1h−1. Sog′xg′−1 = z. So x ∼ z.

(b) If G is abelian, then the only y which will satisfy the relation isy = x, but since ∼ is an equivalence relation, ∼ breaks G apart intodisjoint and exhaustive classes, so G has |G| conjugacy classes of theform {x}.(c) S3 = {e, (12), (23), (13), (123), (132)}. A 3-cycle has a 2-cycleas a conjugate because 3-cycles are even permutations, 2-cycles areodd, so even-cycle times odd-cycle is odd-cycle. Therefore, one classC1 will contain (123), (132). Now, by inspection I find (12)(13)(12)= (12)(123) = (23), so (13) ∼ (23). Also, (23)(12)(23) = (23)(123) =(13), so (12) ∼ (13). So, another class C2 will contain (12), (13), (23).

3 Polynomials

We need definitions, notation explained, the theorems, the proofs; theexercises are already here.

Exercise 3.1. Find all primes p such that x + 2 is a factor of

x4 + x3 + x2 − x + 1 ∈ Zp[x].

Solution. If x + 2 is a factor of x4 + x3 + x2 − x + 1 ∈ Zp[x], then it’sbecause −2 is a root. By evaluating the polynomial at −2, we get 15

25

as the result. Now, 15 = 0 (mod 3) and 15 = 0 (mod 5). Hence wehave p in {3, 5}.

Exercise 3.2. Find all irreducible polynomials of degree 3 in Z3[x].Solution. We can express all polynomials of degree 3 over Z3 by

ax3 + bx2 + cx + d.

The constant term d cannot be zero, or zero is a root. The coefficientscannot sum to zero, or one is a root. Now, to derive a third rule,evaluate x = 2 = −1: −a+ b+−c+d = 0 which implies b+d = a+ c.That is, whenever the sum of the even coefficients equals the sum ofodd coefficients, we have 2 = −1 as a root. Since the degree must be3, then a cannot be 0, so a is either 1 or 2.

Since factorization is unique except for scalar multiple (and order),consider all monic polynomials and multiply each by 2 to notice thatfactorization doesn’t change. So we may consider only the monic poly-nomials; in case we want to distinguish them, we may multiply eachirreducible by 2 to obtain its match, doubling the total of irreduciblepolynomials.

So, with these criteria, we have 2 choices for d, 3 choices for c, 3 choicesfor b, with 1 choice for a. This gives us 18 polynomials p(x). Supposenow that p(x) is reducible; then its factors will be of degree 2 anddegree 1; that is,

p(x) = g(x)h(x) = (x2 + Ax + B)(x + C).

Since d isn’t zero, B and C are not zero. So, C = 1 or C = 2, B = 1 orB = 2. This gives us 6 possibilities for g(x) and 2 possibilities for h(x);that is, 12 factorizations, which are reducible. We take advantage ofthe list to add the irreducibles in the order that they should appear.We list only the coefficients of each polynomial in the order x3, x2,x1, x0.

1 0 0 11 0 0 21 0 1 11 0 1 21 0 2 1 irreducible1 0 2 2 irreducible

1 1 0 11 1 0 2 irreducible

26

1 1 1 11 1 1 2 irreducible1 1 2 1 irreducible1 1 2 2

1 2 0 1 irreducible1 2 0 21 2 1 1 irreducible1 2 1 21 2 2 11 2 2 2 irreducible

We found 8 polynomials. If we multiply each by 2, we would get theirmatches, yielding 16 irreducible polynomials as shown below.

2 0 0 12 0 0 22 0 1 1 irreducible2 0 1 2 irreducible2 0 2 12 0 2 2

2 1 0 12 1 0 2 irreducible2 1 1 1 irreducible2 1 1 22 1 2 12 1 2 2 irreducible

2 2 0 1 irreducible2 2 0 22 2 1 12 2 1 2 irreducible2 2 2 1 irreducible2 2 2 2

Exercise 3.3. Demonstrate that f(x) = x4 − 22x2 + 1 is irreducibleover Q. Solution. By the quadratic formula,

x2 = 11±√

222 − 4/2

= 11± 2√

30

27

So x = ±√

11± 2√

30.

Now we can write f(x) as the product of 4 linear factors in R[x]. Thelinear factors are irreducible polynomials in R[x]; so, the factorizationof f(x) is unique — except for order and scalar multiples. Therefore,if f(x) were to factor in Q[x], we would see rational roots in theselinear factors, which we don’t.

Because the irrationals are not closed under multiplication, we mustalso make sure that no factors of degree 2 will have rational coefficients.We take the linear factors and multiply them in all possible ways.

Let

A = x−√

11 + 2√

30

B = x +√

11 + 2√

30

C = x−√

11− 2√

30

D = x +√

11− 2√

30.

So

AB = x2 − 11− 2√

30

AC = x2 + (−√

11 + 2√

30−√

11− 2√

30)x + 1

AD = x2 + (−√

11 + 2√

30 +√

11− 2√

30)x + 1

BC = x2 + (−√

11− 2√

30 +√

11 + 2√

30)x + 1

BD = x2 + (√

11− 2√

30 +√

11 + 2√

30)x + 1

CD = x2 + 241− 44√

30

No factor with rational coefficients; so f(x) is irreducible in Q[x].

Exercise 3.4. Let φ : G → H be a group homomorphism which isonto. Suppose G has 60 elements and H has 4 elements. Find |ker(φ)|.Let h be in H. How many elements of G map to h?

Solution. By the fundamental theorem of homomorphisms, G/ker(φ)∼= Image(φ). Since φ is onto, H ∼= Image(φ), so G/ker(φ) ∼= H.Since |H| = 4, 60/|ker(φ)| = 4; hence, |ker(φ)| = 15. Let h be in H;so |φ−1[h]| = 15.

28

Exercise 3.5. Show that α =√

2+√

3 is algebraic over Q by findingf(x) in Q[x] such that f(α) = 0. Repeat for α =

√3√

2− i.

Solution.

α2 = 5 + 2√

6

α2 − 5 = 2√

6

α4 − 10α2 + 25 = 24

α4 − 10α2 + 1 = 0

So,√

2 +√

3 is algebraic over Q; so it’s an algebraic number.

For the other, we write

α2 = 3√

2− i

α2 − i = 3√

2

(α2 − i)3 = 2

(α4 − 2iα2 − 1)(α2 − i) = 2

α6 − iα4 − 2iα4 − 2α2 − α2 + i = 2

α6 − 3iα4 − 3α2 + i = 2

α6 − 3α2 − 2 = 3iα4 − i

(α6 − 3α2 − 2)2 = −9α8 + 6α4 + 1

(α6 − 3α2 − 2)2 + 9α8 − 6α4 − 1 = 0

Now the coefficients are rational. Since α is a root of the polynomialabove, and α is in C and C is an extension field ofQ, then α is algebraicover Q, so α is algebraic number.

Exercise 3.6. The complex numbers s + ti and s − ti with s, t in Rare often called conjugates. Show that they satisfy the definition ofconjugates. What happens if t = 0?

Solution. If the coefficients of the polynomial

p(x) = (x− s + ti)(x− s− ti),

are real numbers, then we can pick R = F and C = E to satisfy thedefinition, because C is an extension field of R. Also, p(x) would beirreducible over R because all of its roots are complex numbers. Since

p(x) = x2 − 2sx + (s2 + t2),

29

we have p(x) in R[x] satisfying the definition. If t = 0, then the dis-criminant found in the application of the quadratic formula is greaterthan zero which implies that all roots are real which causes p(x) to bereducible in R failing the definition for the chosen fields here. If wewant to, however, we can pick other fields to satisfy the definition forwhen t = 0.

Exercise 3.7. Let F be the field of rational numbers and let p(x) =x2 +x/3−1. Show that x3 +1, −1/3x2 +x+1 and 10x/9+2/3 are inthe same equivalence class. Show that x4 +x2 +1 and −28x/27+28/9are in the same equivalence class.

Solution. We have x3 +1 = 10x/9+2/3 and −x2/3+x+1 = 10x/9+2/3. This shows they’re all in the same equivalence class. For theother class, we have x4 + x2 + 1 = −28x/27 + 28/9.

Exercise 3.8. Still using Q/〈p(x) = x2 + x/3 − 1〉, find linear poly-nomials equivalent to x2, x3, x4. Use the results to find a linearpolynomial equivalent to 3x4 + x3 − 2/3x2 + x + 1.

Solution. For the first part, we do long division on each. So

x2 = −x/3 + 1

x3 = 10x/9− 1/3

x4 = −19x/27 + 10/9.

For the second part, we use the equivalences above to perform theappropriate substituitions for 3x4 + x3 − 2/3x2 + x + 1. That is, wecan write it as

3(−19x/27) + 10/9 + (10x/9)− 1/3 + (−2/3)− (x/3) + 1 + x + 1,

which is equivalent to (46 + 16x)/9.

Exercise 3.9. What is the multiplicative identity of F [x]/〈p(x)〉?Solution. The multiplicative inverse e(x) is an element of the fieldsuch that k(x)e(x) = e(x)k(x) = k(x). If we take any member of k(x)and multiply it by a constant term, we get a member of k(x), so allconstants are members of e(x).

Exercise 3.10. Let q(x) = x + 1/2; check if q(x)−1

= −4/5x + 2/5is the inverse of q(x) in the field Q[x]/〈x2 + 1〉.

30

Solution. To check, we verify that q(x) q(x)−1

= e(x) and we dividee(x) by x2 + 1. If we get a constant polynomial, it worked. It doesbecause

q(x) q(x)−1

= 1/5− 4x2/5 = 1.

But it could’ve been any constant other than 1, since all constants aremembers of the class 1.

Exercise 3.11. Find the inverse of x in Q[x]/〈x2 + 1〉.Solution. What polynomial that when multiplied by x will yield aconstant mod x2 + 1? The polynomial x itself, because (x)(x) = x2

= −1 mod x2 + 1. So x−1 = x.

Exercise 3.12. Divide θ2 + 1 by tθ + s, express their gcd linearly,find the inverse of s+ ti, and compare with the preceding result of thebook.

Solution. The division algorithm yields

θ2 + 1 = (tθ + s)(θ/t− s/t2) + (1 + s2/t2).

Since (1 + s2/t2) is a constant, it is a gcd(θ2 + 1, tθ + s). We expressit linearly now with

(1 +

s2

t2

)= θ2 + 1−

(tθ + s

)(θ

t− s

t2

)

= θ2 + 1−(

θ

t− s

t2

)(tθ + s

)

We want to express 1 as a combination of θ2 + 1 and tθ + s. So wemultiply the equation by 1/(1 + s2/t2), which yields

1 =[θ2 + 1−

(θ

t− s

t2

)(tθ + s

)]1

1 + s2/t2

=[

11 + s2/t2

(θ2 + 1

)− 1

1 + s2/t2

(θ

t− s

t2

)(tθ + s

)]

=[

t2

t2 + s2

(θ2 + 1

)−

(tθ − s

t2 + s2

)(tθ + s

)]

So

1 + 〈θ2 + 1〉 = − tθ − s

t2 + s2

(tθ + s

)+ 〈θ2 + 1〉.

The inverse of s + ti iss− ti

s2 + t2.

31

The comparison we can make here is that

s− ti

s2 + t2

seems to be the same as

− tθ − s

t2 + s2=

s− tθ

t2 + s2,

where θ plays the role of i.

Exercise 3.13 (7-12).

Exercise 3.14 (7-14).

Exercise 3.15. Find the minimum polynomials over the small fieldof the following elements in the following extensions: (a) i in C : Q,(b) 3√

7 in R : Q, (c) (1 + 5√

)/2 in C : Q, (d)√

5 +√

3 in C : Q.

Solution. We set x = r, where r is the desired root, then we workwith the equation until we get the coefficients over the desired field.For (a), we have x2 +1, for (b) x3−7, (c) x2−x−1, (d) x4−16x2 +4.

Exercise 3.16. Find an irreducible polynomial p(x) in Z3[x] of degree2. Explain why it is irreducible. Contruct the field E = Z3[x]/〈p(x)〉.Factor p(x) into two linear terms in E[x].

Solution. For a prime p, Zp is a field, so it has no zero divisors andit’s closed under the operations, so a polynomial is reducible if andonly if it has a root. In Z3, x2 + 1 has no roots.

Let the roots be caled T and −T = 2T . We attach the roots to Z3,and we compute what other elements will be appear in the field bycomputing the addition and multiplication tables.

+ | 0 1 2 T 2T 1+T 1+2T 2+T 2+2T-----------------------------------------------------------0 | 0 1 2 T 2T 1+T 1+2T 2+T 2+2T1 | 1 2 0 1+T 1+2T 2+T 2+2T T 2T2 | 2 0 1 2+T 2+2T T 2T 1+T 1+2TT | T T+1 T+2 2T 0 1+2T 1 2+2T T2T | 2T 1+2T 2+2T 0 T 1 1+T 2 2+T1+T |1+T 2+T T 1+2T 1 2+2T 2 2T 01+2T |1+2T 2+2T 2T 1 1+T 2 2+T 0 T

32

2+T |2+T T 1+T 2+2T 2 2+2T 0 1+2T 12+2T |2+2T 2T 1+2T 2 2+T 0 T 1 1+T

* | 0 1 2 T 2T 1+T 1+2T 2+T 2+2T-----------------------------------------------------------0 | 0 0 0 0 0 0 0 0 01 | 0 1 2 T 2T 1+T 1+2T 2+T 2+2T2 | 0 2 1 2T T 2+2T 2+T 1+2T 1+TT | 0 T 2T 2 1 2+T 1+T 2+2T 1+2T2T | 0 2T T 1 2 1+2T 2+2T 1+T 2+T1+T | 0 1+T 2+2T 2+T 1+2T 2T 2 1 T1+2T | 0 1+2T 2+T 1+T 2+2T 2 T 2T 12+T | 0 2+T 1+2T 2+2T 1+T 1 2T T 22+2T | 0 2+2T 1+T 1+2T ---- T 1 2 2T

We can now factor every polynomial in Z3(T, 2T ) all the way to linearfactors. The roots of x2 + 1 are T and 2T , so

x2 + 1 = (x + T )(x + 2T ).

4 Algebraic field extensions

Definition 4.1. If every element of a field F is an element of a fieldE and if F has the same field operations as E, then F is a subfield ofE and E is an extension of F . We write F ⊆ E.

Definition 4.2. Let a0 + a1x+ a2x2 + ...+ anxn = 0 be a polynomial

equation of degree n with coefficients ai in a field F , and let E be anextension field of F . If r in E is a root of such an equation, then r issaid to be algebraic over F . If F is Q, then r is an algebraic number.

Definition 4.3. A real number that is not the root of any polynomialequation with rational coefficients is called transcendental.

Knowing that π is transcendental from hearing people say it, makesone wonder if we can show that it is not transcendental. All we haveto do is find a polynomial with rational coefficients which has π as aroot. We can’t consider x− π = 0 because we can’t assume that π isrational. A rational number times an irrational number is irrational,so we will need at least a polynomial of degree 2.

33

Why? Because the idea is: if π isn’t transcendental, then eventuallywe would find some natural n such that πn = r, where r is a rationalnumber. Consider the irrational

√2. If we square it, we get 2; so this

way we can find a polynomial to which√

2 is a root. For example,x2 − 2 = 0. So, if we can apply some power to π and get it tobecome a rational number, then we would be done. If I believe themathematicians, then I have to say that this is not possible and if not,then this is another way to think of transcendental numbers: numberswhich won’t square to a rational number, ever.

Definition 4.4. An algebraic number is called an algebraic integerif and only if it is a root of a monic polynomial equation of degree nwith coefficients in Z.

Every algebraic number r is a root of an equation with coefficients inZ. Why? Because r is a root of an equation with coefficients in Q, bydefinition. So, multiply the equation by the least common multiple ofthe denominators of the rational coefficients and we get an equationwith integral coefficients; the polynomial equation won’t necessarilybe monic.

Theorem 4.1. Let F be a field and let r be in E ⊆ F be algebraicwith respect to F . Then r is a root of a unique monic irreduciblepolynomial p(x) in F [x] and p(x) divides each polynomial in F [x]that has r as a root.

Definition 4.5. Let F be a field. Then r1, r2 in E ⊆ F are conjugatesif they are roots of the same irreducible polynomial in F [x].

Definition 4.6. If every element of an extension E of F is algebraicwith respect to F , then E is an algebraic extension of F .

Definition 4.7. Let F be a field and let p(x) be in F [x] such that p(x)is irreducible in F [x]. We let F [x]/〈p(x)〉 stand for the polynomialswith coefficients in F taken modulo 〈p(x)〉 with equivalence classesdescribed as {r(x) + p(x)q(x), where q(x) is in F [x]}, represented byr(x) mod p(x).

The uniqueness of the remainder of a long division is guaranteed bythe Division Algorithm, so the representation is described in the defi-nition above is also unique. Also, we must check that p(x), from thedefinition above, is really irreducible.

34

Definition 4.8. We may write the representative of an equivalenceclass of polynomials as a polynomial with a bar on top; for example,(10/9)x + 2/3.

Theorem 4.2 (Kronecker). Let F be a field, and let p(x) be irre-ducible in F [x]. Then F [x]/〈p(x)〉 is a field, and contains a subfieldisomorphic to F . The polynomial p(x) has a root in F [x]/〈p(x)〉.

Let’s review some concepts. Let F be a field. Suppose r is algebraicover F which means r is the root of some polynomial over F . If werequire r to be the root of some irreducible polynomial, then there’sonly one polynomial to which r is a root of. We call such polynomial,the minimum polynomial of r over F .

For example, if F = Q and r = 3√

2, then p(x) = x3 − 2. But if F =Q, then 3

√2 implies the existence of p(x) = x− 3

√2. So, it depends on

the field. Now, consider π as the root of some polynomial over Q.

Theorem 4.3 (Kronecker, in new words). We have a polynomial p(x)which is irreducible over F . We can take all polynomials in F [x] andquotient it out with p(x); that is, we can create F [x]/〈p(x)〉; this is afield, with nice properties.

The first property is that the field, mentioned in the theorem of Kro-necker, has F inside it: the constant polynomials. Now x is a rootof p(x). For example, x2 + 2x + 2 is irreducible by E.I.C. Now takeQ[x]/〈p(x)〉 and notice that since p(x) is congruent to zero by defini-tion, x is a root. To check, just plug x in the polynomial.

Another way to think of the previous example is to declare r to be aroot of p(x). Now, we form a field F (r) — read this as “F adjoin r”.Now use the fact that p(r) = 0 to “multiply” things around — whatwould this mean?

Going on; if p(x) has degree n, then every element of F (r) is of theform c0 + c1r + c2r

2 + ... + cn−1ran−1 with ci in F . Let’s look at an

example.

Declare r a root of x2+2x+2. In Q(r), we know that r2+2r+2 = 0 orr2 = −2r−2. Now Q(r) = {x+yr : x, y ∈ Q}. With this informationnow, we write (6 + 2r)(−1− 3r) = −6− 20r − 6r2 = −6− 20r − 6r2

= 6− 8r because r2 = −2r − 2.

Remark 4.1. So, we have a field. But how do we divide things in it?That is, what’s 1/a, for example? We can use the divison algorithm to

35

answer that. The multiplication by conjugates doesn’t always work;that is, in C you can define division in terms of multiplication byinverses, but it seems to me that this works in C and not on everyfield. So, I should be able to find one field in which this does not work,right? This is my task now, because I don’t know one right now, butI will say that I believe that not all the time we get something like i2

= −1 in a field that will answer all of our questions. So I say that Cis a very nice field, but there are some mean ones.

Definition 4.9. A simple extension is an extension which includesonly one element. The degree of an extension F ⊆ E is the numberof elements of E that we need to span E over F .

What does this mean? Take

Q(√

2,√

3) = {a + b√

2 + c√

3 + d√

6},

and notice that it is a degree 4 extension of Q, because we absolutelyneed 1,

√2,√

3, and√

6.

Now, we have to notice that Q(√

2 +√

3) = Q(√

2,√

3), but theyhave different bases and different degrees. Doesn’t this sound notwell-defined? This alters perhaps what we call, in this document, asplitting field. I owe an explanation here.

Theorem 4.4. If r has minimum polynomial p(x) of degree n overF , then F (r) : F is an extension of degree n.

So, for example, take p(x) = x5 + 2x2 + 2x + 6, which is irreducibleover Q, and declare r as a root, so

Q(r) = {a + br + cr2 + dr3 + er4}

has degree 5.

Now, consider x2 + 2x + 2. Here,

Q(r) = {a + br : a, b ∈ Q}.

The rule for multiplication here is r2 = −2r − 2. This is just like thecomplex i such that i2 = −1. Now, we have a root, so we can factorit. We do the long division between x− r and x2 + 2x + 2. We get

x2 + 2x + 2 = (x− r)(x + 2 + r) + 0.

So r and −r − 2 are conjugates in Q(r).

36

Consider now x3−2 which is irreducible over Q. So we extend the fieldby adjoining 3

√2. So a division should happen with zero remainder.

We getx3 − 2 = (x− 3

√2)(x2 +

√2 + 3√

4) + 0.

Now (x2 +√

2 + 3√

4) is irreducible. We then expand the field to getit to be reducible. So Q( 3

√2, 3√

2(−1/2 + i√

3/2)). I wondered if theseadditions to the field would make the field completely factorable, but itdoesn’t; these additions make these polynomials factorable, but othersare still irreducible, so what happens to C? C somehow is completelyfactorable. This is a miracle; it turns out that R[x] mod x2 + 1 makesC a beautiful field.

Definition 4.10. Let p(x) be irreducible over F . Suppose F ⊆ E ⊆ Dare fields. The polynomial “splits” if it factors into linear terms. Wesay D is a spliting field for p(x) if p(x) splits in D[x], and p(x) doesnot split in E[x] for any F ⊂ E ⊂ D.

By the way, all roots of p(x) lie in the splitting field. Why? Becauseif it factors then

p(x) = (x− r1)(x− r2)...(x− rn).

Theorem 4.5. There exists is a unique splitting field. Take F (r1, r2,..., rn) which are the roots of p(x).

Let’s now see an example. Find a splitting field of x4 − 3x2 + 1. Howdo we do it? Find the roots of the polynomial; use the quadratic

formula. So x2 = (3±√5)/2, so x = ±√

(3±√5)/2. Now,

Q(√

(3 +√

5)/2,

√(3−

√5)/2

)

is a splitting field. We’re done. There may be a better way to writethis field, and this is one of the frustrating thing: there is so manyways to write it and it’s not always easy to figure out whether we havethe simplest.

Theorem 4.6. F (r1, r2) = F (r1)F (r2) = F (r2)F (r1) et cetera.

Theorem 4.7. Suppose F ⊆ E ⊆ D with degree E : F = m, degreeD : E = n. Then D : F = mn.

37

An example of this theorem is F = Q and E = Q(√

2) = {a + b√

2: a, b ∈ Q}. Degree E : F = 2. Now, look at E = Q(

√2) and

D = E( 4√

2). The minimum polynomial of 4√

2 over Q(√

2) is x2 −√2so degree D : E = 2 and degree E : Q = 4. I don’t really follow thisexample, so if there’s anything wrong here I wouldn’t be able to tell.

Degree Q(π) : Q = ∞. Now consider Q(π) : Q(π2) which has degree2, and π is root of x2 − π2.

In this section, we’ve studied algebraic numbers and transcendentalnumbers. We have seen irreducible polynomials over a field F . Wehave seen Kronecker’s theorem which claims the existence of a fieldF [x]/〈p(x)〉 which is an extension field of F where p(x), in particular,has a root; we can think of F [x]/〈p(x)〉 as a simple degree n extensionF (r) where p(r) = 0 degree p(x) = n.

We’ve seen minimum polynomials, conjugates of field elements. Onthe next section, we will be looking at roots in which when we permutethem around, we will find nice properties; that is, we’ll be building agroup of permutations of roots.

Exercise 4.1. Prove that 1, i =√−1 form a linear basis for R(i),

where R is the real field. Show that g =√

2− 2i is algebraic relativeto R by finding a polynomial having g as a zero.

Solution. Every element of R(i) can be written as a + bi, so 1 and ispan R(i). To show linear independence, we must show that for all rin R(i), r = a · 1 + b · i = 0 if and only if a = b = 0.

If a = b = 0, then r = 0. Now if r = 0, then

r = a + ib = 0a = −ib

a2 = (−ib)2

= (−i)2b2

= −b2

a2 + b2 = 0

But since a and b are real numbers, a2 and b2 are non-negative, soa = b = 0.

38

To show that g =√

2− 2i is algebraic relative to R, we write

g2 = 2 + 4i2 − 4i√

2

g2 − 2 + 4 = −4i√

2

(g2 + 2)2 = (−4i√

2)2 = −32

(g2 + 2)2 + 32 = 0

Exercise 4.2. Construct a splitting field for p(x) = x4 − 8x2 + 15over Q.

Solution. We factor p(x) into linear terms and see which elementswe’re missing into some field, so that we can build the smallest possiblefield such that p(x) factors completely.

p(x) = (x2 − 5)(x2 − 3)

= (x +√

5)(x−√

5)(x +√

3)(x−√

3)

So p(x) splits into linear factors in Q(√

5,√

3).

Exercise 4.3. Is Q(π) a finite extension? If so, give a basis of Q(π): Q.

Solution. Suppose it is finite. Then we could try 1 and π as a basis,but we need to account for π2, so we could try, 1, π, π2 as a basis, butthen we would have to account for π3 as well, so we could expand thebasis again and so on, so it doesn’t look like this will ever terminate.

Exercise 4.4. Find minimal polynomial of α ∈ Q[x] and degree ofQ(α) : Q.

(a) α =√

3−√6

a2 = 3−√

6

a2 − 3 = −√

6

(a2 − 3)2 = 6

(a2 − 3)2 − 6 = 0

a4 − 6a2 + 3 = 0

Degree Q(α) : Q = 4. Let p = 3, so by E.I.C. it is irreducible.

39

(b) α =√

13

+√

7

α2 =13

+√

7

α2 − 13

=√

7(

α2 − 13

)2

= 7

(α2 − 1

3

)2

− 7 = 0

α4 − 2α2

3− 62

9= 0

9α4 − 6α2 − 62 = 0

Degree Q(α) : Q = 4. Let p = 2, so by E.I.C. it is irreducible.

(c) α =√

2 + i

α2 = 1 + 2i√

2

α2 − 1 = 2i√

2

α4 − 2α2 + 9 = 0

Degree Q(α) : Q = 4. E.I.C. doesn’t seem to help here. What wouldbe another approach? The only thing I can think of is to compute theroots and show they won’t allow reducibility over Q.

The polynomial factors into

(α +√

2 + i)(α +√

2− i)(α−√

2 + i)(α−√

2− i).

Let A,B, C, D be first, second, third, forth factors, respectively. ThenAB has an irrational coefficient, AC has a complex one, AD has acomplex one, BC has a complex one, BD has a complex one and CDhas a complex one; brute force shows no factors over Q are possible.

(d) α =√

2 +√

5

α2 = 2 + 5 + 2√

10

(α2 − 7)2 − 40 = 0

α4 − 14α2 + 9 = 0

Degree Q(α) : Q = 4. Brute force again. The roots are r1 =√

2+√

5,r2 =

√2−√5, r3 = −√2 +

√5, r4 = −√2−√5. So it factors over R

as(x− r1)(x− r2)(x− r3)(x− r4).

40

Let A,B,C, D be the factors in the other they appear above. SoAB, AC, AD, BC, BD and CD have an irrational coefficient. Sofactorization over Q is not possible.

Exercise 4.5. For each α ∈ C and each given field F , classify α aseither algebraic or transcendental over F . If α is algebraic, find thedegree of F (α) : F .

(a) α = i, F = Q.

α2 = i2

α2 + 1 = 0.

So i is algebraic over Q, and degree of Q(i) : Q = 2.

(b) α = π, F = Q. Transcendental.

(c) α = π2, F = Q(π). We have −π2 as a root of p(α) = α − π2, soπ2 is algebraic over Q(π) and the degree of Q(π2) : Q(π) = 1.

(d) α = π2, F = Q(π3). Transcendental.

Exercise 4.6. Find degree and basis for the given field extensions.

(a) Q(√

2) : Q. Basis (1,√

2), degree 2.

(b) Q(√

2,√

3,√

18) : Q. Basis (1,√

2,√

3,√

6), degree 4.

(c) Q( 3√

2,√

3) : Q. Basis (1,√

3, 3√

2, 3√

4,√

3 3√

2,√

3 3√

4), degree 6.

(d) Q(√

2,√

3) : Q(√

2 +√

3). Since√

3 +√

2 is in the field, we alsohave

√6,√

2,√

3, so the fields here are the same, and we have 1 as abasis, so degree is 1.

Exercise 4.7. Find all conjugates for α =√

2 +√

3 over Q. Repeatfor α = 3

√2.

Solution. We look for an irreducible polynomial over Q which has αas a root. So p(x) = x4 − 10a2 + 1 is irreducible over Q and its rootsare α itself, and all of its conjugates which are −√3−√2, −√3+

√2,√

3−√2.

Exercise 4.8. Let σ, τ : F → F be two field automorphisms. Provethat σ−1 and (σ ◦ t) are also field automorphisms.

Proof. Since σ is an automorphism, it is one-to-one and onto and takeseach f ∈ F to some f ′ ∈ F for each f ′ ∈ F . By definition, σ−1 is the

41

one-to-one and onto function such that (σ ◦ σ−1)(f) = (σ−1 ◦ σ)(f) =f for all f ∈ F .

So, if we show that σ−1 is a homomorphism, we would have shownthat it is an isomorphism, and therefore, an automorphism. Let a, bbe in F . Since σ is an automorphism, there are u, v in F with a =σ(u), b = σ(v). So,

σ−1(ab) = σ−1(σ(u)σ(v))

= σ−1(σ(uv))= uv

= σ−1(a)σ−1(b).

Similarly for addition,

σ−1(a + b) = σ−1(σ(u) + σ(v))

= σ−1(σ(u + v))= u + v

= σ−1(a) + σ−1(b).

Now σ ◦ τ . Since σ and τ are field automorphisms, they preserve theoperations. Let a and b be in F . So, for multiplication, we do

(σ ◦ τ)(ab) = σ(τ(ab))= σ(τ(a)τ(b))= σ(τ(a))σ(τ(b))= (σ ◦ τ)(a)(σ ◦ τ)(b).

For addition,

(σ ◦ τ)(a + b) = σ(τ(a + b))= σ(τ(a) + τ(b))= σ(τ(a)) + σ(τ(b))= (σ ◦ τ)(a) + (σ ◦ τ)(b).

So (σ ◦ τ) is a homomorphism. To show that it is onto, let f ′ be in Fand then we must show that there’s an f in F such that (σ ◦ τ)(f) =f ′, for all f ′ in F.

Since τ is an automorphism, then for each f ′ in F , there is an f in Fsuch that τ(f) = f ′. Since σ is also an automorphism, then for eachf ′ in F , there’s τ(f) in F such that σ(τ(f)) = f ′. So for each f ′ inF , there’s an f in F such that σ(τ(f)) = (σ ◦ τ)(f) = f ′, so it’s onto,and hence, an automorphism.

42

Exercise 4.9. Prove that G(C/R) is cyclic with 2 elements.

Proof. By definition, the Galois group G(C/R) is the group of auto-morphisms of C that fix every r ∈ R, where the binary operation iscomposition. One automorphism is the identity map, which is one-to-one and onto, and is the identity of the group.

We will try to find a second one. Define a map σ that maps conjugates;that is, the rule is “map a + bi to a− bi” and notice that every r ∈ Ris of the form “a + b · 0,” so for a real number r, σ(r) = r, so everyr ∈ R is fixed. Now, if σ is an isomorphism, then it is a member ofthe group, and if we can show that σ2 = (σ ◦ s) = e, then we wouldhave shown that there are no more elements in the group, hence thatthe group is cyclic.

Let a + bi, c + di ∈ C. So,

s((a + bi)(c + di)) = s(ac− bd + (ad + bc)i)= ac− bd− (ad + bc)i= (a− bi)(c− di)= s(a + bi)s(c + di).

Also,

s((a + bi) + (c + di)) = s((a + c) + (b + d)i)= a + c− (b + d)i= a− bi + c− di

= s(a + bi)s(c + di)

So σ is a homomorphism. Let a+ bi be in C. We want to show that σis onto; that is, we want to show that there is a c + di in C such thatσ(c + di) = a + bi.

We know that σ(c + di) = c − di, where c and −d are real numbers.The task was to find c and d such that c = a, d = −b or −d = b. Thereare such numbers, pick c = a and d = −b. So σ is an isomorphism.

Now we must show that σ2 = (σ ◦ s) = e. Let a + bi in C. Compute(σ ◦ σ)(a + bi) = σ(σ(a + bi)) = σ(a− bi) = a + bi = e(a + bi); hence,the group is cyclic with two elements, e and σ.

Exercise 4.10. Let F be a field. Define the characteristic of F to bethe smallest n ≥ 1 such that 1 + 1 + ... + 1 = 0 where we are adding1 to itself n times. If no such n exists, we say F has characteristiczero. What is char(Q)? What is char(Z7)? Suppose F has finite

43

characteristic n. Prove that n must be prime. Hint: fields have nozero divisors.

We have char(Q) = 0, and char(Z7) = 7.

Now char(F ) = n. Proof. Let F have finite characteristic n; n mustbe composite or prime. Suppose n is composite. Then there is a primep < n such that n = pq, for some natural q.

Then (1 + 1 + ... + 1)p × (1 + 1 + ... + 1)q = 0, where the index pmeans 1 added p times, and similarly for q. Since fields have no zerodivisors, F isn’t a field if n is composite; hence, n is prime.

5 Galois theory

Definition 5.1. An automorphism σ of a mathematical system Swith an operation defined on S is a one-to-one correspondence fromS onto itself that is an isomorphism with respect to the operation.

Definition 5.2. An automorphism σ of a field F is an automorphismof F that preserves both field operations. An element f in F is fixedby an automorphism σ if σ maps f to itself; that is, σ(f) = f ; it ismoved by σ if σ(f) 6= f .

Theorem 5.1. Let D be a finite extension of F , let E be an inter-mediate field F ⊆ E ⊆ D. Let G(D/F ) be the automorphisms of Dthat fix every f in F . Let G(D/E) be the automorphisms of D thatfix every e in E. Then G(D/F ) is a group under composition andG(D/E) is a subgroup of G(D/F ).

Proof. Suppose that G(D/F ) is a group under composition. Then it’seasy to believe that G(D/E) is a subgroup of G(D/F ) because weknow that we would have a similar set with less automorphisms. Soanyway, we will first prove that G(D/F ) is a group.

We show that we have closure under composition. Let σ1, σ2 be inG(D/F). So any d in D has an image σ2(d) in D because σ2 is anautomorphism of D, and σ1(σ2(d)) is in D. So the composition of twoautomorphisms produces an element in D. Now... oh well, can yougo on?

Definition 5.3. Let E be the splitting field of a polynomial f(x) inF [x]. The group G(E/F ) of automorphisms of E that fix every f in

44

F is the Galois group or group of the polynomial and the group of theequation f(x) = 0.

Theorem 5.2. Let F be a field. Let F (r) : F = n. Any automor-phism σ in G(F (r)/F ) is completely determined by σ(r). It’s truethat |G(F (r)/F )| ≤ n. Let E be the splitting field of an irreduciblepolynomial p(x) in F [x]. The Galois group G(E/F ) is isomorphic to agroup of permutations of the k zeros of p(x) in E, hence to a subgroupof the symmetric group Sk.

Notice the importance of this theorem. It links group theory to thepermutations of the roots of an equation. The theory of groups, as wesee in the Abstract Algebra books today wasn’t very much developedby Galois or Abel. They studied particular groups formed by the per-mutations of the roots of an equation, but in 1870, Kronecker isolatedthe group properties in their own right.

Theorem 5.3 (?). Suppose r is algebraic over F with a minimalpolymonial of degree n. Then |G(F (r)/F )| ≤ n.

Proof. Let σ be in G(F (r)/F ). By definition, the group leaves everyu in F fixed, so there are at most n possible images of r under σ.Since every σ is one-to-one and onto, and since every u in F is alreadytaken, there are at most n possible mappings available because thereat most n roots for the minimal polynomial.

Consider the extension fieldQ(r) : Q, and let the minimum polynomialof r be p(x) = x5 + 2

3x3 + 2x + 1. How do we know that σ(r) mustbe a root? We know that p(r) = 0. So now apply σ to r, so σ(p(r))= σ(0) = 0 because 0 is in Q and so it is fixed. Now, since σ is anautomorphism, we can write

σ(p(r)) = σ(r5 +2r3

3+ 2r + 1)

= σ(r5) +2σ(r3)

3+ 2σ(r) + 1

= σ(r)5 +2σ(r)3

3+ 2σ(r) + 1 = 0

So σ(r) is a root.

So, suppose that σ is in G(E/F ) and p(x) is in F [x], then σ permutesany roots of p(x) which lie in E. We prove this with the argumentabove.

45

Definition 5.4. A field extension E ⊆ F is normal if and only if anyirreducible p(x) in F [x] which has a root in E must factor completelyin E.

Here’s an example of a non-normal extension. Consider Q( 3√

2) : Q,which is not normal; it isn’t because x3 − 2 has a root, but doesn’tfactor completely, and this is a problem for us. We want a bigger fieldwith a Galois group.

E : F is normal if and only if every irreducible p(x) in F [x] with aroot in E splits in E, since “to split” means to factor all the way downinto linear terms.

This is the same as saying that E is the splitting field of some poly-nomial. Or equivalently, {a ∈ E : σ(a) = a for all σ ∈ G(E/F )} = F .Normal extensions will correspond to normal subgroups.

Definition 5.5. A field F is called a field of characteristic zero if Fcontains a subfield isomorphic to the field Q of rational numbers.

Let’s review some stuff. Suppose E ⊂ F is a field extension. The Ga-lois group G(E/F ) = {σ : E → E such that σ is a field automorphismand σ(r) = r for every r in F}.We have to check that field automorphism form a group under com-position. I also have to check that if σ and τ both fix f , then so doσ−1 and σ ◦ τ .

Remark 5.1. Let σ be in G(E/F ), then σ fixes polynomials in F [x],thus σ permutes roots of polynomials in F [x].

Galois group elements take roots to roots, but the roots must belongin the field. Suppose a cubic root is a a root of some polynomial withonly one real solution. Suppose the complex roots do not belong inthe field. Then the only element will be the identity.

Proposition 5.4. Suppose r is algebraic over F with minimal poly-nomial of degree n. Then G(F (r)/F ) has at most n elements.

Proof. Let r have minimal polynomial

p(x) = a0 + a1x + ... + anxn.

46

So p(r) = 0. Let u be in G(F (r)/F ). So

p(u(r)) = a0 + a1u(r) + ... + an(u(r))n

= a0 + a1u(r) + ... + an(u(r)n)= u(a0) + u(a1r) + ... + u(anrn)= u(a0) + a1r + ... + anrn)= u(p(r))= u(0)= 0

The last two equations hold because p(r) = 0 and u(0) = 0 because uis an automorphism, so it maps 0 to 0. So u(r) is also a root. Now, wenotice that every element of G(F (r)/F ) corresponds, in a one-to-oneand onto fashion — because they’re automorphisms —, with a rootof p(x). Since there are at most n roots for p(x), there are at most nautomorphisms in G(F (r)/F ).

A word on degrees. The degree of E:F depends on both E and F .Sure, compare Q(

√2,√

3) : Q with Q(√

2,√

3) : Q(√

2).

By the way, in this document, our definition of splitting field is thesmallest possible for a minimal polynomial. There seems to be otherequivalent definitions.

Proposition 5.5. Suppose E is a splitting field of a polynomial ofdegree k in F [x]. Then G(E/F ) is a subgroup of Sk.

Proof. Let r1, r2, ..., rk be roots of p(x) over E. We know that overE, p(x) splits. Then E = F (r1, r2, ..., rk).

Let σ be in G(F (r1, ..., rk)/F ). To finish the proof, we must show that(1) σ is completely determined by σ(r1), ..., σ(rk). (2) σ(ri) must berj for some j. To conclude, notice that (1) and (2) tell us that theGalois group is isomorphic to some subgroup of Sk.

Definition 5.6. Galois group of p(x) in F [x] is G(E/F ) for E asplitting field of p(x).

Exercise 5.1. Find the Galois group of x4 − 2 over Q.

Solution. We first look for the roots of x4−2. The polynomial factorsas

(x2 +√

2)(x2 −√

2) = (x + 4√

2i)(x− 4√

2i)(x + 4√

2)(x + 4√

2)

47

Let r = 4√

2. So roots are r, −r, ri, −ri. What’s the splitting field?Well, throw, in the basis, the elements r and ri, but notice that weonly need r and i.

What do we need now? Let f be in G(Q(r, i)/Q). To determine f , weneed to know f(r) and f(i). What are the possibilities for f(r)? Allthe roots of x4 − 2. What about f(i)? It goes to i or −i.

So we have 4 possibilities for r, and 2 for i, so 4 × 2 = 8. Thisgroup has at most 8 elements. It is not necessary that all of thesepossibilities will work. We need more theory to show how this couldwork, but they are all technically possible so for.

Let’s list them. As we read the table, notice that σ2(r) = σ(σ(r)) =−r. That’s σ(−r) = −ir. So σ4 is the identity. We introduce τ . Now,we do σ ◦ τ .

automorphism domain range1 r i

σ ir i

σ2 −r i

σ3 −ir i

τ r −i

σ ◦ τ ir −i

σ2 ◦ τ −r −i

σ3 ◦ τ −ir −i

Do you recognize this table? This group is isomorphic to D8, thedihedral group with 8 elements; the symmetries of a square. We candescribe it by τ ◦ σ = σ3 ◦ τ .

Theorem 5.6. Suppose F ⊂ E ⊂ D. Then (1) G(D/E) ≤ G(E/F ).(2) Suppose H ≤ G(D/F ). Let DH = {d ∈ D : σ(d) = d for allσ ∈ H}. Then F ⊆ DH ⊆ D.

Recall that E/F is a normal extension if any of the following 3 equiv-alent conditions hold: (1) E is a splitting field of some polynomialin F [x], (2) any irreducible polynomial p(x) in F [x] with a root in Emust split in E, (3) fixed field of G(E/F ) is F .

Theorem 5.7 (Fundamental Theorem of Galois Theory). Supposethe characteristic of F = 0, and F ⊂ D is a normal extension. Let F ⊆E ⊆ D; then (1) E ←→ G(D/E) gives a one-to-one correspondencebetween intermediate fields E and subgroups of G(D/F ), (2) The

48

degree [D : E] = |G(D/E)|, (3) F ⊆ E1 ( E2 ⊆ D if and only if{e} ≤ G(D/E2) � G(D/E1) ≤ G(D/F ), (4) E is normal over F ifand only if G(E/F ) P G(D/F ).

We now solve a series of exercises which will help us to understandthe claims of the Fundamental Theorem of Galois Theory.

Exercise 5.2. Determine the Galois group of

(x2 − 2)(x2 − 3)(x2 − 5),

over Q.

Solution. The polynomial splits in Q(√

2,√

3,√

5). Now, let σ bearbitrary in G(Q(

√2,√

3,√

5)/Q). Then, the splitting field must alsosplit r’s minimal polynomial which is r2 − 2 = 0, and so σ must take√

2 to√

2 or −√2. The same goes for√

3; σ must take√

3 to√

3 or−√3, and the same for

√5; σ must take

√5 to

√5 or −√5.

So, the automorphisms are

automorphisms√

2√

3√

51√

2√

3√

5σ

√2

√3 −√5

τ√

2 −√3√

5σ ◦ τ

√2 −√3 −√5

γ −√2√

3√

5γ ◦ σ −√2

√3 −√5

γ ◦ τ −√2 −√3√

5γ ◦ σ ◦ τ −√2 −√3 −√5

So the galois group is determined, and it is completely determined byσ :√

5→ −√5, τ :√

3→ −√3 and γ :√

2→ −√2.

Exercise 5.3. Now determine all subfields of the splitting field foundin the previous exercise.

Solution. The subfields of the “second level” of the field diagramwould be Q(

√2,√

3), Q(√

2,√

5), Q(√

3,√

5), Q(√

2,√

15), Q(√

6,√

5),Q(√

6,√

10), Q(√

3,√

10). In the “third level” we would get Q(√

2),Q(√

3), Q(√

5), Q(√

6), Q(√

10), Q(√

15), Q(√

30). So, including thetrivial ones, we get a total of 16 in the diagram.

What if I wanted to know the fixed field of this Galois group? I wouldhave to look at every automorphism and see which elements of the

49

basis never get moved. So I collect these elements and throw them inQ to get the fixed field.

When we look at each of automorphism, we see that

σ(a + b√

2 + c√

3 + d√

5 + e√

6 + f√

10 + g√

15 + h√

30) =

a + b√

2 + c√

3− d√

5 + e√

6− f√

10− g√

15− h√

30),

τ(a + b√

2 + c√

3 + d√

5 + e√

6 + f√

10 + g√

15 + h√

30) =

a + b√

2− c√

3 + d√

5− e√

6 + f√

10− g√

15− h√

30),

σ ◦ τ(a + b√

2 + c√

3 + d√

5 + e√

6 + f√

10 + g√

15 + h√

30) =

a + b√

2− c√

3− d√

5− e√

6− f√

10 + g√

15 + h√

30,

γ(a + b√

2 + c√

3 + d√

5 + e√

6 + f√

10 + g√

15 + h√

30) =

a− b√

2 + c√

3 + d√

5− e√

6− f√

10 + g√

15− h√

30,

γ ◦ σ(a + b√

2 + c√

3 + d√

5 + e√

6 + f√

10 + g√

15 + h√

30) =

a− b√

2 + c√

3− d√

5− e√

6 + f√

10− g√

15 + h√

30,

γ ◦ σ ◦ τ(a + b√

2 + c√

3 + d√

5 + e√

6 + f√

10 + g√

15 + h√

30) =

a− b√

2− c√

3− d√

5 + e√

6 + f√

10 + g√

15− h√

30,

so it looks like that the only field that is fixed by every automorphismis Q itself.

If now I wanted to know each corresponding subgroup of G(Q(√

2,√

3,√

5)/Q),then one way would be to work with Z2 × Z2 × Z2. We assign (0,0,1)to σ because σ moves only

√5, so we consider

√5 as the 1 in the third

component of the 3-tuple. Similarly, we assign (0, 1, 0) with τ , and(1, 0 ,0) with γ, and (0,0,0) with the identity map.

We then work with Z2 × Z2 × Z2 under addition. To get each sub-group now, we can consider (0,0,0), (0,0,1) and notice that it wouldcorrespond to Q(

√(2),

√(3)) because it moves only

√5. The others

would be found similarly.

Exercise 5.4 (8-3). Let F = Q, D = Q(√

2, i), E1 = Q(√

2), E2 =Q(i). Then F has characteristic 0 and D = F (r), where r =

√2 + fi,

f 6= 0 is in Q. Find σ(r) for each automorphism σ ∈ G(D/E1) andτ(r) for each τ ∈ G(D/E2). Find the four conjugates of r = r1, hence,by Theorem 8-2, the four automorphisms of G(D/F ).

They say D = F (r), but isn’t D = Q(√

2, i)? So are they saying thatD can be written as F (r) where r is the root of some polynomial thatwhen adjoined will give us

√2 and i? I think so.

50

Such root is r =√

2 + fi, with f 6= 0, f ∈ Q. Its minimal polynomialis

r2 = 2− f2 + 2fi√

2

r2 − 2 + f2 = 2fi√

2

(r2 − 2 + f2)2 = −8f2

(r2 − 2 + f2)2 + 8f2 = 0

This polynomial has degree 4 and therefore will have at most 4 roots.Since one root is complex, I guess the fundamental theorem of algebraimplies that we will have 4 distinct roots.

Now I must find σ(r) for each automorphism in G(D/E1). Since E1

= Q(√

2), then each automorphism moves only i, so s(r) = i or s(r)= −i. So all of the automorphisms here are

automorphism i

1 i

σ −i

Similarly for τ(r) ∈ G(D/E1). Since E2 = Q(i), then each automor-phism moves

√2 only, so σ(

√2) =

√2 or s(r) = −√2. So all of the

automorphisms here are

automorphism√

21√

2τ −√2

The four conjugates of r1 =√

2 + fi are r2 = −√2 + fi, r3 = −√2−fi, r4 =

√2 − fi. Now I must find, by the theorem above, the 4

automorphisms of G(D/Q). I’m failing to see a clear way to use thetheorem, even though I may have gotten the automorphisms correctlyhere. The exercise suggests that by the theorem, we are able to getall the automorphisms. I guess it must be all permutations of theelements that get moved in G(D/E1) and G(D/E2).

Any automorphism of G(D/Q) fixes Q and moves√

2 and i.

automorphism√

2 i

1√

2 i

σ −√2 i

τ√

2 −i

γ −√2 −i

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The theorem says that this Galois group, in particular, is isomorphicto a group of permutations of 4 zeros of p(x) in Q(

√2, i); hence isomor-

phic to a subgroup of the symmetric group S4. Every automorphismsquares to the identity, so it should be isomorphic to the Klein-4 whichis isomorphic to the flips of a square.

Exercise 5.5 (8-4). Still with F = Q and D = Q(√

2,i). Noticethat G(D/F ) is isomorphic to the Klein-4 group. Find the uniquesubgroup of G(D/F ) that corresponds to each intermediate field E1

= Q(√

2) and E2 = Q(i). Also, find the subfield corresponding to eachsubgroup.

Solution. As proper subgroups of G(D/F ), we have {1, σ}, {1, τ}, {1,γ}, which are isomorphic to subgroups of the Klein-4. The exerciseis asking for the subgroups that correspond to the fields E1 and E2.Which subgroup would correspond to Q(

√2)?

It would be the subgroup that is linked to the maps that do not moveQ(√

2), so it’s {1, τ}. For Q(i), it would be the subgroup that is linkedto the maps that do not move Q(i); so it’s {1, σ}.Here’s a question. The claim iii of the Fundamental Theorem ofGalois Theory seems to say that we have Q(

√2, i) ⊆ Q(i) ⊂ Q(

√2) ⊆

Q if and only if {1} ⊆ G(Q(√

2, i)/Q(i)) ⊂ G(Q(√

2, i)/Q(√

2)) ⊆G(Q(

√2, i)/Q). Q(i) is not an of Q(

√2); it’s not even a subset. Is

this a case in which we don’t satisfy the hypothesis of the claim iii?Or is my interpretation wrong?

The exercise also says “find the subfield corresponding to each sub-group.” Isn’t this the same task above? Once we find the subgroupcorrespondent to the subfield, then we’ve found the subfield corre-spondent to the subgroup.

Exercise 5.6 (8-5). Verify that the claim ii of the Fundamental The-orem of Galois Theory holds for the example we’ve been working with.

Solution. The claim says, for this example in particular, that

[Q(√

2, i) : Q(i)] = |G(Q(√

2, i)/Q(i))|which is true because the degree of the extension is 2 since the basisis a + b

√2 for a, b ∈ Q(i), and for the Galois group we have only two

possible automorphisms; one that takes√

2 to −√2 and the identity.

The Theorem also claims that

[Q(√

2, i) : Q(√

2)] = |G(Q(√

2, i)/Q(√

2))|

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which is almost the same case as above. Here, we have the basis asa + bi, for a, b ∈ Q(

√2). The degree of the extension is 2 and there

are only two possible automorphisms in the Galois group; they are theidentity and another one that takes i to −i.

Also, the theorem claims that

[Q(i) : Q] = [G(Q(√

2, i)/Q) : G(Q(√

2, i)/Q(i))].

I think that this is a link between a degree extension and the numberof left cosets of a subgroup in its group. The degree extension is 2.G(Q(

√2, i)/Q) = G has 4 automorphisms, and G(Q(

√2, i)/Q(i)) =

H has 2. Now, by Lagrange’s theorem, we have that 4 = 2 [G : H],so [G : H] = 2 as well and it matches the degree extension Q(i) : Q.

Also, the theorem claims that

[Q(√

2) : Q] = [G(Q(√

2, i)/Q) : G(Q(√

2, i)/Q(√

2))].

The degree extension is 2,

|G| = |G(Q(√

2, i)/Q)| = 4,

|H| = |G(Q(√

2, i)/Q(√

2))| = 2.

So again, by Lagrange’s theorem, 4 = 2 [G : H], so [G : H] = 2matching the degree extension of Q(

√2) over Q.

Exercise 5.7 (8-6). Still with F = Q and D = Q(√

2, i). Showthat {1} is a proper subset of G(D/E) which is a proper subset ofG(D/F ) for E = E1 = Q(

√2) and for E = E2 = Q(i) but that

G(D/E2) * G(D/E1). Why are the inclusions reversed in order inthe second clause of the third claim of the Fundamental Theorem ofGalois Theory?

Solution. Let E = Q(√

2). Then G(D/E) has two automorphisms {1,σ}, where σ takes i to −i, so {1} is a proper subset of G(D/E). Wealso know that G(D/F ) has 4 automorphisms: {1, σ, τ , γ} as definedin Exercise 5.4. So, evidently, G(D/E) is a proper subset of G(D/F ).

Let E = Q(i). Then G(D/E) has two automorphisms {1, τ}, whereτ takes

√2 to −√2, so {1} is a proper subset of G(D/E). We also

know that G(D/F ) has 4 automorphisms: {1, σ, τ , γ} as defined inExercise 5.4. So, evidently, G(D/E) is a proper subset of G(D/F ).

Is G(D/E2) ⊂ G(D/E1)? No, because σ is in G(D/E2), but not inG(D/E1). Also, τ is in G(D/E1), but not in G(D/E2). That is, myobservations here do not satisfy claim iii.

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Why are the inclusions reversed? They are reversed because the lesselements the Galois group has to permute, the smaller it gets. So ifthe field increases in elements to be fixed, the Galois group decreasesin order.

Exercise 5.8 (8-7). Still working the same D, F , E1, and E2 fields,show that G(D/E1) is normal in G(D/F ) and that G(E1/F ) is iso-morphic to G(D/F )/G(D/E1). Repeat for E2.

Solution. We can show that the left cosets match the right cosets,since we’re working with a particular case. G(D/E1) is {1, τ} = H.So the left cosets are H, and {σ, σ ◦ τ = γ}. But the index is 2; so,it’s normal, by some theorem we’ve seen in the past.

To show that G(E1/F ) is isomorphic to G(D/F )/G(D/E1), we needto find a one-to-one and onto map that preserves the operations be-tween them. That’s easy here because we have only two automor-phisms in G(E1/F ), and identity goes with identity, so the other mapgoes with the other map.

Repeating for E2, we know that G(D/E2) is normal because the in-dex is 2; it’s isomorphic to G(D/F )/G(D/E2) for the same reasondescribed above.

Exercise 5.9. Recall that we determined that the splitting field ofx4 − 2 over was Q(ζ, i) where ζ = 4

√2. Also, we’ve calculated before

that the Galois group was isomorphic to the dihedral group with eightelements: D8 = 〈σ, τ : σ4 = τ2 = e, τ ◦ σ = σ3 ◦ τ〉. Now, let H be thesubgroup {1, σ3 ◦ τ}. Show by explicit calculation that the fixed fieldQ(ζ, i)H is equal to Q((1− i)ζ).

Solution. No idea.

Theorem 5.8 (9-3). Let E0 be a field of characteristic zero containingall the nth roots of unity. Let f ∈ E0 and let E be a splitting field forxn − f over E0. If w is any zero of xn − f in E, then E = E0(w) andG(E/E0) is cyclic.

Exercise 5.10. In Theorem 9-3, let E0 = Q, n = 2, f = 7. Find thesplitting field for x2 − 7 over Q. Show that G(E/E0) has order 2.

Solution. The splitting field for x2 − 7 over Q is Q(√

7) since x2 − 7= (x − √7)(x +

√7). So, in the theorem, E = Q(

√7), and E0 = Q.

We must show that G(Q(√

7)/Q) is cyclic.

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Here, the w of the theorem is either√

7 or −√7, so the theoremsays that we can adjoin

√7 or −√7 and get a cyclic Galois group

G(Q(w)/Q). We let w =√

7. So now we must show that G(Q(w)/Q)has order 2.

Since it’s cyclic of order 2, we must show that every element squaresto the identity. The identity itself does. The only non-trivial elementis σ that takes

√7 to −√7 and squares to the identity as well because

(σ ◦ σ) takes√

7 to −√7 and then back to√

7.

Here, Q(√

7) is a normal extension of Q because (1) Q is the fixed fieldof G(Q(w)/Q) or instead we could justify by saying that (2) Q(

√7) is

the splitting field over Q for x2 − 7 over Q.

Exercise 5.11. In Theorem 9-3, let E0 = Q, n = 2, f = 4. Show thatx2 − f in this case splits in E0 so that G(E/E0) has only the identityautomorphism.

Solution. It’s true that x2 − 4 = (x− 2)(x + 2) which evidently splitsin Q. So, in the Galois group of Q over Q, there’s really “nothing”to permute because the roots of the polynomial were already in Qanyway. The degree of the extension is 1, so we have only one element:the identity.

Exercise 5.12. In Theorem 9-3, let E0 = Q(r) where r is a primitivethird root of unity. Let n = 3, f = 7. Factor x3−f into linear factorsin its splitting field over E0. Show that the Galois group G(E/E0) iscyclic of order 3.

Solution. The roots of unity for n = 3 are

1,−12

+√

3i

2,−12

+√

3i

2.

They form a cyclic group under multiplication. We can’t let r = 1,though, otherwise it wouldn’t generate the group. Let’s let

r =−12

+√

3i

2.

So x3 − 1 splits over E0.

We’re interested in x3 − 7. Its roots are

3√

7,− 3√

72

+√

3 3√

7i

2,− 3√

72−√

3 3√

7i

2.

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So, we really just need 3√

7, because we can express the roots as r 3√

7,r2 3√

7, r3 3√

7 — noticing that r3 = 1, the identity of the cyclic groupof roots of unity. So the factorization can be written as

(x− r3√

7)(x− r2 3√

7)(x− r3 3√

7).

So we adjoin 3√

7 to get E = Q(r, 3√

7). We now consider G(E/E0) andshow that it’s cyclic of order 3.

The Galois group fixes Q(r), and roots must be taken to roots. Letσ( 3√

7) = r 3√

7, so σ2( 3√

7) = σ(σ( 3√

7) = σ(r 3√

7) = r2 3√

7 because σ is,by definition, multiplying its argument by r. Finally, σ3 = r3 3

√7 =

3√

7. That shows that we have only 3 elements and that they’re cyclicwith r 3

√7 as generator.

Exercise 5.13. Show that the Galois group for x3 − 7 over Q isisomorphic to S3 and therefore nonabelian.

Solution. Take a look at the Fundamental Theorem. Notice how manyelements the group has, and that there is a theorem that claims thatthe Galois group would be isomorphic to some subgroup of Sn, buthere we will find 6 elements and this can be seen by using the degreeof the extension field and therefore it will have to be isomorphic to S3

itself, which is nonabelian.

Exercise 5.14. Prove that A5 is not abelian. Why doesn’t the chainS5 B 〈1〉 provide a solvability chain for S5?

Solution. A5 is not abelian because

(1324)(1432) = (234) 6= (123) = (1432)(1324)

and (1324), (1432) are even permutations.

The definition of solvability requires a chain of subgroups G = G0 BG1 B ... B Gt = 〈1〉 to have the property that Gi/Gi+1 is abelian fori = 0, 1, ..., t− 1.

For the chain S5 B 〈1〉 satisfy the definition, we need G0/G1 abelian,but G0/G1 = S5/〈1〉 = S5, and S5 isn’t abelian.

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