1 r r2 3 4 5 - City University of New York · 2017. 9. 5. · 1 r r2 r3 r4 r5 6 1 5 4 3 2 5 6 4 3 2...

1 r r2 r3 r4 r5

16

5

4 3

2

65

4

3 2

1

54

3

2 1

6

43

2

1 6

5

32

1

6 5

4

21

6

5 4

3

s rs r2s r3s r4s r5s

61

2

3 4

5

12

3

4 5

6

23

4

5 6

1

34

5

6 1

2

45

6

1 2

3

56

1

2 3

4

Warmup: Draw the symmetries for the triangle.(1) How many symmetries are there?(2) If we call the move “rotate clockwise” r, what is the order of r? Isthere a way to write r−1 in terms of some positive power of r?(3) If we call the move “flip across a vertical axis” s, what is the order ofs? Is there a way to write s−1 in terms of some positive power of s?(4) Note that rasb means “flip b times and then rotate a times” (readactions right to left like function composition). Now label each of thesymmetries by some rasb. Then label each of the symmetries by somesbra, and then by sbr−a, and compare all three forms.Repeat parts (1)–(4) for the square.

Review: Let G be a set. A binary operation ? on G is a function

? : G×G→ G.

A group is a pair (G, ?) consisting of a set G and a binaryoperation ? on G such that:

1. ? is associative, i.e. (a ? b) ? c = a ? (b ? c);

2. there is an identity element e ∈ G, i.e.

e ? g = g = g ? e for all g ∈ G;

3. every element of G has an inverse; i.e. for all g ∈ G, there isan element g−1 such that gg−1 = e = g−1g.

Favorite examples so far:

1. Zn,Qn,Rn,Cn under addition.

2. Q×,R×,C× under multiplication.

3. Z/nZ under addition.

4. (Z/nZ)× = {a ∈ Z/nZ | a is relatively prime to n} undermultiplication.

Order

The order of a group G, denoted |G|, is the size of the underlyingset.

For any element x ∈ G, if xn = e for some n ∈ Z>0, we say theorder of x is the smallest such n.

Theorem

1. An element x ∈ G has order 1 if and only if x = e.

2. xm = e iff |x| divides m.

1 r r2 r3 r4 r5

16

5

4 3

2

65

4

3 2

1

54

3

2 1

6

43

2

1 6

5

32

1

6 5

4

21

6

5 4

3

s rs r2s r3s r4s r5s

61

2

3 4

5

12

3

4 5

6

23

4

5 6

1

34

5

6 1

2

45

6

1 2

3

56

1

2 3

4

LetD2n = group of symmetries of a regular 2n-gon,

where symmetries means ways to move the 2n-gon so that theoutline ends looking the same, but the vertices have moved.Some properties:

(1) There are always 2n symmetries, i.e. D2n has order 2n.(2) The symmetries are, for example, generated by r =“rotate

clockwise (360/n)◦” and s =“flip over a vertical axis”.(3) The element r has order n, and the element s has order 2.(4) The elements r and s don’t commute, but they do satisfy

rs = sr−1 and sr = r−1s.

Group presentationsA subset of elements S ⊆ G with the property that every elementof G can be written as a finite product of elements of S and theirinverses is called a set of generators of G. We write 〈S〉 = G.

Ex: D2n is generated by S = {r, s}; Z is generated by 1; Z/nZ isgenerated by 1̄.

Any equations that are satisfied in G are called relations.

Ex: The generators S = {r, s} satisfy s2 = rn = 1 and rs = sr−1.

If a set of relations R has the property that any relation in G canbe derived from those in R then those generators and relationsform a presentation of G, written

〈generators | relations〉.

In short, a presentation is everything you need to build the group.Some examples:

D12 = 〈r, s | r6 = e, s2 = e, r−1s = sr〉Z/3Z = 〈1̄ | 1̄3 = e〉 (Yes, weird notation!) Z = 〈1 | ∅〉 = 〈1〉

(If there are no relations, i.e. R = ∅, we write G = 〈S〉.)


Ex: D2n is generated by S = {r, s}

; Z is generated by 1; Z/nZ isgenerated by 1̄.









Ex: D2n is generated by S = {r, s}; Z is generated by 1

; Z/nZ isgenerated by 1̄.














In short, a presentation is everything you need to build the group.

Some examples:D12 = 〈r, s | r6 = e, s2 = e, r−1s = sr〉

Z/3Z = 〈1̄ | 1̄3 = e〉 (Yes, weird notation!) Z = 〈1 | ∅〉 = 〈1〉(If there are no relations, i.e. R = ∅, we write G = 〈S〉.)








D12 = 〈r, s | r6 = e, s2 = e, r−1s = sr〉

Z/3Z = 〈1̄ | 1̄3 = e〉 (Yes, weird notation!) Z = 〈1 | ∅〉 = 〈1〉(If there are no relations, i.e. R = ∅, we write G = 〈S〉.)








D12 = 〈r, s | r6 = e, s2 = e, r−1s = sr〉Z/3Z = 〈1̄ | 1̄3 = e〉 (Yes, weird notation!)

Z = 〈1 | ∅〉 = 〈1〉(If there are no relations, i.e. R = ∅, we write G = 〈S〉.)

Intuition from linear algebraGenerators are like “spanning sets” from linear algebra.

For example, let G = Z2. Then x = (1, 0) generates

x+ x = (2, 0), x+ x+ x = (3, 0), . . . ,

and also

x−1 = (−1, 0) x−1 + x−1 = (−2, 0), . . . .

Throwing in y = (0, 1) you also get

y−1 = (0,−1), x+ y = (1, 1), etc..

So S = {x, y} generates Z2. The only additional information youneed to define the group is that xy = yx. So

Z2 = 〈x, y | xy = yx〉.

A minimum set of generators is like a basis from linear algebra.CAUTION!! Minimum versus minimal: Z = 〈1〉 = 〈2, 3〉.

Intuition from linear algebraGenerators are like “spanning sets” from linear algebra.For example, let G = Z2. Then x = (1, 0) generates

x+ x = (2, 0), x+ x+ x = (3, 0), . . . ,

and also

x−1 = (−1, 0) x−1 + x−1 = (−2, 0), . . . .


y−1 = (0,−1), x+ y = (1, 1), etc..


Z2 = 〈x, y | xy = yx〉.



x+ x = (2, 0), x+ x+ x = (3, 0), . . . ,

and also

x−1 = (−1, 0) x−1 + x−1 = (−2, 0), . . . .


y−1 = (0,−1), x+ y = (1, 1), etc..

So S = {x, y} generates Z2.

The only additional information youneed to define the group is that xy = yx. So

Z2 = 〈x, y | xy = yx〉.



x+ x = (2, 0), x+ x+ x = (3, 0), . . . ,

and also

x−1 = (−1, 0) x−1 + x−1 = (−2, 0), . . . .


y−1 = (0,−1), x+ y = (1, 1), etc..


Z2 = 〈x, y | xy = yx〉.


ExampleLet G be the group

G = 〈a, b | a2 = b2 = e, aba = bab〉

Now a−1 = a and b−1 = b.

Other ways of writing aba = bab:

abab = ba ababa = b ababab = e bbaaaba = bab · · ·

Claim: G has 6 elements.

ExampleLet G be the group

G = 〈a, b | a2 = b2 = e, aba = bab〉Now a−1 = a and b−1 = b.

Other ways of writing aba = bab:

abab = ba ababa = b ababab = e bbaaaba = bab · · ·

Claim: G has 6 elements.

Presentation problem

Question: When are two presentations 〈S1 | R1〉 and 〈S2 | R2〉actually presentations for the same group?

HARD

Claim: If we let a = s and b = rs, then

〈a, b | a2 = b2 = e, aba = bab〉 = 〈r, s | r3 = s2 = e, rs = sr−1〉 = D6

Presentation problem

Question: When are two presentations 〈S1 | R1〉 and 〈S2 | R2〉actually presentations for the same group? HARD

Claim: If we let a = s and b = rs, then

〈a, b | a2 = b2 = e, aba = bab〉 = 〈r, s | r3 = s2 = e, rs = sr−1〉 = D6

The symmetric groupLet X be a finite non-empty set, and let SX be the set of bijectionsfrom the set to itself, i.e. the set of permutations of the elements.

For example, if X = {1, 2, 3} then SX contains

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

SX forms a group under function composition.

I A permutation σ followed by another permutation τ is τ ◦ σ,which is itself a permutation (binary operation)

I Function composition is associative.I The bijection x 7→ x for all x ∈ X serves as the identity.I Every bijection is invertible.

The group SX is called the symmetric group on X.

The symmetric groupLet X be a finite non-empty set, and let SX be the set of bijectionsfrom the set to itself, i.e. the set of permutations of the elements.For example, if X = {1, 2, 3} then SX contains

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3






1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3



I Function composition is associative.

I The bijection x 7→ x for all x ∈ X serves as the identity.I Every bijection is invertible.



1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3



I Function composition is associative.I The bijection x 7→ x for all x ∈ X serves as the identity.

I Every bijection is invertible.



1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3

1

1

2

2

3

3





The symmetric group

When X = [n] = {1, 2, . . . , n} we denote SX by Sn, and call it thesymmetric group of degree n.

Fact: It turns out that SX is essentially the same group as S|X|.(Later: they are isomorphic.)

Proposition

The order of Sn is |Sn| = n!.

Some notationPermutations can be represented in many ways:

σ =

1

1

2

2

3

3

4

4

5

5

6

6

7

7

means σ(1) = 3, σ(2) = 4, etc.

(Cauchy’s) two-line notation:

σ =

(1 2 3 4 5 6 73 4 2 1 7 6 5

)

One-line notation: σ = 3421765Cycle notation:

1

3 2

4 5

6

7

denoted by (1324)(57)(6)

or just (1324)(57)


σ =

1

1

2

2

3

3

4

4

5

5

6

6

7

7

means σ(1) = 3, σ(2) = 4, etc.


σ =

(1 2 3 4 5 6 73 4 2 1 7 6 5

)

One-line notation: σ = 3421765

Cycle notation:

1

3 2

4 5

6

7

denoted by (1324)(57)(6)

or just (1324)(57)


σ =

1

1

2

2

3

3

4

4

5

5

6

6

7

7

means σ(1) = 3, σ(2) = 4, etc.


σ =

(1 2 3 4 5 6 73 4 2 1 7 6 5

)


1

3 2

4 5

6

7

denoted by (1324)(57)(6)

or just (1324)(57)


σ =

1

1

2

2

3

3

4

4

5

5

6

6

7

7

means σ(1) = 3, σ(2) = 4, etc.


σ =

(1 2 3 4 5 6 73 4 2 1 7 6 5

)


1

3 2

4 5

6

7denoted by (1324)(57)(6)

or just (1324)(57)

Multiplication of diagrams:

σ =

1

1

2

2

3

3

4

4

5

5

τ =

1

1

2

2

3

3

4

4

5

5

Cycles

A cycle is a string of integers what represents the element of Snthat cyclically permutes these integers (and fixes all others).

Specifically,

(a1a2 . . . a`) sends

a1 7→ a2a2 7→ a3

...a` 7→ a1

Example: (435) is the permutation in Sn that sends 4 to 3, 3 to5, 5 to 4, and everything else to itself.

Every permutation can be expressed as the product (composition)of cycles, usually in several ways.

Example: (13) = (31) = (23)(12)(23)(draw the permutation diagrams)

Cycles

A cycle is a string of integers what represents the element of Snthat cyclically permutes these integers (and fixes all others).Specifically,

(a1a2 . . . a`) sends

a1 7→ a2a2 7→ a3

...a` 7→ a1

Example: (435) is the permutation in Sn that sends 4 to 3, 3 to5, 5 to 4, and everything else to itself.

Every permutation can be expressed as the product (composition)of cycles, usually in several ways.

Example: (13) = (31) = (23)(12)(23)(draw the permutation diagrams)

Algorithm for writing a permutation in cycles:1. Start with “(1”.2. If a is the last element of the cycle, either:

(i) If σ(a) is the first element of the cycle, close the cycle. Ifthere are any numbers left unused, start a new cycle with theleast available number.

(ii) If σ(a) is the not first element of the cycle, add σ(a) next inthe cycle.

3. Repeat 2 until all numbers 1, . . . , n appear in some cycle.4. Delete any cycles of length 1.

σ =

1

1

2

2

3

3

4

4

5

5

τ =

1

1

2

2

3

3

4

4

5

5

Multiplying cycles:

Algorithm:

1. Start the first cycle of your answer with a = 1.

2. If the last number that you wrote in your answer is a, let x = a.

3. Look for the rightmost cycle with a x in it, and reset x to itssuccessor in that cycle.

4. Always moving left from cycle-to-cycle, look for the next cycle withx in it and replace it with it’s successor.

5. When you run out of cycles, write x as the successor of a in youranswer, unless. . .

6. If x was the first element of the current cycle of your answer, thenclose the cycle, and start a new one with the least number notappearing yet in your answer.

If σ = (1352) and τ = (123)(45) as before, thenστ = (1352)(123)(45) =

Multiplying cycles:Algorithm:

1. Start the first cycle of your answer with a = 1.

2. If the last number that you wrote in your answer is a, let x = a.

3. Look for the rightmost cycle with a x in it, and reset x to itssuccessor in that cycle.

4. Always moving left from cycle-to-cycle, look for the next cycle withx in it and replace it with it’s successor.

5. When you run out of cycles, write x as the successor of a in youranswer, unless. . .

6. If x was the first element of the current cycle of your answer, thenclose the cycle, and start a new one with the least number notappearing yet in your answer.

If σ = (1352) and τ = (123)(45) as before, thenστ = (1352)(123)(45) =

You try:

(a) Write in cycle notation:

σ1 =

1

1

2

2

3

3

4

4

5

5

6

6

7

7

σ2 =

1

1

2

2

3

3

4

4

5

5

6

6

7

7

(b) Draw the maps (like the diagrams above) for

τ1 = (137)(52) τ2 = (12)(34)(56)

(c) Use the cycle notation to compute σ1σ2 and τ2τ1. Check usingthe diagrams (stack σ1 on top of σ2 and resolve).

1 r r2 3 4 5 - City University of New York · 2017. 9. 5. · 1 r r2 r3 r4 r5 6 1 5 4 3 2 5 6 4 3 2...

Documents

Transcript of 1 r r2 3 4 5 - City University of New York · 2017. 9. 5. · 1 r r2 r3 r4 r5 6 1 5 4 3 2 5 6 4 3 2...