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Math Methods ILia Vas

Groups

Introduction of Groups via Symmetry Groups of MoleculesDetermination of the structure of molecules is one of many examples of the use of the group

theory. Electronic structure of a molecule can be determined via its geometric structure. In thiscase, one consider the symmetry of a molecule since it reveals information about its properties (i.e.,structure, spectra, polarity, chirality, etc).

The following diagram represents the above steps.

Geometry of a molecule −→ Symmetry Group

↑ ↓

Structural Properties ←− Group Representation

Symmetry of a molecule is characterized by the fact that it is possible (theoretically) to carry outoperations which will interchange the position of some (or all atoms) and result in the arrangementof atoms that is indistinguishable from the initial arrangements. Thus, the operations we shallconsider are exactly those that we can apply on a model of a molecule so that the resulting moleculeappears the same as the original one.

Operations are:

- rotations - physically possible, are called proper rotations.

- reflections with the respect to a mirror plane or to the center of symmetry - physicallyimpossible, are called improper rotations.

These set of all those operations is called a group of symmetries. The features of such groupthat we are interested in is that

1) a composite of two operation from the group is again an operation from the group.

Let us denote the composite of two operations a and b with a · b or, shorter ab. With this notation,the operations a, b and c of a symmetry group satisfy the associativity law:

2) a(bc) = (ab)c

There is a distinct element of every group of symmetry, called identity element and denoted by1 which corresponds of operation of not moving molecule at all (equivalently, rotation for 0 degrees).Thus, the identity element 1 satisfies

3) a1 = 1a = a for every operation a.

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Finally, for every operation a defines the corresponding operation, denoted a−1 reversing theeffect of a. For example, if a is a rotation for α degrees, then a−1 is the rotation for α degrees in theopposite direction (equivalently, rotation for −α degrees). Thus, the composition of a and a−1 is theidentity operation.

4) aa−1 = a−1a = 1.

These four laws are independent of the chemical setting and, as it turned out, there are manyother situations in which the set of elements considered satisfies these four laws. So, it turned outthat the study of any structure satisfying the above four laws, a group, was of interest for manydisciplines. The study of group became known as the group theory.

History

Historically, group was not defined in the con-text of chemistry and symmetries of molecules.There are three historical roots of group theory:1) the theory of algebraic equations, 2) number

theory and 3) geometry. Euler, Gauss, Lagrange,Abel and Galois were early researchers in the fieldof group theory. Galois is honored as the firstmathematician linking group theory and field the-ory, with the theory that is now called Galois the-ory.

Galois remains an intriguing and unique person in the history of mathematics. The footnotecontains some more information from wikipedia.org. 1

1

Évariste Galois (October 25, 1811 May 31, 1832) was a French mathematician born in Bourg-la-Reine. He was amathematical child prodigy. While still in his teens, he was able to determine a necessary and sufficient conditionfor a polynomial to be solvable by radicals, thereby solving a long-standing problem. His work laid the fundamentalfoundations for Galois theory, a major branch of abstract algebra, and the subfield of Galois connections. He was thefirst to use the word ”group” as a technical term in mathematics to represent a group of permutations. He died in aduel at the age of twenty.

In 1828 he attempted the entrance exam to École Polytechnique, without the usual preparation in mathematics,and failed. He failed yet again on the second, final attempt the next year. It is undisputed that Galois was morethan qualified; however, accounts differ on why he failed. The legend holds that he thought the exercise proposed tohim by the examiner to be of no interest, and, in exasperation, he threw the rag used to clean up chalk marks on theblackboard at the examiner’s head. More plausible accounts state that Galois refused to justify his statements andanswer the examiner’s questions. Galois’s behavior was perhaps influenced by the recent suicide of his father.

His memoir on equation theory would be submitted several times but was never published in his lifetime, due tovarious events. Initially he sent it to Cauchy, who told him his work overlapped with recent work of Abel. Galoisrevised his memoir and sent it to Fourier in early 1830, upon the advice of Cauchy, to be considered for the GrandPrix of the Academy. Unfortunately, Fourier died soon after, and the memoir was lost. The prize would be awardedthat year to Abel posthumously and also to Jacobi.

Despite the lost memoir, Galois published three papers that year, which laid the foundations for Galois Theory.Galois was a staunch Republican, famous for having toasted Louis-Philippe with a dagger above his cup, which

leads some to believe that his death in a duel was set up by the secret police. He was jailed for attending a Bastille

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Mathematical Definition of a GroupLet us give a precise mathematical definition of a group.

Definition. A group is a nonempty set G with operation · such that

A1 The result of operation applied to two elements of G is again an element of G (i.e. if a and bare in G, a · b is also in G). In this case we say that the operation · is closed).

A2 Associativity holds: (a · b) · c = a · (b · c) Operation with this property is said to be associative.

A3 There is identity element 1 so that a · 1 = 1 · a = a for every element a.

A4 Every element a has the inverse a−1 (i.e. a · a−1 = a−1 · a = 1).

As before, we will shorten the notation a · b and write just ab. Also, if G with operation · is agroup, we will say that G is a group under operation ·.

Note that A4 implies the cancellation law: ab = ac or ba = ca imply b = c.This can be verified by multiplying ab = ac from the left with a−1 to get that b = c. Similarly,

multiply ba = ca from the right with a−1 to get b = c.

Examples.

1. Real numbers without 0 under multiplications. The identity element is 1 and 1/a is the inverseof a. Note that rational or complex numbers without 0 under multiplication also form groups.Integers without 0 under multiplication, however, are not a group as A4 is not satisfied forevery integer a. Give an example of this.

Note also that we have to throw out 0 as all real numbers under multiplication are not a group.Indeed, the element 0 does not have an inverse as there is no solution of the equation 0x = 1(i.e. we cannot divide with 0).

2. Real numbers under addition. This is an important example because it tells us that the notationfor operation in specific group might be denoted differently that the one used in the generaldefinition. Note that in this example the identity element is 0 and the inverse of a is −a. Theessential fact is that the laws A1–A4 remain to hold regardless of the change in notation:

A1 a and b are real numbers, a + b is also a real number.

A2 (a + b) + c = a + (b + c).

Day protest in 1831, and was released only 2 days before his death.The night before the duel, supposedly fought in order to defend the honor of a woman, he was so convinced of

his impending death that he stayed up all night writing letters to his Republican friends and composing what wouldbecome his mathematical testament. Hermann Weyl, one of the greatest mathematicians of the 20th century, said of

this testament, ”This letter, if judged by the novelty and profundity of ideas it contains, is perhaps the most substantialpiece of writing in the whole literature of mankind.” In his final papers he outlined the rough edges of some work hehad been doing in analysis and annotated a copy of the manuscript submitted to the academy and other papers. Onthe 30th of May 1832, early in the morning, he was shot in the abdomen and died the following day at ten in theCochin hospital (probably of peritonitis) after refusing the offices of a priest. He was 20 years old. His last words tohis brother Alfred were: ”Don’t cry, Alfred! I need all my courage to die at twenty.”

Galois’ mathematical contributions were finally fully published in 1843 when Liouville reviewed his manuscript anddeclared that he had indeed solved the problem first proposed and also solved by Abel. The manuscript was finallypublished in the October-November 1846.

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if the table is symmetric with respect to the main diagonal. Using this rule, we conclude that thegroup with the above Cayley table is abelian.

Groups with 2 elements. Let us use Cayley table to try to describe all the groups with 2elements a, b. As one of them has to be identity, let us take a = 1 so ab = ba = b and aa = a. As the

result of bb has to be different than ba, bb has to be a. Hence,· a ba a b

b b a

is the resulting table.

Note that this is the only possible way we could fill the table given the condition that a is theidentity. If we chose b to be identity and rearrange the heading column so that the identity is the

first element of the heading column or row, we would end up with the following table.· b ab b aa a b

This table describes the same structure of the group, the only difference is the names we assignedto the elements. Having the correspondence a → b and b → a from first table to the second, wewould end up with the same group. The above correspondence is an example of group isomorphism.The two groups are isomorphic if you can relabel the elements of one group then producing theother. As isomorphic groups are intrinsically the same, mathematicians are often considering themas one. One of the most important questions in group theory is to determine if a given two groupsare isomorphic or not. Note that some chemists refer to isomorphic groups as isomorphous groups(e.g. in Kettle’s Symmetry and Structure).

A group isomorphism preserves all the group properties. For example, if one group is abelianand the other is not, then they cannot be isomorphic. This gives you useful criterion for determiningthat two groups are not isomorphic.

• To demonstrate that two groups are isomorphic, you can match their elements and showthat the matching preserves the Cayley table. We shall see later that two groups having thesame presentation also shows their isomorphism.

• To demonstrate that two groups are not isomorphic, you can note that they do not sharethe same properties. For example, having different number of elements, one being abelian andthe other not, elements having different order etc.

We present further examples of groups isomorphic to the groups above. Let us consider, thegroup of two integers 1 and -1 under the multiplication. This is another group isomorphic to theabove group with elements a and b. Yet another example is the group of remainders when dividingwith 2. Note that when any integer is divided by 2, the remainder is either 0 or 1.

Thus 1 + 0 = 0 + 1 = 1, 0 + 0 = 0. As 1+1=2and 2 has remainder 0 when divided by 2, we have1+1=0. Hence the following table represents this

group.

+ 0 10 0 1

1 1 0

This is another example of a group isomorphic with the above three groups with two elementswe considered. In mathematics, this group is denoted by Z 2. In chemistry, the notation C 2 is used.

We concluded that all the groups with two elements have to be isomorphic to each other. Wecan take C 2 to be the representative of all the groups with two elements.

Groups of 3 elements.

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Let us consider the groups of three elements.As one of the three elements has to be identity,let us denote elements by 1, a and b and let usstart filling the Cayley table.

· 1 a b1 1 a ba ab b

If we put 1 in the first empty place in the second row of the table (again the rows are counted

without the heading elements), then we will have to put b in the last free place in the second row if we do not want to violate the rule for each element in each row appearing exactly once. But then bwill appear twice in the last column, so we cannot fill the table this way.

This mean that the second row has to be a,b, 1and this uniquely determines the last row andhence the entire table. So, the following is thecomplete Cayley table.

· 1 a b1 1 a ba a b 1b b 1 a

Note that this group is isomorphic to the group from example on page 6. Another group of three

elements can be obtained considering remainders when dividing with 3, similarly as in the examplewith C 2. Any integer has a remainder when divided by 3 either 0, 1 or 2. So, we take these threeelements to be the elements of the group.

To obtain the Cayley table, note that the ele-ments add considering the remainder of their sumwhen dividing by 3. For example 2+2=4 whichhas remainder 1 when divided by 3, so 2+2=1.

+ 0 1 20 0 1 21 1 2 02 2 0 1

Note that the identity element here is 0, not 1 since we are using the additive, not multiplica-

tive notation. Analogously to the reminders when dividing by 2, this group is denoted by Z 3 (inmathematics) or C 3 (in chemistry).

As we have seen, there is just one (isomorphism type of a) group with three elements. As thenext section will show, for groups with more elements, the situation can be different.

Groups of 4 elements. As we will see, the situation when describing the groups of 4 ele-ments is more complex. So, before considering the groups with 4 elements, consider some additionalobservations that will help us create different Cayley tables.

The rule A4 has the consequence that ab = 1 implies ba = 1.If a2 = 1 then ab = 1 = ba for all b = a. This means that each non-identity element is either its

own inverse or it pairs up with another element which is its unique inverse creating the two possible

situations:

· ... a ... b ......a 1...b 1

...

· ... a ......a 1...

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When trying to classify the finite groups using the Cayley table, it is possible to rearrange theorder of the labeled columns such that the first columns to appear, reading from left to right, arethose representing self-inverse elements, starting first with the identity element which is always aself-inverse. After all the columns of self-inverses should follow the columns representing pairs of inverses: each pair of inverses should be represented by a pair of adjacent columns. Following thisrule, in the case of groups with four elements 1, a,b, and c we have two possibilities.

· 1 a b c1 1 a b ca a 1b b 1c c 1

· 1 a b c1 1 a b ca a 1b b 1c c 1

These resulting arrangement of identity elements are called the identity skeletons of the Cayleytable. It is usually useful to write the identity skeleton when filling the Cayley table of a group.Groups with different identity skeletons cannot be isomorphic. An identity skeleton boils downsimply to numbers of self-inverses versus number of inverse pairs.

Since the remainder of these two tables can be filled on unique way, we obtained that there areexactly two classes of non-isomorphic groups with four elements and that they are

· 1 a b c1 1 a b ca a 1 c bb b c 1 ac c b a 1

· 1 a b c1 1 a b ca a 1 c bb b c a 1c c b 1 a

The first one is the Cartesian product C 2 × C 2 (set of ordered pairs of elements from C 2). Thecorrespondence here is

1 → (0, 0), a → (1, 0), b → (0, 1) and c → (1, 1).

The second table is isomorphic to the group C 4, the group of remainders when dividing with 4.The correspondence here is

1 → 0, a → 2, b → 1 and c → 3.

Classes of GroupsThere are many different classes of groups. Classifying different groups is one of the largest

problems of group theory. We will mention some important classes of groups.Cyclic groups C n. Cyclic groups are those that are generated with a single element. This

means that every non-identity element of a cyclic group can be obtained from that single element.For example, the set of integers under addition is generated by 1. It turns out that all the infinitecyclic groups are isomorphic to this one.

Using analogous notation as for C 2 and C 3, letus define C n to be the group of remainders whendividing by n. For example, C 5 has the followingCayley table:

+ 0 1 2 3 40 0 1 2 3 41 1 2 3 4 02 2 3 4 0 13 3 4 0 1 24 4 0 1 2 38


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All these groups are cyclic since the element 1 generates the entire group. Even more is true: everycyclic group with n elements is isomorphic to C n. We can illustrate this by the following example.

Consider the cyclic group of 5 elements. Sincethe group is cyclic, there is an element a thatgenerates it. In this case, the group elements are1, a , a2, a3, a4. The element a5 has to be 1 sincethe group would have more than 5 elements oth-erwise. In this case, the following is the Cayley

table of this group.· 1 a a2 a3 a4

1 1 a a2 a3 a4

a a a2 a3 a4 1a2 a2 a3 a4 1 aa3 a3 a4 1 a a2

a4 a4 1 a a2 a3

So, we can see that the above two Cayley tables are isomorphic by

0 → 1, 1 → a, 2 → a2, 3 → a3 and 4 → a4.

One can also see that every cyclic group is abelian.Order of group. Order of group element. If a group G has n elements, we say it has order

n. If an element a is such that an = 1 and am == 1 for any m < n, we say that a has order n.Presentation of Cyclic Groups. Consider the cyclic group of order n. If we denote the

generator by a, the group is uniquely determined by the relation an = 1. For example, if n = 5, wecan determine the above Cayley table simply knowing (1) that there is just one generator a, (2) thatthis generator satisfies the equation a5 = 1. Analogously, any group can be determined by the list of generators and the list of relations these generators satisfy. This is called the group presentation.In a group presentation, the generators are listed followed by relations among them. For example,the group presentation of the cyclic group of order n is

a|an = 1.

As the infinite cyclic group does not have any relations on its generator, the presentation is a. Thisgroup is isomorphic to the group of integers under addition via the isomorphism n → an.

Dihedral Group Dn and its presentation. Dihedral groups are groups of symmetries of regular polygons. The group of symmetries of a regular polygon of n sides is denoted by Dn.

For example, let us consider a square. Thesymmetries of a square are: rotations for 0, 90,180 and 270 degrees, reflections with respect todiagonals and x and y axes (if the square is cen-tered at the origin so that the sides are parallel tox or y axis). Clearly, the rotation for 0 degrees isthe identity, let us denote it with 1. If we denotethe rotation for 90 degrees by a, then the rota-tions by 180 and 270 degrees are a2 and a3 andthen a4 = 1.

Let us denote the reflection with respect to y-axis by b. Then b2 = 1, ab is reflection with respectto the main diagonal, a2b is reflection with respect to x-axis and a3b is reflection with respect to thenon-main diagonal. So, all the symmetries of the square can be written via a and b. This meansthat a and b are generators of D4. Also, note that ba = a

3b.

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Since ba is the reflection with respect to non-main diagonal, it is different than ab, the reflec-tion with respect to main diagonal. So, D4 is notabelian.

The equations a4 = 1, b2 = 1, and ba = a3b onthe generators a and b completely determine the

Cayley table of D4. This means that the groupD4 is given by the presentation

a, b|a4 = 1, b2 = 1, ba = a3b.

Using this presentation, we can fill the table.

· 1 a a2 a3 b ab a2b a3b1 1 a a2 a3 b ab a2b a3ba a a2 a3 1 ab a2b a3b ba2 a2 a3 1 a a2b a3b b aba3 a3 1 a a2 a3b b ab a2bb b a3b a2b ab 1 a3 a2 a

ab ab b a3b a2b a 1 a3 a2

a2b a2b ab b a3b a2 a 1 a3

a3b a3b a2b ab b a3 a2 a 1

Every dihedral group has analogous presentation. The group Dn has two generators a, therotation for 360/n degrees, and b, reflection with respect to y axis, and it satisfies an = 1, b2 = 1,ba = an−1b. Thus,

Dn = a, b|an = 1, b2 = 1, ba = an−1b.

Direct Product of Groups and their presentations. As the plane of real numbers is obtained

by considering ordered pairs of real numbers, a new group can be obtained by considering orderedpairs of elements from two other groups. More precisely, if G1 and G2 are two groups, we can definea new group G = G1 × G2 by considering the elements of G to be the ordered pairs (g1, g2) where g1is from G1 and g2 is from G2. The operation in G is defined on the following way:

(g1, g2) · (h1, h2) = (g1h1, g2h2).

If G1 has n elements and G2 has m elements, then G1 × G2 has mn elements.If we know the presentations of G1 and G2, the presentation of G1×G2 is obtained on the following

way

- generators are all of G1 and of G2

- the relations are all of G1, G2 plus relations that assert that the generators of G1 commutewith the generators of G2

Example 1. The group C 3 × C 2 = a|a3 = 1 × b|b2 = 1 has the presentation

a, b|a3 = b2 = 1, ab = ba.

So, this group has 6 elements 1, a , a3,b,ab,a2b as any other ”word” in two letters a and b can bewritten as one of those 6 using the above relations.

Example 2. As another example, let us look at D3 × C

2. As D

3 = a, b|an = b2 = 1ba = a2b

and C 2 = c|c2 = 1, the presentation of D3 × C 2 is

a,b,c|an = b2 = c2 = 1, ba = a2b,ac = ca, bc = cb.

So, this group has 12 elements

1, a , a2,b,ab,a2b c, ac, a2c,bc,abc,a2bc

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as any other ”word” in three letters a, b and c can be written as one of those 12 using the aboverelations.

If any of the groups G1 or G2 is infinite, then G1 × G2 is infinite. We have seen examples of directproduct of groups. The first example was the group of vectors in real plane under addition. Thisgroup is denoted by R2 and it is the direct product of real numbers R under addition with itself.We can consider a direct product of R2 and R (denoted by R3). This is the real three dimensional

space. It is a group under addition. On page 9, we have seen one example of a direct product of finite groups when considering a group of four elements C 2 × C 2.

Direct product of Cyclic Groups. Let us consider the group C 3 × C 2 with presentationa, b|a3 = b2 = 1, ab = ba from Example 1 and compare it with the cyclic group of order 6 C 6 =c|c6 = 1. We claim that these two groups are isomorphic.

To note that, note that the element ab has order 6. Indeed

(ab)0 (ab)1 (ab)2 (ab)3 (ab)4 (ab)5 (ab)6

1 ab a2 b a a2b 1

Then note that this can be matched with the powers of element c in C 6 = c|c6 = 1.

(ab)0 (ab)1 (ab)2 (ab)3 (ab)4 (ab)5 (ab)6

1 ab a2 b a a2b 1

1 c c2 c3 c4 c5 c6

Comparing the Cayley’s tables shows that the pairing of the elements above really is the isomorphismof the two groups.

C 3 × C 2 1 ab a2 b a a2b

1 1 ab a2 b a a2bab ab a2 b a a2b 1

a2 a2 b a a2b 1 abb b a a2b 1 ab a2

a a a2b 1 ab a2 ba2b a2b 1 ab a2 b a

C 6 1 c c2 c3 c4 c5

1 1 c c2 c3 c4 c5

c c c2 c3 c4 c5 1

c2 c2 c3 c4 c5 1 cc3 c3 c4 c5 1 c c2

c4 c4 c5 1 c c2 c3

c5 c5 1 c c2 c3 c4

On the other hand, the groups C 2 × C 2 and C 4 are not isomorphic. One way to see that is bynoting that the square of all elements in C 2 × C 2 is the identity while the square of the generator of C 4 is not identity.

Another way to see that C 2 × C 2 and C 4 are not isomorphic is to compare the identity skeletonsof the Cayley’s tables.

C 2 × C 2 1 a b ab1 1 a b aba a 1 ab bb b ab 1 a

ab ab b a 1

C 4 1 c c

2

c

3

1 1 c c2 c3

c c c2 c3 1c2 c2 c3 1 cc3 c3 1 c c2

Our observations are just special cases of the following claim.

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C m × C n is isomorphic to C mn if and only if m and n are relatively prime(i.e. the greatest common divisor of m and n is 1).

To prove this claim, you can argue as we did in two examples above:

• If m and n are relatively prime, the element ab will have order mn so you can define theisomorphism by mapping ab → c. Note that this determines the images of the rest of theelements just like in the example with m = 3 and n = 2.

• If m and n have the largest common divisor d > 1, then the group C m × C n does not havean element of order mn (that is, all its elements are of order smaller than mn). On the otherhand, the generator c of C mn has the order mn.

This claim makes possible to determine all the isomorphism classes of abelian groups of certain(finite) order. We illustrate that by the following example.

Example 3. Produce all isomorphism classes of abelian groups of order 24.

Solution. Write 24 as product of powers of prime numbers: 24 = 8 · 3 = 23 · 3. Since 2 and3 are relatively prime, the group C 3 can be combined with any group of C 2 × C 2 × C 2, C 4 × C 2,or C 8, creating isomorphic pairs of groups. Any of these three groups, on the other hand, are notisomorphic because the order of elements do not match. Thus, there are 3 non-isomorphic abeliangroups of order 24

1. C 3 × C 2 × C 2 × C 2 ∼= C 6 × C 2 × C 22. C 3 × C 4 × C 2 ∼= C 12 × C 2 ∼= C 6 × C 43. C 3 × C 8 ∼= C 24

As a large percentage of point groups encountered are cyclic, dihedral, product of two cyclic or

product of a cyclic and a dihedral, let us look more closely to those examples.

Group notation no. of el. presentation

Cyclic (order n) C n n a|an = 1

Product of 2 cyclic C n × C m mn a, b|an = bm = 1, ba = ab

Dihedral Dn 2n a, b|an = b2 = 1, ba = an−1b

Product of Dn and C m Dn × C m 2nm see below

Using the above analysis for getting a presentation of a product, the presentation of Dn × C m, is

a,b,c|an = b2 = cm = 1, ba = an−1b,ca = ac, bc = cb

We will be specially interested in the case when m = 2 both when considering C n × C m andDn × C m.

We should also note the distinction between C n × C 2 and Dn. Both of these two groups have 2nelements. They are not isomorphic as C n × C 2 is abelian, while Dn is not. This can be seen the bestwhen comparing their presentations

a, b|an = b2 = 1, ba = ab and a, b|an = b2 = 1, ba = an−1b.

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4. Symmetric Groups. Let us consider a set {1, 2, 3}. Let us look at all the possible permuta-tions of this set (i.e. one-to-one mappings of this set onto itself). As when the symmetries of polygons were considered, the product of two such mappings is their composition. There are 6such mappings mapping (1, 2, 3) to

(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), and (3, 2, 1).

This groups is called the symmetric group on 3 letters and is denoted by S 3. Analogously, thepermutations on n elements form a group denoted by S n. S n has n! elements.

Symmetric groups are especially important in group theory because of the Cayley’s theoremstating that every group can be represented as a subgroup of some symmetric group.

There is another important class of groups called alternating groups An. They are related toS n. One of the presentations will focus on alternating groups An.

5. Symmetries of Platonic Solids: tetrahedral T d, octahedral Oh and icosahedral group I h.

Platonic or regular solids are convex poly-

hedra with equivalent faces composed of con-gruent convex regular polygons. Euclid provedthat there are exactly five such solids: thecube, dodecahedron, icosahedron, octahedron,and tetrahedron.

Plato 2 related these geometrical shapes to classical elements.

The space limitations that reduce the number of regular three-dimensional solids to only fiveare the following:

1. Each vertex of the solid must coincide with one vertex each of at least three faces.2History. (from wikipedia) The Platonic solids are named after Plato, who wrote about them in Timaeus. Plato

learned about these solids from his friend Theaetetus. Plato conceived the four classical elements as atoms with thegeometrical shapes of four of the five platonic solids that had been discovered by the Pythagoreans (in the Timaeus).These are, of course, not the true shapes of atoms; but it turns out that they are some of the true shapes of packedatoms and molecules, namely crystals: The mineral salt sodium chloride occurs in cubic crystals, fluorite (calciumfluoride) in octahedra, and pyrite in dodecahedra.

This concept linked fire with the tetrahedron, earth with the cube, air with the octahedron and water with theicosahedron. There was intuitive justification for these associations: the heat of fire feels sharp and stabbing (likelittle tetrahedra). Air is made of the octahedron; its minuscule components are so smooth that one can barely feelit. Water, the icosahedron, flows out of one’s hand when picked up, as if it is made of tiny little balls. By contrast,a highly un-spherical solid, the hexahedron (cube) represents earth. These clumsy little solids cause dirt to crumbleand breaks when picked up, in stark difference to the smooth flow of water.

The fifth Platonic Solid, the dodecahedron, Plato obscurely remarks, ”...the god used for arranging the constellationson the whole heaven” (Timaeus 55). He didn’t really know what else to do with it. Aristotle added a fifth element,aithêr (aether in Latin, ”ether” in English) and postulated that the heavens were made of this element, but he hadno interest in matching it with Plato’s fifth solid.

Historically, Johannes Kepler followed the custom of the Renaissance in making mathematical correspondences,and identified the five platonic solids with the five planets Mercury, Venus, Mars, Jupiter, Saturn which themselvesrepresented the five classical elements.

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2. At each vertex of the solid, the total, among the adjacent faces, of the angles betweentheir respective adjacent sides must be less than 360.

3. The angles at all vertices of all faces of a Platonic solid are identical, so each vertex of each face must contribute less than 360/3=120.

4. Regular polygons of six or more sides have only angles of 120 or more, so the commonface must be the triangle, square, or pentagon. And for:

- Triangular faces: each vertex of a regular triangle is 60, so a shape should be possiblewith 3, 4, or 5 triangles meeting at a vertex; these are the tetrahedron, octahedron,and icosahedron respectively.

- Square faces: each vertex of a square is 90, so there is only one arrangement possiblewith three faces at a vertex, the cube.

- Pentagonal faces: each vertex is 108; again, only one arrangement, of three faces at avertex is possible, the dodecahedron.

The tetrahedral group T d is the group of symmetries of the tetrahedron. It has order 24 and

is isomorphic to the group S 4.The icosahedral group I h is the group of symmetries of the icosahedron and dodecahedronhaving order 120, equivalent to the group direct product A5 × C 2 (where A5 is the alternating groupand C 2 is the cyclic group).

The octahedral group Oh is the group of symmetries of the octahedron and the cube. It isisomorphic to S 4 × C 2 and has order 48.

There are many other classes of groups but we concentrated here to those that appear in chemistrywhen considering groups of symmetries of molecules.

Finding a good classification for groups (i.e. finding classes that can describe various types of groups well) and finding a good way to represent various abstract groups are two very difficult tasksof the group theory. Group representation is a subfield of group theory that deals with these

issues.SubgroupsA nonempty subset H of a group G is a subgroup if the elements of H form a group under the

operation from G restricted to H.Every group has a subgroup consisting of identity element alone. This is called the trivial sub-

group. The identity elements is an element of every subgroup of a group.The entire group is a subgroup of itself. This is called the improper subgroup. The more

interesting examples are of nontrivial and proper subgroups.

Examples.

1. Set of even integers is a subgroup of all integers under addition.

2. Set of positive numbers is a subgroup of all real numbers different from 0 under multiplication.

3. Vectors co-linear with x-axis are a subgroup of all vectors in a real plane under addition.

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4. Let determine the subgroups of the two non-isomorphic groups of order 4 G1 and G2 with the

tables

G1 1 a b c1 1 a b ca a 1 c bb b c 1 ac c b a 1

and

G2 1 a b c1 1 a b ca a 1 c bb b c a 1c c b 1 a

.

Consider the left upper 2 × 2 part of the table for G1, we see that it contains just the elements

1 and a.· 1 a1 1 aa a 1

Note that this is a group for itself. So, this is a subgroup of G1. The same

holds for· 1 b1 1 bb b 1

and· 1 c1 1 cc c 1

An non-example is

1 a b1 1 a ba a 1 cb b c 1

as the product of a

and b is c and c is not among the heading elements. So, the set {1, a , b} is not a subgroup of G1. Thus, the list of all the subgroups of G1 is: {1}, {1, a}, {1, b}, {1, c}, and G1.

Turning to G2 we can conclude that there is just one nontrivial and proper subgroup and thisis {1, a}. To prove this, suppose that there is a proper nontrivial subgroup H of G2 containingb (similar proof works if you take c) instead of b. As bb = a, a must be in H as well. But ab = c,so then c must be in H as well and so H = G2 so it is not proper.

In general, determining the number of subgroups of a given can be very difficult.3

Symmetry (Point) Groups of Molecules

A point group is a group of symmetry operations which all leave at least one point unmoved.These groups have the following operations as their basics elements. Note that if two operations arein the group, then their composition is also an element of that group.

Element Operation

Identity E not moving anything (i.e. rotation for 0 degrees)Symmetry plane σ (σh, σv, σd) Reflection with respect to a plane (horizontal, vertical, dihedral)

Inversion center i Reflection with respect to the originProper axis C n Rotation by 360/n degrees

Improper axis S n C n followed by σh

3Another ”unsolvable” problem regarding groups is the so called word problem. The following two paragraphsare taken from wikipedia.

The word problem for groups is the problem of deciding whether two given words of a presentation of a grouprepresent the same element. There exists no general algorithm for this problem, as was shown by Pyotr SergeyevichNovikov. The proof was announced in 1952 and published in 1955. A much simpler proof was obtained by Boone in1959.

The word problem is only concerned with finitely presented groups, i.e. those groups which can be specified byfinitely many generators and finitely many relations among those generators. A word is a product of generators, andtwo such words may denote the same element of the group even if they appear to be different, because by using thegroup axioms and the given relations it may be possible to transform one word into the other. The problem then isto find an algorithm which for any two given words decides whether they denote the same group element.

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The set of all possible symmetries of each molecule constitutes a group. This set of operationsdefine the point group of the molecule.

There is a step-by-step algorithm that assigns a molecule to a point group (some of you maycover it in a chemistry course). In this course, we will be interested both in understanding themathematical structure of point groups as well as the process of assigning one to a given molecule.

Type of point group notation

cyclic C ncyclic with horizontal planes C nh

cyclic with vertical planes C nvnon-axial C i, C sdihedral Dn

dihedral with horizontal planes Dnhdihedral with planes between axes Dnd

improper rotation S 2ncubic groups I, I h, O , Oh, T , T h, T d

linear C ∞, C ∞v, C ∞h, D∞, D∞h

Word of caution: In chemistry, the same letter is used to denote both a group and an elementof a group. For example, a cyclic group of order n is denoted by C n but the generating element isalso denoted by C n. One should keep this is mind always when working with the point groups.

Chem. Math. no. of el. presentation

C n C n n a|an = 1

C nh C n × C 2 2n a, b|an = b2 = 1, ba = ab

C nv Dn 2n a, b|an = b2 = 1, ba = an−1b

C i, C s C 2 2 b|b2 = 1

Dn Dn 2n a, b|an = b2 = 1, ba = an−1b

Dnh C nv × C 2 = Dn × C 2 4n see belowDnd D2n 4n a, b|a

2n = 1, b2 = 1, ba = a2n−1bS 2n C 2n 2n a|a

2n = 1I A5 60 a, b|a

2 = b3 = (ab)5 = 1I h A5 × C 2 120 a,b,c|a

2 = b3 = (ab)5 = 1, ac = ca, bc = cbO S 4 24 a, b|a

2 = b3 = (ab)4 = 1Oh S 4 × C 2 48 a,b,c|a

2 = b3 = (ab)4 = 1, ac = ca, bc = cbT A4 12 a, b|a

2 = b3 = (ab)3 = 1T h A4 × C 2 24 a,b,c|a

2 = b3 = (ab)3 = 1, ac = ca, bc = cbT d S 4 24 a, b|a

2 = b3 = (ab)4 = 1

C ∞ C ∞ = SO(2, R) ∞ no finite presentationC ∞v D∞ ∞ no finite presentationC ∞h C ∞ × C 2 ∞ no finite presentationD∞ D∞ ∞ no finite presentation

D∞h D∞ × C 2 ∞ no finite presentation

Note that some of these groups are isomorphic, so they do not have any differences significant formathematician, but are significantly different from a chemical point of view.

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Let us concentrate first at the first eight groups in the above table. All of them are either dihedral,cyclic, products of two cyclic or products of dihedral and cyclic groups. In the above representations,the element a denotes the rotation and the element b a symmetry or, in the case of C i, inversion i.

• If there is just one generator, the group is cyclic. This is the case for C n, S 2n, C s and C i.

• If there are two generators, a rotation a of order n or 2n and a symmetry b, the group will be

determined by the fact if the generators commute or not.

If they do, then we have a direct product of cyclic group generated with a and C 2, generatedwith b. This will happen in the cases when b is the symmetry with respect to horizontal planebecause then it commutes with the rotation. Note that C nh = C n × C 2.

If a and b do not commute, we have Dn or D2n. This will happen if b is the symmetry withrespect to vertical plane because then ba = a−1b. This is the case for C nv, Dn and Dnd.

• If there are three generators: rotation a, symmetry with respect to a vertical plane b, and asymmetry with respect to a horizontal plane c, these generators satisfy the relations

an

= b2

= c2

= 1, ba = an−1

b,bc = cb, ac = ca

and these relations define the group Dn × C 2 (Dnh in chemical notation).

In practice, not all values of n are possible. In crystallography, the feasible values of n are onlyn = 1, 2, 3, 4, 6, due to the crystallographic restriction theorem. In its basic form, this theorem isthe observation that the rotational symmetries of a crystal are limited to 2-fold, 3-fold, 4-fold, and6-fold. This is strictly true for the mathematical formalism, but in the physical world quasicrystalsoccur with other symmetries, such as 5-fold. (find out more at wikipedia.org). So, there are just 32crystallographic point groups.

• Let us turn our attention to point groups of linear molecules. Although none of these groupshave finite presentation, we can still describe them nicely enough. For example, group C ∞contains all the rotations around a fixed axis. So, we can identify it with the set of all possibleangles of rotations. As the rotation for 360 degrees is the same as rotation for 0 degrees, thisgroup is isomorphic to a group of all the angles represented on a unit circle. The angles areadded on an usual way. In mathematics, this group is known as SO(2, R). To follow notationin chemistry, we denote it by C ∞.

If x is a rotation in C ∞ and b the symmetry in C 2 (so b2 = 1) then either bx = xb (in case b is

a symmetry with respect to a horizontal plane) or bx = x−1b (if b is a symmetry with respectto a vertical plane). If bx = xb, then we have a direct product C ∞ × C 2 = C ∞h. If bx = x

−1b,then we have an infinite version of Dn that is denoted by D∞. Note that C ∞v = D∞.

If we have symmetry both with respect to vertical plane b and horizontal plane c, and x is anyrotation in C ∞, then we have the following relations

c2 = b2 = 1, bx = x−1b,cx = xc.

These relations define the group D∞h = D∞ × C 2. This group is the infinite version of thegroup Dnd.

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Examples.

1. Water H2O. This molecule has the following symmetries: identity E, rotation for 180 degreesC 2, reflections with the respect to the vertical plane σv and their product C 2σv, (which is alsoa reflection but with respect to a vertical plane perpendicular to the one used in σv).

Using the second table, we can conclude that

this group is C 2v. Writing down the Cayley ta-ble, we can see that this group is isomorphicto C 2 × C 2 = a, b|a

2 = b2 = 1, ba = ab. In theabove presentation a corresponds to rotationand b to one (any) of the symmetries.

2. Ethylene C2H4. If we place the molecule in thecoordinate system (lying in the xy-plane), thenthe 3 reflections, with respect to xy, yz and xz -planes, are elements of the point group of this

molecule. The remaining elements of the pointgroup are:

rotations for 180 degrees with respect to all 3 axis (we can denote them by C 2(x), C 2(y) andC 2(z )), inversion i and identity E . So, this group has 8 elements.

If we denote the reflections with a,b, and c, then the three rotations are ab, bc and abc.Inversion is the remaining element ac. Together with identity, this corresponds to the 8 elementspreviously described. So, we have 3 generators, all of order 2, giving us a group C 2 × C 2 × C 2 =C 2v × C 2 = D2 × C 2 = D2h.

3. Boron trifluoride BF3. There are two nontriv-

ial rotations: a rotation for 120 and a2

rotationfor 240 degrees. There are symmetries

with respect to vertical plane b and horizontal plane c. As a and c commute and ba = a2b, wehave D3 × C 2. Similarly as in previous example, this is Dnh-type and in this case n = 3. Thepresentation of this group is

a,b,c|a3 = 1, b2 = 1, c2 = 1, ca = ac, cb = bc, ba = a2b.

4. Bromine Pentafluoride BrF5. Four fluor atomsline in the same plane forming the vertices of asquare. Bromine atom is in the center of thatsquare and the remaining fluor atom is directlyabove the bromine.

Because of that fifth fluor, there are no symmetries with respect to horizontal plane.

Recall that Dn is the group of symmetries of a regular n-tagon. So, the group of symmetriesof a square is D4. We have two generators - rotation for 90 degrees a and the symmetry b with

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respect to (any of the four) vertical planes: two with respect to vertical planes (xz and yz axes)and two with respect to diagonals of the square. Thus, D4 has 8 elements: identity, 3 rotationsa, a2 and a3 and 4 symmetries b,ab, a2b and a3b.

So, the bromine pentafluoride molecules has the same symmetries as a square D4 = C 4v.

5. CHFClBr. All 5 atoms are different so just the

trivial symmetry is present. Thus, the pointgroup is the trivial (one element) group.

6. HClBrC-CHClBr. There is just one nontrivialoperation - the inversion with respect to thecenter. So, the group is C i = C 2 = b|b

2 = 1.

7. Hydrogen chloride HCl. All the rotations for any angle between 0 and 2π with 0 and 2πidentified are the elements of the point group of this molecule. These rotations constitute thegroup denoted C ∞ = SO(2, R). There is also the

symmetry with respect to the vertical planeb = σv (with respect to yz -plane if the moleculestands ”upright”). Since b does not commutewith the rotations, we have C ∞v = D∞.

8. Hydrogen H2. The hydrogen is a linear molecule with two identical atoms.

In addition to transformations from the pre-vious example, the symmetry with respect tohorizontal plane c = σh is also present. Sincec commutes with a and b, the group is D∞h =C ∞v × C 2.

There are many resources on the web detailing step-by-step process for finding the point group forany molecule and multimedia programs that helps you identify the point group of a given molecule.

Many websites also have more examples of point groups. Feel free to explore those resources.Practice Problems.

1. (a) Prove that the set of real numbers different from −13

is a group under the following operation

a ∗ b = a + b + 3ab.

(b) Determine whether it a group if − 13 is included in the set.

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2. Consider 2 × 2 matrices of the form

a b0 c

where a,b,c are real numbers with a = 0 and

c = 0. These matrices are called upper triangular invertible matrices. Show that the set of such matrices with matrix multiplication is a group.

3. Since 5 is a prime number, there is just one abelian group of order 5. Prove that this is the only(non-isomorphic) group of order 5. Hint: Create identity skeleton first. Show that the case

when all the elements are inverse to themselves is not possible – prove that the associativityfails.

4. Write down the Cayley tables for the following groups.

(a) C 2 × C 3, (b) C 2 × C 2 × C 2, (c) D5.

5. Groups of order 8. The groups C 8, C 4 × C 2 and C 2 × C 2 × C 2, D4 have order 8. There isanother group of order 8, called the quaternion group, usually denoted by Q, that can bepresented by

a, b|a4 = 1, a2 = b2, ba = a3b.

(a) Write down the Cayley table for this group and compare. Compare the Cayley tables forthree groups of order 8 generated by two elements: C 4×C 2 = a, b|a

4 = 1, b2 = 1, ba = ab,D4 = a, b|a

4 = 1, b2 = 1, ba = a3b, and Q = a, b|a4 = 1, b2 = a2, ba = a3b.

(b) Demonstrate that five groups of order 8, C 8, C 4 × C 2 and C 2 × C 2 × C 2, D4 and Q, are notisomorphic to each other.

6. Produce all isomorphism classes of abelian groups of order 36.

7. Determine if the following pairs of groups are isomorphic. If they are, produce the isomorphism.If they are not, explain why.

(a) C 3 and D3. (b) C 6 and D3.

(c) S 3 and D3. (d) S n and Dn for n > 3.

8. Describe all the subgroups of (a) D3, (b) C 6 = C 3 × C 2.

9. Describe the point groups of the following molecules. Write down the presentations of the pointgroups. Identify each group element as a symmetry operation.

(a) Ammonia NH3, (b) Chloramine NH2Cl (c) Hydrogen cyanide HCN.

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Solutions.

1. Let us check axioms A1–A4.

A1. If a and b are real numbers different from − 13

, it is clear that the product a ∗ b = a + b + 3abis a real number but you need to check it is different from −13 . Let us examine conditions ona and b that would make this product equal to −1

3.

a ∗ b = a + b + 3ab = −1

3 ⇒ a + b + 3ab +

1

3 = 0 ⇒ a +

1

3 + b(1 + 3a) = 0 ⇒

a + 1

3 + 3b(

1

3 + a) = 0 ⇒ (a +

1

3)(1+3b) = 0 ⇒ a +

1

3 = 0 or 1 + 3b = 0 ⇒ a = −

1

3 or b = −

1

3.

So, if a and b are real numbers different from −13

, then the product a ∗ b is different from −13

as well. Thus, the operation is closed.

A2. Check the associativity.

(a ∗ b) ∗ c = (a + b + 3ab) ∗ c= (a + b + 3ab) + c + 3(a + b + 3ab)c

= a + b + 3ab + c + 3ac + 3bc + 9abc= a + b + c + 3ab + 3ac + 3bc + 9abc

a ∗ (b ∗ c) = a ∗ (b + c + 3bc)= a + (b + c + 3bc) + 3a(b + c + 3bc)= a + b + c + 3bc + 3ab + 3ac + 9abc= a + b + c + 3ab + 3ac + 3bc + 9abc

Thus, the axiom A2 holds.

A3. You are looking for a number x = −13

with the property that a ∗ x = a and x ∗ a = a forevery a = −13 .

a ∗ x = a ⇒ a + x + 3ax = a ⇒ x + 3ax = 0 ⇒ x(1 + 3a) = 0Since a = −13 , 1 + 3a = 0 and we can cancel the equation x(1 + 3a) = 0 to get that x = 0.Thus, a ∗ 0 = a.

Check that 0 ∗ a = 0 + a + 3(0)a = a as well. Thus, the group identity element is 0.

A4. For any a = −13 , you are looking for a number x = −13 with the property that a ∗ x = 0

and x ∗ a = 0.

a ∗ x = 0 ⇒ a + x + 3ax = 0 ⇒ x + 3ax = −a ⇒ x(1 + 3a) = −a ⇒ x = −a

1 + 3a

Note that we can divide by 1 + 3a since a = −13 . Check that x ∗ a = 0 as well. Indeed−a

1+3a ∗ a = −a1+3a + a + 3a

−a1+3a =

−a+(1+3a)a−3a2

1+3a = −a+a+3a

2−3a

2

1+3a = 01+3a = 0.

If we considered all real numbers instead of all numbers different from −13 , we would not get agroup since the axiom A4 would fail. Indeed, the equation −1

3 ∗ x = 0 has no solutions:

−1

3 ∗ x = 0 ⇒

−1

3 + x + 3(

−1

3 )x = 0 ⇒

−1

3 + x − x = 0 ⇒

−1

3 = 0.

Thus, the element −13 does not have an inverse.

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2. Check the four axioms.

A1. We need to show that the product of two upper triangular invertible matrices is again an

upper triangular invertible matrix. Consider the matrices

a b0 c

and

p q 0 r

with a, c, p, r

non-zero.

a b0 c

p q 0 r

=

ap aq + br

0 cr

Thus, the product is again an upper triangular matrix. It is invertible since ap = 0 becauseboth a and p are non-zero, and cr = 0 because both c and r are non-zero.

A2.

a b0 c

p q 0 r

u v0 w

=

ap aq + br

0 cr

u v0 w

=

apu apv + (aq + br)w

0 crw

a b0 c

p q 0 r

u v0 w

=

a b0 c

pu pv + qw

0 rw

=

apu a( pv + qw) + brw

0 crw

The associativity holds since

apv + (aq + br)w = apv + aqw + brw and a( pv + qw) + brw = apv + aqw + brw.

A3. You are looking for an invertible matrix X =

x y0 z

such that AX = A for any invertible

matrix A = a b

0 c

.

AX = A ⇒

a b0 c

x y0 z

=

a b0 c

⇒

ax ay + bz

0 cz

=

a b0 c

This yields the equations ax = a, ay + bz = b and cz = c. Since a = 0 and c = 0, the firstand third equation give us x = 1 and z = 1. The second equation becomes ay + b = b ⇒ ay =

0 ⇒ y = 0 since a = 0. Thus we have that X =

1 00 1

, the identity matrix. The equation

XA = A holds in this case as well.

If you suspected that the identity matrix is the identity element, you could just check that

A

1 00 1

= A and

1 00 1

A = A.

A4. Let I denote the identity matrix

1 00 1

. For any given invertible matrix A =

a b0 c

,

you are looking for an invertible matrix X =

x y0 z

such that AX = I and X A = I .

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AX = I ⇒

a b0 c

x y0 z

=

1 00 1

⇒

ax ay + bz

0 cz

=

1 00 1

This yields the equations ax = 1, ay + bz = 0 and cz = 1. Since a = 0 and c = 0, the first andthird equation give us x = 1

a and z = 1

c. The second equation becomes ay + b 1

c = 0 ⇒ ay =

−b

c ⇒ y = −b

ac since a = 0. Thus we have that X = 1a −bac

0 1c

. You can check that XA = I as

well.

3. Following the hint, assume that the square of all four non-identity elements a, b, c, d is 1. In thiscase, the product ab is either c or d. Assume it is c first. This implies the following Cayley’stable that yields further relations described in the second table.

· 1 a b c d1 1 a b c da a 1 cb b 1

c c 1d d 1

⇒

· 1 a b c d1 1 a b c da a 1 c d bb b 1

c c d 1d d a 1

However, the associativity fails in this case since a(bb) = a · 1 = a and (ab)b = cb = d soa(bb) = (ab)b. You arrive to the similar contradiction assuming that ab = d.

Thus, you have to have two pairs of mutually inverse elements. Assume that a is inverse to dand b to c. The product aa can be b or c in this case. Assume it is b first. This creates thefollowing table and the isomorphism with C 5.

· 1 a b c d

1 1 a b c da a b 1b b 1c c 1d d 1

⇒

· 1 a b c d

1 1 a b c da a b c d 1b b c d 1 ac c d 1 a bd d 1 a b c

∼=

· 1 a a2 a3 a4

1 1 a a

2

a

3

a

4

a a a2 a3 a4 1a2 a2 a3 a4 1 aa3 a3 a4 1 a a2

a4 a4 1 a a2 a3

Assuming that aa = c also creates an isomorphism with C 5.

4. (a) The table for C 2 × C 3 can be found in the section on direct product of cyclic groups.(b) C 2 × C 2 × C 2 = a,b,c|a

2 = b2 = c2 = 1, ab = ba,ac = ca, bc = cb. The Cayley table is alsodisplayed.(c) D5 = a, b|a

5 = 1, b2 = 1, a4b = ba. Thus, this group consists of 10 elements: 1, a , a2, a3,

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a4, b, ab, a2b, a3b, a4b.

1 a b c ab ac bc abc1 1 a b c ab ac bc abca a 1 ab ac b c abc bcb b ab 1 bc a abc c acc c ac bc 1 abc a b ab

ab ab b a abc 1 bc ac cac ac c abc a bc 1 ab bbc bc abc c b ac ab 1 a

abc abc bc ac ab c b a 1

D5 1 a a2 a3 a4 b ab a2b a3b

1 1 a a2 a3 a4 b ab a2b a3b a a a2 a3 a4 1 ab a2b a3b a4b

a2 a2 a3 a4 1 a a2b a3b a4b b a3 a3 a4 1 a a2 a3b a4b b ab a4 a4 1 a a2 a3 a4b b ab a2b b b a4b a3b a2b ab 1 a4 a3 a2

ab ab b a4b a3b a2b a 1 a4 a3

a2b a2b ab b a4b a3b a2 a 1 a4

a3b a3b a2b ab b a4b a3 a2 a 1 a4b a4b a3b a2b ab b a4 a3 a2 a

5. (a) The three Cayley tables for C 4 × C 2, dihedral D4 and the quaternion group Q are below.The first group differs from the latter two in the bottom half of the table. The differencesbetween D4 and Q are in the bottom right part of the table and they are highlighted in the

table for Q.C 4 × C 2 1 a a

2 a3 b ab a2b a3b1 1 a a2 a3 b ab a2b a3ba a a2 a3 1 ab a2b a3b ba2 a2 a3 1 a a2b a3b b aba3 a3 1 a a2 a3b b ab a2bb b ab a2b a3b 1 a a2 a3

ab ab a2b a3b b a a2 a3 1a2b a2b a3b b ab a2 a3 1 aa3b a3b b ab a2b a3 1 a a2

D4 1 a a2

a3

b ab a2

b a3

b1 1 a a2 a3 b ab a2b a3ba a a2 a3 1 ab a2b a3b ba2 a2 a3 1 a a2b a3b b aba3 a3 1 a a2 a3b b ab a2bb b a3b a2b ab 1 a3 a2 a

ab ab b a3b a2b a 1 a3 a2

a2b a2b ab b a3b a2 a 1 a3

a3b a3b a2b ab b a3 a2 a 1

Q 1 a a2

a3

b ab a2

b a3

b1 1 a a2 a3 b ab a2b a3ba a a2 a3 1 ab a2b a3b b

a2 a2 a3 1 a a2b a3b b aba3 a3 1 a a2 a3b b ab a2bb b a3b a2b ab a2 a 1 a3

ab ab b a3b a2b a3 a2 a 1a2b a2b ab b a3b 1 a3 a2 aa3b a3b a2b ab b a 1 a3 a2

(b) Out of five groups of order 8, C 8, C 4 × C 2 and C 2 × C 2 × C 2, D4 and Q, the first three are

abelian and the last two are not so none of the three abelian groups is isomorphic with twonon-abelian groups. Furthermore, no abelian group is isomorphic to any other abelian groupbecause the order of elements do not match: C 8 has an element (four of them in fact) of order 8,the other two do not. C 4 × C 2 has an element (four of them in fact) of order 4 and C 2 × C 2 × C 2does not.

D4 and Q are not isomorphic because D4 has 5 elements of order 2 and just 2 of order 4 andQ has 2 elements of order 2 and 5 elements of order 4.

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6. Write 36 as product of powers of prime numbers: 36 = 4 · 9 = 22 · 32. Since 2 and 3 arerelatively prime, the groups C 3 × C 3 and C 9 can be combined with any of the groups C 2 × C 2,or C 4 creating isomorphic pairs of groups. C 3 × C 3 is not isomorphic to C 9 and C 2 × C 2 is notisomorphic to C 4. Thus, there are 4 non-isomorphic abelian groups of order 36

1. C 3 × C 3 × C 2 × C 2 ∼= C 6 × C 3 × C 2 ∼= C 6 × C 6

2. C 3 × C 3 × C 4 ∼

= C 3 × C 123. C 9 × C 2 × C 2 ∼= C 18 × C 24. C 9 × C 4 ∼= C 36

7. (a) C 3 and D3 are not isomorphic because one has 3 elements and the other has 6 elements.

(b) C 6 and D3 are not isomorphic because one is abelian and the other is not.

(c) S 3 and D3 are isomorphic.

Recall that S 3 has 6 elements representedby mappings that map (1, 2, 3) to (1, 2, 3),(2, 3, 1), (3, 1, 2), (1, 3, 2), (3, 2, 1), and(2, 1, 3). Let us denote these 6 mappingsby f 1 to f 6. Compare that with symmetryoperations of an equilateral triangle.

The map f 1 is the identity, f 2 is rotation for 120 degrees, f 3 is rotation for 240 degreesand so f 22 = f 3. Thus, the order of f 2 is 3. The elements f 4, f 5 and f 6 are symmetrieswith respect to three axes on the figure above. Thus, these elements are of order 2.The product f 2f 4 represents the composition of maps f 2 and f 4 that turns out to be the

mapping f 2f 4 = (3, 2, 1) = f 5. Finally, the product f 22 f 4 = f 2f 5 = f 6.This demonstrates that there is one-to-one mapping of elements of S 3 onto the elementsof D3 that preserves all the relations among the elements. More precisely, note that f 2and f 4 generate the group. Let us denote f 1 = 1, f 2 = a and f 4 = b. Then f 3 = f

22 = a

2,ab = f 2f 4 = f 5 and a

2b = f 22 f 4 = f 6. The relation for commuting the generators isba = f 4f 2 = f 6 = a

2b. Thus, S 3 can be presented by

a, b|a3 = 1, b2 = 1, ba = a2b

which is the presentation of D3 as well. So, the groups are isomorphic.

Writing Cayley tables for these two groups produces the same tables that are a match

further demonstrates the validity of this reasoning.(d)

(e) S n and Dn are not isomorphic for n > 3 because one has 2n and the other n! elements.n! is larger than 2n for n > 3.

8. (a) Besides the identity, D3 has three elements of order 2 (b, ab, and a2b) and two elements of

order 3 (a and a2). This determines the following.

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number of subgroups order of subgroup subgroups1 1 {1}3 2 {1, b}, {1, ab}, {1, a2b}1 3 {1, a , a2}1 6 D3

total number = 6

This represents the complete list because adding any element to any one of the four nontrivialsubgroups, you will end up with the entire group D4. For example, if you add a to {1, b}, thenthe product a2, ab and a2b have to be in this subgroup in order for it to remain closed. If youadd ab to {1, b}, then the products abb = a and bab = a2 have to be in. But in this case a2bhas to be in as well. So, you again arrive to all six elements present. Convince yourself that allthe other scenarios will result in the same conclusion: the list of the subgroups is complete.

(b) Besides the identity, C 6 has two elements of order 6 (a, and a5), two elements order 3 (a2

and a4) and one element of order 2 (a3). This determines the following.

number of subgroups order of subgroup subgroups1 1 {1}1 2 {1, a3}1 3 {1, a2, a4}1 6 C 6

total number = 4

This represents the complete list because adding any element to any of the two nontrivialsubgroups, you will end up with the entire group C 6. For example, if you add a

2 to {1, a3},then the product a2a3 = a5. Moreover, the inverse a4 of a2 has to be in too and so the producta4a3 = a has to be in too. Thus, you end up with all six elements. Similarly, adding a3, for

example, to {1, a2, a4} forces the product a2a3 = a5 and a3a4 = a to be in the subgroup. So itbecomes entire C 6. Thus, the above four subgroups represent the complete list.

9. (a) Ammonia molecule has the same symmetries as the equilateral triangle. It has no sym-metries with respect to the horizontal plane since this molecule is not planar. Thus, thepoint group of NH3, is the dihedral group D3 = C 3v. It has presentation a, b|a

3 = 1, b2 =1, ba = a2b. The elements a and a2 correspond to rotations by 2π3 and

4π3 . The elements

b, ab and a2b correspond to symmetries with respect to 3 vertical axis of symmetries.

(b) Chloramine NH2Cl. The molecule is not planar so the only non-identity group elementis a single symmetry of order 2. So, the point group has two elements and so it is

C 2 = a|a2

= 1.(c) Hydrogen cyanide HCN is a linear molecule. Thus all the rotations for any angle between

0 and 2π are in the point group. There is also symmetry b with respect to the verticalplane (vertical if the molecule stands ”upright”). Since b does not commute with therotations, we obtain C ∞v = D∞.

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point group

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