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MT310 Homework 3

Solutions

Due Friday, February 12 by 5:00 pm

Exercise 1. On a groupG define a relation by: ab if there exists g G such thatb = gag1. Provethat this is an equivalence relation.

Proof. Leta,b, c G.Sincea = eae1, we have a a so the relation is reflexive.Ifa b, theng G such thatgag1 =b. Thena = g1bg, sob a. Hence the relation is symmetric.Ifa b and b c, there are g, h G such that b = g ag1 and c = hbh1. Thenc = hga(hg)1 soa c. Hence the relation is transitive.

Exercise 2. The equivalence classes under the equivalence relation of exercise 1 are called conjugacyclasses. Find the conjugacy classes in S3, D4 and A4.

Solution.

The conjugacy classes inS3 are{e}{(12), (13), (23)}, (2-cycles){(123), (321)}(3-cycles).

ForD4, we number the square as in hw 2:

1 2

34

The congugacy classes in D4 are{e}{(13)(24)}(180 degree rotation),{(14)(23), (12)(34)} (edge reflections),{(13), (24))}(vertex reflections),{(1234), (4321)} (90 degree rotations).

Alternative solution with matrices:

{

1 00 1

}

{

1 00 1

} (180 degree rotation),

{

1 00 1

,

1 00 1

} (edge reflections),

{

0 11 0

,

0 11 0

} (vertex reflections),

{

0 11 0

,

0 11 0

} (90 degree rotations).

The conjugacy classes in A4 are{e},{(12)(34), (13)(24), (23)(14)}, (edge reflections of tetrahedron),{(123), (134), (142), (243))} (face rotations in same direction),{(123), (134), (142), (243))} (face rotations in the other direction).

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Exercise 3. For G= S3, choose one element a in each conjugacy class and compute the order of itscentralizer CG(a). Do the same for G = D4 and G= A4. What relation do you observe between theorderCG(a) and the number of elements in the conjugacy class ofa?

Solution.a CS3(a) |CS3(a)| |conj. class ofa|e S3 6 1

(12) (12) 2 3

(123) (123) 3 2

a CD4(a) |CD4(a)| |conj. class ofa|e D4 8 1

(13)(24) D4 8 1(12)(34) (12), (34) 4 2

(13) (13), (24) 4 2(1234) (1234) 4 2

a CA4(a) |CA4(a)| |conj. class ofa|e A4 12 1

(12)(34) (12)(34), (13)(24) 4 3(123) (123) 3 4(321) (123) 3 4

In all cases we have the relation

|CG(a)| |conj. class ofa|=|G|.

(This holds for any element a of any finite group G, as we will prove later.)

Exercise 4. Let (a1, a2, . . . , ak) be a cycle of length k in Sn, and let Sn be an arbitrary permu-tation. Prove that

(a1, a2, . . . , ak)1 = ((a1), (a2), . . . , (ak)).

Proof. Letbi= (ai), for 1i k. Letc {1, 2, . . . , n}. Ifc / {b1, . . . , bk}, then1c / {a1, . . . , ak},so

(a1, a2, . . . , ak)1(c) = 1(c) = c = (b1, . . . , bk)(c).

Ifc = bi, then1(c) = ai, so reading subscripts modulok , we have

(a1, a2, . . . , ak)1(c) = (a1, a2, . . . , ak)ai = (ai+1) = bi+1 = (b1, . . . , bk)(c).

Hence the permutions (a1, a2, . . . , ak)1 and (b1, . . . , bk) have the same effect on every number in{1, 2, . . . , n}, so they are the same permutation.

Exercise 5. The subset{(12)(34), (13)(24), (14)(23)} S4 is a conjugacy-class. Number the elementsasx1 = (12)(34), x2 = (13)(24), x3 = (14)(23). If S4 is an arbitrary permutation, then x1

1 =xi, for some i {1, 2, 3}. Likewise, x2

1 =xj and x31 =xk, for some j, k {1, 2, 3}. Thus, we

have a permutation

f() =

1 2 3

i j k

.

Prove that this defines a homomorphism f :S4 S3 and compute ker fand im f.

Proof. To see that f is a homomorphism, we take two elements , S4 and compute:

f()(xi) = xi()1 = xi

11,

andf()f()(xi) = f()( xi

1) = xi11,

sof() = f()f().The kernel off consists of those S4 such thatxi1 =xi for all i. In other words,

ker f=CS4(x1) CS4(x2) CS4(x3).

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Each centralizer is isomorphic to D4. In exercise 1 we listed the elements ofCS4(x2). Of these, thosewhich commute with x1 are the 22-cycles and e. These also commute with x3, so we have

ker f={e, (12)(34), (13)(24), (14)(23)}.

The image of f is all of S3. To see this, it suffices to find , in S4 such that f() = (12) andf() = (23). Since

(23)x1(23) =x2, (23)x3(23) =x3, (34)x2(34) =x3, (34)x1(34) =x1,

it follows that f((23)) = (12) and f((34)) = (23), so im f=S3, as claimed.

Exercise 6. LetCbe the group of rigid motions of a cube. Judson [Thm. 4.12] defines an isomorphismg: CS4, whereg() is the permutation of{1, 2, 3, 4}induced by the action of on the four diagonalsof the cube. From the previous exercise, there must also be a homomorphism f : C S3. Give ageometric construction offby finding three things in the cube permuted by C.

Solution. The homomorphism f :CS3 is given by permuting the three perpendicular lines throughopposite faces of the cube. Each line is the axis of a 180 degree rotation, which corresponds to a22-cycle in S4. IfL is one of these lines and x C is 180 degree rotation about L, then any CsendsL to the line L whose 180 degree rotation is x1. So permutes the lines in the same wayit permutes the 22-cycles under conjugation. Hence this is the same homomorphism as in the previous

exercise.

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