Math 3560 HW Set 9

Kara

November 5, 2013

4 2 3 s 3 4 (14.1) The conjugacy classes of D5 are {e}, {r, r }, {r , r }, {s, rs, r , r s, r s}.

(14.4) Conjugacy classes in S6 are formed by permutations of the same cycle structure. There are exactly 11 cycle structures in S6 and all permutations with a given structure form one conjugacy class. Here, there is a list of representatives of 11 possible cycle structures: e, (12), (123), (1234), (12345), (123456), (12)(34), (123)(45), (123)(456), (1234)(56), (12)(34)(56). Using the algorithm in the book(pages 76, 77), we ﬁnd g = (156423) satisfying the equation g(123)(456)g−1 = (531)(264). This g is odd. But this is not the only element satifying the equation above. Note that (531)(264) = (264)(531) and h = (126)(345) is even and satisﬁes h(123)(456)h−1 = (264)(531).(use the same algorithm for (264)(531)).

We can easily ﬁnd h ∈ S8 conjugating α = (12345)(678) to β = (43786)(215), for example h = (14856237). This h is odd, but it is not enough to claim that α and β are not conjugate in A8. There might be other elements conjugating α to β, maybe not all of them are odd. In fact all of them will be odd - but this is not trivial and requires a proof. Assume there is g ∈ A8 such that

gαg−1 = β = hαh−1 where h = (14856237).

This means h−1gα = αh−1g i.e. h−1g, which is an odd permutation being a product of an even and an odd permutations, commutes with α. Now, we will show that any permutation which commutes with α must be even. This will give us a contradiction which means that α and β cannot be conjugate in A8. Assume that γ commutes with α: γα = αγ i.e. γαγ−1 = α Note that: α(γ(6)) = γαγ−1(γ(6)) = γα(6) = γ(7), α(γ(7)) = γαγ−1(γ(7)) = γα(7) = γ(8), α(γ(8)) = γαγ−1(γ(8)) = γα(8) = γ(6). Thus (γ(6), γ(7), γ(8)) is a 3-cycle in α. We have exactly one 3-cycle in α, (678) = (786) =

1 (867). Therefore γ(6) can be 6,7 or 8. If γ(6) = 7, then (γ(6), γ(7), γ(8)) = (786) which means γ contains the 3-cycle (678). If γ(6) = 8, then (γ(6), γ(7), γ(8)) = (867) which implies that γ contains the 3-cycle (687) = (678)2. The remaining case corresponds to the case where γ contains e = (678)3. So, we see that γ contains a 3-cycle which is a power of (678). A similar argument gives that γ also contains a 5-cycle which is a power of (12345). Henece, γ is an even permutation since powers of 3-cycles and 5-cycles are even(so is their product). (14.5) To prove. The 3-cycles in A5 form a single conjugacy class.

Proof. Let x, y be 3-cycles in A5. The 3-cycles in S5 form a single conjugacy class. There- −1 fore, there exists g ∈ S5 such that gxg = y. If g is even, then we are done, so suppose g is odd. Without loss of generality, suppose x = (123). Then (45)x(45) = x because x and (45) are disjoint. Therefore, letting h = g(45),

hxh−1 = g(45)(123)(45)g−1 = g(123)g−1 = gxg−1 = y.

Furthermore, since g and (45) are odd, h is even. Hence, x is conjugate to y.

(14.6) Let S be a set of 8 elements, and suppose {Ui} is a partition of S such that {Ui} has three sets, two of which have 3 elements and the third of which has 2 elements. An element, σ, of S8 preserves the partition if for all j ∈ S, j ∈ Ui if and only if σ(j) ∈ Ui. Then there are 2 · 2 · 1 = 4 elements of S8 which preserve a given partition.

The number of elements of S8 with cyclic structure (12)(345)(678) is equal to the product of the number of ways to partition a set of 8 into a set of 2 and two sets of 3 and the 8 number of elements which preserve a partition. There are 2 ways to choose the set of 2, 6 and from the remaining 6 elements, there are 3 ways to choose another set of 3. However, 1 6 since we do not distinguish between the two sets of 3, there are 2 3 ways to partition a set of 6 into two sets of 3. Therefore, there are 8 16 4 · · = 1120 2 2 3 elements of S8 with cyclic structure (12)(345)(678). (12 . . . n−1) ∈ An, so s(12 . . . n−1) = (12 . . . n − 1)s, which implies s(n) = n and s(i + 1) = s(i) + 1 mod n − 1 for 1 ≤ i < n. s(1)−1 Thus, s = (12 . . . n − 1) . Noting that An−1 can be viewed as a subgroup of An, we can use the same argument to s how s = e.

(14.11) Denote by Fj matrix obtained from I by multiplying j-th row by −1, and by Eij matrix obtained from I by switching i-th and j-th row. Both of this operations will change determinant of I by multiplying it by −1. Thus all these matrices have nonzero determinant, so they are invertible.

2 We are to ﬁnd center of GLn(R). Call it Z. If a matrix A is in Z then in particular it commutes with all the matrices Fj, Eij. As usual denote (i, j)-th entry of A by aij. Note that AFj is matrix A with j-th column multiplied by −1, while Fj A is matrix A with j-th row multiplied by −1. Therefore AFj = Fj A gives that for all k 6= j, −akj = akj. Varying j we get that A must be a diagonal matrix.

Now notice that AEij is matrix A with i and j-th columns switched, while Eij A is matrix A with i and j-th row switched. Analyzing the (i, j)-th entry in AEij and in Eij A gives us equation aii = ajj. Again, varying i, j we obtain that, to commute with all the matrices Fj, Eij, our matrix A needs to be diagonal, with same entry on diagonal, that is A = aI for some a 6= 0 (as we want A to be in GLn(R)). Such matrices commute with every element of GLn(R). Therefore we have proved that center of GLn(R) is {a I | a ∈ R \{0}}.

(15.3) The order of the quaternion group Q is 8 and the order of any subgroup of Q divides 8. Hence, the orders of possible non-proper subgroups of Q are 2 and 4. The only element of Q of order 2 is -1, and all the other elements except the identity have order 4. Hence, H = {1, −1} is the only subgroup of Q with 2 elements. Since both 1 and -1 commute with every element of Q, xH = Hx for every x ∈ Q. Thus, H is normal by Theorem 15.3. All the other subgroups of Q with 4 elements are also normal by Theorem 15.4 since the index of such a subgroup in Q is equal to 2.

(15.6) To prove. If H,K are normal subgroups of a group, and H ∩ K = {e}, then hk = kh for all h ∈ H, k ∈ K.

Proof. Since H and K are normal, kH = Hk and hK = Kh for all h ∈ H, k ∈ K. Then there exist h0 ∈ H, k0 ∈ K such that

kh = h0k, hk−1 = k0h.

This implies khk−1 = h0, khk−1 = kk0h, so combining these, we have h0 = kk0h, or equivalently, h0h−1 = kk0. However, h0h−1 ∈ H, kk0 ∈ K, and H ∩ K = {e}. Therefore, h0h−1 = e, so h0 = h, which implies kh = hk, as desired.

(15.12) A proper normal subgroup of A4 is {e, (12)(34), (13)(24), (14)(23), } (see page 80 in the book). The commutators are: (12345)−1(345)−1(12345)(345) = (245)

3 (12)(34)−1(345)−1(12)(34)(345) = (354) Let H be a normal subgroup of A5. It can contain only even elements so only 5-cycles, 3-cycles, or composition of disjoint transpositions. If it contains a 5-cycle, then we can relabel elements if needed and assume this 5-cycle is a = (12345). Note that a−1 also is in H, and any cojungate of a is in H, and products of these. Therefore its commutator with (345) needs to be in H. Our computations show that it is a 3-cycle. Thus if H contains 5-cycle, it needs to contain a 3-cycle as well. Similarly, if H contains composition of disjoint transpositions, we can assume that it is (12)(34) and use above computations to deduce H contains a 3-cycle - the commutator of (12)(34) and (345). In A5, 3-cycles forms one conjugacy class, therefore H needs to contain all 3-cycles. From theorem 6.5 we know that 3-cycles generate all A5. Thus H = A5.