Mathematics 352 – Abstract Algebra II Solutions for Problem Set 2 January 28, 2008

Section 8.2

22. We prove the result first for finite abelian p-groups for an arbitary prime p. Let G and H be two finite abelian p-groups such that for all m ≥ 1, the numbers of elements of order m in G and H are the same. The m’s involved in this case are powers of p, of course. Since G and H are finite, they have the same number of elements of the same maximal order p^{M} for some M. then writing

N_{i}(G) = |{g ∈ G : |g| = p^{i}}|,

(the number of elements of order p^{i}), we have

|G| = ∑_{i=0}^{M} N_{i}(G) = ∑_{i=0}^{M} N_{i}(H) = |H|.

We will argue by induction on the order of the groups that G and H must be isomorphic under the given hypothesis. If |G| = |H| = 1, then there is nothing to prove since G = {0_{G}} and {0_{H}} = H are both trivial groups hence are isomorphic.

Now assume the result has been proved for p-groups of all orders < p^{k} and consider G, H of order p^{k} such that for all m ≤ p^{k}, the numbers of elements of order m in G and H are the same. By hypothesis, the maximal orders are the same, say p^{M} (M ≤ k), and there are the same numbers of elements in G and H of order p^{M}. Pick one such element a ∈ G and one such element b ∈ H. By Lemma 8.6 (used in the proof of the Structure Theorem), we have G = ⟨a⟩ ⊕ K and H = ⟨b⟩ ⊕ L for some subgroups K, L. Now, K, L are p-groups of order p^{k−M}.

Moreover, we claim that they must also satisfy the hypothesis – K and L must contain the same number of elements of order p^{i} for each p^{i}. We will prove this by a separate induction argument. The base case is i = 0 and the statement is true in that case since there is only one element of order 1 in each group (the 0 elements). Suppose we know that

N_{j}(K) = N_{j}(L), for all 0 ≤ j ≤ i − 1.

Consider the numbers of elements of order p^{i} in K, L, G, and H. By the direct sum decompositions for G, any element of order p^{i} in G must look like b+c where b ∈ ⟨a⟩ and c ∈ K and lcm(|b|, |c|) = p^{i} (that is, at least one of |b|, |c| is p^{i} and the other is ≤ p^{i}). We count how many elements this gives, writing C = ⟨a⟩:

N_{i}(G) = (N_{0}(C) + ⋯ + N_{i}(C))N_{i}(K) + (N_{0}(K) + ⋯ + N_{i−1}(K))N_{i}(C).

Similarly from the direct sum decomposition H = ⟨b⟩ ⊕ L = D ⊕ L we get

N_{i}(H) = (N_{0}(D) + ⋯ + N_{i}(D))N_{i}(L) + (N_{0}(L) + ⋯ + N_{i−1}(L))N_{i}(C).

Now C and D are isomorphic since they are cyclic groups of the same order. This means N_{j}(C) = N_{j}(D) for all j. The induction hypothesis shows that the terms in the second sets of parentheses are equal. Moreover N_{i}(G) = N_{i}(H) by hypothesis. Therefore N_{i}(K) = N_{i}(L) and this finishes the inner induction argument. This shows that K and L satisfy the same hypothesis as G and H.

Hence by the induction hypothesis (for the outer induction), K ≃ L. Since C ≃ D as above, this shows that G ≃ H.

Now, consider the general case, where G, H are finite abelian groups, but not necessarily pgroups. The same reasoning used for p-groups (but applied to elements of all possible orders) shows that |G| = |H|. The hypothesis implies in particular that for any prime p, the number of elements of order p^{i} for each i is the same in both G and H. Letting i range over all integers i ≥ 0, these elements are exactly the elements of the subgroups G(p) and H(p). The above proof shows that G(p) ≃ H(p). This is true for all p dividing the order of G, H. Hence by the Structure Theorem G ≃ H.

Section 8.3

2. (a) Since |S_{4}| = 24 = 2^{3} · 3, Sylow 2-subgroups are subgroups of order 2^{3} = 8. We can get one such subgroup by taking the cyclic subgroup generated by the 4-cycle (1234) and including additional suitable elements of order 2 to make subgroups isomorphic to D_{4}:

H_{1} = {(), (1234), (13)(24), (1432), (24), (13), (12)(34), (14)(23)}

Then, other subgroups of the same order can be generated by conjugating this one. For instance conjugating by (12) we get

H_{2} = {(), (1342), (23)(14), (1243), (14), (23), (12)(34), (24)(13)}.

Conjugating by (23) we get

H_{3} = {(), (1324), (12)(34), (1423), (34), (12), (13)(24), (34)(12)}.

(b) The Sylow 3-subgroups have order 3; they are cyclic, generated by the 3-cycles:

K_{1} = {(), (123), (132)}
K_{2} = {(), (124), (142)}
K_{3} = {(), (134), (143)}
K_{4} = {(), (234), (243)}.

3. Since |A_{4}| = 12 = 2^{2}·3, the Sylow 2-subgroups have order 4. There is exactly one such subgroup:

H = {(), (12)(34), (13)(24), (14)(23)}.

The Sylow 3-subgroups have order 3; they are cyclic, generated by the 3-cycles (this is the same as in S_{4}):

K_{1} = {(), (123), (132)}
K_{2} = {(), (124), (142)}
K_{3} = {(), (134), (143)}
K_{4} = {(), (234), (243)}.

Comment: Note by Sylow 3 that the number of 2-subgroups must be an odd number dividing 12. The possible numbers are 1 and 3, but there are only 3 elements of order a power of 2 in A_{4}.

4. The elementary divisors are 2, 2^{2}, 2^{2}, 3, 3, 5. There is exactly one Sylow 2-subgroup, isomorphic to Z_{2} ⊕ Z_{4} ⊕ ℤ_{4}. There is one Sylow 3-subgroup, isomorphic to ℤ_{3} ⊕ ℤ_{3}. There is one Sylow 5-subgroup, isomorphic to ℤ_{5}.

5. (a) By Sylow theorem 3, the number of 3-subgroups is congruent to 1 mod 3 and divides 72. The number could be 1 or 4. (b) Similarly, the number is congruent to 1 mod 5 and divides 60. Hence the possibilities are 1 or 6.

6. (b) 143 = 11 · 13. By Sylow 3, the number of Sylow 11-subgroups is 1. Similary, the number of 13-subgroups is 1. By Sylow theorem 2, both of these subgroups are normal. Hence G ≃ H × K is cyclic of order 143.

Section 8.4

1 (a) Use the permutation form of D_{4} from 8.3/2a above:

D_{4} ≃ {(), (1234), (13)(24), (1432), (24), (13), (12)(34), (14)(23)} ⊂ S_{4}

to make computing conjugates easier(!) The conjugacy classes are

C_{1} = C_{()} = {()} in the center
C_{2} = C_{(1234)} = {(1234), (1432)}
C_{3} = C_{(13)(24)} = {(13)(24)} also in the center
C_{4} = C_{(12)(34)} = {(12)(34), (14)(23)}
C_{5} = C_{(13)} = {(13), (24)}.

There is no real “trick” to this – just compute!

(c) The conjugacy classes in

A_{4} = {(), (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23)} are

C_{1} = {()}
C_{2} = {(12)(34), (13)(24), (14)(23)}
C_{3} = {(123), (243), (134), (142)}
C_{4} = {(132), (234), (143), (124)}.

(Note in particular that the two elements of order 3 in each Sylow 3-subgroup end up in different conjugacy classes in A_{4}. This is true because the permutations that conjugate (123) to (132) in S_{4}, like (23), are odd – they are not in A_{4}. Hence the conjugacy classes in S_{4} split up into smaller subsets in A_{4}.)