Abstract Algebra Theory and Applications - Computer Science ...

400 HINTS AND SOLUTIONS2. The four Sylow 3-subgroups of S 4 areP 1 = {(1), (123), (132)},P 2 = {(1), (124), (142)},P 3 = {(1), (134), (143)},P 4 = {(1), (234), (243)}.5. Since |G| = 96 = 2 5 · 3, G has either one or three Sylow 2-subgroups by theThird Sylow Theorem. If there is only one subgroup, we are done. If thereare three Sylow 2-subgroups, let H and K be two of them. |H ∩ K| ≥ 16;otherwise, HK would have (32 · 32)/8 = 128 elements, which is impossible.H ∩K is normal in both H and K since it has index 2 in both groups. Hence,N(H ∩K) contains both H and K. Therefore, |N(H ∩K)| must be a multipleof 32 greater than 1 and still divide 96, so N(H ∩ K) = G.8. G has a Sylow q-subgroup of order q 2 . Since the number of such subgroupsis congruent to 1 modulo q and divides p 2 q 2 , there must be either 1, p, or p 2Sylow q-subgroups. Since q̸ | p 2 − 1 = (p − 1)(p + 1), there can be only oneSylow q-subgroup, say Q. Similarly, we can show that there is a single Sylowp-subgroup P . Every element in Q other than the identity has order q or q 2 ,so P ∩ Q = {e}. Now show that hk = kh for h ∈ P and k ∈ Q. Deduce thatG = P × Q is abelian.10. False.17. If G is abelian, then G is cyclic, since |G| = 3 · 5 · 17. Now look at Example 5.23. Define a mapping between the right cosets of N(H) in G and the conjugatesof H in G by N(H)g ↦→ g −1 Hg. Prove that this map is a bijection.26. Let aG ′ , bG ′ ∈ G/G ′ . Then (aG ′ )(bG ′ ) = abG ′ = ab(b −1 a −1 ba)G ′ =(abb −1 a −1 )baG ′ = baG ′ .Chapter 14. Rings1. (a) 7Z is a ring but not a field. (c) Q( √ 2 ) is a field. (f) R is not a ring.3. (a) {1, 3, 7, 9}. (c) {1, 2, 3, 4, 5, 6}.(e){( 1 00 1) ( 1 1,0 1) ( 1 0,1 1) ( 0 1,1 0) ( 1 1,1 0) ( 0 1,1 1)}.4. (a) {0}, {0, 9}, {0, 6, 12}, {0, 3, 6, 9, 12, 15}, {0, 2, 4, 6, 8, 10, 12, 14, 16}.(c) There are no nontrivial ideals.7. Assume there is an isomorphism φ : C → R with φ(i) = a.8. False. Assume there is an isomorphism φ : Q( √ 2 ) → Q( √ 3 ) such thatφ( √ 2 ) = a.