Introduction to Abstract Algebra, Solutions Manual, 4th Edition

9.2 Solvable Groups

1. No. The group S₄ is solvable (Example 4) but Z(S₄) = {ε}.

3. No. S₄ is solvable (Example 4) but is not abelian. Indeed because S₄/A₄ is abelian. Thus , {ε} or K = {ε, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. But S₄/{ε} and S₄/K are not abelian (see Exercise 30 §2.9).

5. If G = A₅ then G is not solvable being nonabelian and simple. Since |A₅| = 60 = 2² · 3 · 5 the Sylow subgroups have orders 4, 3 or 5, and so are abelian.

7. G need not be solvable. If G = A₅ × C₂ then K = {ε} × C₂ is an abelian normal subgroup which is maximal because G/K ≅ A₅ is simple. But G is not solvable because A₅ is not solvable.

8. 1. This is because α[a, b] = [α(a), α(b)] and the fact that G′ consists of products of commutators.

3. Every commutator from H is a commutator from G.

9. By Exercise 14 §8.4, let K G where K ≠ {1}, G. If |G| = p²q let K G, K ≠ {1}, K ≠ G. Then |K| = p, q, p² or pq, so |G/K| = pq, p², q or p. Thus both K and G/K are either abelian or of order pq, and hence are ...