§ 7.6 Solvable Groups: solved problems

9. Let p be a prime and let G be a nonabelian group of order p³. Show that the center Z(G) of G equals the commutator subgroup G' of G.

Solution: Since G is nonabelian, by Exercise 7.2.13 of the text we have |Z(G)| = p. (The center is nontrivial by Theorem 7.2.8, and if |Z(G)| = p², then G/Z(G) is cyclic, and Exercise 3.8.14 of the text implies that G is abelian.) On the other hand, any group of order p² is abelian, so G/Z(G) is abelian, which implies that G' Z(G). Since G is nonabelian, G' {e}, and therefore G' = Z(G).

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10. Prove that D_n is solvable for all n.

Solution: One approach is to compute the commutator subgroup of D_n, using the standard description

D_n = { aⁱ b^j | 0 i < n, 0 j < 2, o(a) = n, o(b) = 2, ba = a^-1b }

Case 1: If x = aⁱ and y = a^j, the commutator is trivial.

Case 2: If x = aⁱ and y = a^jb, then
xyx^-1y^-1 = aⁱa^jba^-ia^jb = aⁱa^jaⁱba^jb = aⁱa^jaⁱa^-jb² = a²ⁱ,
and thus each even power of a is a commutator.

Case 3: If x = a^jb and y = aⁱ, we get the inverse of the element in Case 2.

Case 4: If x = aⁱb and y = a^jb, then
xyx^-1y^-1 = aⁱba^jbaⁱba^jb, and so we get
xyx^-1y^-1 = aⁱa^-jb²aⁱa^-jb² = a^2(i-j),
and again we get even powers of a.

Thus the commutator subgroup D_n' is either < a > (if n is odd) or < a² > (if n is even). In either case, the commutator subgroup is abelian, so D_n'' = {e}.

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11. Prove that any group of order 588 is solvable, given that any group of order 12 is solvable.

Solution: We have 588 = 2² · 3 · 7². Let S be the Sylow 7-subgroup. It must be normal, since 1 is the only divisor of 12 that is 1 mod 7. By assumption, G / S is solvable since | G / S | = 12. Furthermore, S is solvable since it is a p-group. Since both S and G / S are solvable, it follows from Corollary 7.6.8 (b) that G is solvable.

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12. Let G be a group of order 780=2² · 3 · 5 · 13. Assume that G is not solvable. What are the composition factors of G? (Assume that the only nonabelian simple group of order 60 is A₅.)

Solution: The Sylow 13-subgroup N is normal, since 1 is the only divisor of 60 that is 1 mod 13. Using the fact that the smallest simple nonabelian group has order 60, we see that the factor G/N must be simple, since otherwise each composition factor would be abelian and G would be solvable. Thus the composition factors are Z₁₃ and A₅.

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