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§ 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 Dn is solvable for all n.

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

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

We must find all elements of the form xyx-1y-1, for x,y in Dn. We consider the cases x = ai or x = aib and y = aj or y = ajb.

Case 1: If x = ai and y = aj, the commutator is trivial.

Case 2: If x = ai and y = ajb, then
xyx-1y-1 = aiajba-iajb = aiajaibajb = aiajaia-jb2 = a2i,
and thus each even power of a is a commutator.

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

Case 4: If x = aib and y = ajb, then
xyx-1y-1 = aibajbaibajb, and so we get
xyx-1y-1 = aia-jb2aia-jb2 = a2(i-j),
and again we get even powers of a.

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