**9.**
Let p be a prime and let G be a nonabelian group of order p^{3}.
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)| = p^{2}, 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 p^{2} 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}
= { a^{i} b^{j} |
0 i < n,
0 j < 2,
*o*(a) = n,
*o*(b) = 2,
ba = a^{-1}b }

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

Case 2: If x = a^{i} and y = a^{j}b, then

xyx^{-1}y^{-1}
= a^{i}a^{j}ba^{-i}a^{j}b
= a^{i}a^{j}a^{i}ba^{j}b
= a^{i}a^{j}a^{i}a^{-j}b^{2}
= a^{2i},

and thus each even power of a is a commutator.

Case 3: If x = a^{j}b and y = a^{i},
we get the inverse of the element in Case 2.

Case 4: If x = a^{i}b and y = a^{j}b, then

xyx^{-1}y^{-1}
= a^{i}ba^{j}ba^{i}ba^{j}b,
and so we get

xyx^{-1}y^{-1}
= a^{i}a^{-j}b^{2}a^{i}a^{-j}b^{2}
= 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 < a^{2} > (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 = 2^{2} **·** 3 **·** 7^{2}.
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=2^{2} **·** 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 A_{5}.)

*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 **Z**_{13} and A_{5}.

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