Discussion
Let G be a group of prime power order. Then :
Answer: C
The answer is G is solvable.
A group of order \(p^{n}\) has a non-trivial centre, so by induction it has a normal series with abelian (indeed cyclic of order \(p\)) factors: every \(p\)-group is nilpotent, hence solvable.
The other statements fail in general. The quaternion group of order 8 is a non-abelian 2-group. A \(p\)-group of order \(p^{n}\) with \(n \ge 2\) has normal subgroups of every order \(p^{k}\), so it is not simple. And the Sylow \(p\)-subgroup of a \(p\)-group is the whole group, which need not be abelian (again the quaternion group).
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