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This is an example from a book. But I don't understand the solution.

Let $G$ be a group of order $p^n, p$ prime. Then $G$ has a composition series such that all its composition factors are of order $p$.

Solution. Clearly, any composition factor is of order $p^k$ for some $k >0$. But since a group of prime power order has a nontrivial center, and each composition factor is a simple group, it follows that each composition factor is a simple abelian group and therefore, of order $p$.

I don't see how the fact that the group has a nontrivial center, and each composition factor is a simple group leads to the conclution that each composition factor is a simple abelian group. Can anyone elaborate on this?

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Let $C$ be a composition factor. You know it is a simple group, and that $Z(C) \ne \{ 1 \}$. Then you must have $Z(C) = C$. i.e., $C$ is abelian. Now a finite abelian group is only simple when it has prime order, for instance by Cauchy's Theorem.

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Hint:

Theorem: A group $\;G\;$ of order $\;p^n\;$ has normal subgroup of order $\;p^k\;$ , for all $\;0\le k\le n\;$.

Highlights of proof: by induction, the decisive inductive step being to remark that $\;Z(G)\neq1\;$ and thus there exists $\;1\neq z\in Z(G)\;$ and look at the quotient $\;G/\langle z\rangle\;$ ...

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