I don't understand how to solve this problem. I don't think I'm allowed to assume its abelian. Also, I know that I will have to use this theorem: If $p$ is a prime and $G$ is a $p$-group with more than one element, then $Z(G)\neq \{1\}$.
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Sorry, it is a duplicate of A finite $P$-group cannot be simple unless it has order $p.$. The center is always a non-trivial normal subgroup, as you said. If the group has order $p$, it is abelian. This is easy to prove, too (just Lagrange). – Dietrich Burde Nov 05 '17 at 20:25
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@DietrichBurde I don't understand how to go in the forward direction. How does G being a finite p-group and simple mean its order is p? I understand the other direction. – NoMayoPlz Nov 05 '17 at 20:37
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If $G$ is a simple $p$-group, then its center is a proper normal subgroup - a contradiction, except if $G=Z(G)$ (because $Z(G)=1$ is impossible). – Dietrich Burde Nov 05 '17 at 20:39