A Group of order $p^n$ is not simple

Question

Let $G$ be a group that $|G|=p^n$, with $n \geq 2$ and $p$ prime.

Show that G is not simple.

I know that, if $G$ is not abelian, then $Z(G) \not= G$ and $Z(G)$ is a normal subgroup of $G$ with $|Z(G)|= p^m >1$ and $m <n$. And since $Z(G) \lhd G$, we have G being not simple.

But what if G is abelian? That proof would not be possible.

If $G$ is abelian of order $p^n$ with $n\ge 2$, then every subgroup of order $p$ is normal. By Cauchy's theorem there is such a subgroup. — Dietrich Burde, Feb 15 '20 at 21:11

user729424 · Accepted Answer · 2020-02-15T21:31:52.627

1

If $G$ is Abelian, then all subgroups are normal so it's enough to find a non-trivial proper subgroup. If $G$ is order $p^n$ with $n\ge2$, then there is an element $x$ of order $p$. If you let $H=\langle x\rangle$, the subgroup generated by $x$, then $\vert H\vert=p$, so $H$ is a non-trivial proper subgroup.

edited Feb 15 '20 at 21:31

answered Feb 15 '20 at 21:12

user729424

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A Group of order $p^n$ is not simple

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