Let $G$ be a non-abelian group of order $p^3$, where $p$ is an odd prime. Show that $G$ has normal subgroup $N$ such that
$$Z(G)<N<G$$
and $N \cong\Bbb Z_p \times\Bbb Z_p$.
I know the following facts about a non-abelian group $G$ of order $p^3$:
$$|Z(G)| = p \mbox{ and } G/Z(G) \mbox{ is elementary abelian. }$$
Thanks!
Best Answer
Let $p:G\rightarrow G/Z(G)$ the quotient map, for every $x,y\in G$, $p(xyx^{-1}y^{-1})=p(e)$ where $e$ is the neutral element since $G/Z(G)$ is commutative. This implies that $xy=zyx$ where $z\in Z(G)$.
We deduce that $(xy)^p=z^{{p(p-1)}\over 2}x^py^p$ since $p$ is odd, ${{p(p-1)}\over 2}$ is divisible by $p$, since the order of $Z(G)$ is $p$, we deduce that $(xy)^p=x^py^p$ and $f(x)=x^p$ is an homomorphism of groups defined on $G$ whose image is contained in $Z(G)$ since $G/Z(G)$ is elementary abelian.
The kernel of $f$ is a subgroup which contains $Z(G)$ since $Z(G)$ is a $p$ group and its order is strictly superior to $p$ otherwise $|G|\leq p^2$. Let $x$ be an element of $\ker(f)$ which is not in $Z(G)$, the subgroup $N$ generated by $Z(G)$ and $x$ is isomorphic to $\mathbb Z_p\times\mathbb Z_p$.