# Group theory is hard

Let $G$ be a finite group such that $G/Z(G)\simeq \Bbb{Z}/p\Bbb{Z}\oplus\Bbb{Z}/p\Bbb{Z}$, with $p$ prime. Then $p$ divides the order of $Z(G)$.

Let $S$ be a Sylow $p$-subgroup of $G$. The order of $S$ is $p^r$ for some number $r$ greater or equal to $2$. If $S$ and $Z(G)$ have not trivial intersection, we are done. Otherwise, note that $S/Z(G)=G/Z(G)$ (and $r=2$). Then $SZ(G)=G$. But $G/Z(G)$ is abelian, thus for every $a,b \in S$, the commutator $[a,b]=aba^{-1}b^{-1} \in S\bigcap Z(G) = \{e\}$. Thus $S$ is abelian and then $G$ is abelian. A contradiction.