I was wondering if anyone could give me some help on the following problem. Any help would be greatly appreciated!
Let $G$ be a group of order $r^3$ with $r$ an odd prime. Assume $G$ is non-abelian and has exponent $r$. Then $G=N:H \cong \mathbb{Z}_{r}^{2}:\mathbb{Z}_{r}$, where $:$ denotes the semi-direct product. Let $F=GF(p)$ and assume that $r$ divides $p-1$. Prove that $Z(G)=\mathbb{Z}_{r}$ where $Z(G)$ denotes the centre of $G$.
What I have so far is that as the centre is a normal subgroup, so must have order $1$,$r$,$r^{2}$ or $r^{3}$. Clearly it can't have order $r^{3}$ otherwise $G$ would be abelian. Any suggestions from here?