Let $G$ be a non abelian group with 27 elements.
(a) Prove that $|Z(G)|=3$.
(b) Prove that for all $x\in G$ we have $x^3\in Z(G)$.
For (a), note that $Z(G)$ is a subgroup of $G$, therefore it's order must divide 27. The order of $G$ is a prime power, therefore $|Z(G)|\neq 1$, also $|Z(G)|\neq 27$ since $G$ is non abelian, and $|Z(G)|=9$ would give rise to a cyclic quotient $G/Z(G)$ which again contradicts that $G$ is non abelian. Therefore $|Z(G)|=3$.
For (b), I noted that the elements of $x$ are again of order $1,3,9,27$. For order $1,3$ this is clear. $G$ does not contain an element of order 27, since it would be abelian. We are left with the case that the order of $x$ is $9$. I can't see why this should be true, but I bet there is an easy argument. Could someone point me in the right direction?