# Non abelian group with 27 elements and its center

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?

Since the order of $Z(G)$ is $3$, the order of $H= G/Z(G)$ is $9$. Hence the order of the image of an element $x \in G$ in the quotient group $H$ is either $1$, $3$, or $9$. But we know there is no element of order $9$ in $H$, as then $G$ would be abelian. Hence the only possible orders are $1$ and $3$. This implies that any element $x \in G$ is either already in $Z(G)$ or $x^3 \in Z(G)$.