Prove that $|Z(G)|=p$

Question

Let $G$ be a non-abelian group of order $p^3$, $p$ prime. Show that $Z(G)$ is a group of order $p$. Deduce that $G/Z(G)$ is abelian.

From the class equation I can get that $p| |Z(G)|$, so $|Z(G)|=\{p,p^2,p^3\}$ it can't be of the order $p^3 $ because that would give us that $G$ is abelian.

If $|Z(G)|=p^2$ how to continue ? I don't see how a contradiction can appear.

Any hints how to continue ?

Answer 1

Let us use that $G/Z(G)$ is cyclic iff $G$ is abelian. Using this, let us prove the result:

We know that $| Z(G)|\in\{p,p^2,p^3\}$ ($Z(G)\neq 0$ since $p$-groups have non trivial center). By hypothesis $G$ is not abelian, so $|Z(G)|\neq p^3$. If $|Z(G)|= p^2$ then $|G/Z(G)|=p $, and so $G/Z(G)$ is cyclic (a group of prime order is always cyclic), which is a contradiction since $G$ is not abelian. Then $|Z(G)|=p$

For the second part notice that $|G/Z(G)|=p^2 $. Since there are only two groups of order $p^2$, which are $C_{p^2}$ and $C_p\times C_p$, both of them abelian, then $G/Z(G)$ must be abelian.

Answer 2

While one can prove this using the oft-quoted result that if $G/Z(G)$ is cyclic then $G$ is abelian, this fact is not necessary to prove the result.

Assume, for the sake of contradiction, that the order of $Z(G)$ is exactly $p^2$. Since $G\neq Z(G)$, let $x\in G$ be an element with $x\notin Z(G)$.

Then $\langle Z(G),x\rangle$ is strictly larger than $Z(G)$, and hence must have order $p^3$ (since it divides $p^3$ but is strictly larger than $p^2$). Therefore, $\langle Z(G),x\rangle = G$.

But now let $a,b\in Z(G)\cup\{x\}$ be two generators. If at least one of them lies in $Z(G)$, then $ab=ba$. And if this is not the case, then they are both equal to $x$, so again $ab=xx=ba$. Thus, any two elements in the generating set commute, and that implies that $G$ is abelian.

But that yields that $Z(G)=G$ is of order $p^3$, a contradiction. Thus, the order cannot be exacty $p^2$.