Let G be a nonabelian group of order $p^3$, where $p$ is a prime number. Prove that the center of $G$ is of order $p$.
I don't think the proof is correct. The following doesn't make sense to me.
We can also assume that $H$ and $Z(G)$ are disjoint. Otherwise, if there didn't exist a disjoint subgroup of order $p$, then the order of $G$ would be $p^2$.
As pointed out by user10444, typically you apply the statement
If G/Z(G) is cyclic, then $G$ is abelian.
to exclude the case $\lvert Z(G)\rvert = p^2$.
The statement is not hard to show: Let $gZ(G)$ be a generator of the cyclic group $G/Z(G)$. Then $G = \langle \{g\}\cup Z(G)\rangle$. Since each pair of generators of $G$ commutes, $G$ is abelian.
I have a question can somebody please clarify. Why is the possibility that $|Z(G)|=1$ not considered? Thanks!
UPDATE:
Here is my attempt at a solution.
Consider $Z(G)$ center of the group $G$. We know that $Z(G)\leq G$.
By Lagrange's Theorem $|Z(G)|$ must divide $|G|$.
Since $|G|=p^{3}$ the only possibilities are $1, p, p^{2}, p^{3}$.
$|Z(G)|\neq p^{3}$ because otherwise we will have $Z(G)=G$ but $G$ is non-abelian.
$|Z(G)|\neq p^{2}$ also because otherwise we will have the order of the factor group by the center as $|G/Z(G)|=|G|/|Z(G)|= p^{3}/p^{2} = p$.
We have:
$|G/Z(G)|=p \implies G/Z(G)$ is cyclic $\implies G$ is abelian. But $G$ is non-abelian.
Now $|Z(G)|\neq 1$ also because $G$ is a $p-group$ and $p-groups$ have non-trivial center.
(thanks to Tobias Kildetoft for pointing this out).
Thus the only possible order for $Z(G)$ is $p$.
$QED$