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 $p$.
Proof
Since $G$ is not abelian, the order of its center cannot be $p^3$. Since it is a $p$-group, the center cannot be trivial. So the order of $Z(G)$ is either $p^2$ or $p$.
Suppose, for contradiction, that $Z(G) = p^2$. Since $p$ is prime, we can assume that a subgroup $H= \langle p \rangle$ of order $p$ exists in $G$. 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 $p^2$. Since $Z(G)$ is the center, they commute with p. Since they commute with $p$, they must also commute with all powers of $p$. So $G = Z(G) \times H \implies$ G is abelian since $H$ and $Z(G)$ are abelian. So $|Z(G)|=p$.
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Best Answer
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 $p$.
$QED$