Let $G$ be group of order $17^4$. I have to find its center $Z(G)$ and $G/Z(G)$.
$|G|=17^4$.
Since $Z(G)$ is subgroup of $G$, the order of center divides the order of the group: $|Z(G)|=17^a$, where $a\leq4$.
$1^{\circ}$ $a=4$ $\Rightarrow |Z(G)|=|G|$. Since center of the group is Abelian, and the group is non-abelian, we have contradiction.
$2^{\circ}$ $a=3$ $\Rightarrow |G/Z(G)|=17$. This means that $G/Z(g)$ is cyclic, and then $G$ is Abelian. Contradiction.
What should I do with the cases $a=2$ and $a=1$?
And also, is this the right way to do this problem? Thank you.