TITLE: Center of a group of order 77
QUESTION [8 upvotes]: Let $G$ be a group of order $77$. Then center of $G$ is isomorphic to which of the following:
$\mathbb{Z}_1$,
$\mathbb{Z}_7$,
$\mathbb{Z}_{11}$, or
$\mathbb{Z}_{77}$?
I have some difficulty with this problem. Can somebody help? 
Thanks.

REPLY [2 votes]: Before using the theorem in amWhy's post or Ink's coputantional way; you may apply Lagrange's theorem  to get the result as follows:

If $|Z(G)|=7\to |G/Z(G)|=11$ so $G$ is abelian so $7=|Z(G)|=|G|=77$ which is contradiction.
If $|Z(G)|=11\to |G/Z(G)|=7$  so $G$ is abelian so $11=|Z(G)|=|G|=77$ which is contradiction.
If $|Z(G)|=77\to |G/Z(G)|=1$ so $|Z(G)|=|G|$ which is what we were looking for.
If $|Z(G)|=1$ so the group cannot be nilpotent so you may use Ink's post here to see that $G$ is nilpotent. Another contradiction

The following was posted before adding the last line above.

However, we know that $G$ is the direct product of its two normal sylow subgroups of orders $7$ and $11$ respectively since $H_7\cap H_{11}=\{e\}$ and $|H_7H_{11}|=77$ and therefore it is nilpotent ($Z(G)\neq \{e\}$). In fact, $H_7$ is a complement of $H_{11}$ in $G$.