For any group $G$, $|G/Z(G)| \neq 91$.

Question

In Malik's Fundamentals of abstract algebra, one can find the following problem:

Prove that for any group $G$, $\vert G/Z(G)\vert \neq 91$.

This exercise is just ahead of Sylow's theorems.

I've tried a couple of things, like using the class equation to derive a contradiction or treating the order of $G$ as $|G| = 91\cdot|Z(G)| = 7\cdot 13\cdot|Z(G)|$ and applying Sylow's first theorem, but I'm yet unable to prove the statement. Do you have any hints?

Answer 1

By Sylow's theorems, the group must have a normal $7$-subgroup and a normal $13$-subgroup, and so it is the direct product of these two subgroups, and is cyclic.

Then the problem then becomes a special case of this problem.

The upshot is that the only possible size of $G/Z(G)$, if it is cyclic, is $1$.

Answer 2

$G/Z(G)$ can not be cyclic if $G$ is nonabelian and the group of order $91$ is cyclic as $7$ does not divide $13-1$.