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?

Show that the only group of order $91$ is cyclic. Lift a generator of $G/Z(G)$ to some $g\in G$. Then $g$ and $Z(G)$ generate $G$, with $Z(G)$ acting trivially on $\langle{g\rangle}$. — anomaly, Nov 26 '14 at 17:33

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

3

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 $1$.

edited Apr 13 '17 at 12:21

Community

1

answered Nov 26 '14 at 17:33

rschwieb

153,510

both are unique, am I right?, there's only one 7-subgroup and only one 13-subgroup? – Miguelgondu Nov 26 '14 at 17:38
Yes, a Sylow $p$-subgroup is normal iff it is unique. – rschwieb Nov 27 '14 at 00:41

score 1 · Answer 2 · answered Nov 26 '14 at 20:09

1

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

answered Nov 26 '14 at 20:09

mesel

14,862

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

2 Answers2