There is no simple group of order $pq^2$.

Asked Oct 02 '19 at 15:06

Active Oct 02 '19 at 15:13

Viewed 134 times

I want to prove following:

Suppose $p$ and $q$ are distinct primes. Then there is no simple group of order $pq^2$.

If $|G|=pq^2$, then there are two Sylow subgroups of order $p$ and $q$. Then I think I need to use third Sylow theorem. Third Sylow theorem implies that $n_p\mid q$. Thus $n_p=1$ or $n_p=q$. When $n_p=1$ we are done. So suppose $n_p=q$. Then how do I proceed the proof?

edited Oct 02 '19 at 15:13

Shaun

44,997

asked Oct 02 '19 at 15:06

1

Closely related. – Jyrki Lahtonen Oct 02 '19 at 15:14
1

It is also possible that $n_p=q^2$. For example $A_4$ has $4=2^2$ Sylow $3$-subgroups. – Jyrki Lahtonen Oct 02 '19 at 15:25
1

But think about the rest of the Sylow theorems and $n_q$ as well. The Sylow $q$-subgroups are abelian. If any two of them intersect non-trivially, the non-identity elements in there have very large centralizers. – Jyrki Lahtonen Oct 02 '19 at 15:26
1

Close to being a duplicate of this. I'm not sure I want to use my dupehammer vote right away without a chance for someone to air objections. – Jyrki Lahtonen Oct 02 '19 at 15:32

There is no simple group of order $pq^2$.

0 Answers0