Any group of order $p^2q^2$ is not simple for $p,q$ prime

Asked Jul 13 '19 at 04:15

Active Jul 16 '19 at 15:57

Viewed 264 times

I want to know if the following is correct:

If $G$ is abelian, then is trivial since any subgroup is normal and because of Lagrange, it has an element of order $p$.

If $G$ is not abelian, then, I can supose $q > p$, so, Sylow says that $n_p | q^2$, $n_p \equiv 1(p)$, $n_1 | p^2$, $n_q \equiv 1(q)$, so $n_q =1$ or $n_q = p^2$, since if $n_q=p$, it can't happen that $n_q \equiv 1(q)$.

If $n_q = 1$ then $G$ is not simple. Otherwise, $n_q = p^2$, so $p^2 \equiv 1(q)$, so $q | p^2-1 = (p-1)(p+1)$. Since $q$ is prime, this means that $q|p-1$ or $q|p+1$. But it can't be that $q|p-1$ for the same reason as before. So $q|p+1$. But the only primes that can accomplish this are $p=2$ and $q=3$. So I'm left with the case that $|G|=36$, but this is a case I already know from before that is not simple.

Is this correct?

Thanks

edited Jul 16 '19 at 15:57

asked Jul 13 '19 at 04:15

Silkking

Yes $;;;;;$ – TomGrubb Jul 13 '19 at 04:20
You don't even have to deal with the abelian case separately – TomGrubb Jul 13 '19 at 04:21
Why can you suppose $q > p$? What if $q=p$? It's still true in that case, but you need a different proof. – Jul 13 '19 at 04:29
3

"If $n_q=1$ then $G$ is simple" - you mean not simple (the $q$-Sylow subgroup is unique, hence characteristic, hence normal). – Jul 13 '19 at 04:32
Bungo, if q=p, then it's a p-group, then the center is non trivial – Silkking Jul 13 '19 at 21:07
@DaniSeidler Indeed. – Jul 15 '19 at 02:06

Any group of order $p^2q^2$ is not simple for $p,q$ prime

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