10
$\begingroup$

Let $p,q$ be primes and let $G$ be a group of order $p^2q^2$, what's the best way to show $G$ is non-simple?


I know it suffices to show that one of the Sylow-p or Sylow-q subgroup of $G$ is normal, but the counting elements argument doesn't work here since different Sylow subgroups may have non-trivial intersection.

$\endgroup$
11
$\begingroup$

Using a little more group theory allows us to prove something stronger (and avoid the reduction to $|G|=36$):

A group of order $p^2q^2$ has either a normal Sylow $p$-group or normal Sylow $q$-group.

For assume that $p<q$, then there are either $1$ or $p^2$ Sylow $q$-groups in $G$.

If there is $1$, it is normal, and we are done.

If there is $p^2$, then the Sylow $q$-groups are self-normalizing. But any group of order $q^2$ is abelian, so Burnside's transfer theorem implies the Sylow $p$ group is normal.

$\endgroup$
1
  • 4
    $\begingroup$ I LOVE when people prove extra things, especially when they state explicitly that they are doing so. It seems the answerer has vanished, but thanks anyway, ghost person! $\endgroup$
    – The Count
    Sep 29 '16 at 13:35
11
$\begingroup$

We may as well assume $p < q$. The number of Sylow $q$-subgroups is $1$ mod $q$ and divides $p^2$. So it is $1, p$, or $p^2$. We win if it's $1$ and it can't be $p$, so suppose it's $p^2$. But now $q \mid p^2 - 1$, so $q \mid p+1$ or $q \mid p-1$.

Thus $p = 2$ and $q = 3$. The case of order $36$ has been proved in an earlier question:

No group of order 36 is simple

$\endgroup$
2
  • $\begingroup$ How do you arrive at $p=2$ and $q=3$? $\endgroup$ Apr 11 '15 at 0:03
  • $\begingroup$ $p<q$, so we must have $q=p+1$ if $q$ divides either $p-1$ or $p+1$. $\endgroup$
    – Nishant
    Apr 11 '15 at 0:49
0
$\begingroup$

I'll give another proof similar to user29743's answer but avoiding to examine groups of order $36$.

Suppose that $p<q$ and $n_q=p^2$.

  • If $Q_i\cap Q_j=1$ for every two distinct Sylow $q-$subgroups then by simple counting we find that $n_p=1$ hence $G$ is not simple.

  • Let $Q_i,Q_j$ be two Sylow $q-$subgroups with $1\not= I= Q_i\cap Q_j$. Since $|Q_i|=|Q_j|=q^2$ these are abelian groups and $I\lhd Q_i,Q_j \Rightarrow 1\not=I\lhd \langle Q_i,Q_j\rangle=M$. It has to be $|M|>|Q_i|=q^2\Rightarrow |M|=p^2q^2$ or $pq^2$.

  1. If $|M|=p^2q^2=|G|$ then clearly $G$ is not simple $\checkmark$

  2. If $|M|=q^2p$ then $|G:M|=p$ so we have a homomorphism $h:G\to S_p$. If $kerh=1$ we have a contradiction since $p^2q^2\not| p!$ so it has to be $1\not= kerh \leq M\lneq G \ \checkmark$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.