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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.

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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

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  • $\begingroup$ How do you arrive at $p=2$ and $q=3$? $\endgroup$ – Al Jebr 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
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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.

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  • 2
    $\begingroup$ 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

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