Let $G$ be a group such that $|G| = p^2q$, where $p, q$ are primes such that $p < q$.

Consider a Sylow $q$-subgroup of $G$. I want to show that such subgroup is normal in $G$.

Now, by Sylow's Theorems, it will suffice to show that there is only one such subgroup. But the usual divisibility and congruence conditions aren't helping me out here (e.g., I can't seem to derive a contradiction by assuming that the number of Sylow $q$-subgroups is $p$ or $p^2$).

Another approach I've attempted is considering the normalizer of such a group: Call such a subgroup $Q$. It will be sufficient to show that $N_G(Q) = G$. I've got it down to an argument regarding the order of $N_G(Q)$, where I've eliminated all possibilities except for $q$... but, again, I see no reason why $Q$ can't be self-normalizing...

Basically, I've hit this problem with everything I've got and am out of ideas.

  • $\begingroup$ This fails for $p^2q=12$. $\endgroup$ – Angina Seng Oct 27 '18 at 18:21

It's difficult to prove things that aren't true. $A_4$ is a counterexample. It has order $2^2\cdot 3$, and it does not have a normal 3-Sylow.

Order 12 is the only failing case, though. To get a counterexample, you need $p^2\equiv1\pmod q$, since $p<q$ precludes $p\equiv 1 \pmod q$. But that means $q$ divides $p^2-1=(p-1)(p+1)$, so $q=p+1$ and since both are primes we have p=2, q=3.

You can also check that $A_4$ is the only order 12 counterexample. Since any counterexample has 4 3-Sylows, it has 8 elements of order 3, so only one 2-Sylow, which is normal. So it is a semidirect product of a group of order 4 with a group of order 3 with a nontrivial action (since the 3-Sylow is not normal it can't be a direct product). The cyclic group of order 4 has no order 3 automorphism, so the 2-Sylow must be the Klein 4 group. That has only one conjugacy class of order 3 automorphisms, so there is only one such semidirect product, up to isomorphism.

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  • $\begingroup$ Oh, that's right. Incidentally, this actually gets me what I need (as this is part of a larger problem I'm working on). So, thanks! $\endgroup$ – thisisourconcerndude Oct 27 '18 at 18:48

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