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So, going through my qual prep classes, there's a two part question: For $p,q$ distinct primes, prove that a group of order $p^2q$ has a nontrivial proper normal subgroup, then prove that it's solvable. Part 2 is pretty trivial given the first, but I've been struggling with it.

Thoughts so far: Attacking with Sylow theorems, if $n_p=1$ or $n_q=1$ we'd be done, so assume neither are one (either for contradiction or to find a group of order $pq$, since I'll show later that such a group would have to be normal)

So, given that we need $n_p=q$ and $n_q=p$ or $n_q=p^2$, but since we also need $n_p>p$ and $n_q>q$, we must have $p<q<p^2$, and $n_q=p^2$. We also have $q=pk+1$ and $p^2=qj+1$ for some natural numbers $k,j$ due to congruency, but I couldn't seem to find a contradiction out of here.

Now, since $p$ is the smallest prime, if I could find a subgroup of order $pq$, it would have index $p$ and thus be normal by the theorem that groups of the smallest prime index of a group are normal. However, I can't seem to figure out why either that would have to be the case, or why the above statements are a contradiction to force a normal $p^2$ group or a normal $q$ group.

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  • $\begingroup$ Try to get as much as you can out of $q = n_p \equiv 1 \mod p$ and $p^2 = n_q \equiv 1 \mod q$. (This is mostly a number theory question.) $\endgroup$ – darij grinberg Aug 9 '15 at 17:47
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You can use a counting argument. Note that if there are $p^2$ subgroups of order $q$, each of them has $q-1$ distinct elements of order $q$ generating that subgroup. Thus, only $p^2 = p^2q - p^2(q-1)$ elements can have orders $1, p, p^2$. This implies that there can be only one subgroup of order $p^2$, as desired.

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For the contradiction: Since $p^2-1=qj$ then $p^2-1$ is a multiple of $q$. Hence, $q$ divides $p-1$ or $p+1$. Since $q>p$ then $q=p+1$, which implies $q=3$ and $p=2$.

This paper lists all possible groups of order $12$ (and includes a proof that none of them are simple).

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Hint: If there are $q$ Sylow $p$-subgroups, then either two of them intersect non-trivially, or between them the Sylow $p$-subgroups cover $1+q(p^2-1)=p^2q-(q-1)$ elements.

  • In the latter case there is room for only a single Sylow $q$-subgroups.
  • In the former case let $H$ be a non-trivial intersection of two Sylow $p$-subgroups: $H=P_1\cap P_2$. Show that the normalizer of $H$ has more than $p^2$ elements.
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  • $\begingroup$ Initially I got this wrong - I tried Rolf's approach, but didn't quite manage :). Should be ok now. $\endgroup$ – Jyrki Lahtonen Aug 9 '15 at 17:54

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