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Show that the composition factors of a group $G$ of order $p^2q$ where $p$ and $q$ are prime and $p \neq q$ are $C_p$, $C_p$ and $C_q$.

So I know that $|G|=p^2q$ is not a simple group. Not really sure where to go from here?

Any tips would be much appreciated.

My attempt at a solution (this feels pretty incomplete):

$e \triangleleft G_2 \triangleleft G_1 \triangleleft G$

Where $|G_1|=pq$ and $|G_2|=p$. Since $\frac{G_1}{G}$ and $\frac{G_1}{G_2}$ and $G_1$ are all of prime order they are cyclic and simple so this is a valid decomposition.

My question is how do I know that $G_1$ and $G_2$ are subgroups of $G$?

Thank you for your time.

Gareth Foley
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  • How...do you know it isn't simple? Do you have some non'trivial normal subgroup in mind? – DonAntonio Nov 29 '17 at 13:00
  • Exactly! Showing it is not simple is the most difficult part of the proof. – Derek Holt Nov 29 '17 at 13:06
  • I did it a few weeks ago as a separate exercise! (Also looks like it is here: https://math.stackexchange.com/questions/57938/application-of-the-sylow-theorems-to-groups-of-order-p2q)

    So if $|G| = p^2$ by Lagrange any subgroups must have order p, and are therefore cyclic and therefore simple. Is this the correct first step?

     – Gareth Foley Nov 29 '17 at 13:07
  • It is not necessarily true that there exists such a series with $|G_2| = p$. It is possible that you have to choose $G_2$ with $|G_2|=q$. You know that $G$ is not simple, so it has a normal subgroup $N$ that has order $p$, $q$, or $pq$. You need to deal with those cases separately. – Derek Holt Nov 29 '17 at 15:08

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