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Let $G$ be a group of order $p^2q$, where $p$ and $q$ are distinct primes. Let $H$ be a subgroup of $G$ with $|H|=p^2$. Assume that $H$ is not normal in $G$.

(a) Show that $H$ has $q$ conjugates (including $H$).

(b) Show that there is a homomorphism $\phi:G\to S_q$ such that $pq$ divides $\text{Im }\phi$.

Progress:

I have shown (a). I believe the homomorphism I am supposed to take is the following: Index the permutation group $S_q$ using the $q$ conjugates of $H$. Let $g$ be an element of $G$ and let the corresponding permutation $\sigma_g$ be the permutation which takes $aHa^{-1}$ to $gaHa^{-1}g^{-1}$. This map is clearly well-defined and is indeed a homomorphism. I fail to show that final condition that $pq$ divides $\text{Im }\phi$.

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    $\begingroup$ The statement in the title of the post is false for abelian groups of order $p^2q$, which is confusing. $\endgroup$
    – Derek Holt
    Aug 4 '20 at 7:32
  • $\begingroup$ @DerekHolt This is true, but unfortunately I was not able to fit a longer title which conveys the question in a detailed and intelligent way. If anyone has any ideas, I will approve their edits. $\endgroup$
    – LAGC
    Aug 4 '20 at 17:50
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The homomorphism $\phi$ is indeed obtained from the action by conjugacy on the $q$ conjugates of $H$. Since the action is transitive, the image has order divisible by $q$. Clearly the order of the image divides $p^{2} q$.

If the image has order exactly $q$, then its kernel, a normal subgroup of $G$, has order $p^{2}$.

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  • $\begingroup$ Very nice. Thanks. $\endgroup$
    – LAGC
    Aug 4 '20 at 17:48

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