Let $G$ be a group of order $p^2q$. Show that there is a homomorphism $G\to S_q$ such that $pq$ divides $|\text{Im }\phi|$.

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 $$\text{Im }\phi$$.

• The statement in the title of the post is false for abelian groups of order $p^2q$, which is confusing. Aug 4 '20 at 7:32
• @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.
– LAGC
Aug 4 '20 at 17:50

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 $$p^{2}$$.