At Aluffi's algebra chapter 0, Claim 2.16 is :
If $p<q$ are prime integers and $q$ is not congruent to $1$ modulo $p$. Let $G$ be a group of order $pq$. Then $G$ is cyclic.
At first, he showed $G$ has a normal subgroup $H$ of order $p$. and since H is normal, conjugation gives an action of $G$ on H, hence there exist a homomorphism $\gamma$ : $G \rightarrow Aut(H)$.
Now we know $H$ is cyclic of order $p$, so $|Aut(H)|=p-1$.
Here is the point where I can't get: the author says "the order of $\gamma(G)$ must divide both $pq$ and $p-1$."
I don't know why it has to divide $pq$. Could you let me know what the reason is?
Hint: $\gamma$ is a homorphism on $G$. By the fundamental homomorphism theorem, $G$ and $\gamma(G)$ are related how? What does this mean for the relationship between $|G|$ and $|\gamma(G)|$?