Let $G$ be a group where $|G|=p^2q$, such that $p,q$ are prime and $q\nmid Aut(G)$ then $G$ is abelian Let $G$ be a group where $|G|=p^2q$, such that $p,q$ are prime and $q\nmid Aut(G)$ then $G \cong P\times Q$, and moreover $G$ is abelian. I can't seem to figure out how to even start. I know for example that if $p<q$ then $G$ is not simple. I'm reviewing the Sylow theorems in my algebra class and I can't think of any relation I can draw here with respect to the $Aut(G)$ group. Any help would be apriciated. Thanks
 A: Let $\alpha\in G$ of order $q$. Notice that $f:G\to G$ by $f(x)=\alpha $$x\alpha^{-1}$ is an automorphism of order $1$ or $q$. Since $|Aut(G)|$ is not divisible by $q$, $f$ is an trivial isomorphism, that is every, element of order $q$ lies in the center.
Thus, it has unique subgroup $H$ of order $q$ (as it lies in the center it is automatically normal and unique). Let $K$ be a Sylow $p$-subgroup of $G$. Notice that $N_G(K)\geq H,K$. Thus $H$ is also normal. As a result $G\cong H\times K$. It follows that $G$ is abelian as both $H$ and $K$ are abelian.
A: Let $G_p$ be the $p$-subgroup and $G_q$ be the $q$-subgroup.
We have a homomorphism $G_q \to \operatorname{Aut}(G)$ by conjugation.
The image must be trivial, by Lagrange.
Therefore, $G_q \le Z(G)$.
Let $f : G_p \times G_q \to G$ be defined as $f : (g,h) \mapsto gh$.
Injectivity: if $gh=g'h'$, then $g^{-1}g'h'h^{-1} = e$. Since $G_p$ and $G_q$ have trivial intersection, $g^{-1}g' = h'h^{-1} = e$, so $g=g'$ and $h=h'$.
Surjectivity: by counting.
Homomorphism: $ghg'h' = gg'hh'$ because $hg'=g'h$ because $g'hg'^{-1}=h$.
Therefore, $f$ is an isomorphism, i.e. $G_p \times G_q \cong G$.
However, $G_p$ and $G_q$ must be abelian, so $G$ is also abelian.
