Prove that $G$ abelian if $|G|= pq^2$. Let $G$ be a group of order $pq^2$, where $p \neq q$ prime and $p$ does not divide $| Aut (G) |$. Show that $G$ is abelian.
 A: I don't have that reference noted by Patrick, but I think you can use the following hint.
Hint: Use the fact that $|\frac{G}{Z(G)}|$ divides the order of $Aut(G)$.
A: Firstly,we can consider a homomorphism $f:G\rightarrow Aut(G)$ such that:
$f(x)=t_x$,where $t_x:G\rightarrow G$ is defined by $t_x(g)=xgx^{-1}$.
Note that $ker f=Z(G)$,then by first isomorphism theorm, we get:
$G/Z(G)\cong f(G) \le Aut(G)$.So $|G| $can be divided by $|Z(G)||Aut(G)|$.
Hence, $Z(G)$ is divided by $p$.By Cauchy Theorem,$Z(G)$ contains a element of order $p$.
We find that the cyclic subgroup generated by that element is of order $p$ and hence is Sylow $p$ subgroup and since it is in the center, we can conclude that it is the unique Sylow $p$ subgroup,which is also normal.
Let $Q$ be a Sylow $q$ subgroup in $G$.Since $p$ and $q$ are relative prime,we have $P\cap Q=\{1\}$.Besides, $PQ$ is a subgroup of order $pq^2$ since $P$ is normal and $|PQ|=|P||Q|/|P\cap Q|$.We have $PQ=G$.
Now, we only focus on $Q$.From class equation ,we get that $Z(Q)$ is nontrivial.If $Z(Q)$ is of order $q^2$,then $Q$ commutes with its elements, otherwise $|Z(Q)|=q$.
In this case,$Q/Z(Q)$ which is of order $q$ and hence is cyclic, so $Q$ is abelian.
In both case, $Q$ always commutes with its elements.
Since $G=PQ$,$P$ is contained in the centre and $Q$ commutes with its elements,we can conclude that $G$ is abelian.
