Let $p,q$ be prime numbers, $q \neq 3$, and $G$ be a group with order $p^2q^n$. Prove that $G$ has a normal subgroup with order $q^n$. Prove that $G$ is a semidrect product of a group with order $q^n$ and a group with order $p^2$.
My attempt: For $n \geq 2$, $q^n>p^2$ and we have a unique $q$-sylow group, thus, we have our normal subgroup. I'm having some problems to show that the group is normal when $n=1$, but I think I can handle this.
For the semidirect product I don't know what I can do, any hint?
This is not true when $p \gt q$: take $G=S_3 \times C_6$, so a group of order $36=3^2 \cdot 2^2$, hence $p=3, q=2=n$. A Sylow $2$-subgroup of $G$ is not normal.
But it is true when $p \lt q$. Let $Q \in Syl_q(G)$, then $n_q(G) \in \{1,p,p^2\}$.
If $n_q(G)=p$, then $p \equiv 1$ mod $q$, hence $q \leq p-1 \lt p \lt q$, which is absurd.
If $n_q(G)=p^2$, then $q \mid p^2-1=(p-1)(p+1)$, hence by the previous argument $q \mid p+1$, but $p \lt q$, so this can only happen when $q=p+1$, meaning $p=2$ and $q=3$, contradicting $q \neq 3$.
So $Q \lhd G$, and if $P \in Syl_p(G)$, $PQ$ is actually a subgroup, but then $G=PQ$ and $P \cap Q=1$, so $G$ is a semi-direct product $P \ltimes Q$.