Non-abelian solvable group of order $p^2q$ that including no subgroup of order $pq$.

Question

I am working on a non-abelian solvable group of order $p^2q$ that including no subgroup of order $pq $. My problem is this:

Does this group have a normal sylow $p$-subgroup?

In the case that $p>q$, yeah I know it does.

For the case $p<q$, I really have no idea. I tried to show this case won't happen, but I couldn't.

How can I use the solvability of $G$?

Thanks in advanced.

Answer 1

Because $G$ is solvable but not abelian, we know it must have an abelian normal subgroup that is not trivial. Because we are also assuming $G$ has no subgroup of order $pq$ at all, then this normal abelian subgroup must be of order $p$, of order $p^2$, or of order $q$.

If it is $P$ of order $p$, and we let $Q=\langle x\rangle$ be a subgroup of order $q$, then the normality of $P$ tells us that $PQ=QP$, so $PQ$ is a subgroup. What is its order?

Similarly, if it has a normal subgroup $Q$ of order $q$, and we let $y$ be an element of order $p$, then letting $P=\langle y\rangle$ we have $QP=PQ$, hence $PQ$ is a subgroup. What is its order?

Answer 2

The case $p<q$ can occur. The alternating group $A_4$ is non abelian, solvable, has no subgroup of order $6$, and does not have a normal subgroup of order $2$ ($p=2,q=3$)