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.
Best Answer
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?