# Thread: Group of order p^2*q^2 is solvable?

1. ## Group of order p^2*q^2 is solvable?

Hi! Im having an exam in group theory in few days. Ive lost some of my solutions and cant proof this by myself.

There was also a specific example regarding group of order 36=2^2*3^2. With number of sylow 2-subgroups or 3-subgroups being one its simple. But I dont know how proceed when lets say number of sylow 3-subgroups is 4.

If anyone could help me with this problems i would be happy.

-santtu

2. Originally Posted by santtu
Hi! Im having an exam in group theory in few days. Ive lost some of my solutions and cant proof this by myself.

There was also a specific example regarding group of order 36=2^2*3^2. With number of sylow 2-subgroups or 3-subgroups being one its simple. But I dont know how proceed when lets say number of sylow 3-subgroups is 4.

If anyone could help me with this problems i would be happy.

-santtu

You only have to prove the existence of a non-trivial normal subgroup in a group $\displaystyle G\,,\,\,|G|=36$ , since then both this sbgp. and the resulting factor group are solvable and thus also $\displaystyle G$ is (assuming you already know/can prove that any group of order less than 36 is solvable).

Assume there are 4 Sylow 3-sbgps.: if all of them have trivial intersection we have $\displaystyle 4\cdot 8=32$ elements of order 3 or 9, and then the other 4 elements must be the unique Sylow 2-sbgp. which is then normal.

So assume $\displaystyle P\cap Q\neq 1\,,\,\,P,\,Q$ two different Sylow 3-sbgps. It then must be that $\displaystyle |P\cap Q|=3\Longrightarrow$ the normalizer of $\displaystyle P\cap Q$ has at least 15 elements (all the ones in both $\displaystyle P,\,Q$...why?) , so it at least has order 18 (why?), and thus either this normalizer sbgp. or $\displaystyle P\cap Q$ are normal non-trivial sbgps. of $\displaystyle G$ and we're done.

Tonio

Ps. It can be proved, but it's lenghty, that any group of order 36 has at least one Sylow sbgp. normal.