# Prove G is a solvable group

• Mar 16th 2010, 03:47 PM
r45
Prove G is a solvable group
Hi there,

I am trying to solve this problem:

Suppose G is a finite group, and the order of G is \$\displaystyle {p_1}^2 {p_2}^2\$, where \$\displaystyle p_1,p_2\$ are primes such that \$\displaystyle {p_1}^2 < p_2\$. Show that G is a solvable group.

I've tried various approaches through Sylow groups and normal subgroups etc but haven't been able to do it, can anyone lend a hand?

Many thanks!
• Mar 16th 2010, 06:11 PM
Black
Let's look at the number of Sylow \$\displaystyle p_2\$-subgroups. By the Sylow theorems, since \$\displaystyle n_{p_2}=kp_2+1\$ and \$\displaystyle n_{p_2}|p_1^2\$, \$\displaystyle n_{p_2}\$ must be 1. Thus, the Sylow \$\displaystyle p_2\$-subgroup, \$\displaystyle P\$ is normal in \$\displaystyle G\$. Now consider

\$\displaystyle 1 \unlhd P \unlhd G\$.

Any group of order \$\displaystyle p^2\$ (for p prime) is abelian, and since \$\displaystyle |G/P|=p_1^2,\$ \$\displaystyle G/P\$ is abelian.