Suppose that I have a group $G$ of order $p^{2}q$ for two distinct primes $p$ and $q$.
I need to first show that one of its Sylow subgroups is normal.
I start by letting $H$ be a Sylow $p$-subgroup and $K$ be a Sylow $q$-subgroup.
If $K$ is not normal, then letting $r$ denote the number of Sylow $q$-subgroups, I have that $r|p^{2}$ and $r\equiv 1$(mod $q$). I also have by the second Sylow theorem that $r\neq 1$.
So $r = p$ or $p^{2}$.
If $r = p^{2}$, I can write out the Sylow $q$-subgroups, and count their combined elements to show that $H$ is uniquely determined and then again imply Sylow number $2$ to obtain that $H$ is normal.
But if $r = p$, then I cannot see how to proceed. Can anyone give any advice?
Thank you.
( I am happy to provide more detail for the $r = p^{2}$ case if it is desired or appropriate. )
Best Answer
Let $N_q$ be the number of Sylow $q$-subgroups. Then $N_q = 1$ or $p$ or $p^2$.
Suppose $N_q = p$. Since $p \equiv 1$ (mod $q$), $p > q$. Similarly if the number of Sylow $p$-subgroup is $q$, $q > p$. This is a contradiction. Hence there is only one Sylow $p$-subgroup.
Suppose $N_q = p^2$. Then the number of elements of order $q$ is $p^2(q - 1)$. Hence the number of elements of order not equal to $q$ is $p^2q - p^2(q - 1) = p^2$. Hence there is only one Sylow $p$-subgroup.