# Showing that a Sylow subgroup of a group of order $p^{2}q$ is normal.

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. )

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Also look at H not normal, that implies a relation between p and q opposite to the one you already have. The two relations are not compatible. –  Jack Schmidt Oct 10 '12 at 20:23
ok thank you i was trying to go with $K$ not normal $\Rightarrow$ $H$ normal. But I will try to get a contradiction with your suggestion. –  roo Oct 10 '12 at 20:24
got it! thanks very much! –  roo Oct 10 '12 at 20:30

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.
As you noted $n_q=1,p,p^2$. I assume $p<q$. If $n_q=p$ then $n_q=1+kq=p\Longrightarrow p>q$ which is a contradiction. If $n_q=p^2$, let $Q_1, Q_2$ be two distinct $q-$ sylow subgroup. They are both cyclic of order $q$. $Q_1\cap Q_2\le Q_1$ then $|Q_1\cap Q_2|\bigg|q$ so $|Q_1\cap Q_2|=1$ or $=q$. if $|Q_1\cap Q_2|=q$ then $Q_1\cap Q_2=Q_1$ which is wrong so $|Q_1\cap Q_2|=1$. Hence, every two $q-$ sylow subgroup has $1$ in common. Now enumerate all non-trivial element of the group which lie in these $p^2$ subgroups. It is as @Makoto pointed. It is exactly the order of a $p$ sylow subgroup of $G$. Since there is not any member in shared among $p-$ sylow subgroups and $q-$ sylow subgroups so the $p-$ sylow is normal in the group.