Suppose $G$ is a group of order $36$ with a subgroup $H$ of order $9$. Show that either $H$ is normal in $G$ or there is a subgroup $K$ contained in $H$ of order $3$ that is normal in $G$.
This is the question and here is what I have managed:
Since $G$ has order $36=2^2\times3^2$ then any subgroup of order $9$ is a Sylow $3$ subgroup. Let $n_3$ denote the total number of Sylow $3$ subgroups of $G$. Then Sylow tells us that $n_3 \equiv 1\pmod 3$ and $n_3\mid 4$ hence we can conclude $n_3=1$ or $n_3=4$. In the case $n_3=1$ then the subgroup is unique and since Sylow subgroups are conjugate then we have that it is a normal subgroup, so this is the first part of the question.
So it remains to show that if $n_3=4$ then there are normal subgroups of $G$ of order $3$ contained in each of the $4$ Sylow $3$ subgroups. I don't know how to prove this.
Any advice?
$H$ is a $3$-Sylow. If it is not unique, then the intersection of all its conjugates has order $3$ or $1$. But such a subgroup is the normal core of $H$ in $G$, which is also the kernel of the $G$-action by left multiplication on the left quotient set $G/H$, of size $[G:H]=4$. So, if you prove that it can't be trivial, you are done. And this is the case, as otherwise you'd have an embedding $G\hookrightarrow S_4$, which is impossible because $|G|>|S_4|$.