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 $n_3 \equiv 1 \mod {3}$ and $n_3|4$ 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?
Best Answer
For $n_3=4$, if every Sylow $3$-subgroup intersects at identity, there will be $9\times4-1=35$ elements of order $3$ or $9$ but then the Sylow $2$-subgroup won't exist in this case so a contradiction.
So there are $2$ distinct Sylow $3$-subgroups $P$ and $Q$ such that $|P\cap Q|=3$. Let $M=P\cap Q$. Since $P$ and $Q$ are abelian, $P,Q\leq N_G(M)$ hence $|N_G(M)|>9$. By using Sylow's Theorem applied to $N_G(M)$, $|N_G(M)|=3^2k$ where $k\geq 4$. Hence $N_G(M)=G$ and $M\lhd G$.