Prove Sylow subgroups $S_5S_{11}\lhd G$ when $|G|=495$

Question

Prove the sylow subgroups $S_5S_{11}\lhd G$ when $|G|=495$.

My work:

I can show either $S_5$ or $S_{11}$ must be normal.

If $S_5$ is normal, I can conclude $S_5S_{11}\lhd G$.

In the case $S_{11}\lhd G$,

1. Considering $G/S_{11}$, we have $S_3\lhd G$.

   By Lattice thm. $\exists\ K\lhd G: |K|=55$. So $K/S_{11}\cong Z_5\\$.

2. $S_5S_{11}/S_{11}\cong S_5\cong Z_5$

Answer 1

Let $H$ denote a $11$-sylow subgroup, $K$ a $5$-sylow subgroup. Let $s_p$ denote the number of sylow $p$-subgroups. Then $s_{11} = 1$ or $45$, $s_5 = 1$ or $11$.

We first show that $s_{11} = 45$ cannot occur, when this happens, counting elements readily shows $s_5 = 1$. Then $K\lhd G$, so we have a subgroup $HK$ of order $55$. A group of order $55$ has only one subgroup of order $11$, so we have $440$ elements of order $11$ not in $HK$, combined with elements from $HK$ gives exactly $495$ distinct elements of $G$, this leaves no room for order $3$ elements, contradiction.

If $s_{11} = s_{5} = 1$, then trivially $HK\lhd G$.

If $s_{11} =1, s_5 = 11$. Denote the normalizer of $HK$ in $G$ by $N_G(HK)$. Then $H\leq N_G(HK)$ and $N_G(K)\leq N_G(HK)$. Since all eleven $5$-sylow are conjugate to each other, $|N_G(K)| = |G|/11 = 45$. Hence $N_G(HK)$ contains a subgroup of order $11$ and $45$, it follows that $N_G(HK) = G$, so $HK$ is normal in $G$.