Groups of order $360$ have a subgroup of order $10$ I want to prove that groups of order $360$ must have a subgroup of order $10$.
By Sylow's theorem, the number of Sylow $5$-subgroups $n_5 \equiv 1 \pmod 5$ and $n_5\mid 360$. There are three solutions: $1, 6, 36$ (let me know if I missed any).
If $n_5=1$, then the only one is normal, making the product with an element of order $2$ we get a subgroup of order $10$.
If $n_5=36$, then pick any Sylow $5$-subgroup, $[G:N_{G}(P)]=36$. It follows that $N_G(P)$ is a subgroup of order $10$.
But how to deal with the case when $n_5=6$?

If $n_5=6$ and $P$ is a Sylow 5-subgroup, you have $|N_G(P)|=60$. Set $H=N_G(P)$, and we find a subgroup of order $10$ in $H$. Your arguments work. 
In $H$, $n_5$ is either $1$ or $6$. If $n_5=1$, multiply Sylow $5$-subgroup with a cyclic of order $2$. If $n_5=6$ and $Q$ is Sylow $5$-subgroup, then $|N_H(Q)|=10$. So, $N_H(Q)\leq H\leq G$ is a subgroup of order $10$.
