Show there exists a subgroup of order 15

Question

I have a group $|G|=375=5^3*3$ by Sylow analysis, I have shown that $H_5$ is normal, but $H_3$ is not necessarily normal. My question is if I assume $H_3$ is normal, how do I show there is a subgroup of order 15? I know $|G/H_3|=125$ but $15$ does not divide $125$. So where does the subgroup come from?

Answer 1

If you had $H_3$ normal then take a subgroup of order $5$, call it $P$. Then $H_3P=PH_3$ (prove this) which implies $H_3P$ is a subgroup of $G$. Also prove in general if $K$ and $H$ are subgroups $KH$ is a set of order $\frac{|K||H|}{|K\cap H|}$

This means that $H_3P$ is a subgroup, and it has the desired order (since $P\cap H_3$ is trivial by lagrange)