Suppose $H$ is a normal subgroup of a group $G$ of prime index $p$.
I need to show that for any $K \leq G$ we have either:
- $K \leq H$ or
- $G = HK$ and $|K: K \cap H| = p$.
What I've determined so far:
I am able to use the second isomorphism theorem to show that if $G = HK$, then $|K: K \cap H| = p$. This is because we'd have that $$ G/H = HK/H \simeq K/K\cap H $$ and we know that $|G/H| = |G:H| = p$ because $H$ is normal, so $|K / K \cap H| = p$.
I can deduce that if $K \leq H$, then $G \neq HK$. This is because if $K \leq H$, then $HK = H < G$.
If $G$ is finite, then I could use Lagrange's Theorem, but it is not guaranteed that $G$ is finite.
So the last thing I really need to prove is that if $K \not\leq H$, then it must be the case that $G = HK$. I've worked through some simple examples and it has held true, but I'm unable to prove that it will always be true.
Note that $H \leq KH \leq G$, and $[G:KH][KH:H]=[G:H]=p$, which is prime. Since $[KH:H] \gt 1$, the only other possibility is $[KH:H]=p$, which implies $[G:KH]=1$ and $KH=G$.