Proof about groups and subgroups

Question

Let $G$ be a finite group, $H$ a subgroup of $G$ with index $p$, where $p$ is a prime number. If $K \leq G$, then either $K \leq H$ or $[K: K \cap H]=p$.

I don't get it to show that always one of the statements is true. I think it's a short proof or is it more complicate?

Answer 1

I have found this exercise in the textbook I use:

3. Prove that if $H$ is a normal subgroup of $G$ of prime index $p$ then for all $K\leq G$ either (i) $K\leq H$ or (ii) $G=HK$ and $|K:K\cap H|=p$.

Proof. Assume $K\not\leq H$. Since $H\unlhd G$, we have $HK/H\cong K/H\cap K$. Also, we have $H\leq HK\leq G$. Since $|G:H|=p$, we have $HK=H$ or $HK=G$. Suppose $HK=H$. Then $K\leq HK=H$, contradiction. So $HK=G$. Hence $G/H\cong K/H\cap K$. So $|K:H\cap K|=p$.

The exercise is from Dummit and Foote page 101.

In the link, $H=A_{G}$ is a normal subgroup of $S_{G}$, so we can still use the fact.