Using 2nd Isomorphism thm Prove the problem

Question

Let $H \lhd G$ with $[G:H]=p$, where p is a prime $\mathbb{Z}$.

Then $\forall K \leq G$,

either

(1) $K \leq H$

or

(2) $G=HK$ and $[K:K \cap H]=p$

I've reread the textbook and tried it by definition, but none of them seem to work.

Could someone give me a proof?

Answer 1

This is Exercise $3.3.3$ here, and a detailed solution is given. It uses first that $HK$ is a subgroup, and then also the second isomorphism theorem to obtain the conclusion.

Answer 2

Suppose $\;K\rlap{\;\,/}< H\;$ . Observe $\;HK\;$ is a subgroup of $\;G\;$ (why?), and also

$$\;K/(K\cap H)\cong KH/H\le G/H\implies HK=G$$

as a group of order a prime has only two subgroups: the trivial ones. End the proof now.