Let $H\unlhd G$ with $[G:H]=p$, prime. Let $K\le G$ and $K\not\subseteq H$. Then is $G=HK$?

Question

Let $G$ be a group and $H$ be a normal subgroup of index $p$ (a prime); suppose $K$ is a subgroup of $G$ not contained in $H$, then is it true that $G=HK$?

I know that the fact is true if $p=2$, but I don't know for the general case.

Answer 1

By Lagrange's,

$$[G : HK][HK:H] = [G:H] = p.$$

Since $K \not\subset H$, $[HK:H] > 1$ implies $[HK:H] = p,$ so $[G:HK] = 1$ implies $G = HK$.

Answer 2

Let $p:G\rightarrow G/H$ be the projection, there exists $x$ in $K$ not in $H$ this implies that $p(x)$ is a generator of $G/H$ since it is a non trivial element of $Z/p$. Let $y\in G$, $p(y)=p(x)^n=p(x^n)$, this implies that $p(yx^{-n})=1$ and $y=hx^n$, $h\in H$.