I am trying to show that for a group $G$ with normal subgroup $N$ with index $[G:N]=p$ (where p is some prime), we have that $HN=G$ if $H$ is a subgroup of $G$ which is not contained in $N$.
So:
$N \unlhd G$
$[G:N]=p$ for some prime $p$
$H \leq G$
$ H \nsubseteq N$
Want to show that $HN=G$
My shot at a solution is, that since the product of a subgroup and a normal subgroup is again a subgroup, we know that $ HN \leq G .$ Thus, if we can show that $ |HN|=|G|,$ then $HN$ must equal to $G.$
To this end, we could possibly use that $|HN|=\frac{|H||N|}{H \cap N},$ but I didn't succeed with this strategy.
I am aware that $N \unlhd G$ gives us $[G:N]=|G/N|,$ which combined with Lagrange's theorem gives $ |G/N|=\frac{|G|}{|N|}, $ which also might be useful.
Alternatively one could ofcause also try to show that $ HN \geq G,$ but I didn't succeed in doing so either.
Any input will be highly appreciated, thanks in advance! :)
Hint. Given a group $G$, and subgroups $H\leq K\leq G$, the following equality holds: $$[G:H] = [G:K][K:H].$$ Here, $[G:H]$ is the index of $H$ in $G$, which is the number of distinct left (equivalently, right) cosets of $H$ in $G$. This equality holds in the sense of cardinalities (not just in the sense of either finite of $\infty$).
A proof that holds in the sense of cardinalities can be found in this previous answer.