Suppose that $N$ is a normal subgroup of a finite group $G$, and $H$ is a subgroup of $G$. If $|G/N| = p$ for some prime $p$, then show that $H$ is contained in $N$ or that $NH = G$.
I imagine this is related to the fact that $|NH| = |N||H|/|N \cap H|$, but this is not really helping me. I considered the fact that since $N$ is normal, we get that $NH \leq G$, and I then used Largrange, but I'm stuck, and some help would be nice.
Best Answer
Consider the homomorphism $\phi:G\rightarrow G/N$ that sends $x$ to $xN$. Since $\phi(H)\leq G/N$, therefore $|\phi(H)|||G/N|$. Hence $|\phi(H)|=1$ or $|\phi(H)|=p=|G/N|$. In the first case we get $\phi(H)=\{N\}$, thus $H\leq N$. In the other case $\phi(H)=G/N$, we deduce that $\forall x\in G\ \exists h\in H[xN=hN]$, it is easy to show that this implies that $NH=G$.
(Note that we don't need G to be finite )