$H\subseteq G$, $N\triangleleft G$, and showing $|[G:N]|$ is a prime number Let $G$ be a finite group and let $N\triangleleft G$ a normal subgroup.
It is given that if $H\subseteq G$ such that $N\subseteq H\subseteq G$, then $H=N$ or $H=G$.
Show that $|[G:N]|$ is a prime number.
I tried to solve the problem, but it was kind of confusing. I didn't quite understand how to use the given data.
A lame beginning:
By contradiction, say $|[G:N]|$ isn't prime. Then $\frac {|G|} {|N|}=p_1^{m_1}\cdot p_2^{m_2}\dots p_n^{m_n}$ for some primes $p_i$ and natural numbers $m_i$.
Let $|N|=\prod p_i^{m_i}$ , $i\in I$ for some set of indexes such that $|N|\mid|G|$ to satisfy Lagrange.
I guess the solution will be by contradiction, by showing $H\neq N$ and $H\neq G$, but I don't know how get there (how to determine the subgroup $H$?)...
Any hints?
 A: Sketch: Consider the quotient $G/N$.  Observe $|G/N|=[G:N]$.  Show that $G/N$ has no proper nontrivial subgroups iff $[G:N]$ is prime (hint: if the group is cyclic, then the structure of subgroups is known, if the group is not cyclic, then no single element generates the entire group).  If $G/N$ has a proper nontrivial subgroup $H'$, then consider $H=\pi^{-1}(H')$ where $\pi:G\rightarrow G/N$ is the natural quotient.  Then, show that $N\subsetneq H\subsetneq G$.
A: I'm not sure if your start can be made whole. It probably can, but I would advise a different route.
This result follows very nicely from the correspondence theorem, which states that

Given a normal subgroup $N$ of $G$, subgroups of $G$ containing $N$ are in bijection with subgroups of the quotient group $G/N$.

Now, since there are only two subgroups of $G$ containing $N$ (such subgroups are called maximal), there can only be two subgroups of $G/N$: the whole group, and the trivial group.
The only groups with exactly two subgroups are the cyclic groups of prime order (each nonidentity element must generate the entire group), and the result follows.
