# Index of maximal normal subgroup is prime

I would like to prove the following statement.

Let $G$ be a group, $M$ is a maximal subgroup of $G$ and $M \trianglelefteq G$. Then $[G:M]$ is finite and equal to a prime.

This problem was already posted here several times (e.g. If a maximal subgroup is normal, it has prime index). However, all the proofs I found used the correspondence theorem (they are quite intuitive). Is there another nice way to prove the statement without the correspondence theorem?

As the correspondence theorem is not an easy one to prove (at least I think so), I would like to rather not introduce it just for this problem. Do you have any other ideas on how to tackle this problem?

Thank you very much for your help!

Kind regards!

• If $G/M$ is not prime-cyclic, it has a non-trivial proper subgroup $A$, the pre-image of which defies maximality of $M$ – Hagen von Eitzen Nov 11 '17 at 19:05
• You can introduce the correspondence theorem for yourself, not only for a specific problem. Why should one avoid basic important theorems? And moreover the proof is easy. – Dietrich Burde Nov 11 '17 at 20:56
• If $M$ is a maximal normal subgroup then $G/M$ could be any simple group, not necessarily of prime order. So I guess the question is asking about maximal subgroups which happen to be normal. Is the statement false for maximal subgroups which are not normal? I.e. does there exist a maximal subgroup for which $|G/H|$ is not prime? Can anyone provide an example? – Stephen Meskin Nov 12 '17 at 8:44