Under what assumptions will $M \unlhd G$? ($M$ maximal) Let $G$ be a finite group (non-trivial), and let $M$ be a maximal subgroup of $G$.

My question is, what criterion on $M$  would allow us to deduce that $M \unlhd G $? (I mean, will $M \unlhd G$ hold when $M$ is cyclic? Abelian? Nilpotent? Metanilpotent? Solvable? None of the above?)

I know already that if we assume $G$ is nilpotent, every maximal subgroup will be normal, hence our particular $M$ is normal. But if we assume nothing about $G$ (other than that it is finite), I don't know what will happen.
Note: Although I would be willing to accept an answer for the finite case alone, I would also be interested to know if the result would be different for $|G|= \infty$.
 A: This is not a complete answer to your question, but it's certainly an answer to your question, so I'll offer it up.
It is a theorem that every nilpotent group has the property that its maximal subgroups are automatically normal. This does not seem to characterize nilpotence though. Here's a proof.
Claim: Every nilpotent group has the property that it has no self-normalizing proper subgroup (this just means the group is its own normalizer).
Proof of Claim: We do induction on the nilpotency class of the group $G$.
Let $H$ be a proper normal subgroup of $G$ and let $N(H)$ be its normalizer. We prove $H$ is a proper subgroup of $N(H)$. Let $Z(G)$ be the center. Clearly $Z(G) \leq N(H)$ since this holds in any group by definition of the center. If $Z(G)$ is not contained in $H$ (which is the generic case), then of course $H$ cannot be its own normalizer. However, it is possible that $Z(G)$ is a subgroup of $H$. In this case, consider instead $H/Z(G)$ in the group $G/Z(G)$. This reduces the nilpotency class by $1$. We apply the inductive hypothesis then and it follows that $H/Z(G) \neq N_{G/Z(G)}(H/Z(G))$. But this more complicated normalizer is the same thing as $N_G(H)/Z(G)$. This proves $H/Z(G) \neq G/Z(G)$ and so $H \neq N(H)$.
But this condition established in the claim actually implies maximal subgroups are normal! This is easy to see. If $H$ is maximal, then $N(H)$ is either $G$ or $H$. But one of these cases is impossible by the condition!
