Intersection of maximal subgroup with subgroup is maximal in subgroup?

As soon as it looks obvious one way, it looks obvious the other way: if $$G$$ is a group, $$M$$ is maximal subgroup and $$N$$ is subgroup of G, is $$M\cap N$$ maximal subgroup of $$N$$? What if $$N\lhd G$$?

In last case I thought: $$M/(M\cap N)\cong MN/N\le G/N\;$$ ...but I can't continue.

Other condition: same question if we assume $$N\lhd G$$ is of finite index...? Then in the above we have $$G/N\;$$ finite group, so $$\;M/N\le G/N\;$$ also finite...but still stuck.

Any help/direction will be thanked.

• There are pathological counterexamples: Take $N = 1$. Then $M \cap 1 = 1$ will not be maximal in $N$ since maximal subgroups are proper. – ε-δ Jul 13 at 11:28
• Why aren't you ruling out $M=N$ or $N=1$? – user10354138 Jul 13 at 11:28
• @ε-δ I think proper argument is $1$ is properly contained in a proper subgroup $\langle a \rangle$ (say) where $a$ is non-trivial element in $N$ – user710290 Jul 13 at 11:35

Let $$G=S_4$$, $$M=S_3$$ and $$N=V_4$$. Clearly $$M$$ is a maximal subgroup of $$G$$, and $$N$$ is a normal subgroup of $$G$$. Also, $$M\cap N=1$$ is not maximal in $$N$$.
Obvious counterexamples are when $$M\ge N$$, so let's assume that's not the case.
In no case is this true. For a finite example take $$G=S\times S$$ with $$S$$ simple and non-abelian (e.g. $$A_5$$), $$N=S\times\{1\}$$ and $$M$$ the diagonal subgroup $$\{(g,g)|g\in S\}$$.
It's a nice exercise to show $$M$$ is maximal in $$G$$, but $$M\cap N=1$$ is clearly not maximal in $$N$$