4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ There are pathological counterexamples: Take $N = 1$. Then $M \cap 1 = 1$ will not be maximal in $N$ since maximal subgroups are proper. $\endgroup$ – ε-δ Jul 13 at 11:28
  • 1
    $\begingroup$ Why aren't you ruling out $M=N$ or $N=1$? $\endgroup$ – user10354138 Jul 13 at 11:28
  • 1
    $\begingroup$ @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 $N$ $\endgroup$ – user710290 Jul 13 at 11:35
4
$\begingroup$

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 $N$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. Trivial intersection makes the cut in this case, certainly. $\endgroup$ – DonAntonio Jul 13 at 14:32
1
$\begingroup$

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 $N$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. Very nice example. $\endgroup$ – DonAntonio Jul 13 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.