1

This is inspired by my earlier duplicate question seen here: Is conjugation in infinite groups well behaved?

The answer to that question is no. This slight tweak might change the result, but I don't have much intuition for infinite groups (who does?).

Let $G$ be an infinite group, and let $M$ be a maximal subgroup (in the sense that there are no proper subgroups of $G$ containing $M$ that aren't $M$). Is it possible for $gMg^{-1}\subset M$, with the containment being proper?

Community
  • 1
user404127
  • 792
  • 4
  • 7

1 Answers1

3

Let $H$ be $gMg^{-1}$, suppose that $H$ is a proper subgroup of $G$.

Notice that $M=g^{-1}Hg$ is properly contained in $g^{-1}Mg$, so $g^{-1}Mg$ is a proper subgroup of $G$ that contains $M$ properly, contradicting the maximality of $M$.

Alteranatively: Note that conjugation is an automorphism, so by the correspondence isomorphism theorem it sends maximal subgroups to maximal subgroups.

Asinomás
  • 105,651