# example of a group which has no proper subgroup of finite index, but it does have maximal subgroups.

Let G be a group. if G has no proper subgroup of finite index, can we say that it has no maximal subgroup? if it is not true, what's the counterexample for this assertion?

• You probably mean no proper subgroup of finite index. – Derek Holt May 10 '16 at 18:18
• yes. thanks. @DerekHolt – Saeed_T May 10 '16 at 18:19
• Tarski monsters are counterexamples. – Derek Holt May 10 '16 at 18:19
• yes. it's helpfull. thank you. – Saeed_T May 10 '16 at 18:21

As I said in my comment, Tarski monsters are counterexamples, because they are infinite but all subgroups have finite order $p$ for some (large) prime $p$.
But for more comprehensible counterxamples, consider the groups ${\rm PSL}(n,K)$ with $n \ge 2$ and $K$ an infinite field. They are infinite simple groups, so they have no proper subgroups of finite index. But they act $2$-transitively on the set of $1$-dimensional subspaces of $K^n$, and so the point stabilizer (i.e. the stabilizer of a $1$-dimensional subspace) is a maximal subgroup.