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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?

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

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