Let G be a simple group of exponent 2^n for n>1. Is G necessarily finite? If not, what is an example of an infinite simple group of exponent a power of 2?
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6$\begingroup$ No (for $n$ large enough). For $n$ large enough it's known that the free Burnside group $B(2,2^n)$ is infinite, and also known that it's virtually aperiodic. (I call a group aperiodic if its only finite quotient is $\{1\}$). This finite index aperiodic subgroup being finitely generated and nontrivial, it has a simple quotient, necessarily infinite. $\endgroup$– YCorCommented Feb 10, 2020 at 19:17
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For every large enough $n$, there exist infinite simple groups of exponent dividing $n$.
(I call a group aperiodic if it has no nontrivial finite quotient.) Indeed, by the solution to the restricted Burnside problem the Burnside group $B(2,n)$ is known to be virtually aperiodic; let $B(2,n)^\circ$ be its (unique) minimal finite index subgroup, it is thus finitely generated. For $n$ large enough, $B(2,n)$ is known to be infinite. Hence $B(2,n)^\circ$ is infinite too, hence has a simple quotient, also aperiodic and hence infinite. $\Box$
The above proof produces groups of exponents dividing $n$, probably with some more efforts we can get exponent exactly $n$ but I think it's unimportant. I don't know if one can obtain quasi-finite groups in this context, when the exponent is $2^m$ (quasi-finite means infinite with all proper subgroups finite); these are close variants of Tarski monsters. Also probably one can produce infinitely generated countable examples, and uncountable too, but this requires other/additional arguments. Nevertheless:
$\forall n$, there exists no infinite locally finite simple group of exponent $n$. (Nor even in which every element has order dividing some power of $n$.)
For $n=2^k$ (or more generally $p^aq^b$ with $p,q$ prime, these would be locally solvable and indeed Malcev proved that there is no infinite, locally solvable simple group. In general, this is essentially due to Hartley (1995) (Springerlink behind paywall). Namely he proved that for every finite subset $F$ in a simple locally finite group $G$, there exists a finite subgroup $H$ containing $F$, and a normal subgroup $N$ of $H$ such that $H/N$ is simple and $F$ projects injectively into $H/N$. Since (by classification) finite simple groups of given exponent have bounded order, this implies the result. (Hartley also quotes Meierfrankelfeld for the same result.) The statement is explicit in a 2005 paper by Cutolo-Smith-Wiegold (ScienceDirect), Lemma 4.
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1$\begingroup$ To finite group theorists "$X$ is involved in $Y$", usually means$Y$ has subgroups $A,B$ with $B \lhd A$ such that $A/B \cong X$" (probably what you mean by "$X$ is a subquotient of $Y$"). . Hartley wasn't exclusively a finite group theorist, but I knew him, and I think he would have used "involved" in that sense. $\endgroup$ Commented Feb 10, 2020 at 21:20
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$\begingroup$ Thank you for your reply. So for sure I can find an infinite simple groups with exponent a power of 2, as the simple factor of the Burnside group B(2,2^m) for m large enough, right? Do you have any good reference for this? Thanks again. $\endgroup$– marcosCommented Feb 10, 2020 at 22:40
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$\begingroup$ @marcos you missed the point: a priori these simple quotients of $B(2,2^m)$ might be finite. I pass to the aperiodic finite index subgroup and then take a simple quotient and then I'm sure that the quotient is infinite. The cost is that the number of generators is not $2$. Probably it's true that $B(2,2^m)$ has infinite simple quotients too, but would require a further argument. $\endgroup$– YCorCommented Feb 10, 2020 at 22:43
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$\begingroup$ I see, so I need to consider the aperiodic finite index subgroup to conclude. Is there any reference where I can find all of it? Thanks. $\endgroup$– marcosCommented Feb 11, 2020 at 9:09