Does there exist an infinite locally finite group of finite rank and bounded exponent?
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$\begingroup$ @Ersoy: By a group $G$ is of finite rank you mean that there exists an integer $r$ such that every finitely generated subgroup of $G$ can be generated by at most $r$ elements. Am I right? $\endgroup$– Alireza AbdollahiCommented Feb 4, 2013 at 19:49
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4$\begingroup$ So the answer is no by the Restricted Burnside Problem. $\endgroup$– Derek HoltCommented Feb 4, 2013 at 20:46
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$\begingroup$ finite rank and locally finite implies finite, so no. You probably mean by locally finite that proper subgroups are finite. In this case, you can consider the Tarski Monsters. en.wikipedia.org/wiki/Tarski_monster $\endgroup$– Ian AgolCommented Feb 4, 2013 at 20:50
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$\begingroup$ yes, a group has finite rank $r$ if every finitely generated subgroup is generated by at most $r$ elements. $\endgroup$– Kıvanç ErsoyCommented Feb 4, 2013 at 20:55
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1$\begingroup$ @Kivanc: Ok, I understand your question now - the group should be infinitely generated. The terminology "rank" has a different meaning in geometric group theory, denoting the minimal number of generators for the group, which led to my confusion. So you want an infinitely generated group, all of whose finitely generated subgroups are finite and of bounded exponent and (geometric) rank. $\endgroup$– Ian AgolCommented Feb 4, 2013 at 21:54
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2 Answers
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I'll expand on Derek Holt's comment, which answers your question. Suppose one has a group $G$ of the type you describe, so that finitely generated subgroups are generated by $r$ elements and have exponent $n$. Consider a finitely generated subgroup $K< G$. By the restricted Burnside problem, there is a universal constant $R(r,n)$ such that $|K|\leq R(r,n)$. Now, choose the largest size subgroup $K< G$ which is finitely generated. Since $K$ is finite and $G$ is infinite, there exists $g\in G-K$ such that $K < \langle K, g\rangle <G$ is finitely generated, so $\langle K, g\rangle$ must be finite. But since $|K|$ is maximal, we have $K=\langle K,g\rangle$, so $g\in K$, a contradiction.
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A classical 1964 theorem of Hall and Kulatilaka is that every infinite locally finite group has an infinite abelian subgroup.
So, if the exponent is finite, the subgroup has to be abelian. But rank $r$ and exponent $\le n$ for an abelian group obviously implies cardinal $\le n^r$. We deduce a contradiction, and hence that every locally finite group of finite exponent is finite.
(No need of the much harder, and more recent, restricted Burnside problem.)
