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Does there exist an infinite locally finite group of finite rank and bounded exponent?

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@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? – Alireza Abdollahi Feb 4 '13 at 19:49
So the answer is no by the Restricted Burnside Problem. – Derek Holt Feb 4 '13 at 20:46
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. – Ian Agol Feb 4 '13 at 20:50
yes, a group has finite rank $r$ if every finitely generated subgroup is generated by at most $r$ elements. – Kıvanç Ersoy Feb 4 '13 at 20:55
Thank you very much Professor Holt. – Kıvanç Ersoy Feb 4 '13 at 20:55
up vote 4 down vote accepted

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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Thank you very much for this rigorous answer. – Kıvanç Ersoy Feb 4 '13 at 23:19

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