# Locally finite groups of finite rank and bounded exponent

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

• @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. en.wikipedia.org/wiki/Tarski_monster – 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
• I didn't mean "proper subgroups are finite", I mean "every finitely generated subgroup is finite". So, Tarski monsters are not example. Thanks. – Kıvanç Ersoy Feb 4 '13 at 20:57

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.