Locally finite groups of finite rank and bounded exponent - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:41:46Z http://mathoverflow.net/feeds/question/120795 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120795/locally-finite-groups-of-finite-rank-and-bounded-exponent Locally finite groups of finite rank and bounded exponent Kıvanç Ersoy 2013-02-04T19:24:24Z 2013-02-04T22:19:05Z <p>Does there exist an infinite locally finite group of finite rank and bounded exponent?</p> http://mathoverflow.net/questions/120795/locally-finite-groups-of-finite-rank-and-bounded-exponent/120807#120807 Answer by Agol for Locally finite groups of finite rank and bounded exponent Agol 2013-02-04T22:13:49Z 2013-02-04T22:19:05Z <p>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 $K&lt; G$. By the <a href="http://en.wikipedia.org/wiki/Burnside%2527s_problem#Restricted_Burnside_problem" rel="nofollow">restricted Burnside problem</a>, there is a universal constant $R(r,n)$ such that $|K|\leq R(r,n)$. Now, choose the largest size subgroup $K&lt; G$ which is finitely generated. Since $K$ is finite and $G$ is infinite, there exists $g\in G-K$ such that <code>$K &lt; \langle K, g\rangle &lt;G$</code> 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. </p>