5/8 bound in group theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:53:26Z http://mathoverflow.net/feeds/question/91685 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91685/5-8-bound-in-group-theory 5/8 bound in group theory John Mangual 2012-03-20T04:35:12Z 2012-03-20T15:28:06Z <p>The <a href="http://www.ime.usp.br/~rbrito/docs/2318778.pdf" rel="nofollow">odds of two random elements of a group commuting</a> is the number of conjugacy classes of the group</p> <p>$$\frac{ \{ (g,h): ghg^{-1}h^{-1} = 1 \} }{ |G|^2} = \frac{c(G)}{|G|}$$</p> <p>If this number exceeds 5/8, the group is Abelian (I forget which groups realize this bound). </p> <p>Is there a character-theoretic proof of this fact? What is a generalization of this result... maybe it's a result about semisimple-algebras rather than groups?</p> http://mathoverflow.net/questions/91685/5-8-bound-in-group-theory/91686#91686 Answer by Will Sawin for 5/8 bound in group theory Will Sawin 2012-03-20T04:42:48Z 2012-03-20T05:36:06Z <p>If $c(G)> 5|G|/8$, then the average character has a dimension-squared of less than $8/5$, so at least $4/5$ of the characters are dimension $1$ (since the next-smallest dimension-squared is $4$), so the abelianization, which has one element for each 1-dimensional character, is more than half the size of the group, so the commutator subgroup has size smaller than $2$ and so is trivial. </p> http://mathoverflow.net/questions/91685/5-8-bound-in-group-theory/91689#91689 Answer by Igor Pak for 5/8 bound in group theory Igor Pak 2012-03-20T06:01:36Z 2012-03-20T06:08:11Z <p>There is a beautiful generalization due to Guralnick and Wilson, <a href="http://plms.oxfordjournals.org/content/81/2/405.abstract" rel="nofollow">The Probability of Generating a Finite Soluble Group</a>. Their results:</p> <p>1) if the probability that two randomly chosen elements of $G$ generate a solvable group is greater than $\frac{11}{30}$ then $G$ itself is solvable,</p> <p>2) If the probability that two randomly chosen elements of $G$ generate a nilpotent group is greater than $\frac{1}{2}$, then $G$ is nilpotent,</p> <p>3) if the probability that two randomly chosen elements of $G$ generate a group of odd order is greater than $\frac{11}{30}$ then $G$ itself has odd order. </p> <p>Interestingly, these probabilities are best possible. Note also the elementary <a href="http://www.jstor.org/stable/3615961" rel="nofollow">McHale article</a> on probability of commutativity again. </p> http://mathoverflow.net/questions/91685/5-8-bound-in-group-theory/91714#91714 Answer by Geoff Robinson for 5/8 bound in group theory Geoff Robinson 2012-03-20T13:42:13Z 2012-03-20T13:42:13Z <p>One elementary result using character theory, but going in the other direction, which is proved in the paper of R. Guralnick and myself mentioned in my comment above is that if ${\chi_1, \chi_2, \ldots, \chi_c }$ are the complex irreducible characters of $G$, where $c = c(G)$ is the numberof conjugacy classes of $G,$ then by Cauchy-Schwarz, we have <code>$\sum_{i=1}^{c} \chi_i(1) \leq \sqrt{c}\sqrt{|G|}$</code>, so that <code>$\frac{c(G)}{|G|} \geq \left( \frac{\sum_{i=1}^{c} \chi_i(1)}{|G|} \right)^{2}.$</code>. </p>