Given a rational number a/b does there exist a finite group G and an automorphism f s.t. f maps exactly a/b elements of G to their own inverses? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T17:11:33Z http://mathoverflow.net/feeds/question/84842 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84842/given-a-rational-number-a-b-does-there-exist-a-finite-group-g-and-an-automorphism Given a rational number a/b does there exist a finite group G and an automorphism f s.t. f maps exactly a/b elements of G to their own inverses? Anthony Deluca 2012-01-03T22:32:02Z 2012-01-04T00:08:25Z <p>I was helping a friend prepare for his intro abstract final and he mentioned the professor had once asked the question name a group and an automorphism that takes 3/4 of the elements of the group to their own inverses (D4, identity). I tried to figure out how to approach this question in general but can't see how.</p> <p>Further questions:</p> <p>Can we construct the group?</p> <p>Does anything change if we allow infinite groups?</p> http://mathoverflow.net/questions/84842/given-a-rational-number-a-b-does-there-exist-a-finite-group-g-and-an-automorphism/84844#84844 Answer by Anthony Quas for Given a rational number a/b does there exist a finite group G and an automorphism f s.t. f maps exactly a/b elements of G to their own inverses? Anthony Quas 2012-01-04T00:08:25Z 2012-01-04T00:08:25Z <p>This may be a well-known chestnut? (well-known to those that know it well, that is)</p> <p>The fraction can never be between 3/4 and 1. To prove this, suppose $\phi\colon G\to G$ is an automorphism of $G$ that sends more than 3/4 of the elements of $G$ to their inverses. Let $S=\lbrace g\in G\colon \phi(g)=g^{-1}\rbrace$.</p> <p>Notice that if $g$, $h$ and $gh$ all lie in $S$ then on the one hand, $\phi(gh)=(gh)^{-1}=h^{-1}g^{-1}$. On the other hand, $\phi(gh)=\phi(g)\phi(h)=g^{-1}h^{-1}$, so that $g^{-1}$ and $h^{-1}$ commute. It follows that $g$ and $h$ commute.</p> <p>Fix $g\in S$ and consider $A=\lbrace h\colon h\in S\text{ and } gh\in S\rbrace$. There are less than $|G|/4$ $h$'s for which the first condition fails and less than $|G|/4$ $h$'s for which the second condition fails, so that $|A|>|G|/2$. By the above, it follows that the centralizer of $g$ (i.e. the set of $h$'s that commute with $g$) is a superset of $A$. Since the centralizer is a subgroup, by Lagrange's theorem the centralizer of $g$ must be all of $G$. That is $g$ lies in the center of the group. Now we have that more than $1/2$ of the group lies in the center (which is again a subgroup), so that $G$ is Abelian.</p> <p>Now since $S$ is more than half of the group, the subgroup generated by $S$ must be all of $G$, so that every element of $G$ is a product of elements of $S$. Now $g\in G$, write $g=s_1\ldots s_n$. Then $\phi(g)=\phi(s_1)\ldots\phi(s_n)=s_1^{-1}\ldots s_n^{-1}=s_n^{-1}\ldots s_1^{-1}=g^{-1}$, so that $\phi(g)=g^{-1}$ for all $g\in G$.</p>