Homomorphism more than 3/4 the inverse - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:42:51Z http://mathoverflow.net/feeds/question/38 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38/homomorphism-more-than-3-4-the-inverse Homomorphism more than 3/4 the inverse Richard Dore 2009-09-29T23:01:26Z 2009-10-01T19:36:49Z <p>Suppose G is a finite group and f is an automorphism of G. If f(x)=x<sup>-1</sup> for more than 3/4 of the elements of G, does it follow that f(x)=x<sup>-1</sup> for all x in G?</p> <p>I know the answer is "yes," but I don't know how to prove it. <hr> Here is a nice solution posted by administrator, expanded a bit:</p> <p>Let S = { x in G: f(x) = x<sup>-1</sup> }.</p> <p>Claim: For x in S, S&cap;x<sup>-1</sup>S is a subset of C(x), the centralizer of x. <br>Proof: For such y, f(y) = y<sup>-1</sup> and f(xy) = (xy)<sup>-1</sup>. Now x<sup>-1</sup> y<sup>-1</sup> = f(x)f(y) = f(xy) = (xy) <sup>-1</sup> = y<sup>-1</sup>x<sup>-1</sup>. So x and y commute.</p> <p>Since S&cap;x<sup>-1</sup>S is more than half of G, so is C(x). So by Lagrange's Theorem, C(x) = G, and x is in the center of G. Thus S is a subset of the center, and it is more than half of G. So the center must be all of G, that is G is commutative. Once G is commutative the problem is easy.</p> http://mathoverflow.net/questions/38/homomorphism-more-than-3-4-the-inverse/48#48 Answer by solbap for Homomorphism more than 3/4 the inverse solbap 2009-10-01T04:35:25Z 2009-10-01T04:35:25Z <p>I think the point of this whole 3/4 business is the following. If G_1 is the set of elements such that f(x) = x^{-1}, then if we look at left multiplication on G by an element of G_1, more than half the elements have to make back into G_1. </p> <p>Combining this with what we know about f it should follow that any g \in G_1 commutes with more than 1/2 the elements of G, so if you say Langrange's thm enough times it should follow that G is abelian and G_1 generates G, which together imply the result.</p> http://mathoverflow.net/questions/38/homomorphism-more-than-3-4-the-inverse/50#50 Answer by Jonathan Wise for Homomorphism more than 3/4 the inverse Jonathan Wise 2009-10-01T05:47:04Z 2009-10-01T05:47:04Z <p>(My girlfriend explained this to me.) After Anton's observation, it's sufficient to show that <code>f = id</code> if <code>f</code> fixes more than half of <code>G</code>. But the elements of <code>G</code> fixed by an automorphism form a group and this group has index less than <code>2</code> by assumption, hence is all of <code>G</code>.</p>