Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose $G$ is a finite group and $f$ is an automorphism of $G$. If $f(x)=x^{-1}$ for more than $\frac{3}{4}$ of the elements of $G$, does it follow that $f(x)=x^{-1}$ for all $x$ in $G\ ?$

I know the answer is "yes," but I don't know how to prove it.


Here is a nice solution posted by administrator, expanded a bit:

Let $S = \{ x \in G: f(x) = x^{-1}\}$.

Claim: For $x$ in $S$, $S\cap x^{-1}S$ is a subset of $C(x)$, the centralizer of $x$.

Proof: For such $y$, $f(y) = y^{-1}$ and $f(xy) = (xy)^{-1}$. Now $$x^{-1} y^{-1} = f(x)f(y) = f(xy) = (xy)^{-1} = y^{-1}x^{-1}.$$ So $x$ and $y$ commute.

Since $S\cap x^{-1}S$ 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.

share|improve this question
3  
If f does not invert more than 3/4 of the elements of G, then the result is false. Take Q={+-1,+-i,+-j,+-k} the order 8 quaternion group, and let f(i)=-i and f(j)=-j (this determines f since i and j generate Q). Then f sends +-1, +-i, and +-j to their inverses (thats 6 out of 8, which is 3/4), but does not send k to its inverse. –  Anton Geraschenko Sep 30 '09 at 2:29
1  
An observation: since f\circ f must be the identity on more than half of the elements of G, it must be the identity. –  Anton Geraschenko Oct 1 '09 at 3:28
add comment

2 Answers 2

up vote 12 down vote accepted

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.

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.

share|improve this answer
    
+1 Awesome ! –  Anton Geraschenko Oct 1 '09 at 14:20
add comment

(My girlfriend explained this to me.) After Anton's observation, it's sufficient to show that f = id if f fixes more than half of G. But the elements of G fixed by an automorphism form a group and this group has index less than 2 by assumption, hence is all of G.

share|improve this answer
    
I don't follow. f does not fix more than half of G, it sends more than 3/4 of the elements to their inverses. It looks like this proves that f\circ f is the identity because it is an automorphism that fixes more than half of G. –  Anton Geraschenko Oct 1 '09 at 14:30
2  
Sorry -- I misunderstood your comment above about f o f. (In defense of my girlfriend, I told her the wrong problem.) –  Jonathan Wise Oct 1 '09 at 17:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.