# Homomorphism more than 3/4 the inverse

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.

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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
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

(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.