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.



Homomorphism more than 3/4 the inverse
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 Lagrange's theorem enough times it should follow that $G$ is abelian and $G\_1$ generates $G$, which together imply the result.


