Question

Suppose G is a finite group and f is an automorphism of G. If f(x)=x^-1 for more than 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.

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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∩x^-1S 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^-1x^-1. So x and y commute.

Since S∩x^-1S 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.

Answer 1

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.

Answer 2

(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.