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## 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 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∩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.

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Maybe we should ask questions as questions, rather than exam instructions. :-) – Scott Morrison Sep 30 2009 at 0:19
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 2009 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 2009 at 3:28

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.

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 +1 Awesome ! – Anton Geraschenko♦ Oct 1 2009 at 14:20

(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.

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 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 2009 at 14:30 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 2009 at 17:37