Automorphism sends more than three-fourths of elements to inverses implies abelian

Statement
Let $$G$$ be a finite group, and $$\sigma$$ be an automorphism of $$G$$, such that the subset:

$$S := \{ g \mid \sigma(g) = g^{-1} \}$$

has size more than three-fourths the size of $$G$$. Then, $$G$$ is an proves property satisfaction of::abelian group (in particular, a proves property satisfaction of::finite abelian group) and $$\sigma$$ is the inverse map.

Similar facts

 * Inverse map is automorphism iff abelian
 * Square map is endomorphism iff abelian
 * Cube map is automorphism implies abelian
 * Endomorphism sends more than three-fourths of elements to squares implies abelian

Tightness
There exist examples of non-abelian groups with automorphisms that send exactly 3/4 of the elements to their inverses. For instance:


 * In the eight-element particular example::quaternion group with elements $$i,j,k$$ of order four, the map sending $$i,j$$ to their negatives but sending $$k$$ to itself, sends exactly 3/4 of the elements to their inverses.
 * In the eight-element particular example::dihedral group:D8, the identity automorphism sends three-fourths of the elements to their inverses, because $$5$$ of the non-identity elements have order two.

Facts used

 * 1) The set of elements that commute with any fixed element of the group, is a subgroup: the so-called centralizer of the element.
 * 2) The set of elements that commute with every element of the group, is a subgroup: the so-called center of the group
 * 3) uses::Subgroup of size more than half is whole group
 * 4) uses::Abelian implies universal power map is endomorphism, in particular the inverse map is an endomorphism. Since the inverse map is also obviously bijective, this also yields that it is an automorphism.
 * 5) The image of a generating set under a homomorphism completely determines the homomorphism.

Proof details
Given: A finite group $$G$$, an automorphism $$\sigma$$ of $$G$$. $$S$$ is the subset of $$G$$ comprising those $$g$$ for which $$\sigma(g) = g^{-1}$$. We are given that $$\! |S| > (3/4) |G|$$.

To prove: $$G$$ is abelian and $$S = G$$, i.e., $$\sigma(g) = g^{-1}$$ for all $$g \in G$$.

Proof: We initially focus on a single element $$x \in S$$ and try to show that its centralizer is the whole of $$G$$. We then shift focus to $$S$$ as a set, show that it is contained in the center, and since the center is a subgroup, it is forced to be all of $$G$$.

Steps (6) and (8) clinch the proof.