Endomorphism sends more than three-fourths of elements to squares implies abelian

Statement
Suppose $$G$$ is a finite group and $$\sigma$$ is an endomorphism of $$G$$. Suppose the subset $$S$$ of $$G$$ given by:

$$S := \{ g \in G \mid \sigma(g) = g^2\}$$

has size more than $$3/4$$ of the order of $$G$$. Then, $$G$$ is an proves property satisfaction of::abelian group (in particular, a proves property satisfaction of::finite abelian group) and $$\sigma$$ sends every element of $$g$$ to its square.

Related facts

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

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
 * 5) The image of a generating set under a homomorphism completely determines the homomorphism.

Proof details
Given: A finite group $$G$$, an endomorphism $$\sigma$$ of $$G$$. $$S$$ is the subset of $$G$$ comprising those $$g$$ for which $$\sigma(g) = g^2$$. We are given that $$\! |S| > (3/4) |G|$$.

To prove: $$G$$ is abelian and $$S = G$$, i.e., $$\sigma(g) = g^2$$ 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.