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

This article is about a result whose hypothesis or conclusion has to do with the fraction of group elements or tuples of group elements satisfying a particular condition.The fraction involved in this case is 3/4
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## 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 abelian group (in particular, a finite abelian group) and $\sigma$ sends every element of $g$ to its square.

## 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. Subgroup of size more than half is whole group
4. Abelian implies universal power map is endomorphism
5. The image of a generating set under a homomorphism completely determines the homomorphism.

## Proof

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### 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 $G$.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation Commentary
1 Suppose $x,y \in G$ are such that $x,y,xy$ are all in $S$. Then $xy = yx$ -- Definition of $S$ as the set of elements mapped to their squares, $\sigma$ is an endomorphism [SHOW MORE] Convert local property of elements going to squares to local property of pairs commuting.
2 For any $x \in S$, define $x^{-1}S := \{ x^{-1}s : s \in S \}$. We have $S \cap x^{-1}S \subseteq C_G(x)$, where $C_G(x)$ is the centralizer of $x$ in $G$. Step (1) [SHOW MORE] Toward establishing a large centralizer for fixed but arbitrary element of $S$ -- use local property of elements going to squares.
3 For any $x \in S$, the intersection $S \cap x^{-1}S$ has size greater than $|G|/2$ $|S| > (3/4)|G|$ [SHOW MORE] Toward establishing a large centralizer for fixed but arbitrary element of $S$ -- use size constraints and combinatorial formula.
4 For any $x \in S$, the subgroup $C_G(x) = \{ y \in G \mid xy = yx \}$ is equal to the whole group $G$ Facts (1), (3) Steps (2), (3) [SHOW MORE] Complete establishing that a fixed but arbitrary element of $S$ has centralizer the whole group -- use that big enough subgroups must contain everything.
5 $S$ is contained in the center $Z(G)$ of $G$ Step (4) [SHOW MORE] Shift focus to $S$ as a set.
6 $G$ equals its own center, hence is abelian Facts (2), (3) $|S| > (3/4)|G|$ Step (5) [SHOW MORE] Clinch -- use that big enough subgroups must contain everything.
7 $S$ is a generating set for $G$. Fact (3) $|S| > (3/4)|G|$ -- [SHOW MORE]
8 $S = G$, i.e., $\sigma(g) = g^2$ for all $g \in G$. Facts (4), (5) -- Steps (6), (7) [SHOW MORE]

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