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
View other such statements

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.

Related facts

Facts used

The set of elements that commute with any fixed element of the group, is a subgroup: the so-called centralizer of the element.
The set of elements that commute with every element of the group, is a subgroup: the so-called center of the group
Subgroup of size more than half is whole group
Abelian implies universal power map is endomorphism
The image of a generating set under a homomorphism completely determines the homomorphism.

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

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] If $x,y,xy\in S$ , we have $\sigma (xy)=\sigma (x)\sigma (y)$ , yielding $(xy)^{2}=x^{2}y^{2}$ which after cancellation of terms gives $xy=yx$ .	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] Suppose $y\in S\cap x^{-1}S$ . In particular, we get that $y\in S$ , and there exists $s\in S$ such that $y=x^{-1}s$ , which rearranges to give $xy=s$ , so $xy\in S$ . Thus, $x,y,xy$ are all in $S$ , so by step (1), $xy=yx$ , yielding $y\in S\cap x^{-1}S$ .	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] Note first that $\|x^{-1}S\|=\|S\|$ . A simple counting formula (the baby case of inclusion-exclusion) yields: $\|S\cap x^{-1}S\|+\|S\cup x^{-1}S\|=\|S\|+\|x^{-1}S\|$ Substituting in the above that $\|S\|=\|x^{-1}S\|>(3/4)\|G\|$ , and $\|S\cup x^{-1}S\|\leq \|G\|$ yields that $\|S\cap x^{-1}S\|>\|G\|/2$ .	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] By step (2), $S\cap x^{-1}S\subseteq C_{G}(x)$ . Step (3) yields that $\|S\cap x^{-1}S\|$ has size greater than $\|G\|/2$ . Thus, $\|C_{G}(x)\|>\|G\|/2$ . By fact (1), $C_{G}(x)$ is a subgroup, so by fact (3), we get $C_{G}(x)=G$ .	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] By step (4), every element of $S$ has centralizer equal to the whole of $G$ . In other words, every element of $S$ is in the center of $G$ , so $S\subseteq Z(G)$ .	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] $Z(G)$ is a subgroup of $G$ , and by step (5), it is a subgroup of size more than $(3/4)\|G\|$ . Thus, fact (3) yields that $Z(G)=G$ , showing that $G$ is abelian.	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] Since $\|S\|>(3/4)\|G\|$ , the subgroup $\langle S\rangle$ generated by $S$ in $G$ also has size at least $(3/4)\|G\|$ , hence by fact (3), must be all of $G$ .
8	$S=G$ , i.e., $\sigma (g)=g^{2}$ for all $g\in G$ .	Facts (4), (5)	--	Steps (6), (7)	[SHOW MORE] By step (6) and fact (4), that the map sending every element to its square is an endomorphism for $G$ . Call that endomorphism $\tau$ . $\sigma$ and $\tau$ are thus endomorphisms of $G$ that agree on the generating set $S$ . This forces $\sigma =\tau$ , so that $\sigma$ is the square map on all of $G$ , so $S=G$ .