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

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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.

Related facts

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.