Cube map is surjective endomorphism implies abelian

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This article describes an easy-to-prove fact about basic notions in group theory, that is not very well-known or important in itself
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Statement

Verbal statement

If the Cube map (?) on a group is an automorphism, or more generally a surjective endomorphism, then the group is an abelian group.

Statement with symbols

Let G be a group such that the map \sigma:G \to G defined by \sigma(x) = x^3 is an automorphism, or more generally, a surjective endomorphism. Then, G is an abelian group.

Related facts

Similar facts for cube map

Similar facts for other values

Weaker facts for other values

Opposite facts

The statement breaks down if we remove the assumption of surjectivity:

Frattini-in-center odd-order p-group implies p-power map is endomorphism: In particular, for p = 3, we can obtain non-abelian groups of order p^3 = 27, such as prime-cube order group:U(3,3) and semidirect product of Z9 and Z3, where the cube map is an endomorphism. In the former case, the cube map is the trivial endomorphism. In the latter, it is a nontrivial endomorphism.

Facts used

  1. Group acts as automorphisms by conjugation: For any g \in G, the map c_g = x \mapsto gxg^{-1} is an automorphism of G.
  2. nth power map is surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center
  3. Square map is endomorphism iff abelian

Proof

Hands-on proof using fact (1)

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Given: A group G such that the map sending x to x^3 is a surjective endomorphism of G.

To prove: G is abelian.

Proof: We denote by c_g the map x \mapsto gxg^{-1}.

Step no. Assertion Facts used Given data used Previous steps used Explanation
1 g^2h^3 = h^3g^2 for all g,h \in G, i.e., every square commutes with every cube Fact (1) Cube map is an endomorphism -- [SHOW MORE]
2 g^2x = xg^2 for all g,x \in G Cube map is surjective Step (1) [SHOW MORE]
3 g^2x^2 = xgxg for all g,x \in G Cube map is an endomorphism -- [SHOW MORE]
4 We get gx = xg for all g,x \in G. Steps (2), (3) [SHOW MORE]

Hands-off proof (using facts (2) and (3))

Given: A group G such that the map sending x to x^3 is a surjective endomorphism of G.

To prove: G is abelian.

Proof: By fact (2), we conclude that the square map must be an endomorphism of G. By fact (3), we conclude that therefore G must be abelian.

Difference from the corresponding statement for the square map

In the case of the square map, we can in fact prove something much stronger:

(xy)^2 = x^2y^2 \iff xy = yx

In the case of the cube map, this is no longer true. That is, it may so happen that (xy)^3 = x^3y^3 although xy \ne yx. Thus, to show that xy = yx we need to not only use that (xy)^3 = x^3y^3 but also use that this identity is valid for other elements picked from G (specifically, that it is valid for their cuberoots).

References

Textbook references

External links

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