Cube map is surjective endomorphism implies abelian
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 be a group such that the map defined by is an automorphism, or more generally, a surjective endomorphism. Then, is an abelian group.
Related facts
Similar facts for cube map
- Cube map is endomorphism iff abelian (if order is not a multiple of 3)
- Cube map is endomorphism implies class four for 2-divisible group
Similar facts for other values
Weaker facts for other values
- nth power map is endomorphism iff abelian (if order is relatively prime to n(n-1))
- nth power map is automorphism implies (n-1)th power map is endomorphism taking values in the center
- nth power map is endomorphism implies every nth power and (n-1)th power commute
- (n-1)th power map is endomorphism taking values in the center implies nth power map is endomorphism
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 , we can obtain non-abelian groups of order , 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
- Group acts as automorphisms by conjugation: For any , the map is an automorphism of .
- nth power map is surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center
- Square map is endomorphism iff abelian
Proof
Hands-on proof using fact (1)
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Given: A group such that the map sending to is a surjective endomorphism of .
To prove: is abelian.
Proof: We denote by the map .
Step no. | Assertion | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | for all , i.e., every square commutes with every cube | Fact (1) | Cube map is an endomorphism | -- | [SHOW MORE] |
2 | for all | Cube map is surjective | Step (1) | [SHOW MORE] | |
3 | for all | Cube map is an endomorphism | -- | [SHOW MORE] | |
4 | We get for all . | Steps (2), (3) | [SHOW MORE] |
Hands-off proof (using facts (2) and (3))
Given: A group such that the map sending to is a surjective endomorphism of .
To prove: is abelian.
Proof: By fact (2), we conclude that the square map must be an endomorphism of . By fact (3), we conclude that therefore 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:
In the case of the cube map, this is no longer true. That is, it may so happen that although . Thus, to show that we need to not only use that but also use that this identity is valid for other elements picked from (specifically, that it is valid for their cuberoots).
References
Textbook references
- Topics in Algebra by I. N. Herstein, ^{More info}, Page 48, Exercise 24