Cube map is endomorphism iff abelian (if order is not a multiple of 3)
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
Consider a finite group whose order is not a multiple of 3. Then, the Cube map (?) (viz the map sending each element of the group to its cube) is an endomorphism if and only if the group is abelian.
Statement with symbols
Let be a finite group whose order is not a multiple of 3. Then, the map defined as is an endomorphism if and only if is abelian.
Related facts
Cube map is automorphism implies abelian
Stronger facts for other values
Weaker facts for other values
Breakdown when the order is a multiple of three
- Frattini-in-center odd-order p-group implies p-power map is endomorphism
- Frattini-in-center odd-order p-group implies (p plus 1)-power map is automorphism
Facts used
- Abelian implies universal power map is endomorphism: In an abelian group, the power map is an endomorphism for all .
- Cube map is surjective endomorphism implies abelian
- kth power map is bijective iff k is relatively prime to the order
Proof
Abelian implies cube map is endomorphism
This is a direct consequence of fact (1).
Cube map is endomorphism implies abelian
Given: A finite group whose order is relatively prime to , and such that is an endomorphism of .
To prove: is abelian.
Proof:
Step no. | Assertion | Given data used | Facts used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | The cube map is an automorphism, and in particular, a surjective endomorphism, of . | The order is not a multiple of 3, and the cube map is an endomorphism. | Fact (3) | -- | Since the order is not a multiple of , the order is relatively prime to , so fact (3) yields that the cube map is bijective. Since we already know that the cube map is an endomorphism, this yields that the cube map is an automorphism of . |
2 | is abelian. | -- | Fact (2) | Step (1) | Since the cube map is an automorphism, must be abelian. |
Difference from the corresponding statement for the square map
In the case of the square map, we 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