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
View other elementary non-basic facts
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this
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.
We say that a group is a n-abelian group if the power map is an endomorphism. Here are some related facts about -abelian groups.
- n-abelian iff (1-n)-abelian
- The set of for which is -abelian is termed the exponent semigroup of . It is a submonoid of the multiplicative monoid of integers.
- abelian implies n-abelian for all n
- n-abelian implies every nth power and (n-1)th power commute
- n-abelian implies n(n-1)-central
- n-abelian iff abelian (if order is relatively prime to n(n-1))
- nth power map is surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center
- (n-1)th power map is endomorphism taking values in the center implies nth power map is endomorphism
- Frattini-in-center odd-order p-group implies p-power map is endomorphism
- Frattini-in-center odd-order p-group implies (mp plus 1)-power map is automorphism
- Characterization of exponent semigroup of a finite p-group
- Alperin's structure theorem for n-abelian groups
|Value of (note that the condition for is the same as the condition for )||Characterization of -abelian groups||Proof||Other related facts|
|2||abelian groups only||2-abelian iff abelian||endomorphism sends more than three-fourths of elements to squares implies abelian|
|-1||abelian groups only||-1-abelian iff abelian|
|3||3-abelian group means: 2-Engel group and derived subgroup has exponent dividing three||Levi's characterization of 3-abelian groups||cube map is surjective endomorphism implies abelian, cube map is endomorphism iff abelian (if order is not a multiple of 3), cube map is endomorphism implies class three|
|-2||same as for 3-abelian||(based on n-abelian iff (1-n)-abelian)|
- 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
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.
|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).