Kth power map is bijective iff k is relatively prime to the order
Contents
Statement
Suppose is a finite group and is a natural number. Then, the map is bijective from to if and only if is relatively prime to the order of .
Note that the power map (also termed a universal power map) is not necessarily an endomorphism of . If it is also an endomorphism, then its being bijective is equivalent to its being an automorphism.
Related facts
Applications
- Cube map is endomorphism iff Abelian (if order is not a multiple of 3)
- nth power map is endomorphism iff Abelian (if order is relatively prime to n(n-1))
- nth power map is automorphism implies Abelian (if order is relatively prime to n-1)
Facts used
- Order of element divides order of group, or equivalently, Exponent divides order: For any finite group and any element , we have that the order of divides the order of . In particular, is the identity element. Equivalently, the exponent of , which is defined as the least common multiple of the orders of all its elements, divides the order of .
- Cauchy's theorem: If is a prime divisor of the order of , then contains an element of order . Equivalently, exponent of a finite group has precisely the same factors as order.
Proof
Relatively prime to the order implies power map is bijective
Given: A finite group , a natural number relatively prime to the order of .
To prove: The map is a bijection from to itself.
Proof: Let be the order of . Since and are relatively prime, the subgroup of generated by and is the whole group . In particular, we can find integers such that .
Consider the map . We now have:
.
By fact (1), is the identity element, so we get:
which is the identity map on . Similarly, is the identity map on . Thus, both and are bijective maps, completing the proof.
Power map is bijective implies relatively prime to the order
Given: A finite group , a natural number that is not relatively prime to the order of .
To prove: The map is not a bijection.
Proof: Since and are not relatively prime, there exists a common prime divisor of both and . By fact (2), there exists a non-identity element such that is the identity element. Thus, we get that , where is the identity element. Thus, , contradicting injectivity.
References
Textbook references
- Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, ^{More info}, Page 61, Exercise 25, Section 2.3 (Cyclic groups and cyclic subgroups)