Kth power map is bijective iff k is relatively prime to the order

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Statement

Suppose G is a finite group and k is a natural number. Then, the map g \mapsto g^k is bijective from G to G if and only if k is relatively prime to the order of G.

Note that the power map (also termed a universal power map) is not necessarily an endomorphism of G. If it is also an endomorphism, then its being bijective is equivalent to its being an automorphism.

Related facts

Applications

Facts used

  1. Order of element divides order of group, or equivalently, Exponent divides order: For any finite group G and any element g \in G, we have that the order of g divides the order of G. In particular, g^{|G|} is the identity element. Equivalently, the exponent of G, which is defined as the least common multiple of the orders of all its elements, divides the order of G.
  2. Cauchy's theorem: If p is a prime divisor of the order of G, then G contains an element of order p. 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 G, a natural number k relatively prime to the order of G.

To prove: The map \alpha: g \mapsto g^k is a bijection from G to itself.

Proof: Let n be the order of G. Since k and n are relatively prime, the subgroup of \mathbb{Z} generated by k and n is the whole group \mathbb{Z}. In particular, we can find integers a,b \in \mathbb{Z} such that ak + bn = 1.

Consider the map \beta:g \mapsto g^a. We now have:

\alpha \circ \beta = g \mapsto (g^k)^a = g^{ka} = g^{1 - bn} = g.(g^n)^{-b}.

By fact (1), g^n is the identity element, so we get:

\alpha \circ \beta = g \mapsto g

which is the identity map on G. Similarly, \beta \circ \alpha is the identity map on G. Thus, both \alpha and \beta are bijective maps, completing the proof.

Power map is bijective implies relatively prime to the order

Given: A finite group G, a natural number k that is not relatively prime to the order of G.

To prove: The map \alpha:g \mapsto g^k is not a bijection.

Proof: Since k and n are not relatively prime, there exists a common prime divisor p of both k and n. By fact (2), there exists a non-identity element g such that g^p is the identity element. Thus, we get that g^k = e, where e is the identity element. Thus, \alpha(g) = g^k = e = e^k = \alpha(e), 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)