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

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.

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

 * 1) uses::Order of element divides order of group, or equivalently, uses::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) uses::Cauchy's theorem: If $$p$$ is a prime divisor of the order of $$G$$, then $$G$$ contains an element of order $$p$$. Equivalently, uses::exponent of a finite group has precisely the same factors as order.

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.

Textbook references

 * , Page 61, Exercise 25, Section 2.3 (Cyclic groups and cyclic subgroups)