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.

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.