I am currently reading this article in which there is a proof that groups $H,K$ with relatively prime order we have $\text{Aut}(H \times K) \equiv \text{Aut}(H) \times \text{Aut}(K)$. In this proof, the order of $H$ is $n$ and the order of $K$ is $m$. The set $\{k^n \vert k \in K\}$ is constructed. For some automorphism $\omega \in \text{Aut}(H \times K)$, the morphism $\gamma: K \to H$ is constructed by $\gamma(k) = \pi_H(\omega(1_H, k))$, where $\pi_H$ is the projection of $H \times K$ onto $H$ and $1_H$ is the identity element in $H$. I can see that the set $\{k^n \vert k \in K\}$ is a subset of the kernel of $\gamma$, however the authors say
Also, since $m$ and $n$ are relatively prime, the set $\{k^n \vert k \in K\}$ consists of $m$ elements.
I do not get this claim: in case we are working with abelian groups, I can see that this is true, for if we consider two different elements in $K$ such that their $n$th power is equal, we have an element of order a divisor of $n$ (other then $1$). But what in the case of non-abelian groups?
If $n$ and $m$ are relatively prime, $n$ is invertible in $ℤ/mℤ$, say with an inverse represented by $c ∈ ℤ$, that is: $nc \equiv 1 \bmod mℤ$. Then, consider the maps $$α \colon K → K,~k ↦ k^n\quad\text{and}\quad β \colon K → K, ~k↦k^c.$$ For all $k ∈ K$, $(k^c)^n = k^{nc} = k^1 = k$ as $nc \equiv 1 \bmod mℤ$ and $k^m = 1$ in $K$ by Lagrange. Hence $α$ is a left inverse to $β$, hence $α$ is surjective, hence $\{k^n;~k ∈ K\} = \operatorname{img} α = K$.