I have no trouble proving the following statement. Let $G$ be a cyclic group of order $n$ and let $k$ be an integer relatively prime to $n$. Prove the map $x \mapsto x^k$ is surjective. It is clear by $<x>=<x^{k}>$ by $\text{gcd}(k,n)=1$.
However, I fail to see why this is also surjective for any finite group of order $n$ though I can see $x^{n}=1$ for any $x$ in the group. Where is the surjectivity coming from in this context?
By Bezout identity, we can find $a,b\in\mathbb Z$ such that $ak+bn=1$. If $x\in G$, since $x^n=1$, $$x=x^{ak+bn}=(x^a)^k,$$ hence $y\mapsto y^k$ is surjective.
Since $\;(n,k)=1\;$ there exist integers $\;x,y\;$ s.t.
$$xn+yk=1\implies\;\forall\,g\in G\;,\;\;g=g^1=g^{xn+yk}=(g^x)^n\cdot(g^y)^k=1\cdot (g^y)^k=(g^y)^k$$
Let $y \in G$, then $y = x^m$ for some $m \in \{1,2,..,n\}$. Write $m = kp + q$. Thus: $y = x^m = x^{kp+q} = (x^k)^p\cdot x^q$. This shows $x^q \in G = <x^k>$ with$0 \leq q < k$. Thus $x^q = e$, and $q = 0$, and $y = x^m = (x^p)^k = f(x^p)$. So $f$ is onto.
Suppose $G=\langle g\rangle$. What you want to show is that $g=x^k$, for some $x\in G$. Since $x=g^r$, for some $r$, you want to solve $$ g=g^{rk} $$ where the unknown is $r$. This amounts to saying $rk\equiv 1\pmod{n}$. Since $k$ is coprime with $n$, you know there are integers $r$ and $s$ such that $rk+sn=1$.
On each cyclic subgroup $C$ of $G$, the map given $\phi(x)=x^k$ is a homomorphism $C \to G$.
The map $\phi$ is injective: Let $x\in\ker\phi$. Then $x^k=1$. Let $m$ be the order of $x$. Since $x^n=1$, we have $m$ divides both $k$ and $n$ and so $m$ must be $1$. This means that $x=1$.
Since $\phi$ actually maps $C$ to $C$, we conclude that $\phi$ is a bijection of $C$. In particular, $\phi$ is surjective as a map $C \to C$.
Now let $g\in G$ and consider $C=\langle g \rangle$. We have seen that $\phi$ is surjective. In particular, $g=\phi(h)=h^k$, which proves that $x \mapsto x^k$ is surjective as a map $G \to G$.
Alternatively, note that the $k$th power map is a homomorphism. Now, the only elements that satisfy $g^k=1$ are those whose orders divide both $k$ and $n$, so only those of order $1$, i.e., just the identity. Thus, this homomorphism is injective, and therefore surjective since $G$ is finite.
