"Prove that $g \mapsto g^a$ for any $a \in \mathbb{Z}$ that is prime to the group order is a Permutation on any finite group G."
Can somebody please help me?
My idea was to show that $g,h \in G$ with $g^a=h^a$ implies $g=h$, but I struggle to show that.
Let $n$ be the order of $G$. SInce $(n,a)=1$, by Bezout's lemma there are $s,t\in \mathbb{Z}$ such that $sn+ta=1$.
Notice the map $h\mapsto h^{t}$ is the inverse of the provided map in the problem. To see that notice that \begin{align*} (g^a)^t&=g^{at}\\ &=g^{at}g^{sn} \quad\quad\text{(since $g^n=1$)}\\ &=g^{at+sn}\\ &=g. \end{align*} The other way around compostion give you similar result.