Consider the following task:
Let $G$ be a group and let $a$ be a fixed element of G. Show that the map $\lambda_a:G \to G$, given by $\lambda_a(g) = ag$ for $g \in G$, is a permutation of the set $G$.
While I feel fairly positive that what I have written down is correct, I am insecure as to how valid the argument is for the proof itself:
We know that $\lambda_a:G \to G$ and $g \mapsto ag$. Since $a, g \in G$, then $ag \in G$, by the definition of a group. Assume, for the sake of simplicity, that $G$ is finite. We have that $\bigl(\begin{smallmatrix}
g_1 & g_2 & g_3 & \cdots & g_{n-1} & g_{n-1} \\
ag_1 & ag_2 & ag_3 & \cdots & ag_{n-1} & ag_n
\end{smallmatrix}\bigr)$. We must have that $ag_x = g_y$ for some $x, y$ by the definition of a group. Thus, the map $\lambda_a: G \to G$ is a permutation. Note that $G$ does not need to be finite for the argument to hold.
Best Answer
You don't even need to assume that $G$ is finite. You know it's surjective since $a(a^{-1}g)=g$. Moreover, if $ag=ah$, then $g=h$ by cancellation. Therefore you have a (set) automorphism of $G$.