If $G$ is finite group how to prove that $f(g)=ag$, $a \in G$, is a bijection for all $g\in G$? 
If $G$ is finite group, how to prove that $f(g)=ag$, $a \in G$, is a bijection for all $g \in G$? Here $ag$ is $a \cdot g$, where $\cdot$ is the operator from the group $G$.

This is what I've tried so far:
$f(g)=ag
\\
a^{-1}g(g)=g. 
$
Since the inverse has been dotted so $f$ is a bijection, I would like to know if there's a misstep on that.

Though the claim holds true for any group, you can take advantage of the assumed finiteness of $G$ to get immediately:

*

*the injectivity holds by the left cancellation law;

*since $G$ is finite, the surjectivity follows from 1.

