$G$ is a group. Prove that $f(x)=ax$ is bijective Let $(G,\circ)$ be a group and $f: G\to G$ be defined by $f(x)=a\circ x$. Prove $f$ is bijective.

I know how to prove injectivity, I need to show that if $f(a)=f(b)$, then $a=b$. Let $x,y\in G$ such that $f(x)=f(y)$. Then  $ax=by\implies x=y$.
How can I show the surjectivity?
Help?
 A: By far the easiest proof is by asking what is the inverse? Since a function is bijective if and only if it is invertible. Let
$$f':G\to G$$
$$f'(x)=a^{-1}x$$
We will show that $f'$ is the inverse of $f$. Indeed
$$f(f'(x))=f(a^{-1}x)=aa^{-1}x=x$$
$$f'(f(x))=f(ax)=a^{-1}ax=x$$
Since $f$ is invertible then it is bijective.

So this is a very important construction in general. For any $a\in G$ define
$$f_a:G\to G$$
$$f_a(x)=ax$$
and
$$\mbox{inv}(G)=\{f:G\to G\ |\ f\mbox{ is invertible}\}$$
Then we have a function
$$\theta:G\to\mbox{inv}(G)$$
$$\theta(a)=f_a$$
So we've already shown that this function is well defined (i.e. $f_a$ is invertible). But you can easily check that $\mbox{inv}(G)$ is a group (with function composition as a multiplication) and $\theta$ is an injective group homomorphism (a.k.a. the Cayley's theorem). In particular the following holds
$$f_{a}^{-1}=f_{a^{-1}}$$
$$f_{ab}=f_{a}\circ f_{b}$$
and what I did in the first part of the answer is showing the first equality.
A: I am assuming here that $G$ is a multiplicative group and that $a \in G$. In this case, noting that $a^{-1} \in G$ $\forall a \in G$ by the group axioms:
Proof. Infectivity and well definition: 
$$
f(x) = f(y) \iff ax = ay \iff a^{-1}(ax) = a^{-1}(ay) \iff (a^{-1}a)x = (a^{-1}a)y \iff 1x = 1y \iff x=y.
$$
Surjectivity:
$$
x \in G \implies a^{-1}x \in G  
$$
but $f(a^{-1}x) = a(a^{-1}x) = (aa^{-1})x = 1x = x$.
