Proving a function is in the symmetric group [closed]

Let $G$ be a group and let $g\in G$. The function $f:G\rightarrow G$ is defined by $f(x)=gx$ for all $x\in G$. Prove that $f\in S_G$, where $S_G$ is the symmetric group on the set $G$.

I don't know how to prove such a statement. Clearly we have that $f$ acts on $G$ to itself, but beyond that I'm unsure how to prove the claim

Solution 1:

Since

$$\begin{align} h:G&\to G, \\ x&\mapsto g^{-1}x \end{align}$$

is an inverse of $f$, we have that $f$ is a bijection on $G$. Thus $f\in S_G$.

Solution 2:

Hint: Prove $ f:G\rightarrow G $ defied by $ f(x)=gx$ is a bijection from $G$ onto itself.

I think you can easily show that the function is bijective.