Proof that G is Abelian given $xy=zx\implies y=z$ for $x,y,z \in G$

Question

Suppose G is a group and that for all $x,y,z\in G$, we have that $xy=zx \implies y=z.$ Prove $G$ is Abelian.

“Proof”:

Since $xy=zx \implies y=z$, $\forall x,y,z\in G$, we have that for $y=z=t, xt=tx$. Hence any two arbitrary elements of $G$ commute. Thus, we see by definition, that $G$ is Abelian.

QED

Is this proof correct and rigorous?

Answer 1

This is just Ihf's proof simplified:

By the given condition:

$$x(yx)=(xy)x \Rightarrow yx=xy$$

Answer 2

Here is a valid proof: $$ xy = xyx^{-1}y^{-1}yx \implies y = xyx^{-1}y^{-1}y \implies e = xyx^{-1}y^{-1} \implies yx = xy $$

Answer 3

Let $x,g \in G$, then $xGx^{-1}=G$ trivially, Hence $xgx^{-1} = g'$ for some $g' \in G$. But then $xg=g'x$ implies $g=g'$. Hence $xg=gx$. Since $x,g$ are arbitrary this completes the proof.