Let $G$ be a group with the following property
$$\forall a, b, c \in G,\quad ab = ca \implies b = c.$$
Show that $G$ is abelian.
I know this hints towards the elements of the set being commutative but not sure.
I understand that this property hints at the set of left cosets being equal to the set of right cosets proving $G$ is abelian but not sure how to prove it fully.
Argue by contradiction: suppose there exists $a,b$ such that $ab\ne ba$, then $b\ne a^{-1}ba$. Let $c = a^{-1}ba$, then $ba = ac$ and so $b = c = a^{-1}ba$ by assumption, a contradiction.
The conjugation operation of $G$ on $G$, given by $b\mapsto aba^{-1}$ then is the identity, i.e., $aba^{-1}=b$ for all $a,b\in G$. Hence $G$ is abelian.
The property implies that the conjugacy class of any $g\in G$ is exactly $\{g\}$, which is equivalent to being abelian.
$xyx = xyx$ therefore with $a=x, b=yx, c=xy$ we get $ab = ca$, so $b=c$, that is $xy=yx$
