TITLE: Left inverse implies right inverse in a finite ring
QUESTION [12 upvotes]: Let $R$ be a finite ring with identity $1$, and assume $\exists x,y\in R$ such that $ xy=1$. How can I show it implies $yx=1$?

REPLY [5 votes]: Let $f_y\colon:R\rightarrow R,\ z\mapsto yz$ then:
$$f_y(z)=f_y(t)\iff yz=yt\Rightarrow x(yz)=x(yt)\Rightarrow (xy)z=(xy)t\Rightarrow z=t$$
hence $f_y$ is one to one. Now since $R$ is finite then the map $f_y$ is bijective hence there's a unique $z\in R$ s.t. $f_y(z)=yz=1$ so $x(yz)=(xy)z=z=x$ and conclude.