Suppose that instead of the property $ab=ba=e$ a group G has the condition that for every element $a$ there exists an element $b$, such that $ab=e$. Prove that $ba=e$. Is the following a valid proof?
Since $ab=e$ then under the condition of the group there exists an element $k$ such that $bk=e$ for some $k$ in the group.
Now $bk=e$ so $abk=ae$ therefore $(ab)k=a$ and finally $ek=a$ and $k=a$.
Is this a valid proof?