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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?

Stefan
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I assume that you mean if for SOME $a,b$ (not every) $ab = e$ then $ba = e$. Your proof is valid, but you could write it much easier without playing with $k$. $$ab = e \Rightarrow bab = b.$$ If you already know the cancellation law, then we are done. Otherwise you may continue by writing $baba = ba$, so that $(ba)^2 =ba$. Just note that the only idempotent in a group is $e$.

Amin
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    The cancellation law is a consequence of the existence of unique inverses (as is the fact that $e$ is the only idempotent). Without expanding the proofs of one of those statements, this is just hiding the desired statement rather than proving it. – Milo Brandt Oct 26 '15 at 15:45
  • @MiloBrandt Cancellation law has nothing to do with "uniqueness". Ones you have a right (left) inverse then you have the right (left) cancellation law. In this problem we have $G$ is a group, so by definition we have right and left inverses. – Amin Oct 26 '15 at 16:13
  • I think after: $$ab = e \Rightarrow bab = b.$$ We can probably just conclude by associativity $(ba)b=b$, and since the identity is unique then $ba=e$ from there? – Stefan Oct 26 '15 at 17:15
  • @StefanT. You should be careful here. As Milo mentioned, using uniqueness is not allowed here unless you prove it. Here you can write $(ba)b = eb$ and use the right cancellation. – Amin Oct 26 '15 at 17:46