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Suppose $R$ is a ring with unity $1$ and for some $a\in R$ there exists more than one left inverse of $a$ in $R$. Show that $R$ has infinitely many left inverses of $a$.

I am trying to define a map $f:X\to X$ which is one-to-one but not onto or vice versa where $X=\{x\in R\mid xa=1\}$. That will directly imply that cardinality of $X$ can't be finite. But I don't have exact idea of proceeding further.

Edit:Given Ring is non commutative.

Arpit Kansal
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    That's not possible. If $xa=1=x'a$ then by commutativity $ax'=x'a$ and we can see that

    $$x=x(x'a)=x(ax')=(xa)x'=x'$$

     – Adam Hughes Sep 11 '14 at 21:31
  • Presumably you mean non-commutative ring? – Mathmo123 Sep 11 '14 at 21:32
  • Also, I think your "finite ring" tag is misplaced, since if the ring is finite, no subset is infinite, in particular left-inverses of $a$ cannot be an infinite set. – Adam Hughes Sep 11 '14 at 21:33
  • I know such a statement, but it doesn't refer to left inverses, but rather the "converse", namely zero divisors. – Martin Brandenburg Sep 11 '14 at 21:33
  • It also looks like the question has been dealt with in the case of right inverses: http://math.stackexchange.com/questions/156238/kaplanskys-theorem-of-infinitely-many-right-inverses-in-monoids which of course is immediately adaptable to left-inverses. It's not in one of the answers, it's quoted in the question. – Adam Hughes Sep 11 '14 at 21:35
  • @AdamHughes Yes. The Kaplansky-Jacobson Theorem. See also http://www.jams.or.jp/scm/contents/Vol-9-5/9-48.pdf – almagest Sep 11 '14 at 21:38

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