I'm in the beginning stages of this proof. My question here I guess is that by definition a monoid has the properties that it is associative and has an identity. So: $(ab)c=a(bc)$ and $de=ed=e$ where $e$ is the identity. If I can prove that the identity is unique, does that prove the inverse is unique?

  • $\begingroup$ No, proving the identity is unique would only prove just that. To prove that inverses are unique when they exist, you can suppose that some $a$ in your monoid has inverses, say $b$ and $c$, and show that $b=c$. $\endgroup$ – Jonas Meyer Jan 27 '12 at 5:01

Of course the identity is unique, $e_1=e_1\circ e_2=e_2$. Now suppose $a,b$ are both inverses of $x$. Then

$$a = a\circ e = a\circ(x\circ b) = (a\circ x)\circ b = e\circ b = b.$$

The fact that the identity is unique plays no role, really, though the proof uses the same method of switching between the left and right perspectives of an expression and substituting with identity.

  • $\begingroup$ Thanks for clearing that up! $\endgroup$ – user23793 Jan 27 '12 at 5:16
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    $\begingroup$ Of course, if the identity were not unique, one would have to specify with respect to what identity the inverse should be interpreted. So showing the identity is unique is necessary to think about inverses. $\endgroup$ – Myself Jan 27 '12 at 19:43
  • $\begingroup$ @Myself: How is it necessary? Just let $e$ be any identity (not that there are any more than one but let's pretend we don't know this) and the line given here is still valid - without assuming or otherwise showing uniqueness of the identity. $\endgroup$ – anon Jan 27 '12 at 19:56
  • $\begingroup$ @anon: sure, your proof remains valid! My remark was that the question becomes invalid, because it is not well defined what 'having an inverse' means. $\endgroup$ – Myself Jan 27 '12 at 20:04
  • $\begingroup$ @Myself: Oh, sorry. Good point. $\endgroup$ – anon Jan 27 '12 at 20:07

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