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?
Prove if an element of a monoid has an inverse, that inverse is unique
6
$\begingroup$
abstract-algebra
monoid
-
0No, 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 $b=c$. – 2012-01-27
1 Answers
6
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.
-
0Thanks for clearing that up! – 2012-01-27
-
1Of 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. – 2012-01-27
-
0@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. – 2012-01-27
-
0@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. – 2012-01-27
-
0@Myself: Oh, sorry. Good point. – 2012-01-27