# Inverse element in associative structure not necessarily unique [duplicate]

I read from the linkhere proving that the inverse element in associative structure should be unique. However, it's only proving that should there exist a left and a right inverse element, then they should be identical. And I remember reading examples in non-commutative rings (so there should also be a similar case in monoids), an element has multiple multiplicative left inverse elements. But I'm not quite sure anymore. Could anybody reproduce such an example?

Also, interestingly, even for non-commutative groups, the inverse element is unique. Does all of this have to do with the fact that every element in the group is assigned with an inverse element?

## marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 22 '17 at 12:46

• Let $E$ be a set and let $S$ be the semigroup of all maps $f:E\to E.$ If $f$ is injective but not surjective, then $f$ has more than one left inverse but no right inverse. If $f$ is surjective but not injective, then $f$ has more than one right inverse but no left inverse. – bof Apr 22 '17 at 2:20
• However, it's only proving that should there exist a left and a right inverse element, then they should be identical. Well, yes, if you have that then it's obvious there's only one left inverse. If $a'$ is any left inverse of $a$, then $a'a=1$, and right multiplying with $a^{-1}$ you get $a'=a^{-1}$. – rschwieb Apr 22 '17 at 12:48
Or, take the monoid $E$ of functions $\Bbb N\to\Bbb N$ under composition. Now, e.g. $s:n\mapsto n+1$ has infinitely many left inverses, as it has freedom on what to assign to $0\notin{\rm im}(s)$.
If you want, you can make a ring out of this by adding elements (with integer [or whatever] coefficients) freely - denoted by $\Bbb Z[E]\$ [or $R[E]$ for arbitrary coefficient ring $R$].
For another example, along the same line, we can find a linear operator corresponding to $s$ on an infinite dimensional vector space (e.g. separable Hilbert space).