# a generalization of unit elements

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• September 12th 2012, 01:56 AM
ymar
a generalization of unit elements
If I have a monoid $S$, and an element $u\in S$, then the following equivalence holds:

$u$ is a unit iff $f(x)=xu$ and $g(x)=ux$ are bijections onto $S$.

Proof. Suppose $u$ is a unit. Let $x,y\in S.$ Then we have

1. $x=xu^{-1}u=f(xu^{-1});$
2. if $f(x)=f(y),$ then $xu=yu$ so $x=xuu^{-1}=yuu^{-1}=y.$

Conversely, if $f$ and $g$ are bijections onto $S,$ then there exists $x\in S$ such that $xu=f(x)=1$ and $y\in S$ such that $ux=g(x)=1.$ Thus $u$ has a left inverse and a right inverse, which is known to imply that $u$ is a unit.

Now let's say $S$ is a semigroup without an identity element. Can there be an element $u\in S$ such that $f(x)=xu$ and $g(x)=ux$ are bijections? Is it possible when $S$ is finite? (Then it's equivalent to asking whether a semigroup without ideantity can have a cancellable element.) Is it possible when $S$ is infinite? Is it possible when we only demand that at least one of $f,g$ be a bijection onto $S?$
• September 12th 2012, 08:02 AM
Deveno
Re: a generalization of unit elements
a partial answer:

if such bijections x→xu and x→ux exists for some element u, then for some element x:

xu = u, and since for ANY y in S we have y = ua (for some a): xy = x(ua) = (xu)a = ua = y. thus x is a left-identity for S.

similarly, there must be some z in S with uz = u, and thus (writing y = bu), yz = (bu)z = b(uz) = bu = y, so z is a right-identity for S.

but then x = xz = z, so S possesses an identity element, which we can call e.

so in THAT case, we have xu = e and uy = e for some x,y in S, so u is indeed a unit (and x = y: x = xe = x(uy) = (xu)y = ey = y).

note that above, we only require that x→ux, x→xu be surjective to show that a right- or left-identity exists, uniqueness of this identity follows (for one of left- or right-) if one of those maps is injective as well (if S is finite, then surjective = bijective) (if x→ux is injective, then we have a unique right-identity, if x→xu is injective, we have a unique left-identity).

i am unsure of how you would define a unit without the presence of a two-sided identity.
• September 13th 2012, 04:12 AM
ymar
Re: a generalization of unit elements
Thanks!
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