a generalization of unit elements

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

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

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

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

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

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

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.
• Sep 13th 2012, 04:12 AM
ymar
Re: a generalization of unit elements
Thanks!