# A non-commutative cyclic semigroup

Allowing exponents of variables to reach every transfinite numbers we can obtain counterexamples to consolidated algebraic results.
Let $\Omega$ be the ordered set of all transfinite numbers.

#### A.Vitali’s semigroup: a non commutative cyclic semigroup:

Consider the set $\{ { x }^{ \alpha }: \alpha \in \Omega \}$. It satisfies the following properties:
1) it is a semigroup with unity, a monoid:
closure: ${ x }^{ \alpha }{ x }^{ \beta }={ x }^{ \alpha +\beta}$
associativity: it comes from additive associativity of transfinite numbers
unity: ${ x }^{0}$ is a good unity
2) it is cyclic: by the very definition every element is a power of x !!! But
3) it is not commutative; ${ x }^{ \omega }x = { x }^{ \omega + 1}$ while $x{ x }^{ \omega } = { x }^{1+ \omega}={ x }^{ \omega }$
4) it admits right but not left cancellation law: it is a consequence of analogue additive properties of transfinite numbers
Remark: in reality $\Omega$ itself is such an example but we have introduced x to have a multiplicative form.

### A.Vitali’s ring: a polinomial ring in one variable in which the powers of the variable don’t commute:

Consider the usual polynomial ring $k[x]$ in one indeterminate x commuting with the elements of a field k. But now we allow the exponents of x to go through all $\Omega$. Then we have an extended polynomial ring ${k}_{V}[x]$i in one variable x in which, by the preceeding example, the powers of the indeterminate don’t commute !!!
It contains as a subring $k[x]$ with its usual properties.

## A.Vitali’s transfinite power series ring:

We can also consider the ring of transfinite powers series ${k}_{V}[[x]]$ with the usual meaning that contains as a subring $k[[x]]$ the ring of infinite power series in x.
It is an easy exercise to see that here again $\sum _{ \alpha \in \Omega }^{ }{ { x }^{ \alpha } }$ is right  inverse of (1-x): $(1-x)(1+x+{x}^{2}\cdot\cdot {x}^{\omega}+{x}^{\omega +1}+{x}^{\omega +2}+\cdot\cdot {x}^{\omega 2}+{x}^{\omega 2+1}\cdot)= 1-x+x-{x}^{2}\cdot+{x}^{\omega}+{x}^{1+\omega}+{x}^{\omega +1}-{x}^{1+\omega +1}\cdot\cdot.=1+({x}^{\omega}-{x}^{\omega}+{x}^{\omega +1}-{x}^{\omega +1})+( )+( )+\cdot\cdot=1$
But this is not the only right inverse: also the traditional in k[x] is, and those with all zero from the second w-segment too, and so on. A right inverse is given by the chose of a subset of these w-segments that constitute $\Omega$.The first w-segment, that of natural numbers must be present. The successive w-segments are indexed, by little thought, by $\Omega$ itself. To distinguish we call it ${\Omega}_{s}$. In definitive (1-x) has at least $|P({\Omega}_{s})|={2}^{|{\Omega}_{s}|}={2}^{|\Omega|}$ right inverses such as $(1+x+{x}_{2}\cdot\cdot)+ { x }^{ \omega 2 }(1+x+{x}_{2}\cdot\cdot)+{ x }^{ \omega 5 }(1+x+{x}_{2}\cdot\cdot)+{ x }^{ {\omega}^{2} }(1+x+{x}_{2}\cdot\cdot)+\cdot\cdot$, and in general $(1+x+{x}_{2}\cdot\cdot)+\sum _{ \alpha \in \beta \in P({ \Omega}_{s}) }^{ }{ { x }^{ \alpha }(1+x+{x}_{2}\cdot\cdot) }$.
It is easy to see that there many others: if we also permit coefficients to the powers, because of the well ordering of $\Omega$, it is necessary and sufficient, guided by the preceeding calculations, that all consecutive terms inside the same w-segment cancel two by two, For this it is necessary and sufficient that all coefficients inside the w-segment be equal. The first coefficient of the first w-segment must be one while the first of the others can be arbitrary. Then the most general right inverse of (1-x) is of the form: $(1+x+{x}_{2}\cdot\cdot)+\sum _{ \alpha \in \beta \in P({ \Omega}_{s}) }^{ }{ {k}_{\alpha}{ x }^{ \alpha }(1+x+{x}_{2}\cdot\cdot) }$.
However: $\sum _{ \alpha \in \omega }^{ }{ { x }^{ \alpha } }=1+x+{x}^{2}\cdot\cdot$ is still left inverse of (1-x), because we can think the calculation in $k[[x]]$, but $\sum _{ \alpha \in \Omega }^{ }{ { x }^{ \alpha } }$ is no more: $(1+x+{x}^{2}\cdot\cdot {x}^{\omega}+{x}^{\omega +1}+{x}^{\omega +2}+\cdot\cdot {x}^{\omega 2}+{x}^{\omega 2+1}\cdot)(1-x) = 1-x+x-{x}^{2}\cdot\cdot +{x}^{\omega}-{x}^{\omega +1}+{x}^{\omega +1}-{x}^{\omega +2 +1}\cdot\cdot.=1+{x}^{\omega}+{x}^{\omega 2}+{x}^{\omega 3}+\cdot\cdot = \sum _{ \alpha \in \Omega }^{ }{ { x }^{\omega \alpha } }\neq 1$
The same calculations with the most general right inverse $(1+x+{x}^{2}\cdot\cdot)+\sum _{ \alpha \in \beta \in P({ \Omega}_{s}) }^{ }{ {k}_{\alpha}{ x }^{ \alpha }(1+x+{x}^{2}\cdot\cdot) }$ gives as a result $1+ \sum _{ \alpha \in \beta \in {\Omega}_{s} }^{ }{{k}_{\alpha} { x }^{\omega \alpha } }\neq 1$.
It is easy to see that $\sum _{ \alpha \in \omega }^{ }{ { x }^{ \alpha } }=1+x+{x}^{2}\cdot\cdot$ is the only left inverse of (1-x). In fact dividing the transfinite powers of this left inverse $\sum _{ \alpha \in \Omega}^{ }{{k}_{\alpha} { x }^{\alpha } }$ in w-segments as usual we see that:
1) the first w-segment must be $(1+x+{x}^{2}\cdot\cdot)$
2) Inside the same w.segment to cancel two by two we must have the same coefficient
3) the leading term of the w-segment survives and hence must have coefficient equal to zero.

References
H. Bachmann       Transfinite Zahlen    Springer Verlag 1967
J. Donald Monk   Introduction to set theory   Mc Graw Hill  New York 1969