Infinite ordinals and cardinals are beautiful mathematical objects, extending in a natural way and sharing most of its properties. I have always wondered whether it would be possible to generalize the constructions of or to get transfinite integers or rationals. In general, putting a group structure on classes extending Ord or Card is not possible as I indicated some years ago (fr). Basically, we would have 1 + 0 = 0 (and also 1 + ω = ω ) which would imply 1 = 0 . However, we can consider a slightly modified problem with weaker constraints, as follows. First we assume we have a class of cardinals C Card containing zero and stable by addition.

(1) { 0 C κ , λ C , κ + λ C

We define the class Z C = C C , where C = { κ κ C , κ 0 } is a representation of "negative" cardinals. As we have seen, the class Z C need not be a group. Nevertheless, we can try to find an equivalence class R over Z C such that G C = Z C R is a group. Moreover, we want this equivalence class to be compatible with addition and opposite i.e.

(2) κ , λ C , κ + λ ¯ = κ ¯ + λ ¯ κ C , κ ¯ = κ ¯

Of course this implies that 0 ¯ is the identity element 0 of the expected group. Similarly, we can settle the same problem for a class C Ord and + is now understood as the ordinal addition. Can we follow this schema to contruct interesting infinite algebraic structures?

For C Card , we have for any infinite κ C the relation κ ¯ = κ + κ ¯ = κ ¯ + κ ¯ so κ ¯ = 0 . Hence G C ω = G C and the initial problem is reduced to the case where C ω contains only finite cardinals (= finite ordinals) and so the problem is still not really exciting. What about the case C Ord ? In general it is possible to have β 1 + γ = β 2 + γ without β 1 , β 2 being equal and so for any α such that β 1 α , β 2 α , γα C , we get that β 1 α ¯ = β 2 α ¯ . In particular for any infinite ordinal γ and α C such that γα C we obtain α ¯ = 0 because 1 + γ = γ (if this is not obvious to you, a more general statement is proved below). As a consequence, more general assumptions on the class C strongly limit the structure of the group. For example if C is closed under an infinite sum ( C = Ord for example) then G C is trivial. Similarly, if we require C to be stable by multiplication (in order to define a ring structure for example) then G C is trivial whenever C contains an infinite ordinal γ (hence the remaining case is again C ω ). If G C is not trivial, we denote α 0 C the least element such that α 0 ¯ 0 (in particular α 0 > 0 ).

One additional natural hypothesis should be added. We know that for any ordinal α < β there exists a unique ordinal γ such that α = β + γ . We would very like γ ¯ to match the difference " β ¯ + α ¯ ". For that purpose, we only require γ to belong to C . Hence we assume that

(3) α , β C , γ Ord , α = β + γ γ C

Let's come back to the case C ω with the new assumption (3) and suppose that G C is not trivial. Then α 0 < ω and any finite m C and be written m = α 0 q + r with r < α 0 . By (3), r C and because C is stable by finite sums, m ¯ = α 0 ¯ q + r ¯ = α 0 ¯ q . Thus G C is the monogenic group generated by α 0 ¯ .

What about the general case? Unfortunately, it turns out that if G C is not trivial it is still a monogenic group generated by α 0 ¯ . To prove this, we need to recall some equalities on ordinals. First, for any α > 0 , we have 1 + ω α = ω α . This is clearly true for α = 1 and we prove the general case by induction on α : 1 + ω α + 1 = 1 + ω ω α = 1 + ( 1 + ω ) ω α = ( 1 + ω α ) + ω α + 1 = ω α + ω α + 1 = ( 1 + ω ) ω α = ω ω α = ω α + 1 and 1 + ω λ = sup 0 < α < λ ( 1 + ω α ) = sup 0 < α < λ ω α = ω λ . Now, if we have two ordinals α < β and if γ > 0 is such that β = α + γ we have ω α + ω β = ω α ( 1 + ω γ ) = ω α ω γ = ω β . Finally, if we have two ordinals γ 1 < γ 2 we can write their Cantor Normal Forms:

(4) γ 1 = ω β 1 1 k 1 1 + ω β 2 1 k 2 1 + + ω β n 1 k n 1 γ 2 = ω β 1 2 k 1 2 + ω β 2 2 k 2 2 + + ω β n 2 k m 2

where β 1 2 β 1 i > β 2 i > > β 2 i and k j i are positive integers. Using the equality ω α + ω β = ω β for α < β it is clear that γ 1 + γ 2 = γ 2 if β 1 2 > β 1 1 .

Now our element α 0 C can be written α 0 = ω β k + γ where ω β k is the first term of its Cantor Normal Form. For any δ C , we can write its euclidean division by α 0 : δ = α 0 l + ε where ε < α 0 . By the previous discussion, if l ω then we can write the Cantor Normal Forms of the two elements α 0 < δ as in (4) and see that β 1 2 > β 1 1 . Hence α 0 + δ = δ and so α 0 ¯ = 0 , which is a contradiction. So l < ω and because C is stable by finite sums, α 0 l C and by the property (3) we get ε C . This means that δ ¯ = α 0 ¯ l + ε ¯ = α 0 ¯ l and hence G C is generated by α 0 ¯ as claimed above.

As a conclusion, if C Card satisfies the properties (1), (2) above and G C = Z C R is a group then C ω Ord . If C Ord satisfies properties (1), (2), (3) above and G C = Z C R is a group then G C is isomorphic to or to n . Conversely, we can build these groups from C = ω by defining the relation R as the equality (respectively the equality modulo n ). But we do not get any new groups...