Re: Group on arbitrary ordinal


hagman wrote:
Saurav schrieb:

Arturo Magidin wrote:
The way I usually construct Z from N is as a set of equivalence
classes of pairs of elements of N. There is no induction at all in
that definition.

We defined Z without induction, I admit it, but to show that it is a
commutative group, didn't we use induction?

No.
Let N be *any* non-empty set with a commutative associative operation
+.
Let Z = (NxN)/~ where the equivalence ~ is given by
(a,b) ~ (a',b') iff a+b'=a'+b.
Define + on Z by
[(a,b)] + [(c,d)] := [(a+c,b+d)].
This is well-defined:
if (a,b) ~ (a'b') and (c,d)~(c',d'), then
(a+c)+(b'+d')=(a+b')+(c+d')=(a'+b)+(c'+d)=(a'+c')+(b+d),
hence
(a+c,b+d) ~ (a'+c',b'+d').

Then (Z,+) is a group:
Associativity is clear from associativity in (N,+).
If a is an element of N, then [(a,a)] is neutral because (a+x,a+y) ~
(x,y): (a+x)+y=x+(a+y).
If [(x,y)] in Z, then [(y,x)] is its inverse because (x+y,y+x)~(a,a):
(x+y)+a=a+(y+x)
Commutativity is again clear from commutativity of (N,+).
No induction here.

-hagman

Correct! I am absolutely sorry that I wrote a himalayan blunder here!
It would possibly be the construction of the monoid on omega that
needed induction; I think this time I have not gone off my head.
Thanks.

.