# Making N (set of all positive integers) a group [closed]

Can anybody please give me an example of a binary operation under which N forms a group? More generally, how to find some operations to make possibly any set a group?

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Take your favorite bijection of N with Z (integers under addition) and transfer the group structure via this bijection. –  David Carchedi May 23 '10 at 10:59

## closed as off topic by Ryan Budney, Dan Petersen, Qiaochu Yuan, Steven Landsburg, Yemon ChoiApr 10 '12 at 7:08

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As explained by Arturo, a simple example of a group structure on N is the operation a ⊕ b = f-1(f(a) + f(b)) where f:NZ is defined by f(2n) = n and f(2n+1) = -n.

The statement that any nonempty set admits a group structure is equivalent to the Axiom of Choice! This is explained in this answer.

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Right: when I assume the given set has a cardinality and that we can find a bijection from $S$ to $\mathbb{Z}^{(|S|)}$, I am invoking choice a few times. –  Arturo Magidin May 23 '10 at 1:36
(If you prefer N to include 0, define f(2n-1) = -n instead.) –  François G. Dorais May 23 '10 at 1:36
@Arturo: More precisely, you're invoking the fact that every infinite S has the same size as the family of all its finite subsets. –  François G. Dorais May 23 '10 at 1:39
@François G. Dorais, nice to learn it. Thanks. –  dexter May 23 '10 at 2:21

Perform the usual addition of natural numbers in decimal representation, but let's forget about carrying the extra digit in the next column. So e.g. 356+75=321 and 123+987=0 . Note: This is very particular, as it is linked to base 10; but makes the example very concrete and quick. The poster asked for an example, and in my view, for an example, simplicity is prior to generality.

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I didn't vote you down, but I think your answer would be improved if you replaced the entire last sentence with a description of the resulting group structure (i.e., an infinite sum of order $m$ cyclic groups). –  S. Carnahan May 24 '10 at 1:47
You are certainly right, but I assumed that that point must be clear for everybody here. But say that asking for a + deserves a - , even if it was a joke. So now I am doubtful about changing the last sentence, as it is somehow instructive as it is :) / :( –  Pietro Majer May 24 '10 at 3:38
Actually my last sentence didn't express well what I meant, so I rewrite it (and incidentally dropped the request of +1). –  Pietro Majer May 24 '10 at 6:34
@Zsbán: It is associative: 16+6=12 –  Sune Jakobsen May 24 '10 at 11:10
Well for me 0 is a natural numer so I'd keep it in there if you don't mind ;) –  Pietro Majer May 25 '10 at 20:49

A useful one is the "nim sum".

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–  François G. Dorais May 23 '10 at 2:38
this is my favorite answer, however –  Pietro Majer May 24 '10 at 11:19

There are groups of any size (except $0$): for finite ones, you can always take the cyclic groups. For infinite ones, to get a group of cardinality $\kappa\geq\aleph_0$, just take the direct sum of $\kappa$ copies of $\mathbb{Z}$. Given a set $S$ of cardinality $\lambda$ (finite or infinite), pick a bijection $f$ from $S$ to a group $G$ of cardinality $\lambda$, and define the operation on $S$ by $a*b = f^{-1}(f(a)f(b))$.

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Well, except there are no groups of cardinality $0$, of course. –  Pete L. Clark May 23 '10 at 1:21
But we still need to exhibit a group $G$ for each infinite cardinality $\lambda$. I don't know anything about set theory (as will become clear, I don't even know what the word "cardinality" means), but here's something: For the countably infinite cardinality, take $\mathbb{Z}$ with addition; for uncountably infinite cardinalities $\lambda$, assuming that $\lambda$ is the cardinality of the power set of some other smaller cardinality, then the group of automorphisms of a set of that smaller cardinality should have cardinality $\lambda$. But I don't know if my assumption is justified. –  Kevin H. Lin May 23 '10 at 1:24
First: I did exhibit a group $G$ for each infinite cardinal $\lambda$: the direct sum of $\lambda$ copies of $\mathbb{Z}$, endowed with the usual structure, has cardinality $\lambda$ (the direct sum is the set of all $\lambda$-tuples, all but finitely many equal to $0$. Second: in ZFC you cannot prove that every infinite cardinal is the cardinal of a power set. If the Continuum Hypothesis is false, for example, then there would be no set $S$ such that $P(S)$ has cardinality $\aleph_1$. And if you assume the Generalized Continuum Hypothesis, then no set would have cardinality $\aleph_{\omega}$. –  Arturo Magidin May 23 '10 at 1:30
@Kevin: I assume, by the way, that by "automorphisms of a set" you mean all bijections from the set to itself. –  Arturo Magidin May 23 '10 at 1:31
@Pete: quite right; I edited the answer to reflect the exception. –  Arturo Magidin May 23 '10 at 1:33
One can argue that the nim sum (strictly speaking on $\lbrace 0 \rbrace \cup \mathbb{N}$) gives a natural group structure since the nim sum of two numbers is recursively defined to be the smallest number that it could possibly be. –  Robin Chapman May 24 '10 at 9:04