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I am interested in finding an operation on the set of natural numbers by which $\mathbb{N}$ becomes a group. I am not able to think of any example.

Curious student
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  • Have a look at : https://en.wikipedia.org/wiki/Grothendieck_group#Examples:_the_integers,_the_Grothendieck_group_of_a_manifold_and_of_a_ring and: https://en.wikipedia.org/wiki/Integer#Construction –  Jun 02 '18 at 08:36
  • That means with any operation we cannot from it as group – Curious student Jun 02 '18 at 08:39
  • I don't know of any binary operation by which the natural numbers become a group. –  Jun 02 '18 at 08:41
  • Multiplication of nimbers, maybe? – Oscar Lanzi Jun 02 '18 at 08:43
  • The positive, negative integers and zero form a group under addition. 0 is the identity element. The non-negative integers by themselves form a semi-group under both addition and multiplication.. – Simone Masiero Jun 02 '18 at 09:13
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    Lookup transport of structure, e.g. see here and many more examples. – Bill Dubuque May 29 '19 at 15:26
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    @jack , the confusion of this post is really the special structure of the natural number. By making more emphasis on it, isnt that make the post more confusing? – Arctic Char Jun 01 '19 at 17:41
  • @ArcticChar: The Wikipedia link may be unnecessary. I have done my part of editing anyway. And yes, one should pay attention that one is talking about the set $\mathbb{N}$ of natural number, not any other extra "structure" on the set $\mathbb{N}$. –  Jun 01 '19 at 17:46

4 Answers4

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There are many of them. my favourite is the nimber addition, also known as bitwise exclusive or. In fact, there’s even a corresponding multiplication operation that makes the finite nimbers into a quadratically closed field.

Of course, there are also a bunch of boring ones based on bijections between ℕ and any group of size $\aleph_0$, such as $\mathbb{Z}$ or $\mathbb{Q}$ or $\mathbb{A}$.

anomaly
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Simone Masiero
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Every group with countable elements induces a group structure on $\mathbb{N}$. Asume that $(A,+,0,-)$ is a group and $A=\{a_1, a_2, \ldots\}$ is countable, w.l.o.g $a_1=0$ and $f(a_n)=n$ and $f^{-1}(n)=a_n$ then we define an addition $\oplus$ on $\mathbb{N}$ by

$$n\oplus m:=f(a_n+a_m)$$ and a neutral element $$\circledcirc:=f(a_1)$$ and an inverse $$\ominus n:=f(-a_n)$$

So for example $(\mathbb{Z}, +, 0, -)$ is a group. $$f(1)=0\\ f(2n)=n\\ f(2n+1)=-n$$

so for the $\mathbb{N}$ we have the following

The neutral element is $1$ and so $$n\oplus1=n$$ Te addition is defined by $$2m \oplus 2n=2(m+n)$$ $$(2m+1)\oplus(2n+1)=2(m+n)+1$$ $$(2m+1)\oplus2n=2(n-m), \text{if} \;n>m$$ $$(2m+1)\oplus2n=2(m-n)+1, \text{if} \;n\le m$$ we have $$\ominus (2n)=2n+1$$ $$\ominus (2n+1)=2n$$

miracle173
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The precise answer depends on whether the natural numbers include zero, but such an operation exists either way. If the natural numbers include zero, then as Simone says nimber addition works; the inversion operation is just the identity in this case. If we adopt a definition that does not include zero, then nimber multiplication qualifies. See here for a set theory based definition of both operations.

Oscar Lanzi
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There is a bijection between the natural numbers and the integers, for example:

$0, 1, -1, 2, -2, 3, -3$....

And the integers are a group for addition.

peterh
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