I want to get a binary operation on $\mathbb{N}_0$. The set $\mathbb{N}_0$ denotes the set of natural numbers including zero. Need a binary operation * such that the ($\mathbb{N}_0$, * ) forms a commutative group.

I am trying but not getting. Please help me.


2 Answers 2


HINT: If $A$ and $B$ have the same cardinality, and $f\colon A\to B$ is a bijection, then whenever $R$ is a relation on $B$ there is some $R'$ which is a relation on $A$ such that $f$ is an isomorphism between $(A,R')$ and $(B,R)$.

This $R'$ is defined in the most obvious way. If $R$ is a set of $k$-tuples then we define: $$R'=\{\langle a_1,\ldots,a_k\rangle\mid\langle f(a_1),\ldots,f(a_k)\rangle\in R\}.$$

Now recall that an binary operation is merely a binary function, or a trenary relation. Find some countably infinite group $(G,\cdot)$ and define $*$ accordingly.

  • $\begingroup$ I liked the hint. This much more impressive. I will try to get other binary operations on it. However I am more enthusiastic now. Would it be possible to get ring or field structure on the set? $\endgroup$
    – Fukuzita
    May 26, 2014 at 10:58
  • 1
    $\begingroup$ Yes. Note that this is not limited just for one relation, you can transport any structure like that. (This why this is called "transport of structure!".) It is important to note that more often than not, any structure you may transport onto the natural numbers will not be compatible with their usual addition/multiplication/ordering. But that's okay. $\endgroup$
    – Asaf Karagila
    May 26, 2014 at 11:02

Define $$ H(n) = \begin{cases} n/2 & \text{if $n$ is even} \\ -(n+1)/2 & \text{if $n$ is odd} \end{cases} $$ This sends the naturals-with-zero to the integers. You can even write down an inverse map, $K$, which I'll leave to you. (It involves doubling things, more or less).

Now define

$$ n \text{#} m = K( H(n) + H(m) ) $$

This new operation converts the naturals-with-zero into an abelian group.


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