# Binary operation on $\mathbb N$

Does there exist a binary operation such that set of all-natural numbers forms a group? I saw a similar question which specifically says it needs a binary operation on the set of natural numbers forms a commutative group.

I was wondering if I can find a non-commutative Group of natural numbers.

I tried defining an operation such that $$a*b = a$$. But can't find an identity.

Yes, just take you favourite bijection $$f \colon {\mathbb N} \to {\mathbb Z}$$ and then take as operator $$a \circ b = f^{-1}(f(a) + f(b)).$$ The identity then is $$f^{-1}(0)$$.

• If you want a non-commutative group, you can generalise this approach by replacing $(\Bbb{Z}, +)$ with any countably infinite group $(G, +)$ (e.g. the product of $\Bbb{Z}$ and $S_3$). Apr 7, 2020 at 6:56

I suggest you have a look at: Why do the rationals, integers and naturals all have the same cardinality?

and

What is an example of function $f: \Bbb{N} \to \Bbb{Z}$ that is a bijection?

Since you said the set of all natural numbers, I take that as including $$0$$. And since you have taken out the old operators of $$\mathbb{N}$$ and intending to build a new group from the elements from it. We could think of $$\mathbb{N}$$ here being the same as $$\mathbb{Z}$$. And since there there is a group operation on $$\mathbb{Z}$$, there is a group operation on $$\mathbb{N}$$ as well.

One option is as follows: You write both numbers in base $$6$$, if necessary writing the smaller one with leading zeroes so that the length matches the larger one. Then you map each digit to one of the permutations in $$S_3$$, mapping the digit $$0$$ to the identity permutation (how you map the others is not relevant, as long as you stay consistent). Then you multiply the permutations for corresponding digits of the two numbers, and turn the result back into a digit. Finally you interpret the result again as base $$6$$ representation of an integer.

For example, assume the following mapping: \begin{align} 0 &\to \operatorname{id}\\ 1 &\to (12)\\ 2 &\to (13)\\ 3 &\to (23)\\ 4 &\to (123)\\ 5 &\to (132) \end{align}

Now consider for example $$16*2$$. Rewriting in base 6 gives $$24_6*02_6$$. Now you go digit-wise: \begin{align} 2*0 &\to (13)\operatorname{id} = (13) \to 2\\ 4*2 &\to (123)(13) = (23) \to 3 \end{align} Finally we find that $$23_6 = 15$$. Therefore $$16*2=15$$

This is a group where the neutral element is $$0$$, and the inverse element is obtained by replacing each $$4$$ by $$5$$ and vice versa in the base $$6$$ representation. It is also non-commutative, as seen by the fact that $$2*16 = 13 \ne 16*2$$.