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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.
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.
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.
