Abelian group which is not one of these

Question

Im struggling to find a finite abelian (commutative , associative) group $(G,\circ)$ with some specific conditions:

• $a\circ b$ isn't naive addition $a+b$ for $a,b\in G$

• $G$ is a subset of $\mathbb{Z}$

• $a\circ b$ doesnt hold multiplication ($a*b$) or third or bigger degree polynomial ($a^3+b^3$), etc...

• $a\circ b$ isnt binary or any boolean manipulation

Examples of non accepted groups

• $(\mathbb{Z},+)$, $(\mathbb{Z},a+b+ab)$, $(\mathbb{Z},a^2+b)$, $(\mathbb{Z}_2,a \text{ xor } b)$

Examples of candidate groups

• $(\mathbb{Z}/10\mathbb{Z},3x+2b+5)$, $(\mathbb{Z},(x+b)\%10)$

in case of non existence I wish to see any proof of that.

Answer 1

For any finite subset $G\subset \mathbb N$ and map $\circ\colon G\times G\to \mathbb N$, we can find a polynomial $f\in\mathbb Q[X,Y]$ such that $f(a,b)=a\circ b$ for all $a,b\in G$.

Answer 2

about finite groups , my last researches came with an expression which fits an abelian cyclic group $(\mathbb{G},\circ)$

• Cyclic:

$x\circ y=(nx+n'y)\mod l $

$x\circ y=(n(x \mod l)+n'(y \mod l))\mod l $

$x\circ y=(n(kl+x_0)+n'(k'l+y_0))\mod l $

$x\circ y=(nx_0+n'y_0)\mod l= x_0\circ y_0$ with $x_0=x \mod l$ and $y_0=y \mod l$

• Commutative:

$x\circ y=y\circ x$

$nx+n'y-ny+n'x=k l$

$(x-y)(n-n')=k l$

$$l/(n-n')$$

• Associative:

Regarding commutativity , many tests would be skipped in next stage:

$ (x \circ z )\circ y =(x \circ y )\circ z $

$ n(nx+n'z)+n'y - n(nx+n'y)+n'z =k l$

$ n²x+nn'z+n'y - n²x-nn'y-n'z =k l$

$ (n-1)n'(y-z) =k l$

$$l/(n-1)n'$$

$ (x \circ z )\circ y =(y \circ z )\circ x x$

$ n(nx+n'z)+n'y - n(ny+n'z)+n'x =k l$

$ n²x+nn'z+n'y - n²y-nn'z-n'x =k l$

$ (n²-n')(x-y) =k l$

$$l/(n²-n')$$

• Identity element:

$x \circ x_0 = x$

$nx+n'x_0-x = kl$

$nx+n'x_0-x = kl$

$n'x_0 = kl-(n-1)x$

this equation is true for specific $k$ when

$$l/(n-1)$$

$x_0=0$

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from al these , and supposed $n+n'>l$ to guarantee cyclic operations :

$n=kl+1$ , $n'=kl+1$

• example:

$(n,n',l)=(8,8,7)$

$\mathbb{G}={0,1,2,3,4,5,6}$ , $x\circ y= (8x+8y)%7$

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• • Inverse element:

$x \circ x^{-1} = 0$

$nx+n'x^{-1} = kl$

$n'x^{-1} = kl-nx$

$n'x^{-1}+x = kl-nx+x = kl -(n-1) x = k'l$ since $l/(n-1)$

remplacing $x^{-1}$ with $-x$

$n'x^{-1}+x = x-n'x=-(n'-1)x=kl$ because $l/(n'-1)$

since the group is cyclic $x^{-1}=l-x$