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I was thinking of the Peano definition of multiplication on $\mathbb{N}$:

$a \times 0 = 0 \\ a \times S(b) = (a \times b) + a$

and wondering if this is possible to generalise this kind of construction to arbitrary groups / monoids? E.g. given a group (monoid) $(G,+)$ with identity $e$, we can define a multiplication $\times$ on $G$ such that for all $g,h \in G$:

$g \times e = e\\ g \times S(h) = (g \times h) + g$

for some suitable (surjective on $G \setminus \{ 0 \}$?) "successor function" $S:G \to G$, and such that $G$ is a (semi)ring under $+$ and $\times$.

It seems this can be done for cyclic groups (monoids): suppose $G = \langle g \rangle$, and let $S:G \to G$ be defined $S(h) = h+g$. Then, $S$ is sufficient (?) to allow us to define $\times$ on $G$ as above. However, as every cyclic group is isomorphic to $\mathbb{Z}$ or $\mathbb{Z}/n\mathbb{Z}$ for some $n$, it seems I have just recovered the usual multiplication of integers (mod $n$ for finite $G$). So, any way to do it for more general abelian (and non-cyclic) groups?

Jordan Mitchell Barrett
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  • What this boils down to for finite groups: You pick an arbitrary bijective map $f\colon G\to \Bbb Z/n\Bbb Z$ with and it turns out that $a\times b=f(b)\cdot a$. The inifinite case will work only for countably infinite $G$ and with a bijective map $G\to \Bbb N_0$ that sends $e\mapsto 0$. – Hagen von Eitzen Jul 03 '19 at 06:42
  • @HagenvonEitzen: it seems your construction may not guarantee that $G$ has a ring structure under $+$ (old operation) and $\times$ (induced operation). E.g. associativity of $\times$: $(a \times b) \times c = [f(c)f(b)] \cdot a$, while $a \times (b \times c) = f(f(c) \cdot b) \cdot a$ (I am assuming you use $n \cdot a$ to mean $a + a + a + \cdots$, $n$ times?).

    I suppose associativity would follow if $f$ was "additive", i.e. $f(n \cdot b) = nf(b)$ for all $n \in \mathbb{Z}, b \in G$.

     – Jordan Mitchell Barrett Jul 03 '19 at 08:14
  • @HagenvonEitzen: just tried this construction with the Klein four-group $V = \langle a,b \mid a^2 = b^2 = (ab)^2 = e \rangle$, and the map $f: V \to \mathbb{Z}_4$ defined $e \mapsto 0, a \mapsto 1, b \mapsto 2, ab \mapsto 3$. It seems then that $a$ and $ab$ are both (right-)multiplicative identities, contradicting the uniqueness of $1$ in a ring... – Jordan Mitchell Barrett Jul 03 '19 at 08:24
  • If the resulting (semi)rings are supposed to be unital, then $S(g)=1\times S(g)=(1\times g)+1=g+1$ and so the choice of $1$ uniquely determines $S$. And there are abelian groups (or more generally cancellative semigroups) where no such choice can be made. Although the case of arbitrary (commutative) semigroup is unclear. – freakish Jul 03 '19 at 15:08

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Suppose we have a successor function $S\colon G\to G$ such that the recursive definition $$\tag1 g\times e:=e,\qquad g\times S(h):=g\times h+g$$ makes sense. Define a map $\phi\colon \Bbb N_0\to G$ by $\phi(0)=e$, $\phi(n+1)=S(\phi(n))$. Then $\phi$ must be onto (and in particular, $G$ must be countable) because otherwise $(1)$ would not define $\times$ completely. By induction, we obtain $$\tag{2}g\times \phi(n)=n\cdot g.$$

If $\phi$ is injective, it is bijective and we obtain an inverse bijection $\psi\colon G\to\Bbb N_0$.

If $\phi$ is not injective, there is a minimal $m$ with $f(m)=f(k)$ for some $k<m$. By surjectivity, it must be the case that $m=|G|$. Then as $g\times\phi(|G|)=|G|\cdot g=e$, we may as well adjust $S$ so that $\phi(|G|)=e$, i.e., (re-)define $S(\phi(|G|-1)):=e$. This allows us to view $\phi$ as map $\Bbb Z/|G|\Bbb Z\to G$, which is then a bijection and we get an inverser bijection $\psi\colon G\to \Bbb Z/|G|\Bbb Z$.

At any rate, we arrive at $$\tag 3g\times h=\psi(g)\cdot g $$ for $\psi$ as defined above.

Under what conditions will this be a semiring?

We have $$ (a\times b)\times c=\psi(\psi(a)\cdot b)\cdot c$$ and $$ a\times(b\times c)=\psi(a)\psi(b)\cdot c.$$ In general, these will differ. A sufficient condition for equality is that $\psi$ is additive, i.e., $(G,+)$ is isomorphic to one of $(\Bbb N_0,+)$, $(\Bbb Z/n\Bbb Z,+)$.

We have $$ (a+b)\times c=\phi(c)\cdot(a+b)=\phi(c)\cdot a+\phi(c)\cdot b=a\times c+b\times c.$$ But $$a\times (b+c)=\phi(b+c)\cdot a $$ will in general not be equal to $$a\times b+a\times c=\phi(b)\cdot a+\phi(c)\cdot a=(\phi(b)+\phi(c))\cdot a.$$ Again, a sufficient condition would be that $\psi$ is an isomorphism.

Hagen von Eitzen
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