Inducing a Peano-style multiplication on a group? 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?
 A: 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.
