Ringoid over symmetric group?

Question

Is there a ringoid whose domain and additive operation is that of a symmetric group?

I know it cannot be a field or a ring because symmetric groups are generally not abelian, so what about a near-ring?

Answer 1

I will take a “near-ring” structure on a group $G$ (which psychologically will be additive) to be a map $\ell: G \to End(G)$, where $\ell(g)$ is the map given by “left-multiplying” by $g$. The condition that the “multiplication” be left-distributive is the same as requiring that $\ell(gh) = \ell(g)\ell(h)$, where we take pointwise products. The condition that the multiplication be right-distributive is the same as requiring each $\ell(g)$ to be a homomorphism.

The pointwise product of two homomorphisms $\phi, \psi: G \to G$ is a homomorphism if and only if $$\phi(gh)\psi(gh) = \phi(g)\psi(g)\phi(h)\psi(h).$$ Since $\phi$ and $\psi$ are homomorphisms, this is the same as requiring $\psi(g)$ and $\phi(h)$ to commute in $G$ for all $g$ and $h$. If $G$ is to be the “additive” group of a near-ring with multiplicative 1, this implies $G$ must be abelian. Even if not, the image of the multiplication must lie in an abelian subgroup of $G$ and thus vanish on the commutator subgroup $[G,G]$. In the case $G= S_n$, it is well-known that $[S_n, S_n] = A_n$, which tells us that the near-ring structures on $S_n$ factor through $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. So there is only one nonzero product, and its image has order two in $S_n$. We deduce that the near-ring structures on $S_n$ correspond to elements of order $1$ or $2$; when we go modulo zero-divisors, the resulting ring is $\mathbb Z/2\mathbb Z$.