When can an operator $\omega : G \to A$ for a group $G \subset A$ be extended to a map $A \to A$? First, a word on notation, I define $\vec{v}_\kappa$ where $\kappa$ is an ordinal as $\{ v_{\alpha} : \alpha \in \kappa \}$.
I'm going to consider a slight generalization of a group with operators.
Let $G \subset A$ be a group and $\Omega$ be a set of operators and let $\omega$ be an otherwise propertyless element of $\Omega$.
Ordinarily, $\omega$ is a map $G \to G$. Let's consider a slight extension of that and make it a map $G \to A$ satisfying the following rule:
$$ g^\omega h^\omega = (gh)^\omega $$
I'm wondering whether $\omega$ can always be extended to a map from $A$ to $A$. And, if not, under what circumstances it cannot be extended.

What follows is my attempt to prove this statement myself. However, this proof is not valid because it makes at least one unjustified assumption, namely that I can get a minimal generating set of a group of a specified cardinality. I'm also not sure whether specifying the definition of an operator on all elements of a minimal generating set and inverses of such elements is sufficient to uniquely define the operator. I'm additionally uncertain about whether a group always has a minimal generating set, since the generating set is required to generate any group element using only finitely many factors.

This is my attempted proof.
Suppose $G = A$, then we're done because $\omega$ is already a map from $A$ to $A$.
Suppose $G$ is a proper subset of $A$.
Let $\vec{a}_\lambda$ where $\lambda$ is some ordinal be a minimal generating set for $A$. Let $\vec{a}_{\lambda_0}$ where $\lambda_0 \in \lambda$ be a minimal generating set for $G$.
Let $\vec{b}_\lambda$ be a minimal generating set for $A$ that is potentially different from $\vec{a}_\lambda$. Let $\vec{b}_\kappa$ where $\kappa \in \lambda$ be a minimal generating set for $G^\omega$. Let us further impose the constraint that $\kappa \le \lambda_0$.
I define $\xi$, the extension of $\omega$ as follows.
For $a_\mu$ where $\mu \in \lambda_0$, let $(a_\mu)^\xi = (a_\mu)^\omega $. Anything inside $\vec{a}_{\lambda_0}$ gets sent wherever $\omega$ would have sent it normally.
For $(a_\mu)^{-1}$ where $\mu \in \lambda_0$, let $((a_\mu)^{(-1)})^\xi$ be $((a_\mu)^{(-1)})^\omega$.
For $a_\mu$ where $\mu \not\in \lambda_0$, let $(a_\mu)^\xi = b_\mu$. Anything outside $\vec{a}_{\lambda_0}$ gets send to the corresponding element of the minimal generating set $\vec{b}$, $(b_\mu)^{(-1)}$.
For $a_\mu$ where $\mu \not\in \lambda_0$, let $((a_\mu)^{(-1)})^\xi$ be $(b_\mu)^{(-1)}$.
Thus, for any group element $g$ in $A$, $\xi$ is determined. Any group element can be written as a finite product of generators and inverses of generators.
 A: No, it is not always possible to extend an operator from a subgroup to the full group.
As an example, take $A$ to be dihedral of order 8, and $G$ to be a non-cylic subgroup of order 4. Then $G$ has an operator $\omega$ which cyclically permutes the three non-identity elements of $G$. However, no endomorphism of $A$ restricts to $\omega$.
In finite groups this is sometimes called "Does G control its own fusion in A?" and so I took a standard counterexample for that.
In abelian groups, you are more asking if $S$ is an injective module (divisible abelian group). For example, if $\omega:G\to A:n \mapsto \tfrac12 n$ where $G=2\mathbb{Z}$ and $A=\mathbb{Z}$ are additively-written groups, then $\omega$ cannot be extended to $A\to A$ since then $\omega(1) = \tfrac12 \notin A$  (more formally, $\omega(1)+\omega(1) = \omega(2) = 1$, and $A$ has unique "square roots" (when thinking multiplicatively), that is, it has unique division by 2 inside $\mathbb{Q}$).
When? In abelian groups, you can get a nice answer for when this is possible (which $S$ work for all $G$? divisible groups, many results of Baer; which $G$ for all $S$? I think that one is direct summand, though check if pure subgroup is correct). In finite groups there are some nice sufficient conditions for control of fusion (choose $G$ abelian Sylow $p$-subgroup, for example), but I don't think the results are as clean as for abelian groups $S$, and the setup I know is a little different than your question.
