Is there an algebraic characterization of disjoint support in some group action? Let $\phi:G\hookrightarrow \text{Sym}(X)$ be a faithful $G$-action on a set $X$. Consider $g,h\in G$ s.t. $\text{supp}(\phi(g))\cap\text{supp}(\phi(h))=\varnothing$. It is easy to show that $\phi(g)$ and $\phi(h)$ commute and $\langle \phi(g)\rangle\cap\langle\phi(h)\rangle=\{\text{id}_X\}$. These are purely algebraic properties that can be brought back by $\phi^{-1}$ so $g$ and $h$ must commute and $\langle g\rangle\cap\langle h\rangle=\{e\}$. So we took a particular $G$-action and learned purely algebraic information about $G$ by looking at the support.
My question is whether we can go the other way and define a purely algebraic conditions on elements $g,h$ for there being a faithful $G$-action s.t. $g$ and $h$'s action on the set have disjoint support. It's obvious not the case that if this holds for one faithful action, then it holds for all faithful actions, e.g. the only pairs of elements with disjoint support in the faithful action of $G$ on itself by left multiplication is the identity with some other element.
My intuition is that $g$ and $h$ commuting and $\langle g\rangle\cap\langle h\rangle=\{e\}$ is a necessary and sufficient condition for there being a faithful $G$-action s.t. $g$ $h$ have disjoint support. My reasoning is that this implies that $\langle g,h\rangle$ is the internal direct sum of $\langle g\rangle$ and $\langle h\rangle$. In my mind, this means that $g$ and $h$ are independent in some sense. However, I have not had any luck constructing a specific $G$-action given elements $g$ and $h$ that meet my criteria. Is there some way to prove this or am I missing a condition? Also, is there any literature related to this topic?
 A: (I'm going to call the two elements $a$ and $b$, instead of $g$ and $h$.)
Two elements $a$ and $b$ of $G$ have disjoint support if and only if they have disjoint support on every orbit of $G$.
Now, choose a particular orbit $O_i$, and let $H_i$ be the stabiliser of a point in this orbit.. The induced action on $O_i$ is equivalent to the action by multiplication on the cosets of $H_i$. The criterion for $a$ and $b$ to have disjoint support on $O_i$ is that,
(1) for every $x\in G$, at least one of $a^x$ or $b^x$ must be contained in $H_i$.
Finally, for the whole action to be faithful, we need that
(2) the core of the intersection of the $H_i$'s in $G$ is trivial.
(In other words, the only normal subgroup of $G$ contained in the intersection is trivial).
This is an if and only if: if you start with $a$ and $b$ and a collection of $H_i$'s satisfying (1) and (2), you can reverse the construction and take the action on the union of the sets of cosets of the $H_i$'s, and $a$ and $b$ will have disjoint support.
So (1)+(2) is what you were looking for. The question is then whether this is equivalent to your condition.
The answer is no: $G=A_4$ is a counterexample.
Let $a,b,c$ be the three double transpositions, and let $g$ be an element of order $3$. $a$ and $b$ have the property you require (they commute and generate subgroups which intersect trivially), but I claim that there is no faithful action of $A_4$ where they have disjoint support.
Let us first determine which subgroups of $A_4$ satisfy (1). Clearly,  $A_4$ and the Klein subgroup do, but not the trivial subgroup or a subgroup of order $3$. Finally, taking $x=1$ shows that $\langle c\rangle$ does not, and taking $x=g$ and then $x=g^{-1}$ shows that $\langle a\rangle$ and $\langle b\rangle$ do not either.
So all possibilities for $H_i$'s contain the Klein subgroup, which is normal, so the action cannot be faithful.
