If groups $G$ and $H$ act on $X$, does $G\times H$ act on $X$?

Question

Suppose two groups $G$ and $H$ act on a set $X$.

What is the group action of $G\times H$ on $X$?

From the actions there a homomorphisms $\varphi\colon G\to S_X$ and $\psi\colon H\to S_X$. So this induces a homomorphism from the coproduct $\Phi\colon G\times H\to S_X$, so there should be an action.

Identifying $G$ and $H$ with the subgroups $\{(g,e):g\in G\}$ and $\{(e,h):h\in H\}$ by the inclusions into the coproduct, it seems $(g,e)$ should act as $g$, and $(e,h)$ should act as $h$, as $\Phi(g,e)=\varphi(g)$ and $\Phi(e,h)=\psi(h)$. Since $(g,h)=(g,e)(e,h)$, it seems like $(g,h)$ should act as $g\cdot (h\cdot x)$.

My concern was that $$ ((g_1,h_1)(g_2,h_2))\cdot x=(g_1g_2,h_1h_2)\cdot x=g_1g_2\cdot(h_1h_2\cdot x) $$ but $$ (g_1,h_2)\cdot((g_2,h_2)\cdot x)=g_1\cdot(h_1\cdot(g_2\cdot (h_2\cdot x)) $$

If the actions of $G$ and $H$ "commute," this seems like it would be okay, since $$ \psi(h)\varphi(g)=\Phi(e,h)\Phi(g,e)=\Phi((e,h)(g,e))=\Phi(g,h)=\Phi((g,e)(e,h))=\varphi(g)\psi(h) $$

but this makes it seem like any two group actions on $X$ commute, which doesn't sound like it should be true. Have I made an error somewhere?

Edit: I made a mistake and used the incorrect coproduct. But it seems like there is at least an action if the actions of $G$ and $H$ commute.

Motivation: I have $G$ and $H$ acting on a ring $X$, and I'm looking at the fixed points $X^{G\times H}$, but I'm not sure what the action of $G\times H$ should be.

Answer 1

Suppose we have a set $X$ equipped with actions coming from two different groups:

$$\begin{array}{ll} \alpha:G_1\to{\rm Sym}(X) \\ \beta:G_2\to {\rm Sym}(X) \end{array} $$

Denote $\ker\alpha=K_1$ and $\ker\beta=K_2$. Automatically we get induced faithful actions

$$\overline{\alpha}:G_1/K_1\to{\rm Sym}(X) \\ \overline{\beta}:G_1/K_1\to{\rm Sym}(X) $$

At this juncture we can define the group $H=\overline{\alpha}(G_1/K_1)\cap \overline{\beta}(G_2/K_2)$. Since $\overline{\alpha}$ and $\overline{\beta}$ are injective by construction, we have maps $\overline{\alpha}^{-1}:H\to G_1/K_1$ and $\overline{\beta}^{-1}:H\to G_2/K_2$. Then our two actions induce an action of the amalgamated free product $G_1/K_1 *_H G_2/K_2$. We could have just said we get an action of the free product $G_1*G_2$, but the elements of $K_1$ and $K_2$ inside act trivially, and then the images of $h\in H$ in $G_1/K_1$ act the same as their images in $G_2/K_2$ so we might as well encode that fact too.

If $[\alpha(G_1),\beta(G_2)]=1$ within ${\rm Sym}(X)$ then we can say there is an induced action of $G_1\times G_2$.

Answer 2

There is no group action unless the actions of $G$ and $H$ commute, in which case there is a unique natural action. The key point is associativity in the definition/characterization of "action", namely, that $$ (g_1\times h_1)\cdot \Big((g_2\times h_2)\cdot x\Big) \;=\; (g_1g_2\times h_1h_2)\cdot x $$ When the actions commute, defining $(g\times h)\cdot x= g\cdot(h\cdot x)$ and or $h\cdot (g\cdot x)$ gives associativity.

When the actions do not commute, that means for some $g$ and $h$ and $x$ we have $g\cdot (h\cdot x)\not=h\cdot (g\cdot x)$. This would contravene associativity, because, while $(g\times 1)$ and $(1\times h)$ commute in $G\times H$, $$ g\cdot (h\cdot x)\;=\; (g\times 1)\cdot \Big((1\times h)(x)\Big) \;=\; \Big((g\times 1)(1\times h)\Big)(x) \;=\; \Big((1\times h)(g\times 1)\Big)(x) \;=\; \ldots \;=\; h\cdot(g\cdot x) $$ contradiction.

(The uniqueness is easy once we say clearly what the requirement is.)