Defining group action formula for cartesian products

Question

I am trying to understand @Martin Brandenburg's this proof. He said that

"The group $H \times K$ acts on the set $HK \subseteq G$ via $(h,k) x := hxk^{-1}$"

However I could not understand how he reached "$(h,k) x := hxk^{-1}$". Sorry for my poor knowledge here.

I think that it is used because of it satify group action rules. I see the first rule by $(e,e)x=exe^{-1}=x$,but I could not prove the second condition "$(g1 g2 )(x) = g1 (g2 x)$ for all $x ∈ X$ and all $g1 , g2 ∈ G$" for $H \times K$.

Secondly, what if the group action would be "$H_1 \times H_2 \times... \times H_n$ acts on the set $H_1H_2...H_n \subseteq G$. Is there any method for that one like $(h,k) x := hxk^{-1}$ ?

Answer 1

Moved (and elaborated) from the comments.

They reached that because, firstly, it is an action and, secondly, because it smartly supports their aim. In fact:

1. $x∈HK⟹hxk^{−1}∈HK$ (good definition);
2. $(e,e)x=x$ for every $x∈X$ (as you noted already);
3. for every $(h,k),(h',k')\in H\times K$ and $x\in X$: $((h,k)(h′,k′))x=$ $(hh′,kk′)x\color\red{:=}$ $(hh′)x(kk′)^{−1}=$ $h(h′xk′^{−1})k^{−1}\color\red{=:}$ $h((h′,k′)x)k^{−1}\color\red{=:}$ $(h,k)((h′,k′)x)$.

With this at hands, the following are sensible objects: \begin{alignat}{1} \operatorname{Orb}(e) &= \{(h,k)e, (h,k)\in H\times K\} \\ &= \{hk^{-1}, (h,k)\in H\times K\} \\ &= HK \end{alignat}

and

\begin{alignat}{1} \operatorname{Stab}(e) &= \{(h,k)\in H\times K\mid(h,k)e=e\} \\ &= \{(h,k)\in H\times K\mid hek^{-1}=e\} \\ &= \{(h,k)\in H\times K\mid hk^{-1}=e\} \\ &= \{(h,k)\in H\times K\mid h=k\} \\ &= \{(a,a)\in H\times K\mid a\in H\cap K\} \\ &\cong H\cap K \end{alignat} whence: $$\left|\operatorname{Orb}(e)\right|=|HK|\space\space\text{ and }\space\space\left|\operatorname{Stab}(e)\right|=|H\cap K|$$ As the orbit-stabilizer theorem does hold for every group action, in this case we get: $$|HK|\cdot|H\cap K|=|H\times K|=|H||K|$$ from which $$|HK|=\frac{|H||K|}{|H\cap K|}$$ follows.