What is the meaning of "$H$-orbits on $G/H$?" I am trying to solve some problem and it phrases as "...if and only if the number of $H$-orbits on $G/H$ is $2$."
I have only learned the orbit of some element, not the group itself. So if I treat $H$ as an element of some larger group(maybe G), it is implied that this larger group $G$ is acting on $G/H$. Am I correct? What is the action like? Is it necessarily a left multiplication? Can I assume that $H$ is a normal subgroup? Or can I treat $G/H$ as just a set of left cosets?
Also, is there any obvious implication that the number of this orbits is $2$?
 A: I guess $H$ is a subgroup of $G$; hence $G/H$ is the set of, say, the left cosets of $H$ (this is not necessarily a group, if $H$ is not normal in $G$). Then, $(h,gH)\mapsto h\cdot gH:=(hg)H$ is indeed an action of $H$ on $G/H$. The point-wise orbit of $gH$ is the set:
\begin{alignat}{1}
O(gH) &= \{h\cdot gH, h\in H\} \\
&=\{(hg)H, h\in H\} \\
\end{alignat}
and the point-wise stabilizer of $gH$ is the subgroup of $H$:
\begin{alignat}{1}
\operatorname{Stab}(gH) &= \{h\in H\mid h\cdot gH=gH\} \\
&= \{h\in H\mid hgH=gH\} \\
&= \{h\in H\mid g^{-1}hgH=H\} \\
&= \{h\in H\mid g^{-1}hg\in H\} \\
\end{alignat}
Now, in case $H\unlhd G$, we get $\operatorname{Stab}(gH)=H$ for every $g\in G$, and hence (Orbit-Stabilizer theorem):
$$|O(gH)||\operatorname{Stab}(gH)|=|H|, \forall g\in G \Longrightarrow |O(gH)|=1, \forall g\in G$$
So, if there are $2$ orbits only, we have in this case $|G/H|=2$.
I'm not sure if this answers any of your questions, but hopefully it helps in some way.
A: The subgroup $H$ acts on $G/H$ by induced multiplication. Taking a equivalence class $[g]\in G/H,$ the action is defined by $h\cdot[g] = [hg]$.
Any group action on a set $X$ is an equivalence relation given by $x R y$ if and only if exists $g\in G$ such that $y=g\cdot x$. So this does partitionate $X$ into equivalence classes. In the case above, it means that there are only two equivalence classes for the action of $H$ on $G/H$.
