Subgroups and an union of orbits I have to prove or disprove the following statement:

If a group $G$ acts on a set $X$, then every subgroup $H$ of $G$ acts on the set $X$ as well, and every orbit of the action $G$ on $X$ is an union of orbits of the action $H$ on $X$.'

But I have absolutely no clue what they mean with this question. The question is translated from Dutch, so I hope that I didn't make mistakes while translating it. What I don't understand is what the action $G$ on $X$ means and what a union of orbits should be.
 A: In general, an action of a group $G$ on a set $X$ is a group homomorphism from $G$ to the group $S_X$ of permutations of the set $X$. This means that we send each group element $g \in G$ to some permutation of the elements in $X$, so each group element "acts" on $X$ by permuting its elements in some way. Usually, we write $g \cdot x$ to denote the element of $X$ that $x$ is sent to by the permutation in $S_X$ that corresponds to $g \in G$.
Then the orbit of a point $x \in X$ is the set of all points of $X$ that we can reach by applying elements of $g$ to $x$. In formula: $\text{Orb}(x) = G(x) = \{g \cdot x: g \in G\}$. This is a subset of $X$.
You are given a group $G$ with a subgroup $H$ and some action $G \to S_X$. Does this also give an action of $H$ on $X$? Yes, it does, because the composition of the homomorphisms $H \to G$ (inclusion) and $G \to S_X$ gives a homomorphism $H \to S_X$. Loosely speaking, since we can apply elements of $G$ to elements of $X$ we can do the same with elements of $H \subset G$. Since $H$ is a group under the group operation of $G$, this gives a well defined action of a group on a set.
Now we have two actions on $X$: one from $G$ and one from $H$. We also have orbits from both actions, partitioning $X$. (So we have two partitions of $X$: one into the orbits of the action of $G$, and one into the orbits of the action of $H$.) The question asks whether it is true that any orbit of $G$ (as a subset of $X$) is a union of orbits of $H$ (which are subsets of $X$ as well).
For instance, the dihedral group $D_4$ acts naturally on the vertices of a square (a set $X$ with $4$ points). $D_4$ has a cyclic subgroup $H$ with two elements, corresponding with reflecting in a (let's say horizontal) side of the square. The action of $D_4$ is transitive, so there is one orbit: $X$ itself. The action of $H$ has two orbits, one consisting of the upper two vertices and one consisting of the lower two vertices of the square. So, in this case, any $G$-orbit is a union of $H$-orbits: the only $G$-orbit $X$ is the union of the two $H$-orbits. The question is whether this holds for all group actions. 
I hope this makes the question clear to you!
And to give you a hint to solve it: because $X$ is partitioned both by the $H$- and the $G$-orbits, it suffices to prove that any $H$-orbit is contained in only one $G$-orbit.
A: It is a two step proof:

*

*Show that an element belonging to a certain orbit is an equivalence relation. Refer


*Show that equivalence relation partitions set into disjoint smaller sets whose total union unions give back original set. Refer
A: As for the second part, recall that every $g\in G$ lays in some right coset of $H$ in $G$. So, denoted with $R\subseteq G$ a complete set of coset representatives, for $x\in X$ we get:
\begin{alignat}{1}
\operatorname{Orb}_G(x) &= \{g\cdot x, g\in G\} \\
&= \{(hg_i)\cdot x, g_i\in R \text{ and } h\in H\} \\
&= \{h\cdot(g_i\cdot x), g_i\in R \text{ and } h\in H\} \\
&= \{h\cdot y_i, y_i\in Y(x)\subseteq \operatorname{Orb}_G(x) \text{ and } h\in H\} \\
&= \bigcup_{y_i\in Y(x)}\{h\cdot y_i, h\in H\} \\
&= \bigcup_{y_i\in Y(x)}\operatorname{Orb}_H(y_i) \\
\end{alignat}
where $Y(x):=\{g_i\cdot x, g_i\in R\}$.  So, the claim is true.
