# Show that the orbit of $H$ containing $x$ is equal to the right coset $Hx$

Let $G$ be a group and let $H$ be a subgroup of $G$. Let $X$ be the set of elements of $G$. Let $\ast : H \times X \to X$ be given by $$h \ast x = hx (h \in H, x \in X)$$.

QUESTION:

Let $x \in X$. Show that the orbit of $H$ containing $x$ is equal to the right coset $Hx$.

ATTEMPT:

Firstly I know that the right coset $Hx = \{ hx | h \in H\}$ And I'm supposed to show that if $A$ is the orbit of orbit of $H$ containing $x$, $A \subset Hx$ and $Hx \subset A$.

But my problem is I have no idea what "the orbit of $H$ containing $x$" means.

Can someone clarify this for me? I can picture orbit of single elements in a group but I can't imagine the orbit of an entire group.

• Let G act on $\Omega$ and $a\in \Omega$ then the orbit containing $a$ is $G.a=\{ga|g\in G\}$. since $1\in G$ then $a\in Ga$. – mesel Apr 2 '14 at 20:22
• As a result orbit of $H$ containing $x$ is $Hx$. – mesel Apr 2 '14 at 20:26

By "orbit of $H$" they mean "orbit of the action of $H$." In terms of language, orbits belong to both elements of $X$ and to the group $H$, so we could say "orbit of $x$" for elements $x\in X$ or we could also say "orbit of $H$" to mean one of the orbits of the action of $H$ on $X$.