The Orbit and Stabilizer of a Point in a Group Acting on a Set

Recall from the Group Actions of a Group on a Set page that if $(G, \cdot)$ is a group and $A$ is a (nonempty) set then a group action of the group $G$ on the set $A$ is a multiplication $G \times A \to A$ denoted for all $g \in G$ and for all $a \in A$ by $(g, a) \to ga$ that satisfies the following properties:

• 1) $g_1(g_2a) = (g_1 \cdot g_2)a$ for all $g_1, g_2 \in G$ and for all $a \in A$.

• 2) $ea = a$ for all $a \in A$

Where $e \in G$ denotes the identity. We now have some additional definitions to go through.

Definition: Let $(G, \cdot)$ be a group acting on a (nonempty) set $A$. For each $a \in A$, the Orbit of $a$ Under $G$ is the set $Ga = \{ b \in A : b = ga \: \mathrm{for \: some \:} g \in G \}$.

For example, let $G$ be a group and let $H \subseteq G$ be a subgroup of $G$. As we saw on the The Left and Right Regular Group Actions of a Group on Itself page, $H$ acts on $G$ through the operation defined on $G$, i.e., the group action is the operation on $G$.

Let $g \in G$. Then the orbit of $g$ under the group (subgroup) $H$ is:

(1)

\begin{align} \quad Hg = \{ g' \in G : g' = hg \: \mathrm{for \: some \:} h \in H \} \end{align}

That is, the orbit of $g$ under $H$ is the right coset of $Hg$.

Definition: Let $(G, \cdot)$ be a group acting on a (nonempty) set $A$. For each $a \in A$, the Stabilizer of $a$ in $G$ is the set $G_a = \{ g \in G : ga = a \}$.

From the example above, if $G$ is a group and $H \subseteq G$ is a subgroup of $G$ then the group $H$ acts on the set $G$ by the operation on $G$.

Let $g \in G$. Then the stabilizer of $g$ under the group (subgroup) $H$ is:

(2)

\begin{align} \quad H_g = \{ h \in H : hg = g \} \end{align}

Since $H$ is a subgroup of $G$, $h, g \in G$ and we see that $hg = g$ implies that $hgg^{-1} = gg^{-1}$, i.e., $h = e$ where $e$ denotes the identity in $G$. Thus the stabilizer of $g \in G$ in $H$ is $\{ e \}$.

The following proposition tells us that if a group $(G, \cdot)$ is acting on a set $A$, then for each $a \in A$, the stabilizer of $a$ in $G$, $G_a$, will be a subgroup of $G$.

Proposition 1: Let $(G, \cdot)$ be a group acting on a set $A$. Then for each $a \in A$, the stabilizer of $a$ under $G$, $G_a$, is a subgroup of $G$.

• Proof: Fix $a \in A$ and let $G_a = \{ g \in G : ga = a \}$. Let $g_1, g_2 \in G_a$. Then $g_1a = a$ and $g_2a = a$ So by the first axiom of a group action we have that:

(3)

\begin{align} \quad (g_1 \cdot g_2)a = g_1(g_2a) = g_1a = a \end{align}

• So $g_1g_2 \in G_a$. Thus $G_a$ is closed under the operation on $G$.

• Now observe that by the the second axiom of a group action we have that $ea = a$. So $e \in G_a$.

• Lastly, let $g \in G_a$. Then $ga = a$. Then observe that by the first and second axioms of a group action that:

(4)

\begin{align} \quad g^{-1}a = g^{-1}(ga) = (g^{-1} \cdot g)a = ea = a \end{align}

• Thus $g^{-1} \in G_a$.

• So $G_a$ is a subgroup of $G$. $\blacksquare$