# Group Actions and Normal Subgroups

Let $G$ be a group acting transitively on a set $X$. If $H \le G$, then there is a natural action of $H$ on $X\text.$ For $x \in X$, let
\begin{equation*}
Hx = \{\, hx \mid h \in H \,\}
\end{equation*}

denote the orbit of $x$ under this action of $H$, and let $X/H$ be the set of all such orbits.

In general, the action of $G$ on $X$ does not naturally induce an action of $G$ on $X/H$.

**Example.** Consider the group $G = S_3 = \{ 1, r, r^2, s, rs, r^2s \}$, where $r = (123)$ and $s = (12)$, acting on the set $X = \{1, 2, 3\}$. Let $H = \langle s \rangle = \{1, s\}$. Then
\begin{equation*}
X/H = \left\{ \bar 1 = \{1,2\}, \bar 3 = \{3\} \right\}.
\end{equation*}

Now, $r \bar 1 = \{r1, r2\} = \{2, 3\} \notin X/H$.

**Theorem.** If $N \unlhd G$, then for every transitive $G$-set $X$, $G$ has a natural action on $X/N$ defined by $g(Nx) = N(gx)$.
*Proof.* $gN = Ng$. $\qquad\square$

**Example.** Let $G = D_4 = \{1, r, r^2, r^3, s, rs, r^2 s, r^3 s\}$, where $r = (1234)$ and $s = (13)$. $G$ acts transitively on the set $X = \{1, 2, 3, 4\}$. Let $N = \{1, r^2 = (13)(24)\}$. Then
\begin{equation*}
X/N = \left\{ \bar 1 = \{1, 3\}, \bar 2 = \{2, 4\} \right\}.
\end{equation*}

Now,
$r \bar 1 = \{r1, r3\} = \{2, 4\} = \bar 2$,
$r \bar 2 = \{r2, r4\} = \{1, 3\} = \bar 1$,
$s \bar 1 = \{s1, s3\} = \{3, 1\} = \bar 1$,
$s \bar 1 = \{s2, s4\} = \{2, 4\} = \bar 2$.
Thus, the action of $G$ on $X/N$ is well-defined.

**Theorem.** If $N \le G$, then $N \unlhd G$ only if $G$ has a well-defined natural action on $X/N$, given by $g \cdot (Nx) = N(gx)$, for every transitive $G$-set $X$.
*Proof.* Consider $G$ acting on its own underlying set by translation. Then $G/N$ is the set of right cosets of $N$ in $G$. If the action of $G$ on $G/N$ given in the statement is well-defined, then $gN = g \cdot (N1) = N(g1) = Ng$, for all $g \in G$. Hence, $N \unlhd G$. $\qquad\square$