Quotient of Transformation Group acts Effectively

Theorem

Let $G$ be a transformation group (which may or may not be effective) acting on $X$.

Then the quotient group $G / G_0$, where $G_0$ is the kernel, does act effectively on $X$.

Proof

Let $g G_0 \in G / G_0$.

Let $x \in X$.

Let $G / G_0$ act on $X$ by the rule:

$g G_0 * x := g x$

Then by definition of effective transformation group:

$G / G_0$ acts effectively on $X$ if and only if:

$g G_0 * x = x \implies g G_0 = G_0$

Suppose $g G_0 * x = g x = x$.

We want to show $g G_0 = G_0$.

By definition of kernel of group action of $G$ on $X$:

$g * x = gx = x \iff g \in G_0$

So $gG_0 = G_0$ by closure of subgroups.

Hence the result.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 53 \gamma$