Stabilizer is Subgroup

Theorem

Let $\struct {G, \circ}$ be a group which acts on a set $X$.

Let $\Stab x$ be the stabilizer of $x$ by $G$.

Then for each $x \in X$, $\Stab x$ is a subgroup of $G$.

Corollary

Let $G$ be a group whose identity is $e$.

Let $G$ act on a set $X$.

Let $x \in X$.

Then:

$\forall g, h \in G: g * x = h * x \iff g^{-1} h \in \Stab x$

Proof

From the Group Action Axiom $\text {GA} 2$:

$e * x = x \implies e \in \Stab x$

and so $\Stab x$ cannot be empty.

Let $g, h \in \Stab x$.

$\ds g, h$

$\in$

$\ds \Stab x$

$\ds \leadsto \ \ $

$\ds g * x$

$=$

$\ds x$

Definition of Stabilizer of $x$ by $G$

$\, \ds \land \, $

$\ds h * x$

$=$

$\ds x$

Definition of Stabilizer of $x$ by $G$

$\ds \leadsto \ \ $

$\ds g * \paren {h * x}$

$=$

$\ds x$

$\ds \leadsto \ \ $

$\ds \paren {g \circ h} * x$

$=$

$\ds x$

Group Action Axiom $\text {GA} 1$

$\ds \leadsto \ \ $

$\ds g \circ h$

$\in$

$\ds \Stab x$

Definition of Stabilizer of $x$ by $G$

Let $g \in \Stab x$.

Then:

$x = \paren {g^{-1} \circ g} * x = g^{-1} * \paren {g * x} = g^{-1} * x$

Hence $g^{-1} \in \Stab x$.

Thus the conditions for the Two-Step Subgroup Test are fulfilled, and $\Stab x \le G$.

$\blacksquare$

Sources

1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.6$. Stabilizers
1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 54$
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: An Introduction to Groups: Exercise $5$
1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.3$: Group actions and coset decompositions
1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.9$