= Stabilizer subgroup
{title2=$G_x=\operatorname{Stab}_G(x)$}
= Isotropy group
{synonym}
= Stabiliser subgroup
{synonym}
= Point stabilizer
{synonym}
= Stabilizer
{synonym}
For a <group action> of a <group> $G$ on a <set> $X$, the <stabilizer subgroup> of $x\in X$ consists of the transformations that leave $x$ fixed:
$$
G_x=\operatorname{Stab}_G(x)=\{g\in G:g\cdot x=x\}\leq G.
$$
It is a <subgroup>: the <identity element> fixes $x$, products of transformations fixing $x$ fix $x$, and the <inverse element> of a transformation fixing $x$ also fixes $x$. A <stabilizer subgroup> is a <subset> of $G$, whereas the <orbit of a group action> is a <subset> of $X$. See https://people.math.harvard.edu/~nate/teaching/UPenn/2007/fall/math_371/lectures/week_1/lecture_1/lecture_1.pdf[Ackerman's algebra notes, definition 0.4.1.2].
For the <square> <group actions> defined under <group action>, direct calculation gives
$$
(D_4)_A=\{e,s\},\qquad (D_4)_{AB}=\{e,rs\},\qquad (D_4)_{AC}=\{e,r^2,s,r^2s\},\qquad (D_4)_O=D_4.
$$
The <reflection> $s$ fixes $A$ and $C$ individually. The <reflection> $rs$ preserves the unoriented <side of a polygon> $AB$ by exchanging its two <vertices of a polygon>. The half-turn <rotation> $r^2$ exchanges the two ends of the unoriented <diagonal of a polygon> $AC$ while preserving that <diagonal of a polygon>. Thus fixing a <side of a polygon> or <diagonal of a polygon> as an object does not mean fixing every point on it; the latter condition uses a <pointwise stabilizer>. The four <stabilizer subgroups> have <group orders> $2,2,4,8$, matching <group orbit> sizes $4,4,2,1$ through the <orbit-stabilizer theorem>.
A <stabilizer subgroup> records the transformations still available after imposing a constraint. Fixing a <vertex of a polygon> of the <square> leaves two choices; fixing its <centroid> leaves all eight. In a <faithful group action>, a <stabilizer subgroup> can still contain a nonidentity transformation, as $s\in(D_4)_A$ shows: a <faithful group action> requires that only the <identity element> fix every point of $X$ simultaneously, not that every individual <stabilizer subgroup> be trivial. The two-point <cyclic group> example under <group action> has both <stabilizer subgroups> equal to the <group kernel> $\{0,2\}$ of its associated <group homomorphism>.
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