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The centralizer of an element $x$ of a group $G$ consists of all elements that commute with $x$:It is the stabilizer subgroup of $x$ under the conjugation action, hence a subgroup of $G$. In conjugacy classes of the symmetric group on three letters, the three representative computations areFor the transposition $(12)$, a commuting permutation must preserve the exchanged pair $\{1,2\}$ and fix $3$; only $e$ and $(12)$ do so. For $r=(123)$, a commuting permutation is determined by where it sends $1$: the equation $g(r(i))=r(g(i))$ forces the images of $2$ and $3$, leaving exactly $e,r,r^2$. For $e$, every permutation commutes. These centralizer sizes $6,2,3$ give conjugacy class sizes $1,3,2$.

The centralizer is useful because it measures how much freedom remains after fixing one operation. Its index of a subgroup counts the distinct results of conjugation, and the intersection of all element centralizers is the center of a group. A centralizer need not be an abelian group: $C_G(e)=G$ even when $G$ is a nonabelian group. In an abelian group, all element centralizers equal $G$. These observations also explain why the surrounding group matters: the centralizer of $(123)$ inside the alternating group $A_3$ is the whole $A_3$, whereas its centralizer inside $S_3$ is a proper subgroup. See McKernan, the centralizer definition preceding lemma 23.6.

For an arbitrary subset $A\subseteq G$, the centralizer of a subset fixes every element of $A$ individually under the conjugation action:This is a pointwise stabilizer of the elements of $A$, not of the vertices of a polygon in a geometric realization. In particular, $r=(123)$ belongs to $C_{S_3}(r)$ although its rotation moves every vertex of a polygon of the equilateral triangle. If $A$ generates a subgroup $H$, commuting with every element of $A$ implies commuting with every product of those elements and their inverse elements; hence $C_G(A)=C_G(H)$. For the triangle rotation subgroup $H=⟨r⟩$, this gives $C_{S_3}(H)=C_{S_3}(r)=H$.

To see the commutation test directly on the equilateral triangle, compare a rotation $r$ with the reflection $s=(12)$. Their permutations satisfyThe two orders send vertex of a polygon $1$ to different places: $rs(1)=3$, but $sr(1)=1$. Thus $s\notin C_{S_3}(r)$. By contrast $r$ and $r^2$ commute, with $r{r}^2=r^{2}r=e$. The reflection reverses the chosen rotation while preserving the subgroup of all triangle rotations, so $s\in N_{S_3}(H)$. The distinction is visible before any counting: commuting preserves an operation itself; membership in the normalizer permits exchanging it with another operation in the same subgroup.

The natural actions of these five sporadic simple groups have degrees 11, 12, 22, 23 and 24, and respective transitivity degrees 4, 5, 3, 4 and 5. They arise from exceptional finite designs and codes.

For two distinct points of the Riemann sphere, conjugate by a Möbius transformation sending them to zero and infinity. A transformation fixing those two points has the form $z\mapsto \lambda z$, $\lambda \ne 0$. Composition multiplies the parameter, so the pointwise stabilizer is isomorphic to the complex multiplicative group. For zero and one, conjugation by $z/(1-z)$ gives $z\mapsto \lambda z/[1+(\lambda -1)z]$.

The Möbius group consists of maps $z\mapsto (az+b)/(cz+d)$ with $ad-bc\ne 0$, acting on the Riemann sphere $\hat{\mathbb{C}}=\mathbb{C}\cup \{\infty \}$. Proportional matrices represent the same map. Thus it is ${\mathrm{PGL}}_2(\mathbb{C})$, equivalently ${\mathrm{PSL}}_2(\mathbb{C})$ after normalizing the determinant. A zero denominator gives the image $\infty$, while $\infty$ maps to $a/c$ when $c\ne 0$ and stays at infinity when $c=0$.

Fixing both zero and infinity forces $b=c=0$, leaving $z\mapsto \lambda z$ with $\lambda \ne 0$. Composition multiplies the parameters, giving a bijective group homomorphism from the multiplicative group of nonzero complex numbers to this pointwise stabilizer.

For the other pair, take $T(z)=z/(1-z)$, which sends zero to zero and one to infinity. Conjugate the scaling subgroup by this Möbius transformation:Every map fixing zero and one arises this way. Composition again multiplies $\lambda$, proving both subgroups are isomorphic to ${\mathbb{C}}^{\ast }$. This is the Möbius pointwise stabilizer of two points; conjugation changes the selected fixed points without changing its group structure.

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: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 Ackerman's algebra notes, definition 0.4.1.2.

For the square group actions defined under group action, direct calculation givesThe 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.