= Centralizer
{title2=$C_G(x)$}
= Centraliser
{synonym}
= Commutant
{synonym}
The <centralizer> of an element $x$ of a <group> $G$ consists of all elements that commute with $x$:
$$
C_G(x)=\{g\in G:gx=xg\}=\{g\in G:gxg^{-1}=x\}.
$$
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 are
$$
C_{S_3}(e)=S_3,\qquad C_{S_3}((12))=\{e,(12)\},\qquad C_{S_3}((123))=\{e,(123),(132)\}.
$$
For 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 https://ocw.mit.edu/courses/18-703-modern-algebra-spring-2013/96221ce2d5b1077b3e5d3c5e08c88afe_MIT18_703S13_pra_l_23.pdf[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>:
$$
C_G(A)=\{g\in G:ga=ag\text{ for all }a\in A\}=\bigcap_{a\in A}C_G(a).
$$
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> $H=\langle r\rangle$, 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> satisfy
$$
rs=(13),\qquad sr=(23),\qquad srs^{-1}=r^{-1}.
$$
The 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 $rr^2=r^2r=e$. 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>.
\Image[/triangle-permutation-operation-order.png]
{title=Commuting triangle <rotations> and noncommuting <rotation>–<reflection> products}
{description=Colored labels track the original <vertices of a polygon>. Products act from right to left. The top row gives the same labeled triangle for $rr^2$ and $r^2r$; the bottom row gives different labeled triangles for $rs$ and $sr$. All six <permutations> in $S_3$ are included in the <centralizer> and <normalizer> membership table.}
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