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If an abelian subgroup $A$ has index $p^a$, the centralizer of any $1\ne g\in A$ contains $A$, so the class size of $g$ divides $p^a$. The prime-power conjugacy-class obstruction to simplicity rules this out. If $A=1$, the group is a p-group and has nontrivial center of a group, also ruling out nonabelian simplicity.

For an adjoint vacuum expectation value proportional to $\mathrm{diag}(1,1,-2)$, the centralizer is $(SU(2)\times U(1))/{\mathbb{Z}}_2$. Four gauge bosons remain massless and four acquire a common mass. This differs from Fundamental-Higgs breaking of SU(3) to SU(2).

Except in degree six, the $3$-cycles are characterized among order-three elements by the largest centralizer. Their cyclic subgroups correspond to three-element supports. Two different such subgroups commute exactly for disjoint supports; with one common point their generator products have order five, and with two common points those products have order two or three. Thus an automorphism induces an automorphism of $J(n,3)$. The maximal cliques of a Johnson graph reconstruct a permutation of the points. After undoing that permutation, an automorphism fixes every cyclic $3$-subgroup. Product orders force its choices of generator inversion to be consistent on the connected graph, and inversion of every $3$-cycle would reverse a noncommuting product. Hence it fixes all $3$-cycles, which generate $A_n$.

The center of a group consists of the elements that commute with every element of that group:where $C_G(g)$ is the centralizer of $g$. These are operations whose position in any product can be changed freely. The center is an abelian group and a characteristic subgroup, hence a normal subgroup. A group is an abelian group precisely when $Z(G)=G$.

For the symmetric group $S_3$, the conjugacy classes areAn element belongs to the center precisely when its conjugacy class has one element, so $Z(S_3)=\{e\}$. Yet commutator subgroup of the symmetric group on three letters gives $S_3^{\prime }=A_3$, with three elements. Thus belonging to the commutator subgroup does not mean commuting with everything. The alternating group $A_3$ is an abelian group as a subgroup, but its nonidentity elements do not commute with the transpositions in the surrounding $S_3$.

For the dihedral group $D_8$ of group order eight in abelianization of the square symmetry group, the center is $\{e,r^2\}$. A rotation $r^i$ commutes with $s$ precisely when $r^i=r^{-i}$, which requires $i$ even. No reflection $r^{i}s$ commutes with $r$, since $(r^{i}s)r=r^{i-1}s$ whereas $r(r^{i}s)=r^{i+1}s$. ConsequentlyThe half-turn can be performed before or after any square symmetry with the same result. Its appearance both here and in the commutator subgroup is a feature of this example, rather than a general equality. See Conrad, Dihedral groups I, theorem 3.4 and example 3.5.

The center and the commutator subgroup answer different useful questions: which operations already commute with everything, and which operations must disappear for the entire group to commute. In an abelian group these answers are $Z(G)=G$ and $G^{\prime }=\{e\}$. Another way to use the center is through conjugation: the group homomorphism $g\mapsto (h\mapsto gh{g}^{-1})$ has group kernel $Z(G)$, so the first isomorphism theorem gives $G/Z(G)\cong \mathrm{Inn}(G)$, the inner automorphism group. The quotient group by the center thus records the distinct conjugation operations; it need not be an abelian quotient.

A nontrivial finite p-group always has a nontrivial center. In the class equation, each noncentral conjugacy class has size divisible by $p$, so $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. The identity element ensures $Z(G)$ is nonempty, hence it has at least $p$ elements. See McKernan, MIT 18.703, lemmas 23.6–23.7 and definition-lemma 23.10.

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.

Both constructions are stabilizer subgroups for the conjugation action. The centralizer $C_G(x)$ fixes an individual element $x$; the normalizer $N_G(H)$ preserves a subgroup $H$ as a whole, while possibly moving its elements. The centralizer of a subset $C_G(H)$ fixes every member of $H$ individually, soFor a cyclic subgroup $H=⟨r⟩$, commuting with the group generator $r$ is equivalent to commuting with all its powers, giving $C_G(H)=C_G(r)$.

In the symmetric group $S_3$, let $r=(123)$ and $H=⟨r⟩=\{e,(123),(132)\}$. ThenEvery transposition sends $r$ to $r^{-1}$ under conjugation. In particular, for $t=(12)$,Thus $t$ belongs to the normalizer of $H$ but not to the centralizer of $r$. Preserving the collection of powers allows the two possible directions of its group generator to be exchanged.

This distinction lets one ask separately which operations leave a chosen operation unchanged and which operations preserve the subgroup it generates. The normalizer also supplies a group homomorphism to the automorphism group of $H$, by $g\mapsto (h\mapsto gh{g}^{-1})$; its group kernel is $C_G(H)$. In this example the group homomorphism has two possible outputs, fixing $r$ or replacing it by $r^{-1}$. The quotient group $N_{S_3}(H)/C_{S_3}(H)$ is therefore a cyclic group of group order two. Milne develops the normalizer as a stabilizer subgroup in Group Theory (v4.01) by James S. Milne.

Identify a regular orbit with the group using a chosen base point. A permutation commuting with all left translations satisfies $c(x)=xc(1)$. Thus the centralizer consists of right translations. Defining $R_g(x)=x{g}^{-1}$ makes $R_g{R}_h=R_{gh}$ and gives an isomorphism to $G$, rather than an unmentioned reversal of multiplication.

The centralizer of a subset $A$ of a group consists of the elements commuting with every member of $A$. It is the intersection of the element centralizers, hence a subgroup. Directly, identity commutes with every element, products of two elements that each commute with every member of $A$ still have that property, and the same is true of inverses. For $A=G$ this is the centre of a group. For an empty subset the commutation condition is vacuous and the centralizer is all of $G$.

For a finite group $G$, choose one representative $x_i$ from each conjugacy class with more than one element. The class equation iswhere $Z(G)$ is the center of a group and $C_G(x_i)$ is the centralizer of $x_i$. Each central element contributes its own one-element conjugacy class. Every remaining conjugacy class contributes its size, given by the orbit-stabilizer theorem. The sum counts each element exactly once. See McKernan, Group actions and automorphisms, lemma 23.6.

For conjugacy classes of the symmetric group on three letters, take $x_1=(12)$ and $x_2=(123)$. Their centralizers have sizes two and three, and the center of a group has size one, givingFor an abelian group, every centralizer is the whole group, every conjugacy class has size one, and the sum over noncentral representatives is empty.

The power of this counting identity is that the sizes of conjugacy classes constrain group structure. For a nontrivial finite p-group with group order $p^n$, every noncentral conjugacy class has size a positive power of $p$, by Lagrange's theorem. The class equation therefore gives $|Z(G)|\equiv 0(\mathrm{mod}p)$. Since the identity element lies in the center of a group, the center of a group must have at least $p$ elements. This obtains an element commuting with the entire group without constructing one. See McKernan, lemma 23.7.

Historically, these ideas belong to the nineteenth-century development of permutation groups and normal subgroups. Milne's historical opening records Galois's introduction of normal subgroups in 1832 and Jordan's systematic treatment of permutation groups in 1870. This is context for the conjugation action and the class equation, rather than a claim that either author first stated the modern formula.^[ref]