Conjugacy class (${\mathrm{Cl}}_G(x)$)

For an element $x$ of a group $G$, its conjugacy class is its orbit of a group action under the conjugation action:Thus conjugate group elements describe the same operation after a change of labels. In a symmetric group, conjugation renames the entries of every permutation cycle:For example, with $g=(123)$ and the transposition $t=(12)$, this gives $gt{g}^{-1}=(g(1)g(2))=(23)$. Products use function composition of permutations, with the rightmost factor applied first. The complete six-element example is conjugacy classes of the symmetric group on three letters.

The conjugacy classes form a partition of a set of $G$, since conjugacy is an equivalence relation. In a symmetric group, two permutations lie in the same conjugacy class precisely when they have the same cycle type: renaming preserves the lengths of permutation cycles, and matching equal-length permutation cycles constructs a renaming in the other direction. This compresses many labelled permutations into a few structural types. See McKernan, Conjugation in Sn, lemma 6.2 and proposition 6.6.

The stabilizer subgroup of $x$ under the conjugation action is its centralizer $C_G(x)$. Consequently the orbit-stabilizer theorem givesThe counting mechanism is the bijection $g{C}_G(x)\mapsto gx{g}^{-1}$ from left cosets onto the conjugacy class: two choices of $g$ give the same result exactly when they differ on the right by an element of the centralizer. A large centralizer means many choices of labels leave the operation unchanged, hence a small conjugacy class. A conjugacy class has size one precisely when its element belongs to the center of a group. The class equation counts these orbits of a group action together.

For a finite group $G$, write $k(G)$ for its number of conjugacy classes. Exactly the elements of the center of a group $Z(G)$ have singleton conjugacy classes; every other conjugacy class has at least two elements. Since the conjugacy classes form a partition of a set $G$, $k(G)-|Z(G)|\le (|G|-|Z(G)|)/2$. If $G$ is a nonabelian group, the cyclic quotient by the center criterion and the fact that every group of group order at most three is a cyclic group give $[G:Z(G)]\ge 4$, hence $k(G)\le 5|G|/8$. The quaternion group attains this bound, with five conjugacy classes among eight elements.

A conjugacy class is real when it is stable under inversion of conjugacy classes. Equivalently, one, and therefore every, member is conjugate to its inverse. Members need not themselves be involutions.

For a group $G$, inversion sends the conjugacy class of $g$ to that of $g^{-1}$, because $(xg{x}^{-1}{)}^{-1}=x{g}^{-1}{x}^{-1}$. It therefore defines an involution on the collection of conjugacy classes, even in a noncommutative group where inversion is not a group homomorphism.

The symmetric group $S_3$ has exactly six permutations, written using permutation cycles:Here $e$ is the identity element; $(12)$, $(13)$ and $(23)$ are transpositions; and $(123)$ and $(132)$ are permutation cycles of length three. With $t=(12)$ and $r=(123)$, rightmost-first function composition gives $r^2=(132)$ and $r^{-1}=r^2$. The conjugacy classes and representative centralizers are

An explicit check of the conjugation $gt{g}^{-1}$ with $g=r=(123)$ sends $1\mapsto 3\mapsto 3\mapsto 1$, $2\mapsto 1\mapsto 2\mapsto 3$, and $3\mapsto 2\mapsto 1\mapsto 2$, where each chain applies $g^{-1}$, then $t$, then $g$. Thus $1$ is fixed and $2,3$ are exchanged: $gt{g}^{-1}=(23)$. Relabelling also produces $(13)$, so the three transpositions share one conjugacy class. A transposition reverses the length-three permutation cycle, since $tr{t}^{-1}=(213)=(132)=r^{-1}$, putting both length-three permutation cycles in one conjugacy class.

Checking each of the six possible permutations against $gx=xg$ gives the displayed centralizers. Only $e$ has a one-element conjugacy class, so the center of a group here is $Z(S_3)=\{e\}$. The class equation is $6=1+3+2$. The second diagram in centralizer and normalizer uses the same $r$ and $t$ to show how preserving a cyclic subgroup differs from fixing its group generator.

A discrete group is ICC when every nonidentity element has an infinite conjugacy class. The trivial group satisfies this condition vacuously. It characterizes when the group von Neumann algebra is a Von Neumann factor.

If an infinite group has a single nonidentity conjugacy class, it is a torsion-free group. Otherwise every nonidentity element has the same finite order $n$. Powers show $n$ is prime. For $n\ge 3$, choose $xg{x}^{-1}=g^2$; then $x^n=1$ gives $g=g^{2^n}$, impossible by Fermat's little theorem. For $n=2$, the group has exponent two and is abelian, so every conjugacy class is a singleton and the hypothesis would force the group to have only two elements. Infinitude excludes this final case.

A shortest conjugacy representative is a word having the fewest written letters among all words representing elements of a fixed conjugacy class. This minimum equals the least group word length of an element in the class; a longer spelling of that element is not itself a shortest representative. Such a word exists because lengths are nonnegative integers. It is freely and cyclically reduced: free cancellation shortens the same representative, while removing mutually inverse first and last letters shortens a conjugate. Every cyclic rotation has the same minimal length.

A nonempty shortest conjugacy representative is a cyclically reduced word. If $w=sv{s}^{-1}$ with first and last letters inverse, $v$ is a shorter word representing a conjugate. Consequently $w^n$ has length exactly $n|w|$ as a freely reduced word, and any segment of the periodic word of length at most $|w|$ fits in a cyclic rotation of $w$. This observation turns a Dehn shortening segment crossing copy boundaries into a shortening of a conjugacy representative.

If $N⊴G$ has prime index $p$, a conjugacy class of $G$ lying in $N$ splits into one or $p$ conjugacy classes of $N$. Conjugation by $G$ acts transitively on these smaller classes, while $N$ fixes every one. This factors through $G/N$, and the orbit-stabilizer theorem makes the number of classes a divisor of $p$.

For an even permutation $g$, its symmetric group class splits into two equal alternating group classes exactly when its symmetric-group centralizer contains no odd permutation. If an odd commuting element exists, an odd conjugator can be multiplied by it to become even; otherwise the index-two subgroup has two conjugation orbits. In cycle notation the splitting types are precisely distinct odd cycle lengths, counting fixed points as length-one cycles: rotating odd cycles is even, whereas an even cycle or the exchange of two equal odd cycles supplies an odd centralizer element.

For an odd $n$-cycle $g$, an inverting permutation fixes one point and swaps $(n-1)/2$ pairs, so its sign is $(-1{)}^{(n-1)/2}$. Its other inverting permutations differ by elements of $C_{S_n}(g)=⟨g⟩$, all even. Thus inversion is possible inside $A_n$ exactly when that sign is positive. In particular $A_n$ is not an ambivalent group when $n\equiv 3(\mathrm{mod}4)$.

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]

For a finite group with trivial centre of a group and a prime divisor $p$ of its order, some nonidentity conjugacy class has size coprime to $p$. Otherwise the class equation would give $|G|\equiv 1(\mathrm{mod}p)$. Primality is essential: in the symmetric group $S_3$, neither nonidentity class size is coprime to the composite divisor $6$.