Conjugation action

A group acts on itself and on its subgroups by conjugation: $g\cdot x=gx{g}^{-1}$ and $g\cdot H=gH{g}^{-1}$.

For a normal subgroup $K$ of $G$, each $c_h:k\mapsto h^{-1}kh$ is a group automorphism of $K$. Its inverse is $c_{h^{-1}}$, but the dependence on $h$ is a group antihomomorphism under ordinary function composition, because $(gh{)}^{-1}k(gh)=h^{-1}(g^{-1}kg)h$. The assignment $h\mapsto (k\mapsto hk{h}^{-1})$ instead is a group homomorphism to the automorphism group. Equivalently the first convention is a right group action, while the second is a left group action.

For a proper subgroup $B$ of a finite group $G$, the number $m$ of its conjugates is $[G:N_G(B)]\le [G:B]$, because its normalizer contains $B$. All conjugates contain the identity, so their union has size at most $1+m(|B|-1)\le |G|-[G:B]+1<|G|$. Thus some element is not conjugate to any element of $B$. Finiteness is essential to this cardinal counting argument.

Conjugation by $g$ is the group automorphism $x\mapsto gx{g}^{-1}$. Its orbits are the conjugacy classes.

For a subset $S$ of a group and an element $g$, its conjugate subset isConjugation is a bijection and a group automorphism, so it preserves cardinality and every property expressed using the group operation.

Two group elements $x$ and $y$ are conjugate when $y=gx{g}^{-1}$ for some group element $g$, equivalently when they belong to the same conjugacy class. Conjugate elements have the same order of a group element.

Two permutations in the same symmetric group are conjugate if $\tau =\pi \sigma {\pi }^{-1}$ for some permutation $\pi$. Conjugation relabels every permutation cycle, so it preserves cycle type. Conversely, equal cycle types permit a relabelling between corresponding cycles, establishing conjugacy.

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 $6$.

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.

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.

Embed SU(2) into SU(3) by $U\mapsto \mathrm{diag}(U,1)$. Its defining representation decomposes as an irreducible doublet and a trivial representation. The Schur lemma forces every commuting matrix to be $\mathrm{diag}(z{I}_2,w)$. Unitarity and unit determinant imply $|z|=1$ and $w=z^{-2}$, proving the formula. With Hermitian generator $X=\mathrm{diag}(1,1,-2)$, the defining representation restricts as $2_1\oplus 1_{-2}$ and the Adjoint representation as $3_0\oplus 2_3\oplus 2_{-3}\oplus 1_0$. The U(1) charges in the Adjoint representation are differences of defining charges. The generated connected subgroup is $(SU(2)\times U(1))/{\mathbb{Z}}_2$, since the two factors meet in $\mathrm{diag}(-I_2,1)$.

For the block embedding $SU(m)\times SU(n)\subset SU(m+n)$ with $m,n\ge 2$, the centralizer is the block-scalar subgroup $C=\{\mathrm{diag}(z{I}_m,w{I}_n):|z|=|w|=1,z^m{w}^n=1\}$. Schur lemma makes the diagonal blocks scalar and kills the off-diagonal intertwiners, since the two blocks are inequivalent representations of the product group.

Put $d=\gcd (m,n)$, $a=m/d$, $b=n/d$. The map $(z,w)\mapsto z^a{w}^b$ has image the $d$th roots of unity and connected kernel $(z,w)=(e^{ib\theta },e^{-ia\theta })$. A Bezout identity completes $(a,b)$ to an invertible integer coordinate change on the two-dimensional torus, proving $C\cong U(1)\times {\mathbb{Z}}_d$. Thus the full centralizer is a circle group exactly when $m,n$ are coprime. For $m=n=2$, $(zw{)}^2=1$ gives two circles, a useful counterexample to the unrestricted connectedness claim.

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 $G$.

Let a Hermitian matrix have distinct real eigenvalues with multiplicities $n_1,\dots ,n_k$. A unitary matrix commuting with it must preserve each eigenspace: if $Av=\lambda v$, then $A(Uv)=UAv=\lambda Uv$. Conversely, any unitary transformation acting independently on these mutually orthogonal eigenspaces commutes with $A$. The centralizer in $U(n)$ is therefore ${\prod }_{j}U(n_j)$. Imposing total determinant one gives the displayed subgroup inside the special unitary group $SU(n)$.

For multiplicities $(2,1)$ in SU(3), it is $S(U(2)\times U(1))\cong U(2)$, embedded by $U\mapsto \mathrm{diag}(U,(\det U{)}^{-1})$. Three distinct eigenvalues leave the diagonal maximal torus $U(1{)}^2$. This determines the unbroken group for a diagonal adjoint scalar field vacuum, including its global determinant constraint.

Commutation with a fixed-point-free involution preserves its partner relation. A commuting permutation can permute the $m$ pairs and independently swap each pair. This gives the permutation wreath product with order $2^{m}m!$.

A permutation commutes with a transposition exactly when it preserves its two-element support as a set. It may swap those two points and independently permute the remaining points. Thus this centraliser has order $2(n-2)!$.

For a subgroup $H$ of a group $G$, the normalizer isIt is the stabilizer subgroup of $H$ under the conjugation action on subgroups. Its elements can permute members of $H$; they need not commute with them. For $r=(123)$ in the symmetric group $S_3$, the cyclic subgroup $H=⟨r⟩$ has normalizer $S_3$, although the centralizer of $r$ has only three elements, as illustrated in centralizer and normalizer.

Every element of $H$ preserves $H$ under conjugation, so $H\subseteq N_G(H)$. Moreover, $H$ is a normal subgroup of $N_G(H)$, and this is the largest subgroup of $G$ with that property. Indeed, any subgroup in which $H$ is a normal subgroup consists entirely of elements preserving $H$ under conjugation. In particular, $N_G(H)=G$ exactly when $H$ is a normal subgroup of $G$.^[ref]

The orbit-stabilizer theorem counts distinct conjugate subsets of a subgroup byFor $H=⟨(123)⟩$ in $S_3$, there is just one, since $[S_3:N_{S_3}(H)]=6/6=1$. For the subgroup $K=⟨(12)⟩=\{e,(12)\}$, preserving $K$ forces its unique nonidentity element to stay fixed. Hence $N_{S_3}(K)=C_{S_3}((12))=K$, and its three distinct conjugate subsets are $\{e,(12)\}$, $\{e,(13)\}$ and $\{e,(23)\}$. This is the subgroup counterpart of counting a conjugacy class with a centralizer: the object being stabilized determines which denominator belongs in the count.^[ref]

The equality in the definition is essential. For a subgroup $H$ that is a finite group, an inclusion $gH{g}^{-1}\subseteq H$ already gives equality by equal cardinalities. For an infinite subgroup, that inference can fail: conjugation might carry it onto a proper subgroup. The normal subgroup criterion is therefore, with the explicit hypothesis $H\le G$,No finiteness or commutativity assumption is needed. A normal subgroup is preserved as a set; its elements need not lie in the center of a group.

There is a second useful count, distinct from counting conjugate subgroups. The normalizer acts on $H$ itself by group automorphisms, giving the group homomorphismIts group kernel is $C_G(H)$, so the centralizer of a subset is a normal subgroup of $N_G(H)$ and the first isomorphism theorem givesThis quotient group records the different rearrangements of $H$ induced by the surrounding group. It gives a faithful group action on $H$, but need not be the entire automorphism group of $H$. For example, take the ambient group $G=H=⟨(123)⟩$. Since $G$ is an abelian group, $N_G(H)=C_G(H)=G$, so the quotient group is trivial; nevertheless $r\mapsto r^{-1}$ is a nonidentity group automorphism of $H$. Returning to the ambient symmetric group $S_3$ makes that group automorphism available through a reflection, and $N_{S_3}(H)/C_{S_3}(H)\cong C_2$. Its two cosets are the three rotations and the three reflections.

The subgroup $H$ itself need not be contained in $C_G(H)$: their intersection is the center of a group $Z(H)$. It is contained precisely when $H$ is an abelian group. Thus $N_G(H)/H$, which exists because $H$ is a normal subgroup of its normalizer, and $N_G(H)/C_G(H)$ answer different questions and should not be substituted for one another.

Over the reals or a finite field with more than two elements, the normalizer of all invertible diagonal matrices in $G{L}_2$ consists of monomial matrices and has quotient $C_2$ over that diagonal subgroup. Over ${\mathbb{F}}_p$ its size is $2(p-1{)}^2$ for $p>2$. For $p=2$ the diagonal subgroup is trivial, so its normalizer is all of $G{L}_2({\mathbb{F}}_2)$, of order six.