core27 min studyLesson 138 in Mathematics

Conjugacy classes and the class equation

Use the conjugation action to organize a finite group and prove why finite prime-power groups have nontrivial centers.

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What you’ll understand

  • Compute a conjugacy class using its centralizer.
  • Separate the center from the centralizer of one element.
  • Derive the class equation and apply it to finite p-groups.

Changing coordinates can change the written form of an operation without changing its structural role. Inside a group, conjugation performs this relabeling: replace x by gxg inverse. Unlike commutation, this is not a claim that the result equals x. The conjugation action partitions a group into conjugacy classes, and orbit–stabilizer turns those classes into a practical counting tool. The resulting class equation exposes central elements that a multiplication table can easily hide.

Conjugation is an action on the group

Let G act on its underlying set by g·x=gxg inverse. The identity acts trivially, and conjugating first by h and then by g equals conjugating by gh. The orbit of x is its conjugacy class. For permutations, conjugation relabels every entry in each disjoint cycle by g, so cycle lengths are preserved. In the full symmetric group, equal cycle type also suffices for conjugacy: choose a relabeling that sends each cycle of one permutation to a cycle of the other with the same length.

The stabilizer of x under this action is the centralizer C_G(x), the subgroup of elements commuting with x. The center Z(G) consists of elements commuting with every element of G. These concepts are different: the centralizer of one x can be large even if the center is small. An element is central exactly when its conjugacy class has size one. In particular, every element of an abelian group forms its own singleton class; conjugacy supplies no additional grouping in that case.

Cl⁡(x)={gxg−1:g∈G},∣Cl⁡(x)∣=[G:CG(x)]\operatorname{Cl}(x)=\{gxg^{-1}:g\in G\},\qquad |\operatorname{Cl}(x)|=[G:C_G(x)]
Apply orbit–stabilizer to the conjugation action; no normality of the centralizer is required.↔ Scroll wide equations horizontally.

Summing the conjugacy classes

Choose one representative from each conjugacy class in a finite group. Since orbits partition the underlying set, summing their sizes gives |G|. Separate the singleton classes, which are exactly the center, and express each remaining size as a centralizer index. This is the class equation. Its force comes from combining two descriptions of the same finite set: an additive partition into classes and multiplicative divisibility from subgroup indices. Always include one term per class, rather than one term per noncentral element.

Conjugation also explains a useful normality criterion. A subgroup N is normal exactly when it is a union of conjugacy classes: membership must survive every conjugation. This does not mean every arbitrary union of classes is a subgroup; closure and inverses still need checking. When searching for normal subgroups in a small finite group, their possible orders must both divide |G| and be expressible as one plus the sizes of some nonidentity classes. These necessary restrictions can sharply narrow the search without determining every multiplication rule.

∣G∣=∣Z(G)∣+∑i=1r[G:CG(xi)]|G|=|Z(G)|+\sum_{i=1}^{r}[G:C_G(x_i)]
The representatives x_i run over the noncentral conjugacy classes exactly once.↔ Scroll wide equations horizontally.

Why a finite p-group has central elements

Suppose |G| is a positive power of a prime p. Every centralizer has prime-power order by Lagrange’s theorem. Consequently every noncentral conjugacy class has size a positive power of p larger than one, hence divisible by p. The class equation shows that |Z(G)| is also divisible by p. Since the identity is central, the center cannot be empty; its order is at least p. This is an existence proof from arithmetic, not a procedure for locating a particular nonidentity central element.

The vectors lab offers only a geometric analogy: a planar rotation changes a vector’s coordinates while preserving its magnitude. Adjust angle at fixed magnitude and observe the projections change. Conjugation preserves group-theoretic data such as order, but the lab does not calculate conjugates, centralizers, or classes. In applications, use the actual group operation to compute those sets. Invariants help rule out conjugacy; in a general group, sharing an invariant such as element order does not by itself prove two elements are conjugate.

The complete class equation of S3

Partition the six permutations of three symbols into conjugacy classes and use the result to identify the center.

  1. The identity is its own class. The three transpositions (12), (13), and (23) have the same cycle type and form one class. The two three-cycles (123) and (132) form the remaining class.

  2. For a transposition such as (12), the centralizer is {e,(12)}, of order two, so its class has size 6/2=3. The centralizer of (123) is the cyclic subgroup it generates, of order three, yielding class size 6/3=2.

  3. The class sizes sum as 6=1+3+2. Only the identity class is a singleton, so Z(S3)={e}. The three transpositions alone do not form a subgroup, even though they form a whole conjugacy class.

∣S3∣=1+3+2,Z(S3)={e}|S_3|=1+3+2,\qquad Z(S_3)=\{e\}
A class size is a centralizer index, not the order of a representative element.↔ Scroll wide equations horizontally.
The result

S3 has three conjugacy classes and a trivial center. Being a union of classes is a normality criterion only after the subset is known to be a subgroup.

Explore the model

Interactive lab

Interactive lab · vectors

An invariant under changed orientation

Change angle while keeping magnitude fixed. This model displays a single planar vector and its projections, so the vector length stays invariant under the illustrated rotations.

An invariant under changed orientationPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forPreserved length is an analogy for an invariant; the model neither computes conjugacy classes nor proves that equal-order group elements are conjugate.

Check your understanding

A finite group G has order 27. What does the class equation guarantee about its center?

Try it yourself

Prove that every group of order p squared, for p prime, is abelian. You may use that if G/Z(G) is cyclic, then G is abelian.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The nontrivial center has order p or p squared. In the first case, consider the quotient order.

Compare with the solution

The p-group argument gives a nontrivial center, and Lagrange restricts its order to p or p squared. If it has order p squared, the whole group is central. If it has order p, the quotient G/Z(G) has prime order and is cyclic. To justify the supplied fact, write any two elements as g to the powers a and b times central elements z and w. Their products agree in either order because powers of g commute and z,w commute with everything. Thus G is abelian in this case too; in fact its center must then be all of G.

Take it with you

  • Conjugacy classes are orbits with centralizers as their stabilizers.
  • The class equation separates the center from noncentral classes.
  • Prime-power divisibility forces every finite p-group to have a nontrivial center.

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