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Conjugacy, the Class Equation, and p-Groups

Apply orbit-stabilizer to conjugation and a single equation falls out that organizes the entire internal structure of a finite group. We read off the center, prove that p-groups always have nontrivial centers, and classify groups of order p².

Conjugation as an action

Let G act on itself by conjugation: g·x = gxg⁻¹. The orbits are the conjugacy classes; two elements are conjugate when one is a “relabelling” of the other by some symmetry of the group. In the symmetric group S_n, conjugacy classes are exactly cycle types — a beautifully concrete instance. The stabilizer of x under conjugation is the centralizer C_G(x) = { g : gx = xg }, the elements that commute with x. By the orbit-stabilizer theorem, the size of the conjugacy class of x is [G : C_G(x)] = |G| / |C_G(x)|.

When is a conjugacy class a single point? Exactly when gxg⁻¹ = x for all g, i.e. x commutes with everything. Those x form the center Z(G), the heart of how non-abelian a group is. An element is central precisely when its conjugacy class is a singleton. (The closely related normalizer N_G(H) plays the same role for a subgroup H instead of an element — it is the stabilizer of H under conjugation of subsets.)

The class equation

Since the conjugacy classes partition G, summing their sizes gives |G|. Splitting off the singleton classes (the center) from the rest produces the class equation: |G| = |Z(G)| + Σ [G : C_G(x_i)], where the sum runs over one representative x_i from each non-central conjugacy class. Every term in that sum is a divisor of |G| strictly greater than 1. This little bookkeeping identity is shockingly powerful — it converts arithmetic constraints on |G| into structural facts about G.

Class equation of S_4  (|S_4| = 24).
Conjugacy classes = cycle types. Class size = 24 / |centralizer|.

  cycle type    representative   class size
  e             ()                 1     <- central
  (a b)         (1 2)              6
  (a b)(c d)    (1 2)(3 4)         3
  (a b c)       (1 2 3)            8
  (a b c d)     (1 2 3 4)          6

Check:  24 = 1 + 6 + 3 + 8 + 6.   (Indeed 1+6+3+8+6 = 24.)

Reading off structure:
- Z(S_4) = {e}, since only the identity class is a singleton.
- The classes of size 1 + 3 = 4 elements that are even and 'square-symmetric'
  ( e and the three (a b)(c d) ) form the normal subgroup V_4.
- Even permutations: classes 1 + 3 + 8 = 12 elements -> the subgroup A_4.
The class equation of S_4 read off from cycle types, revealing the center and normal subgroups.

p-groups have nontrivial center

A p-group is a finite group whose order is a power of a prime p. Here the class equation pays off immediately. Each term [G : C_G(x_i)] in the sum is a divisor of |G| = pⁿ that is greater than 1, hence divisible by p. Also |G| is divisible by p. So in the equation |Z(G)| = |G| − Σ [G : C_G(x_i)], the right-hand side is divisible by p. Therefore p divides |Z(G)|. Since Z(G) contains the identity it is nonempty, so |Z(G)| ≥ p > 1: every nontrivial p-group has a nontrivial center. This single fact is the engine behind almost everything we can prove about p-groups.

Cash it in: any group of order p² is abelian. If G has order p², its center Z(G) has order p or p². If |Z(G)| = p, then G/Z(G) has order p, hence is cyclic — but a standard lemma says G/Z(G) cyclic forces G abelian, contradicting |Z(G)| = p. So |Z(G)| = p², i.e. G = Z(G) is abelian. Combined with the structure theorem you will meet in guide 5, every group of order p² is isomorphic to Z/p²Z or to Z/pZ × Z/pZ — a complete classification, obtained from the class equation plus one lemma.