The Class Equation

The class equation is the entry point to the structure theory of finite $p$-groups, which in turn underlies the Sylow Theorems (though those are typically proved via slightly different orbit-counting arguments on cosets rather than conjugacy alone, the philosophy is identical). It is the direct ancestor of the proof that every group of order $p^2$ is abelian, and of induction arguments showing $p$-groups are nilpotent. In representation theory, the number of conjugacy classes of a finite group equals the number of irreducible complex representations, so understanding conjugacy class sizes (exactly what the class equation tracks) is foundational to character theory. In chemistry and physics, conjugacy classes of point groups and crystallographic space groups classify physically indistinguishable symmetry operations, directly generalizing the group-theoretic notion here. In computational group theory and cryptography, algorithms for computing conjugacy classes and centralizers (e.g. in GAP or Magma) are basic primitives, and centralizer computations feature in some group-based cryptographic protocols. Finally, the technique of "counting an orbit decomposition and extracting divisibility" recurs throughout combinatorics (Burnside's lemma/orbit-counting), making the class equation a template argument students see again and again.

:::quiz[In the class equation $|G| = |Z(G)| + \sum [G:C_G(g_i)]$, what does each term $[G:C_G(g_i)]$ represent?]

• The order of the centralizer of $g_i$
• The size of the conjugacy class of $g_i$ ✓
• The number of conjugacy classes in $G$
• The order of $G$ divided by $|Z(G)|$

Ask the tutor: Can you show why every group of order $p^2$ must be abelian, using the nontrivial-center result as a starting point?

Picture a finite group as a room full of objects, and imagine two objects as "the same shape" if one can be rotated (conjugated) into the other. Conjugacy classes are the shape-equivalence classes: all elements in one class are algebraically interchangeable from the group's internal perspective, related by an inner automorphism. An element sits alone in its class exactly when nothing in the group can "rotate" it into something different — that is precisely the defining property of the center. The size of a conjugacy class is governed by the Orbit-Stabilizer Theorem: the class of $g$ has size $[G:C_G(g)]$, the index of its centralizer. Elements with large centralizers (lots of commuting partners) have small conjugacy classes; elements with small centralizers (few commuting partners) get flung around into large classes. Central elements are the extreme case: centralizer equal to the whole group, class size $1$. The class equation is simply bookkeeping: add up $|Z(G)|$ (the size-1 classes) and the sizes of all the bigger classes, and the total must be $|G|$. The magic happens when $|G|$ is a prime power. Each noncentral class size $[G:C_G(g_i)]$ is a proper divisor of $p^n$ greater than $1$, hence is itself divisible by $p$. That means $|G| - \sum[G:C_G(g_i)]$ is divisible by $p$ — and that quantity is exactly $|Z(G)|$. Since $|Z(G)| \geq 1$ (it always contains $e$) and is divisible by $p$, it must be at least $p$, so it cannot be trivial.

Example 1: The dihedral group $D_5$ of order 10. Let $D_5 = \langle r, s \mid r^5 = s^2 = e,\ srs = r^{-1} \rangle$, the symmetries of a regular pentagon, with $|D_5| = 10$. Note $10 = 2 \cdot 5$ is not a prime power, so this example tests the class equation itself (Part 1) without expecting the $p$-group conclusion (Part 2) to apply. We compute conjugacy classes directly. The elements are $\{e, r, r^2, r^3, r^4, s, sr, sr^2, sr^3, sr^4\}$. Center: Suppose $r^k \in Z(D_5)$ for $1 \le k \le 4$. Then $s r^k s^{-1} = r^k$ is required, but $srs^{-1} = r^{-1}$, so $sr^ks^{-1} = r^{-k}$. This equals $r^k$ iff $r^{2k} = e$ iff $5 \mid 2k$ iff $5 \mid k$ (since $\gcd(2,5)=1$), impossible for $1 \le k \le 4$. So no nontrivial rotation is central. No reflection is central either, since reflections don't commute with $r$ (as just shown, $r$ and $s$ don't commute: $rs \ne sr$ because that would force $r=r^{-1}$, i.e. $r^2=e$, false since $r$ has order 5). Hence $Z(D_5) = \{e\}$, so $|Z(D_5)| = 1$. Conjugacy class of $r$: Conjugates of $r$ by rotations: $r^j r r^{-j} = r$. Conjugates by reflections: $sr^j \cdot r \cdot (sr^j)^{-1} = s r s^{-1} = r^{-1}$ (using that all reflections conjugate $r$ the same way, since $D_5$'s rotation subgroup is abelian and normal — a direct computation confirms $(sr^j) r (sr^j)^{-1} = s r^j r r^{-j} s^{-1} = srs^{-1} = r^{-1} = r^4$). So $\mathrm{cl}(r) = \{r, r^4\}$, size $2$. Centralizer: $C_G(r)$ has index $2$ by orbit-stabilizer, so $|C_G(r)| = 5$; indeed $C_G(r) = \langle r \rangle$ since rotations commute with $r$ but no reflection does. Conjugacy class of $r^2$: By the same computation, conjugating $r^2$ by a reflection gives $r^{-2} = r^3$. So $\mathrm{cl}(r^2) = \{r^2, r^3\}$, size $2$, centralizer also $\langle r \rangle$ of order $5$. Conjugacy class of $s$: Conjugates of $s$: by $r^j$, $r^j s r^{-j} = r^j s r^{-j}$. Using $sr = r^{-1}s$ repeatedly (equivalently $rs = sr^{-1}$), one computes $r^j s r^{-j} = r^{2j} s$. As $j$ ranges over $0,1,2,3,4$, $r^{2j} \bmod r^5$ takes all values $r^0, r^2, r^4, r^1, r^3$ (since $\gcd(2,5)=1$), so $\{r^{2j}s\} = \{s, r^2s, r^4s, rs, r^3s\}$ — all five reflections. Conjugating by reflections stays inside this set as well (conjugation by an element of the class of $s$ times a rotation reduces to the same orbit by transitivity of the orbit already computed). So $\mathrm{cl}(s) = \{s, rs, r^2s, r^3s, r^4s\}$, all five reflections in one class of size $5$. Centralizer of $s$ has index $5$, so $|C_G(s)| = 2$; indeed $C_G(s) = \{e, s\}$ since $s$ commutes with itself and $e$, but not with any rotation or other reflection (a short direct check: $r^j s (r^j)^{-1}=r^{2j}s \ne s$ for $j\neq 0$). Class equation check:

This matches $|D_5| = 10$ exactly, confirming the class equation. Note $|Z(D_5)| = 1$: since $10$ is not a prime power, Part 2 of the theorem simply does not apply, and indeed the center is trivial here — illustrating that the $p$-group hypothesis is doing real work.

Example 2: A group of order $p^3 = 27$ — the Heisenberg group mod 3. Let $p = 3$ and let

the group of upper unitriangular $3\times 3$ matrices over the field $\mathbb{F}_3 = \{0,1,2\}$, under matrix multiplication. This is a standard nonabelian group of order $27 = 3^3$. Write $M(a,b,c)$ for the matrix above. Direct matrix multiplication gives

with all entries reduced mod $3$. Center: We seek $M(a,b,c)$ commuting with every $M(a',b',c')$. Compute both products:

These agree for all $a',c'$ iff $ac' = a'c$ for all $a', c' \in \mathbb{F}_3$. Taking $a'=1,c'=0$ forces $0 = c \cdot 1 \cdot 0$ (automatically true) and taking $a'=0,c'=1$ forces $a \cdot 1 = 0$; more carefully, setting $c'=1,a'=0$: $a\cdot 1 = 0 \cdot c$, i.e. $a=0$; setting $a'=1,c'=0$: $a\cdot 0 = 1\cdot c$, i.e. $c=0$. So $a=c=0$ is required (and clearly sufficient, since then $ac'=a'c=0$ always). Hence

which has order $3$. This confirms Part 2's guarantee ($3 \mid |Z(G)|$, and indeed $|Z(G)| = 3 > 1$): the center is nontrivial, exactly as the theorem forces for any group of order $3^3$. A conjugacy class and its centralizer (illustrating the class equation numerically): take $g = M(1,0,0)$ (so $a=1,b=0,c=0$), a noncentral element since $a \ne 0$. We compute $C_G(g)$: we need $M(a',b',c')$ with $M(1,0,0)M(a',b',c') = M(a',b',c')M(1,0,0)$. Using the multiplication rule with $(a,b,c)=(1,0,0)$:

These agree iff $b' + c' = b'$, i.e. $c' = 0$. So

which has order $9$ (free choice of $a', b' \in \mathbb{F}_3$, each with $3$ choices). By the Orbit-Stabilizer Theorem,

consistent with Part 2's guarantee that noncentral class sizes are divisible by $p=3$ (here the class size is exactly $3$, i.e. $p^1$). One can carry out the analogous computation for representatives of all conjugacy classes; the full class equation for this group reads

since $27 - 3 = 24 = 8 \times 3$, and one checks (by an argument parallel to the one above, varying $g$ over noncentral representatives) that all noncentral classes here indeed have size exactly $3$. This is consistent with, and a concrete instance of, Part 1 and Part 2 of the theorem simultaneously.