nLab class equation

Idea

The term class equation (or class formula, or orbit decomposition formula) refers to a basic type of counting argument that comes about by decomposing a finite G-set as a union of its orbits. This has a number of fundamental applications in group theory.

Statement

Let $G$ be a group and let $A$ be a G-set (given by a homomorphism $G \to \hom_{Set}(A, A)$ of monoids, with which is associated an action $\alpha: G \times A \to A$). Recall that $A$ is connected in the category of G-sets if $A$ is inhabited and the action is transitive; in this case, choosing an element $a \in A$, there is a surjection of $G$-sets $G \to A$ sending $1 \mapsto a$, and this induces an isomorphism $G/Stab(a) \cong A$ where $Stab(a)$ is the stabilizer of $a$ and $G/Stab(a)$ is the $G$-set consisting of left cosets of $Stab(a)$.

More generally, each $G$-set $A$ admits a canonical decomposition as a coproduct of its connected components; the components are usually called the orbits of the action. Choosing a representative element $a_x$ in each orbit $x$, this means we have an isomorphism of $G$-sets

$$A \cong \sum_{orbits x} G/Stab(a_x).$$

By taking $G$ and $A$ to be finite and counting elements, we get an equation of the form

$${|A|} = \sum_{orbits x} \frac{{|G|}}{{|Stab(a_x)|}}.$$

We call an instance of this equation a class equation. By judicious choice of groups $G$ and $G$-sets $A$, often in combination with number-theoretic arguments, one can derive many useful consequences; some sample applications are given below.

Notice that reading the class equation equivalently as

$$\sum_{orbits x} \frac{{1}}{{|Stab(a_x)|}} = \frac{|A|}{|G|}$$

it expresses the groupoid cardinality of the action groupoid of $G$ acting on $A$.

Applications

Centers of $p$-groups

Let $p$ be a prime; recall that a $p$-group is a finite group whose order is a power of $p$. A basic structural result is the following.

Proposition

A non-trivial $p$-group $G$ has a nontrivial center $Z(G)$.

Proof

Let $G$ act on itself by the conjugation action $G \times G \to G$, $(g, h) \mapsto g h g^{-1}$. In this case an orbit $Orb(h)$ is usually called the conjugacy class of $h$, and $Orb(h)$ is trivial (consists of exactly one element $h$) iff $h$ belongs to $Z(G)$. In any case ${|Orb(h)|} = \frac{{|G|}}{{|Stab(h)|}}$ divides ${|G|} = p^n$, and therefore $p$ divides ${|Orb(h)|}$ if $h$ is noncentral. In this case the class equation takes the form

$${|G|} = {|Z(G)|} \; + \sum_{nontrivial\; orbits x} \frac{{|G|}}{{|Stab(a_x)|}}$$

and now since $p$ divides ${|G|}$ as well as each term in the sum over nontrivial orbits, it must also divide ${|Z(G)|}$. In particular, $Z(G)$ has more than one element.

It follows by induction that $p$-groups are solvable, since the center is a normal subgroup and the quotient $G/Z(G)$ is also a $p$-group. Since a group obtained from an abelian group by repeated central extensions is nilpotent, $p$-groups are in fact nilpotent.

Number of fixed points

An elementary observation that is frequently useful is that the number of fixed points of an involution on a finite set $S$ has the same parity as $S$. This is a statement about $\mathbb{Z}/(2)$-sets; we generalize this to a statement about $G$-sets for general $p$-groups $G$. (Again, a fixed point of a $G$-set is an element whose orbit is a singleton.)

Proposition

If $G$ is a $p$-group acting on a set $A$, then

• ${|A|} \equiv {|Fix(A)|} \; mod p$.

As special cases, if there is just one fixed point, then ${|A|} \equiv 1 \; mod p$, and if $p$ divides ${|A|}$, then $p$ divides ${|Fix(A)|}$.

Proof

The class equation takes the form

$${|A|} = {|Fix(A)|} \; + \sum_{nontrivial\; orbits x} \frac{{|G|}}{{|Stab(a_x)|}}$$

where $p$ divides each summand over nontrivial orbits on the right, since $G$ is a $p$-group. Now reduce mod $p$.

Wedderburn’s theorem

In the theory of finite projective planes, an important result is that a projective plane is Pappian if it is Desarguesian. The purely algebraic version of this is Wedderburn’s theorem:

Theorem

A finite division ring $D$ is commutative.

Proof

(Witt) The center of $D$ is a field; if $D$ is of characteristic $p \gt 0$, then the center $F$ has $q = p^f$ elements for some $f$. We may regard $D$ as an $F$-vector space of dimension $n$, whence the number of elements of its multiplicative group $D^\times$ is $q^n-1$.

The conjugation action of $D^\times$ on itself yields a decomposition

$$D^\times \cong F^\times \; + \sum_{nontrivial\; orbits x} \frac{{|D^\times|}}{{|Stab(a_x)|}}$$

where again the elements of the center $F^\times = Z(D^\times)$ correspond to the trivial orbits. The stabilizer of any $a_x$, together with $0$, forms a division ring (strictly) intermediate between $F$ and $D$; usually this is called the centralizer $C(a_x)$ of $a_x$. Putting $d_x = dim_F(C(a_x))$, the division ring $C(a_x)$ has $q^{d_x}$ elements, and notice $d_x$ divides $n$ because $n/d_x$ is just the dimension of $D$ seen as a vector space (module) over $C(a_x)$. Thus ${|Stab(a_x)|} = q^{d_x} - 1$, and we have a class equation

$$q^n - 1 = q - 1 + \sum_x \frac{q^n - 1}{q^{d_x} - 1}.$$

Now let $\zeta \in \mathbb{C}$ be any primitive $n^{th}$ root of unity. Since $z - \zeta$ divides each polynomial $z^n-1$ and $\frac{z^n - 1}{z^{d_x} - 1}$, so does $\prod_{prim.\; \zeta} (z-\zeta)$. It follows that the algebraic integer $\prod_{prim.\; \zeta} (q - \zeta)$ divides each of the integers $q^n-1$ and $\frac{q^n - 1}{q^{d_x} - 1}$, and hence divides $q-1$ according to the class equation. But also ${|q - \zeta|} \geq {|q-1|}$ for any root of unity $\zeta$. Thus ${|q - \zeta|} = {|q-1|}$ and $n = 1$, i.e., $F = Z(D)$ is all of $D$ as was to be shown.

Sylow theorems

Let $G$ be a finite group of order $n$, and suppose that $p$ is a prime that divides $n$; say $n = p^f u$ where $p$ does not divide $u$. Recall that a Sylow p-subgroup is a $p$-subgroup of maximal order $p^f$.

A fundamental fact of group theory is that Sylow $p$-subgroups exist and they are all conjugate to one another; also the number of Sylow $p$-subgroups is $\equiv 1 \; mod p$.

Existence of Sylow $p$-subgroups can be proven by exploiting the same type of argument as in the proof of Proposition :

Theorem

If $G$ has order $n$ and $p^k$ is a prime power dividing $n$, then there is a subgroup of $G$ of order $p^k$.

Proof

First we show that Sylow subgroups exist. We start by observing that if a group $H$ has a $p$-Sylow subgroup $P$, then so does any subgroup $G$. To prove this, first note that if we let $G$ act on $H/P$ by left translation, then the stabilizer of any element $h P$ is $G \cap h P h^{-1}$, a $p$-group since $h P h^{-1}$ is. Then note that since $H/P$ has cardinality prime to $p$, so must one of its connected components $G/Stab(a_x)$ in its $G$-set decomposition

$$H/P \cong \sum_{orbits\; x} G/Stab(a_x),$$

and this makes $Stab(a_x)$ a $p$-Sylow subgroup of $G$.

Then, if $G$ is of order $n$, apply this observation to the embedding

$$G \stackrel{Cayley}{\hookrightarrow} Perm({|G|}) \cong S_n \hookrightarrow GL_n(\mathbb{Z}/(p)) = H$$

where we embed the symmetric group $S_n$ via permutation matrices into the group $H$ of $n \times n$ invertible matrices over $\mathbb{Z}/(p)$. The group $H$ has order $(p^n - 1)(p^n - p)\ldots (p^n - p^{n-1})$, with maximal $p$-factor $p^{n(n-1)/2}$. It thus has a $p$-Sylow subgroup given by unitriangular matrices, i.e., upper-triangular matrices with all $1$'s on the diagonal. Therefore $p$-Sylow subgroups $P$ exist for any finite group $G$.

Finally, note that by Proposition , $P$ is solvable and therefore has a composition series

$$\{1\} = P_0 \subset P_1 \subset \ldots \subset P$$

where each $P_k$ has order $p^k$.

Theorem

If $H$ is a $p$-subgroup of $G$ and $P$ is a Sylow $p$-subgroup, then $g^{-1} H g \subseteq P$ for some $g \in G$. In particular, all Sylow $p$-subgroups are conjugate to one another.

Proof

$G$ acts on the set of cosets $G/P$ as usual by left translation, and we may restrict the action to the $p$-subgroup $H$. By maximality of $P$, we see ${|G/P|}$ is prime to $p$, and so by Proposition , ${|Fix_H(G/P)|}$ is also prime to $p$. In particular, $Fix_H(G/P)$ has at least one element, say $g P$. We infer that $h g P = g P$ for all $h \in H$, or that $g^{-1} h g P = P$ for all $h \in H$, and this implies that $g^{-1} H g \subseteq P$.

Theorem

The number of Sylow $p$-subgroups of $G$ is $\equiv 1 \; mod p$.

Proof

Let $Y$ be the set of Sylow $p$-subgroups; $G$ acts on $Y$ by conjugation. As all Sylow $p$-subgroups are conjugate, there is just one orbit of the action, and the stabilizer of an element $P \in Y$ is just the normalizer $N_G(P)$ (by definition of normalizer). Thus $Y \cong G/N_G(P)$ as $G$-sets.

Restrict the action to the subgroup $P$. Of course the element $P \in Y$ is a fixed point of this restricted action, and if $Q$ is any other fixed point, it means $x Q x^{-1} = Q$ for all $x \in P$, whence $P \subseteq N_G(Q)$. Now: $P, Q$ are both Sylow $p$-subgroups of $N_G(Q)$ and are therefore conjugate to each other (as seen within the group $N_G(Q)$). But $Q$ is already fixed by the conjugation action in its stabilizer $N_G(Q)$, so we conclude $P = Q$. We conclude $Fix_P(Y)$ has exactly one element. From ${|Y|} \equiv {|Fix_P(Y)|} \; mod p$ (Proposition ), the theorem follows.

Corollary

If $G$ is a group of order $n = p^k m$ where $p$ is prime to $m$, then the number $n_p$ of $p$-Sylow subgroups divides $m$.

Proof

Because $G$ acts transitively on $p$-Sylow subgroups (Theorem ) , the number $n_p$ divides ${|G|} = p^k m$. From Theorem , we have $a n_p + b p^k = 1$ for some integers $a, b$. Since $n_p$ divides both terms of the left side of $a n_p m + b p^k m = m$, it divides $m$.

The Sylow theorems are routinely used throughout group theory. As a sample application: if $p, q$ are distinct primes, with $p^2 \nequiv 1\; \mod q$ and $q \nequiv 1\; \mod p$, then any group of order $p^2 q$ is abelian. (For example, a group of order $2023 = 7 \cdot 17^2$ must be commutative.)

Proof

We have $n_p|q$ by Corollary , but $n_p \neq q$ (using Theorem and $q \nequiv 1\; \mod p$), so $n_p = 1$. Arguing similarly we have $n_q|p^2$ but $n_q \neq p$ and $n_q \neq p^2$, so $n_q = 1$. The $p$-Sylow subgroup $P$ of order $p^2$ and the $q$-Sylow subgroup $Q$ of order $q$ are both abelian. $P \cap Q = \{1\}$ since $p, q$ are relatively prime, and $P, Q$ are normal subgroups of $G$ since $n_p, n_q$ are both $1$. It follows that $P Q$ is a subgroup of order $p^2 q$, hence $P Q = G$. Thus to prove $G$ abelian, it suffices to show that if $x \in P$ and $y \in Q$, then $x$ and $y$ commute, i.e., $x y x^{-1} y^{-1} = 1$. But by normality of $Q$, the element $(x y x^{-1}) y^{-1}$ belongs to $Q$; similarly, the element $x(y x^{-1} y^{-1})$ belongs to $P$, and so $x y x^{-1} y^{-1} \in P \cap Q = \{1\}$. The result follows.

Related concepts

• class function

• Klein geometry