MATH 101A: ALGEBRA I PART A: GROUP THEORY 114. p-groups and the class formulaThe class formula is used to prove that p-groups are nilpotent. Theclass formula in turn comes from the orbit-stabilizer formula.4.1. orbit-stabilizer formula. This f ollows from the following theo-rem.Theorem 4.1. Suppose that G acts on a set X and x ∈ X. Then thereis a bijection between the orbit πG(x) of x and the set of left cosets ofthe stabilizer Gxof x. The bijectionφ : G/Gx≈−→ πG(x)is given by φ(gGx) = gx.When X is finite, we get a numerical formula.Corollary 4.2 (orbit-stabilizer formula). If X is a G-set and x ∈ X,the size of the orbit of x is equal to the index of the stabilizer of x.Recall that X is a disjoint union of orbits:X =!xirepπG(xi)where xi∈ X are representatives of the orbits (i.e., one element fromeach orbit).Corollary 4.3 (orbit-sum formula). The number of elements in a finiteG-set X is given by|X| ="xirep|πG(xi)| ="xirep|G : Gxi|.4.2. actions of p-groups. One example of the orbit-sum formula isgiven by the action of p-groups on finite sets. If P is a p-group, with|P | = pkthen every subgroup has order a power of p. If P acts on afinite set X then the size of each orbit is also a power of p:|πP(x)| = |P : Px| =|P ||Px|=pkpj= pk−j.Notice that this is divisible by p except when it is equal to 1. The sizeof X is given by the orbit-sum formula:|X| ="xi|P : Pxi| ="xipk−ji.Now we want to separate the s ummands which are equal to 1 and thosewhich are greater than 1. If |P : Px| = 1 then Px= P and x is a fixed12 MATH 101A: ALGEBRA I PART A: GROUP THEORYpoint of the action. This means g · x = x for all g ∈ P . The otherorbits have more than one element and therefore the size of these orbitsis divisible by p. This gives the follow ing theorem which we need later.Theorem 4.4. Suppose that P is a p-group acting on a finite set X.Then the number of elements in X is congruent modulo p to the numberof fixed points of the action.The key use of this formula is the following.Corollary 4.5. Every nontrivial finite p-group has a nontrivial center.Proof. Suppose that P is a p-group with pk> 1 elements. Then Pacts on P by conjugation. The theorem says that the number of fixedpoints of this action is congruent to |P | = pkmodule p. In other words,p divides the number of fixed points. But g ∈ P is a fixed point of theconjugation action if and only if g ∈ Z(P ). Therefore, p divides |Z(P )|which implies that Z(P ) has at least p elements. !4.3. class formula. This is another example of the orbit-sum formula.Take a finite group G acting on the set G by conjugation. Then theorbit of xi∈ G is the conjugacy class of xiand the stabilizer is thecentralizer CG(xi) of xi. This gives the following formula.|G| ="xi|γG(xi)| ="xi|G : CG(xi)|This is not the class formula. We need to separate the summands whichare equal to 1. By the orbit-stabilizer formula,|γG(x)| = |G : Gx| = |G : CG(x)| = 1This means CG(x) = G. In other words, x is central, or x ∈ Z(G).You can also just look at the definition:γG(x) = {gxg−1| g ∈ G}.This is {x} if and only if gxg−1= x, i.e., x ∈ Z(G).Every central element contributes 1 to the sum and every noncentralelement contributes a number |G : CG(x)| $= 1. This gives the following.Theorem 4.6 (class formula). The number of elements in any finitegroup G is given by|G| = |Z(G)| +"xi|γG(xi)| = |Z(G)| +"xi|G : CG(xi)|where the xiare representatives of the conjugacy classes in G whichcontain more than one element.This formula is used to prove Corollary 4.5 but we already did that.MATH 101A: ALGEBRA I PART A: GROUP THEORY 135. Nilpotent groupsThere are two definitions of nilpotent groups. I don’t rememberwhether I proved that they are equivalent but we can do that here.Definition 5.1. A group G is nilpotent if there is a normal towerG = G0" G1" G2" · · · " Gn= {e}with the following properties for all i.(1) Gi# G and(2) Gi−1/Gi≤ Z(G/Gi).A normal tower satisfying these conditions will be called a central se-ries.The smallest possible value of n is called the nilpotency class of G.Thus G has nilpotency class 1 if and only if it is abelian (and nontrivial).The other definition that I gave used iterated centers Z(i)(G) definedrecursively as follows.(1) Z(1)(G) := Z(G).(2) Z(n+1)(G) is the unique normal subgroup of G which containsZ(n)(G) and so thatZ(k+1)(G)Z(k)= Z#GZ(k)(G)$Notice that this recursive definition implies that(5.1)Z(n)(G)Z(G)= Z(n−1)#GZ(G)$.Proposition 5.2. A group G is nilpotent if and only if G = Z(n)(G)for some n. Furthermore, the smallest such n is equal to the nilpotencyclass of G.Proof. If G = Z(n)(G) then the towerG = Z(n)(G) ≥ Z(n−1)(G) ≥ · · · ≥ Z(G) ≥ {e}is a central series. Therefore, G is nilpotent of class c ≤ n.Conversely, suppose that G is nilpotent of class c. Then we have acentral series:G = G0" G1" · · · " Gc−1" Gc= {e}Claim: Gc−k≤ Z(k)(G) for all k.If this is true then, putting k = c, we get the that G = Z(c)(G),which means that c ≥ n where n is the smallest number satisfying14 MATH 101A: ALGEBRA I PART A: GROUP THEORYG = Z(n)(G). Since c ≥ n and c ≤ n we conclude that c = n. So, itsuffices to prove the claim.The claim holds for k = 1 by the assumption that Giform a centralseries. Suppose by induction on k that Gc−k≤ Z(k)(G) and we havethe quotient map:φ : G/Gc−k$ G/Z(k)(G).Any epimorphism has the property that it sends central elements intocentral elements. Therefore,φ#Gc−k−1Gc−k$≤ φ#Z#GGc−k$$≤ Z#GZ(k)(G)$=Z(k+1)(G)Z(k).Since φ is the quotient map this implies that Gc−k−1≤ Z(k+1)(G).Therefore, the claim and thus the proposition holds. !Corollary 5.3. A nontrivial group G is nilpotent of class c if and onlyif G/Z(G) is nilpotent of class c − 1.Proof. This follows from the proposition and Equation (5.1). G is nilpo-tent of class ≤ c if and only if Z(c)(G) = G. By (5.1) this is equivalentto saying that Z(c−1)(G/Z(G)) = G/Z(G). By the proposition thelast statement is equivalent to saying that G/Z(G) is nilpotent of class≤ c − 1. Putting these together we get the corollary. !Corollary 5.4. Finite p-groups are nilpotent.Proof. If P is a p-group then |P | = pkfor some k. If k = 1 then P iscyclic and therefore abelian making it nilpotent of class 1. If k > 1 thenP has a nontrivial center Z(P ) and P/Z(P ) is nilpotent by induction.By the previous corollary this implies that P is nilpotent.