Milne - Group Theory.. - Free

38 4 GROUPS ACTING ON SETS
Proof. Let x 0 ∈ X. Then
Ker(G → Sym(X)) = ⋂
Stab(gx 0 ) 4.4
= ⋂ g · Stab(x 0 ) · g −1 .
Stab(x) = ⋂
x∈X g∈G
Hence, the proposition is a consequence of the following lemma.
Lemma 4.10. For any subgroup H of a group G, ⋂ g∈G gHg−1 is the largest normal
subgroup contained in H.
Proof. Note that N 0 = df
⋂g∈G gHg−1 , being an intersection of subgroups, is itself
a subgroup. It is normal because
g 1 N 0 g −1
1 = ⋂ g∈G(g 1 g)N 0 (g 1 g) −1 = N 0
— for the second equality, we used that, as g runs over the elements of G, soalso
does g 1 g. Th us N 0 is a normal subgroup of G containedin1H1 −1 = H. If N is a
second suchgroup, then
N = gNg −1 ⊂ gHg −1
for all g ∈ G, andso
N ⊂ ⋂ gHg −1 = N 0 .
The class equation
When X is finite, it is a disjoint union of a finite number of orbits:
X =
m⋃
i=1
O i
(disjoint union).
Hence:
Proposition 4.11. The number of elements in X is
#X =
m∑
#O i =
i=1
m∑
(G :Stab(x i )), x i in O i .
i=1
When G acts on itself by conjugation, this formula becomes:
Proposition 4.12 (Class equation).
(G :1)= ∑ (G : C G (x))
(x runs over a set of representatives for the conjugacy classes), or
(G :1)=(Z(G) :1)+ ∑ (G : C G (y))
(y runs over set of representatives for the conjugacy classes containing more than one
element).