MATH 817 Notes
JD Nir
[email protected]
www.math.unl.edu/∼jnir2/817.html
October 2, 2015
• Final version of problem set 5 will be posted soon
• Exam?
LOIS: If G acts on X, ∀x ∈ X,
[G : Gx ] = #(G · x)
Cor If #G < ∞, G acts on X, then ∀x ∈ X, #(G · x) | #G
Pf Follows from LOIS and Lagrange:
[G : Gx ] · #(Gx ) = #G
L1
L2
L3
Ex G = D12
G acts on {L1 , L2 , L3 } = X
The action is transitive ⇒ just one orbit, of “length” 3.
[G : GL1 ] = 3
⇒ #GL1 = 4
GL1 = e, s, r3 , sr3
Ex G = group of rotational symmetries of a
G acts on the 6 faces {f1 , . . . , f6 } = X.
#Gf1 = 4 (4 rotationas about a line, by 0, 90, 180, 270)
[G : Gf1 ] = #(G · f1 ) = 6
⇒ #G = #[G : Gf1 ] · #Gf1
= 6 · 4 = 24
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MATH 817
JD Nir
[email protected]
Ex G̃ = group of all symmetries, including orientation reversing ones, of
G̃ acts on X, transitively. (G ≤ G̃)
∴ [G̃ : G̃f1 ] = 6
#G̃ = 6 · #(G̃f1 )
#G̃f1 = 8 (at least intuitively)
∴ #G̃ = 48
If G is any group, H ≤ G any subgroup, then G acts on the set X of all left cosets of H in G,
X = {gH | g ∈ G}, by (g 0 · (gH) := (g 0 g)H
Special case: H = {e}: G acts on G via left multiplication.
For any H ≤ G this action is transitive: the orbit of H ∈ X is all of X.
GH = {g ∈ G | gH = H} = H
LOIS: # of left cosets = [G : H]
Cayley’s Theorem If #G = n, then G is isometric to a subgroup of Sn .
Pf: Let G act on itself by left multiplication. This action determins a group homomorphism
ρ : G → Perm(G) where ρ(g) is the permutation of G.
Sending x ∈ G to g · x:
ker ρ = {g | g · x = x, ∀x ∈ G} = {e}
∴ G∼
= im ρ ≤ Perm(G) ∼
= Sn .
H ≤ G, G acts on X = {gH | g ∈ G}, action is transitive.
We get a group homomorphism
ρ : G → Perm(X) ∼
= Sm , m = [G : H]
(Assume m < ∞)
ker ρ = {g | gxH = xH, ∀x ∈ G}
= g | x−1 gx ∈ H, ∀x ∈ G
= g | g ∈ xHx−1 , ∀x ∈ G
\
=
xHx−1 ⊆ H
x∈G
Theorem #G < ∞, H ≤ G, [G : H] = p, where p is the smallest prime divisor of #G. Then H E G.
Proof Let X = {gH | g ∈ G}, G acts on X in the usual way, and the action is transitive. #X = p.
So we get a group homomorphism ρ : G → Sp
∼ im ρ ≤ Sp
1st isomorphism theorem ⇒ G/ker ρ =
So, # (G/ker ρ) | p! but # (G/ker ρ) | |G|
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MATH 817
JD Nir
[email protected]
gcd(#G, (p − 1)!) = 1. ∴ # (G/ker ρ) = 1 or p.
Also, recall ker ρ ≤ H.
# (G/ker ρ) = 1 ⇒ ker ρ = G ⇒⇐.
# (G/ker ρ) = p. But [G : H] = p and ker ρ ⊆ H ⇒ ker ρ = H ⇒ H E G.
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