MATH 817 Notes
JD Nir
[email protected]
www.math.unl.edu/∼jnir2/817.html
September 4, 2015
G = group, X ⊆ G a subset
• CG (X) = {g ∈ G | gx = xg , ∀x ∈ X} centralizer of X in G.
| {z }
m
gxg −1 =x
• NG (X) = g ∈ G|gXg −1 = X , where gXg −1 = gxg −1 | x ∈ X normalizer of X in G.
NG (H) = G
m
Usually, X ≤ G. A normal subgroup is a subgroup H ≤ G s.t.
∀g ∈ G, gHg −1 = H.
CG (X) ≤ G and NG (X) ≤ G
Proof: 1 e · X · e−1 = X Xe ∈ NG (X)
2 a, b ∈ NG (X)
(ab)X(ab)−1 = a(bXb−1 )a−1
= aXa−1 = X
∴ ab ∈ NG (X)
3 a ∈ NG (X), ∴ aXa−1 = X.
NTS a−1 X(a−1 )−1 = X
a−1 X(a−1 )−1 = a−1 Xa
aXa−1 = X ⇒ a−1 (aXa−1 ) = a−1 X
⇒ eXa−1 = a−1 X
⇒ Xa−1 = a−1 X
⇒ Xa−1 a = a−1 Xa
⇒ X = a−1 Xa X
Note In general, NG (X) 6= g ∈ G | gxg −1 ∈ X, ∀x ∈ X
But we do have = if
• X≤G
or
• #<∞
Ex
• CG (G) = Z(G) = center
• CG ({y}) = {g ∈ G | gy = yg}
• CG (X) ≤ NG (X)
• CG ({y}) = NG ({y})
1
JD Nir
[email protected]
MATH 817
• NG (G) = G
(H ≤ G ⇒ H ≤ NG (H) ≤ G)
If G acts on a set S, for s ∈ S,
Gs = {g ∈ G | g · s = s}
= stabalize of s (for the given action)
Then Gs ≤ G.
• e · s = s ⇒ e ∈ Gs .
• x, y ∈ Gs ⇒ x· (y · s) = (xy) · s
Z, Y, W ⊆ G ⇒
-
Z(Y W ) = (ZY )W
x ∈ Gs ⇒ x · s = s ⇒
⇒
•
⇒
⇒


x−1 · (x · s) = x−1 · s
e · s = x−1 · s
s = x−1 · s
x−1 ∈ Gs
Let S = G and let G act on S by conjugation:
g · s := gsg −1
Gs = g | gsg −1 = s
Then
= CG ({s})
Also, X ⊆ G ⇒\
CG (X) =
CG ({x})
x∈X
\
=
Gx ≤ G
x∈X
Note: For S = G + action is conjugation,
NG (X) =?
NG (X) = {g ∈ G | g · X = X}
Ex Recall D2n acs on {1, 2 . . . , n}
1
n
2
3
(D2n )1 = {e, s} ≤ D2n
(D2n )j = e, the unique relation that fixes j th number, srn−j+1
Ex Sn acts on {1, 2, . . . , n} in the following way
2
MATH 817
JD Nir
[email protected]
Pick i, 1 ≤ i ≤ n. (Sn )i = {σ ∈ Sn | σ(i) = i} ≤ Sn
(Sn )i ∼
= Sn−1
If ϕ : G → H is any group homomorphism, then ker ϕ := {g ∈ G | ϕ(g) = e} is a subgroup of G +
im(ϕ) ≤ H
ker ϕ ≤ G :
• ϕ(e) = e ⇒ e ∈ ker ϕ
• ϕ(x · y) = ϕ(x) · ϕ(y) ⇒ if x, y ∈ ker ϕ then x · y ∈ ker ϕ
• ϕ(x−1 ) = ϕ(x)−1
(ϕ(x−1 · ϕ(x)) = ϕ(x−1 · x) = ϕ(e) = e ∴ ϕ(x−1 ) = ϕ(x)−1 )
If x ∈ ker ϕ, then x−1 ∈ ker ϕ
Note: In fact, ker(ϕ) is a normal subgroup of G.
Ex: f = field, e.g. F = R
• SLn (F ) = {A ∈ GLn (F ) | det(A) = 1}
SLn (F ) ≤ GLn (F ) because SLn (F ) = ker(det : GLn (F ) → F x )
•
An = {σ ∈ Sn | σ can be written as a product of an even number of transpositions}
= ker(sgn : Sn → ({±1} , ·)) ≤ Sn
Def If G is a group and x ∈ G,
hxi := {xn | n ∈ Z}
= subgroup of G generated by x
G is cyclic if G = hxi for some x ∈ G
Ex 1 G = e, r, r2 , . . . , rn−1 ≤ D2n
G = hri = hrj i, if gcd(j, n) = 1
2 (Z, +) = h−1i = h1i
3 Z/n = 0, 1, . . . , n − 1
i+j =r
i+j =q·n+r 0≤r <n
Z/n
= h1i = hji if gcd(j, n) = 1
4 for n ∈ N
µn = {z ∈ Cx | z n = 1} ≤ Cx
n
o
j
= e2πi· n | j = 0, 1, . . . , n − 1
2πi
µn = he n i
j
= he2πi n i if gcd(j, n) = 1
| {z }
primitive nth root of 1
3
JD Nir
[email protected]
MATH 817
e
2πi
7
e2πi = 1
6
e2πi· 7
4