MATH 817 Notes
JD Nir
[email protected]
www.math.unl.edu/∼jnir2/817.html
September 14, 2015
H ≤ G.
A left coset of H in G is a subset of G of the form xH = {xh | h ∈ H} for some x ∈ G.
A right coset of H in G is a subset of G of the form Hx = {hx | h ∈ H} for some x ∈ G.
Lemma H ≤ G.
1 For x, y ∈ G, (xH = yH or xH ∩ yH = ∅) and (Hx = Hy or Hx ∩ Hy = ∅)
2 xH = yH ⇔ y −1 x ∈ H ⇔ x−1 y ∈ H
3 Hx = Hy ⇔ xy −1 ∈ H ⇔ yx−1 ∈ H
Pf 2 2nd “⇔” holds since H ≤ G
If xH = yH, then x = yh, some h ∈ H, and so y −1 x = h ∈ H X
If y −1 x = h ∈ H, then xh0 = yhh0 ∈ yH ∀h0 ∈ H ∴ xH ⊆ yH
By symmetry, yH ⊆ xH
3 Similar
1 If xH ∩ yH 6= ∅, then xh = yh0 , some h, h0 ∈ H, and so y −1 x = h0 h−1 ∈ H. By 2 xH = yH.
2nd statement is similar.
So, H ≤ G determines two equivalence relations on G:
left
def
x ∼ y ⇔ xH = yH ⇔ y −1 x ∈ H
H
right def
x ∼ y ⇔ Hx = Hy ⇔ xy −1 ∈ H
H
left
right
H
H
Do ∼ and ∼ respect multiplication?
left
H
left
⇒
x·z ∼y·z
??
??
x∼y
m
yz −1 xz ∈ H
m
m
y −1 x ∈ H
H
⇒
(z −1 (y −1 x)z ∈ H
No. Unless H is normal in G.
Prop H ≤ G The following are equivalent
1 xH = Hx ∀x ∈ G
left
right
H
H
2 ∼ and ∼ coincide
3
xHx−1
= H ∀x ∈ G
4 xHx−1 ⊆ H ∀x ∈ G
1
MATH 817
5
6
JD Nir
[email protected]
left
∼ respects multiplication
H
right
∼ respects multiplication
H
If these hold, we say H is a normal subgroup of G. Notation H E G
I’ll show 4 ⇒ 3
Assume xHx−1 ⊆ H, ∀x ∈ G
Pick h ∈ H. For any x,
x−1 h(x−1 )−1 = x−1 hx ∈ H
So x−1 hx = h0 ∈ H and h = xh0 x−1 ∈ xHx−1
∴ ∀x H ⊆ xHx−1
Rest: Exercise
Recall If ∼ respects multiplication then G/∼ is a group under
[x] · [y] = [xy]
left
right
H
H
So, if H E G, then ∼ = ∼ =: ∼ and G/∼
is a group
H
G/H
:=
G/∼
G/H
:=
H
=
Notation If H E G,
H
{xH | x ∈ G} = {Hx | x ∈ G}
(xH)(yH) = (xy)H
or
is a group via
(Hx)(Hy) = H(xy)
The canonical map π : G → G/H is a ssurjective group homomorphism, where π(x) = xH = Hx
Ex 1 Define ∼ on Z by
i ∼ j ⇔ i ≡ j (mod n)
⇔ i − j ∈ hni ≤ Z
∴ ∼ is the equivalence relation associated to hni E Z
Z/hni
=: Z/n is a quotient group
2 {±1} E Q8
(Z(G) E G)
Z/2 × Z/2 ∼
= Q8/{±1} = {{±1} , {±i} , {±j} , {±k}}
{±1} · {±i} = {±i}
{±i} · {±j} = {±k} etc.
{±i}2 = {±1}
3 Fix n. hri E D2n
k
{1,r,...,rn−1 }
sri s−1 = r−i
∴ s ∈ ND2n (H) := g | gHg −1 = H
2
MATH 817
JD Nir
[email protected]
r ∈ ND2n (H)
But ND2n (H) ≤ D2n
hs, ri = D2n ∴ ND2n (H) = D2n ⇒ H E G.
D2n/H
is a group with 2 elements:
H and sH = s, sr, . . . , srn−1
D2n/H
∼
= Z/2
4 hx1 , . . . , xm | R1 , . . . , Rn i :=
F
N
:=
(free group on x1 , x2 , . . . , xm )
N
N = unique smallest normal subgroup of G that contains R1 , . . . , Rn
5 If ϕ : G → G0 is a group homomorphism, then
ker ϕ := {g ∈ G | ϕ(g) = e}
is a normal subgroup of G.
Pf: h ∈ ker ϕ, x ∈ G
⇒ ϕ(xhx−1 ) = ϕ(x)ϕ(h)ϕ(x)−1
= ϕ(x)ϕ(x)−1
= e
⇒ xhx−1 ∈ ker ϕ
∴ ∀x x · (ker ϕ)x−1 ⊆ ker ϕ ⇒ ker ϕ E G.
:=
5’ SLn (R) E GLn (R)
ker(det:GLn (R)→R× )
Question If H E G what’s the kernal ofthe canonical group
ker π : G → G/H ?
ker π = {g ∈ G | gH = H}
= H
Prop Given a subgroup H of G,
H is normal ⇔ H is the kernal of some group homomorphism ϕ : G → G0
Pf (⇐) Done
(⇒) Use π : G → G/H .
3