MATH 817 Notes
JD Nir
[email protected]
www.math.unl.edu/∼jnir2/817.html
September 21, 2015
1st Isomorphism Theorem ϕ : G → H a group homomorphism ⇒
G/ker ϕ
∼
= im ϕ
via g · (ker ϕ) 7→ ϕ(g)
2nd Isomorphism Theorem G group, H ≤ G, K ≤ G, H ⊆ NG (K) (i.e. hK −1 h = K, ∀h ∈ H) then
a HK ≤ G
b K E HK
c H ∩K EH
d
HK/K
hk · K →
7
h · (H ∩ K)
∼
= H/H∩K via
h · K ←[ h · (H ∩ K)
G
HK
E
K
H
E
H ∩K
d : Two ways to intepret H and K, and they are ismorphic
K 6⊆ H
• make H bigger H 7→ HK
or
• make K smaller: K 7→ H ∩ K
Proof a already proven
b H ⊆ NG (K) and K ⊆ NG (K) ⇒ hH, Ki ≤ NG (K) and hH, Ki = HK
∴ K E HK
c Pick h ∈ H. If x ∈ H ∩ K, hxh−1 ∈ H ∩ K, so, H ∩ K E H.
d Define ϕ : H → HK/K by ϕ(h) = h · K ∈ HK/K
ϕ is the composition of H ,→ HK inc.
can.
HK/K
and so ϕ is a group homomorphism
1
JD Nir
[email protected]
MATH 817
It’s onto since hk · K = h · K = ϕ(h)
= {h ∈ H | h · K = K}
ker ϕ = h | h · K = eHK/K
= H ∩K
By 1st Isomorphism Theorem
H/H∩K
∼
= HK/K
via h · (H ∩ K) 7→ h · K.
,→
=
=
=
=
inclusion
1-1
onto
isomorphism
−→
=
isomorphism
∼
=
(
Ex G = GLn (R), K = SLn (R) E G, H = Dn :=
r1 0
..
.
0 rn
!
)
| ri ∈ R×


0  1
(1st row of X)
det
X


1
.
exercise

  2nd row of X 
HK/K ∼
=
GLn X = 
= H/H∩K HK


..

..
. 
.
0
1



r1
0









..


.
×
H ∩K = 
 | r1 , . . . , rn−1 ∈ R




rn−1





 0
1
r1 ···rn−1

GLn
SLn
∴
2nd IT
↓
∼
=

r1











0
..
.
rn−1
det X
Dn


0






×
 : r1 , . . . , rn−1 ∈ R





1
r1 ···rn−1
∼
=
R×
3rd Isomorphism Theorem Let G be a group, H E K E G and H E G then K/H E G/H and
/ ∼
= G/K
G/H K
/H
gH 7→ gK
where gH = coset of the subgroup of K/H of G/H represented by the element
gH ←[ gK
gH ∈ G/H
via
Notes 1 H E K E G 6⇒ H E G
2
K/H
= {kH | k ∈ K} ⊆ G/H
Proof: Define ϕ : G/H → G/K by ϕ(gH) = gK.
To see ϕ is a well-defined group homomorphism, consider π : G → G/K , the canonical map.
ker π = K ⊇ H and H E G
2
JD Nir
[email protected]
MATH 817
∴ UMP ⇒ ∃ϕ : G/H → G/K such that ϕ is a group homomorphism and ϕ(gH) = π(g) = gK.
Moreover, ϕ is onto.
ker phi = {gH | gK = K} = {gH | g ∈ K} = K/H .
∴ By 1st Isomorphism Theroem, G/H/K/H ∼
= G/K
Ex SLn (R) E Pn E GLn (R) SLn (R) E GLn (R)
Pn = {X ∈ GLn (R) | det X > 0}
Pn ≤ GLn (R), SLn (R E Pn
X ∈ Pn , Y ∈ GLn
det(Y XY −1 ) = det(X) > 0 ∴ Pn E GLn (R)
/
3rd Isom
∼
=
GLn/SLn P
n/SLn
Pn/SLn
∼
=
R×
≤
Reality check:
≤
GLn/SLn
GLn/Pn
∼
= (R>0 , ·)
GLn/SLn
Pn/SLn
ϕ : GLn → ({±1} , ·) by ϕ(X) =
GLn ∼
∼
= R×/R>0 ∼
= ({±1} , ·)
= ({±1} , ·) ∴
Pn
det(X)
| det(X)|
3