Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-I
19
( G )
N G
( H )
≅
H
N
Define
f : G N
by
Na
Ha
is well-defined, since Na = Nbfor
we get
. Since N ⊆ H,
thus gives
and so
Ha = Hb. f is a homomorphism:
the identity of
⇔ Ha = H
⇔ a ∈ H
Hence ker . As ker f ∆ G ,
N so H ∆G
N N
The fundamental homomorphism theorem for groups implies that
G
N
ker
≅ Im f = G
f
N
( ) ) ( ) ( ) ( ) ( H ) f ( Nb)
∈
∀f G
⇔ker
−1
,b
G,
= H
HN
Na f
Ha
∩,∀
, ∈Na
bf
Nker
a
= Nb ∆
f = , ⇔f
fNab
( ) = HabH
, = HaHb = f Na
a b
H
≅
∈Nab
HN
∩=
NNaNb
= f a f b ∀a,
b∈
N ( H ∩ N ) G
⇔ a ∈ N and a ∈ H∴
N ≅ G
H H
⇔ a ∈ H ∩ N
N
Theorem 6 (Second Isomorphism Theorem)
Let G be a group, and let N ∆G,
let H be any subgroup of g. Then HN is a subgroup of G,
and
Proof:
Define f:H
by a
is a homomorphism since
a ∈ ker f ⇔ f ( a) = N,
the identity element of HN and a
H
So