Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-I
47
Remark :
If G is abelian, then
abelian group.
Solution :
(i) Given gug − 1
∈U ∀ g ∈G
, ∀ u ∈U . To show U ∃ ∆ G
Θ
is the subgroup generated by U.
4
could be simplified to ( a ) ( a ) 5
, but this may not be true in the non
in U}
= {all finite products of integral powers of
Θ∃ Θ 3 5 7
ax
gxg
n n
k
i 1) 11−
2 1 n n
u1 u2
(........
a2
gu
1 n nk
U( ∀G xyG′
yxG′ ∃−
gux ( = ′ = ) u
( a 2
iy xG ∈g G U′ ∃ ′ −1
y −
x yx x)
= , i
yx U∈
U)
G........
1)
∀′
i ∈
1 2k
u
I
G′
k
= {all finite products of integral powers of element in U}
Let x ∈U
∃ ,
, ui U i ∈ Z −1 n1 −1
n2 1
g = gu
g g g
nk
−1
1 u
g.......
g g
2
uk
=
−1 n1
− 1 n2
−1 n
( gu1g
) ( gu2g
) ................. ( gu k
g ) k
∈U
∃
because
Hence gxg
∈U∃
∀g
∈G
⇒ U ∃ ∆ G (i.e. G ∆ G )
(ii) U =
− −
{ xy x y x,
y ∈G}
The Commutator subgroup of G
From (1) U ∃ = G′∆ G .
l
(iii) G G′ = xG'| x ∈Gq, To show
We must show xG′ yG′ =
i.e.
L.H.S.
=
=
=
(Θ
abelian,
1
2
− −
is a commutator and so y x yxG′ = G′
1 1
)
−1 −1
= ( xyy x ) yxG′ = yxG′
= yG′ xG′ = R.H.S.
(iv)
G N
abelian ⇔ ′ ⊂
To show
G
N
−1 −1 −1 −1
−1 −1 −1 −1
−1 −1 −1 −1
⇒ x Ny N = y Nx N x y N = y x N xyx y N = N xyx y ∈N
i.e. every commutator to a group N, hence all finite products of integral powers of commutators are in
N. ∴ G ′ ⊂ N.