Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-I
35
integers x and y. Note that x and y both cannot be +ve integers because if 1 in R.H.S. Let x be a +ve integer
and y be -ve integer. Hence
mx−ny
ab = a b
−F
= a a
mx−ny
mx ny
HG I K J
-ny mx
b b
mx
{ d i d i }
y
a
n a n mx
-y
-y -y
b a b G
RST d i d i UVW d −
Θ , ∈ i
= mx -y
n
-y
n
a a b b
mx
a
Claim: g g
Consider
Caution:
m n n
1 2 2
= g g m v 1 g g ∈G
v
1 2
We can not write mx times, if x ∈ N , x is -ve integer. Here mx is a+ve integer as
d1 d1 2 1i i − 1 1 2
2 1 id 2= 1gi g dg d2 ig1 ig2
1 d2 1 i d2 1 i −
1 2 i d3d i2 1di
2 1 i
g
=
mx + ny = 1 mx − ny = 1, x, y ∈ N, the set of natural numbers.
m ∴
m, m m n n m mx m n m = n m g − − − g n m n
g mx 1
n
g 1 g m− −n− −m g g mx ng
m
1
x
g m n m mx −
= ny n m
mx−ny
g m g g= g g
mx
1
gn n ∈ g n g
m
2
= N.
2 1=
where mx ∈ N times n
d i
m
Also g g
as ny ∈ N.
d i
m x
= g g m −m n m
1 3 g1 , where g 3 = g2 g1
∈G
d i
2
d i
x
= g m m −m m m m
g g Θ a m b = b a v a, b ∈G
3 1 1
dg g i
{ }
n
−ny
m n ny
1 2
=
1 2
d i
= g n g
∴ g g =
−1
m ny 1
{ 2 1 } − from above
m n
−ny
n m
−
d g g
ny
1 2i d2 1 i (2)
Hence from (1) and (2) we get
g
(1)
v g1, g2
∈ G,
v
(3)