Page 35 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Solution: Let us see if o is a binary operation on T.
Now f o f (x , y) = f (x + c, y + d) = (x + c + a, y + d + b)
c,d
a,b
a,b
= f a+c, b+d (x,y) for any (x, y)  R .
2
 f of = f a+c, b+d  T.
c,d
a,b
Thus, o is a binary operation on T.
Now, f o f = f  f  T.
0,0
a,b
a,b
a,b
Therefore, f is the identity element.
0,0
Also, f o f -a, -b = f  f  T.
0,0
a,b
a,b
Therefore, f is the inverse of f  T.
a,b
-a,’
Thus, (T, o) satisfies G1’, G2' and G3', and hence is a group.
Note that f o f = f o f  f f  T. Therefore, (T, o) is abelian.
,
a,b c,d c,d a,b a,b c,d
2.3 Properties of Groups
Before understanding the properties of group lets first give notational conventions.
Convention: Let us, we will denote a group (G, *) by G, if there is no danger of confusion. We
will also denote a * b by ab, for a, b  G, and say that we are multiplying a and b. The letter e will
continue to denote the group identity.
Now let us discuss a simple theorem.
Theorem 4: Let G be a group. Then

(a) (a ) = a for every a  G.
–1 –1
(b) (ab) = b a for all a, b  G.
–1
–1
–1
Proof: (a) By the definition of inverse,
(a ) (a ) = e = (a ) (a ) .
-1
-1
-1 -1
-1 -1
But, a a = a a = e also, Thus, by Theorem 1 (b), (a ) = a.
-1
-1 -1
-1
(b) For a, b  G, ab  G. Therefore, (ab)  G and is the unique element satisfying (ab) (ab) -1
-1
= (ab) (ab) = e.
-l
However, (ab) (b a ) = ((ab) b ) a -1
–1
-1
-1
= (a (b b ) a )
-1
-1
= (a e) a -1
= aa -1
=e
Similarly, (b a ) (ab) = e.
-1
-1
Thus, by uniqueness of the inverse we get (ab) = b a .
–1
–1
-1
Note that, for a group G, (ab) = a b  a, b  G only if G is abelian.
-1
-1
-1
You know that whenever ba = ca or ab = ac for a, b, c in R*, we can conclude that b = C. That is,
we can cancel a. This fact is true for any group.
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