Page 37 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes and that  and  are commutative and associative. A is also commutative since A A B

= B  A  A, B  (S).

Also,  is the identity element since A A  = A  A  (S).
Further, any element is its own inverse, since A A A =   A  (S).

Thus, ((S). A) is an abelian group.
For A, B in ((S), A) we want to solve Y A A = B. But we know that A is its own inverse. So, by
Theorem 6, Y = B A A = B A A is the unique solution. What we have also proved is that (B A A)
-1
A A = B for any A, B in  (S).
Definition: Let G be a group. For a  G, we define
(i) a = e.
0
(ii) a = a . a, if n > 0
n-1
n
(iii) a = (a )n, if n > 0.
-1
–n
n is called the exponent (or index) of the integral power a of a.
n
Thus, by definition a = a, a = a . a, a = a . a, and so on.
2
1
2
3
n
Notes When the notation used for the binary operation is addition, a becomes
na. For example, f a any a  Z,
na = 0 if a = 0,
na = a + a+ ... +a (n times) if n > 0,
na = (–a) + (–a) + .... + (–a) (–n times) if n < 0.

Let us now prove some laws of indices for group elements.
Theorem 7: Let G be a group. For a " G and m, n " Z,

(a) (a ) = a = (a ) , (b) a , a = a m+n , (c) (a ) = a mn
n
n -1
-n
-1 n
m n
m
Proof: We prove (a) and (b), and leave the proof of (c) to you.
(a) If n = 0, clearly (a ) = a = (a ) .
n -l
-0
-1 n
Now suppose n > 0. Since aa = e, We see that
-1
e = e = (aa )
–1 n
n
= (aa ) (aa ) .... (aa ) (n times)
-1
-1
-1
= a (a ) , since a and a compute.
-1 a
n
-1
-1 n
n -1
 (a ) = (a ) .
Also, (a ) = a , by definition.
-n
-1 n
 (a ) = (a ) = a when n > 0.
n -1
-1 a
-a
If n < 0, then (–n) > 0 and
(a ) = [a ]
-(n) -1
n -1
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