Page 36 - DMTH403_ABSTRACT_ALGEBRA

Unit 2: Groups

Theorem 5: For a, b, c in a group G, Notes
(a) ab = ac  b = c. (This is known as the left cancellation law.)

(b) ba = ca  b = c. (This is known as the right cancellation law.)
Proof: We will prove (a) and leave you to prove (b).
(a) Let ab = ac. Multiplying both sides on the left hand side by a " G, we get
–I
a (ab) = a (ac)
–1
-1
 (a a) b = (a a) c
–1
-1
 eb = ec, e being the identity element.
 b = c.
Now let us prove another property of groups.

Theorem 6: For elements a. b in a group G, the equations ax = b and ya = b have unique solutions
in G.
Proof: We will first show that these linear equations do have solulions in G, and then we will
show that the solutions are unique.
For a, b  G, consider a b  G. We find that a(a b) = (aa ) b = eb = b. Thus, a b satisfies the
-1
-1
-1
-1
equation ax = b, i.e., ax = b has a solution in G.
But is this the only solution? Suppose x , x are two solutions of ax = b in G. Then ax, = b = ax . By
2
1
2
the left cancellation law, we get x = x . Thus, a b is the unique solution in G.
-1
l
2
Similarly, using the right cancellation law, we can show that ba is the unique solution of
-1
ya = b in G.
Now we will illustrate the property given in Theorem 6.
Example: Consider A    2 3  ,B    1 5  in GL, (R)
 1 2   0 4 
Find the solution of AX = B.
Solution: From Theorem 6, we know that X = A B. Now,
-1
2   3
A =  
–1
  1 2 
2   2
–1
 A B =    X.
  1 3 
In the next example we consider an important group.

Example: Let S be a non-empty set. Consider  (S) with the binary operation of
symmetric difference A, given by
A  B=(A\B)  (B\A)  A, B  (S).

Show that ((S), A) is an abelian group. What is the unique solution for the equation Y A=B?
Solution: A is an associative binary operation. This can be seen by using the facts that
A\B=A  B , (A  B) = A  B , (A  B) = A  B C
C
C
C
C
C
C
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