Theorem 1 (Left & Right Cancellation Laws)
If G is a group with binary operation *, then the left and right cancellation laws hold in G, that is
a*b = a*c implies b = c, and
b*a = c*a implies b = c for all a,b,c in G
Similarly, we can prove the right cancellation law.
Theorem 2 (Unique Solution)
If G is a group with binary operation *, and if a & b are any element of G, then the linear equations
b*a = c*a implies b = c for all a,b,c in G
Proof:
(To prove the left cancellation law)
To show if a*b = a*c then b = c
Let a*b = a*c
Since G is a group, G3 is satisfied. Then, for all a' in G such that,
(To prove the left cancellation law)
To show if a*b = a*c then b = c
Let a*b = a*c
Since G is a group, G3 is satisfied. Then, for all a' in G such that,
a'*(a*b) = a'*(a*c) --G3
(a'*a)*b = (a'*a)*c --G1,* associative
e*b = e*c --by def. of a' in G3
b = c
(a'*a)*b = (a'*a)*c --G1,* associative
e*b = e*c --by def. of a' in G3
b = c
Similarly, we can prove the right cancellation law.
Theorem 2 (Unique Solution)
If G is a group with binary operation *, and if a & b are any element of G, then the linear equations
a*x = b and y*a =b
have unique solution x and y in G.
Proof :
Let a,b are any element in G
(To show the existence of at least one solution of a*x = b)
Suppose (a'*b) is one of the solution.
Then,
a*(a'*b)
= (a*a')*b --by G1
= e*b --by def. of G3
= b -- by def. of G2
Thus, (a'*b) is a solution of a*x = b
(Next, to show y = b*a' is a solution of y*a = b)
a*(b*a')
= a*(a'*b) --* is commutative in G
= (a*a')*b -- by G1
= e* b -- by def. of G3= b -- by def. of G2
Thus, (b*a') is a solution of y*a = b
(To show the solution are unique)
Let x' & x" are solution of a'*x = b
Then,
Hence,
Proof :
Let a,b are any element in G
(To show the existence of at least one solution of a*x = b)
Suppose (a'*b) is one of the solution.
Then,
a*(a'*b)
= (a*a')*b --by G1
= e*b --by def. of G3
= b -- by def. of G2
Thus, (a'*b) is a solution of a*x = b
(Next, to show y = b*a' is a solution of y*a = b)
a*(b*a')
= a*(a'*b) --* is commutative in G
= (a*a')*b -- by G1
= e* b -- by def. of G3= b -- by def. of G2
Thus, (b*a') is a solution of y*a = b
(To show the solution are unique)
Let x' & x" are solution of a'*x = b
Then,
a*x' = b
a*x" = b
a*x" = b
Hence,
a*x' = a*x"
x' = x"
(by left cancellation law)
Finally, the uniqueness of y follows similarly.
x' = x"
(by left cancellation law)
Finally, the uniqueness of y follows similarly.
sources : http://www.maranausd.org/images/pages/N3555/math1.gif
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