Page 33 - DMTH502_LINEAR_ALGEBRA

Unit 1: Vector Space over Fields

This theorem can be generalised as: Notes
if a, b, c, … k, l, m G, then
–1
–1
–1
–1
(a o b o c o … k o l o m) –1 = m o l o k o. .c o b o a .
–1
–1
Theorem 5: Cancellation laws hold good in a group, i.e., if a, b, c, are any elements of G, then
a o b = a o c b = c (left cancellation law)
and b o a = c o a b = c (right cancellation law)
Proof: Let a G. Then
–1
a G a G such that a o a = e
–1
= a o a , where e is the identity element
–1
Now, let us assume that
a o b = a o c
then a o b = a o c a o (a o b) = a o aoc
–1
–1
–1
(a oa) ob = (a oa) oc (by associative law)
–1
–1
eob = eoc (because a oa = e)
b = c.
Similarly, b o a = c o a

(boa) o a = (coa) o a –1
–1
–1
bo (ao a ) = co (ao a )
–1
boe = coe
b = c.

Theorem 6: If G is a group with binary operation o and if a and b are any elements of G, then the
linear equations
aox = b and yoa = b

have unique solutions in G.
–1
Proof: Now a G a G,
–1
–1
and a G, b G a ob G.
Substituting a ob for x in the equation aox = b, we obtain
–1
a o (a o b) = b
–1
(a o a ) o b = b
–1
e o b = b
b = b [because e is identity]

–1
Thus x = a ob is a solution of the equation aox = b.
To show that the solution is unique let us suppose that the equation aox = b has two solutions
given by

x = x and x = x
1 2

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