Page 32 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes = a (because e is identity)
–1
–1
–1
Also, (a o a) o a = e o a (because a o a = e)
= a (because e is identity)
–1
–1
But a o (ao a ) = (a oa) o a as in a group composition is associative
a –1 = a .
–1 –1
–1
–1
Theorem 3: If the inverse of a is a then the inverse of a is a, i.e., (a ) = a.
Proof: If e is the identity element, we have
–1
a o a = e (by definition of inverse)
–1 –1
(a ) o (a o a) = (a ) o e
–1 1
–1
[because a G (a ) G]
–1 –1
–1
–1 –1
[(a ) o a ] o a = (a )
–1
–1 –1
[because Composition in G is associative and e is identity element]
–1 –1
e o a = (a )
–1 –1
a = (a )
–1 –1
(a ) = a.
Theorem 4: The inverse of the product of two elements of a group G is the product of the inverse
taken in the reverse order i.e.,
–1
–1
(a o b) –1 = b o a a, b G.
Proof: Let us suppose a and b are any two elements of G. If a and b are inverses of a and b
–1
–1
respectively, then
–1
a o a = e = a o b (e being the identity element)
–1
–1
and b o b = e = b o b –1
–1
–1
Now, (a o b) o b o a –1 = [(a o b) ob ] o a (by associativity)
–1
–1
= [a o (b o b )] o a (by associativity)
–1
= (a o e) o a [because b o b = e]
–1
–1
–1
= a o a [because a o e = a]
–1
= e [because a o a = e]
–1
Also (b o a ) o (aob) = b o [a o (a o b)] (by associativity)
–1
–1
–1
= b o [(a o a) ob]
–1
–1
–1
–1
= b o (e o b) [because a o a = e]
–1
= b o b [because e o b = b]
= e.
Hence, we have
–1
–1
(b o a o (a o b) = e = (a o b) o (b o a )
–1
–1
Therefore, by definition of inverse, we have
–1
(a o b) –1 = b o a –1
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