Properties of groups Algebraic structures
(Monday, May 3, 2010)
Theorem 1:
In a group áG, *ñ, a left identity element is also a right identity element, and a left inverse of an element is also its right inverse.
Proof:
Let e be a left identity in group áG, *ñ. Then, for all a e G, e*a =a e G.
Also let a' e G be the left inverse of an element of a e G. (1)
Also let a'' e G be the left inverse of a'
\a' * a = e (2)
and a'' * a' = e (3) (by G4)
Then (a' *a) * a' = e * a' = a' by (1)
or a' * (a * a' ) = a' (by G2)
Again a'' * (a' * (a * a' ))=a'' * a'=e by (3)
or (a'' * a )* (a * a') = e (by G2)
or e * (a * a') = e by (3)
or a' * (a * a' ) = a' (by G2)
\a * a' = e (4)
Hence, a' is also the right inverse of the element of a e G
Now, a * e = a * (a' * a)
= (a' * a) * a
= e * a
= a
Hence, e is also the right identity element in G. Similarly it can be proved that in a group áG, *ñ, a right identity is also a left identify, and the right inverse of an element in G is also its left inverse.
Thus, an identity element and an inverse element in a group are both-sided.
Theorem 2:
The identity element and the inverse element in a group áG, *ñ are both unique
Proof:
(i) If possible, let e and e' be two identity elements in áG, *ñ. The identity elements are both sided.
Then e * e' = e'(e being the left identity)
Also e*e'=e (e' being the right identity)
Hence, it follows that e'=e, which shows the uniqueness of identity element in a group.
(ii) If possible, let a' and a'' be two inverses of an element a in áG, *ñ. Note that a' and a'' are both sided.
Then, we have
a'=a'*e=a'*(a * a'')=(a'*a)*e''=e*a'‘=a'‘
Hence, the result.
Theorem 3
For any three elements a, b, c e áG, *ñ,
(i) If a*b=a*c, the b=c (left cancellation
law)
(ii) If a*c=b*c, the a=b (right
cancellation law)
Proof:
Let e be the identity element and a-1,
b-1, c-1, be the inverse elements of a, b, c respectively in áG,*ñ. Note that e, a-1, b-1, and c-1 are all unique.
Now, when a*b=a*c, we have,
a-1 * (a * b) = a-1 * (a * c)
or (a-1 * a) * b = (a-1 * a) * c
or e*b=e*c
\ b = c
Again, if a * c = b * c, then we have
(a * c) * c-1 = (b * c) * c-1
or a * (c * c-1) = b * (c * c-1)
or a * e = b * e
\a = b
Hence, the theorem.
Theorem 4:
For any two elements a, b e áG, *ñ
(i) (a-1)-1 =a
(ii) (a * b)-1 = b-1 * a-1
Proof:
(i) Let e be an element in áG, *ñ. Then
a * a-1 = a-1 * e and
(a-1)-1 *a-1=a-1 *(a-1)-1 = e
As a-1 * a = e and a-1 * (a-1)-1 = e for a, a-1 e G, by the left cancellation law it follows that (a-1)-1 = a
(ii) We have (a * b)-1 * (a * b) = e, where a * b and (a * b)-1 e G.
Now, (b-1 * a-1)*(a * b) = b-1 * {(a-1 * a)*b}
= b-1 * (e * b)
= b-1 * b = e
\ (a * b)-1 = b-1 * a-1, the inverse of an element is unique.
Theorem 4:
In a group áG, *ñ the equations a * x = b and y * a = bhave unique solutions
x=a-1 * b and y= b *a-1, where a, b e G.
Proof:
We have a *(a-1 *b) = (a * a-1) *b
= e * b
= b
and
We have (b *a-1) *a) = b * ( a-1 *a )
= b * e
= b
Where e is the identity element in G. Thus, as a * x = b and a * (a-1 *b) = b, it follows that x=a-1 * b.
Further, y * a = b and (b * a-1) * a = b, it follows that y=b * a-1.
By the virtue of uniqueness of inverse and identity elements, it follows that the solution are also unique.
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