Page 25 - DMTH502_LINEAR_ALGEBRA

Unit 1: Vector Space over Fields

Solution: Notes
(i) A B = {a} = {a} = B
Similarly, A C = C, A D and A A = A.
Also, B B = B, B C = {a} {a, b} = {a, b} = C,

B D = {a} {a, b, c} = {a, b, c} = D
C C = C, C D = {a, b} {a, b, c} = {a, b, c} = D
Hence is a binary operation on S.
(ii) Again, A A = A, A B = {a} = = A

A C = A, A D = A
and B B = B, B C = {a} {a, b} = {a} = B
B D = {a} {a, b, c} = {a} = B
C C = C, C D = {a, b} {a, b, c}

= {a, b} = C.
Hence is a binary operation on S.

Self Assessment

1. Show that multiplication is a binary operation on the set A = {1, –1} but not on B = {1, 3}.
2. If A = {1, –1} and B = {1, 2}, then show that multiplication is a binary operation on A but not
on B.
3. If S = {A, B, C, D} where A = , B = {a, b}, C = {a, c}, D = {a, b, c} show that is a binary
operation on S but is not.

Group
Definition: An algebraic structure (G, o) where G is a non-empty set with a binary operation ‘o’ defined
on it is said to be a group, if the binary operation satisfies the following axioms (called group axioms).
(G ) Closure Axiom: G is closed under the operation o, i.e., a o b G, for all a, b G.
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(G ) Associative Axiom: The binary operation o is associative, i.e.,
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(a o b) o c = a o (b o c) a, b, G.
(G ) Identity Axiom: There exists an element e G such that
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e o a = a o e = a a G.

The element e is called the identity of ‘o’ in G.
(G ) Inverse Axiom: Each element of G possesses inverse, i.e., for each element a G, there
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exists an element b G such that
b o a = a o b = e.
–1
–1
The element b is then called the inverse of a with respect to ‘o’ and we write b = a . Thus a is an
element of G such that
–1
a o a = a o a = e.
–1

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