Page 39 - DMTH502_LINEAR_ALGEBRA

Unit 1: Vector Space over Fields

Solution: Let us prepare a composition table as given below: Notes

+ (mod 5) 0 1 2 3 4
0 0 1 2 3 4
1 1 2 3 4 0
2 2 3 4 0 1
3 3 4 0 1 2
4 4 0 1 2 3

Closure Property: All the entries in the composition table are elements of the set G. Hence G is
closed under addition modulo 5.

Associative Property: Addition modulo 5 is associative always.
Identity: 0 G is the identity element.
Inverse: It is clear from composition table.
Element — 0 1 2 3 4

Inverse — 0 4 3 2 1
Inverse exists for every element of G.
Commutative Law: The composition is commutative as the corresponding rows and columns is
G are 5.
Hence {G, + (mod 5)} is a finite abelian group of order 5.

Example 11: Prove that the set G = {1, 2, 3, 4, 5, 6} is a finite abelian group of order 5 with
respect to multiplication modulo 7.
Solution: Let us prepare the following composition table:

X 7 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 1 3 5
3 3 6 2 5 1 4
4 4 1 5 2 6 3
5 5 3 1 6 4 2
6 6 5 4 3 2 1

Closure Property: All the entries in the table are elements of G. Therefore G is closed with
respect to multiplication modulo 7.
Associative Property: Multiplication modulo 7 is associative always.

Identity: Since first row of the table is identical to the row of elements of G in the horizontal
border, the element to the left of first row in vertical border is identity element, i.e., 1 is identity
element in G with respect to multiplication modulo 7.
Inverse: From the table it is obvious that inverses of 1, 2, 3, 4, 5, 6 are 1, 4, 5, 2, 3 and 6 respectively.
Hence inverse of each element in G exists.

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