Page 7 - ALGEBRA STRUCTURE cyclic group BY MIFTAHUL JANNAH (4193311004) MESP2019

(8) = {1,3,5,7} with the multiplication operation modulo 8 is not a cyclic group, because

< 1 > = {1}

< 3 > = {3,1}

< 5 > = {5,1}

< 7 > = {7,1}

It can be seen that there isn’t a ∈ = <a> so that is not cyclic group. From the definition of
8
8
the order of an element obtained the order of 3 or (3) = 2, (5) = 2 and (7) = 2

Example 4

(10) = {1,3,7,9}with the multiplication operation modulo 10 is a group, is (10) is a cyclic

group, if yes, specify the generators.

Solution :

By using Cayley table, it can be shown that U(10) is group.

*(10) 1 3 7 9

1 1 3 7 9

3 3 9 1 7

7 7 1 9 3

9 9 7 3 1

1) The first axiom (closed property) is satisfied because the all result of the operation are on

set U(10).
2) The second axiom (associative property) of multiplication modulo 10 is satisfied with

integers, therefore, U(10) is also satisfied.
3) The third axiom (identity element) is satisfied because U(10) is an identity element because

U(10) is fulfilled a*1 = 1*a = a.

4) The fourth axiom (inverse element) is satisfied, that is
1 the inverse is 1; 3 the inverse is 7; 7 the inverse is 3; 9 the inverse is 9

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