Page 16 - ALGEBRA STRUCTURE cyclic group BY MIFTAHUL JANNAH (4193311004) MESP2019
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15
10
25
5
5
10
〈a 〉 = {e, a , a , a , a , a } order 6
〈a 〉 = {e, a , a , a , a } order 5
6
12
18
6
24
10
10
20
〈a 〉 = {e, a , a } order 3
15
〈a 〉 = {e, a } order 2
15
〈a 〉 = {e} order 1
30
In general, if 〈a〉 has order n and k divisor of n then 〈 n/k 〉 is a single sub group of order K
Due to Theorem B-4 : (Sub group of )
For each positive k of n, the set <n/k> is a single sub group of Z or order k.
n
Example 11
Define the sub group of Z
30
Based on the due to theorem B-4, the subgroup of Z are
30
< 1 >= {0,1,2, … ,29} order 30
< 2 > = {0,2,4, … ,28} order 15
< 3 >= {0,3,6, … ,27} order 10
< 5 >= {0,5,10,15,20,25} order 6
< 6 >= {0,6,12,18,24} order 5
< 10 >= {0,10,20} order 3
< 15 >= {0,15} order 2
< 30 >= {0} order 1
15
