Page 15 - ALGEBRA STRUCTURE cyclic group BY MIFTAHUL JANNAH (4193311004) MESP2019

1
Let H ≠ {e}, so H must contain elements in the form a with t > 0
m
Let m is the smallest positive integer that satisfies a ∈ H

m
As a result H = < a >
Take any b ∈ H, then b = a for every k ∈ Z
k
With division algorithm ∃! q, r ∈ Z ∋ k = mq + r with 0 ≤ r < m

k
r
k
so that b = a = a mq+r with 0 ≤ r < m or b = a = a mq+r = a mq . a ∈ H,
k
r
m −
k
a = a −mq . a , a = b ∈ H and a −mq = (a ) ∈ H
m
But m is a smallest positive integer, a ∈ H and 0 ≤ r < m so r = 0
k
m q
m
k
Then a −mq . a = e. b = a = (a ) ∈ < a >
Because |< a >| = n
m n
So it is obtained (a ) = (a ) = e = e
n
n m
Let k be the divisor of n
) ≠ e, t < k 〈a
) = a = a = e and (a
(a n/k k n m n/k t n/k 〉 have order k, we can write n = mq +
r, 0 ≤ r < m

n
r
m −q
It is obtained e = a = a mq+r that satisfies a = a −mq = (a ) ∈ H
n
m
Because r = 0 and n = mq so that k = |H| = |〈a 〉| =
m
n
m
As a result, m = and H = 〈a 〉 = 〈a n/k 〉
k
Example 10

Let |〈a〉| = 30

The positive divisor of 30 is 1,2,3,5,6,10,15,30 so that according to theorem B-4 〈a〉 has sub group,

namely :

29
2
〈a〉 = {e, a, a , … , a } order 30
2
〈a 〉 = {e, a , a , …, a } order 15
2
28
4
〈a 〉 = {e, a , a , …, a } order 10
27
3
3
6
14