Page 13 - ALGEBRA STRUCTURE cyclic group BY MIFTAHUL JANNAH (4193311004) MESP2019

0
3
5
4
= {0,1,2,3,4,5} = {5 , 5 , 5 , 5 , 5 , 5 } =< 5 >
2
1
6
Based on the classification of the cyclic group part 2 it can be seen that | | = |< 5 >| = 6
6
6
18
Then 5 = 5 12 = 5 it is because 6| (12 − 6) also 6| (18 − 12)
Due to theorem B-2
k
Let G group, a ∈ G with |a| = n. If a = e, then n divides k

Proof :

Given a = e = 0 , we know from the classification of the subdivided cyclic groups that n is the
k
divisor of k-0.

Example 7

= {0,1,2,3,4,5} in example 5 is a cyclic group with generator 5 or = {0,1,2,3,4,5} =
6
6
4
5
2
3
{5 , 5 , 5 , 5 , 5 , 5 } = < 5 > and |〈5〉| = 6
0
1
By using the result of theorem B-2 we get 5 12 = e = 0 then 6 | 12

Theorem B-3

Let G = < a > is a cyclic group, then G =< a > if and only if g c d (k,n) = 1
k

Proof :

It will be proven

k
(1) G = < a > is n ordered cyclic group, if g c d (k,n) = 1 then G =< a >
(2) G = < a > is n ordered cyclic group, if G =< a > then g c d (k,n) = 1
k

Proof I

Gcd (k,n) = 1 by using the concept of linear combination so it can be written that there is u, v ∈

Z ∋ 1 = ku + nv

k u
1
ku nv
Take any a ∈ G then a = a = a ku+nv = a a = a ku = (a )
12