Page 14 - ALGEBRA STRUCTURE cyclic group BY MIFTAHUL JANNAH (4193311004) MESP2019

k
It shows that G = < a >

Proof II (It will be proven by contraposition)

Suppose gcd (k,n) ≠ 1 or gcd (k,n) = d > 1 it means k = td and n = sd

k
sd t
n t
td s
k s
Pay attention to (a ) = (a ) = (a ) = (a ) = with |a | ≤ s ≤ n
k
k
It shows that a is not generator of G and G ≠ < a >
Example 8
From the previous example U(10) = {1,3,7,9} with multiplication operation of modulo 10 is a
2
1
0
3
cyclic group with generator 3 or U(10) = {3 , 3 , 3 , 3 }
|< 3 >| = 4 then according to theorem 3 above, 3 is a generator because gcd (3,4) = 1, likewise
3
1
2
3 is a generator because gcd (1,4) = 1 but 3 is not generator because gcd (2,4) ≠ 1
Due to Theorem B-3

An integer k ∈ Z is a generator of if and only if gcd (k,n) = 1

Example 9

= {0,1,2,3,4,5,6,7} with addition operation of modulo 8, it can be shown that is a cyclic
8
8
group. According to the due to theorem B-3, then has generator 1,3,5, and 7.
8

Theorem B-4

If |< a >| = n then the order of any subgroup of < a > is a factor of n and for every positive

divisor k of n, the group < a > has exactly one sub group order k called < a n/k >

Proof

Let G = < a > and H is a sub group of G, it will be shown that H is a cyclic group.

If H = {e}; it is certain that H is cyclic, because it is built by e itself.

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