Page 10 - ALGEBRA STRUCTURE cyclic group BY MIFTAHUL JANNAH (4193311004) MESP2019

r
a = a n−qm
n
n −q
= a (a ) ∈ H
r
So, a ∈ H with 0 ≤ r < m

Suppose r ≠ 0, then 0 < r < 0 means there is a positive integer r < m so that a ∈ H or m is not
r
m
the smallest positive number so a ∈ H….
A contradiction arises between (A) and (B)

So the reliability is wrong, the correct one is r = 0

m q
n
m
If r = 0 this means n = qm so b = a = a qm = (a ) ∈ H, ∀b ∈ H or H =< a >
Proven that H is cyclic subgroup.

Example 5

In example 4 before U(10) = {1,3,7,9} with multiplication operation modulo 10 is a cyclic group

with generator 3, we can choose {1, 9} is a subgroup of U(10) and {1, 9} is a cyclic subgroup with

generator 9.

Classification of cyclic group

1. If G is a cyclic group with an infinite number of elements, then G has the following

properties: a = a → n divide (k-h)
k
h
2. If G is a cyclic group with a finite elements (n element), then G has the following properties
k
h
: a = a → n divide (k-h)
Proof I

The statement above can be interpreted as :

k
h
G =< a > and |G| = infinite → a = a → k = h
Proof

In logic we have equivalence : → ≅ ∩
̅
̅
̅
̅
̅
̅
̅
̅
̅

h
ℎ
Suppose : = → = ℎ, this means (a = a ) ∩ (k ≠ h)
k
Let k > h then a k−h = e, with k − h > 0
9