Page 9 - ALGEBRA STRUCTURE cyclic group BY MIFTAHUL JANNAH (4193311004) MESP2019

So that 3 and 7 are generator for U(10).

Definition B-4

Division algorithm

If n, m ∈ Z, m > 0 then ∃ ! q, r ∈ Z ∋ n = qm + r, 0 ≤ r < m

Example 1 :

n = 38, m = 7 then ∃ q = 5 and r = 3 so that 38 = 5 . 7 + 3

Example 2

n = −36, m = 8 then ∃ q = −5 and r = 4 so that − 36 = −5 . 8 + 4

Theorem B-2 : Classification Subgroup of Cyclic Group

Every subgroup of cyclic group are cyclic.

Proof :

Subgroup of cyclic group are cyclic.

Let G = < a > is a cyclic group, and H ≤ G (read H subgroup of G).

It will be shown that H is a cyclic group.

p
G = < a >, because H ≤ G then the elements in H must be in formed a with p ∈ Z.
p
If a ∈ H then a −p ∈ H (remember H is subgroup of G)
Case I : If H = {e} then H = < e > cyclic group

Case II : If H ≠ {e} then H must be contained elements that form a with p > 0
p

m
Let m = smallest positive integers ∋ a ∈ H…… (A)
Take any b ∈ H then b = a for a n ∈ Z
n
n
With division algorithm then ∃ ! , ∈ ∋ = + with 0 ≤ < so that b = a = a qm+r
with 0 ≤ r < m

or

n
a ∈ H
b = a = a qm r
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