Page 11 - ALGEBRA STRUCTURE cyclic group BY MIFTAHUL JANNAH (4193311004) MESP2019

m
Let m = smallest positive integers so that a = e

Take any b ∈ G =< a > then = , for a n ∈ Z, according to the division algorithm then
∃ ! q, r ∈ Z ∋ n = mq + r, with 0 ≤ r < m, so that it is obtained

r
q r
n
m q r
b = a = a qm+r = (a ) a = e a = a
So ∀ b ∈ G ∃ r ∈ Z ∋ b = a with 0 ≤ r < m
r
2
1
0
So that the G elements can be written : {a , a , a , … , a m−1 }
The contradiction arises that G has infinite elements
k
h
So the right reliance is wrong a = a → k = h

Example 6

= {0,1,2,3,4,5} with addition operation of modulo 6.
6
Is a cyclic group? If yes, specify the generator. By using Cayley table it can be shown that is
6
6
a cyclic group with generator 1 and 5

Proof

Cayley table in set
6
+ 0 1 2 3 4 5
6

0 0 1 2 3 4 5

1 1 2 3 4 4 0

2 2 3 4 5 0 1

3 3 4 5 0 1 2

4 4 5 0 1 2 3

5 5 0 1 2 3 4

By looking the table above, so it is obtained :

1. The first axiom (closed property) is satisfied because all operation result are in set .
6

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