Question

In Z12,find the subgroups <2>,<3>,<4>,<5> and <6>

Answer 1

$$Solution$$

In order to find the subgroups we need to understand that given,

$Zn,$ we need to show that $〈[a]n〉 ⊆ 〈[d]n〉and〈[d]n〉 ⊆ 〈[a]n〉.$

To show that $〈[a]n〉 ⊆ 〈[d]n〉$ note that $a=dq$ for some $q∈Z$ since $d$ is a divisor of $a$. Then $[a]n=q[d]n$ showing $[a]n∈ 〈[d]n〉$

For $〈[d]n〉 ⊆ 〈[a]n〉$ ,we can say $d=ma+nq$ for some $m, q∈Z$ since the greatest common divisor of $a$ and $n$ can be written as a linear combination of $a$ and $n$ .Then $d≡ma(modn), and\ so [d]n=m[a]n∈ 〈[a]n〉.$ We can thus solve this problem as below.

We need to find$<[a]>$ for all congruence classes$[a]\in\Z_{12}$ .$〈[1]〉=Z12\\ 〈[2]〉={[0],[2],[4],[6],[8],[10]}\\ 〈[3]〉={[0],[3],[6],[9]}\\ 〈[4]〉={[0],[4],[8]}\\ 〈[5]〉={[0],[5],[10],[3],[8],[1],[6],[11],[4],[9],[2],[7]}\\ 〈[6]〉={[0],[6]}$

Therefore, you can see that the subgroups of $Z_{12}$ correspond to the divisors of 12.