In order to find the subgroups we need to understand that given,
"Zn," we need to show that "\u3008[a]n\u3009 \u2286 \u3008[d]n\u3009and\u3008[d]n\u3009 \u2286 \u3008[a]n\u3009."
To show that "\u3008[a]n\u3009 \u2286 \u3008[d]n\u3009" note that "a=dq" for some "q\u2208Z" since "d" is a divisor of "a". Then "[a]n=q[d]n" showing "[a]n\u2208 \u3008[d]n\u3009"
For "\u3008[d]n\u3009 \u2286 \u3008[a]n\u3009" ,we can say "d=ma+nq" for some "m, q\u2208Z" since the greatest common divisor of "a" and "n" can be written as a linear combination of "a" and "n" .Then "d\u2261ma(modn), and\\ so [d]n=m[a]n\u2208 \u3008[a]n\u3009." We can thus solve this problem as below.
We need to find"<[a]>" for all congruence classes"[a]\\in\\Z_{12}" ."\u3008[1]\u3009=Z12\\\\\n\u3008[2]\u3009={[0],[2],[4],[6],[8],[10]}\\\\\n\u3008[3]\u3009={[0],[3],[6],[9]}\\\\\n\u3008[4]\u3009={[0],[4],[8]}\\\\\n\u3008[5]\u3009={[0],[5],[10],[3],[8],[1],[6],[11],[4],[9],[2],[7]}\\\\\n\u3008[6]\u3009={[0],[6]}"
Therefore, you can see that the subgroups of "Z_{12}" correspond to the divisors of 12.
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