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Math Help - Subgroups

  1. #1
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    Subgroups

    How many distinct subgroups does Z/36Z have? How would I begin solving this?
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  2. #2
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    Re: Subgroups

    things to remember with cyclic groups:

    1) every subgroup is cyclic.

    2) if G is cyclic of order n, and k divides n, G has a UNIQUE subgroup of order k (so it has one, and only one, for each divisor k of n).

    in particular, if G = <a>, with |a| = n, and k|n, so that n = kd, then the subgroup of order k is <ad>.

    for ABELIAN groups, ad is usually written da. since Z/36Z is generated by 1+36Z = [1], we need only consider <d[1]> = <[d]> for each divisor d of 36.

    the divisors of 36 are:

    1,2,3,4,6,9,12,18, and 36.

    your turn.
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  3. #3
    Super Member ILikeSerena's Avatar
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    Re: Subgroups

    Hi lovesmath!

    The size of a subgroup divides the size of the group.
    How many dividers of 36 can you find?
    Pick one. Is it a subgroup?
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  4. #4
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    Re: Subgroups

    Okay, so the divisors are 1, 2, 3, 4, 6 9, 12, 18, 36. When the order is 36, the generator is <1>. When the order is 18, the generator is <2>.
    So, we have the following:
    order 36, <1> = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 0}
    order 18, <2> = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 0}
    order 12, <3> = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 0}
    order 9, <4> = {4, 8, 12, 16, 20, 24, 28, 32, 0}
    order 6, <6> = {6, 12, 18, 24, 30, 0}
    order 4, <9> = {9, 18, 27, 0}
    order 3, <12> = {12, 24, 0}
    order 2, <18> = {18, 0}
    order 1, <0> = {0}

    Am I on the right track?
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  5. #5
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    Re: Subgroups

    those are them!
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  6. #6
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    Re: Subgroups

    But is each one of those subgroups distinct?
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  7. #7
    Senior Member jakncoke's Avatar
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    Re: Subgroups

    yes, there is no other subgroups of those sizes which are different.
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