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Math Help - Finding Subgroups in Z

  1. #1
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    Finding Subgroups in Z

    I'm a little confused by exactly what this question is asking and am thinking I might be over simplifing it.

    The question reads "Find all subgrounps of {Z}_{12}".

    Does this simply mean list
    {null, {0}, {1}, ... , {0, 1}, {0, 2}, ..., {0,1,2,3,4,5,6,7,8,9,10, 11}} as all of the subgrounds?

    There is no operation defined on the group, so I'm not sure what else to do?

    Thanks!
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  2. #2
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    Quote Originally Posted by apsis View Post
    I'm a little confused by exactly what this question is asking and am thinking I might be over simplifing it.

    The question reads "Find all subgrounps of {Z}_{12}".

    Does this simply mean list
    {null, {0}, {1}, ... , {0, 1}, {0, 2}, ..., {0,1,2,3,4,5,6,7,8,9,10, 11}} as all of the subgrounds?

    There is no operation defined on the group, so I'm not sure what else to do?

    Thanks!
    This is the set \{ [0]_{12}, [1]_{12}, ... , [11]_{12} \} with binary operation "addition modulo 12".
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  3. #3
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    How do I count the subgroups then? If I say that [0], [1], ... , [11] are the subgroups its not correct, because the closure property is not satisfied, i.e., [1] + [1] is not a member of [1] in {Z}_{12}, and so [1] is not a subgroup. So I would have to expand that group to include [2], but then I run into the same issue of [1] + [2]. Eventually I would include all elements and rebuild the group {Z}_{12}.

    The same argument applies for all elements of the group (except [0]). So does that mean the only proper subgroup is [0]? Or am I thinking of this wrong?


    Thanks
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  4. #4
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    The subgroup of a cyclic group is cyclic. Therefore, any subgroup must be generated by an element.
    Take for example [2]_{12} then the subgroup generated by it is \{ [0]_{12},[2]_{12},[4]_{12},[6]_{12}, [8]_{12},[10]_{12} \}.
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  5. #5
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    So if I'm thinking of this right then the subgroups are

    [0] -> {[0]}
    [1] -> {Z}_{12}
    [2] -> {[0], [2], [4], [6], [8], [10]}
    [3] -> {[0], [3], [6], [9]}
    [4] -> {[0], [4], [8]}
    [6] -> {[0], [6]}

    Any other generator just results in the set {Z}_{12}, which would be repetitive.
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  6. #6
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    Quote Originally Posted by apsis View Post
    So if I'm thinking of this right then the subgroups are

    [0] -> {[0]}
    [1] -> {Z}_{12}
    [2] -> {[0], [2], [4], [6], [8], [10]}
    [3] -> {[0], [3], [6], [9]}
    [4] -> {[0], [4], [8]}
    [6] -> {[0], [6]}

    Any other generator just results in the set {Z}_{12}, which would be repetitive.
    Yes ! In fact for each k|12, 1\leq k\leq 12, we can generate a distinct subgroup. Furthermore, \left < [k_1]_{12} \right> \subseteq \left< [k_2]_{12}\right> if and only if k_1|k_2.

    You want to try doing a similar problem for \mathbb{Z}_{60} by finding all the subgroups using the above result.
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  7. #7
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    wicked thanks! those results make it much easier to determine the subgroups so for {Z}_{60} the generators of distinct subgroups are the integers that divide 60.

    But for your second result showing the subgroups of subgroups, isn't that backwards? In my list I have <6> \subseteq <3>. Not the other way around as your result would predict?
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  8. #8
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    Quote Originally Posted by apsis View Post
    But for your second result showing the subgroups of subgroups, isn't that backwards? In my list I have <6> \subseteq <3>. Not the other way around as your result would predict?
    Yes! Sorry about that. It should say k_2|k_1 and not k_1|k_2.
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