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Math Help - A couple more subgroup questions

  1. #1
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    A couple more subgroup questions

    1. Find all the subgroups of Q8. (Q8 = {I, -I, J, -J, K, -K, L, -L}. Different notation can be used as well). Show that Q8 is an example of a nonabelian group with the property that all its proper subgroups are cyclic.

    2. a) Let G be a cyclic group of order n. Show that if m is a positive integer, then G has an element of order m IFF m divides n.
    b) Let G be a cyclic group of order 40. List all the possibilities for the orders of elements of G.

    3. Let G = <x> be a cyclic group of order 144. How many elements are there in the subgroup <x^26>?

    4. Construct a nonabelian group of order 16, and one of order 24.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Janu42 View Post
    1. Find all the subgroups of Q8. (Q8 = {I, -I, J, -J, K, -K, L, -L}. Different notation can be used as well). Show that Q8 is an example of a nonabelian group with the property that all its proper subgroups are cyclic.
    using lagrange's thm, subgroups of Q8 is of order 1, 2, 4 and 8.. get all subgroups then..

    Quote Originally Posted by Janu42 View Post
    2. a) Let G be a cyclic group of order n. Show that if m is a positive integer, then G has an element of order m IFF m divides n.
    b) Let G be a cyclic group of order 40. List all the possibilities for the orders of elements of G.
    a) for ( \Rightarrow)  G is cyclic, then G=<a> and a^n=e. Let b\in G such that b^m=e. then b=a^k for some natural number k. thus b^m = a^{km} = e = a^n you can continue from here..

    on the other direction, if m|n, then n=mk for some integer k. thus e=a^n=a^{mk}=(a^k)^m...

    Quote Originally Posted by Janu42 View Post
    3. Let G = <x> be a cyclic group of order 144. How many elements are there in the subgroup <x^26>?
    \frac{\mbox{lcm }\{144, 26\}}{26}

    Quote Originally Posted by Janu42 View Post
    4. Construct a nonabelian group of order 16, and one of order 24.
    i think, Q_8\times \mathbb{Z}_2 is non-abelian..
    S_4 is non-abelian
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    I'm still confused on #1 and #4....

    Can someone explain to me how #1 works exactly?
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  4. #4
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    Quote Originally Posted by Janu42 View Post
    4. Construct a nonabelian group of order 16, and one of order 24.
    In general if for any even number >4 there exists an non-abelian group. Just construct D_n - the dihedral group of order 2n.
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    The dihedral group? I don't know if we've learned that yet....

    What exactly is the structure of the dihedral group?
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    Quote Originally Posted by Janu42 View Post
    What exactly is the structure of the dihedral group?
    A geometric way to think of it is all the symmetries of a regular n-gon.
    There are n rotations which perserve the polygon.
    And there are 2 reflections through each vertex.
    So there are a total of 2n symmetries, and these symmetries form the dihedral group.
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  7. #7
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    i think, Q_8\times \mathbb{Z}_2 is non-abelian..
    S_4 is non-abelian
    Is Q_8\times \mathbb{Z}_2 a group of order 16? And the group S_4 is order 24?

    And for #1, the subgroups of Q8, if I'm not mistaken, are the subgroups {I}, {I, -I}, {I, -I, J, -J}, {I, -I, K, -K}, and {I, -I, L, -L}? However, I don't know if these subgroups are cyclic, because the question asks us to show that all the subgroups of Q8 are cyclic, so I don't know, I'm rather confused on this whole question.
    Last edited by mr fantastic; February 23rd 2009 at 12:41 AM. Reason: Merged posts
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  8. #8
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Janu42 View Post
    And for #1, the subgroups of Q8, if I'm not mistaken, are the subgroups {I}, {I, -I}, {I, -I, J, -J}, {I, -I, K, -K}, and {I, -I, L, -L}? However, I don't know if these subgroups are cyclic, because the question asks us to show that all the subgroups of Q8 are cyclic, so I don't know, I'm rather confused on this whole question.
    the subgroups of Q_8 are {I}, {I, -I}, {I, -I, J, -J}, {I, -I, K, -K}, and {I, -I, L, -L} and they are all cyclic..

    Quote Originally Posted by Janu42 View Post
    Is Q_8\times \mathbb{Z}_2 a group of order 16? And the group S_4 is order 24?
    if (a,b) \in Q_8\times \mathbb{Z}_2, then there will be 8 choices for a and 2 choices for b which make it 16 choices for the element (a,b).. since Q_8 is non-abelian, so is Q_8\times \mathbb{Z}_2

    |S_n|= n!
    Last edited by mr fantastic; February 23rd 2009 at 12:40 AM. Reason: Merged posts
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