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Thread: 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 ($\displaystyle \Rightarrow$) $\displaystyle G$ is cyclic, then $\displaystyle G=<a>$ and $\displaystyle a^n=e$. Let $\displaystyle b\in G$ such that $\displaystyle b^m=e$. then $\displaystyle b=a^k$ for some natural number $\displaystyle k$. thus $\displaystyle b^m = a^{km} = e = a^n$ you can continue from here..

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

    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>?
    $\displaystyle \frac{\mbox{lcm }\{144, 26\}}{26}$

    Quote Originally Posted by Janu42 View Post
    4. Construct a nonabelian group of order 16, and one of order 24.
    i think, $\displaystyle Q_8\times \mathbb{Z}_2$ is non-abelian..
    $\displaystyle S_4$ 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 $\displaystyle >4$ there exists an non-abelian group. Just construct $\displaystyle D_n$ - the dihedral group of order $\displaystyle 2n$.
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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 $\displaystyle n$-gon.
    There are $\displaystyle n$ rotations which perserve the polygon.
    And there are $\displaystyle 2$ reflections through each vertex.
    So there are a total of $\displaystyle 2n$ symmetries, and these symmetries form the dihedral group.
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    i think, $\displaystyle Q_8\times \mathbb{Z}_2$ is non-abelian..
    $\displaystyle S_4$ is non-abelian
    Is $\displaystyle Q_8\times \mathbb{Z}_2$ a group of order 16? And the group $\displaystyle S_4$ 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; Feb 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 $\displaystyle Q_8$ 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 $\displaystyle Q_8\times \mathbb{Z}_2$ a group of order 16? And the group $\displaystyle S_4$ is order 24?
    if $\displaystyle (a,b) \in Q_8\times \mathbb{Z}_2$, then there will be 8 choices for $\displaystyle a$ and 2 choices for $\displaystyle b$ which make it 16 choices for the element $\displaystyle (a,b)$.. since Q_8 is non-abelian, so is $\displaystyle Q_8\times \mathbb{Z}_2$

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