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Thread: quaternions and Diheral group

  1. #1
    ux0
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    quaternions and Diheral group

    Prove that the quaternions $\displaystyle \bold Q$ and the Dihedral group $\displaystyle D_8$ are non-isomorphic groups of order 8.
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    An isomorphism preserves the order of an element. Therefore if, for example, there are strictly more elements of order $\displaystyle 2$ in $\displaystyle D_4$ than in $\displaystyle \bold Q,$ then theese groups are not isomorphic.
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  3. #3
    ux0
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    ok well i know that there is only one element in $\displaystyle \bold Q$ of order 2, which is $\displaystyle -I$, and i know it is normal, how do I figure out how many subgroups of order 2 are in $\displaystyle D_8 $?
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    The dihedral group of order 8 can be seen as the group of symmetries of the square. There are two diagonal symetries, which are involutions, therefore $\displaystyle D_8$ (with your notation) contains at least 2 elements of order 2; that is sufficient to prove what you want.

    $\displaystyle D_8$ has exactly 5 elements of order 2 (two diagonal symmetries, 2 lateral symmetries and one central symmetry) and 2 elements of order 4 (one rotation of $\displaystyle \frac{\pi}{2}$ and its inverse).
    Last edited by clic-clac; Oct 27th 2009 at 06:22 AM. Reason: involutions and not idempotent...
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by clic-clac View Post
    There are two diagonal symetries, which are idempotent...
    Maybe I am missing something here, but the only idempotent in a group is the identity - an idempotent is an element $\displaystyle a$ such that $\displaystyle a^2=a$. In a group we can cancel this to get that $\displaystyle a=1$...
    Last edited by Swlabr; Oct 27th 2009 at 07:56 AM.
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    You're not missing anything, I wrote idempotent thinking of involution... My bad
    Thanks for pointing out the error.
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