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Thread: cosets

  1. #1
    Senior Member sfspitfire23's Avatar
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    cosets

    hi all, just an interesting coset problem I'm having trouble with -

    Find the right and left cosets in G of the subgroups H and K of G.

    $\displaystyle G=A_4$; $\displaystyle H=\{e,(1 2)(3 4),(1 3)(2 4), (1 4)(2 3)\}$; $\displaystyle K=<(1 2 3)>$


    What do you guys think?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    hi all, just an interesting coset problem I'm having trouble with -

    Find the right and left cosets in G of the subgroups H and K of G.

    $\displaystyle G=A_4$; $\displaystyle H=\{e,(1 2)(3 4),(1 3)(2 4), (1 4)(2 3)\}$; $\displaystyle K=<(1 2 3)>$


    What do you guys think?
    Firstly, you know precisely how many right/left cosets there are for each subgroup, this is just $\displaystyle |G|/|A|$ where $\displaystyle A$ is your subgroup.

    Now, for left cosets we know $\displaystyle xA=yA \Leftrightarrow x^{-1}y$, and we know that cosets all have equal size (they have size $\displaystyle |A|$). Thus, you need to find however many sets of 4 elements not in $\displaystyle H$ such that $\displaystyle x^{-1}y \in H$ for all 4 elements.

    These will be your left cosets. For your right cosets, you apply a similar formula ($\displaystyle xy^{-1}$).

    Remember that left and right cosets are equal if and only if your subgroup is $\displaystyle Normal$. I'm afraid I can't tell you off hand whether $\displaystyle K$ is normal, although I am pretty positive it is. $\displaystyle H$ is definitely normal. So, hopefully, in each case your left and right cosets will be equal.
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  3. #3
    Senior Member sfspitfire23's Avatar
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    Would e*k={e, (123),(132)}?

    I think this would be correct...but why is the (132) in there?


    thanks
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Would e*k={e, (123),(132)}?

    I think this would be correct...but why is the (132) in there?


    thanks
    This is correct - the coset $\displaystyle eH$ is just $\displaystyle H$, as when you multiply every element in the subgroup $\displaystyle H$ by the identity you get the identity. Similarly, $\displaystyle hH=H$ for all $\displaystyle h \in H$. $\displaystyle (132)$ is in the coset because $\displaystyle K=\{e, (123), (132)\}=<(123)>$. The $\displaystyle <X>$ brackets mean "the subgroup generated by the set $\displaystyle X$". Have you come across this notation before?
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  5. #5
    Senior Member sfspitfire23's Avatar
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    ah indeed! I see now!
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