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Thread: Homomorphisms

  1. #1
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    Homomorphisms

    Hello all. I am stuck trying to figure out if these are homomorphisms and what there images/kernels are if that is the case. Any help would be great! James


    $\displaystyle G=\Sigma_5, X=\Sigma_5$. Then the homomorphism takes $\displaystyle \sigma\rightarrow (3 5 2 4 1)*\sigma*(3 1 4 2 5)$. I know that this is a homomorphism, I am just stuck on the kernel/image


    Working in $\displaystyle \Sigma_8$. Then $\displaystyle G=<(1 2 3)(8 4 6 7 5)>$. If $\displaystyle \sigma=(1 2 3)(8 4 6 7 5) $, the homomorphism maps to $\displaystyle \mathbb{Z}_5$ by $\displaystyle \sigma^i\rightarrow i+5\mathbb{Z}$. I'm not sure what to do from here

    Thanks
    Last edited by alittletouched; May 19th 2008 at 01:39 PM.
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    Quote Originally Posted by alittletouched View Post
    Hello all. I am stuck trying to figure out if these are homomorphisms and what there images/kernels are if that is the case. Any help would be great! James


    $\displaystyle G=\Sigma_5, X=\Sigma_5$. Then the homomorphism takes $\displaystyle \sigma\rightarrow (3 5 2 4 1)*\sigma*(3 1 4 2 5)$. I know that this is a homomorphism, I am just stuck on the kernel/image


    Working in $\displaystyle \Sigma_8$. Then $\displaystyle G=<(1 2 3)(8 4 6 7 5)>$. If $\displaystyle \sigma=(1 2 3)(8 4 6 7 5) $, the homomorphism maps to $\displaystyle \mathbb{Z}_5$ by $\displaystyle \sigma^i\rightarrow i+5\mathbb{Z}$. I'm not sure what to do from here

    Thanks
    How is $\displaystyle \Sigma_n$ defined?
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  3. #3
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    $\displaystyle \Sigma_n$ is the symmetric group of n letters

    As examples

    $\displaystyle \Sigma_2=\left\{1_{\Sigma_2},(1 2)\right\}$
    and
    $\displaystyle \Sigma_3=\left\{1_{\Sigma_3},(1 2),(2 3),(1 3),(1 2 3),(1 3 2)\right\}$

    Thank! James
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alittletouched View Post
    Hello all. I am stuck trying to figure out if these are homomorphisms and what there images/kernels are if that is the case. Any help would be great! James


    $\displaystyle G=\Sigma_5, X=\Sigma_5$. Then the homomorphism takes $\displaystyle \sigma\rightarrow (3 5 2 4 1)*\sigma*(3 1 4 2 5)$. I know that this is a homomorphism, I am just stuck on the kernel/image
    Oh! i always used the notation $\displaystyle S_n$. anyway, the Kernel of the homomorphism is only contains the identity.

    note that $\displaystyle (3~5~2~4~1)^{-1} = (3~1~4~2~5)$

    to get an output of $\displaystyle (1)$ we need $\displaystyle \sigma * (3~1~4~2~5) = (3~5~2~4~1)^{-1}$

    clearly $\displaystyle \sigma$ has to be the identity permutation
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    Quote Originally Posted by alittletouched View Post
    Working in $\displaystyle \Sigma_8$. Then $\displaystyle G=<(1 2 3)(8 4 6 7 5)>$. If $\displaystyle \sigma=(1 2 3)(8 4 6 7 5) $, the homomorphism maps to $\displaystyle \mathbb{Z}_5$ by $\displaystyle \sigma^i\rightarrow i+5\mathbb{Z}$. I'm not sure what to do from here

    Thanks
    as for this one. if we are mapping to $\displaystyle \mathbb{Z}_5$ then we should be mapping to equivalence classes mod 5, right?

    so you mean $\displaystyle \sigma^i \mapsto [i + 5 \mathbb{Z}]$

    the order of $\displaystyle \sigma$ is 8, so that $\displaystyle 0 \le i \le 8$, and then everything repeats for higher values. thus, to find the kernel (which i suppose is what you want), it would be $\displaystyle \{ \sigma^i \mid i + 5 \mathbb{Z} \text{ is a multiple of 5}\}$
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    Quote Originally Posted by Jhevon View Post
    as for this one. if we are mapping to $\displaystyle \mathbb{Z}_5$ then we should be mapping to equivalence classes mod 5, right?

    so you mean $\displaystyle \sigma^i \mapsto [i + 5 \mathbb{Z}]$

    the order of $\displaystyle \sigma$ is 8, so that $\displaystyle 0 \le i \le 8$, and then everything repeats for higher values. thus, to find the kernel (which i suppose is what you want), it would be $\displaystyle \{ \sigma^i \mid i + 5 \mathbb{Z} \text{ is a multiple of 5}\}$
    just a couple of points:

    1) the order of $\displaystyle \sigma$ is 15 not 8.

    2) it is important to note that $\displaystyle \sigma^i \rightarrow [i + 5\mathbb{Z}]$ is basically well-defined. that's because 15 (the order of $\displaystyle \sigma$) is divisible by 5.

    3) then the map is obviously homomorphism and onto and its kernel is $\displaystyle \{(1), (1 \ 2 \ 3), (1 \ 3 \ 2) \}.$
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    just a couple of points:

    1) the order of $\displaystyle \sigma$ is 15 not 8.
    yes, you're right. i added when i should have multiplied
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    That's great thanks

    How would I find the images for the homomorphisms? I expect that the image of the second homomorphism is merely the integers but Im not sure about the first
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