Results 1 to 8 of 8

Math Help - Homomorphisms

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    8

    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


    G=\Sigma_5, X=\Sigma_5. Then the homomorphism takes \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 \Sigma_8. Then G=<(1 2 3)(8 4 6 7 5)>. If \sigma=(1 2 3)(8 4 6 7 5) , the homomorphism maps to \mathbb{Z}_5 by \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 02:39 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    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


    G=\Sigma_5, X=\Sigma_5. Then the homomorphism takes \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 \Sigma_8. Then G=<(1 2 3)(8 4 6 7 5)>. If \sigma=(1 2 3)(8 4 6 7 5) , the homomorphism maps to \mathbb{Z}_5 by \sigma^i\rightarrow i+5\mathbb{Z}. I'm not sure what to do from here

    Thanks
    How is \Sigma_n defined?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2008
    Posts
    8
    \Sigma_n is the symmetric group of n letters

    As examples

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

    Thank! James
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    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


    G=\Sigma_5, X=\Sigma_5. Then the homomorphism takes \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 S_n. anyway, the Kernel of the homomorphism is only contains the identity.

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

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

    clearly \sigma has to be the identity permutation
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by alittletouched View Post
    Working in \Sigma_8. Then G=<(1 2 3)(8 4 6 7 5)>. If \sigma=(1 2 3)(8 4 6 7 5) , the homomorphism maps to \mathbb{Z}_5 by \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 \mathbb{Z}_5 then we should be mapping to equivalence classes mod 5, right?

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

    the order of \sigma is 8, so that 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 \{ \sigma^i \mid i + 5 \mathbb{Z} \text{ is a multiple of 5}\}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Jhevon View Post
    as for this one. if we are mapping to \mathbb{Z}_5 then we should be mapping to equivalence classes mod 5, right?

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

    the order of \sigma is 8, so that 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 \{ \sigma^i \mid i + 5 \mathbb{Z} \text{ is a multiple of 5}\}
    just a couple of points:

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

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

    3) then the map is obviously homomorphism and onto and its kernel is \{(1), (1 \ 2 \ 3), (1 \ 3 \ 2) \}.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by NonCommAlg View Post
    just a couple of points:

    1) the order of \sigma is 15 not 8.
    yes, you're right. i added when i should have multiplied
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    May 2008
    Posts
    8
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 5th 2011, 07:19 PM
  2. Homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: February 6th 2011, 12:39 PM
  3. No of Homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 29th 2010, 05:01 AM
  4. Homomorphisms to and onto
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: April 23rd 2009, 10:21 AM
  5. Homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: April 23rd 2007, 12:53 PM

Search Tags


/mathhelpforum @mathhelpforum