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Math Help - Homomorphisms & the isomorphism thm.

  1. #1
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    Homomorphisms & the isomorphism thm.

    P: G to H is a group homomorphism, L = Image(P) = {h in H | h= P(g) for some g in G}.

    A. Show L is a subgroup of H.

    Now assume G= Z_{12} under addition and H= S_{3}.

    B. Show that P can't be onto. Which subgroups of S_{3} are possible for L?
    C. Find a specific example of some group homomorphism P from G to H which isn't the trivial map.

    my thoughts:
    for part A, I had that P(g)P(h) = P(gh) for any g, h in G, P( 1_G)= 1_H and P( G^{-1})= P (G)^{-1}, by the definition of a hom, but was told that this is wrong because I need to discuss L. What am I missing here?
    for B, I know the possible subgroups of H are <1>, <(123)>, <(12)>,<(23)>, <(13)>, and H, all of order 1, 2, 3 or 6, which all divide the order of G -- from there I don't know how to show there is no onto hom.
    for (C), I have no idea how to find a mapping that will work!

    Thank you so much for any help!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kimberu View Post
    P: G to H is a group homomorphism, L = Image(P) = {h in H | h= P(g) for some g in G}.

    A. Show L is a subgroup of H.
    Is it closed under mult.
    Spoiler:
    \theta(g),\theta(g')\in\theta(G)\implies \theta(g)\theta(g')=\theta(gg')\in\theta(G) since g,g'\in G\implies gg'\in G


    Does it contain the identity element?
    Spoiler:
    e_G\in G and \theta(e_G)=e_H


    Does it contain inverses?
    Spoiler:
    If \theta(g)\in G then g\in G and so g^{-1}\in G and \theta(g^{-1})\in \theta(G) and \theta(g)\theta(g^{-1})=\theta(gg^{-1})=\theta(e_G)=e_H

    Now assume G= Z_{12} under addition and H= S_{3}.

    B. Show that P can't be onto. Which subgroups of S_{3} are possible for L?
    Hint:
    Spoiler:
    If G is abelian and \theta:G\to G' is an epimorphism (a surjective homomorphism) then clearly a,b\in G' means that a=\theta(g),b=\theta(g') for some g,g'\in G and so ab=\theta(g)\theta(g')=\theta(gg')=\theta(g'g)=\th  eta(g')\theta(g)=ba........so

    C. Find a specific example of some group homomorphism P from G to H which isn't the trivial map.
    Why don't you think about this one for a little more.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post

    Hint: If G is abelian and \theta:G\to G' is an epimorphism (a surjective homomorphism) then clearly a,b\in G' means that a=\theta(g),b=\theta(g') for some g,g'\in G and so ab=\theta(g)\theta(g')=\theta(gg')=\theta(g'g)=\th  eta(g')\theta(g)=ba........so
    Does this mean that, accordingly, L = the image is an abelian subgroup, so the only possibilities are <1>, <(123)>, <(12)>,<(23)>, <(13)>, and not all of H, so P is not onto?

    Then, for part (C), does the following map work?
    P(0) --> e
    P(1, 3, 5, 7, 9, 11) ---> (12)
    P(2,4,6, 8,10) ----> e

    If so, the way I found it was basically by guessing...I guess what I'm asking is, is there a systematic way to find this sort of thing dealing with the orders of elements? I tried to find a map onto the group of <(123)>, but I couldn't figure that one out.

    Thank you!!
    Last edited by kimberu; April 12th 2010 at 07:37 PM.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kimberu View Post
    Does this mean that, accordingly, L = the image is an abelian subgroup, so the only possibilities are <1>, <(123)>, <(12)>,<(23)>, <(13)>, and not all of H, so P is not onto?

    Then, for part (C), does the following map work?
    P(0) --> e
    P(1, 3, 5, 7, 9, 11) ---> (12)
    P(2,4,6, 8,10) ----> e

    If so, the way I found it was basically by guessing...I guess what I'm asking is, is there a systematic way to find this sort of thing dealing with the orders of elements? I tried to find a map onto the group of <(123)>, but I couldn't figure that one out.

    Thank you!!
    I'm not quite sure what you're asking. The reason that the map in your example can't be surjective is that it would imply that S_3 is abeilan since \mathbb{Z}_{12} is.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    I'm not quite sure what you're asking. The reason that the map in your example can't be surjective is that it would imply that S_3 is abeilan since \mathbb{Z}_{12} is.
    That makes sense, thanks. Just to confirm, does that mean that <1>, <(123)>, <(12)>,<(23)>, <(13)> are all possibilities for L, as they are not abelian?
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