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Group theory question

  1. Nov 29, 2007 #1
    1. The problem statement, all variables and given/known data
    If G is a group, is it true that right multiplication by a given element is a homomorphism but left multiplication is not? That does not really make sense to me because aren't right and left multiplication "symmetric"?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 29, 2007 #2

    morphism

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    In general neither is a homomorphism.
     
  4. Nov 29, 2007 #3

    Office_Shredder

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    Note to be a homomorphism, it needs to map the identity to the identity. Only a select few elements can multiply with the identity and give the identity back
     
  5. Nov 29, 2007 #4
    I think I asked the wrong question.
    Let G = {g_1, ..., g_n} be a group. Let X be a graph with vertices {g_1, ..., g_n}. Let [tex] \alpha_{g}[/tex] be the permutation of the vertices defined by [tex] \alpha_{g}(g_i) = g_i*g [/tex]

    Given a bijection between the elements of G and the elements of the group of \alpha_g, I want to show that the group of \alpha_{g} and G are isomorphic.

    So, let [tex] h(g_i) = \alpha_{g_i} [/tex].

    h is not a homomorphism, correct?
     
  6. Nov 30, 2007 #5
    does anyone understand this question?
     
  7. Nov 30, 2007 #6

    quasar987

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    The group of alpha_g? I assume you mean the subgroup of [tex]S_G[/tex] obtained from the map h you defined lastly?

    If so, then that's just Cayley's theorem.

    In Cayley's theorem, we show that every group G is isomorphic to a subgroup of [tex]S_G[/tex] (the group of permutation of G.)

    To do this, we show that the map [tex]h:G\rightarrow S_G[/tex] defined as you did is a monomorphism. Then the restriction of its codomain to h(G) is an isomorphism.
     
  8. Nov 30, 2007 #7
    The group I speak of is [tex]\{\alpha_{g} : g \in G\}[/tex] under composition of functions.

    I didn't realize that but you're right. So my question really is then, to prove Cayley's Theorem can you define [tex] \alpha_{g}(g_i) = g_i*g [/tex] or do you need to define
    [tex] \alpha_{g}(g_i) = g*g_i [/tex]?

    I think that we need the latter, but I'm afraid I am making a stupid mistake.
     
  9. Nov 30, 2007 #8

    quasar987

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    IT works either way. As you said, it's "symetric"

    1° For all g, [tex] \alpha_{g}(g_i) = g_i*g [/tex] defines a permutation of G, because 1) if g1*g=g2*g, then discover that g1=g2 by multiplying by g^-1 from the right. That's injectivity. 2) let g' be in G, then [tex]g_i=g'*g^{-1}[/tex] is sent to g' by [tex] \alpha_{g}[/tex]. That's surjectivity.

    2° Show h is an homomorphism. That's routine.

    3° Let g be in Ker(h). Then [tex]\alpha_{g}=id_G[/tex]. So for any g' in G, g'*g=g'. This can only be so if g=1. Hence the kernel is trivial, hence h is a monomorphism.
     
  10. Nov 30, 2007 #9

    quasar987

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    Funny!

    2° [tex]h(g*g')(g_i)=\alpha_{g*g'}(g_i)=g_i*(g*g')=(g_i*g)*g'=\alpha_{g'}(g_i*g)=\alpha_{g'}(\alpha_{g}(g_i))=(\alpha_{g'}\circ \alpha_{g})(g_i)=[h(g')\circ h(g)](g_i)[/tex]

    so it seems you were right in worrying! It's not a morphism if we multiply from the right!
     
  11. Nov 30, 2007 #10
    I claim that h is not a homomorphism in this case:

    [tex] \alpha_{g_1*g_2}(g_i) = g_i*g_1*g_2 [/tex]
    but
    [tex] \alpha_{g_1}\circ \alpha_{g_2}(g_i) = \alpha_{g_1}(\alpha_{g_2}(g_i)) = g_i*g_2*g_1[/tex]

    EDIT: you beat me
     
    Last edited: Nov 30, 2007
  12. Nov 30, 2007 #11

    quasar987

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    It works if we define [tex]\alpha_g[/tex] to act on gi by multiplication from the right by the inverse of g instead:

    [tex]\alpha_g(g_i)=g_ig^{-1}[/tex].

    Because now

    [tex]\alpha_{g*g'}(g_i)=g_i(g*g')^{-1}=g_ig'^{-1}g^{-1}=(\alpha_{g}\circ\alpha_{g'})(g_i)[/tex].
     
    Last edited: Nov 30, 2007
  13. Nov 30, 2007 #12

    quasar987

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    Yes, morphism and homomorphism are interchangeable terms.
     
  14. Nov 30, 2007 #13
    I just realized that and deleted the post. But thanks. Everything is clear now.
     
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