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Homomorphism of a cyclic subgroup is a cyclic subgroup ?

  1. Mar 20, 2012 #1

    Leb

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    1. The problem statement, all variables and given/known data

    hom.jpg

    Let [itex]\alpha:G \rightarrow H [/itex] be a homomorphism and let x[itex]\in[/itex]G

    Prove [itex]\alpha(<x>) =<\alpha(x)> [/itex]

    2. Relevant equations

    α(<x>) = α({x^{r}: r ∈ Z}) = {α(x^{r}) : r ∈ Z} = {α(x)^{r}: r ∈ Z} = <α(x)>.


    I do not understand how can we take out the 'r' out of a(x^{r}) to get (a(x))^{r} ? What would a(x)^2 say mean ? a(x) (some binary operation defined in H) a(x) or would it be (a°a)(x) i.e. a(a(x)) ?
     
  2. jcsd
  3. Mar 20, 2012 #2

    tiny-tim

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    Hi Leb! :smile:

    (try using the X2 button just above the Reply box :wink:)
    No, that would be ar(x).

    This is (a(x))r, the product of r elements of group H. :wink:
     
  4. Mar 20, 2012 #3

    Leb

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    Thank you for your reply tiny-tim !
    However, I am still not sure what will happen...

    (a(x))r = a(x) * a(x) *.....*a(x) r-times ? But what is * then ? Is it defined in H or in G ?
     
  5. Mar 20, 2012 #4

    Well, you have that [itex]a[/itex] is homomorphism, so [itex]a\bigl(x^2\bigr)=a(x\star x)=a(x)*a(x)=\bigl(a(x)\bigr)^2[/itex], same for [itex]a(x^{r})[/itex], where [itex]\star[/itex] is binary operation in [itex]G[/itex] and [itex]*[/itex] in [itex]H[/itex].
     
  6. Mar 20, 2012 #5

    tiny-tim

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    in H

    x is in G, a(x) is in H …

    so x can only undergo G's operations,

    and a(x) can only undergo H's operations :wink:
     
  7. Mar 20, 2012 #6

    Leb

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    Ah, OK, now it is clear. Thanks guys !
     
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