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Groups :Ismorphism

  1. Aug 11, 2011 #1
    To prove that : f : U[itex]_{s}[/itex] (st) [itex]\rightarrow[/itex] U(t) is an onto map.

    Note that

    Us(st)= {x [itex]\in[/itex] U(st): x= 1 (mod s)}

    Let x [itex]\in[/itex]U(t)

    then (x, t)=1 and 1<x< t

    How to proceed beyond point ?
     
  2. jcsd
  3. Aug 12, 2011 #2

    Fredrik

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    I don't understand what you're saying. How do you define the sets (groups?) U(t) and U(st) and? Is s an integer? How is the function f defined?
     
  4. Aug 13, 2011 #3
    1) U ( n) is a multiplication group modulo n, for any integer n.

    ( U(n) is group contains all the non zero units of Zn, that is , all the integers belonging to Zn that are relatively prime to n.

    2) s and t are integers such that (s, t) =1.( rel prime)

    3) f (x) = x mod s, where x belongs to Us(st)
     
    Last edited: Aug 13, 2011
  5. Aug 14, 2011 #4

    HallsofIvy

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    But why are you limiting yourself like that? For any group, G, the function [itex]f_s(t)= st[/itex] (or in "additive" notation, [itex]f_s(t)= s+ t[/itex]) is an isomorphism- and so both one-to-one and onto.

    To prove it is onto, suppose y is a given member of the group. You merely need to show that there exist x such that sx= y- and that is easy.
     
  6. Aug 14, 2011 #5

    Stephen Tashi

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    This might be a question about a mapping between the multiplicative group of integers mod st to the multiplicative group of integers mod t rather than a question about mapping the multiplicative of integers into itself. In fact, from the definition of [itex] U_s[/itex] in the original post, it might be a question about mapping subsets of the group of integers mod st onto the group of integers mod t.
     
  7. Aug 14, 2011 #6
    Do you know U(st) is isomorphic to U(s)xU(t) though the map x->(x mod s, x mod t)?

    Then Us(st) is a subgroup of U(st). Images of subgroups are subgroups, and in this case the image is {1} x U(t) which is isomorphic to U(t).
     
    Last edited: Aug 14, 2011
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