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Mapping Question

  1. Apr 11, 2010 #1
    Can an element of a quotient group G/N in an isomorphism f:G/N--H map back onto itself if it does not have a corresponding element in H? The example I am looking at is the quotient group of the group of symmetries of an n-gon, G, where n is an even number and N equal to the normal subgroup containing the identity and a 180 rotation. H represents another n-gon with an even n, but is less than the order of G.

    I have found that the quotient group G/N contains every permutation in G (which I think is incorrect) because for f to be an isomorphism, G contains rotations that don't map to H.

    What's going on here?
     
  2. jcsd
  3. Apr 12, 2010 #2
    Are you referring to one of the isomorphism theorems?

    Let's do a small example, with the dihedral group [itex]D_4[/itex] (the group of symmetries of a square). The 8 elements are [itex]\{ e, h, v, d_1, d_2, r_{90}, r_{180}, r_{270} \}[/itex]. If [itex]N=\{ e, r_{180} \}[/itex] then what are the elements of [itex]D_4/N[/itex]? (From Lagrange's theorem, we expect only 4 elements in this group.)

    [itex]eN = r_{180}N = N[/itex]
    [itex]vN = hN = \{ h, v \}[/itex]
    [itex]d_1N = d_2N = \{ d_1, d_2 \}[/itex]
    [itex]r_{90}N = r_{270}N = \{ r_{90}, r_{270} \}[/itex]

    Notice that the elements of [itex]D_4/N[/itex] are essentially _cosets_ of N; so I am not sure what you mean by "G/N contains every permutation."
     
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