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Is the Mod2-Reduction Map Onto?

  1. Aug 2, 2011 #1
    Hi, Algebraists:

    The modN reduction map r(N) from a matrix group (any group in which the elements
    are matrices over Z-integers) over the integers, in which r is defined by

    r(N) : (a_ij)-->(a_ij mod N) is not always commutative, e.g.:

    r(6) :Gl(2,Z) --Gl(2,Z/N)

    is not onto, since Gl(2,Z) is unimodular over Z, but Gl(2,Z/N) is not, e.g., we can

    take units of Z/N that are not 1modN , so that the determinants do not match up;

    e.g., for N=10 , take the unit , say, 7 in Z/10Z ; then the matrix M with a_11=7

    a_22 =0 and 0 otherwise, is in Gl(2,Z/10) (use, M' with a_11'=3 , and a_22'=1 )

    but it is not the image of any matrix in Gl(2,Z), since the determinants do not match

    up.

    ** question **:

    Anyone know any results re when the mod2 reduction map r(2) is onto?

    Thanks.
     
  2. jcsd
  3. Dec 28, 2011 #2
    Ughh I'm coming across a similar problem..
     
  4. Dec 30, 2011 #3

    Bacle2

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    Science Advisor

    Sorry; I still don't have a good answer. I will keep looking into it.
     
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