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Proving that a subgroup is normal.

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data

    How can we prove that a subgroup H of [itex]Gl_2(Z_3)[/itex] is normal?

    These are the elements of H:

    [tex]\begin{pmatrix}1&1\\1&2 \end{pmatrix}[/tex]

    [tex]\begin{pmatrix}1&2\\2&2 \end{pmatrix}[/tex]

    [tex]\begin{pmatrix}2&1\\1&1 \end{pmatrix}[/tex]

    [tex]\begin{pmatrix}2&2\\2&1 \end{pmatrix}[/tex]

    [tex]\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}[/tex]

    [tex]\begin{pmatrix} -1 & 0 \\0 & -1 \end{pmatrix}[/tex]

    [tex]\begin{pmatrix} 0 & 2 \\1& 0 \end{pmatrix}[/tex]

    [tex]\begin{pmatrix} 0 & 1 \\2 & 0 \end{pmatrix}[/tex]

    So the determinant of the elements is 1 and the trace is zero...and H additionally contains the identity element and the -identity element.


    2. Relevant equations



    3. The attempt at a solution

    I don't think I can find the left and right cosets and show that they are equal, because there are too many elements in [itex]Gl_2(Z_3)[/itex]. There is a theorem in our textbook that states,

    Let H be a subgroup of G. Then H is normal in G if and only if there is a group structure on the set G/H of left cosets of H with the property that the canonical map π : G → G/H is a homomorphism. If H is normal in G, then the group structure on G/H which makes π a homomorphism is unique: we must have gH · gH = ggH for all g, g ∈ G. Moreover, the kernel of π : G → G/H is H. Thus, a subgroup of G is normal if and only if it is the kernel of a homomorphism out of G.

    So I'm trying to use this and show that [itex]f: Gl_2(Z_3) \rightarrow Gl_2(Z_3)/H[/itex] is a homomorphism...but I'm stuck...

    We know that [itex]f(a)f(b)=aHbH[/itex] and that [itex]f(ab)=abH[/itex]. But how can we show that aHbH=abH? Can anybody please give me a hint?

    Thanks in advance
     
  2. jcsd
  3. Mar 9, 2013 #2

    jbunniii

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    All the elements of ##H## have determinant ##1##, so H is a subgroup of ##SL_2(Z_3)##. How many subgroups of order 8 does ##SL_2(Z_3)## have?
     
  4. Mar 10, 2013 #3
    Thanks, but I didn't read about [itex]SL_2(Z_3)[/itex] yet.
     
  5. Mar 10, 2013 #4

    jbunniii

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    ##SL_2(Z_3)## is the subgroup of ##GL_2(Z_3)## consisting of the matrices with determinant 1. What I have in mind is to try the following argument: (1) ##H## is characteristic in ##SL_2(Z_3)##; (2) ##SL_2(Z_3)## is normal in ##GL_2(Z_3)##; (3) therefore...?

    In order to make that argument, you need to know the orders of ##GL_2(Z_3)## and ##SL_2(Z_3)##. Can you calculate these?
     
    Last edited: Mar 10, 2013
  6. Mar 11, 2013 #5
    Alright, thanks :smile:
     
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