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Abstract Algebra: Finding Conjugates

  1. Apr 30, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider this group of six matrices:

    Let G = {I, A, B, C, D, K}, Matrix Multiplication>

    [tex]I =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}[/tex] [tex]A =\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}[/tex] [tex]B =\begin{bmatrix}0 & 1\\-1 & -1\end{bmatrix}[/tex]

    [tex]C =\begin{bmatrix}-1 & -1\\0 & 1\end{bmatrix}[/tex] [tex]D =\begin{bmatrix}-1 & -1\\1 & 0\end{bmatrix}[/tex] [tex]K =\begin{bmatrix}1 & 0\\-1 & -1\end{bmatrix}[/tex]

    Operation Table for this group:

    _|I A B C D K
    I |I A B C D K
    A|A I C B K D
    B|B K D A I C
    C|C D K I A B
    D|D C I K B A
    K|K B A D C I

    Define [tex] f:G\rightarrow[/tex] [tex]\left\langle\(R^{*}, \bullet\right\rangle[/tex] by f(x) = det(x) for any Matrix x [tex]\in[/tex] G.


    Finally, the question. Haha.

    Find all the conjugates of A:
    Find all the conjugates of B:

    2. Relevant equations
    N/A

    3. The attempt at a solution

    Now I know this isn't that hard of a concept and I understand what a conjugate is but I don't know how to find them. Any help on how I would go about finding them is greatly appreciated.
     
    Last edited: Apr 30, 2010
  2. jcsd
  3. Apr 30, 2010 #2
    For conjugates of A,

    G={I, A, B, C, D, K}

    I^-1*A*I=A
    B^-1*A*B=D*C=K
    C^-1*A*C=C*B=K
    D^-1*A*D=B*K=C
    K^-1*A*K=K*D=C

    So K and C are conjugates of A.

    Do same procedure to find conjugates of B.
     
  4. Apr 30, 2010 #3
    Or given the fact that you've worked out the multiplication table, you can see by inspection that it's a nonabelian group of of order 6. (You don't need to check the associative law because that's taken care of by the elements being real matrices under multiplication.)

    That means it's isomorphic to S3 and the conjugacy classes are
    (i) the identity
    (ii) the elements of order 2
    (iii) the rest

    The identity is I of course.
    The elements of order 2 can be read off by finding I on the main diagonal, so A,C,K.
    And the rest is B,D.

    Of course you need to have proved that there is only one nonabelian group of order 6, and determined the conjugacy classes in S3 first.
     
  5. May 2, 2010 #4
    Ok I understand that. Thanks. One question though, can A or B be considered a conjugate of either x-1Ax or x-1Bx respectively?
     
  6. May 2, 2010 #5
    Yes. For take the identity element, I^-1*A*I=A So A is a conjugate of A, and likewise B is a conjugate of itself.
     
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