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    The following are prime decompositions in \Bbb{Z}_7[x]:

    x^8+1= (x^2-x-1)(x^2+x-1)(x^2+3x-1)(x^2+4x-1)

    x^4+1= (x^2+3x+1)(x^2+4x+1)

    (a) Give representatives for the conjugacy classes of elements of order dividing 16 in Gl_2(\Bbb{Z}_7) and give the order of each.

    (b) Show that there are no elements of order 32 in Gl_2(\Bbb{Z}_7).

    (a) We can write the rational canonical forms as representatives.

    x^2-x-1 \rightarrow \begin{bmatrix} 0 & 1 \\1 & 1 \end{bmatrix}

    x^2+x-1  \rightarrow  \begin{bmatrix} 0 & -1 \\1 & 1 \end{bmatrix}

    x^2+3x-1 \rightarrow \begin{bmatrix} 0 &  -3 \\1 & 1 \end{bmatrix}

    x^2+4x-1 \rightarrow \begin{bmatrix} 0 &  -4 \\1 & 1 \end{bmatrix}

    x^3+3x+1 \rightarrow \begin{bmatrix} 0 &  -3 \\1 & -1 \end{bmatrix}

    x^2+4x+1 \rightarrow \begin{bmatrix} 0 &  -4 \\1 & -1 \end{bmatrix}

    x^2+1 \rightarrow \begin{bmatrix} 0 &  0 \\1 & -1 \end{bmatrix}, but since the determinant is zero, it is not in Gl_2(\Bbb{Z}_7).

    The first four matrices have order 16, since they satisfy x^8+1=0 so x^8=-1 \implies (x^8)^2 = 1. The last two have order 8, since x^4=-1 \implies x^8=1.

    (b)We don’t know the factorization of x^{16}+1, so we can’t check whether or not the matrices that satisfy it actually belong to Gl_2(\Bbb{Z}_7). I wasn’t able to tell whether this was even reducible in Z_7[x] to begin with, so I was kind of stuck here, and I was wondering if anybody could give me a hint...

    Thanks in advance
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    Hey there. While you're waiting for a reply to your post, we thought we'd give your thread a timely bump.

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    (Original post by Artus)
    (b)We don’t know the factorization of x^{16}+1,
    To start just replace x with x^2 in the factorisation of x^8+1. It's somewhat messier after that.
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    (Original post by ghostwalker)
    To start just replace x with x^2 in the factorisation of x^8+1. It's somewhat messier after that.
    Thanks
 
 
 
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