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Math Help - Prove that any element of GL2(Z)

  1. #1
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    Prove that any element of GL2(Z)

    Prove that any element of GL2(Z) of finite order has order 1,2,3,4, or 6.
    Extend this to GL3(Z), and show that it fails in GL4(Z).

    Thanks for your help!
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  2. #2
    Senior Member
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    Quote Originally Posted by dabien View Post
    Prove that any element of GL2(Z) of finite order has order 1,2,3,4, or 6.
    Extend this to GL3(Z), and show that it fails in GL4(Z).

    Thanks for your help!
    Here is for the first part:

    Let X be an element of finite order in GL_2(Z) such that X=\begin{bmatrix}<br />
a& b\\<br />
c & d\\ <br />
\end{bmatrix} with X^n=I_2.

    Then the characteristic polynomial of X for eigenvalues \lambda_1, \lambda_2 is \lambda^2-(a+d)\lambda + (ad - bc)=0. Since \lambda^k is an eigenvalue of X^k for a positive integer k, we see that \lambda^n=1 for \lambda=\lambda_1 and \lambda_2.

    Case 1. Both \lambda_1 and \lambda_2 are reals. In this case, \lambda_i for i=1,2 are +1 or -1. Thus it has order 1 or 2.

    For order 1, X=\begin{bmatrix}<br />
1& 0\\<br />
0 & 1\\ <br />
\end{bmatrix}.
    For order 2, X=\begin{bmatrix}<br />
1& 0\\<br />
0 & -1\\ <br />
\end{bmatrix}, \begin{bmatrix}<br />
-1& 0\\<br />
0 & 1\\ <br />
\end{bmatrix}, \begin{bmatrix}<br />
-1& 0\\<br />
0 & -1\\ <br />
\end{bmatrix}.

    Case 2. \lambda_1 and \lambda_2 are complex numbers. Since they are roots of quadratic equations with real coefficients, they are conjugate to each other. Since they are complex roots of unity, \lambda_1 + \lambda_2=0, 1, -1.

    If \lambda_1 + \lambda_2=0, then \lambda_1=-\lambda_2 with \lambda_k = +i or -i for k=1,2. Thus the order of X is 4 (verify this). If \lambda_1 + \lambda_2=-1, then characteristic polynomial is \lambda^2+\lambda + 1=0. In this case, X can be X=\begin{bmatrix}<br />
0& 1\\<br />
-1 & -1\\ <br />
\end{bmatrix}. It has an order 3.
    If \lambda_1 + \lambda_2=1, then characterist polynomial is \lambda^2-\lambda + 1=0. In this case, X can be X=\begin{bmatrix}<br />
0& 1\\<br />
-1 & 1\\ <br />
\end{bmatrix}. It has an order 6.

    Thus any element of GL2(Z) of finite order has order 1,2,3,4, or 6.
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  3. #3
    MHF Contributor

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    the possible orders of elements in GL_3(\mathbb{Z}) is again 1, 2, 3, 4 and 6. however, the order of A=\begin{pmatrix}0 & 0 & 0 & -1 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{pmatrix} \in GL_4(\mathbb{Z}) is equal to 5. note that A is the companion matrix of

    \Phi_5(x)=x^4+x^3+x^2+x+1, the cyclotomic polynomial of order 5. more examples are B=\begin{pmatrix}0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \end{pmatrix} \in GL_4(\mathbb{Z}) and C=\begin{pmatrix}0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \in GL_4(\mathbb{Z}). the orders of

    B and C are 10 and 12 respectively. also, B and C are the companion matrices of \Phi_{10}(x) and \Phi_{12}(x) respectively. finally, note that \varphi(5)=\varphi(10)=\varphi(12)=4, where \varphi is the

    Euler totient function! extending this idea, we see that for any integer n \geq 2 there exists A \in GL_{\varphi(n)}(\mathbb{Z}) with order n.
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