Prove that any element of GL2(Z)

**dabien** · November 9th 2009, 01:59 PM

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!

**aliceinwonderland** · November 9th 2009, 08:49 PM

Originally Posted by dabien

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 $X=\begin{bmatrix} a& b\\ c & d\\ \end{bmatrix}$ 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, $X=\begin{bmatrix} 1& 0\\ 0 & 1\\ \end{bmatrix}$ .
For order 2, $X=\begin{bmatrix} 1& 0\\ 0 & -1\\ \end{bmatrix}$ , $\begin{bmatrix} -1& 0\\ 0 & 1\\ \end{bmatrix}$ , $\begin{bmatrix} -1& 0\\ 0 & -1\\ \end{bmatrix}$ .

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 $X=\begin{bmatrix} 0& 1\\ -1 & -1\\ \end{bmatrix}$ . 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 $X=\begin{bmatrix} 0& 1\\ -1 & 1\\ \end{bmatrix}$ . It has an order 6.

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

**NonCommAlg** · November 9th 2009, 11:45 PM

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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