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This question arose during discussions about interesting examples of "orders of group elements" for a group theory course.

Definition: $GL(2,\mathbb{R})$ is the group of $2 \times 2$ matrices with real number entries, with non-zero determinant. The binary operation for this group is matrix multiplication.

Question: What is $\{\mathrm{ord}(M):M \in GL(2,\mathbb{R})\}$?

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Hint: consider rotations. – Robert Israel Mar 19 '12 at 21:21
More interesting is the $GL(2,\mathbb{Z})$ case. – user641 Mar 19 '12 at 21:22
Well...$GL(2, \mathbb{R})$ is a pretty big group. It contains elements of every possible order. An element of infinite order is $\left(\begin{array}{cc}1&0\\2&1\end{array}\right)$, and along with its pair $\left(\begin{array}{cc}1&2\\0&1\end{array}\right)$ they generate the free group on two generators, while one can use primitive roots of unity to find elements of any given finite order. – user1729 Mar 19 '12 at 21:23
In order to have an element of $GL(2,{\mathbb Z})$ with order $n$ (an integer $\ge 3$), its eigenvalues would have to be primitive $n$'th roots of unity, but they are also roots of a quadratic polynomial over $\mathbb Z$. Since the $n$'th cyclotomic polynomial is irreducible over the rationals, the only possibilities are $n=3,4,6$ which are the cases where the cyclotomic polynomials are quadratic. – Robert Israel Mar 19 '12 at 21:36
The "group-theoretic" way to see what Robert Israel said is to write $GL(2,\mathbb{Z})$ as an amalgamated product of $D_8$ and $D_{12}$, so that every element of finite order lives in (a conjugate of) $D_8$ or $D_{12}$. – user641 Mar 19 '12 at 22:33
up vote 12 down vote accepted

It can be any number. Take a matrix that represents a rotation with an angle $\phi=\frac{2\pi}{n}$ ($n\in \mathbb{Z}$), that is $$\left(\begin{array}{cc}\cos\phi & \sin \phi\\-\sin\phi & \cos\phi\end{array}\right)$$ Its order is clearly $n$.

EDIT: Just for completeness, it can of course also be infinite. An example would be $$\left(\begin{array}{cc}1 & 1\\ 0 & 1\end{array}\right)$$ which is invertible and satisfies $$\left(\begin{array}{cc}1 & 1\\ 0 & 1\end{array}\right)^n=\left(\begin{array}{cc}1 & n\\ 0 & 1\end{array}\right)$$

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Thanks for that. That very succinctly answers my question. – Douglas S. Stones Mar 19 '12 at 21:29

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