Show that an element in $E(2)$ of the form $(A, x)$, where $x$ is not equal to $0$, has infinite order. I'm trying to show that an element in $E(2)$ of the form $(A, x)$, where $x \neq 0$, has infinite order. And $E(2)$ is an euclidean group. $A\in O(2), x\in\mathbb{R}^2.$ However, I think the statement is wrong. I think every element in $E(2)$ has finite order. Can anyone help me?
 A: I came across the same problem (guess I'm using the same text book). I also think the statement is wrong. Consider this element $(A,x)$ where
$$A=\begin{pmatrix}-1&0\\
0&-1\end{pmatrix}\,,\,x=\begin{pmatrix}1\\0\end{pmatrix}$$
(a rotation of 180º counterclockwise and the translation defined by $x$). Then $(A,x)\in E(2)$ and $x\neq 0$ but it has finite order: 
$$(A,x)^2=(A^2,Ax+x)=(I,-x+x))=(I,0)$$
A: For a general matrix $A$ and vector $x$ you have that
$$(A,x)^n = (A^n, \sum_0^{n-1} A^jx).$$
If this element has finite order then $A^n=I$ and $x\in \ker(\sum_0^{n-1}A^j)$. The matrices in $O(2)$ are just rotations and reflections. Suppose first that $A$ is a rotation with angle $2\pi\theta$, so its eigenvalues are $e^{\pm i2\pi \theta}$ and it is diagonalizable. It follows that $B=\sum_0^{n-1}A^j$ is also diagonalizable and its eigenvalues are exactly the polynomial $\sum_0^{n-1}x^j$ applied to $e^{\pm 2\pi i \theta}$ which is exactly $\frac{1-e^{\pm n 2 \pi i\theta}}{1-e^{\pm 2 \pi i\theta}}$ which can be zero only if $\theta$ is rational (and not an integer in which case $A=I$) and then both eigenvalues are zero. Since $B$ is diagonalizable, you get that it is either zero for rational $\theta$ (which is not a multiple of or invertible. In the first case you get that $(A,x)^n=(I,0)$ so it has finite order, and otherwise the order is infinite. 
If $A$ is a reflection, then $A^2=I$ and its eigenvalues are $\pm 1$. Every vector $x$ we can decompose to $x=x_+ + x_-$ such that $Ax=x_+ - x_-$ and then
$$(A,x)^2=(A^2,Ax+x)=(I,2x_+)$$ which is the identity if $x_+ =0$ and otherwise has infinite order.
