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I have this exercise, I only need help with d.

a. Show that $SO_2(\mathbb{R})=\{R_\theta=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}| \theta \in \mathbb{R} \}$

is an abelian subgroup of $GL_2(\mathbb{R})$

b). Define $\phi:\mathbb{R}\rightarrow SO_2(\mathbb{R})$, with $\phi(\theta)=R_\theta$. Find the kernel to $\phi$. Identify $SO_2(\mathbb{R})$ with a factor-group of the real numbers.

c) Show that $R_\theta \in SO_2(\mathbb{R})$if and only if there are real umbers m,n such that $\theta=\frac{2\pi m}{n}$.

d) Explain why there are elements $g,h \in GL_2(\mathbb{R})$, both of order 2, such that $gh$ does not have finite order.

Any hints for d)? I assume that since the entire exercise has to do with rotations, there is going to be a rotation in d) aswell. If an element has order two as a rotation, it must rotate a multiple of $\pi$? So we may put $g=R_\pi$?, but what about h? I thought that h might be flips around the cordinate axis for instance. Then h has order 2, but I allways get that $gh$ also have order 2(not infinite), no matter which I try. Any hints?

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It doesn't necessarily has to do with rotation. Only invertible its is the requirement.

Consider $$A=\begin{pmatrix}0&1\\1&0 \end{pmatrix} \quad B= \begin{pmatrix}0&2\\1/2 & 0 \end{pmatrix}$$

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  • $\begingroup$ Thanks, how did you figure this example out? $\endgroup$ – user119615 Mar 2 '15 at 18:09
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Consider $$ g=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \qquad\text{and}\qquad h=\begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}. $$ The orders of $g$ and $h$ are finite, but the order of $gh$ is infinite. Namely, we have $g^4=I$ (while $g^2 \neq I$; here $I$ is the identity matrix) and $h^3 = I$, i.e. $\text{ord}(g)=4$ and $\text{ord}(h)=3$. But $$ gh=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $$ and for any $n \in \mathbb{N}$ $$ (gh)^n= \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \neq I. $$

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