Is it true that (for each $n$) $\mathbb{Z}_n$ is isomorphic to a subgroup of $GL(2,R)$?

Question

Is it true that $GL(2,R)$ has elements of all order ? If so then it’s done, but don't know how to exhibit this.

Answer 1

Identify $\Bbb R^2$ with $\Bbb C$, then the multiplication with a nonzero $x \in \Bbb C$ can be identified with an element in $GL(2, \Bbb R)$.

If you take $x$ to be a primitive root of $X^n$ then this is corresponding to and element of order $n$ in $GL(2, \Bbb R)$.

If you instead choose a generic ring $R$, it is not true. A simple argument for this can be found for a finite ring $R$, since the group $GL(2, R)$ is finite. You could also look at the Jordan normal forms and show that there is not always such an element for $R= \Bbb Q$.

Edit:

Let $\omega$ be a primitive $n$-th root of unity. Then the minimal polynomial of $\omega$ over $\Bbb R$ is given by $x^2-(\omega+ \bar{\omega})x+1$.

Now consider the ring $$ \Bbb R[X] / \langle X^2-(\omega+ \bar{\omega})X+1 \rangle $$ it is a real algebra with basis $(1,X)$ and $X$ has order $n$. What you are looking for is the matrix representing multiplication by $X$. Now $1$ is mapped to $X$ and $X$ is mapped to $X^2=(\omega+ \bar{\omega})x-1$, so in this basis you have the representing matrix $$\begin{bmatrix}0&-1\\1& \omega+ \bar{\omega}\end{bmatrix}. $$ Using Euler formula one choice for $\omega + \bar{\omega}$ is $2\cos (2\pi /n)$. In the case $n=3$ we don't need this since the sum of the $n$-th roots of unity is zero and this gives us $\omega +\bar{\omega}+1=0$ and thus the matrix simplifies to $$\begin{bmatrix}0&-1\\1& -1\end{bmatrix} .$$

What about $\Bbb Q$? By Jordan normal form you know that the desired matrix $M$ has eigenvalue $\omega$. Since $\omega$ can not be real and the characteristic polynomial is real $\bar{\omega}$ is also an eigenvalue of $M$. But for $n=p$ prime the minimal polynomial of $\omega$ has degree $p-1$, thus for $n$ prime and $n>3$ there is no such matrix over $\Bbb Q$.