Problem 18.1.10 in Dummit and Foote's Abstract Algebra, third edition:
Prove that $\text{GL}_2(\mathbb{R})$ has no subgroup isomorphic to
$Q_8$. [EA: The quaternion group]. [This may be done by direct
computation using generators and relations for $Q_8$. Simplify these
calculations by putting one generator in rational canonical form.]
What I've done so far: $Q_8$ has presentation $\langle -1, i, j, k \mid (-1)^2 = e, i^2 = j^2 = k^2 = ijk=-1 \rangle.$ The element $-1$ should probably have RCF
$$\left( \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right) \text{ and not }\left( \begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix} \right)$$ or else it would make it difficult for it to be in the center. Not really sure where to go from here, or what they're looking for.
Best Answer
The minimal polynomial of an element of order $4$ has to be $x^2+1$. By replacing the generators by conjugates you can assume that $i$ is in rational canonical form. That is, $i = \left(\begin{array}{rr}0&-1\\1&0\end{array}\right)$. Let $j = \left(\begin{array}{rr}a&b\\c&d\end{array}\right)$. Now use the equations $k=ij$, $j^2=k^2=i^2=\left(\begin{array}{rr}-1&0\\0&-1\end{array}\right)$ to get some equations involving $a,b,c,d$ and derive a contradiction. It's not too hard.