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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.

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    $\begingroup$ I would prove this in a completely different way: A faithful representation of $Q_8$ on a real vector space $V$ makes $V$ into a quaternionic vector space, so the dimension of $V$ has to be divisible by $4$. $\endgroup$ – Andreas Cap Mar 30 '15 at 7:15
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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.

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it is clear that $Q_8$ has the 3-cycle $(i,j,k)$ as an outer automorphism. i suspect, but don't know how to prove, that the only non-trivial outer automorphism (modulo inner automorphisms) of a finite matrix subgroup of $GL(2,\mathbb{R})$ is the involution obtained by the permutation $(a_{11} a_{22})(a_{12} a_{21})$

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  • $\begingroup$ The Klein $4$-group $C_2 \times C_2$ is a subgroup of ${\rm GL}(2,\mathbb{R})$ and has an outer automorphism of order $3$. $\endgroup$ – Derek Holt Mar 28 '15 at 11:31
  • $\begingroup$ @DerekHolt thx , Derek. i see now that the matrices corresponding to $\pm 1, \pm i$ in the standard $GL(2,\mathbb{R})$ representation for $\mathbb{C}$ will give a $V_4$ $\endgroup$ – David Holden Mar 28 '15 at 12:10

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