TITLE: Show that $Q_8$ can't be embedded in $M_{2 \times 2}(\mathbb{R})$ as a group.
QUESTION [6 upvotes]: So, suppose that we're working in a field $F$. Consider the ring $M_{n \times n} (F)$ which is the set all $n \times n$ matrices with entries in $F$. Is it possible to determine whether a matrix polynomial equation has solutions? If yes, is it possible to find the solutions for low degree polynomials?
My knowledge of mathematics is very limited as an undergraduate. I'm sure that there are many aspects of this question that are possibly beyond my knowledge at this level, but my interest in this question comes from this problem:
Show that the group $Q_8 = \{ \pm 1, \pm i, \pm j,\pm k\}$ under multiplication defined on Quaternions can not be embedded in the group of invertible elements of $M_{2\times 2}(\mathbb{R})$ under matrix multiplication.
So, because the cardinality of $M_{2 \times 2}(\mathbb{R})$ is infinite, I thought it might be very difficult to try group homomorphisms from $Q_8$ to $M_{2 \times 2}(\mathbb{R})$ and show that they can't have trivial kernel. Especially because if we replace $\mathbb{R}$ by $\mathbb{C}$ the problem becomes false. So, I thought that I should carefully study the number of solutions in each group and find some contradictions. For example, is it possible to show that the equation $A^4=I$ has less than $8$ solutions in $M_{2 \times 2}(\mathbb{R})$?

REPLY [5 votes]: By the earlier answers, all of the elements of order $4$ all have minimal polynomial $x^2+1$, so you can assume that one of them is $A = \left(\begin{array}{cc}0&1\\-1&0\end{array}\right)$.
Suppose $B = \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$ is another matrix of order $4$ in the image. Then, from the structure of $Q_8$, we have $AB=-BA$, giving $b=c$, but $B^2 = -I$ gives $a^2+bc= -1$, so $a^2+b^2 = -1$, which is not possible in ${\mathbb R}$.