Show that $Q_8$ can't be embedded in $M_{2 \times 2}(\mathbb{R})$ as a group.

Question

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 $M_{2 \times 2}(\mathbb{R})$?

Answer 1

Usually the way to approach matrix polynomial equations (of one variable) is to figure out what the minimal and/or characteristic polynomial of the solution matrices would be. Note that any conjugate of a solution to $p(A)=0$ is also a solution, so if there is one solution there will be infinitely many (except in cases where the only solutions are scalars). So you won't be able to show there are less than 8 solutions.

If $A$ satisfies $A^4 = I$, then the minimal polynomial of $A$ is a divisor of $x^4-1$. Since $x^4-1$ has no repeated roots, the minimal polynomial of $A$ is equal to the characteristic polynomial of $A$, which has degree 2. Therefore the characteristic polynomial is a real degree 2 factor of $x^4 - 1$, so it must be $x^2-1$ or $x^2+1$. By the Cayley-Hamilton theorem, this is also a sufficient condition for a matrix to satisfy $A^4 = I$.

So: A necessary and sufficient condition for a 2 by 2 real matrix $A$ to satisfy $A^4 = I$ is that its characteristic polynomial is $x^2 \pm 1$. This is equivalent to $A$ having trace 0 and determinant $\pm 1$, so this gives a pretty concrete form of what any such $A$ will look like. I think, from here, you should be able to work out the rest of the proof (that there is no embedding from quaternion group into $GL(2,\mathbb{R})$), though I haven't worked out the details.

(In the original answer, I used $A^2 = -I$ instead, but that assumes that -1 maps to $-I$ in any prospective embedding, which need not be the case.)

Answer 2

There are many solutions of $A^4=I$ in $GL(2,\Bbb R)$, because of conjugation. However it is fairly easy to see that one can nevertheless not embed $Q_8$ in $GL(2,\Bbb R)$. Assuming one has such an embedding, then since $Q_8$ is finite one can find a positive definite bilinear form that is invariant under all elements of $Q_8$ (this a standard averaging trick: transform the standard inner product by all elements of $Q_8$ and take their average). Changing basis to an orthogonal basis for this invariant form, we get an embedding of $Q_8$ into the group $O(2,\Bbb R)$ of orthogonal $2\times 2$ matrices. But in $O(2,\Bbb R)$ there are only two elements of order$~4$.

Answer 3

Suppose that the quaternion group can be embedded into $M_2(\mathbb{R})$. Then it is isomorphic to a finite subgroup of $GL_2(\mathbb{R})$. Since all finite subgroups of $GL_2(\mathbb{R}$ have a faithful real character of degree $2$, this would apply to the quaternion group, too. But this is not the case, see here, a contradiction. Actually, the question has been answered implicitly here: Finite Subgroups of GL(n,R).

Answer 4

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 ${\mathbb R}$.