$GL_2(R)$ and $\mathbb{A}_4$

Question

So, i have to prove that $\mathbb{A}_4$ is not isomorphic to a subgroup $G$ of $GL_2(R)$.

Here is a "hint" (are previuos part of the same problem that i was thinking, so i supose should help):

If $H$ is a subgroup of $G$ of $GL_2(R)$isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$, there would be 2 matrices $A,B$ such that $AB=BA$, $A^2=B^2=I$, in wich case $A,B$ are simultaneosly diagonalizables, so $H$ is the conjugated of a subgroup of diagonal matrices...wich i dont know how to use to prove the affirmation about hte isomorphism

Answer 1

The observation that $A$ and $B$ are simultaneously diagonalizable does help. Proffering the following route.

Assume that a subgroup $G\le GL_2(\Bbb{R})$ isomorphic to $A_4$ would exist. Because we can replace $G$ by any of its conjugates, we may as well assume that the matrices $$ A=\left(\begin{array}{rr}-1&0\\0&1\end{array}\right)\qquad\text{and}\qquad B=\left(\begin{array}{rr}1&0\\0&-1\end{array}\right) $$ are both elements of $G$ (ask, if you need details for this). Then comes the hint.

1. Let $K$ be the subgroup generated by $A$ and $B$. Let $N=N_{GL_2}(K)$ be the normalizer of $K$ in $GL_2(\Bbb{R})$. Show that $N$ consists of the so called monomial matrices $$ N=\left\{\left(\begin{array}{rr}a&0\\0&b\end{array}\right)\,\big\vert\, a,b\in\Bbb{R}^*\right\}\cup \left\{\left(\begin{array}{rr}0&a\\b&0\end{array}\right)\,\big\vert\, a,b\in\Bbb{R}^*\right\}. $$
2. Show that there are no elements of order $3$ in $N$.
3. Show that $G$ cannot exist.