## Return to Answer

13 Added the nilpotent case

(Edit: the arguments below build on observations by several other people in this thread.)

Case: $2 = 0$ in $R$. R$. As has been observed elsewhere, in this case$G = \text{SL}_3(\mathbb{F}_2)$embeds into$\text{SL}_3(R)$. Let$g \in \text{SL}_2(R)$be an element of order dividing$4$and trace$r$. Then$g^4 = 1$and$g^2 = rg + 1$, hence$g^4 = r^2 g^2 + 1$and$r^2 g^2 = 0$. This gives$r^4 g^4 = 0$, hence$r^4 = 0$. But$r^2 g^2 = r^3 g + r^2 = 0$, hence$r^4 g = r^3 = 0$, hence$r^3 g = r^2 = 0$, and this is both necessary and sufficient. The elements of$R$which square to zero form a nilpotent ideal$I$. It follows that if$G$embeds into$\text{SL}_2(R)$, the image of any elements of order$4$in$\text{SL}_2(R/I)$must, by the above computation, have order dividing$2$. In particular the image of$G$is not isomorphic to$G$, so it must be trivial since$G$is simple. Hence the image of$G$in$\text{SL}_2(R)$consists only of matrices congruent to the identity$\bmod I$. But any such matrix has trace squaring to zero, hence order dividing$4$, which contradicts the existence of elements of order$7$in$G$. So no such embedding exists. Case:$2$is nilpotent in$R$. It is still true in this case that$SL_3({\mathbb F}_2)$embeds into$\text{SL}_3(R)$, because the two complex 3-dimensional irreducible representations of$SL_3({\mathbb F}_2)=\text{PSL}_2({\mathbb F}_7)$are realisable over${\mathbb Z}_2$(they need$(1\pm\sqrt{-7})/2$which are in${\mathbb Z}_2$), and$R$is a${\mathbb Z}_2$-algebra. However, as explained in the follow-up question,$SL_3({\mathbb F}_2)$is not a subgroup of$\text{SL}_2(R)$for any$R$. Case:$2$is not a zero divisor in$R$. R$. As has been observed elsewhere, in this case $G = S_4$ embeds into $\text{SL}_3(R)$. Let $g \in \text{SL}_2(R)$ be an element of order dividing $2$ and trace $r$. Then $g^2 = 1$ and $g^2 - rg + 1 = 0$, hence $rg = 2$. Squaring gives $r^2 = 4$, so $r$ is also not a zero divisor. It follows that $g$ must be a scalar multiple of the identity, hence central. But $S_4$ contains elements of order $2$ which are not central. (As does $S_3$. Hence this argument also shows, as in Tim Dokchitser's answer, that $S_3$ does not embed into $\text{SL}_2(R)$ in this case.)

In particular the above arguments cover the case that $2$ is invertible, as well as the case that $R$ is an integral domain. The remaining case is that $2 \neq 0$but , and it is a zero divisor but is not nilpotent.

One may hope that the reduction to Artinian rings as in the follow-up question and the fact that we know the answer both when $2$ is invertible and for ${\mathbb Z}_2$-algebras can actually finish this off.

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(Edit: the arguments below build on observations by several other people in this thread.)

Case: $2 = 0$ in $R$. As has been observed elsewhere, in this case $G = \text{SL}_3(\mathbb{F}_2)$ embeds into $\text{SL}_3(R)$. Let $g \in \text{SL}_2(R)$ be an element of order dividing $4$ and trace $r$. Then $g^4 = 1$ and $g^2 = rg + 1$, hence $g^4 = r^2 g^2 + 1$ and $r^2 g^2 = 0$. This gives $r^4 g^4 = 0$, hence $r^4 = 0$. But $r^2 g^2 = r^3 g + r^2 = 0$, hence $r^4 g = r^3 = 0$, hence $r^3 g = r^2 = 0$, and this is both necessary and sufficient.

The elements of $R$ which square to zero form a nilpotent ideal $I$. It follows that if $G$ embeds into $\text{SL}_2(R)$, the image of any elements of order $4$ in $\text{SL}_2(R/I)$ must, by the above computation, have order dividing $2$. In particular the image of $G$ is not isomorphic to $G$, so it must be trivial since $G$ is simple. Hence the image of $G$ in $\text{SL}_2(R)$ consists only of matrices congruent to the identity $\bmod I$. But any such matrix has trace squaring to zero, hence order dividing $4$, which contradicts the existence of elements of order $7$ in $G$. So no such embedding exists.

Case: $2$ is not a zero divisor in $R$. As has been observed elsewhere, in this case $G = S_4$ embeds into $\text{SL}_3(R)$. Let $g \in \text{SL}_2(R)$ be an element of order dividing $2$ and trace $r$. Then $g^2 = 1$ and $g^2 - rg + 1 = 0$, hence $rg = 2$. Squaring gives $r^2 = 4$, so $r$ is also not a zero divisor. It follows that $g$ must be a scalar multiple of the identity, hence central. But $S_4$ contains elements of order $2$ which are not central. (As does $S_3$. Hence this argument also shows, as in Tim Dokchitser's answer, that $S_3$ does not embed into $\text{SL}_2(R)$ in this case.)

In particular the above arguments cover the case that $2$ is invertible, as well as the case that $R$ is an integral domain. The remaining case is that $2 \neq 0$ but it is a zero divisor.

One quick observation. $S_3$ can embed into $\text{SL}_2(R)$ when $2$ is a zero divisor in $R$. For example, in $\mathbb{Z}/6\mathbb{Z}$, the embedding

$$(12) \mapsto \left[ \begin{array}{cc} 1 & 3 \\ 0 & 1 \end{array} \right], (23) \mapsto \left[ \begin{array}{cc} 1 & 0 \\ 3 & 1 \end{array} \right]$$

works. More generally, this embedding works with $3$ replaced by $x$ whenever $x \neq 0$ and $2x = x^3 + x = 0$ in $R$.

11 deleted 4 characters in body

(Edit: the arguments below build on observations by several other people in this thread.)

Case: $2 = 0$ in $R$. As has been observed elsewhere, in this case $G = \text{SL}_3(\mathbb{F}_2)$ embeds into $\text{SL}_3(R)$. Let $g \in \text{SL}_2(R)$ be an element of order dividing $4$ and trace $r$. Then $g^4 = 1$ and $g^2 = rg + 1$, hence $g^4 = r^2 g^2 + 1$ and $r^2 g^2 = 0$. This gives $r^4 g^4 = 0$, hence $r^4 = 0$. But $r^2 g^2 = r^3 g + r^2 = 0$, hence $r^4 g = r^3 = 0$, hence $r^3 g = r^2 = 0$, and this is both necessary and sufficient.

The elements of $R$ which square to zero form a nilpotent ideal $I$. It follows that if $G$ embeds into $\text{SL}_2(R)$, the image of any elements of order $4$ in $\text{SL}_2(R/I)$ must, by the above computation, have order dividing $2$. In particular the image of $G$ is not isomorphic to $G$, so it must be trivial since $G$ is simple. Hence the image of $G$ in $\text{SL}_2(R)$ consists only of matrices congruent to the identity $\bmod I$. But any such matrix has trace squaring to zero, hence order dividing $4$, which contradicts the existence of elements of order $7$ in $G$. So no such embedding exists.

Case: $2$ is not a zero divisor in $R$. As has been observed elsewhere, in this case $G = S_4$ embeds into $\text{SL}_3(R)$. Let $g \in \text{SL}_2(R)$ be an element of order dividing $2$ and trace $r$. Then $g^2 = 1$ and $g^2 - rg + 1 = 0$, hence $rg = 2$. Squaring gives $r^2 = 4$, so $r$ is also not a zero divisor. It follows that $g$ must be a scalar multiple of the identity, hence central. But $S_4$ contains elements of order $2$ which are not central. (As does $S_3$. Hence this argument also shows, as in Tim Dokchitser's answer, that $S_3$ does not embed into $\text{SL}_2(R)$ in this case.)

In particular the above arguments cover the case that $2$ is invertible, as well as the case that $R$ is an integral domain. The remaining case is that $2 \neq 0$ but it is a zero divisor.

One quick observation. $S_3$ can embed into $\text{SL}_2(R)$ when $2$ is not a zero divisor in $R$. For example, in $\mathbb{Z}/6\mathbb{Z}$, the embedding

$$(12) \mapsto \left[ \begin{array}{cc} 1 & 3 \\ 0 & 1 \end{array} \right], (23) \mapsto \left[ \begin{array}{cc} 1 & 0 \\ 3 & 1 \end{array} \right]$$

works. More generally, this embedding works with $3$ replaced by $x$ whenever $x \neq 0$ and $2x = x^3 + x = 0$ in $R$.

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