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Edit: I think the answer to Question 2 is yes. Let $R = \mathbb{Z}[x, y, z]/I$ where

$$I = (2x, 2y, 2z, z - xz^2, x - x^2 z, z - yz^2, y - y^2 z).$$

Then I think the embedding

$$(12) \mapsto \left[ \begin{array}{cc} 1 & x \\ 0 & 1 \end{array} \right], (23) \mapsto \left[ \begin{array}{cc} 1 & 0 \\ z & 1 \end{array} \right], (34) \mapsto \left[ \begin{array}{cc} 1 & y \\ 0 & 1 \end{array} \right]$$

works. I have not yet verified that $I$ is as small as I think it is, though. I think would be enough to verify that $I$ does not contain $xz, yz$, or $x - y$.

Here's what I know about Question 1 from tinkering with the linked question for awhile.

• If $R$ is an algebraically closed field of characteristic zero then the classification is the same as the classification of the finite subgroups of $\text{SL}_2(\mathbb{C})$: the cyclic groups, the dicyclic groups, and the binary polyhedral groups. Hence if $R$ is an integral domain of characteristic zero then any finite subgroup must be on this list.

• If $R$ is an integral domain of characteristic $2$ then $\text{SL}_2(R)$ has no nontrivial elements of order $2$. This rules out any finite group of even order. Whoops: what I meant to say is that $\text{SL}_2(R)$ has no nontrivial elements of order $4$. Actually it's enough that $R$ is reduced. This rules out any finite group with such an element. Slightly more generally, if $2 = 0$ and $I$ is the ideal of $R$ consisting of elements squaring to zero, then an element of $\text{SL}_2(R)$ has order $4$ if and only if its trace lands in $I$, so there are no elements of order $4$ in $\text{SL}_2(R/I)$. If $G$ is a non-abelian simple group in $\text{SL}_2(R)$ with an element of order $4$, then its image in $\text{SL}_2(R/I)$ is not faithful, so must be trivial, but that means every element of $G$ has trace landing in $I$, hence order $4$; contradiction.

• If $2$ is not a zero divisor in $R$ (in particular, if $2 \neq 0$ and $R$ is an integral domain) then any element of order $2$ in $\text{SL}_2(R)$ is a scalar multiple of the identity, in particular central. This rules out many finite groups, including non-abelian simple groups by Feit-Thompson.

• In general, let $I$ be the ideal of $R$ consisting of the elements annihilated by $2$. The above bullet point shows that any element of order $2$ in the quotient $\text{SL}_2(R/I)$ must be a scalar multiple of the identity, in particular central, in this quotient. Hence if $G$ is a non-abelian simple subgroup of $\text{SL}_2(R)$, it must land in the kernel of the above map: the subgroup congruent to the identity $\bmod I$.

I strongly suspect that $\text{SL}_2(R)$ can't contain a non-abelian simple group in general but have not yet been able to prove it. It just seems as if there is not enough room to be "too noncommutative." For example, the first three cases above all rule out $\text{SL}_3(\mathbb{F}_2) \sim \text{PSL}_2(\mathbb{F}_7)$.

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Edit: I think the answer to Question 2 is yes. Let $R = \mathbb{Z}[x, y, z]/I$ where

$$I = (2x, 2y, 2z, z - xz^2, x - x^2 z, z - yz^2, y - y^2 z).$$

Then I think the embedding

$$(12) \mapsto \left[ \begin{array}{cc} 1 & x \\ 0 & 1 \end{array} \right], (23) \mapsto \left[ \begin{array}{cc} 1 & 0 \\ z & 1 \end{array} \right], (34) \mapsto \left[ \begin{array}{cc} 1 & y \\ 0 & 1 \end{array} \right]$$

works. I have not yet verified that $I$ is as small as I think it is, though. I think would be enough to verify that $I$ does not contain $xz, yz$, or $x - y$.

Here's what I know about Question 1 from tinkering with the linked question for awhile.

• If $R$ is an algebraically closed field of characteristic zero then the classification is the same as the classification of the finite subgroups of $\text{SL}_2(\mathbb{C})$: the cyclic groups, the dicyclic groups, and the binary polyhedral groups. Hence if $R$ is an integral domain of characteristic zero then any finite subgroup must be on this list.

• If $R$ is an integral domain of characteristic $2$ then $\text{SL}_2(R)$ has no nontrivial elements of order $2$. This rules out any finite group of even order. Whoops: what I meant to say is that $\text{SL}_2(R)$ has no nontrivial elements of order $4$. Actually it's enough that $R$ is reduced. This rules out any finite group with such an element. Slightly more generally, if $2 = 0$ and $I$ is the ideal of $R$ consisting of elements squaring to zero, then an element of $\text{SL}_2(R)$ has order $4$ if and only if its trace lands in $I$, so there are no elements of order $4$ in $\text{SL}_2(R/I)$. If $G$ is a non-abelian simple group in $\text{SL}_2(R)$ with an element of order $4$, then its image in $\text{SL}_2(R/I)$ is not faithful, so must be trivial, but that means every element of $G$ has trace landing in $I$, hence order $4$; contradiction.

• If $2$ is not a zero divisor in $R$ (in particular, if $2 \neq 0$ and $R$ is an integral domain) then any element of order $2$ in $\text{SL}_2(R)$ is a scalar multiple of the identity, in particular central. This rules out many finite groups, including non-abelian simple groups by Feit-Thompson.

• In general, let $I$ be the ideal of $R$ consisting of the elements annihilated by $2$. The above bullet point shows that any element of order $2$ in the quotient $\text{SL}_2(R/I)$ must be a scalar multiple of the identity, in particular central, in this quotient. Hence if $G$ is a non-abelian simple subgroup of $\text{SL}_2(R)$, it must land in the kernel of the above map: the subgroup congruent to the identity $\bmod I$.

I strongly suspect that $\text{SL}_2(R)$ can't contain a non-abelian simple group in general but have not yet been able to prove it. It just seems as if there is not enough room to be "too noncommutative." For example, the first three cases above all rule out $\text{SL}_3(\mathbb{F}_2) \sim \text{PSL}_2(\mathbb{F}_7)$.

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Edit: I think the answer to Question 2 is yes. Let $R = \mathbb{Z}[x, y, z]/I$ where

$$I = (2x, 2y, 2z, z - xz^2, x - x^2 z, z - yz^2, y - y^2 z).$$

Then I think the embedding

$$(12) \mapsto \left[ \begin{array}{cc} 1 & x \\ 0 & 1 \end{array} \right], (23) \mapsto \left[ \begin{array}{cc} 1 & 0 \\ z & 1 \end{array} \right], (34) \mapsto \left[ \begin{array}{cc} 1 & y \\ 0 & 1 \end{array} \right]$$

works. I have not yet verified that $I$ is as small as I think it is, though.

Here's what I know about Question 1 from tinkering with the linked question for awhile.

• If $R$ is an algebraically closed field of characteristic zero then the classification is the same as the classification of the finite subgroups of $\text{SL}_2(\mathbb{C})$: the cyclic groups, the dicyclic groups, and the binary polyhedral groups. Hence if $R$ is an integral domain of characteristic zero then any finite subgroup must be on this list.

• If $R$ is an integral domain of characteristic $2$ then $\text{SL}_2(R)$ has no nontrivial elements of order $2$. This rules out any finite group of even order. Whoops: what I meant to say is that $\text{SL}_2(R)$ has no nontrivial elements of order $4$. Actually it's enough that $R$ is reduced. This rules out any finite group with such an element. Slightly more generally, if $2 = 0$ and $I$ is the ideal of $R$ consisting of elements squaring to zero, then an element of $\text{SL}_2(R)$ has order $4$ if and only if its trace lands in $I$, so there are no elements of order $4$ in $\text{SL}_2(R/I)$. If $G$ is a non-abelian simple group in $\text{SL}_2(R)$ with an element of order $4$, then its image in $\text{SL}_2(R/I)$ is not faithful, so must be trivial, but that means every element of $G$ has trace landing in $I$, hence order $4$; contradiction.

• If $2$ is not a zero divisor in $R$ (in particular, if $2 \neq 0$ and $R$ is an integral domain) then any element of order $2$ in $\text{SL}_2(R)$ is a scalar multiple of the identity, in particular central. This rules out many finite groups, including non-abelian simple groups by Feit-Thompson.

• In general, let $I$ be the ideal of $R$ consisting of the elements annihilated by $2$. The above bullet point shows that any element of order $2$ in the quotient $\text{SL}_2(R/I)$ must be a scalar multiple of the identity, in particular central, in this quotient. Hence if $G$ is a non-abelian simple subgroup of $\text{SL}_2(R)$, it must land in the kernel of the above map: the subgroup congruent to the identity $\bmod I$.

I strongly suspect that $\text{SL}_2(R)$ can't contain a non-abelian simple group in general but have not yet been able to prove it. It just seems as if there is not enough room to be "too noncommutative." For example, the first three cases above all rule out $\text{SL}_3(\mathbb{F}_2) \sim \text{PSL}_2(\mathbb{F}_7)$.

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