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edited Nov 30 2010 at 17:26 |
(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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edited Nov 29 2010 at 20:59 |
(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 $R$.
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edited Nov 29 2010 at 12:32 |
(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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edited Nov 29 2010 at 12:25 |
(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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edited Nov 29 2010 at 12:12 |
(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.
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edited Nov 28 2010 at 17:35 |
(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.
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.
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edited Nov 28 2010 at 17:08 |
(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 $r^2 g^2 = r^3 g^2 g + r^2 = 0$, hence $r^4 g^2 g = r^3 = 0$, hence $r^3 g^2 g = r^2 = 0$, 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 $2$ 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.
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.
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edited Nov 28 2010 at 16:26 |
(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 $r^2 g^2 = r^3 g^2 + r^2 = 0$, hence $r^4 g^2 = r^3 = 0$, hence $r^3 g^2 = r^2 = 0$, 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 $2$ 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.
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.
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edited Nov 28 2010 at 16:06 |
Some CW (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 $4$ 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 is impossible if gives $R$ is an integral domainr^4 g^4 = 0$, hence $r^4 = 0$. Kevin Buzzard shows But $r^2 g^2 = r^3 g^2 + r^2 = 0$, hence $r^4 g^2 = r^3 = 0$, hence $r^3 g^2 = r^2 = 0$, and this when 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 characteristic 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 $2$ 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 commentsidentity $\bmod I$. If But any such matrix has trace squaring to zero, hence order $4$, which contradicts the characteristic is existence of elements of order $2$, then we proceed as follows7$ in $G$. So no such embedding exists.If 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 $g \in \text{SL}_2(R)$ satisfies be an element of order $g^4 2$ and trace $r$. Then $g^2 = 1$ then and $(g g^2 - 1)^4 rg + 1 = 0$, hence $rg = 2$. Squaring gives $r^2 = 4$, so the characteristic polynomial of $r$ is also not a zero divisor. It follows that $g$ must be $g^2 = 1$a scalar multiple of the identity, hence central. We conclude that But $\text{SL}_2(R)$ has no S_4$ contains elements of order $4$. But 2$ which are not central. In particular the element $\left[ \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right]$ (for example) in $\text{SL}_3(R)$ has order above arguments cover the case that $4$. If 2$ is invertible, as well as the subring generated by case that $1 \in R$ is an integral domain. The remaining case is that $\mathbb{Z}/n\mathbb{Z}$ one might want to look at 2 \neq 0$ but it is a Borel subgroup of $\text{SL}_3(\mathbb{Z}/n\mathbb{Z})$, which seems too large to fit into $\text{SL}_2(R)$ to mezero divisor.
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edited Nov 28 2010 at 15:17 |
Some CW observations.
This is impossible if $R$ is an integral domain. Kevin Buzzard shows this when the characteristic is not $2$ in the comments. If the characteristic is $2$, then we proceed as follows. If $g \in \text{SL}_2(R)$ satisfies $g^4 = 1$ then $(g - 1)^4 = 0$, so the characteristic polynomial of $g$ is $g = 1$ or must be $g^2 = 1$. We conclude that $\text{SL}_2(R)$ has no elements of order $4$. But the element $\left[ \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right]$ (for example) in $\text{SL}_3(R)$ has order $4$.
If the subring generated by $1 \in R$ is $\mathbb{Z}/n\mathbb{Z}$ one might want to look at a Borel subgroup of $\text{SL}_3(\mathbb{Z}/n\mathbb{Z})$, which seems too large to fit into $\text{SL}_2(R)$ to me.
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edited Nov 28 2010 at 14:06 |
Some CW observations.
This is impossible if $R$ is an integral domain. Kevin Buzzard shows this when the characteristic is not $2$ in the comments. If the characteristic is $2$, then we proceed as follows. If $g \in \text{SL}_2(R)$ satisfies $g^4 = 1$ then $(g - 1)^4 = 0$, so the characteristic polynomial of $g$ is $g = 1$ or $g^2 = 1$. We conclude that $\text{SL}_2(R)$ has no elements of order $4$. But the element $\left[ \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right]$ (for example) in $\text{SL}_3(R)$ has order $4$.
If the subring generated by $1 \in R$ is $\mathbb{Z}/n\mathbb{Z}$ one might want to look at a Borel subgroup of $\text{SL}_3(\mathbb{Z}/n\mathbb{Z})$, which seems too large to fit into $\text{SL}_2(R)$ to me.
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edited Nov 27 2010 at 0:06 |
Some CW observations.
This is impossible if $R$ is an integral domain. Kevin Buzzard shows this when the characteristic is not $2$ in the comments. If the characteristic is $2$, then we proceed as follows. If $g \in \text{SL}_2(R)$ satisfies $g^4 = 1$ then $(g - 1)^4 = 0$, so the characteristic polynomial of $g$ is either $g = 1$ or $g^2 = 1$. We conclude that $\text{SL}_2(R)$ has no elements of order $4$. But the element $\left[ \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right]$ (for example) in $\text{SL}_3(R)$ has order $4$.
If the subring generated by $1 \in R$ is $\mathbb{Z}/n\mathbb{Z}$ one might want to look at the a Borel subgroup of upper triangular matrices with $1$s on the diagonal in $\text{SL}_3(R)$. In particular this group has an element of order $n^2$, \text{SL}_3(\mathbb{Z}/n\mathbb{Z})$, which I think is seems too large to fit into $\text{SL}_2(R)$.\text{SL}_2(R)$ to me.
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answered Nov 26 2010 at 20:32 |
Some CW observations.
This is impossible if $R$ is an integral domain. Kevin Buzzard shows this when the characteristic is not $2$ in the comments. If the characteristic is $2$, then we proceed as follows. If $g \in \text{SL}_2(R)$ satisfies $g^4 = 1$ then $(g - 1)^4 = 0$, so the characteristic polynomial of $g$ is either $g = 1$ or $g^2 = 1$. We conclude that $\text{SL}_2(R)$ has no elements of order $4$. But the element $\left[ \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right]$ (for example) in $\text{SL}_3(R)$ has order $4$.
If the subring generated by $1 \in R$ is $\mathbb{Z}/n\mathbb{Z}$ one might want to look at the subgroup of upper triangular matrices with $1$s on the diagonal in $\text{SL}_3(R)$. In particular this group has an element of order $n^2$, which I think is too large to fit into $\text{SL}_2(R)$.
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