$\newcommand{\op}[1]{\operatorname{#1}}\def\GL{\op{GL}}$I am interested in the subgroups of $\GL(n,K)$ of the form $\GL(n/s, E)$ for $E$ some extension field of $K$ of degree $s$. These arise in the following manner: If $V = K^n$ then $V \cong E^{n/s}$ as $K$-vector spaces, so $\GL(n/s, E)$ acts on $V$. Since an $E$-linear action is in particular $K$-linear, this amounts to an embedding of $\GL(n/s, E)$ in $\GL(n, F)$.

Said a bit less clumsily, the assertion is that if $E$ is a subfield of $\op{End}(V)$ of degree $s$ over $K$ then $\op{Aut}_E(V)$ (the invertible elements of the centralizer $C_E(V)$) is isomorphic to $\GL(n/s, E)$, and $E$ can be recovered as the center of $\op{Aut}_E(V)$. Thus embedded copies of $\GL(n/s, E)$ in $\GL(n, K)$ are parameterized by homomorphisms $E \to \op{End}(V)$.

(There may be other subgroups of $\GL(n, K)$ abstractly isomorphic to general linear groups, but we ignore those: we want only the naturally embedded ones.)

My question is about this lattice of subgroups. Specifically: $\def\Aut{\op{Aut}}$

What can we say about $\langle\op{Aut}_E(V), \op{Aut}_F(V)\rangle$ inside $\op{Aut}_{E \cap F}(V)$?

For example, there are many copies of $\GL(1, \mathbf{C})$ embedded in $\GL(2,\mathbf{R})$: the only restriction is that the matrix representing $i$ should have square $-I$. Let $E$ and $F$ be the embedded copies of $\mathbf{C}$ in $\op{End}(\mathbf{R}^2)$ corresponding to $i \mapsto (0,-1;1,0)$ and $i\mapsto(-1,-2;1,1)$, respectively. Then one can check that $\Aut_E(V)$ and $\Aut_F(V)$ generate $\GL^+(2,\mathbf{R})$, the group of all matrices with positive determinant, which has index 2 in $\Aut_{E\cap F}(V) = \GL(2,\mathbf{R})$.

On the other hand if we replace $\mathbf{R}$ with $\mathbf{Q}$ and $\mathbf{C}$ with $\mathbf{Q}(i)$, then $\Aut_E(V)$ and $\Aut_F(V)$ are both contained in the subgroup of matrices having determinant of the form $a^2+b^2$. I suspect that they generate this subgroup.

Optimistically, are these determinant-type obstructions the only obstructions?

Assume $E \cap F = K$. Is it true that $\langle \Aut_E(V), \Aut_F(V)\rangle$ contains $\op{SL}(n,K)$?

I am actually most interested in the case of finite fields, and in this case I actually know a proof that $\langle \Aut_E(V), \Aut_F(V)\rangle = \Aut_{E\cap F}(V)$. But the proof is fairly non-elementary and it doesn't feel very natural to me, so I'm hoping to find a better one.

In the case of finite fields, is there a simple proof that $\langle \Aut_E(V), \Aut_F(V) \rangle = \Aut_{E\cap F}(V)$?

Any comments (even if it's just a different way of looking at the problem, an interesting special case, a relevant reference, etc.) are much appreciated!

$\newcommand{\op}[1]{\operatorname{#1}}\def\GL{\op{GL}}$I am interested in the subgroups of $\GL(n,K)$ of the form $\GL(n/s, E)$ for $E$ some extension field of $K$ of degree $s$. These arise in the following manner: If $V = K^n$ then $V \cong E^{n/s}$ as $K$-vector spaces, so $\GL(n/s, E)$ acts on $V$. Since an $E$-linear action is in particular $K$-linear, this amounts to an embedding of $\GL(n/s, E)$ in $\GL(n, K)$.

Said a bit less clumsily, the assertion is that if $E$ is a subfield of $\op{End}(V)$ of degree $s$ over $K$ then $\op{Aut}_E(V)$ (the invertible elements of the centralizer $C_E(V)$) is isomorphic to $\GL(n/s, E)$, and $E$ can be recovered as the center of $\op{Aut}_E(V)$. Thus embedded copies of $\GL(n/s, E)$ in $\GL(n, K)$ are parameterized by homomorphisms $E \to \op{End}(V)$.

(There may be other subgroups of $\GL(n, K)$ abstractly isomorphic to general linear groups, but we ignore those: we want only the naturally embedded ones.)

My question is about this lattice of subgroups. Specifically: $\def\Aut{\op{Aut}}$

What can we say about $\langle\op{Aut}_E(V), \op{Aut}_F(V)\rangle$ inside $\op{Aut}_{E \cap F}(V)$?

For example, there are many copies of $\GL(1, \mathbf{C})$ embedded in $\GL(2,\mathbf{R})$: the only restriction is that the matrix representing $i$ should have square $-I$. Let $E$ and $F$ be the embedded copies of $\mathbf{C}$ in $\op{End}(\mathbf{R}^2)$ corresponding to $i \mapsto (0,-1;1,0)$ and $i\mapsto(-1,-2;1,1)$, respectively. Then one can check that $\Aut_E(V)$ and $\Aut_F(V)$ generate $\GL^+(2,\mathbf{R})$, the group of all matrices with positive determinant, which has index 2 in $\Aut_{E\cap F}(V) = \GL(2,\mathbf{R})$.

On the other hand if we replace $\mathbf{R}$ with $\mathbf{Q}$ and $\mathbf{C}$ with $\mathbf{Q}(i)$, then $\Aut_E(V)$ and $\Aut_F(V)$ are both contained in the subgroup of matrices having determinant of the form $a^2+b^2$. I suspect that they generate this subgroup.

Optimistically, are these determinant-type obstructions the only obstructions?

Assume $E \cap F = K$. Is it true that $\langle \Aut_E(V), \Aut_F(V)\rangle$ contains $\op{SL}(n,K)$?

I am actually most interested in the case of finite fields, and in this case I actually know a proof that $\langle \Aut_E(V), \Aut_F(V)\rangle = \Aut_{E\cap F}(V)$. But the proof is fairly non-elementary and it doesn't feel very natural to me, so I'm hoping to find a better one.

In the case of finite fields, is there a simple proof that $\langle \Aut_E(V), \Aut_F(V) \rangle = \Aut_{E\cap F}(V)$?

Any comments (even if it's just a different way of looking at the problem, an interesting special case, a relevant reference, etc.) are much appreciated!

$\newcommand{\op}[1]{\operatorname{#1}}\def\GL{\op{GL}}$I am interested in the subgroups of $\GL(n,K)$ of the form $\GL(n/s, E)$ for $E$ some extension field of $K$ of degree $s$. These arise in the following manner: If $V = K^n$ then $V \cong E^{n/s}$ as $K$-vector spaces, so $\GL(n/s, E)$ acts on $V$. Since an $E$-linear action is in particular $K$-linear, this amounts to an embedding of $\GL(n/s, E)$ in $\GL(n, F)$.

Said a bit less clumsily, the assertion is that if $E$ is a subfield of $\op{End}(V)$ of degree $s$ over $K$ then $\op{Aut}_E(V)$ (the invertible elements of the centralizer $C_E(V)$) is isomorphic to $\GL(n/s, E)$, and $E$ can be recovered as the center of $\Aut_E(V)$. Thus embedded copies of $\GL(n/s, E)$ in $\GL(n, K)$ are parameterized by homomorphisms $E \to \op{End}(V)$.

(There may be other subgroups of $\GL(n, K)$ abstractly isomorphic to general linear groups, but we ignore those: we want only the naturally embedded ones.)

My question is about this lattice of subgroups. Specifically: $\def\Aut{\op{Aut}}$

What can we say about $\langle\op{Aut}_E(V), \op{Aut}_F(V)\rangle$ inside $\op{Aut}_{E \cap F}(V)$?

For example, there are many copies of $\GL(1, \mathbf{C})$ embedded in $\GL(2,\mathbf{R})$: the only restriction is that the matrix representing $i$ should have square $-I$. Let $E$ and $F$ be the embedded copies of $\mathbf{C}$ in $\op{End}(\mathbf{R}^2)$ corresponding to $i \mapsto (0,-1;1,0)$ and $i\mapsto(-1,-2;1,1)$, respectively. Then one can check that $\Aut_E(V)$ and $\Aut_F(V)$ generate $\GL^+(2,\mathbf{R})$, the group of all matrices with positive determinant, which has index 2 in $\Aut_{E\cap F}(V) = \GL(2,\mathbf{R})$.

On the other hand if we replace $\mathbf{R}$ with $\mathbf{Q}$ and $\mathbf{C}$ with $\mathbf{Q}(i)$, then $\Aut_E(V)$ and $\Aut_F(V)$ are both contained in the subgroup of matrices having determinant of the form $a^2+b^2$. I suspect that they generate this subgroup.

Optimistically, are these determinant-type obstructions the only obstructions?

Assume $E \cap F = K$. Is it true that $\langle \Aut_E(V), \Aut_F(V)\rangle$ contains $\op{SL}(n,K)$?

I am actually most interested in the case of finite fields, and in this case I actually know a proof that $\langle \Aut_E(V), \Aut_F(V)\rangle = \Aut_{E\cap F}(V)$. But the proof is fairly non-elementary and it doesn't feel very natural to me, so I'm hoping to find a better one.

In the case of finite fields, is there a simple proof that $\langle \Aut_E(V), \Aut_F(V) \rangle = \Aut_{E\cap F}(V)$?

Any comments (even if it's just a different way of looking at the problem, an interesting special case, a relevant reference, etc.) are much appreciated!

$\newcommand{\op}[1]{\operatorname{#1}}\def\GL{\op{GL}}$I am interested in the subgroups of $\GL(n,K)$ of the form $\GL(n/s, E)$ for $E$ some extension field of $K$ of degree $s$. These arise in the following manner: If $V = K^n$ then $V \cong E^{n/s}$ as $K$-vector spaces, so $\GL(n/s, E)$ acts on $V$. Since an $E$-linear action is in particular $K$-linear, this amounts to an embedding of $\GL(n/s, E)$ in $\GL(n, F)$.

Said a bit less clumsily, the assertion is that if $E$ is a subfield of $\op{End}(V)$ of degree $s$ over $K$ then $\op{Aut}_E(V)$ (the invertible elements of the centralizer $C_E(V)$) is isomorphic to $\GL(n/s, E)$, and $E$ can be recovered as the center of $\op{Aut}_E(V)$. Thus embedded copies of $\GL(n/s, E)$ in $\GL(n, K)$ are parameterized by homomorphisms $E \to \op{End}(V)$.

(There may be other subgroups of $\GL(n, K)$ abstractly isomorphic to general linear groups, but we ignore those: we want only the naturally embedded ones.)

My question is about this lattice of subgroups. Specifically: $\def\Aut{\op{Aut}}$

What can we say about $\langle\op{Aut}_E(V), \op{Aut}_F(V)\rangle$ inside $\op{Aut}_{E \cap F}(V)$?

For example, there are many copies of $\GL(1, \mathbf{C})$ embedded in $\GL(2,\mathbf{R})$: the only restriction is that the matrix representing $i$ should have square $-I$. Let $E$ and $F$ be the embedded copies of $\mathbf{C}$ in $\op{End}(\mathbf{R}^2)$ corresponding to $i \mapsto (0,-1;1,0)$ and $i\mapsto(-1,-2;1,1)$, respectively. Then one can check that $\Aut_E(V)$ and $\Aut_F(V)$ generate $\GL^+(2,\mathbf{R})$, the group of all matrices with positive determinant, which has index 2 in $\Aut_{E\cap F}(V) = \GL(2,\mathbf{R})$.

On the other hand if we replace $\mathbf{R}$ with $\mathbf{Q}$ and $\mathbf{C}$ with $\mathbf{Q}(i)$, then $\Aut_E(V)$ and $\Aut_F(V)$ are both contained in the subgroup of matrices having determinant of the form $a^2+b^2$. I suspect that they generate this subgroup.

Optimistically, are these determinant-type obstructions the only obstructions?

Assume $E \cap F = K$. Is it true that $\langle \Aut_E(V), \Aut_F(V)\rangle$ contains $\op{SL}(n,K)$?

I am actually most interested in the case of finite fields, and in this case I actually know a proof that $\langle \Aut_E(V), \Aut_F(V)\rangle = \Aut_{E\cap F}(V)$. But the proof is fairly non-elementary and it doesn't feel very natural to me, so I'm hoping to find a better one.

In the case of finite fields, is there a simple proof that $\langle \Aut_E(V), \Aut_F(V) \rangle = \Aut_{E\cap F}(V)$?

Any comments (even if it's just a different way of looking at the problem, an interesting special case, a relevant reference, etc.) are much appreciated!