Does $SL_2(K) \simeq SL_2(L)$ imply $K\simeq L$?

Question

Let $K$ and $L$ be two fields. Assume characteristics are not 2. I can show in a quite elementary way that if the statement $SL_2(K) \simeq SL_2(L) \implies K \simeq L$ holds, then for $n \geq 2$, the statement $SL_n(K) \simeq SL_n(L) \implies K \simeq L$ holds. But I do not know how to prove this for $n=2$ in its full generality. We can of course assume the groups i.e. the fields are infinite, otherwise counting the number of elements should be enough.

On the other hand by using any non-central diagonal element as a parameter, one can define the field $K$ in the group $SL_2(K)$ as follows. Let $t_0$ be one such element. Let $T=C_{SL_2(K)}(t_0) \simeq K^*$ (torus). We may regard $T$ as the group of diagonal matrices with determinant 1. There are exactly two abelian subgroups $H$ of $SL_2(K)$ of the form $\langle h^T\cup\{1\} \rangle$ for any $1\neq h \in H$ and with the property that $H \cap Z(SL_2(K))=1$ , the strictly upper and lower triangular matrices, say $U$ and $V$ (unipotent) respectively. (Because $x = (1+x/2)^2 - 1^2 - (x/2)^2$ for any $x\in K$, see below.) They are both isomorphic to the addive group of $K$. Choose one of them, say $U$. The choice does not matter as the automorphism "transpose inverse" interchanges them fixing $T$. Denote the elements of $T$ by $t(x)$ where $x\in K^*$ and elements of $U$ by $u(y)$ where $y\in K$. Then $T$ acts on $U$ as follows $u(y)^{t(x)} = u(x^2y)$. Thus we get the subfield of $K$ generated by the squares. But since $x = (1+x/2)^2 - 1^2 - (x/2)^2$ for any $x\in K$, the subfield generated by the squares is $K$ itself. Thus the field $K$ is definable with one parameter, namely $t_0$. (Except that the group does not know the unit element 1 of the field, we only get an affine version of a field; to fix 1 of the field $K$ we need one more parameter, but this is irrelevant to us). It follows that in the group $SL_2(L)$ both fields $K$ and $L$ are definable.

In particular if the automorphism takes a non-central diagonalizable element of $SL_2(K)$ to a non-central diagonalizable element of $SL_2(L)$, then we will necessarily have $K\simeq L$. This will be so if we can distinguish diagonalizable elements of $SL_2(K)$ from its non-diagonalizable semisimple elements (i.e. diagonalizable in the algebraic closure) in a group theoretic way. If $K$ and $L$ are algebraically closed, all the semisimple elements will be diagonalizable, so there will be no problem, in this case $K$ will be isomorphic to $L$.

Answer 1

I will write the argument in my comment in more detail. (I have now edited it to avoid citing Clifford's Theorem.)

Let $G_K = {\rm SL}(2,K)$ for a field $K$, and let $T_K$ and $U_K$ be respectively the subgroups of diagonal matrices and of upper unitriangular matrices. Then $B_K = T_KU_K$ is a Borel subgroup.

Let $\phi:G_K \to G_L$ be an isomorphism for fields $K,L$. I claim that $\phi(B_K)$ is conjugate in $G_L$ to $B_L$. Then, when the characteristics of $K$ and $L$ are not 2, we can use the argument in the post to conclude that $K \cong L$.

First observe that, since $G_K = B_K \cup B_KxB_K$ for any $x \in G_K \setminus B_K$ (or, equivalently,$G_K$ acts doubly transitively on on the cosets of $B_K$), $B_K$ is a maximal subgroup of $G_K$, so $\phi(B_K)$ is maximal in $G_L$.

Now $B_K$ is the stabilizer in $G_K$ of a 1-dimensional subspace of $K^2$ in the natural action. Any reducible subgroup of $G_K$ must fix a 1-dimensional subspace of $K^2$, and $G_K$ acts transitively on the set of all such subspaces, and so there is a unique conjugacy class in $G_K$ of reducible maximal subgroups, and $B_K$ is one of these.

So, if $\phi(B_K)$ is reducible, then it is conjugate to $B_L$, and we are done.

So suppose, for a contradiction, that $\phi(B_K)$ is irreducible. We will prove that $B_K$ has an abelian subgroup of index 2, which is false, at least when $|K| > 5$, and we can assume that $K$ and $L$ are both infinite.

Now $\phi(U_K)$ is an abelian normal subgroup of $\phi(B_K)$. Suppose first that $\phi(U_K)$ is reducible over $L$. Then under the action of $\phi(U_K)$, $L^2$ has an invariant $1$-dimensional subspace $U$. Since $\phi(B_K)$ is irreducible, there exists $g \in \phi(B_K)$ with $Ug \ne U$ and then $Ug$ is also invariant under $\phi(U_K)$ and $L^2 = U \oplus Ug$.

If $U$ and $Ug$ were isomorphic as $\phi(U_K)$-modules, then the action of $\phi(U_K)$ would be scalar, and $\phi(U_K)$ would be central in $\phi(K)$, which it isn't. So they are not isomorphic.

Then $U$ and $Ug$ are the only $\phi(U_K)$ invariant subspaces of $L^2$, since any other such subspace would have the form $\{(u,\tau(u)g) : u \in U \}$ where $\tau:U \to Ug$ is a $\phi(U_K)$-module isomorphism. So $U$ and $Ug$ are fixed or interchanged by all elements of $\phi(B_K)$, and then the subgroup of $\phi(U_K)$ that fixes them is an abelian subgroup of index 2, contradiction. (The above argument is just Clifford's Theorem, but I cannot apply that directly because it is usually only stated for normal subgroups of finite index.)

So $\phi(U_K)$ is irreducible over $L$. Choose $1 \ne g \in \phi(U_K)$, and let $\lambda^{\pm 1}$ be the eigenvalues of $g$ over the algebraic closure $\bar{L}$ of $L$.

If $\lambda=\lambda^{-1}$ then $\lambda = \pm 1 \in L$, and so $g$ fixes a necessarily unique subspace of $L^2$, which is then fixed by all of $\phi(U_K)$, contradicting its irreducibility over $L$.

So $\lambda \ne \lambda^{-1}$, but then $g$ fixes exactly two 1-dimensional subspaces of $\bar{L}^2$ (i.e. the eigenspaces associated with $\lambda$ and $\lambda^{-1}$), which are fixed or interchanged by all of $\phi(B_K)$, and then the above argument shows that $\phi(B_K)$ has an abelian subgroup of index 2, contradiction.