Suppose $n > 1$ is a natural number. Suppose $K$ and $L$ are fields such that the general linear groups of degree $n$ over them are isomorphic, i.e., $GL(n,K) \cong GL(n,L)$ as groups. Is it necessarily true that $K \cong L$?


I'm also interested in the corresponding question for the special linear group in place of the general linear group.


NOTE 1: The statement is false for $n = 1$, because $GL(1,K) \cong K^\ast$ and non-isomorphic fields can have isomorphic multiplicative groups. For instance, all countable subfields of $\mathbb{R}$ that are closed under the operation of taking rational powers of positive elements have isomorphic multiplicative groups.


NOTE 2: It's possible to use the examples of NOTE 1 to construct non-isomorphic fields whose additive groups are isomorphic *and* whose multiplicative groups are isomorphic.



Isomorphic general linear groups implies isomorphic fields?
The answer is "yes", see below.


Dieudonné in his book "La géométrie des groupes classiques"
considers the abstract group $SL\_n(K)$ for a field $K$, not necessarily commutative,
and writes $PSL\_n(K)$ for $SL\_n(K)$ modulo the center.
In Ch. IV, Section 9, he considers the question whether $PSL\_n(K)$ can be isomorphic to $PSL\_m(K')$ for $n\ge 2,\ m\ge 2$.
He writes that they can be isomorphic only for $n=m$, except for $PSL\_2(\mathbb{F}\_7)$ and $PSL\_3(\mathbb{F}\_2)$.
If $n=m>2$, then the isomorphism is possible only if $K$ and $K'$ are isomorphic or anti-isomorphic.
The same is true for $m=n=2$ if both $K$ and $K'$ are commutative, except for the case $K=\mathbb{F}\_4$,
$K'=\mathbb{F}\_5$. Dieudonné gives ideas of proof and references to Schreier and van der Waerden (1928), to his paper "On the automorphisms of classical groups"
 in Mem. AMS No. 2 (1951) and to the paper of Hua L.-K. and Wan in J. Chinese Math. Soc. 2 (1953), 1-32.


This answers affirmatively the question for $SL\_n$, because if $SL\_n(K)\cong SL\_n(K')$, then $PSL\_n(K)\cong PSL\_n(K')$. In the case $n=2$, $K=\mathbb{F}\_4$,
$K'=\mathbb{F}\_5$, the orders $|SL\_2(\mathbb{F}\_4)|=60$ and $|SL\_2(\mathbb{F}\_5)|=120$ are different, and therefore these groups are not isomorphic.


This also answers affirmatively the question for $GL\_n$, because $SL\_n(K)$ is the commutator subgroup of $GL\_n(K)$, except for $GL\_2(\mathbb{F\_2})$, see Dieudonné, Ch. II, Section 1. In the case $n=2$, $K=\mathbb{F}\_2$, we have $|GL\_2(\mathbb{F}\_2)|=6$ , which is less than $|GL\_2(\mathbb{F}\_q)|=q(q-1)(q^2-1)$ for any $q=p^r>2$, hence 
$GL\_2(\mathbb{F}\_2)\not\cong GL\_2(\mathbb{F}\_q)$ for $q>2$.


