TITLE: Isomorphic general linear groups implies isomorphic fields?
QUESTION [47 upvotes]: 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.

REPLY [40 votes]: 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 $q>2$.