Isomorphic general linear groups implies isomorphic fields? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T15:16:21Z http://mathoverflow.net/feeds/question/106838 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106838/isomorphic-general-linear-groups-implies-isomorphic-fields Isomorphic general linear groups implies isomorphic fields? Vipul Naik 2012-09-10T19:45:22Z 2012-09-11T15:30:23Z <p>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\$?</p> <p>I'm also interested in the corresponding question for the special linear group in place of the general linear group.</p> <p>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.</p> <p>NOTE 2: It's possible to use the examples of NOTE 1 to construct non-isomorphic fields whose additive groups are isomorphic <em>and</em> whose multiplicative groups are isomorphic.</p> http://mathoverflow.net/questions/106838/isomorphic-general-linear-groups-implies-isomorphic-fields/106841#106841 Answer by Jim Humphreys for Isomorphic general linear groups implies isomorphic fields? Jim Humphreys 2012-09-10T20:05:53Z 2012-09-11T13:06:43Z <p>The answer to the question is yes, though I don't have all the old literature at my fingertips. This kind of question for various classes of linear groups has a long history in the study of homomorphisms and isomorphisms of classical groups and then other algebraic groups (van der Waerden, Dieudonne, ...) The most comprehensive treatment was given by Borel and Tits in their Ann. of Math 97 (1973) paper, but emphasizing simple types rather than general reductive groups. Anyway, for general linear groups the ideas occur much earlier and also involve the uniqueness of <code>\$n\$</code>. (As you point out, the case <code>\$n=1\$</code> has a different flavor.) I'll check the sources, but you could also work back from the references in Borel-Tits.</p> <p>P.S. Note that any isomorphism (of abstract groups) between two general linear groups induces an isomorphism of the derived groups. Given <code>\$n&gt;1\$</code>, these are <em>special linear groups</em> and fit well into the older or newer sources I mentioned. (Probably there is enough detail in the 1928 Hamburg paper by Schreier and van der Waerden to settle your question, but I confess I've never gone back that far in the literature.)</p> <p>ADDED: One relatively modern reference I should point out is <em>Lectures on Linear Groups</em> by O.T. O'Meara, CBMS 22, Amer. Math. Soc., 1974. While O'Meara's own research interest at the time was in the direction of linear groups over various rings of interest, these lecture notes also incorporate older work over fields. See in particular his Sections 5.5-5.6 for theorems most relevant to the question asked here. </p> <p>Roughly speaking, the central concern in these isomorphism theorems is what happens to unipotent elements (classically, transvections and the like). In the setting of classical matrix groups, the original techniques rely on the underlying geometry of the situation. But in the broader treatment by Borel-Tits the structure theory of reductive groups (Jordan decomposition, Bruhat decomposition, etc.) plays the most prominent role. For general or special linear groups, it's hard for me to judge what approach is really "simplest".</p> http://mathoverflow.net/questions/106838/isomorphic-general-linear-groups-implies-isomorphic-fields/106916#106916 Answer by Mikhail Borovoi for Isomorphic general linear groups implies isomorphic fields? Mikhail Borovoi 2012-09-11T14:22:06Z 2012-09-11T15:30:23Z <p>The answer is "yes", see below.</p> <p>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.</p> <p>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.</p> <p>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\$.</p>