We prove that if $G$ is a finite simple group which is the unit group of a
ring, then $G$ is isomorphic to either (a) a cyclic group of order 2; (b) a
cyclic group of prime order $2^k -1$ for some $k$; or (c) a projective special
linear group $PSL_n(\mathbb{F}_2)$ for some $n \geq 3$. Moreover, these groups
do (trivially) all occur as unit groups. We deduce this classification from a
more general result, which holds for groups $G$ with no non-trivial normal
2-subgroup

Davis, Christopher

Occhipinti, Tommy

English

arXiv

We prove that if G is a finite simple group which is the unit group of a ring, then G is isomorphic to either (a) a cyclic group of order 2; (b) a cyclic group of prime order 2^k −1 for some k; or (c) a projective special linear group PSLn(F2) for some n ≥ 3. Moreover, these groups do all occur as unit groups. We deduce this classification from a more general result, which holds for groups G with no non-trivial normal 2-subgroup

Davis, Christopher James

Copenhagen University Research Information System

u n i ve r s i t y  o f  co pe n h ag e n  Københavns UniversitetWhich finite simple groups are unit groups?Davis, Christopher James; Occhipinti, TommyPublished in:Journal of Pure and Applied AlgebraDOI:10.1016/j.jpaa.2013.08.013Publication date:2014Document versionEarly version, also known as pre-printCitation for published version (APA):Davis, C. J., & Occhipinti, T. (2014). Which finite simple groups are unit groups? Journal of Pure and AppliedAlgebra, 218(4), 743-744. https://doi.org/10.1016/j.jpaa.2013.08.013Download date: 02. Feb. 2020WHICH FINITE SIMPLE GROUPS ARE UNIT GROUPS?CHRISTOPHER DAVIS AND TOMMY OCCHIPINTIAbstract. We prove that if G is a finite simple group which is the unit group of a ring, then G is isomorphicto either (a) a cyclic group of order 2; (b) a cyclic group of prime order 2k−1 for some k; or (c) a projectivespecial linear group PSLn(F2) for some n ≥ 3. Moreover, these groups do all occur as unit groups. Wededuce this classification from a more general result, which holds for groups G with no non-trivial normal2-subgroup.Throughout this paper, rings will be assumed to be unital, but not necessarily commutative, and ringhomomorphisms send 1 to 1. The finite groups G of odd order which occur as unit groups of rings weredetermined in [3]. We will prove similar results for a more general class of groups; the description of thisclass of groups uses the following.Definition 1. For a finite group G, the p-core of G is the largest normal p-subgroup of G. We denote thissubgroup by Op(G). It is the intersection of all Sylow p-subgroups of G.We now state the main result. The authors1 are most grateful to the anonymous referee for our earlierpaper [2], who recognized that one of the results proved in that paper could be strengthened into the following.Theorem 2. Let G denote a finite group such that O2(G) = {1} and such that G is isomorphic to the unitgroup of a ring R. ThenG ∼= GLn1(F2k1 )× · · · ×GLnr (F2kr ).Before proving Theorem 2, we record the following corollary.Corollary 3. The finite simple groups which occur as unit groups of rings are precisely the groups(a) Z/2Z,(b) Z/pZ for a Mersenne prime p = 2k − 1,(c) PSLn(F2) for n ≥ 3.Proof. If G is a finite simple group, then either O2(G) = {1} or O2(G) = G. If O2(G) = G, then G is a2-group, and because we are assuming G is simple, we must have G ∼= Z/2Z, which for instance is isomorphicto the unit group of Z.Hence assume G is a finite simple group which is isomorphic to the unit group of a ring and further assumeO2(G) = {1}. By Theorem 2, we knowG ∼= GLn1(F2k1 )× · · · ×GLnr (F2kr ).These groups all occur as unit groups of the corresponding products of matrix rings, so we are reduced todetermining which of them are simple; this forcesG ∼= GLn(F2k).If n > 1 and k > 1, then the subgroup of invertible scalar matrices forms a nontrivial normal subgroup.Hence two possibilities remain. If n = 1, then GL1(F2k) is cyclic of order 2k − 1; such a group is simple ifand only if its order is prime. If k = 1, then GLn(F2) = PSLn(F2). For the case k = 1, n = 2, we havePSL2(F2) ∼= S3 (see for example [4, Section 3.3.1]); this group is not simple. For the cases k = 1, n ≥ 3, it iswell-known that PSLn(F2) is simple (see for example [4, Section 3.3.2]). This completes the proof. Date: August 6, 2013.1The authors also thank Colin Adams, John F. Dillon, Dennis Eichhorn, Noam Elkies, Kiran Kedlaya, Charles Toll, andRyan Vinroot for many useful discussions.1Remark 4. The simple groups A8 and PSL2(F7) also occur as unit groups. This follows immediately fromthe exceptional isomorphismsA8 ∼= PSL4(F2) and PSL2(F7) ∼= PSL3(F2).See for instance [4, Section 3.12].Having recorded the above consequences of the main result, we now gather the preliminary results usedin its proof. We begin with the following observation.Lemma 5. Let G denote a finite group with O2(G) = {1}, and let R denote a ring with R× ∼= G. Then Rhas characteristic 2.Proof. The elements 1 and −1 are units in R and are in the center of R, hence are in the center of R×. Bythe assumption O2(G) = {1}, the center of G cannot contain any elements of order 2. Hence 1 = −1. Lemma 6. Keep notation as in Lemma 5, and fix an isomorphism R× ∼= G. Because R has characteristic 2,we have a natural mapϕ : F2[G]→ Rextending the fixed embedding of G into R. The image of ϕ is a ring with unit group isomorphic to G.Proof. Write S for the image of ϕ. On one hand, we have that S× ⊆ R× ∼= G. On the other hand, theinduced map ϕ : G→ S× → R× is surjective. This shows that the unit group of S is isomorphic to G. Lemma 7. Let R denote a finite ring of characteristic 2. If J ⊆ R is a two-sided ideal such that J2 = 0,then 1 + J is a normal elementary abelian 2-subgroup of R×.Proof. Note that for any j, k ∈ J and r ∈ R×, we have• (1 + j)2 = 1 + j2 = 1;• (1 + j)(1 + k) = 1 + j + k + jk = 1 + j + k = (1 + k)(1 + j);• r(1 + j)r−1 = 1 + rjr−1 ∈ 1 + J .The first of these calculations shows that 1 + J is a subset of R×, and the three calculations together showthat it is a normal elementary abelian 2-group. We now use these preliminary results to prove our main theorem.Proof of Theorem 2. By Lemma 6, we may assume R is a finite ring (and is in particular artinian) and hascharacteristic 2. Let J denote a two-sided ideal of R such that J2 = 0. By Lemma 7, the set 1 + J is anormal 2-subgroup of R×, and so by the assumption O2(G) = {1}, we have J = {0}. Thus the ring R hasno non-zero two-sided ideals J with J2 = 0, and hence R has no non-zero two-sided nilpotent ideals. By [1,Theorem 5.4.5], the artinian ring R is semisimple. By Wedderburn’s Theorem [1, Theorem 5.3.4], we haveR ∼= Mn1(D1)× · · · ×Mnr (Dr)for some n1, . . . , nr ≥ 1 and some division algebras D1, . . . , Dr. Our ring R is finite and hence each Di isfinite. By another theorem of Wedderburn [1, Theorem 3.8.6], we have that each Di is a finite field. Finally,because the ring R has characteristic 2, each field Di has characteristic 2. This completes the proof. References[1] P. M. Cohn. Algebra. Vol. 2. John Wiley & Sons Ltd., Chichester, second edition, 1989.[2] Christopher Davis and Tommy Occhipinti. Which alternating and symmetric groups are unit groups?http://arxiv.org/abs/1304.0143.[3] S. Z. Ditor. On the group of units of a ring. Amer. Math. Monthly, 78:522–523, 1971.[4] R. Wilson. The Finite Simple Groups. Graduate Texts in Mathematics. Springer, 2009.University of California, Irvine, Dept of Mathematics, Irvine, CA 92697Current address: University of Copenhagen, Dept of Mathematical Sciences, Universitetsparken 5, 2100 København Ø,DenmarkE-mail address: davis@math.ku.dkUniversity of California, Irvine, Dept of Mathematics, Irvine, CA 92697Current address: Carleton College, Dept of Mathematics, Northfield, MN 55057E-mail address: tocchipinti@carleton.edu2

Which finite simple groups are unit groups?

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