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The length and depth of algebraic groups
The length and depth of algebraic groups
Burness, Timothy; Liebeck, Martin; Shalev, Aner
2018-06-06 00:00:00
Math. Z. https://doi.org/10.1007/s00209-018-2101-6 Mathematische Zeitschrift 1 2 Timothy C. Burness · Martin W. Liebeck · Aner Shalev Received: 10 January 2018 / Accepted: 12 April 2018 © The Author(s) 2018 Abstract Let G be a connected algebraic group. An unreﬁnable chain of G is a chain of subgroups G = G > G > ··· > G = 1, where each G is a maximal connected 0 1 t i subgroup of G . We introduce the notion of the length (respectively, depth) of G,deﬁned i −1 as the maximal (respectively, minimal) length of such a chain. Working over an algebraically closed ﬁeld, we calculate the length of a connected group G in terms of the dimension of its unipotent radical R (G) and the dimension of a Borel subgroup B of the reductive quotient G/R (G). In particular, a simple algebraic group of rank r has length dim B +r, which gives a natural extension of a theorem of Solomon and Turull on ﬁnite quasisimple groups of Lie type. We then deduce that the length of any connected algebraic group G exceeds dim G. We also study the depth of simple algebraic groups. In characteristic zero, we show that the depth of such a group is at most 6 (this bound is sharp). In the positive characteristic setting, we calculate the exact depth of each exceptional algebraic group and we prove that the depth of a classical group (over a ﬁxed algebraically closed ﬁeld of positive characteristic) tends to inﬁnity with the rank of the group. Finally we study the chain difference of an algebraic group, which is the difference between its length and its depth. In particular we prove that, for any connected algebraic group G with soluble radical R(G), the dimension of G/R(G) is bounded above in terms of the chain difference of G. The third author acknowledges the hospitality and support of Imperial College, London, while part of this work was carried out. He also acknowledges the support of ISF Grant 1117/13 and the Vinik chair of mathematics which he holds. B Timothy C. Burness t.burness@bristol.ac.uk Martin W. Liebeck m.liebeck@imperial.ac.uk Aner Shalev shalev@math.huji.ac.il School of Mathematics, University of Bristol, Bristol BS8 1TW, UK Department of Mathematics, Imperial College, London SW7 2BZ, UK Institute of Mathematics, Hebrew University, 91904 Jerusalem, Israel 123 T. C. Burness et al. Mathematics Subject Classiﬁcation Primary 20E32, 20E15 ; Secondary 20G15, 20E28 1 Introduction The length of a ﬁnite group G, denoted by l(G), is the maximum length of a chain of subgroups of G. This interesting invariant was the subject of several papers by Janko and Harada [10,13,14] in the 1960s, culminating in Harada’s description of the ﬁnite simple groups of length at most 7 in [10]. In more recent years, the notion of length has arisen naturally in several different contexts. For example, Babai [1] considered the length of the symmetric group S in relation to the computational complexity of algorithms for ﬁnite permutation groups (a precise formula for l(S ) was later determined by Cameron, Solomon and Turull in [6]). Motivated by applications to ﬁxed-point-free automorphisms of ﬁnite soluble groups, Seitz, Solomon and Turull studied the length of ﬁnite groups of Lie type in a series of papers in the early 1990s [21,23,24]. Let us highlight one of their main results, [24, Theorem A*], which states that if G = G (p ) is a ﬁnite quasisimple group of Lie type and k is sufﬁciently large (with respect to the characteristic p), then l(G) = l(B) + r, (1) where B is a Borel subgroup of G and r is the twisted Lie rank of G. The dual notion of the depth of a ﬁnite group G, denoted by λ(G), is the minimal length of a chain of subgroups G = G > G > ··· > G > G = 1, 0 1 t −1 t where each G is a maximal subgroup of G . This invariant was studied by Kohler [15] i i −1 for ﬁnite soluble groups and we refer the reader to more recent work of Shareshian and Woodroofe [22] for further results in the context of lattice theory. In [4] we proved several results on the depth of ﬁnite simple groups and we studied the relationship between the length and depth of simple groups (see [5] for further results on the length and depth of ﬁnite groups). For instance, [4, Theorem 1] classiﬁes the simple groups of depth 3 (it is easy to see that λ(G) 3 for every non-abelian simple group G)and [4, Theorem 2] shows that alternating groups have bounded depth (more precisely, λ(A ) 23 for all n, whereas l(A ) n n tends to inﬁnity with n). Upper bounds on the depth of each simple group of Lie type over F are presented in [4, Theorem 4]; the bounds are given in terms of k,where q = p with p a prime. In this paper, we extend the above notions of length and depth to connected algebraic groups over algebraically closed ﬁelds. Let G be a connected algebraic group over an alge- braically closed ﬁeld of characteristic p 0. An unreﬁnable chain of length t of G is a chain of subgroups G = G > G > ··· > G > G = 1, 0 1 t −1 t where each G is a maximal closed connected subgroup of G (that is, G is maximal i i −1 i among the proper connected subgroups of G ). We deﬁne the length of G, denoted by i −1 l(G), to be the maximal length of an unreﬁnable chain. Similarly, the depth λ(G) of G is the minimal length of such a chain. Notice that we impose the condition that the subgroups in an unreﬁnable chain are connected, which seems to be the most natural (and interesting) deﬁnition in this setting. In the statements of our main results, and for the remainder of the paper, we assume that the given algebraic group is connected and the underlying ﬁeld is algebraically closed (unless 123 The length and depth of algebraic groups stated otherwise). Also note that our results are independent of any choice of isogeny type. Our ﬁrst result concerns the length of an algebraic group. Theorem 1 Let G be an algebraic group and let B be a Borel subgroup of the reductive group G = G/R (G).Then l(G) = dim R (G) + dim B + r, where r is the semisimple rank of G. Corollary 2 Let G be a simple algebraic group of rank r and let B be a Borel subgroup of G. Then l(G) = dim B + r. Moreover, every unreﬁnable chain of G of maximum length includes a maximal parabolic subgroup. The last sentence of the corollary is justiﬁed in Remark 3.1. By Lemma 2.2, the solubility of B implies that l(B) = dim B, so Corollary 2 is the algebraic group analogue of the aforementioned result of Solomon and Turull [24, Theorem A*] for ﬁnite quasisimple groups (see (1) above). Next, we relate the length of arbitrary algebraic groups G to their dimension. We clearly have l(G) dim G. Theorem 3 Let G be an algebraic group. (i) l(G)> dim G. (ii) l(G) = dim G if and only if G/R(G) A for some t 0,where R(G) is the soluble radical of G. The lower bound in part (i) of Theorem 3 is essentially best possible. For example, if G = C is a symplectic group of rank r 1, then Corollary 2 implies that l(G) 1 3 1 = + → as r →∞. dim G 2 4r + 2 2 We now turn to the depth of simple algebraic groups. Our ﬁrst result shows that simple algebraic groups in characteristic zero have bounded depth. Theorem 4 Let G be a simple algebraic group in characteristic zero. Then 3if G = A ⎪ 1 5if G = A (r 3, r = 6), B , D or E r 3 r 6 λ(G) = ⎪ 6if G = A 4 in all other cases. Our next result shows that the depth of simple groups in the positive characteristic setting is rather different. In particular, the depth can be arbitrarily large. To state this result, we need some additional notation. Given a prime p, deﬁne a sequence e (p) (n ∈ N) as follows: e (p) = p,and for l > 1, e (p) e (p) = p . l+1 Now deﬁne a function ψ : R → N by ψ (x ) = min (l : e (p) x ) . p l 123 T. C. Burness et al. Theorem 5 Let G be a simple algebraic group in characteristic p > 0 with rank r. (i) If G is an exceptional group then λ(G) 9, with equality if and only if G = E and p = 2. (ii) If G is a classical group, then λ(G) 2(log r ) + 12. (iii) For any G, we have λ(G) ψ (r ). In particular, λ(G) →∞ as r →∞. Part (i) of Theorem 5 is an immediate corollary of Theorem 3.3, which gives the exact depth of each exceptional algebraic group. For parts (ii) and (iii), see Theorems 3.4 and 3.10, respectively. We also give an example (Example 3.11) to show that the lower bound ψ (r ) in (iii) is of roughly the correct order of magnitude. By a well-known theorem of Iwasawa [12], the length and depth of a ﬁnite group G are equal if and only if G is supersoluble. This result does not extend directly to algebraic groups. However, it follows from our results on length and depth that the only simple algebraic group with λ(G) = l(G) is G = A (see Lemma 3.12). More generally, we prove the following result on arbitrary connected groups with this property, which can be viewed as a partial analogue of Iwasawa’s theorem. Theorem 6 Let G be an algebraic group satisfying λ(G) = l(G) and let R(G) be the radical of G. Then either G is soluble, or G/R(G) A . In fact it follows from our arguments that λ(G) = l(G) if and only if λ(G) = dim G. This is proved in Sect. 3.7. Note that the converse is false: for example, if G = UA , a semidirect product where U is a nontrivial irreducible KA -module, then A is maximal 1 1 in G,so λ(G) 1 + λ(A ) = 4, while l(G) = dim U + l(A )> 4. On the other hand, 1 1 if G = U × A ,orif G is a nonsplit extension of the irreducible module U by A ,then 1 1 λ(G) = l(G). More generally, we can consider the chain difference of G, which is deﬁned by cd(G) = l(G) − λ(G). This invariant was studied for ﬁnite simple groups. See [3,11,19] for the study of ﬁnite simple groups of chain difference one, and Corollary 9 in [4], where we bound the length of a ﬁnite simple group in terms of its chain difference. For algebraic groups we prove a stronger result, without assuming simplicity. Theorem 7 Let G be an algebraic group and set G = G/R(G).Then dim G (2 + o(1)) cd(G), where o(1) = o (1).Moreprecisely, cd(G) dim G 2cd(G) + 40 400 + 2cd(G) + 800. This result will be proved in Sect. 3.8. We also consider the chain ratio cr(G) = l(G)/λ(G) of an algebraic group G,and show in Sect. 4 that in contrast to the chain difference, the dimension of G/R(G) is not in general bounded in terms of cr(G). 2 Preliminaries As stated in Sect. 1, for the remainder of the paper we assume G is a connected algebraic group over an algebraically closed ﬁeld (unless stated otherwise). We start with the following elementary observation. 123 The length and depth of algebraic groups Lemma 2.1 Let G be an algebraic group and let N be a connected normal subgroup. (i) λ(G) l(G) dim G. (ii) l(G) = l(N ) + l(G/N ). (iii) λ(G/N ) λ(G) λ(N ) + λ(G/N ). Proof Parts (i) and (iii) are obvious, and part (ii) is proved just as [6, Lemma 2.1]. Recall that if U is a connected unipotent algebraic group, then the Frattini subgroup (U ) of U is the intersection of the closed subgroups of U of codimension 1 (see [9]). Lemma 2.2 If G is a soluble algebraic group, then λ(G) = l(G) = dim G. Proof It is sufﬁcient to show that any connected maximal subgroup M of G has codimension 1. Write G = UT,where U = R (G) and T is a maximal torus. If U M then M = US, where S is a connected maximal subgroup of T , and the result follows since dim S = dim T − 1. Now assume U M,so M = (M ∩ U )T and M ∩ U is a maximal T -invariant subgroup of U.Now (U ) M, so by factoring out (U ) we can assume that (U ) = 1. Then U K ,an n-dimensional vector space over the underlying algebraically closed ﬁeld K (see [9, Proposition 1]). Moreover, T acts linearly on U , and since T is diagonalisable on V , a maximal T -invariant subspace has codimension 1 (this is proved in much greater generality in [18, Theorem B]). Hence M has codimension 1 in G in this case also. Lemma 2.3 Let G be an algebraic group and let m ∈{1, 2, 3}.Then λ(G) = m if and only if dim G = m. Proof The case m = 1isobvious,soassume m ∈{2, 3}.If λ(G) = 2then G has a maximal T or U subgroup and clearly G is soluble, so dim G = 2 by Lemma 2.2.Conversely, if 1 1 dim G = 2then G is soluble and once again the result follows from Lemma 2.2. Now assume dim G = 3. By the m = 2 case already proved, λ(G) 3. If G is soluble, then Lemma 2.2 implies that λ(G) = 3, otherwise G = A and λ(G) = 3since A > U T > T > 1 1 1 1 1 is unreﬁnable (here we write U for a unipotent group of dimension k, and similarly T for k k a k-dimensional torus). Finally, suppose λ(G) = 3 and assume that G is insoluble. If G is reductive, it has a maximal connected subgroup of depth 2, hence of dimension 2, and the only possibility is G = A . Otherwise, let U = R (G) be the unipotent radical of G.Since G 1 u is insoluble, the previous sentence implies that G/U = A .But G has a maximal subgroup of depth 2, which must be soluble of dimension 2. This is clearly not possible. Remark 2.4 Notice that the conclusion of Lemma 2.3 does not extend to integers m > 3. For example, if r 2 then the symplectic group C has a maximal A subgroup in characteristic r 1 0, so there are depth 4 algebraic groups of arbitrarily large dimension. Lemma 2.5 Let G = U.L be an algebraic group, where U and L are nontrivial connected subgroups of G, with U normal. Then λ(G) 1 + λ(L). Proof By Lemma 2.1(iii), λ(G) λ(L). Assume for a contradiction that λ(G) = λ(L) = t, and let G = G > G > ··· > G > G =1(2) t t −1 1 0 123 T. C. Burness et al. be an unreﬁnable chain. If L = G U/U L,thenwemusthave L < L for all i,since i i i i +1 otherwise the depth of L would be less than t. This means that G = G U > G U > ··· > G U > G U = U t t −1 1 0 is a chain of connected subgroups. But then if we choose i minimal such that U G ,we have G < G U < G , contradicting the unreﬁnability of (2). i i i +1 3 Proofs 3.1 Proof of Theorem 1 Let G be an algebraic group. The proof goes by induction on dim G. For dim G = 1, the result is obvious. Write U = R (G). By Lemmas 2.1(ii) and 2.2, l(G) = l(U ) + l(G/U ) = dim U + l(G/U ). If U = 1 then the conclusion follows by induction, so we assume that U = 1; that is, G is reductive. Write G = G ··· G Z, a commuting product with each G simple and 1 t i Z = Z (G) ,and let B be a Borel subgroup of G . Note that B = B ··· B Z is a Borel i i 1 t subgroup of G.Let r be the rank of G ,so r = r is the semisimple rank of G.By i i i Lemma 2.1(ii) we have l(G) = l(G ) + dim Z . If t > 1or Z = 1 we can apply induction to deduce that l(G ) = l(B ) + r for each i,and i i i hence l(G) = dim B + r + dim Z = dim B + r, i i i i as required. Hence we may assume that G is simple of rank r. By considering an unreﬁnable chain passing through B, noting that l(B) = dim B by Lemma 2.2, it follows that l(G) dim B + r. (3) Our goal is to show that equality holds. Let M be a maximal connected subgroup of G with l(M ) = l(G) − 1. By [2, Corollary 3.9], M is either parabolic or reductive. Suppose ﬁrst that M is reductive and let B be a Borel subgroup of M. By induction, l(M ) = dim B + rank(M ). But it is easy to see that dim B < dim B − 1 and thus l(G)< dim B + r. This contradicts (3), so we have reduced to the case where M is a maximal parabolic subgroup. Write M = QL where Q = R (M ) and L is a Levi subgroup. By induction, l(M ) = dim Q + dim B + rank(L ), where B is a Borel subgroup of L.Since B = QB is a Borel subgroup of G, and rank(L ) = L L r − 1, it follows that l(G) = l(M ) + 1 = dim B + r, as required. This completes the proof of Theorem 1. 123 The length and depth of algebraic groups Remark 3.1 Let G be a simple algebraic group. By the proof of Theorem 1, it follows that every unreﬁnable chain of G of maximum length includes a maximal parabolic subgroup. This gives Corollary 2. 3.2 Proof of Theorem 3 First consider (i). In view of Lemma 2.2, the bound is clear if G is soluble, so assume otherwise. Write G/R(G) = G ··· G , where each G is simple. By applying Corollary 2, 1 t i it is easy to see that l(G )> dim G for each i, so by combining Lemmas 2.1(ii) and 2.2 i i we get 1 1 l(G) = l(R(G)) + l(G )> dim R(G) + dim G dim G i i 2 2 i i as required. An entirely similar argument establishes (ii), noting that l(G ) = dim G if and i i only if G = A (this is easily deduced from Corollary 2; see Lemma 3.12). i 1 3.3 Proof of Theorem 4 Let G be a simple algebraic group over an algebraically closed ﬁeld of characteristic 0. The maximal connected subgroups of G were determined by Dynkin [7,8] and we repeatedly apply these results throughout the proof. To begin with, let us assume G is a classical group of rank r. By Lemma 2.3, λ(G) 3 with equality if and only if G = A ,sowemay assume r 2. As observed by Dynkin [8], G = C has an irreducible maximal subgroup of type A and r 1 thus λ(G) = 4. Similarly, λ(G) = 4if G = B and r = 3. The group G = B needs r 3 special attention because it does not have a maximal A subgroup (indeed, an irreducibly embedded A is contained in a G subgroup of G). One checks that dim M > 3for every 1 2 maximal connected subgroup M of G, so Lemma 2.3 implies that λ(G) 5. In fact, we see that equality holds since B > G > A > U T > T > 1 3 2 1 1 1 1 is an unreﬁnable chain of length 5. Next assume G = D ,so r 3 by simplicity. Here λ(G) 5since G does not have a maximal A subgroup. But G does have an unreﬁnable chain of length 5: D > B > A > U T > T >1if r = 4 r r −1 1 1 1 1 D > A > A > U T > T > 1 4 2 1 1 1 1 and thus λ(G) = 5. To complete the proof for classical groups, suppose G = A and r 2. Here G has a maximal A subgroup if and only if r = 2, so we get λ(G) = 4if r = 2, otherwise λ(G) 5. It is easy to see that λ(G) = 5if r 3and r = 6. Indeed, we have the following unreﬁnable chains: A > B > A > U T > T >1if r 4even, r = 6 r r/2 1 1 1 1 A > C > A > U T > T >1if r 3 odd. r (r +1)/2 1 1 1 1 Note that if r = 6 then the ﬁrst chain is reﬁnable (as noted above, A is non-maximal in B ) 1 3 and we get λ(G) 6via A > B > G > A > U T > T > 1. 6 3 2 1 1 1 1 123 T. C. Burness et al. Table 1 The depth of low rank pA A B G A B C A B C D F 1 2 2 2 3 3 3 4 4 4 4 4 simple algebraic groups 23 6 5 5 6 6 6 9 7 7 7 8 34 5 5 5 6 5 6 5 6 7 6 545 6555656 74 54 56 5 11 4 4 5 > 11 4 We claim that λ(G) = 6 in this case. To see this, let M be a maximal connected subgroup of G.By[8], either M = B or M is a parabolic subgroup of the form U A T , U A A T or 3 6 5 1 10 4 1 1 U A A T (here U denotes a normal unipotent subgroup of dimension k). If M is parabolic, 12 3 2 1 k then Lemma 2.1(iii) implies that λ(M ) λ(A ) for some k ∈{3, 4, 5} and thus λ(M ) 5 by our above work. Since λ(B ) = 5, the claim follows. Finally, let us assume G is an exceptional group. By [7], G has a maximal A subgroup if and only if G = E ,so λ(G) = 4 in these cases. For G = E we have λ(G) 5and 6 6 equality holds since G has a maximal G subgroup (see [7]) and so there is an unreﬁnable chain E > G > A > U T > T > 1. 6 2 1 1 1 1 This completes the proof of Theorem 4. 3.4 Proof of Theorem 5(i) Let G be a simple algebraic group of rank r over an algebraically closed ﬁeld K of charac- teristic p > 0. In this subsection we determine the precise depth of G in the case where G is of exceptional type. We start with a preliminary lemma, which gives the precise depth of the simple algebraic groups of rank at most 4. In Table 1, if the ﬁnal entry c in a column occurs in the row corresponding to p = ,then λ(G) = c for all p . For example, Table 1 indicates that ⎨ 7if p = 2 λ(C ) = 6if p ∈{3, 5, 7} 4if p 11. Lemma 3.2 Let G be a simple algebraic group of rank r 4 in characteristic p > 0.Then λ(G) is given in Table 1. Proof First recall that λ(G) 3, with equality if and only if G = A . Now assume r 2 and let G > M > M > ··· > M = 1 1 t be an unreﬁnable chain of minimal length. Recall that M is either parabolic or reductive. If G is an exceptional group, then the possibilities for M have been determined by Liebeck and Seitz (see [17, Corollary 2(ii)]). Similarly, if G is a classical group with natural module V , then [16, Theorem 1] implies that either M stabilises a proper nontrivial subspace of V,or a tensor product decomposition of the form V = U ⊗ W,or M is simple and V | is an 123 The length and depth of algebraic groups irreducible KM-module with p-restricted highest weight. We refer the reader to [25,Table 5] for a convenient list of the relevant reductive maximal connected subgroups of G. It will be useful to observe that λ(M ) 5if M is a maximal parabolic subgroup of G (this follows immediately from Lemma 2.5). Suppose G = A .If p 3then G has a maximal A subgroup, so λ(G) = 4. On the 2 1 other hand, if p = 2then M = U A T and M ∈{A T , U A } are the only possibilities, 2 1 1 1 1 1 2 1 so λ(G) = 2 + λ(M ) = 6. Next let G = B or C .If p 5then A is a maximal subgroup and thus λ(G) = 4. If 2 2 1 p ∈{2, 3} then dim M > 3 and thus λ(G) 5. In fact, equality holds since there is a chain B > A A > A > U T > T > 1, 2 1 1 1 1 1 1 where A < A A is diagonally embedded. Next assume G = G . In the usual manner, we 1 1 1 2 deduce that λ(G) = 4if p 7, so let us assume p ∈{2, 3, 5}.Heredim M > 3 and thus λ(G) 5. Since G > A A > A > U T > T > 1 2 1 1 1 1 1 1 is unreﬁnable, we conclude that λ(G) = 5. Now assume G = A .Heredim M > 3, so λ(G) 5. If p 3then A > A A > A > U T > T > 1 3 1 1 1 1 1 1 is unreﬁnable and thus λ(G) = 5. Suppose p = 2, so either M = B or M is parabolic. Since λ(B ) = 5 as above, and λ(M ) 5when M is parabolic, it follows that λ(G) = 6. Next suppose G = B ,so λ(G) 5since dim M > 3. Now G has a maximal G 3 2 subgroup, so λ(G) λ(G ) + 1 and thus λ(G) = 5if p 7, otherwise λ(G) 6. By considering the various possibilities for M, it is easy to check that λ(M ) 5when p ∈{2, 3, 5} (we can assume M is reductive, so the possibilities are listed in [25,Table 5]) and we conclude that λ(G) = 6. The case G = C is similar. If p 7then M = A 3 1 and λ(G) = 4. Now assume p ∈{2, 3, 5},so λ(G) 5. If p ∈{3, 5} then we can take M = A A , which gives λ(G) = 5. Finally, if p = 2 then one checks that λ(M ) 5, with 1 1 equality if M = G , hence λ(G) = 6. Suppose G = A .Here λ(G) 5since dim M > 3. If p 5then A > B > A > U T > T > 1 4 2 1 1 1 1 is unreﬁnable and thus λ(G) = 5. If p = 3 then either M = B or M is a parabolic subgroup, whence λ(G) = 6since λ(B ) = 5 as above. Finally, suppose p = 2. Here every maximal connected subgroup of G is parabolic, so we need to consider the depth of P = U A T 1 4 3 1 and P = U A A T . The Levi factor of P is maximal, so P > A T > A is unreﬁnable 2 6 2 1 1 1 1 3 1 3 and thus λ(P ) 8since λ(A ) = 6. Now Lemma 2.5 gives λ(P ) 2 + λ(A ) = 8and 1 3 1 3 λ(P ) 3 + λ(A ) = 9, so λ(P ) = 8and λ(G) = 9. 2 2 1 Next assume G = D .Onceagain, λ(G) 5. If p 5then λ(G) = 5since D > A > A > U T > T > 1 4 2 1 1 1 1 is unreﬁnable. Now assume p ∈{2, 3}.Here λ(G) λ(B ) + 1 = 7 and we claim that equality holds. To see this, ﬁrst observe that M is either a parabolic subgroup, or M = B , A B (p = 3), A or A (p = 2). Note that λ(B ) = 6and λ(A ) = 6 (with p = 2inthe 1 2 2 3 2 latter case). By applying Lemma 2.5, it is also easy to see that λ(M ) 6 in the remaining cases. For example, if M = U A T then λ(M ) 1 + λ(A T ) 7. This justiﬁes the claim. 6 3 1 3 1 123 T. C. Burness et al. Next consider G = B . First observe that λ(G) = 4 if and only if p 11. If p ∈{3, 5, 7} then we can take M = A A , which gives λ(G) = 5. Now assume p = 2. We claim that 1 1 λ(G) = 7. Certainly, λ(G) 7 since there is a chain B > B B > B > A A > A > U T > T > 1. 4 2 2 2 1 1 1 1 1 1 To establish equality, we need to consider the possibilities for M.If M is reductive, then M = D , A B or B B . By our earlier work, λ(D ) = 7and λ(A B ) 1 + λ(B ) = 7. 4 1 3 2 2 4 1 3 3 Similarly, λ(B B ) = 6 and one checks that λ(M ) 6if M is parabolic. The claim follows. 2 2 Now assume G = C . As in the previous case, λ(G) = 4 if and only if p 11. If p ∈{3, 5, 7} then C > A > A A > A > U T > T > 1 4 1 1 1 1 1 1 is unreﬁnable and thus λ(G) 6. In fact, it is easy to see that λ(M ) 5 for every connected maximal subgroup M of G and thus λ(G) = 6. Finally, suppose p = 2. Here λ(G) 7via the chain C > C C > C > A A > A > U T > T > 1. 4 2 2 2 1 1 1 1 1 1 We claim that λ(G) = 7. To see this, we need to show that λ(M ) 6 for every connected maximal subgroup M of G.If M is reductive, then M = C C , A C or D . By combining 2 2 1 3 4 Lemma 2.5 with our earlier work, we have λ(A C ) 1 + λ(C ) = 7and λ(D ) = 7. It is 1 3 3 4 also easy to see that λ(C C ) = 6. It is routine to verify the claim when M is parabolic. For 2 2 instance, if M = U A T then λ(M ) 1 + λ(A T ) 2 + λ(A ) = 8. 10 3 1 3 1 3 To complete the proof of the lemma, we may assume G = F .Here G has a maximal A 4 1 subgroup if and only if p 13, so we may assume p < 13. If p ∈{7, 11} then λ(G) = 5 via the chains F > B > A > U T > T >1if p = 11 4 4 1 1 1 1 F > G > A > U T > T >1if p = 7. 4 2 1 1 1 1 Next assume p ∈{3, 5}.Here λ(G) 6 since there is a chain F > A A > A > A > U T > T > 1. 4 2 2 2 1 1 1 1 By considering the various possibilities for M and using Lemma 2.5 and our earlier work, it is easy to show that λ(M ) 5 and thus λ(G) = 6. Finally, let us assume p = 2. First observe that λ(G) 8 via the chain F > C > C C > C > A A > A > U T > T > 1. 4 4 2 2 2 1 1 1 1 1 1 We claim that λ(G) = 8. To see this, we need to show that λ(M ) 7 for every maximal connected subgroup M of G.If M is reductive, then M = C , B or A A .Asabove,we 4 4 2 2 have λ(B ) = λ(C ) = 7and λ(A A ) 1 + λ(A ) = 7 and the result follows. If M = UL 4 4 2 2 2 is a parabolic subgroup, with unipotent radical U , then Lemma 2.5 gives λ(M ) 1 + λ(L) and it is straightforward to see that λ(L) 6. For example, if M = U A A T then Lemma 20 1 2 1 2.5 yields λ(L) 2 + λ(A ) = 8. The result follows. We are now in a position to prove our main result for exceptional groups. In particular, part (i) of Theorem 5 is an immediate corollary of the following result. In Table 2, we adopt the same conventions as in Table 1. Theorem 3.3 Let G be an exceptional algebraic group in characteristic p > 0.Then λ(G) is given in Table 2. 123 The length and depth of algebraic groups Table 2 The depth of pG F E E E 2 4 6 7 8 exceptional algebraic groups 2 5 8 689 3 5 6 677 5 5 6 555 74 5 5 5 11 5 5 5 13 4 5 5 17 4 5 19 5 > 19 4 Proof In view of Lemma 3.2, we may assume G = E , E or E .Let 6 7 8 G > M > M > ··· > M = 1 1 t be an unreﬁnable chain of minimal length. Recall that the possibilities for M are determined in [17]. First assume G = E and note that λ(G) 5since dim M > 3. If p 5, then G has a maximal A subgroup and λ(A ) = 4 by Lemma 3.2,so λ(G) = 5. Now assume p ∈{2, 3}. 2 2 Here G < G is maximal and thus λ(G) λ(G ) + 1 = 6. We claim that λ(G) = 6. If M 2 2 is reductive, then [17] implies that M ∈{A A , A , G , C (p = 3), F , A G } 1 5 2 4 4 2 2 and it is easy to check that λ(M ) 5. By applying Lemma 2.5, we see that the same conclusion holds when M is parabolic. This justiﬁes the claim. Next assume G = E .Here λ(G) = 4 if and only if p 17. If 5 p 13 then E > A A > A > U T > T > 1 7 1 1 1 1 1 1 is unreﬁnable and thus λ(G) = 5. Now assume p = 3. First observe that λ(G) 7 via the chain E > A G > A A > A A > A > U T > T > 1. 7 1 2 1 2 1 1 1 1 1 1 We claim that λ(G) = 7. To see this, let M be a maximal connected subgroup of G. Suppose M is reductive, in which case M ∈{A D , A , A A , A G , A F , G C }. 1 6 7 2 5 1 2 1 4 2 3 Now λ(D ) 5and λ(A ) 5 (neither group has a maximal A subgroup) and one can 6 5 1 readily check that λ(A ) 6 (the only reductive maximal connected subgroups are C , D 7 4 4 and A C ). Therefore, by applying Lemmas 2.5 and 3.2 we deduce that λ(M ) 6. Similarly, 1 3 one checks that the same bound holds if M is parabolic and the claim follows. Now assume G = E and p = 2. Here there is an unreﬁnable chain E > G C > G G > G > A A > A > U T > T > 1 7 2 3 2 2 2 1 1 1 1 1 1 and thus λ(G) 8. By essentially repeating the above argument for p = 3, it is straight- forward to show that λ(M ) 7 for every maximal connected subgroup M of G, whence λ(G) = 8. 123 T. C. Burness et al. To complete the proof of the theorem, we may assume G = E .If p 23 then λ(G) = 4 since G has a maximal A subgroup. If 5 p 19 then λ(G) = 5 via the chain E > B > A > U T > T > 1. 8 2 1 1 1 1 Now assume p = 3. Here λ(G) 7 since there is a chain E > A > A A > A > A > U T > T > 1. 8 8 2 2 2 1 1 1 1 To see that λ(G) = 7, we need to show that λ(M ) 6. If M is reductive then [17, Corollary 2(ii)] implies that M ∈{D , A E , A , A E , A A , G F } (4) 8 1 7 8 2 6 4 4 2 4 and it is straightforward to show that λ(M ) 6 (recall that λ(A ) = λ(F ) = λ(E ) = 6 4 4 6 and λ(E ) = 7). For example, if M = A then either M is parabolic and λ(M ) 5, or 7 8 1 1 M ∈{B , A A } and λ(M ) = 5. As usual, the bound λ(M ) 6 is easily checked when 1 4 2 2 1 M is parabolic. Finally, let us assume G = E and p = 2. There is an unreﬁnable chain E > D > B > B B > B > A A > A > U T > T > 1 8 8 4 2 2 2 1 1 1 1 1 1 and thus λ(G) 9. To establish equality, we need to show that λ(M ) 8. If M is reductive then (4) holds and we consider each possibility in turn. The bound is clear if M = A E , 1 7 A A or G F since λ(E ) = λ(F ) = 8and λ(A ) = 9. Similarly, λ(A E ) 8since 4 4 2 4 7 4 4 2 6 λ(A ) = λ(E ) = 6. If M = A and M is reductive, then M = A A is the only option 2 6 8 1 1 2 2 and thus λ(M ) 1 + λ(A ) = 7. It is easy to see that λ(M ) 7if M is parabolic, so 1 2 1 1 λ(A ) 8 as required. Similarly, one checks that λ(D ) 8. Finally, if M is a maximal 8 8 parabolic subgroup with Levi factor L, then Lemma 2.5 gives λ(M ) 2 + λ(L ) and it is straightforward to show that λ(L ) 6. The result follows. 3.5 Proof of Thereom 5(ii) Now assume G is a simple classical algebraic group. The next result establishes the upper bound on λ(G) in part (ii) of Theorem 5. Theorem 3.4 If G is a simple classical algebraic group of rank r, then λ(G) 2(log r ) + 12. We partition the proof into a sequence of lemmas, starting with the case where G is a symplectic group. Note that λ(B ) = λ(C ) when p = 2. r r a a 1 k Lemma 3.5 Let G = C and write r = 2 + ··· + 2 with a > a > ··· > a 0.Then r 1 2 k λ(G) 4k − 1 + 2 a 2(log (r + 1))(log r ) + 5 2 2 and the conclusion of Theorem 3.4 holds. Proof In view of Lemma 3.2, we may assume r 5. First observe that a log r and k log (r + 1),so 4k − 1 + 2 a 4k − 1 + 2(k log r − k(k − 1)/2) = 2k log r − k + 5k − 1 2 2 2(log (r + 1))(log r ) + 5. 2 2 123 The length and depth of algebraic groups Therefore, it sufﬁces to show that λ(G) 4k − 1 + 2 a . (5) We proceed by induction on k. We will repeatedly use the fact that if H is a symplectic group with natural module V , then the stabiliser in H of any proper nondegenerate subspace of V is a maximal connected subgroup of H . First assume k = 1, so r = 2 and a 3. Let M be the stabiliser in G of a nondegenerate r-space, so M = C a −1C a −1.Now M has a diagonally embedded maximal subgroup of 1 1 2 2 type C a −1 , so there is an unreﬁnable chain C > C a −1C a −1 > C a −1 . 2 1 1 1 2 2 2 By repeating this process, we can descend from G to C in 2a steps and thus 1 1 λ(G) 2a + λ(C ) = 2a + 3. 1 1 1 This establishes the bound in (5)when k = 1. Now assume k > 1. Let M be the stabiliser in G of a nondegenerate 2 -space, so a a M = C C a . By Lemma 2.1,wehave 1 2 k 2 2 +···+2 λ(M ) λ(C a ) + λ(C a a ) 1 2 k 2 2 +···+2 so by induction we get λ(G) 1 + (3 + 2a ) + 4(k − 1) − 1 + 2 a (6) 1 i i =2 and thus (5) holds. Lemma 3.6 If G = A ,then λ(G) 2(log r ) + 4. Proof By Lemma 3.2, we may assume that r 5. If r is odd then C is a maximal (r +1)/2 connected subgroup of G, so Lemma 3.5 implies that λ(G) 1 + λ(C ) 1 + 2(log ((r + 3)/2))(log ((r + 1)/2)) + 5 2(log r ) + 4 (r +1)/2 2 2 2 as required. Similarly, if r is even then A > U A T > A T > A > C r r r −1 1 r −1 1 r −1 r/2 is an unreﬁnable chain and a further application of Lemma 3.5 yields λ(G) 4 + λ(C ) 4 + 2(log (r/2 + 1))(log (r/2)) + 5 2(log r ) + 4. r/2 2 2 2 The result follows. Lemma 3.7 Suppose G = D ,where r 3.Then λ(G) 2(log r ) + 11. Proof As usual, we may assume r 5. First assume p = 2and let M = B be the r −1 stabiliser of a nonsingular 1-space. Since λ(B ) = λ(C ) when p = 2, Lemma 3.5 r −1 r −1 implies that λ(G) 1 + λ(M ) 1 + 2(log (r − 1))(log r ) + 5 < 2(log r ) + 6. 2 2 2 For the remainder, we may assume p = 2. 123 T. C. Burness et al. a a 1 k Suppose r is even and write r = 2 + ··· + 2 ,where a > a > ··· > a 1. We 1 2 k claim that the upper bound on λ(G) in (5) holds, in which case λ(G) 2(log (r + 1))(log r ) + 5. 2 2 To prove this, we use induction on k. Note that if M is the connected component of the stabiliser in G of a nondegenerate -space of the natural module, with 1 r and = 2, then M is a maximal connected subgroup of G (if = 2then M = T D is the Levi factor 1 r −1 of a parabolic subgroup of G). First assume k = 1, so r 8. We can construct an unreﬁnable chain D > D D > D > ··· > D > D D > D a −1 a −1 a −1 r 1 1 1 8 4 4 4 2 2 2 of length 2a − 4, so λ(G) 2a − 4 + λ(D ) 2a + 3. 1 1 4 1 a a a Now assume k > 1 (and continue to assume r is even). Then M = D D is a 1 2 k 2 2 +···+2 maximal closed connected subgroup of G,so a a a λ(G) 1 + λ(M ) 1 + λ(D ) + λ(D ) 1 2 k 2 2 +···+2 and by induction we deduce that (6) holds. The result follows. Finally, let us assume r 5 is odd. If r = 5then λ(G) 6since B is a maximal subgroup of G, so we can assume r 7. Let M be the connected component of the stabiliser in G of a nondegenerate 6-space. Then M = D D is a maximal connected subgroup of 3 r −3 G, so by the previous result for even rank, we get λ(G) 1 + λ(D D ) 6 + λ(D ) 6 + 2(log (r − 2))(log (r − 3)) + 5 3 r −3 r −3 2 2 and this yields λ(G) 2(log r ) + 11 as required. The next lemma completes the proof of Theorem 3.4. Lemma 3.8 Suppose G = B ,where r 3.Then λ(G) 2(log r ) + 12. Proof This is an immediate corollary of Lemma 3.7 since D is a maximal connected sub- group of G. 3.6 Proof of Theorem 5(iii) Let G = Cl(V ) be a classical algebraic group of rank r over an algebraically closed ﬁeld K of characteristic p > 0, with natural module V.If M is a maximal connected subgroup of G,thenby[16, Theorem 1], one of the following holds: (i) M is the connected stabiliser of a proper nontrivial subspace U of V that is either totally singular, nondegenerate, or a nonsingular 1-space (the latter only when G is orthogonal and p = 2); (ii) M is the connected stabiliser Cl(U ) ⊗ Cl(W ) of a tensor product decomposition V = U ⊗ W ; (iii) M ∈ S(G), the collection of maximal connected simple subgroups of G such that V is a p-restricted irreducible KM-module. Lemma 3.9 Let G be as above, let M ∈ S(G) and suppose M is of classical type. Then rank(M)> log r. 123 The length and depth of algebraic groups Proof Let k = rank(M ). Using Weyl’s character formula, it is easy to see that the p-restricted irreducible KM-module of largest dimension is the Steinberg module, which has dimension N 2 p ,where N is the number of positive roots in the root system of M.Since N k ,it follows that dim V p . The conclusion follows, as r < dim V . Now we prove Theorem 5(iii). As in the statement, deﬁne e (p) = p,and e (p) = 1 l+1 e ( p) p for l > 1, and for x ∈ R set ψ (x ) = min (l : e (p) x ) . p l Note that e (p) = log e (p),and for x > p we have l l+1 ψ (x ) = 1 + ψ log x . (7) p p Theorem 3.10 If G is a simple algebraic group of rank r in characteristic p > 0,then λ(G) ψ (r ). 2 p Proof The proof proceeds by induction on r.If r < e (p) = p ,then ψ (r ) 2, so the 3 p conclusion holds. Now assume that r e (p). Then certainly r > 8, so G is classical. Choose a maximal connected subgroup M of G such that λ(M ) = λ(G) − 1. Then M is as in one of the possibilities (i)-(iii) above, and in case (iii) we have rank(M)> log r, by Lemma 3.9.In cases (i) and (ii), M has a simple quotient Cl(U ) with dim U dim V . Hence in any case, there is a simple connected group H of rank at least log r, such that λ(M ) λ(H ).By induction, λ(H ) ψ ( log r ), andsoby(7)wehave λ(G) = 1 + λ(M ) 1 + ψ ( log r ) = ψ (r ). p p This completes the proof by induction. The proof of Theorem 5 is now complete. We conclude with an example showing that the lower bound ψ (r ) in Theorem 5(iii) is of roughly the correct order of magnitude. Example 3.11 Fix a prime p 5 and consider the series of embeddings of odd dimensional orthogonal groups via their Steinberg modules: B < B < ··· < B , r r r 0 1 k where r = 1, r = (p − 1)/2and r = (p − 1)/2for l 1. By [20], each term in this 0 1 l+1 series is maximal in the next, so λ(B ) k + 3. 3.7 Proof of Theorem 6 We begin by classifying the simple algebraic groups G with λ(G) = l(G). Lemma 3.12 The only simple algebraic group G satisfying λ(G) = l(G) is G = A . 123 T. C. Burness et al. Proof First observe that λ(A ) = l(A ) = 3, by Lemma 2.3. Conversely, suppose G is 1 1 simple of rank r > 1and λ(G) = l(G). We know that l(G) = l(B) + r by Corollary 2.If r 4or G is exceptional, this contradicts Lemma 3.2 or Theorem 3.3.And if r 5, then Theorem 5(ii) gives a contradiction. Now we prove Theorem 6.Let G be a connected algebraic group over an algebraically closed ﬁeld. Suppose λ(G) = l(G) and G is insoluble. Set G = G/R(G) and note that ¯ ¯ ¯ λ(G) = l(G). Write G = G ··· G , where each G is simple. Then λ(G ) = l(G ) for each 1 t i i i i. By Lemma 3.12, this implies that G A for all i.Since λ(A A ) = 4 < l(A A ) = 6, i 1 1 1 1 1 we must have t = 1, so G = A , as in Theorem 6. 3.8 Proof of Theorem 7 Let G be an algebraic group and recall that cd(G) = l(G) − λ(G) is the chain difference of G. Lemma 3.13 If N is a connected normal subgroup of G, then cd(G) cd(N ) + cd(G/N ). Proof By Lemma 2.1 we have l(G) = l(N ) + l(G/N ) and λ(G) λ(N ) + λ(G/N ).The conclusion follows. The analogous result for ﬁnite groups is [3, Lemma 1.3]. We now state some immediate consequences of the above lemma. Corollary 3.14 (i) If N is a connected normal subgroup of G, then cd(G/N ) cd(G). (ii) If 1 = G G ··· G G = G is a chain of connected subgroups of G, then t t −1 1 0 cd(G) cd(G /G ). i −1 i (iii) If G = G × ··· × G ,where each G is connected, then cd(G) cd(G ). 1 t i i The next result bounds dim G in terms of cd(G) when G is simple. Proposition 3.15 Let G be a simple algebraic group in characteristic p 0.Then 2cd(G) +3if p = 0 dim G 2cd(G) + 40 for any p. Proof Let r be the rank of G. Using Corollary 2 and its notation we obtain cd(G) = dim B + r − λ(G). (8) Suppose ﬁrst that p = 0, and let c be the value of λ(G) as in Theorem 4.Thenwehave cd(G) = dim B + r − c dim B − 2. Therefore dim G 2dim B − 1 2cd(G) + 3, as required. Suppose now that p > 0. Applying (8)and Theorem 5 we obtain dim G = 2dim B − r 2cd(G) +
24 − 3r + 4(log r ) , and the right hand side is at most 2cd(G) + 40. 123 The length and depth of algebraic groups The next result is of a similar ﬂavour, dealing with certain semisimple groups. Proposition 3.16 Let G = S where k 2 and S is a simple algebraic group in character- istic p 0.Then 2cd(G) +2if p = 0 dim G 2cd(G) + 28 for any p. Proof By considering a series of diagonal subgroups, we can construct an unreﬁnable chain k k−1 S > S > ··· > S of length k − 1, so λ(S ) k − 1 + λ(S). This yields cd(G) = k · l(S) − λ(S ) k · l(S) − (k − 1 + λ(S)) = k(l(S) − 1) − λ(S) + 1. Corollary 2 shows that if B is a Borel subgroup of S,and r = rank(S),then cd(G) k(dim B + r − 1) − λ(S) + 1, and so k dim B cd(G) − k(r − 1) + λ(S) − 1. Since dim G = k dim S k(2dim B − 1),wesee that dim G 2(cd(G) − k(r − 1) + λ(S) − 1) − k = 2cd(G) + a, where a = 2λ(S) − 2k(r − 1) − 2 − k = 2(λ(S) − 1) − k(2r − 1). If p = 0 then it is easy to check using Theorem 4 that a 2 in all cases, as required. Now suppose p > 0. Then Theorem 5 yields 2 2 a
2 2(log r ) + 11 − k(2r − 1)
4(log r ) − 4r + 24, 2 2 which is at most 28. Lemma 3.17 Let S ,..., S be simple algebraic groups that are pairwise non-isomorphic. 1 n Then dim S n . Proof Let d = dim S , and assume d d ···.For any r, the number of distinct types i i 1 2 of simple algebraic groups of rank at most r is at most 4r. Hence rank(S ) ,and so n n 1 1 2 3 d i > n . 16 48 i =1 i =1 This is greater than n provided n > 48. For n 48, the conclusion can readily be checked by computation. We are now ready to prove Theorem 7.Let G be a connected algebraic group. If G is soluble then the conclusion holds trivially, so suppose G is insoluble. Let R(G) be the radical of G and write G = G/R(G) = S , i =1 123 T. C. Burness et al. where the S are pairwise non-isomorphic simple algebraic groups. By Corollary 3.14, cd(G) cd(S ), and hence Propositions 3.15 and 3.16 imply i i 1 1 cd(G) (k dim S − 40) = dim G − 20n. (9) i i 2 2 i =1 1 2 Now Lemma 3.17 gives cd(G) (n − 40n), and it follows that n 20 + 400 + 2cd(G). Therefore by (9), dim G 2cd(G) + 40n 2cd(G) + 40 400 + 2cd(G) + 800. This completes the proof of Theorem 7. Remark 3.18 Let G be an algebraic group in characteristic p 0and set G = G/R(G). For a simple group G, it is easy to see that cd(G) = 1 if and only if G = A and p = 2. In the general case, by arguing as in the proof of Theorem 6, one can show that cd(G) = 1 only ¯ ¯ if G = A ,or p = 2and G = A . For example, if G = UA , a semidirect product where 1 2 1 U is the natural module for A ,then l(G) = 5and λ(G) = 4. 4Chain ratios In this ﬁnal section we consider the chain ratio cr(G) of an algebraic group G,which is deﬁned by cr(G) = l(G)/λ(G). First we show that if G is simple, then its dimension is bounded in terms of its chain ratio. Proposition 4.1 Let G be a simple algebraic group in characteristic p 0.Then (i) dim G < 12 cr(G) if p = 0. (ii) dim G <(1 + o(1)) · cr(G) · (log cr(G)) if p > 0,where o(1) = o (1). cr(G) Proof Set d = dim G and note that l(G)> d/2 by Theorem 3. Therefore d < 2l(G) = 2λ(G)cr(G). Note that this holds for any algebraic group. Assuming G is simple and p = 0, we have λ(G) 6 by Theorem 4, proving part (i). Now suppose p > 0and let r be the rank of G.Then λ(G) 2(log r ) + 12 by Theorem 2 1 2 5.Since d > r we obtain λ(G)< (log d) + 12, so d <((log d) + 24) cr(G). This easily implies the conclusion of part (ii). In contrast to this result, we shall exhibit a sequence of algebraic groups G for which dim G/R(G) is not bounded above in terms of the chain ratio cr(G). To show this we need the following result. Lemma 4.2 If S is a simple algebraic group and k ∈ N,then λ(S ) k + 2. 123 The length and depth of algebraic groups Proof The proof goes by induction on k, the case k = 1 being clear. Suppose k > 1, write G = S and let π : G → S be the projection to the i-th factor. Let M be a maximal connected subgroup of G with λ(M ) = λ(G) − 1. If π (M ) = M < S for i i k−1 k−1 some i,then M = M × S ,and so λ(M ) λ(S ) k + 1 by induction, giving the conclusion. Now assume π (M ) = S for all i.Then M is a product of diagonal subgroups of various k 2 k−2 subsets of the simple factors of S , and maximality forces M = diag(S ) × S ,where 2 2 k−1 diag(S ) denotes a diagonal subgroup of S . Hence M = S and the conclusion again follows by induction. Now, ﬁx a simple algebraic group S and let G = S for k 1. Since l(G) = k · l(S) and λ(G) k + 2 by Lemma 4.2, it follows that cr(G) = l(G)/λ(G)< k · l(S)/k = l(S). Letting k tend to inﬁnity, we see that cr(G) is bounded, while dim G/R(G) = dim G tends to inﬁnity. 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