Right Engel elements of stability groups of series in vector spaces

Journal of Pure and Applied Algebra 220 (2016) 2701–2710

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Journal of Pure and Applied Algebra www.elsevier.com/locate/jpaa

Right Engel elements of stability groups of series in vector spaces B.A.F. Wehrfritz School of Mathematical Sciences, Queen Mary University of London, London E1 4NS, England, United Kingdom

a r t i c l e

i n f o

Article history: Received 10 July 2015 Available online 8 January 2016 Communicated by S. Donkin MSC: 20F45; 20F19; 20H25

a b s t r a c t We study the relationships of the subsets of right and bounded right Engel elements with certain terms of the upper central series of the stability groups of series of subspaces in vector spaces over fields and division rings. We focus mainly, but not exclusively, on ascending series and descending series. The left and bounded left Engel elements of these stability groups were studied in an earlier paper of the author, see [5] below. © 2015 Elsevier B.V. All rights reserved.

1. Introduction Let V be a vector space over a division ring D and L a series of subspaces of V containing {0} and V , usually either ascending or descending. Let S = Stab(L) be the (full) stability group of this series in GL(V ). In [1] Casolo and Puglisi prove that if L is ascending, and often if it is descending, then the Hirsch–Plotkin radical HP (S) of S and the Fitting subgroup Fitt(S) of S are equal and in [4] Traustason shows that in the descending case they need not be equal. In [5] the author proves that for ascending series and often for descending series the set L(S) of left Engel elements of S and the set L− (S) of bounded left Engel elements of S are equal to Fitt(S), as indeed are the Gruenberg radical σ(S) and the Baer radical σ − (S) of S. Here we consider the sets R(S) and R− (S) of right Engel and bounded right Engel elements of S and ask whether analogous results exist. The brief answer is yes they do. In this context the right analogue of HP (S) is the hypercentre ζ(S) of S and the right analogue of Fitt(S) is the ω-th term ζω (S) of the upper central series of S. In some ways replacing left by right interchanges ascending with descending. For example, for descending series ζ(S) and ζω (S) are always equal, but usually they are not for ascending series, in fact only if the length λ of L is finite (the trivial case), or if λ is a limit ordinal, or if λ = ω + 1. (See [6], where the exact central heights for all ascending and all descending series are computed.) Further we also use below the analogue of σ(S), which is   ρ(S) = x ∈ S : for all g ∈ S the subgroup g is an ascendant subgroup of g, xS  E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jpaa.2015.12.006 0022-4049/© 2015 Elsevier B.V. All rights reserved.

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and the analogue of σ − (S), which is   ρ− (S) = x ∈ S : for some k ≥ 1 and all g ∈ S, g is subnormal in g, xS  in at most k steps . By [3], 7.31, for any group G we have that ρ(S) and ρ− (S) are normal subgroups of G with ζ(G) ≤ ρ(G) ⊆ R(G) and ζω (G) ≤ ρ− (G) ⊆ R− (S). Theorem 1.1. Let L = {Vα : 0 ≤ α ≤ λ} be a strictly ascending series of subspaces of the vector space V over the division ring D indexed by the ordinal numbers α ≤ λ, where {0} = V0 and Vλ = V . Set λ = μ + n, where μ is zero or a limit ordinal and n is a non-negative integer, and set S = Stab(L). a) b) c) d)

R− (S) = ρ− (S) = ζω (S). R(S) = ρ(S) = ζ(S). If n = 0, then R(S) = {1}. ζω (S) = ζ(S) if and only if either n = 0 or λ ≤ ω + 1.

N.B. 1.1b) and [6], 1.2, immediately imply 1.1c) and 1.1d). If μ = 0, then S is nilpotent and the claims are trivial. Thus our proofs below concentrate on the case where μ is at least ω. Theorem 1.2. Let L = {Vα : 0 ≤ α ≤ λ} be a strictly descending series of subspaces of the vector space V over the division ring D indexed by the ordinal numbers α ≤ λ, where {0} = Vλ and V0 = V . Set λ = μ + n, where μ is zero or a limit ordinal and n is a non-negative integer, and set S = Stab(L). a) R− (S) = ρ− (S) = ζω (S) = ζ(S). b) If dimD V is countable or if V has an L-basis, then R(S) = ρ(S) = ζω (S). c) If n = 0 and if V satisfies the hypotheses of b) above, then R(S) = {1}. An L-basis of V is a basis B of V such that B ∩ L is a basis of L for all L in L. (This is equivalent to assuming that B is a basis of V such that for every jump Λα /Vα of L, the set B ∩ (Λα \Vα ) is a basis of Λα modulo Vα .) Note that 1.2b) and [6], 1.4, immediately imply 1.2c). Again the claims are trivial if μ = 0 and hence our proofs assume that μ ≥ ω. We are able to say something about series in general. For the definition and basic properties of general series see [3], Section 1.2. We use 1.3 in the proofs of both 1.1 and 1.2 (in order to deduce that R(S) and R− (S) are actually subgroups of S). Theorem 1.3. Let L = {(Λα , Vα ) : α ∈ A} be a series of subspaces of the vector space V over the division ring D running from {0} to V . Set S = Stab(L). a) R− (S) = ρ− (S). b) If L(S) = Fitt(S), e.g. if V and L satisfy the hypotheses of [5], Theorem B, then R(S) = ρ(S). 2. General series 2.1. For any group G we have R(G) ∩ Fitt(G) ⊆ ρ(G). (R(G) ∩ Fitt(G) is always a subgroup of G by [3], 7.34, Corollary 2, but we make no use of this here.) Proof. Let x ∈ R(G) ∩ Fitt(G). Then X = xG  is nilpotent. Let g ∈ G and h ∈ gX. Then h = g i k for some integer i and for some k in X. Hence [h, r g] = [k, r g] for all r ≥ 1. Now gX is nilpotent-by-cyclic, so

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R(gX) = ρ(gX) = ζ(gX) by [3], 7.34, Corollary 1 and Theorem 1.2iii) of [2]. Clearly X is generated by elements of R(G). Consequently X ≤ ζ(gX), so gX = ζ(gX) and g is ascendant in gX. This is for all g in G and therefore x ∈ ρ(G). 2.1 follows. 2 Corollary 2.2. If L(G) = Fitt(G), then R(G) = ρ(G). For by [3], 7.11, we have R(G) ⊆ L(G)−1 and 2.1 applies. Let V be a left vector space over the division ring D and L the general series {Λα , Vα : α ∈ A}, where  V \{0} = α∈A Λα \Vα . Set S = Stab(L). 2.3. R− (S) = ρ− (S). For R− (S)−1 ⊆ L− (S) by [3], 7.11. Hence R− (S) ⊆ Fitt(S) by [5], Theorem A. Consequently R− (S) ⊆ ρ− (S) by Theorem 1.6 of [2]. Always ρ− (S) ⊆ R− (S). 2.4. If L is an ascending series, or if dimD V is countable, or if V has an L-basis, then R(S) = ρ(S). For in each case L(S) = Fitt(S) by Theorem B of [5] and then 2.2 applies. 3. Ascending series In this section V is a left vector space over the division ring D and L denotes the strictly ascending series {0} = V0 < V1 < . . . < Vα < . . . < Vλ = V of subspaces of V indexed by the ordinal numbers α ≤ λ, where λ = μ + n, μ is a (non-zero) limit ordinal and n is a non-negative integer. Let S = Stab(L). Set H = HomD (V /Vμ , Vμ ) made into a right S-module via the diagonal action. 3.1. Let θ ∈ H be such that for some positive integer r we have [θ, r g] = 0 for all g ∈ C = CS (V /Vμ ). Then V θ ≤ Vr . Proof. Suppose w ∈ V with wθ ∈ / Vr . Pick vi ∈ Vi \Vi–1 for 1 ≤ i ≤ r. Since L is an ascending series there is an L-basis B of V containing the vi and wθ. Hence there exists g in S with wθ(g − 1) = vr and vi (g − 1) = vi–1 for 1 < i ≤ r and bg = b for all other b in B. Then [θ, g] = g −1 θg − θ = θ(g–1) since g ∈ C, so w[θ, r g] = wθ(g − 1)r = v1 = 0. This contradicts the hypothesis on r. Consequently V θ ≤ Vr . 2 Let R denote the set of all bounded right S-Engel elements of H; that is θ in H lies in R if and only if for some integer r the element θ satisfies the hypothesis of 3.1. Then 3.1 implies that: 3.2. For all θ in R there exists a positive integer r with V θ ≤ Vr .  Let Hi = HomD (V /Vμ , Vi ) for 0 ≤ i ≤ ω and Hω− = i<ω Hi ≤ Hω . Then R ≤ Hω− by 3.2 and {0} = H0 < H1 < . . . < Hi < . . . is a series of D–S submodules of H. Now Hi+1 /Hi is D–S isomorphic to HomD (V /Vμ , Vi+1 /Vi ), = Li say, and Li is C-trivial. Also if Lij = HomD (V /Vμ+j , Vi+1 /Vi ), then Li = Li0 ≥ Li1 ≥ . . . ≥ Lin = {0} and Lij /Li,j+1 is S-isomorphic to the S-trivial S-module HomD (Vμ+j+1 /Vμ+j , Vi+1 /Vi ). Thus Hω− is S-hypercentral with C-central height, and hence S-central height, at most ω (in fact exactly ω by [6], 3.5). Consequently we have proved that: 3.3. R = Hω− = ζω (H S ) ∩ H, where H S denotes the split extension of the S-module H by S.

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Theorem 3.4. If λ = μ > 0, then R(S) = {1}. We break the proof of 3.4 into several parts. 3.4a). Suppose μ = ω and Vi = Vi−1 ⊕ Dei for 1 ≤ i < ω. We use standard matrix notation for GL(V ), so {ei,j } is its standard basis relative to the ei . Set g =1+



ei+1,i .

i≥1

Then g ∈ S, with g −1 = 1 +



(−1)i−j ei,j .

i>j≥1

Consider x = 1 + for xx = 1 +

 k≥t

 k≥t

ak ek,1 ∈ S, where ak ∈ D and t ≥ 2. If x = 1 +

ak ek,1 +

 k>t

ak−1 ek,1

and xx g = g +



 k>t

ak ek,1 +

k≥t

ak−1 ek,1 , then x = [x, g −1 ];



ak ek+1,1 = gx.

k≥t

Thus, if at = 0, then [x, k g −1 ] = 1 for all k ≥ 0. Consequently x ∈ / R(S). 3.4b). If μ = ω, then R(S) = {1}. Proof. Let x ∈ R(S)\{1}. Now by [3], 7.11 and [5], Theorem B, we have x ∈ L(S)−1 = Fitt(L), which here just means that V (x − 1) ≤ Vj for some j with 1 ≤ j < ω. We choose j minimal. Then there exists some w in V with w(x − 1) ∈ Vj \Vj−1 . Also w ∈ Vk \Vk−1 for some k > j. The stabilizer S1 of the subseries {0} ≤ Vj−1 < Vj < Vk < Vk+1 < . . . ≤ Vω = V embeds into S and S1 maps onto the stability group of the series {0} < Vj /Vj−1 < Vk /Vj−1 < Vk+1 /Vj−1 < . . . ≤ V /Vj−1 of V /Vj−1 . Thus to obtain our contradiction we may assume that j = 1 and k = 2. In particular we now have V (x − 1) ≤ V1 .  Set v1 = w(x − 1), v2 = w and choose any vi ∈ Vi \Vi−1 for 2 < i < ω. Set W = j≥1 Dv i and  Wi = Dv . Then W ∩ V = W = W ⊕ Dv for each i. Now {v : i ≥ 1} is a subset of some j i i i−1 i i j≤i L-basis B of V , so there exists g ∈ S satisfying vi (g − 1) = vi−1 for i ≥ 2 and bg = b for all other b in B. Now V1 = U ⊕ Dv 1 for some subspace U and U , indeed all of V1 , is centralized by S. Therefore S acts on V /U . Also V (x − 1) ≤ V1 , so vi x and vi x−1 both lie in the subset V1 + vi ≤ V1 + W = U ⊕ W . Also v2 (x − 1) = v1 ∈ / U . Thus x, g ≤ S acts on U ⊕ rW/U and 3.4a) yields that for each k ≥ 1 the element [x, k g −1 ] acts non-trivially on U ⊕ W/U . Consequently x ∈ / R(S), a contradiction that completes the proof of 3.4b). 2 3.4c). The completion of the proof of 3.4 Let x ∈ R(S). There exists a maximal α with x centralizing Vα . Clearly 1 ≤ α ≤ μ. Suppose that α < μ. Then α + ω ≤ μ. Applying 3.4b) to the series {0} < Vα < Vα+1 < . . . < Vα+i < . . . ≤ Vα+ω

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implies that x centralizes Vα+ω . This contradiction of the choice of α shows that α = μ and hence that x = 1. Therefore R(S) = {1}. Theorem 3.5. R− (S) = ρ− (S) = ζω (S). Proof. Already we have R− (S) = ρ− (S) ≥ ζω (S), see 2.3. Set R = R− (S) and T = CC (Vμ ); recall C = CS (V /Vμ ). We have the standard S-isomorphism of T onto H. Let Tω− denote the inverse image of Hω− in T , see 3.4 and its preamble. Then R∩T = Tω− ≤ ζω (S). By 3.4 we have that R ≤ CS (Vμ ). Pick a subspace W of V with V = W ⊕ Vμ . Let S1 = CS (W ) and S2 = NS (W ) ∩ CS (Vμ ). Then [S1 , S2 ] = 1, S = S1 T S 2 , C = S1 T and R ≤ CS (Vμ ) = T S 2 , see [6], where the notation is exactly the same. Let x ∈ S2 and t ∈ T with xt ∈ R, say r ≥ 1 with [xt, r g] = 1 for all g in S. If g ∈ S1 , then since [S1 , S2 ] = 1, we have [xt, g] = [t, g] and [t, r g] = [xt, r g] = 1. This holds for all g in S1 and hence for all g in S1 T = C. If θ is the image of t in H, then V θ ≤ Vr by 3.1, so θ ∈ Hω− and t ∈ Tω− ≤ ζω (S) ⊆ R. Consequently x ∈ R ∩ S2 and therefore we may choose t = 1. / Vμ . Choose vi ∈ Vi \Vi−1 for 1 ≤ i ≤ r. Suppose x = 1. Then there exists w in V \Vμ with w(1 − x−1 ) ∈ There exists an L-basis B of V containing a basis of W and v1 , v2 , . . . , vr and an element g of S1 with vi (g − 1) = vi−1 for 1 < i ≤ r and bg = b for all other b in B. Choose θ in Hr with w(1 − x−1 )θ = vr . Then w(1 − x−1 )θ(g − 1)r−1 = v1 = 0. Let t be the inverse image of θ under our canonical isomorphism of T onto H. Then [x, gt] = [x, t] since [x, g] = 1, recall g ∈ S1 , so [x, r gt] = [x, t, r−1 g]. Now x centralizes Vμ and g centralizes V /Vμ . Hence the image of [x, t] in H is (1 − x−1 )θ and of [x, t, r−1 g] is (1 − x−1 )θ(g − 1)r−1 . Consequently [x, r gt] = 1, a contradiction of the choice of r. Therefore R ∩ S2 = {1} and so R ≤ Tω− ≤ ζω (S) ≤ R. The proof of 3.4 is complete. 2 Theorem 3.6. R(S) = ρ(S) = ζ(S). Note that Theorem 1.1 follows immediately from, in order, 3.6, 3.5, 3.4 and the remarks in the introduction. Proof. Let Hα = HomD (V /Vμ , Vα ) for all α ≤ μ, regarded as a submodule of HomD (V, Vμ ) as usual. If  ν ≤ μ is a limit ordinal set Hν− = α<ν Hα ≤ Hν . Now each Hα+1 /Hα is S-hypercentral, see [6], 3.2. Let θ ∈ Hν \Hν− . Then for each α < ν there exists v ∈ V with vθ ∈ Vν \Vα . / Vα(1) . Then v1 θ ∈ Vβ(1) \Vβ(1)−1 for some β(1) with α(1) ≤ Choose α(1) < ν and v1 in V with v1 θ ∈ β(1) − 1 < β(1) < ν. Since θ ∈ / Hν− there exists v2 in V and an α(2) with β(1) < α(2) < ν and v2 θ ∈ / Vα(2) . Then v2 θ ∈ Vβ(2) \Vβ(2)−1 , where α(1) < β(1) < α(2) < β(2). We keep going; in this way we construct vi , α(i) and β(i) for all 1 ≤ i < ω with vi θ ∈ Vβ(i) \Vβ(i)−1 and with α(1) < β(1) < . . . < α(i) < β(i) < . . . . There exists an L-basis B of V containing all the vi θ. Define g ∈ S by setting vi θ(g−1) = vi−1 θ for 1 < i < ω  and bg = b for all other b in B. (Note that g −1 does exists and is given by vi θg −1 = j≤i (−1)i−j vj θ.) Now V θ ≤ Vμ , so g centralizes V /Vμ and vi+1 [θ, g] = vi θ; more generally vj+1 [θ, j g] = v1 θ = 0 for 1 ≤ j < ω. Therefore [θ, j g] = 0 for all 1 ≤ j < ω. In other words, no element of Hν \Hν− is right Engel with respect to S. As before we set T = CS (V /Vμ ) ∩ CS (Vμ ), which is isomorphic to H in the usual way. Again we write Tα for the inverse image of Hα in T (including Tν− for Hν− ). The above shows that R(S) ∩ Tν = R(S) ∩ Tν− for all limit ordinals ν ≤ μ. Since each Tα+1 /Tα is S-hypercentral, we have that R(S) ∩T ≤ ζ(S) ∩T ≤ R(S) ∩T . Now R(S) is a normal subgroup of S (by 2.4), R(S) is a subgroup of CS (Vμ ) by 3.4 and S/CS (V /Vμ ) is nilpotent of class n − 1. Hence  R(S), n−1 S ≤ R(S) ∩ CS (V /Vμ ) ∩ CS (Vμ ) = R(S) ∩ T ≤ ζ(S). Therefore R(S) ≤ ζ(S) ≤ R(S). The proof of 3.6 is complete.

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4. Descending series In this section V is a left vector space over the division ring D and L denotes the strictly descending series V = V0 > V1 > . . . > Vα > . . . > Vλ = {0} of subspaces of V indexed by the ordinal numbers α ≤ λ, where λ = μ + n, μ is a (non-zero) limit ordinal and n is a non-negative integer. Set S = Stab(L). 4.1. Let U be a subspace of V such that for every g in CS (Vω ) there is a positive integer m with V (g−1)m ≤ U . If λ ≥ ω, assume that V has an M-basis B for M the subseries V = V0 > V1 > . . . > Vα > . . . > Vω ≥ {0} of L. Then there is an integer i with Vi ≤ U . If dimD (V /Vω ) is countable, then V always has an M-basis, see [5], 1.7 (and its proof). Proof. If λ is finite the claim is trivial. Hence assume that λ ≥ ω. For each 1 ≤ i < ω choose an element ei of B ∩ (Vi−1 \Vi ). Consider first the case where λ = ω and each Vi−1 = Vi ⊕ Dei . Then using standard matrix notation for GL(V ), so {ei,j ) is the standard basis of Dω×ω relative to the basis B, set g =1+



ei,i+1 +

i≥1



e2j,2j+2 .

j≥1

Then g ∈ S, with g −1 = 1 −

 i≥1

ei,i+1 +



e2j−1,2j+1 .

j≥1

Now e2i (g − 1) = e2i+1 + e2i+2 and e2i−1 (g − 1) = e2i . Thus Vi (g − 1) = Vi+1 and V (g − 1)i = Vi for each i ≥ 0. By hypothesis there exists m > 0 with V (g − 1)m ≤ U . Consequently Vm ≤ U . Now consider the general case where λ = ω. Define the linear map g of V into itself by setting e2i g = e2i + e2i+1 + e2i+2 and e2i−1 g = e2i−1 + e2i for each i ≥ 1 and setting bg = b for all other elements b of B. Then g ∈ S, so V (g − 1)m ≤ U for some m, m depending of course on our choice of g and the ei . Set  Wi = j>i Dej for i ≥ 0. Then by the previous case applied to W0 we have Wm ≤ U . Suppose that no Vi for 1 ≤ i < ω is contained in U . Let Xi be the subspace of V spanned by B ∩(Vi \Vi+1 ).  Then Vi = Xi ⊕ Vi+1 and, since B is an M-basis (= L-basis here) of V , Vi = j≥i Xj for all i ≥ 0. Then for each i there exists j ≥ i with Xj not contained in U . Consequently there is an infinite sequence 1 ≤ j(1) < j(2) < . . . < j(k) < . . . of integers such that no B ∩ Xj(k) is contained in U . But now we may assume that we have initially picked the ei ∈ B ∩ (Vi−1 \Vi ) so that no ej(k)+1 lies in U . But then the argument of the previous paragraph yields an m with ej(m)+1 ∈ U . This contradiction completes the proof of the case where λ = ω. Finally suppose λ > ω. Apply the case where λ = ω to V /Vω . Hence we are given an m with 1 ≤ m < ω and Vm ≤ U + Vω . To complete the proof of 4.1 it suffices to prove that Vω ≤ U . Let w ∈ Vω and set k(i) = i(i + 1)/2 for each i ≥ 1. Define h ∈ CS (Vω ) by ek(i)+r (h − 1) = ek(i)+r+1 for 0 ≤ r < i, ek(i)+i (h − 1) = w and bh = b for all other b in B. Then ek(i) (h − 1)i+1 = w. But by hypothesis

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there exists m ≥ 1 with V (h − 1)m ≤ U . Consequently w = ek(m) (h − 1)(h − 1)m ∈ U . This proves that Vω ≤ U , which completes the proof of 4.1. 2 Theorem 4.2. R− (S) = ρ− (S) = ζ(S) = ζω (S). Proof. Note first that R = R− (S) = ρ− (S) by 2.3, so R is a normal subgroup of S. Also ζ(S) = ζω (S) by [6], 1.4, and always ζω (S) ⊆ R− (S). Thus it remains to prove that R− (S) ≤ ζ(S). If λ is finite, then S is nilpotent and the claims are trivial. Thus assume that λ is infinite. Suppose first that n = 0, so λ = μ ≥ ω. Let x ∈ R\1. Then x−1 ∈ L− (S) ([3], 7.11), so x ∈ Fitt(L) by [5], Theorem A; that is, x stabilizes a finite subseries M of the completion L∗ of L. Suppose M is V = U0 > U1 > . . . > Up > {0}, where the Ui lie in L∗ and we choose p minimal. Hence Up (x − 1) = {0} but Up−1 (x − 1) = {0}. Choose v1 ∈ Up−1 \Up with v2 = v1 (x − 1) = 0. Then v2 ∈ Vβ \Vβ+1 for some β < μ. (Recall μ = λ is a limit ordinal.) Clearly Up ≥ Vβ > Vβ+1 > Vμ = {0}. For j > 2 choose vi for 2 < i ≤ j with vi in Vβ+i−2 but not in Vβ+i−1 . Consider the following subseries N of L∗ ; N : V = U0 > U1 > . . . > Up ≥ Vβ > Vβ+1 . . . > Vβ+j−2 > {0}. Since this is a series of finite length with each jump covering at most one of the vi , there is an N-basis B of V containing the set {v1 , v2 , . . . , vj }. Define g ∈ GL(V ) by vi (g − 1) = vi+1 for 1 ≤ i < j and bg = b for all other elements b of B. Then g ∈ S. Clearly v1 (x − 1) = v2 and vi x = vi for 1 < i ≤ j. Then by 1.1 of [5] we have v1 [x, j−1 g] = vj = 0. Since this is for all j > 2 we have a contradiction of x ∈ R− (S). Consequently R = 1 whenever λ = μ. Now assume that n > 0. We use the notation of [6]. Thus T = CS (V /Vμ ) ∩ CS (Vμ ) and T is isomorphic as S-module to H = HomD (V /Vμ , Vμ ), which we equate to the set of θ in HomD (V, Vμ ) with Vμ θ = {0}. Let K denote the set of all θ in H such that for some i < ω we have Vi θ = {0}. Then K is a D–S bisubmodule of H and 4.2 of [6] yields that K is the S-hypercentre of H and H has S-central height exactly ω. Let θ ∈ H\K. Then Vj θ = {0} for all j < ω. Fix j with 1 ≤ j < ω and pick w ∈ Vj with wθ = 0. For 0 ≤ i < j choose vi ∈ Vi \Vi+1 and set vj = w. There exists x in S with vi (x − 1) = vi+1 for 0 ≤ i < j and with Vj (x − 1) = {0}. Then  vi θ, x−1 = vi xθx−1 − vi θ = vi (x − 1)θ = vi+1 θ for 0 ≤ i < j since V θ ≤ Vμ ≤ Vj . Therefore  v0 θ, j x−1 = v0 (x − 1)j θ = vj θ = wθ = 0. Hence K is equal to 

 θ ∈ H : for some integer j and all x in S we have [θ, j x] = {0} .

Let TK denote the inverse image of K in T under our canonical isomorphism of T onto H. Then TK = T ∩ ζ(S) = T ∩ ζω (S) and the above yields that TK = T ∩ R. The case above where λ is a limit ordinal applied to S/CS (Vμ ) (and [5], 1.5) yields that R ≤ CS (Vμ ). Also S/CS (V /Vμ ) is nilpotent of class n − 1. Therefore [R, n−1 S] ≤ T ∩ R ≤ ζ(S); recall R is a normal subgroup of S. Consequently R ≤ ζ(S) and the proof of 4.2 is complete. 2 4.3. Suppose V =

 1≤i<ω

Dei , Vi =

 ji

Dej and λ = ω. Then R(S) = {1}.

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Proof. We identify GL(V ) with the matrix ring Dω×ω , so {ei,j } is its standard basis. Let g =1+



ei,i+1 +

i≥1



e2j,2j+2 .

j≥1

   Then g ∈ S with g −1 = 1 − i≥1 ei,i+1 + j≥1 e2j−1,2j+1 . Consider y = 1 + 2≤r≤2t ar e1,r , where the ar lie in D with t ≥ 1 and a2t = 0. Now

[y, g] =

1−

 r

 ar e1,r g

−1

1+

 s

 as e1,s g = 1 +



as g −1 e1,s g −

s

 r

ar e1,r −



ar as e1,r g −1 e1,s g.

r,s

But e1,s g = e1,s + e1,s+1 + d, where d = 0 if s is odd and d = e1,s+2 if s is even. Also g −1 e1,s g = e1,s g and e1,r g −1 e1,s g = 0; recall r ≥ 2 and s ≥ 2. The following is a consequence of this. 4.3a). Let x = 1 + ae1,2t with a ∈ D{0} and t ≥ 1 and let k ≥ 1. Then [x, k g] = 1; hence x does not lie in R(S). A very similar calculation yields the following. 4.3b). Let x = 1 + ae1,2t+1 with a ∈ D\{0} and t ≥ 1 and let k ≥ 1. Then [x, k g −1 ] = 1; hence x does not lie in R(S). Set T = Fitt(S). By Theorem B of [5] we have T = Fitt(L); thus T is just the McLain group defined by the series L = {Vi : i = 0, 1, 2, . . .}, see [3], Section 6.2. We claim that 4.3c). CS (T ) = 1.  For suppose g = i≤j gi,j ei,j ∈ CS (T ). Then 1 + em,n ∈ T , where m < n, and em,n g = gem,n . Hence   j gn,j em,j = i gi,m ei,n . Thus gn,j = 0 if j = n and gi,m = 0 if i = m. Always gm,m = 1 = gn,n since g ∈ S. Therefore g = 1 and 4.3c) is proved. We now complete the proof of 4.3. Recall R(S) = ρ(S) by 2.4 and hence R(S) is a normal subgroup of S. If ρ(S) ∩ T = 1, then ρ(S) ≤ CS (T ) = 1 by 4.3c). Suppose ρ(S) ∩ T = 1. Then, cf. Theorem 6.2iii) of [3], there exists an x = 1 + aem,n ∈ ρ(S) for some non-zero a in D and some 1 ≤ m < n. Let φ : S → GL(Vm−1 ) be the obvious restriction map. Then Sφ ≤ GL(Vm−1 ) is the (full) stability group of the series {Vi : m − 1 ≤ i < ω}, see [5], 1.5. Clearly xφ ∈ R(S)φ ⊆ R(Sφ). But this contradicts either 4.3a) or 4.3b) applied with V = Vm−1 , depending on whether n is even or odd. Consequently ρ(S) ∩ T = 1 and hence R(S) = {1}. 2 4.4. Suppose λ = ω and suppose V has an L-basis B (e.g. suppose dimD V is countable, see [5], 1.7). Then R(S) = {1}. Proof. Let x ∈ R(S)\{1}. Then x−1 ∈ L(S) = Fitt(L) by [3], 7.11 and [5], Theorem B. Hence x stabilizes a finite subseries of L∗ that we can take to be V = V0 > V1 > . . . > Vk > {0}. Taking k minimal we have 1 ≤ k < ω, Vk (x − 1) = {0} and Vk−1 (x − 1) = {0}. There exists vk ∈ B ∩ (Vk−1 \Vk ) with vk (x − 1) ∈ Vk \{0}, say vm = vk (x − 1) ∈ Vm−1 \Vm , where k < m. There exists an element g in GL(V ) fixing every element of B with one exception c ∈ Vm−1 \Vm where cg = vm . Clearly g normalizes each Vi and hence normalizes S. Thus we may assume that vm as well as  vk lies in B. For each i = k, m choose any element vi of B ∩ (Vi−1 \Vi ). Set W = i≥k Dv i ≤ V , and M = {W ∩ L : L ∈ L}. If h ∈ SW = Stab(M), then setting bh = b for all b ∈ B\W extends the action of h and more generally of SW to V . In this way regard SW as a subgroup of S. Now vk x = vk + vm and vi x = vi for all i > k. Thus x|W ∈ SW . Also x|W ∈ R(SW ). Then x|W = 1 by 4.3. This contradiction completes the proof that R(S) = 1. 2

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4.5. Suppose that λ = μ ≥ ω and assume that either V has an L-basis or has countable dimension. Then R(S) = {1}. Proof. Let x ∈ R(S)\{1}. Then as in the proof of 4.4 the element x stabilizes a finite subseries of L∗ . Hence there exist β < β  < μ such that Vβ (x − 1) = {0}, Vβ (x − 1) ≤ Vβ  and Vβ  (x − 1) = {0}. Pick γ maximal subject to Vβ (x − 1) ≤ Vγ ∈ L∗ . Then β  ≤ γ, so Vγ (x − 1) = {0}. Also Vβ (x − 1) is not contained in Vγ+1 , so there exists v ∈ Vβ with v(x − 1) ∈ Vγ \Vγ+1 . Further γ + ω ≤ μ. Set W = Vβ and U = Vγ+ω . Consider W/U and its series M : W/U = Vβ /U > Vγ /U > Vγ+1 /U > . . . > Vγ+i > . . . > Vγ+ω /U = {0}. Then S acts on W/U in the obvious way and S/CS (W/U ) contains Stab(M) = SM say. Clearly y = x|W/U ∈ SM \1, so in fact y ∈ R(SM ). Further if B is an L-basis of V , then {b + U : b ∈ B ∩ (W \U )} is an M-basis of W/U and if dimD V is countable, then W/U has an M-basis by [5], 1.7. But then 4.4 implies that y = 1, a contradiction that completes the proof of 4.5. From now on assume that μ ≥ ω and that n > 0. Set C = CS (Vμ ) and T = CC (V /Vμ ). There is an obvious S-isomorphism of T onto H = HomD (V /Vμ , Vμ ), an S-module which we equate with its obvious image in HomD (V, Vμ ). Let   K = θ ∈ H : there exists i < ω with Vi θ = {0} and let TK denote the inverse image of K in T under our canonical isomorphism of T to H. Then by Section 4 of [6] we have that TK = ζ(S) = ζω (S). Let M denote the subseries V = V0 > V1 > . . . > Vi > . . . > Vω ≥ {0} of L∗ and N the series {W/Vω : W ∈ M} of V /Vω . Suppose C is an N-basis of V /Vω , suppose D is a subset of V mapping one-to-one onto C under the canonical map of V onto V /Vω and suppose that E is any basis of Vω . Then B = D ∪ E is an M-basis of V . 2 4.6. In the notation above, if V /Vω has an N-basis, e.g. if dimD (V /Vω ) is countable, then T ∩ R(S) = TK = ζ(S) = ζω (S). Proof. Let θ ∈ H. If g ∈ C, then [θ, g −1 ] = gθg −1 − θ = (g − 1)θ and more generally [θ, k g −1 ] = (g − 1)k θ for all k ≥ 0. If θ is the image in H of some t ∈ T ∩ R(S), then [θ, m g −1 ] = 0 for some m > 0 and consequently (g − 1)m θ = 0 and V (g − 1)m ≤ ker θ. By 4.1 there an integer i with Vi ≤ ker θ. This means that θ ∈ K and t ∈ TK . Thus T ∩ R(S) ≤ TK = ζ(S) ≤ T ∩ R(S). This completes the proof of 4.6. 2 Theorem 4.7. Suppose V has an L-basis or dimD V is countable. Then R(S) = ζ(S) = ζω (S). Proof. We are assuming here that μ ≥ ω, so we can apply 4.5 to V /Vμ . Thus we have that R(S) ⊆ CS (V /Vμ ). Also R(S) = ρ(S), which is a normal subgroup of S and S/C is nilpotent of class n − 1. Therefore  R(S), n−1 S ≤ C ∩ CS (V /Vμ ) ∩ R(S) = T ∩ R(S) ≤ ζ(S), the latter by 4.6. Consequently we have R(S) ≤ ζ(S) = ζω (S) ≤ R(S). The proof is complete. 2 5. Remarks In [4] Traustason constructs an example V with a descending series L of length ω for which Fitt(S) = HP (S). In fact the same example satisfies the stronger result Fitt(S) = P H(S), see [5]. (P H(S) denotes

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subgroup of S that is the union of all the normal subgroups of S that are hypercentral in their own right.) In view of 1.2 we would like an example V with a descending series L satisfying R− (S) = R(S). In view of the above Traustason’s example would be a prime candidate for this. Unfortunately it is not such an example.  To see this let U = i≥1 F ui and W = i≥1 F wi . (F is any field and the ui and wi are all non-zero of  course). Set Uj = i>j F ui and Wj = i>j F wi ≤ W . Then put V = U ⊕ W and Vi = Ui ⊕ Wi for each i.

Let L denote the series {Vi : i ≥ 0} of V and note that Vi = {0} and L is a descending series of length ω. Set S = Stab(L). We prove that R(S) = {1}. Now this V does not have an L-basis, so 4.4 cannot apply here. However we can modify the proof of 4.4 to cover this case. Let x ∈ R(S)\{1}. As in the proof of 4.4 there exist 1 ≤ k < m < ω and a vk in Vk−1 \Vk such that  Vk (x − 1) = {0} and vk (x − 1) = vm ∈ Vm−1 \Vm . For i = k, m, let vi = ui and set X = i≥k F v i ≤ Vk−1 . Let SX denote the stability group (in GL(X)) of the series {Vi ∩ X : i ≥ 0}. We need to extend the action of SX to the whole of V .  Let Bi be a basis of Vi−1 modulo Vi such that B = i≤m Bi contains all the vi for i ≤ m. (For example,  let Bi = {ui , wi } if i = k, m and Bi = {ui , vi } or {vi , wi } if i = k, m.) Set Y = k≤i≤m F v i , B = B\Y  and Z = b∈B F b. Then V = Y ⊕ Vm ⊕ Z = Y ⊕ Um ⊕ Wm ⊕ Z = X ⊕ Wm ⊕ Z. Extend the action of SX on X to an action on V by making SX act trivially on Wm ⊕ Z. Let g ∈ SX . Suppose i ≥ m. Then Ui ≤ Vi ∩ X = (Ui ⊕ Wi ) ∩ X = Ui ⊕ (Wi ∩ X) = Ui , since i ≥ m. Hence Vi = (Vi ∩ X) ⊕ Wi and so Vi (g − 1) ≤ Vi+1 ∩ X. Now suppose that i < m. Then Vi = (Vi ∩ Y ) ⊕ (Vi ∩ Z) ⊕ Um ⊕ Wm = (Vi ∩ Y ) ⊕ Um ⊕ (Vi ∩ Z) ⊕ Wm , = (Vi ∩ X) ⊕ (Vi ∩ Z) ⊕ Wm . Therefore again Vi (g − 1) ≤ Vi+1 ∩ X. Thus g stabilizes the series L and in this way we regard SX as a subgroup of S. Now vk x = vk + vm and vi x = vi for all i > k. Thus x|X ∈ SX and hence x|X ∈ R(SX ). But R(SX ) = {1} by 4.3 and yet x|X = 1. This contradiction completes the proof that R(S) = {1}. References [1] [2] [3] [4] [5] [6]

C. Casolo, O. Puglisi, Hirsch–Plotkin radical of stability groups, J. Algebra 370 (2012) 133–151. K.W. Gruenberg, The upper central series in soluble groups, Ill. J. Math. 5 (1961) 436–466. D.J.S. Robinson, Finiteness Conditions and Generalized Soluble Groups, Springer-Verlag, Berlin, 1972. G. Traustason, On the Hirsch–Plotkin radical of stability groups, J. Algebra 425 (2015) 31–41. B.A.F. Wehrfritz, Stability groups of series in vector spaces, J. Algebra 445 (2016) 352–364. B.A.F. Wehrfritz, The central heights of stability groups of series in vector spaces, Czechoslov. Math. J. (2016), in press.