1 Introduction
In 1957, G. Higman [18] investigated finite groups in which every element has prime power order (later they were called the CP-groups).
He gave a description of solvable CP-groups by showing that any such group is a p-group, or Frobenius, or 2-Frobenius, and its order has at most two distinct prime divisors.
Concerning a nonsolvable group G with the same property, he proved that G has the following structure:
(1.1)1⩽K<H⩽G,
where the solvable radical K of G is a p-group for some prime p, H/K is a unique minimal normal subgroup of G/K and is isomorphic to some nonabelian simple group S, and G/H is cyclic or generalized quaternion.
Later, Suzuki, in his seminal paper [34], where the new class of finite simple groups (now known as the Suzuki groups) was presented, found all nonabelian simple CP-groups.
The exhaustive description of CP-groups was completed by Brandl in 1981 [3].
It turns out that there are only eight possibilities for nonabelian composition factors S in (1.1):
L2(q), q=4,7,8,9,17,
L3(4),
Sz(q), q=8,32;
the solvable radical K must be a 2-group (possibly trivial), and there is only one CP-group with nontrivial factor G/H, namely, M10, an automorphic extension of A6.
In the middle of the 1970s, Gruenberg and Kegel invented the notion of the prime graph of a finite group (nowadays, it is also called the Gruenberg–Kegel graph) and noticed that, for groups with disconnected prime graph, very similar results to Higman’s can be proved.
Recall that the prime graphGK(G) of a finite group G is a labelled graph whose vertex set is π(G), the set of all prime divisors of |G|, and in which two different vertices labelled by r and s are adjacent if and only if G contains an element of order rs.
So, according to this definition, G is a CP-group if and only if GK(G) is a coclique (all vertices are pairwise nonadjacent).
Gruenberg and Kegel proved that a solvable finite group G with disconnected prime graph is Frobenius or 2-Frobenius and the number of connected components equals 2 (cf. Higman’s result), while a nonsolvable such group has again a normal series (1.1), where the solvable radical K is a nilpotent π1-group (here π1 is the vertex set of the connected component of GK(G) containing 2), H/K is a unique minimal normal subgroup of G/K and is isomorphic to some nonabelian simple group S, and G/H is a π1-group.
The above results were published for the first time by Gruenberg’s student Williams in [46].
There, he also started the classification of finite simple groups with disconnected prime graph that was completed by Kondrat’ev in 1989 [23] (see [29, Tables 1a–1c] for a revised version).
Though many nonabelian simple groups, for example, all sporadic ones, have disconnected prime graph, there is a bulk of classical and alternating simple groups which do not enjoy this property.
Nevertheless, as we will see below, if an arbitrary finite group has the set of orders of elements as a nonabelian simple group, then its structure can be described as Higman’s and Gruenberg–Kegel’s theorems do.
For convenience, we refer to the set ω(G) of the orders of elements of a finite group G as its spectrum and say that groups are isospectral if their spectra coincide.
It turns out that there are only three finite nonabelian simple groups L, namely, L3(3), U3(3), and S4(3), that have the spectrum as some solvable group [48, Corollary 1] (again, the latter must be Frobenius or 2-Frobenius).
It is also known that a nonsolvable group G isospectral to an arbitrary nonabelian simple group L has a normal series (1.1) with the only nonabelian composition factor H/K (see, e.g., [15, Lemma 2.2]).
Here we are interested in the nilpotency of the solvable radical K of G.
Theorem 1.
Let L be a finite nonabelian simple group distinct from the alternating group A10.
If G is a finite nonsolvable group with ω(G)=ω(L), then the solvable radical K of G is nilpotent.
Observe that, as shown in [28, Proposition 2] (see [33] for details), there is a nonsolvable group having a non-nilpotent solvable radical and isospectral to the alternating group A10.
Theorem 1 together with the aforementioned results gives the following.
Corollary 2.
Let L be a finite nonabelian simple group distinct from L3(3), U3(3), and S4(3).
If G is a finite group with ω(G)=ω(L), then there is a nonabelian simple group S such that S⩽G/K⩽AutS, with K being the largest normal solvable subgroup of G, is nilpotent provided L≠A10.
As in the case of CP-groups, a thorough analysis of groups isospectral to simple ones allows us to say more.
Although there are quite a few examples of finite groups with nontrivial solvable radical which are isospectral to nonabelian simple groups (see [26, Table 1]), in general, the situation is much better.
In order to describe it, we refer to a nonabelian simple group L as recognizable (by spectrum) if every finite group G isospectral to L is isomorphic to L, and as almost recognizable (by spectrum) if every such group G is an almost simple group with socle isomorphic to L.
It is known that all sporadic and alternating groups, except for J2, A6, and A10, are recognizable (see [30, 11]), and all exceptional groups excluding D43(2) are almost recognizable (see [42, 50]).
In 2007, Mazurov conjectured that there is a positive integer n0 such that all simple classical groups of dimension at least n0 are almost recognizable as well.
Mazurov’s conjecture was proved in [15, Theorem 1.1] with n0=62.
Later, it was shown [32, Theorem 1.2] that we can take n0=38.
It is clear that this bound is far from being final, and we conjectured that the following holds [15, Conjecture 1].
Conjecture 3.
Suppose that L is one of the following nonabelian simple groups:
Un(q), where n⩾5 and (n,q)≠(5,2),
S2n(q), where n⩾3, n≠4 and (n,q)≠(3,2),
O2n+1(q), where q is odd, n⩾3, n≠4 and (n,q)≠(3,3),
O2nε(q), where n⩾4, ε∈{+,-}, and (n,q,ε)≠(4,2,+),(4,3,+).
Then every finite group isospectral to L is an almost simple group with socle isomorphic to L.
In order to prove the almost recognizability of a simple group L, one should prove the triviality of the solvable radical K of a group isospectral to L.
It is not surprising that establishing the nilpotency of K is a necessary step toward that task (see, e.g., [16]).
Thus our main result, besides everything, provides a helpful tool for the verification of the conjecture.
2 Preliminaries
As usual, [m1,m2,…,mk] and (m1,m2,…,mk) denote respectively the least common multiple and greatest common divisor of integers m1,m2,…,ms.
For a positive integer m, we write π(m) for the set of prime divisors of m.
Given a prime r, we write (m)r for the r-part of m, that is, the highest power of r dividing m, and (m)r′ for the r′-part of m, that is, the ratio m(m)r.
If ε∈{+,-}, then, in arithmetic expressions, we abbreviate ε1 to ε.
The next lemma is well known (see, for example, [19, Chapter IX, Lemma 8.1]).
Lemma 2.1.
Let a and m be positive integers, and let a>1.
Suppose that r is a prime and a≡ε(modr), where ε∈{+1,-1}.
If r is odd, then (am-εm)r=(m)r(a-ε)r.
If a≡ε(mod4), then (am-εm)2=(m)2(a-ε)2.
Let a be an integer.
If r is an odd prime and (a,r)=1, then e(r,a) denotes the multiplicative order of a modulo r.
Define e(2,a) to be 1 if 4 divides a-1 and to be 2 if 4 divides a+1.
A primitive prime divisor of am-1, where |a|>1 and m⩾1, is a prime r such that e(r,a)=m.
The set of primitive prime divisors of am-1 is denoted by Rm(a), and we write rm(a) for an element of Rm(a) (provided that it is not empty).
The following well-known lemma was proved in [2] and independently in [49].
Lemma 2.2 (Bang–Zsigmondy).
Let a and m be integers, |a|>1 and m⩾1.
Then the set Rm(a) is not empty, except when
(a,m)∈{(2,1),(2,6),(-2,2),(-2,3),(3,1),(-3,2)}.
Lemma 2.3.
Let k, m, and a be positive integers numbers, where a>1.
Then Rmk(a)⊆Rm(ak).
If, in addition, (m,k)=1, then Rm(a)⊆Rm(ak).
Proof.
It easily follows from the definition of Rm(a) (see, e.g., [14, Lemma 6]).
∎
The largest primitive divisor of am-1, where |a|>1, m⩾1, is the number
km(a)={∏r∈Rm(a)|am-1|rifm≠2,∏r∈R2(a)|a+1|rifm=2.
The largest primitive divisors can be written in terms of cyclotomic polynomials Φm(x).
Lemma 2.4.
Let a and m be integers, |a|>1 and m⩾3.
Suppose that r is the largest prime divisor of m and l=(m)r′.
Then km(a)=|Φm(a)|(r,Φl(a)).
Furthermore, (r,Φl(a))=1 whenever l does not divide r-1.
Proof.
This follows from [31, Proposition 2] (see, e.g., [39, Lemma 2.2]).
∎
Recall that ω(G) is the set of the orders of elements of G.
We write μ(G) for the set of elements of ω(G) that are maximal under divisibility.
The least common multiple of the elements of ω(G) is equal to the exponent of G and denoted by exp(G).
Given a prime r, ωr(G) and expr(G) are respectively the spectrum and the exponent of a Sylow r-subgroup of G.
Similarly, ωr′(G) and expr′(G) are respectively the set of the orders of elements of G that are coprime to r and the least common multiple of these orders.
A coclique of a graph is a set of pairwise nonadjacent vertices.
Define t(G) to be the largest size of a coclique of the prime graph GK(G) of a finite group G.
Similarly, given r∈π(G), we write t(r,G) for the largest size of a coclique of G containing r.
It was proved in [35] that a finite group G with t(G)⩾3 and t(2,G)⩾2 has exactly one nonabelian composition factor.
Below, we provide the refined version of this assertion from [38].
Lemma 2.5 ([35, 38]).
Let L be a finite nonabelian simple group such that t(L)⩾3 and t(2,L)⩾2, and suppose that a finite group G satisfies ω(G)=ω(L).
Then the following holds.
There is a nonabelian simple group S such that S⩽G¯=G/K⩽AutS, with K being the largest normal solvable subgroup of G.
If
ρ
is a coclique of GK(G) of size at least
3
, then at most one prime of
ρ
divides |K|⋅|G¯/S|.
In particular, t(S)⩾t(G)-1.
If r∈π(G) is not adjacent to
2
in GK(G), then r does not divide |K|⋅|G¯/S|.
In particular, t(2,S)⩾t(2,G).
The next lemma summarizes what we know about almost recognizable simple groups (see [15, 39, 32]).
Lemma 2.6.
Let L be one of the following nonabelian simple groups:
a sporadic group other than J2,
an alternating group An, where n≠6,10,
an exceptional group of Lie type other than D43(2),
Ln(q), where n⩾27 or q is even,
Un(q), where n⩾27, or q is even and (n,q)≠(4,2),(5,2),
S2n(q),O2n+1(q), where either q is odd and n⩾16, or q is even, n≠2,4 and (n,q)≠(3,2),
O2n+(q), where either q is odd and n⩾19, or q is even and (n,q)≠(4,2),
O2n-(q), where either q is odd and n⩾18, or q is even.
Then every finite group isospectral to L is isomorphic to some group G with L⩽G⩽AutL.
In particular, there are only finitely many pairwise nonisomorphic finite groups isospectral to L.
Now we list the spectra of some groups of low Lie rank and give some lower bounds on the exponents of exceptional groups of Lie type.
Throughout the paper, we repeatedly use, mostly without explicit reference, the description of the spectra of simple classical groups from [5] (with corrections from [13, Lemma 2.3]) and [4], as well as the adjacency criterion for the prime graphs of simple groups of Lie type from [43] (with corrections from [44]).
Also, we use the abbreviations Lnε(q) and E6ε(q), where ε∈{+,-}, that are defined as follows:
Ln+(q)=Ln(q),
Ln-(q)=Un(q),
E6+(q)=E6(q), and
E6-(q)=E62(q).
Lemma 2.7 ([4]).
Let q be a power of an odd prime p, and let L=L4ε(q).
The set ω(L) consists of all divisors of the following numbers:
(q2+1)(q+ε)(4,q-ε),
q3-ε(4,q-ε),
q2-1,
p(q2-1)(4,q-ε),
p(q-ε),
In particular, expp′(L)=(q4-1)(q2+εq+1)2.
Lemma 2.8 ([4]).
Let q be a power of an odd prime p, and let L=L6ε(q).
The set ω(L) consists of all divisors of the following numbers:
(q3+ε)(q2+εq+1)(6,q-ε),
q5-ε(6,q-ε),
q4-1,
p(q4-1)(6,q-ε),
p(q3-ε),
p(q2-1),
In particular,
expp′(L)=(q6-1)(q5-ε)(q2+1)q-ε.
Lemma 2.9 ([34]).
Let u=22k+1⩾8.
Then ω(B22(u)) consists of all divisors of the numbers 4, u-1, u-2u+1, and u+2u+1.
In particular, we have
exp(B22(u))=4(u2+1)(u-1).
Lemma 2.10.
Let u be a power of a prime v.
Then ω(G2(u)) consists of all divisors of the numbers u2±u+1, u2-1, v(u±1) together with all divisors of
In particular, expv′(S)=u6-1(3,u2-1).
Furthermore, if a Sylow r-subgroup of G2(u) is cyclic, then r divides u2+u+1 or u2-u+1.
Lemma 2.11.
If S and f(u) are as follows, then exp(S)>f(u).
| S | E8(u) | E7(u) | E6±(u) | F4(u) | F42(u) | G22(u) |
| f(u) | 2u80 | 3u48 | u26 | 3u16 | 5u10 | 2u4 |
Proof.
If S≠F4(u), G22(u), the assertion is proved in [16, Lemma 3.6].
It follows from [45] that exp(G22(u))=9(u3+1)(u-1)4, and so
exp(G22(u))>9(u4-u3)4>2u4.
Let S=F4(u), and let u be a power of a prime v.
Using [8] and Lemma 2.1, it is not hard to see that expv(F4(u))⩾13 and expv′(F4(u))=(u12-1)(u4+1)(2,u-1)2, so we have the desired bound.
∎
Lemma 2.12.
Let S be a finite simple group of Lie type.
If r,s,t∈π(S) and rt,st∈ω(S), but rs∉ω(S), then a Sylow t-subgroup of S is not cyclic.
Proof.
Let S be a group over the field of order u.
Observe that S is not L2(u), B22(u), S4(u) since every connected component of the prime graphs of these groups is a clique [25].
In particular, we may assume that t is coprime to u.
It is well known that Sylow 2-subgroups of a nonabelian simple group cannot be cyclic, so we may also assume that t≠2.
It is sufficient to show that S contains an elementary abelian group of order t2, and to do so, we use [6] and the information about tori in exceptional groups collected in [21].
For brevity, in this proof, we denote a direct product of two cyclic groups of order a by a×a.
Suppose that S=Lm(u) with m⩾4.
Applying [36, Lemma 2.4 (ii)] if r and s are coprime to u and [43, Proposition 3.1] otherwise, and interchanging r and s if necessary, we may assume that e(r,u)>n2.
If we have e(t,u)>n2, then by [36, Lemma 2.4 (iii)], we conclude that e(r,u)=e(t,u).
By [43, 44], the neighbors in GK(S) of a prime w with e(w,u)>2 depend only on the number e(w,u), so t is adjacent to s if and only if r is, a contradiction.
Thus e(t,u)⩽n2, and hence S contains a subgroup of the form (ue(t,u)-1)×(ue(t,u)-1).
The same argument goes for the groups Um(u), S2m(u), O2m+1(u), and O2m±(q), where m⩾4, with the function φ(⋅,S) defined in [36, p. 328] in place of e(⋅,u).
For the groups L3τ(u), where τ∈{+,-}, S6(u), O7(u) and exceptional groups, we use the compact form of their prime graphs given in [44, 24] to determine possible values of t.
If S=L3τ(u), then t divides u-τ and t≠3 if (u-τ)3=3.
The lemma follows since L3τ(u) contains a subgroup of the form u-τ(3,u-τ)×u-τ(3,u-τ).
Also, we see that S≠G22(u) because t≠2.
All the remaining groups contain a subgroup of the form (u±1)×(u±1), so we may assume that t does not divide u2-1.
Then S is one of E6±(u), E7(u), E8(u) and either t∈R3(u)∪R6(u), or t∈R4(u) and S≠E6±(u).
All these groups contain subgroups of the form (q2±q+1)×(q2±q+1), and E7(u), E8(u) contain a subgroup of the form (q2+1)×(q2+1).
∎
The next five lemmas are tools for calculating the orders of elements in group extensions.
Most of them are corollaries of well-known results (such as the Hall–Higman theorem).
Lemma 2.13.
Suppose that G is a finite group, K is a normal subgroup of G and w∈π(K).
If G/K has a noncyclic Sylow t-subgroup for some odd prime t≠w, then tw∈ω(G).
Proof.
Let W be a Sylow w-subgroup of K, and let T be a Sylow t-subgroups of NG(W).
By the Frattini argument, G=KNG(W), and so T is also noncyclic.
By the classification of Frobenius complements, T cannot act on W fixed-point-freely; therefore, tw∈ω(G).
∎
Lemma 2.14.
Suppose that G is a finite group, K is a normal w-subgroup of G for some prime w and G/K is a Frobenius group with kernel F and cyclic complement C.
If (|F|,w)=1 and F is not contained in KCG(K)/K, then we have w|C|∈ω(G).
Proof.
See [27, Lemma 1].
∎
Lemma 2.15.
Let v and r be distinct primes, and let G be a semidirect product of a finite v-group U and a cyclic group 〈g〉 of order r.
Suppose that [U,g]≠1 and G acts faithfully on a vector space W of positive characteristic w≠v.
Then either the natural semidirect product W⋊G has an element of order rw, or the following holds:
v=2 and r is a Fermat prime.
Proof.
See [36, Lemma 3.6].
∎
Lemma 2.16.
Let G be a finite group, let N be a normal subgroup of G, and let G/N be a simple classical group over a field of characteristic v.
Suppose that G acts on a vector space W of positive characteristic w, r is an odd prime dividing the order of some proper parabolic subgroup of G/N, the primes v, w, and r are distinct, and v,r∉π(N).
Then the natural semidirect product W⋊G has an element of order rw.
Proof.
Let S=G/N.
We may assume CG(W)⊆N since, otherwise, CG(W) has S as a section and the lemma follows.
Let P be the proper parabolic subgroup of S containing an element g of order r, and let U be the unipotent radical of P.
By [10, Statement 13.2], it follows that [U,g]≠1.
Since both v and r are coprime to N, there is a subgroup of G isomorphic to U⋊〈g〉, and this subgroup acts on W faithfully.
Applying Lemma 2.15, we see that either rw∈ω(W⋊G), or v=2, r is a Fermat prime, U is nonabelian, and CU(g)≠1.
Suppose that the latter case holds.
If S≠Un(u), then the conditions v=2 and 2r∈ω(S) imply that r divides the order of a maximal parabolic subgroup of S with abelian unipotent radical (for example, the order of the group P1 in notation of [22]), and we can proceed as above with this parabolic subgroup instead of P.
Let S=Un(u).
Writing k=e(r,-u), we have that k⩽n-2 and k divides t-1.
Since r is a Fermat prime, it follows that k=1 or k is even, and so r divides u2-1 or uk-1.
If n⩾4, then S includes a Frobenius group whose kernel is a v-group and complement has order r, and we again can apply Lemma 2.15.
Let n=3.
Then r divides u+1.
Since |S|r=(u+1)r2 and expr(S)=(u+1)r, Sylow r-subgroups of S are not cyclic, and applying Lemma 2.13 completes the proof.
∎
Lemma 2.17.
Let G=L2(v), where v>3 is a prime.
Suppose that G acts on a vector space W of positive characteristic w and w∉π(G).
If r and s are two distinct odd primes from π(G), then {r,s,w} is not a coclique in GK(W⋊G).
Proof.
Clearly, we may assume that G acts on W faithfully.
If one of r and s, say r, divides v-12, then applying Lemma 2.15 to a Borel subgroup of G yields wr∈GK(W⋊G).
If both r and s divide v+12, then rs∈ω(G).
Thus we are left with the case where one of r and s is equal to v, while another divides v+12.
We prove that either an element of order v or an element of order v+12 has a fixed point in W.
We may assume that W is an irreducible G-module.
The ordinary character table of L2(v) is well known.
We use the result and notation due to Jordan [20].
Define ϵ to be +1 or -1 depending on whether v-1 is even or odd.
Also, let ζ be some fixed not-square in the field of order v.
We denote by μ and ν, respectively, the conjugacy classes containing the images in L2(v) of the matrices (1101) and (1ζ01).
We denote by Sb with 1⩽b⩽v-ϵ4 the conjugacy class containing the image of the b-power of some fixed element of order v+12.
Then the values of nontrivial irreducible characters of G on elements of orders dividing v and v+12 are as follows:
| 1 | μ | ν | Sb |
| χ0 | v | 0 | 0 | -1 |
| χ± | v+ϵ2 | ϵ±ϵv2 | ϵ∓ϵv2 | -(1-ϵ)(-1)b2 |
| χu | v+1 | 1 | 1 | 0 |
| χt | v-1 | -1 | -1 | -tb-t-b |
Here u and t are the roots (except ±1) of the respective equations u(v-1)/2=1 and t(v+1)/2=1.
If g∈G and χ is the character of the representation on W, then CW(g)≠0 if and only if the sum
(2.1)∑h∈〈g〉χ(h)
is positive.
If g∈μ, then the sum is equal to χ(1)+(v-12)⋅(χ(μ)+χ(ν)), and so it is clearly positive unless χ=χ± and ϵ=-1, or χ=χt.
Let g∈S1.
If ϵ=-1, then (2.1) is equal to
χ(1)+χ(S(v+1)/4)+2∑1⩽b⩽v-34χ(Sb).
Taking χ=χ±, we have
v-12-(-1)(v+1)/4-2∑1⩽b⩽v-34(-1)b=v-12-∑1⩽b⩽v+12(-1)b=v+12.
Similarly, if χ=χt, then we have
v-1-2t(v+1)/4-2∑1⩽b⩽v-34(tb+t-b)=v-1-2∑1⩽b⩽v-12tb=v+1.
If ϵ=+1 and χ=χt, then (2.1) is equal to
χt(1)+2∑1⩽b⩽v-14χt(Sb)=v-1-2∑1⩽b⩽v-12tb=v+1.
The proof is complete.
∎
We conclude the section with a lemma from [16, Lemma 2.9].
We give it with the proof because we will use variations of this proof further.
Lemma 2.18.
Let G be a finite group, and let S⩽G/K⩽AutS, where K is a normal solvable subgroup of G and S is a nonabelian simple group.
Suppose that, for every r∈π(K), there is a∈ω(S) such that π(a)∩π(K)=∅ and ar∉ω(G).
Then K is nilpotent.
Proof.
Otherwise, the Fitting subgroup F of K is a proper subgroup of K.
Define G~=G/F and K~=K/F.
Let T~ be a minimal normal subgroup of G~ contained in K~, and let T be its preimage in G.
It is clear that T~ is an elementary abelian t-group for some prime t.
Given r∈π(F)∖{t}, denote the Sylow r-subgroup of F by R, its centralizer in G by Cr, and the image of Cr in G~ by C~r.
Since C~r is normal in G~, it follows that either T~⩽C~r or C~r∩T~=1.
If T~⩽C~r for all r∈π(F)∖{t}, then T is a normal nilpotent subgroup of K, which contradicts the choice of T~.
Thus there is
r∈π(F)∖{t} such that C~r∩T~=1.
If CG~(T~) is not contained in K~, then it has a section isomorphic to S.
In this case, ta∈ω(G) for every a∈ωt′(S), contrary to the hypothesis.
Thus
CG~(T~)⩽K~.
Choose a∈ωr′(S) such that π(a)∩π(K)=∅ and ra∉ω(G), and let x∈G~ be an element of order a.
Then x∉CG~(T~); therefore, [T~,x]≠1, so [T~,x]⋊〈x〉 is a Frobenius group with complement 〈x〉.
Since C~r∩T~=1, we can apply Lemma 2.14 and conclude that ra∈ω(G), contrary to the choice of a.
∎
3 Reduction
In this section, we apply known facts concerning Theorem 1 to reduce the general situation to a special case.
Let L be a finite nonabelian simple group, and let G be a nonsolvable finite group with ω(G)=ω(L).
Since the spectrum of a group determines its prime graph, it follows that GK(G)=GK(L).
In particular, if GK(L) is disconnected, then so is GK(G).
In this case, G satisfies the hypothesis of the Gruenberg–Kegel theorem; hence the solvable radical K of G must be nilpotent (see [46, Theorem A and Lemma 3]).
Thus, proving Theorem 1, we may assume that GK(L) is connected.
If L is sporadic, alternating, or an exceptional group of Lie type, Lemma 2.6 says that L is either almost recognizable or one of the groups J2, A6, A10, D43(2).
In the former case, the solvable radical K is trivial, while in the latter case, if we exclude A10, then K is nilpotent because GK(L) is disconnected [29, Tables 1a–1c].
Thus we may suppose that L is a classical group.
Let p and q be the characteristic and order of the base field of L, respectively.
If L is one of the following groups:
Lnε(q), where n⩽3,
S2n(q),O2n+1(q), where n=2,4, U4(2), U5(2), S6(2), and O8+(2),
then GK(L) is disconnected [29].
Together with Lemma 2.6, this shows that we may assume that q is odd, and n⩾4 for L=Lnε(q), n⩾3 for L∈{S2n(q),O2n+1(q)}, and n⩾4 for L=O2nε(q).
Furthermore, applying information on the sizes of maximal cocliques and 2-cocliques from [43, 44], we obtain t(L)⩾3 and t(2,L)⩾2, so the conclusion of Lemma 2.5 holds for G.
In particular, G has a normal series
(3.1)1⩽K<H⩽G,
where K is the solvable radical of G, H/K is isomorphic to a nonabelian simple group S, and G/K is isomorphic to some subgroup of Aut(S).
The group S is neither an alternating group by [40, Theorem 1] and [41, Theorem 1], nor a sporadic group by [40, Theorem 2] and [41, Theorem 2].
If S is a group of Lie type over a field of characteristic p, then S≃L due to [41, Theorem 3] and [15, Theorem 2].
Then [12, Corollary 1.1] yields that either K=1 or L=L4ε(q).
In the latter case, K must be a p-group in view of [47, Lemma 11].
Thus we may assume that S is a simple group of Lie type over the field of order u and characteristic v≠p.
Finally, applying Lemma 2.18, we derive the following assertion.
Lemma 3.1.
Let q be odd, and let L be one of the simple groups Lnε(q), where n⩾4, S2n(q), where n⩾3, O2n+1(q), where n⩾3, or O2nε(q), where n⩾4.
Suppose that G is a finite group such that ω(G)=ω(L), and K and S are as in (3.1).
Then either K is nilpotent, or one of the following holds:
L=O2nε(q), where n is odd, q≡ε(mod8), and Rn(εq)∩π(S)⊆π(K);
L=Lnε(q), where 1<(n)2<(q-ε)2, and Rn-1(εq)∩π(S)⊆π(K);
L=Lnε(q), where (n)2>(q-ε)2 or (n)2=(q-ε)2=2, and
Rn(εq)∩π(S)⊆π(K).
Proof.
If L≠L4ε(q), then this follows from [16, Lemma 4.1].
Let L=L4ε(q).
We show that either, for every r∈π(K), there is a∈ω(S) satisfying π(a)∩π(K)=∅ and ar∉ω(G), in which case K is nilpotent by Lemma 2.18, or one of (ii) and (iii) holds.
If (q-ε)2=4, then the elements of R4(εq)∪R3(εq) are not adjacent to 2 in GK(G), so we can take a=r4(εq) if r is coprime to m4=(q2+1)(q+ε)(4,q-ε), and a=r3(εq) otherwise.
Let (q-ε)2>4.
If r is coprime to m4, then a=r4(εq).
If r divides m4 and there is s∈(π(S)∩R3(εq))∖π(K), then a=s.
If (q-ε)2<4, the same argument goes through, but with m4, r4(εq) and R3(εq) replaced by m3=q3-ε(4,q-ε), r3(εq) and R4(εq), respectively.
Therefore, (ii) or (iii) holds, and the proof is complete.
∎
4 General case
Let G and L be as in Theorem 1, and suppose that the solvable radical K of G is not nilpotent.
According to the previous section, we may assume that L is a classical group over a field of odd characteristic p and order q with connected prime graph, G has a normal series
1⩽K<H⩽G,
where S=H/K is a simple group of Lie type over a field of characteristic v≠p and order u, and G/K is a subgroup of Aut(S).
Moreover, we may assume that one of the assertions (i)–(iii) of the conclusion of Lemma 3.1 holds.
Before we proceed with the proof of Theorem 1, we introduce some notation which allows us to deal with the cases described in (i)–(iii) of Lemma 3.1 simultaneously.
We define z to be the unique positive integer such that Rz(εq)⊆π(L) and each r∈Rz(εq) is not adjacent to 2 in GK(L).
We also define y to be the unique positive integer such that Ry(εq)⊆π(L) and each r∈Ry(εq) is adjacent to 2 but not to p.
Both z and y are well-defined for all groups L in (i)–(iii).
Indeed, y=n, z=2n-2 in (i); y=n-1, z=n in (ii); and y=n, z=n-1 in (iii).
Moreover, the last containments of all three assertions in Lemma 3.1 can be written uniformly as
(4.1)Ry(εq)∩π(S)⊆π(K).
Observe that rz(εq) is not adjacent to both 2 and p, and hence {p,rz(εq),ry(εq)} is a coclique in GK(L).
Also, observe that z,y⩾3, so if r∈Rz(εq)∪Ry(εq), then r⩾5.
It is not hard to check that every number in ω(L) that is a multiple of rz(εq) or ry(εq) has to divide the only element of μ(L) which we denote by mz or my, respectively.
Namely, mz=(qn-1+1)(q+ε)4, my=qn-ε4 in (i);
and in (ii), (iii),
{mz,my}={qn-1(q-ε)(n,q-ε),qn-1-ε(n,q-ε)}.
The numbers mz and my are coprime, so rz(εq) and ry(εq) have no common neighbors in GK(L).
Also, we note that my is even.
The definition of the number z and Lemma 2.5 (ii) imply that
Rz(εq)∩(π(K)∪π(G/H))=∅,
and so kz(εq)∈ω(S).
Hence there is a number k(S)∈ω(S) such that kz(εq) divides k(S) and every r∈π(k(S)) is not adjacent to 2 in GK(S).
Using, for example, [43, Section 4], one can see that this number is uniquely determined and of the form kj(u) for some positive integer j provided that S≠L2(u) and S is not a Suzuki or Ree group.
If k(S)≠kj(u), then k(S)=v when S=L2(u), and k(S)=f+(u) or f-(u), where f±(u) is as follows, when S is a Suzuki or Ree group.
| S | f±(u) |
| B22(u) | u±2u+1 |
| G22(u) | u±3u+1 |
| F22(u) | u2±2u3+u±2u+1 |
Lemma 4.1.
Ry(εq)⊆π(K).
Proof.
Suppose that r∈Ry(εq)∖π(K).
Then (4.1) yields r∈π(G/H)∖π(H).
Since S is a simple group of Lie type, r∉π(S) and r⩾5, there is a field automorphism φ of S of order r lying in G/K and u=u0r.
Denote by S(u0) the simple group of the same Lie type as S over the field of order u0.
The centralizer of φ in S has a section isomorphic to S(u0); therefore, r⋅π(S(u0))⊆ω(G).
Since r∈π(G/H), we have that p∈π(S) by Lemma 2.5.
By the above, it follows that p∉π(S(u0)), and so e(p,u0)≠e(p,u).
Then the equality u=u0r yields e(p,u0)=re(p,u).
Since e(p,u0) divides p-1, we see that r divides p-1, contrary to the fact that y⩾3.
∎
Lemma 4.2.
Let F be the Fitting subgroup of K.
If s∈π(G) and (s,my)=1, then Sylow s-subgroups of G are cyclic.
Proof.
Suppose that K is not nilpotent.
We will use the notation from the proof of Lemma 2.18, which means the following.
The Fitting subgroup F of K is a proper subgroup of K, and G~=G/F, K~=K/F.
We choose T~ to be a minimal normal subgroup of G~ contained in K~ and let T be its preimage in G.
It is clear that T~ is an elementary abelian t-group for some prime t.
Given r∈π(F)∖{t}, denote the Sylow r-subgroup of F by R, its centralizer in G by Cr, and the image of Cr in G~ by C~r.
Since F is a proper subgroup of F, there is r∈π(F)∖{t} such that C~r∩T~=1, and we fix some r enjoying this condition.
Suppose first that CG~(T~)⩽K~.
Then r divides mz; otherwise, due to the standard arguments from the proof of Lemma 2.18, we have rrz(εq)∈ω(G), which is not the case.
By Lemma 4.1, we have that Ry(εq)⊆π(K).
If w∈Ry(εq) and W is a Sylow w-subgroup of K, then R⋊W is a Frobenius group because (my,mz)=1.
It follows that W is cyclic.
Thus NG(W)/CG(W) must be abelian.
The Frattini argument implies that NG(W) contains a nonabelian composition section S, and so does CG(W).
Then wrz(εq)∈ω(G), a contradiction.
Thus we may assume that CG~(T~)⩽̸K~.
Then CG~(T~) contains a nonabelian composition factor isomorphic to S, and in particular, t is adjacent to every prime in π(S).
Since rz(q) divides |S|, it follows that t divides mz and
Ry(εq)∩π(S)=∅.
If Ry(εq)∩(π(K)∖π(F))≠∅ and W~ is a Sylow w-subgroup of K~ for some prime w from this intersection, then the group T~⋊W~ is Frobenius, and we get a contradiction as above.
Thus Ry(εq)⊆π(F).
Let w∈Ry(εq), and let W be a Sylow w-subgroup of F, while P be a s-subgroup of G for some s∈π(G) such that (s,my)=1.
Since w and s are nonadjacent in GK(G), the group W⋊P is Frobenius.
Therefore, P is cyclic.
∎
Lemma 4.3.
We may assume that L=L6ε(q) and z=5, or L=L4ε(q).
Proof.
Suppose that L≠L6(εq), L4(εq).
One can easily check using [44] that t(L)⩾4 and there is a coclique of size 4 in GK(L) that contains an element of the form ry(εq) and does not contain p.
Let ρ be such a coclique and ρ′=ρ∖Ry(εq).
Lemma 4.2 (i) and Lemma 2.5 imply that
{p}∪ρ′⊆π(S)∖(π(K)∪π(G/H)).
On the other hand, p is not adjacent to at most one element of ρ′, an element of the form rz(εq).
Hence p is adjacent to at least two elements of ρ′.
By Lemma 2.12, a Sylow p-subgroup of S is not cyclic.
This contradicts Lemma 4.2 (ii).
Suppose that L=L6ε(q) and z=6.
Since
{p,r6(εq),r5(εq)} and {r3(εq),r4(εq),r5(εq)}
are cocliques in GK(L), it follows that
{p}∪R3(εq)∪R4(εq)⊆π(S)∖(π(K)∪π(G/H)).
However, p is adjacent to r3(εq) and r4(εq) in GK(L), and we derive a contradiction as above.
∎
5 Small dimensions
In this section, we handle the remaining case L=L6ε(q) and z=5, or L=L4ε(q).
If L is one of the groups L4(3), U4(3), L4(5), and U6(5), then L has disconnected prime graph, so K is nilpotent.
If L=U4(5) or L6(3), then K=1 (see [37] and [7], respectively).
If L=L6(5) or L=U6(3), then z=6.
Thus we may assume that q>5, and in particular expp(L)⩽q.
We begin with some more notation and two auxiliary lemmas.
Choose x∈{1,2} such that 2∉Rx(εq).
By definition and Lemma 2.2, Rx(εq) is not empty and consists of odd primes.
Furthermore, it is not hard to see that x=2 if z=n and x=1 if z=n-1.
If x=2, then it is clear that rx(εq) is adjacent to rz(εq) but not to ry(εq) in GK(L).
If x=1, then r∈Rx(εq) is adjacent to rz(εq) if and only if (q-ε)r>(n)r and not adjacent to ry(εq) if and only if (q-ε)r⩾(n)r (see, for example, [43, Propositions 4.1 and 4.2]).
Since n=6 or 4, we conclude that rx(εq) is always adjacent to rz(εq) but not to ry(εq) unless n=6, rx(εq)=3, and (q-ε)3=3 in which case rx(εq) is not adjacent to both rz(εq) and ry(εq).
Lemma 5.1.
Let L=L6ε(q), z=5, and σ=π(L)∖R6(εq), and suppose that S≠D43(u) is an exceptional group of Lie type such that k5(εq)∈ω(S)⊆ω(L).
Then the following hold:
if S≠G2(u),
B22(u), then expσ(L)<exp(S);
if S=G2(u),
B22(u),
η=σ∖R1(εq),
(q-ε,5)=1, then expη(L)<exp(S).
Proof.
Write k=k5(εq), and recall that k(S) is the number in ω(S) such that k divides k(S) and every r∈π(k(S)) is not adjacent to 2 in GK(S).
By Lemma 2.4, we have k=q5-ε(q-ε)(5,q-ε).
(i) It is easy to see using Lemmas 2.8 and 2.1 that
expσ(L)=expp(L)⋅(3,q+ε)(q5-ε)(q2+εq+1)(q2+1)(q+ε).
If ε=+, then k>q45 and
expσ(L)<q⋅3⋅q5⋅q3-1q-1⋅q4-1q-1<3q11(qq-1)2⩽3q11⋅(76)2=49q1112.
So q4<5k(S) and
expσ(L)<49⋅(5k(S))11/412<342(k(S))11/4.
Similarly, if ε=-, then k>7q440 and
expσ(L)<q⋅3⋅(q5+1)(q2-q+1)(q2+1)(q-1)<3q11.
It follows that
expσ(L)<3⋅(40k(S)7)11/4<363(k(S))11/4.
In either case, we have expσ(L)<exp(S) unless
(5.1)exp(S)<363(k(S))11/4.
Observe that 211/4<7.
Let S=E8(u).
Then exp(S)>2u80 by Lemma 2.11 and k(S)<2u8.
It follows from (5.1) that 2u80<363(2u8)11/4, and so u58<182⋅7.
Similarly, if S=E7(u), then exp(S)>3u48 by Lemma 2.11 and k(S)<2u6, and we derive that 3u48<363(2u6)11/4, or equivalently, u63/2<121⋅7.
In both cases, we have a contradiction.
Let S=E6±(u).
Then exp(S)>u26 and k(S)⩽2u6, so u26<363(2u6)11/4, or equivalently, u19<3632⋅211/2.
The last inequality yields u=2.
If S=F4(u), then exp(S)>3u16 and k(S)⩽u4+1.
Thus 3u16<363(u4+1)11/4, and so again u=2.
In either case, we have 7q440<k<k(S)<128, contrary to the fact that q>5.
If S=F42(u), then we have exp(S)>5u10 and k(S)⩽2u2.
It follows that 5u10<363(2u2)11/4, whence u9/2<73⋅7.
This is a contradiction since u⩾8.
If S=G22(u), then exp(S)>2u4 and k(S)⩽u+3u+1<2u.
We have 2u4<363(2u)11/4, whence u5<(3632)4⋅211, and so u=33 or u=35.
In fact, using u+3u+1 instead of 2u, one can check that (5.1) does not hold for u=35.
It follows that k<2⋅33, but we saw above that this is false.
(ii) Since z=5, it follows that (q-ε)2=2, and so the R1(εq)-part of expσ(L) is equal to (q-ε)(5,q-ε)(3,q-ε)2.
Hence
expη(L)⩽2expσ(L)q-ε=2expp(L)⋅(3,q+ε)(q5-ε)(q2+εq+1)(q2+1)(q+ε)q-ε.
If ε=+, then k>q4 and
expη(L)<6q⋅q5-1q-1⋅q3-1q-1⋅q4-1q-1⩽6q10⋅(76)3<10q10<10k5/2.
If ε=-, then k>7q48 and
expη(L)<6qq5+1q+1⋅(q2-q+1)(q2+1)(q-1)⩽6q10<6(8k7)5/2<9k5/2.
In either case, expη(L)<exp(S) unless
(5.2)exp(S)<10(k(S))5/2.
If S=G2(u), then
exp(S)⩾7(u6-1)(3,u2-1) and k(S)⩽u2+u+1.
So (5.2) yields 7(u6-1)<30(u2+u+1)5/2, or equivalently,
7(u2-1)(u2-u+1)<30(u2+u+1)3/2.
The last inequality is not valid if u⩾16, and hence u⩽13, which implies that k⩽183.
This is a contradiction since k⩾7q48.
If S=B22(u), then exp(S)=4(u2+1)(u-1) and k(S)⩽u+2u+1.
So we have 4(u2+1)(u-1)<30(u+2u+1)5/2, or equivalently,
2(u-2u+1)(u-1)<15(u+2u+1)3/2.
It follows that u=23 or u=25, and therefore, k⩽41, a contradiction.
∎
Lemma 5.2.
Let L=L4ε(q), σ=π(L)∖Ry(εq) and suppose that S≠D43(u) is an exceptional group of Lie type such that kz(εq)∈ω(S)⊆ω(L).
Then the following holds:
if S≠G2(u),
B22(u), then expσ(L)<exp(S);
if S=G2(u),
B22(u),
η=σ∖Rx(εq), then expη(L)<exp(S).
Proof.
Write k=kz(εq).
(i) If z=4, then k=q2+12>q22 and
expσ(L)=expp(L)⋅(3,q-ε)(q4-1)2<3q52<3(2k)5/22<9k5/2.
Let z=3.
If ε=+, then k=q2+q+1(3,q-1)>q23 and
expσ(L)=expp(L)⋅(q2+q+1)(q2-1)<7q56<7(3k)5/26<19k5/2.
If ε=-, then k=q2-q+1(3,q+1)>7q224 and
expσ(L)=expp(L)⋅(q2-q+1)(q2-1)<q5<(24k7)5/2<22k5/2.
Thus expσ(L)<exp(S) unless
(5.3)exp(S)<22(k(S))5/2.
Since (5.3) is stronger than (5.1), we may assume that S is one of the groups E6±(2), F4(2), G22(33).
It is easy to check that (5.3) does not hold for all these groups.
(ii) If z=4, then x=2 and the Rx(εq)-part of exp(L) is equal to q+ε2.
So
expη(L)=expp(L)⋅(3,q-ε)(q2+1)(q-ε)<7q42<7(2k)22=14k2.
Similarly, if z=3 and ε=+, then
expη(L)=2expp(L)⋅(q2+q+1)(q+1)<2q4(76)2<2⋅(3k)2⋅(76)2<25k2.
And if z=3 and ε=-, then
expη(L)=2expp(L)⋅(q2-q+1)(q-1)<2q4<2⋅(24k7)2<24k2.
Thus expη(L)<exp(S) unless
(5.4)exp(S)<25(k(S))2.
If S=G2(u), then arguing as in the proof of Lemma 5.1, we derive that (5.4) yields
7(u2-1)(u2-u+1)<75(u2+u+1),
whence u⩽4.
But then k⩽u2+u+1⩽21.
Since k⩾19, we see that u=4 and k divides 21.
This contradicts the fact that k is coprime to 3.
Similarly, if S=B22(u), then 4(u-2u+1)(u-1)<25(u+2u+1).
It follows that u⩽8, and so k⩽13, which is a contradiction.
∎
The next step in our proof is the following lemma.
In fact, this lemma remains valid without the assumption that L=L6ε(q), L4ε(q), but we need it only in this section.
Lemma 5.3.
If r∈π(K), (r,my)=1 and r does not divide the order of the Schur multiplier M(S) of S, then r∉π(S) and expr(K)⋅ω(S)⊆π(G).
Proof.
By Lemma 4.2 (ii), Sylow r-subgroups of G are cyclic.
In particular, if R is a Sylow r-subgroup of K, then R is cyclic.
By the Frattini argument, we derive that C=CG(R) has a composition factor isomorphic to S.
A Hall r′-subgroup A of the solvable radical of C centralizes the Sylow r-subgroup of this radical, so A is normal in C.
Factoring out C by A, we get a central extension of S by an r-subgroup.
Denote this central extension by C~.
The derived series of C~ terminates in a perfect central extension of S by an r-group.
By the hypothesis, this perfect central extension is isomorphic to S, and therefore C~ includes a subgroup isomorphic to R×S.
Now the lemma follows.
∎
Now we are ready to complete the proof of Theorem 1.
Applying Lemmas 2.5, 4.1 and 4.2, we conclude that p and rz(εq) lie in π(S)∖(π(K)∪π(G/H)) and the corresponding Sylow subgroups of S must be cyclic.
Suppose that S is a classical group.
Let w∈Ry(εq).
Then
w∈π(F)∖π(H/F),
and writing G~=G/Ow′(K), K~=K/Ow′(K), and W~=Ow(K~), we have that N=K~/W~ is a w′-subgroup.
By the Hall–Higman lemma [17, Lemma 1.2.3], it follows that CK~(W~)⩽W~.
Let r∈π(N)∩π(my), and let R be a Sylow r-subgroup of N.
By the Frattini argument, there is an element g∈NG~/W~(R) of order rz(εq).
Since rrz(εq)∉ω(G), we have that R⋊〈g〉 is a Frobenius group.
Applying Lemma 2.14 yields wrz(εq)∈ω(G), which is not the case.
Thus we have π(N)∩π(my)=∅.
In particular, 2∉π(N).
Furthermore, by Lemma 5.3, it follows that π(N)∩π(S)⊆π(M(S)).
Assume that rz(εq)≠v.
At least one of the numbers rz(εq) and p, denote this number by r, divides the order of a proper parabolic subgroup of S (cf. [36, Lemma 3.8]).
By Lemma 2.16, we see that rw∈ω(G) unless v∈π(N).
In this case, by the results of the preceding paragraph, v is odd and v divides |M(S)|.
It follows that v=3 and S is one of the groups L2(9), U4(3), and S6(3).
But then the v-subgroup of S is not cyclic, contrary to the fact that v is coprime to my and Lemma 4.2.
If rz(εq)=v, then S=L2(v) since rz(εq) is not adjacent to 2 and the corresponding Sylow subgroup is cyclic.
Furthermore, in this case π(N)∩π(S)=1.
Applying Lemma 2.17, we see that {p,rz(εq),w} is not a coclique in GK(G), a contradiction.
If S=D43(u), then rz(εq) divides u4-u2+1, and so p divides u6-1.
This implies that Sylow p-subgroups of S are not cyclic (see the structure of maximal tori of D43(q) in [9]).
Thus we may assume that S is an exceptional group of Lie type other than D43(u).
Let L=L6ε(q) and z=5.
Since R6(εq)∩π(S)=∅, it follows from Lemma 5.1 (i) that S is B22(u) or G2(u).
We claim that R1(εq)∩π(S)=∅.
Recall that z=5 yields 2∉R1(εq).
Assume that r∈R1(εq)∩π(S).
Since r does not divide m6(εq), the corresponding Sylow subgroup of G is cyclic.
This, in particular, implies that r≠v,3.
If r∈π(K), then, by Lemma 5.3, we have r∈π(M(S)).
This is a contradiction since the Schur multiplier of B22(u) or G2(u) is either trivial, or a 2-group, or a 3-group.
Suppose that r∈π(G/H).
Since r is odd, it follows that u=u0r for some u0 and G/K contains an automorphism φ of S of order r such that CS(φ) is B22(u0) or G2(u0), respectively.
It is not hard to check using Lemmas 2.9 and 2.10, that there is t∈π(k(S))∖π(CS(φ)); for example, if
S=G2(u) and k(S)=u2+τu+1 for someτ∈{+,-},
then we can take t=r3r(τ0).
Observe that t∈R1(εq)∪R5(εq) since k(S) divides m5(εq).
Let T be a Sylow t-subgroup of S.
By the Frattini argument, there is an element g∈NG/K(T) of order r.
The choice of t implies that T⋊〈g〉 is a Frobenius group.
By the result of the previous paragraph, both r and t are coprime to |K|, so we may assume that this Frobenius group acts on the Sylow w-subgroup of K for some w∈R6(εq).
Applying Lemma 2.14, we see that either tw or rw lies in π(G), a contradiction.
Thus if r∈R1(εq)∩π(S), then r∉π(K)∪π(G/H) and r≠3.
It follows that r is adjacent to both rz(εq) and p in GK(S), while rz(εq) and p are not adjacent.
Lemma 2.12 yields that a Sylow r-subgroup of S cannot be cyclic, a contradiction.
Therefore, R1(εq)∩π(S)=∅.
To apply Lemma 5.1 (ii) and derive a final contradiction for L6ε(q), it remains to show that (5,q-ε)=1.
Suppose that 5∈R1(εq).
Then 5∉π(S), and therefore S=G2(u).
Furthermore, since k(S) divides m5(εq) and the ratio m5(εq)k5(εq)=q-ε is coprime to |S|, it follows that
k5(εq)=k(S)=u2+τu+1 for someτ∈{+,-}.
Also, by Lemmas 2.1 and 2.10, we have that exp5(L)=(m5(εq))5=5(q-ε)5, and so
k5(εq)⋅exp5(L)∈ω(G),but p⋅exp5(L)∉ω(L).
Assume that exp5(K)<exp5(G).
Then 5∈π(G/H), G/K contains a field automorphism φ of S of order 5, and u2+εu+1∈ω(CS(φ)).
As we remarked previously, this is not the case.
Thus exp5(K)=exp5(G).
Applying Lemma 5.3 yields p⋅exp5(L)∈ω(G)∖ω(L).
This completes the proof for L6ε(q).
The proof for L4ε(q) follows exactly the same lines with Rx(εq) in place of R1(εq) and Lemma 5.2 in place of Lemma 5.1.
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,
Mariya A. Grechkoseeva
