Isomorphic Subgroups

G. Stroth Martin-Luther-Universität Halle-Wittenberg Fachbereich Mathematik und Informatik Institut für Algebra und Geometrie 06099 Halle, Germany

Dedicated to Professor Dieter Held on his 60th Birthday

In this paper we are going to analyze the following situation.

(∗) Let G be a finite group. If H, K ≤ G with H ≅ K then there is some g ∈ G with H^{g} = K.

There are several papers dealing with situations similiar to (∗). For example in [FeS] W. Feit and G. Seitz showed that a finite group G in which any isomorphism between two subgroups extends to an inner automorphism of G are Σ_{1}, Σ_{2} and Σ_{3}. Furthermore Cherlin and Felgner [CheFe1], [CheFe2] and independently for solvable groups P. Neumann classified finite groups G in which any isomorphism of subgroups extends to an isomorphism of G. Related to this is the paper of Zhang Jiping [J1], where the groups where any isomorphism between cyclic subgroups can be extended to an automorphism of G is treated. There are quite a few papers which relate the order of subgroups to the question of the existence of inner automorphisms or isomorphisms [J2], [W], [BW] [Gr]. But these assumptions are much too strong as they immediately restrict the structure of G very heavily. Nevertheless this paper depends on the results of E.E. Shult [Sh1], [Sh2] , G. Higman [BlHu1; VIII(7.9)] and E.G. Bryakhanova [Bry] where p - groups whose automorphism group acts transitively on subgroups of order p are determined. The aim of this paper is to prove the following theorem.

Theorem. Let G be a finite group in which (∗) holds. Then

F^{∗}(G) ≅ X_{1} × ⋯ × X_{k}

where | X_{i} | are pairwise coprime and X_{i} is one of the following

(i) X_{i} ≅ L_{2}(5), SL_{2}(5), L_{2}(8), Sz(8)

(ii) X_{i} is an elementary abelian p-group of order p^{2}

(iii) X_{i} is isomorphic to a quaternion group of order 8, a Sylow 2-subgroup of Sz(8) or a Sylow 2-subgroup of U_{3}(4)

(iv) X_{i} is elementary abelian of order p^{3}

(v) X_{i} is a cyclic p-group.

If P ∈ Syl_{p}(F^{∗}(G)) then either P is cyclic or P ∈ Syl_{p}(G).

If there is X_{i} of type (i) then there is H ≤ G with H = X_{i} × I C_{G}(X_{i}) and gcd(| X_{i} |, | I C_{G}(X_{i}) |) = 1. Further G = H or X_{i} ≅ Sz(8) and | G : H |= 3. In all cases Sylow 3-subgroups of G are cyclic.

If there is some X_{i} as in (ii) then H = I C_{G}(X_{i}) = X_{i}×Y_{i}, gcd(| X_{i} |, | Y_{i} |) = 1 and one of the following holds

(1) G/H ≅ SL_{2}(5), p ∈ {11, 19, 29, 59}

(2) G/H ≅ SL_{2}(5) × ℤ_{7}, p = 29

(3) G/H ≅ SL_{2}(5) × ℤ_{29}, p = 59

(4) G/H ≅ SL_{2}(3), p ∈ {5, 11}

(5) G/H ≅ SL_{2}(3) × ℤ_{5}, p = 11

(6) G/H ≤ ΓL_{1}(p^{2}) and acts transitively on the 1-spaces of X_{i}.

Furthermore at most one of (1)-(6) occurs.

If there is some X_{i} as in (iii) then there is H ≤ G, H = X_{1} × Y_{1}, | Y_{1} | odd and one of the following holds

(1) X_{i} is quaternion of order eight and G/H ≅ ℤ_{3}

(2) X_{i} is a Sylow 2-subgroup of Sz(8) and G/H ≅ F_{21}

(3) X_{i} is a Sylow 2-subgroup of U_{3}(4) and G/H ≅ ℤ_{15}

If X_{i} is as in (iv) then G/ I C_{G}(X_{i}) is a subgroup of ΓL_{1}(p^{3}) which is transitive on the 1-subspaces of X_{i}.

We have that there is at most one X_{i} as in (i) − (iii).

There are a few remarks to the theorem. The group in general is not a direct product of groups of the type X_{i}H_{i}, where H_{i} ≅ G/ I C_{G}(X_{i}).

For example let X_{1} be elementary abelian of order 27 and X_{2} be elementary abelian of order 79^{2}. Let H be a group of order 80 in GL_{2}(79) acting transitively on the 1-spaces of X_{i} and K ≤ Z(GL_{2}(79)), | K |= 13. We let H × K act on X_{2} as a subgroup of GL_{2}(79) and let H act trivially on X_{1} and K as a subgroup of GL_{3}(3). Then (X_{1} × X_{2})(H × K) satisfies (∗) but it is not a direct product. One of the results of the theorem is that such things cannot occur if X_{1} and X_{2} are elementary abelian of rank two.

Furthermore if G satisfies (∗) we can always get a larger group G_{1} which also satisfies (∗) as follows. Let X be cyclic, gcd (| X |, | G |) = 1. Set G_{1} = G × X.

It is interesting to ask what happens if one drops the assumption that G be finite. The paper [SaW] seems to suggest that there is no hope to classify these groups. On the other hand I do not know very many infinite groups satisfying (∗). Of course the Prüfer groups do have property (∗). So a very ambitious assertion is that in general G looks as follows:

There is X ≤ G, I C_{G}(X) ⊆ X and X = X_{1} × … × X_{k}, where X_{i} are as in the theorem or X_{i} ≅ ℤ(p^{∞})

Finally for any X_{i} ≅ ℤ̸(p^{∞}) we have that G/ I C_{G}(X_{i})X_{i} is as in the theorem and any group X_{i} with X_{i} ≅ ℤ(p^{∞}) is normal in G.

For the proof of the theorem we first show that E(G) is quasisimple if nontrivial. Then application of the classification of finite simple groups gives (i). For the structure of F (G) the main tool is the classification of linear groups acting transitively on 1-spaces, which can be obtained from [LPS].

The claim in the infinite case will be supported by the way we prove the theorem in the finite case. First of all if V is an infinite dimensional vectorspace over GF (p), then no automorphism group can act transitively on isomorphic subspaces, as there are W_{1}, W_{2} ≤ V, W_{1} ≅ W_{2} and | V : W_{1} |= p, | V : W_{2} |= p^{2}. This means that we can never have an infinite product of groups X_{i} ≅ ℤ(p^{∞}) for the same prime p. If we have X_{1} × … × X_{n}, n > 1, X_{1} ≅ … ≅ X_{n} ≅ ℤ(p^{∞}), then we have a subgroup Y_{1} × … × Y_{n} = Y, Y_{1} ≅ ℤ_{p^{2}}. But an automorphism group acting transitively on cyclic subgroups of order p^{2} in Y has order divisible by p. So probably we will get some p-element in G \ (X_{1} × … × X_{n}), which should not be the case.

§1 Preliminaries

(1.1) Lemma. Let V be a vector space of dimension n > 1 over GF (p). Let H ≤ GL(V). Suppose that whenever W_{1} and W_{2} are subspaces of V of the same dimension, then there is some g ∈ H with g(W_{1}) = W_{2}. Then one of the following holds :

(i) SL(V) ≤ H

(ii) n = 4, p = 2, H ≅ A_{7}

(iii) n ≤ 3, K ≤ H, K ≅ ℤ_{(p^{n}−1)/(p−1)} and H is isomorphic to a subgroup of ΓL_{1}(p^{n})

(iv) n = 2, p ∈ {11, 19, 29, 59}, K ≤ H, K ≅ SL_{2}(5)

(v) n = 2, p ∈ {5, 7, 11, 23}, K ≤ H, K ≅ SL_{2}(3)

Proof. We may assume SL(V) ⊈ H. First of all H acts transitively on the 1-dimensional subspaces of V. Let A be the stabilizer of such a subspace in Aut(V). Then Aut(V) = HA. As A is a maximal parabolic in Aut(V) and H ≤ M, M maximal in Aut(V), we have Aut(V) = MA, where M and A are maximal subgroups of Aut(V). Now application of [LPS] to the images of H and A in P GL(V) shows that there is K ≤ M and one of the following holds

(i) K ≅ SL_{a}(p^{b}), ab = n, a > 1

(ii) K ≅ P Sp_{n}(p), n even, n ≥ 4

(iii) n = 4, p = 2 and 15| | K |

(iv) n = 2, p ∈ {11, 19, 29, 59}, K ≅ SL_{2}(5)

(v) n = 2, p ∈ {5, 7, 11, 23}, K ≅ SL_{2}(3)

(vi) n ≤ 3 and K ≅ ℤ_{(p^{n}−1)/(p−1)}.

Suppose first n > 3. Then the same group H is also transivite on the 2subspaces and so let B be the maximal parabolic stabilizing a 2- subspace we get BM = Aut(V) again. Application of [LPS] now shows that there is K_{1} ≤ M such that one of the following holds

(a) n = 5, p = 2, | K_{1} |= 31 · 5

(b) n = 4, p = 2, K_{1} is 4-homogeneous on the eight letters on which A_{8} acts. Hence by [BlHu2,XII(6.7) and (6.13)] K_{1} ≅ A_{7}.

So we see that for n > 3, we have n = 4, p = 2 and H ≅ A_{7}.

Assume next n = 3. Then all we get is case (vi) and so H is isomorphic to a subgroup of ΓL_{1}(p^{3}).

Let finally n = 2. If we have (vi) we have the assertion. Assume (iv) or (v). Now we get with the subgroup list of SL_{2}(p) from [Hu,II 8.27] that either K ≤ H or we have H ≤ ΓL_{1}(p^{2}).

(1.2) Lemma. Let G be a group satisfying (∗) and N ≤ G. Let x, y ∈ G with o(xN) = o(yN) = p^{f}, p prime. Then ⟨xN⟩ ∼ ⟨yN⟩ in G/N. Furthermore if x ∈ G \ N is a p-element, then ⟨x⟩ ∩ N is a cyclic subgroup of N of maximal order.

Proof. (see also the proof of [J1;(1.1)].) We may assume o(x) = p^{m}, o(y) = p^{n}. If n = m we have the assertion as ⟨x⟩ ∼ ⟨y⟩ in G. So assume m > n. Then f ≤ n. Choose ⟨z⟩ < ⟨x⟩ ∩ N such that | ⟨x⟩ |= p^{t}, t = min{m − f, n}. Let ⟨u⟩ ≤ ⟨y⟩, | ⟨u⟩ |= p^{t}. Then ⟨u⟩ ⊈ N. But this contradicts ⟨u⟩ ∼ ⟨z⟩ in G.

(1.3) Lemma. Let G be quasisimple such that 2 | | Z(G) |. If Aut(G) acts transitively on the involutions of G then G ≅ SL_{2}(q) q odd, 2·A_{7}, 6·A_{7} or 6·A_{6}.

Proof. According to [Go;1.46] we have that Out(G) is solvable. Hence a Sylow 2-subgroup S of G admits a solvable group which is transitive on the involutions. By [BlHu1;VIII(7.3)] S is a Suzuki 2-group, homocyclic or quaternion. As G/Z(G) is simple we see that either G/Z(G) possesses a dihedral Sylow 2subgroup or an abelian Sylow 2-subgroup. Now application of [Be1] and [Be2] implies that G/Z(G) ≅ L_{2}(q), A_{7} or J_{1}. As 1 ≠ Z(G) is a subgroup of the Schur multiplier of G we get the assertion with [Go; Table 4.1 and Prop. 4.233].

(1.4) Lemma. Let G be a quasisimple group. Suppose that for each pair H, K ≤ G with H ≅ K, there is some automorphism α of G with α(H) = K, then G ≅ L_{2}(q), SL_{2}(q), q odd, ^{2}G_{2}(3^{n}), L_{2}(8) or Sz(8).

Proof. Let Z(G) ≠ 1 and p be a prime with p | | Z(G) |. If p ≠ 2, then there is some ν ∈ Z(G), o(ν) = p, and some ω ∈ G \ Z(G), o(ω) = p, a contradiction as α(⟨ν⟩) = ⟨ω⟩ with some α ∈ Aut (G). So p = 2 and the assertion follows with (1.3). Hence we may assume G to be simple.

Suppose G ≅ A_{n}, n > 6. Let i = (123) and j = (123)(456). Then there is no automorphism α of G with α(i) ∈ ⟨j⟩.

Let G be sporadic. Then G possesses exactly one class of involutions. Hence by [CCNPW] we have G ≅ J_{1}, J_{3}, M_{11}, M_{22}, M_{23}, F_{3}, McL, LyS, O′N. Furthermore G possesses just one class of subgroups of order three. Again [CCNPW] shows G ≅ J_{1}, M_{11}, M_{22}, M_{23}. If G ≅ M_{22}, there are E_{1} ≅ E_{16} ≅ E_{2}, with N_{G}(E_{1})/E_{1} ≅ A_{6} and N_{G}(E_{2})/E_{2} ≅ Σ_{5}. Hence E_{1} ≁ E_{2} in Aut(G). If G ≅ M_{23} there are E_{1} ≅ E_{16} ≅ E_{2}, N_{G}(E_{1})/E_{1} ≅ A_{7} and N_{G}(E_{2})/E_{2} ≅ (ℤ_{3} × A_{5}) · 2, so E_{1} ≁ E_{2} in G. If G ≅ J_{1} there is a maximal subgroup D_{6} × D_{10} by [CC-NPW]. Hence there are subgroups isomorphic to D_{6} which are not conjugate. So let G ≅ M_{11}. Let T ∈ Syl_{3}(G). Then T is elementary abelian of order nine and N_{G}(T)/T is semidihedral of order sixteen. As Aut(T) ≅ GL_{2}(3), and Aut(M_{11}) = M_{11}, we see that there are two cyclic groups Z_{1}, Z_{2} of order four in N_{G}(T) with I C_{N_{G}(T)}(Z_{1}) = Z_{1} and I C_{N_{G}(T)}(Z_{2}) > Z_{2} such that T Z_{1} ≁ T Z_{2} in Aut(G) but T Z_{1} ≅ T Z_{2} as both groups are conjugate in T Aut(T).

Let G be a group of Lie type. If we have roots of different length, the corresponding root groups are not conjugated in Aut(G) provided we do not have diagram automorphism. Using [Bour] this shows G ≅ L_{n}(q), E_{n}(q), D_{n}(q), F_{4}(2^{n}), 2F_{4}(q), Sz(q),^{2} G_{2}(q), Sp_{4}(2^{n}), G_{2}(3^{n}) or U_{3}(q).

Let G ≅ L_{n}(q), n ≥ 3. If q is odd, there are ī, j̄ ∈ G which are homorphic images of

i = ( \begin{matrix}1 \\ & 1 \\ & 1 & 1 \\ & & & 1 \\ & & & & . & . & . \\ & & & & & & & 1 \end{matrix} ) j = ( \begin{matrix}1 \\ 1 & 1 \\ & 1 & 1 \\ & & & 1 \\ & & & & . & . & . \\ & & & & & & & 1 \end{matrix} )

such that ⟨ī⟩ and ⟨j̄⟩ are not conjugate under Aut (G).

If q is even and n ≥ 4, then

i = ( \begin{matrix}1 \\ 1 & 1 \\ & & 1 \\ & & 1 & 1 \\ & & & & 1 \\ & & & & & . & . & . \\ & & & & & & & & 1 \end{matrix} ) and j = ( \begin{matrix}1 \\ 1 & 1 \\ & & 1 \\ & & & . & . & . \\ & & & & & & 1 \end{matrix} )

are not conjugate under Aut(G). Let n = 3, q = 2^{m}. We may assume q > 2. Let E ≤ G, | E |= q^{2}, E elementary abelian. Then if F_{1}, F_{2} ≤ E, | F_{1} |=| F_{2} |, then F_{1} ∼ F_{2} in N_{Aut(G)}(E). So we may apply (1.1) with dim V ≥ 4. We have that N_{Aut(G)}(E)/ I C_{Aut(G)}(E) has a normal subgroup L_{2}(q) and so we have a contradiction with (1.1).

If G ≅ D_{n}(q), n ≥ 4, F_{4}(q), ^{2}F_{4}(q) or E_{n}(q), then by [AschS], [Iw] and [Dieu] or [Asch; chapt. 15] in case of D_{n}(q), q odd, G has at least two classes of involutions which are not conjugate under Aut(G).

Let G ≅ Sp_{4}(2^{n}), n > 1. By [AschS; chapt.7] there are three classes of involutions which cannot be fused under Aut(G).

Let G ≅ G_{2}(3^{n}). By [Iw] there is just one class of involutions. Let i be such an involution. Then I C_{G}(i) = (X_{1}X_{2})⟨j⟩ where [X_{1}, X_{2}] = 1, X_{1} ≅ X_{2} ≅ SL_{2}(3^{n}), X_{1} ∩ X_{2} = ⟨i⟩ and j ∈ N_{G}(X_{k}) for k = 1, 2. Let ω_{k} ∈ X_{k}, o(ω_{k}) = 3. Then ⟨iω_{1}⟩ ≅ ℤ_{6} ≅ ⟨iω_{1}ω_{2}⟩. But it is obvious that both groups are not conjugate in Aut(G) as they cannot be conjugate in I C(i).

If G ≅ U_{3}(q) q odd, we get a contradiction as in the case G ≅ L_{3}(q). So let G ≅ U_{3}(q), q = 2^{n}. Now there is a subgroup of order (q + 1)/d, d = gcd(3, q + 1) which acts fixed point freely on the natural module. If p | (q + 1)/d, p prime p ≠ 3, then G contains an elementary abelian subgroup of order p^{2}. Hence there is some subgroup of order p which has fixed points on the natural module. This implies that not all subgroups of order p are conjugate. So we are left with q + 1 is a 3-power, i.e. q = 8. Now in SU_{3}(8) there is a subgroup isomorphic to ℤ_{9} × L_{2}(8) whose image in U_{3}(8) is isomorphic to ℤ_{3} × L_{2}(8). So there are 3-elements ρ ∈ U_{3}(8), which correspond to elements of order 3 in SU_{3}(8) and 3-elements ω which correspond to elements of order 9 in SU_{3}(8). This shows that ⟨ρ⟩ and ⟨ω⟩ cannot be conjugate in Aut(U_{3}(8)) as any automorphism of U_{3}(8) lifts to an automorphism of SU_{3}(8).

Let G ≅ Sz(q). Let S ∈ Syl_{2}(G), T = Ω_{1}(S). Then | T |= q and T is elementary abelian. Now all elementary abelian subgroups of T of the same order are conjugate in N_{Aut(G)}(S), which is solvable. So by (1.1) we get q = 8.

(1.5) Lemma. Suppose G satisfies (∗). Let H ≤ G where H is quasisimple or simple. Then H ≅ L_{2}(5), SL_{2}(5), L_{2}(8) or Sz(8).

Proof. By (1.4) we have H ≅ L_{2}(q), SL_{2}(q), q odd, ^{2}G_{2}(3^{n}), L_{2}(8) or Sz(8). Let first H ≅ SL_{2}(q) or L_{2}(q), q odd. By (∗) we have that Z(H) ∈ Syl_{2}( I C_{G}(H)). Let x ∈ G\H and assume that x is a 2-element. Choose x of minimal order. Then x^{2} ∈ H. But now there is y ∈ H, o(y) = o(x), and there is no way to conjugate ⟨x⟩ to ⟨y⟩ in G. So we have that G induces just inner × field automorphisms on H.

Set M = H/Z(H). Hence M ≅ L_{2}(q). If q ≢ 3, 5(8), then M contains two fours group which are not conjugate under G/Z(H). As both preimages are quaternion of order eight these groups are not conjugate in G. So we have q ≡ 3, 5(8). Let S ∈ Syl_{2}(M). Then | S |= 4. Let ε = ±1, where 4 | q + ε. If q ≠ 5, then 1 ≠ x = (q + ϵ)/4. We have some ω ∈ M, o(ω) = x. Further N_{M}(⟨ω⟩) ≅ ℤ_{2} × D_{2x}, where D_{2x} is dihedral of order 2x. So we have two dihedral groups of order (q +ϵ)/2 in N_{M}(⟨ω⟩), which are not conjugate under G/Z(H). As the preimages of both groups in G are isomorphic we see that G cannot satisfy (∗).

Let now H ≅ ^{2}G_{2}(3^{n}). Then we see that | G : H | is odd. Now H possesses a subgroup H_{1} ≅ ℤ_{2} × L_{2}(3^{n}) and G induces just field automorphisms on L_{2}(3^{n}). So as before H contains to subgroups isomorphic to ℤ_{2} × D_{2s}, s = (3^{n} + 1)/2, which are not conjugate in G.

(1.6) Lemma. Let H be a group with H/O_{2}(H) ≅ L_{3}(2) and | O_{2}(H) |= 8, O_{2}(H) the natural module for H/O_{2}(H). Then H \O_{2}(H) contains involutions.

Proof. Let u ∈ H such that uO_{2}(H) is of order 4. Set z = u^{4}. If z = 1 we are done. Otherwise as ⟨z⟩ = I C_{O_{2}(H)}(u) = [⟨u^{2}⟩, O_{2}(H)], there is some v ∈ O_{2}(H) such that [u^{2}, v] = z. Now u^{2}v is an involution not in O_{2}(H).

§2 Proof of the theorem

Throughout this section we assume (∗) without any further notice.

(2.1) Lemma. If E(G) ≠ 1, then E(G) ≅ L_{2}(5), SL_{2}(5), L_{2}(8) or Sz(8).

Proof. Let E(G) = E_{1} ∗ ⋯ ∗ E_{k}, E_{i} quasisimple. Suppose k > 1. Let x ∈ E_{1} \ Z(E(G)) and y ∈ E_{2} \ Z(E(G)) with x^{2}, y^{2} ∈ Z(E(G)). Then also xy ∉ Z(E(G)), but (xy)^{2} ∈ Z(E(G)). Hence by (1.2) xZ(E(G)) is conjugated to xyZ(E(G)) in G/Z(E(G)), a contradiction. So we have k = 1 and the assertion follows with (1.5).

(2.2) Lemma. If E(G) ≠ 1, then there is X ≤ G such that | X | is coprime to | E(G) | and I C_{G}(E(G))E(G) = E(G) × X. Further G = I C_{G}(E(G))E(G) or E(G) ≅ Sz(8) and | G/ I C_{G}(E(G))E(G) |= 3.

Proof. We have I C_{G}(E(G)) ∩ E(G) = Z(E(G)). Furthermore there is y ∈ E(G) \ Z(E(G)) with y^{2} ∈ Z(E(G)). So application of (1.2) shows that for x ∈ I C_{G}(E(G)) with x^{2} ∈ Z(E(G)) we have x ∈ Z(E(G)) and so Z(E(G)) ∈ Syl_{2}( I C_{G}(E(G)). In particular I C_{G}(E(G)) possesses a normal 2-complement X. By (∗) |E(G) | is coprime to | X |. Furthermore E(G) contains all 2-elements of G. This shows that G = E(G) × X or E(G) ≅ Sz(8) and G/(E(G) × X) ≅ ℤ_{3}.

(2.3) Lemma. Let P ∈ Syl_{p}(F (G)), p odd. Then P is elementary abelian or cyclic. If P is not cyclic, then P ∈ Syl_{p}(G).

Proof. We have P < G. Furthermore G acts transitively on the set of subgroups of order p in G. Hence by [Sh1] [Sh2] P is homocyclic. Let K = I C_{G}(P) and H/K = O_{p}(G/K). Let Q ∈ Syl_{p}(H). Then the Frattini argument implies that N_{G}(Q) acts transitively on the subgroups of order p in Q. Hence again by [Sh1, Sh2] Q is abelian. This implies H = K.

Suppose now that Q is not cyclic and p| | G : K |. If there is U ≤ G/K, U fixed point free on Ω_{1}(Q), then G contains a subgroup of order p not in K, a contradiction.

On the other hand we have that G/K is one of the groups of (1.1). From this we see that G/K contains X ≅ SL_{n}(p), | Ω_{1}(Q) |= p^{n}, n odd. Now X and so G/K contains cyclic subgroups of order p which are not conjuagted, as they have different canonical form on Ω_{1}(Q). But this contradicts (1.2).

So we have that P ∈ Syl_{p}(G) if P is not cyclic. Suppose now that Q is neither cyclic nor elementary abelian. Then the number of cyclic subgroups of order p^{2} in Q is divisible by p and so by [Hu;IV 2.5] p | | G : K |, a contradiction. This is the assertion.

(2.4) Lemma. Let P ∈ Syl_{2}(F (G)). Then P is cyclic, elementary abelian of order at most eight, a quaternion group of order eight, or a Sylow 2-subgroup of Sz(8) or U_{3}(4). If P is not cyclic then P ∈ Syl_{2}(G). If P is isomorphic to a Sylow 2-subgroup of Sz(8), then G induces F_{21} on P while G induces ℤ_{15} on P if P is isomorphic to a Sylow 2-subgroup of U_{3}(4).

Proof. Suppose first that P has exactly one involution. Then P is cyclic or quaternion. Suppose P is quaternion. As G acts transitively on the subgroups of order 4 in P, we get | P |= 8. Now a Sylow 2-subgroup P_{1} of G is quaternion. As Ω_{2}(P_{1}) = P, we get P_{1} = P.

So assume now that | Ω_{1}(P) |≥ 4. Then G/ I C_{G}(Ω_{1}(P)) is one of the groups in (1.1). Suppose 2^{n} =| Ω_{1}(P) |≥ 16. Then L_{n}(2) or A_{7} is induced. Now we have at least two classes of elements of order three in G/ I C_{G}(Ω_{1}(P)), contradicting (1.2).

So we have | Ω_{1}(P) |≤ 8. Let | Ω_{1}(P) |= 8. Then by (1.1) P has an automorphism of order 7. Application of [BlHu1; VIII(7.9)] gives that P is a Suzuki 2-group and so either | P |= 2^{6} or 2^{9} or P is homocyclic. If | Ω_{1}(P) |= 4, we get | P |= 2^{6} or P is homocyclic by [BlHu1; VIII (7.9)] again.

Let first P be not homocyclic. Then application of [BlHu1; VIII (7.9)] yields P ∈ Syl_{2}( I C_{G}(Ω_{1}(P))). We next show P ∈ Syl_{2}(G).

If | Ω_{1}(P) |= 4, there is ω ∈ G with ω^{3} ∈ I C(Ω_{1}(P)) and I C_{Ω_{1}(P)}(ω) = 1. Let T ∈ Syl_{3}(G), then G = I C_{G}(Ω_{1}(P))N_{G}(T) and 2 |̸ | N_{I C_{G}(Ω_{1}(P))}(T) | . As G/ I C_{G}(Ω_{1}(P)) ≲ Σ_{3} and all involutions of G are in P, we see that | G : I C_{G}(Ω_{1}(P)) |= 3 and so P ∈ Syl_{2}(G).

Let | Ω_{1}(P) | = 8 and 2 | | G : I C_{G}(Ω_{1}(P)) |. Then G/ I C_{G}(Ω_{1}(P)) ≅ L_{3}(2). This contradicts [Bry]. So we have shown that P ∈ Syl_{2}(G) if P is nonabelian.

Let next | P/Ω_{1}(P) |= 64. We have 63 || G : I C_{G}(P)P |. Furthermore G/ I C_{G}(P)P is a subgroup of odd order in L_{6}(2). Let ω ∈ I C_{G}(Ω_{1}(P)) \ I C_{G}(P) with ω^{7} ∈ I C_{G}(P). Then there is P_{1} < P, |P_{1}| = 64 and [P_{1}, ω] = P_{1}. But then all elements in P_{1} \ Ω_{1}(P_{1}) have the same square, i.e. [P_{1}, ω] possesses exactly one involution, a contradiction. So we have that 7 |̸ | I C_{G}(Ω_{1}(P))|. Let ν ∈ G\I C_{G}(P) with ν^{7} ∈ I C_{G}(Ω_{1}(P)). Then [Ω_{1}(P), ⟨ν⟩] = Ω_{1}(P) and [P, ⟨ν⟩] = P. Now [ν, I C_{G}(Ω_{1}(P))] ≤ I C_{G}(P). Let now ω ∈ I C_{G}(Ω_{1}(P)). Then I C_{P/Ω_{1}(P)}(ω) is ν - invariant. So we have | I C_{P/Ω_{1}(P)}(ω)| = 1, 2^{3}, or 2^{6}. If | I C_{P/Ω_{1}(P)}(ω)| = 2^{3}, we see that |[P/Ω_{1}(P), ω]| = 2^{3} and so o(ω) = 7, a contradiction. So we have that | I C_{P/Ω_{1}(P)}(ω) |= 1 for ω ∈ I C_{G}(Ω_{1}(P)) and [P, ω] ≠ 1. This gives that I C_{G}(Ω_{1}(P))/ I C_{G}(P) is a subgroup of a cyclic group of order 9. As | N_{G}(Ω_{1}(P))/ I C_{G}(Ω_{1}(P)) | | 21, we get that | G : I C_{G}(P)P |= 63 or 3 · 63.

Now we count the number of isomorphism classes in P of groups U with Ω_{1}(P) < U and | U : Ω_{1}(P) |= 4. Let Ω_{1}(P) = ⟨z_{1}, z_{2}, z_{3}⟩ and U = ⟨a, b, Ω_{1}(P)⟩. If U is abelian, we have U ≅ ℤ_{4} × ℤ_{4} × ℤ_{2}. Let U be nonabelian. Assume ϕ(U) = ⟨e_{1}⟩. Then U ≅ Q_{8} × ℤ_{2} × ℤ_{2}. Let ϕ(U) = ⟨e_{1}, e_{2}⟩. Then we may assume a^{2} = e_{1}, b^{2} = e_{2}. Now the isomorphism type is determined by [a, b]. We have [a, b] ∈ ⟨e_{1}, e_{2}⟩. If [a, b] = e_{1}e_{2}, then (ab)^{2} = 1, a contradiction. So [a, b] = e_{1} or e_{2}. But these two groups are isomorphic. Hence up to now we have three isomorphism types. Let finally ϕ(U) = Ω_{1}(P). Then we have a^{2} = e_{1}, b^{2} = e_{2} and [a, b] ∈ Ω_{1}(P) \ ⟨e_{1}, e_{2}⟩ and all these groups are isomorphic. Hence we have exactly four types.

There are 3 · 7 · 31 such subgroups in P. None of them is normalized by an element of order 7. As 3 · 7 · 31/4 > 63, we must have at least one orbit of length 3^{3} · 7. Hence there are 66 · 7 further subgroups. As 66 · 7/3 = 22 · 7 > 63, we get another regular orbit and so 39 · 7 subgroups are left. As 39 · 7/2 > 63, we have a further regular orbit. So for the last orbit we have 12 · 7 subgroups. But 12·7 does not divide 63·3, a contradiction. So we have | P |= 64 if P is nonabelian.

Let now | Ω_{1}(P) |= 4, then | P/Ω_{1}(P) |= 16 and N(P)/P acts transitively on the involutions of P/Ω_{1}(P). Hence 5 || G : I C_{G}(P)P |. So G/ I C_{G}(P)P contains a cyclic group of order 15. As this group is maximal of odd order in L_{4}(2), we have G/N_{G}(P)P ≅ ℤ_{15}. Now it is easy to see that P is isomorphic to a Sylow 2-subgroup of U_{3}(4) [Bei, Satz 3]. Let next | Ω_{1}(P) |= 8 and | P/Ω_{1}(P) |= 8. Then G/N_{G}(P)P ≅ ℤ_{7} or F_{21}. Let U, V ≤ P, U ≅ V ≅ ℤ_{4} × ℤ_{2} such that ϕ(U) = ϕ(V). There is some g ∈ G with U^{g} = V. Hence g ∈ I C(ϕ(U)). This shows G/N_{G}(P)P ≅ F_{21}. So the lemma has been proved in case of P nonabelian.

Let now P be homocyclic but not cyclic. Suppose P not to be elementary abelian. Let Q be a Sylow 2-subgroup of I C_{G}(Ω_{1}(P)). Then by [BlHu1;VIII(7.9)] Q is homocyclic or a Suzuki 2-group. As P is not elementary abelian, all elements of order four in G are in P, which shows that Q is homocyclic. Let now |Ω_{1}(P) |= 4 =| Ω_{1}(Q) |. As the number of cyclic subgroups of order four in Q is even, we have that 2 || G : I C_{G}(Ω_{1}(P)) | and so G/ I C_{G}(Ω_{1}(P) ≅ Σ_{3}. But now there are involutions in G\P, a contradiction. So let | Ω_{1}(P) |= 8. Again we have that 2 || G : I C_{G}(Ω_{1}(P)) | and so G/ I C_{G}(Ω_{1}(P)) ≅ L_{3}(2). By (1.6) there is some y ∈ G \ I C_{G}(Ω_{1}(P)) with y^{2} ∈ I C_{G}(Ω_{1}(P)) and o(y) = o(t) for some t ∈ P. But this contradicts ⟨y⟩ ∼ ⟨t⟩ in G.

So we have that P is elementary abelian and | P |≤ 8. Next we show P ∈ Syl_{2}(G). Let Q ∈ Syl_{2}( I C_{G}(P)). Then Q is either homocyclic or a Suzuki 2-group by [BlHu1;VIII(7.9)].

Let Q be homocyclic. By [Hu; IV 2.5] and the Frattini argument any two cylic groups of order 4 in Q are conjugated under N_{G}(Q). But as the number of cyclic subgroups of order four in Q is even, we get a contradiction as before. So let Q be a Suzuki 2-group. Then we see that either |P | = 8 and there is some ν ∈ G with ν^{7} ∈ I C_{G}(P) and [ν, Q] = Q, or |P | = 4 and |Q| = 64. Suppose the latter. As there are no involutions in G \ P, we see that |G/ I C_{G}(P)| = 3 and so Q ∈ Syl_{2}(G). Now let Q_{1}, Q_{2} ≤ Q with P ≤ Q_{1}∩Q_{2} and |Q_{1} : P | = 2 = |Q_{2} : P |. Then by [Hu; IV 2.5] Q_{1} ∼ Q_{2} in N_{G}(Q). This shows that there is some ν ∈ G with ν^{5} ∈ I C_{G}(Q) and [ν, Q] = Q. In any case Q ∉ Syl_{2}(F (G)) and so there is some R ∈ Syl_{p}(F (G)), p odd, such that I C_{Q}(R) = P. By (2.3) and (1.1) we see | R |≤ p^{3} and so Q/P ≤ GL_{3}(p). But this is not possible, as now E_{8} · ℤ_{7} or E_{16} · ℤ_{5} would be a subgroup of GL_{3}(p), which is not the case.

So we have Q = P in any case. Let P ∉ Syl_{2}(G), then G/ I C_{G}(P) ≅ Σ_{3} or L_{3}(2) and we get involutions in G \ I C_{G}(P) using (1.6) again. This contradiction proves the lemma.

(2.5) Lemma. Let X ∈ Syl_{p}(F (G)), X elementary abelian of order p^{2}. Then one of the following holds :

(a) G/ I C_{G}(X) ≅ SL_{2}(5), p ∈ {11, 19, 29, 59}

(b) G/ I C_{G}(X) ≅ SL_{2}(5) × ℤ_{7}, p = 29

(c) G/ I C_{G}(X) ≅ SL_{2}(5) × ℤ_{29}, p = 59

(d) G/ I C_{G}(X) ≅ SL_{2}(3), p ∈ {5, 11}

(e) G/ I C_{G}(X) ≅ SL_{2}(3) × ℤ_{5}, p = 11

(f) G/ I C_{G}(X) is isomorphic to a subgroup of ΓL_{1}(p^{2}) which acts transitively on the subgroups of order p in X.

Proof. We know that G/ I C_{G}(X) acts transitively on the subgroups of order p in X. Now with (1.1) we get either (f) or there is a normal subgroup H ≅ SL_{2}(p), SL_{2}(5) or SL_{2}(3) in G/ I C_{G}(X). As X ∈ Syl_{p}(G), we have H ≅ S̸L_{2}(p). So we have H ≅ SL_{2}(5), p ∈ {11, 19, 29, 59} or H ≅ SL_{2}(3) and p ∈ {5, 7, 11, 23}. If p = 7 or 23, then there are 8 or 24 subgroups of order p in X. Hence G/ I C_{G}(X) ≅ SL_{2}(3) · 2. But now there are cyclic subgroups of order two or four (depending whether G/ I C_{G}(X) splits over SL_{2}(3) or not) in G/ I C_{G}(X) which are not conjugate, contradicting (1.2). Now H contains all elements of order four in G/ I C_{G}(X) and so | G/ I C_{G}(X) : H | is odd. Hence G/ I C_{G}(X) = H × I C(H) and so gcd(| H |, | I C(H) |) = 1. Furthermore | I C(H) || p − 1. This shows (a)-(e).

(2.6) Lemma. Let X ∈ Syl_{2}(F (G)), X a quaternion group of order eight, or isomorphic to a Sylow 2-subgroup of Sz(8) or U_{3}(4). Then we have

(a) G/ I C_{G}(X)X ≅ ℤ_{3} if X is quaternion

(b) G/ I C_{G}(X)X ≅ F_{21} if X is isomorphic to a Sylow 2-subgroup of Sz(8)

(c) G/ I C_{G}(X)X ≅ ℤ_{15} if X is isomorphic to a Sylow 2-subgroup of U_{3}(4).

Proof. As Aut(Q_{8})/Inn(Q_{8}) ≅ Σ_{3} and X ∈ Syl_{2}(G) by (2.4), we have that G/ I C_{G}(X)X ≅ ℤ_{3} for X ≅ Q_{8}. If X is isomorphic to a Sylow 2-subgroup of Sz(8), then by (2.4) G/ I C_{G}(X) contains H ≅ F_{21}. As this group is maximal in Aut(X/ϕ(X)) and X ∈ Syl_{2}(G), we have H = G/ I C_{G}(X)X.

Let finally X be isomorphic to a Sylow 2-subgroup of U_{3}(4). Then by (2.4) there is H ≅ ℤ_{15} in G/ I C_{G}(X)X. Furthermore | G/ I C_{G}(X)X | is odd. But a cyclic group of order 15 is a maximal odd order subgroup in A_{8}, the assertion.

(2.7) Lemma. Let X ∈ Syl_{p}(F (G)), X elementary abelian of order p^{3}. Then G/ I C_{G}(X) is a subgroup of ΓL_{1}(p^{3}) which is transitive on the subgroups of order p in X.

Proof. The assertion follows with (1.1) as X ∈ Syl_{p}(G) by (2.3).

(2.8) Lemma. Let F^{∗}(G) = E(G) × F (G). If E(G) ≠ 1, then Sylow p-subgroup of F (G) are cyclic or elementary abelian of order p^{3}. If E(G) = 1, then there is at most one Sylow p-subgroup of F (G) which is neither cyclic nor elementary abelian of order p^{3}.

Proof. Let E(G) ≠ 1. Then by (2.2) I C_{G}(E(G)) = E(G)×X, where | X | is coprime to | E(G) |. In particular | X | is odd. Now by (2.3)-(2.5) we see that if S ∈ Syl_{p}(F (X)), S neither cyclic nor elementary abelian of order p^{3}, then | S |= p^{2} and G/ I C_{G}(S) is a subgroup of ΓL_{1}(p^{2}) which is transitive on the subgroups of S. So p + 1 || G : I C_{G}(S) | and so by (2.2) (p + 1)/gcd(p + 1, 3) || I C(E(G)) |. But p is odd and so 2 || X |, a contradiction.

So let E(G) = 1. If there is a noncyclic Sylow 2-subgroup U of F (G), then | G : I C_{G}(U)U | is odd by (2.4) and I C_{G}(U)U = U × O_{2′}(I C_{G}(U)). So we may argue as before.

So we may assume there is A ∈ Syl_{p}(F (G)) with | A |= p^{2}, A noncyclic and B ∈ Syl_{q}(F (G)) with | B |= q^{2}, B noncyclic, p, q odd, p ≠ q.

Let W = I C_{G}(A × B), W_{A} = I C_{G}(A), W_{B} = I C_{G}(B). Then G/W_{A} and G/W_{B} are given in (2.5). We have G/W ≲ G/W_{A} × G/W_{B}. If x, y ∈ G \ W with x^{2} ∈ W ∋ y^{2}. Then we have xW ∼ yW by (1.2). By (2.5) | G/W_{A} | and | G/W_{B} | are even and so Sylow 2-subgroups of G/W have just one involution. If SL_{2}(3) or SL_{2}(5) is involved in G/W_{A} or G/W_{B} we see that a Sylow 2-subgroup of G/W is quaternion of order 8. If SL_{2}(5) is involved now p, q ∈ {11, 19, 29, 59} while p, q ∈ {5, 11} if SL_{2}(3) is involved. In both cases (p + 1)(q + 1) is divisible by 8. As there are (p + 1)(q + 1) subgroups of order pq in A × B which are all conjugate in G and one of them is normalized by an involution we see that | G/W | is divisible by 16, a contradiction.

So we have G/W_{A} is a subgroup ΓL_{1}(p^{2}) and G/W_{B} is a subgroup of ΓL_{1}(q^{2}). Let 2^{a}⊤p + 1 and 2^{b}⊤q + 1, a ≤ b. Let S ∈ Syl_{2}(G/W) and S_{1} ≤ S maximal such that S_{1} acts on each subgroup of order q in A × B. Then | S : S_{1} |= 2^{b} as S is cyclic. So S_{1} also acts on each subgroup of order p. In particular S_{1} normalizes some subgroup of order pq. But 2^{a+b} divides the number of subgroups of order pq in A × B and so 2^{a+b} ≤ 2^{b}, a contradiction. This proves the lemma.

(2.9) Proposition. The theorem holds.

Proof. The structure of F^{∗}(G) follows from (2.1), (2.2), (2.3) and (2.4). That Sylow p-subgroups of F^{∗}(G) are Sylow p-subgroups of G if not cyclic follows from (2.3) and (2.6). The remainder follows from (2.5)-(2.8).

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