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Orders of real elements in finite groups
Accepted Manuscript Orders of real elements in finite groups
Hung P. Tong-Viet
PII: DOI: Reference:
S0021-8693(19)30173-5 https://doi.org/10.1016/j.jalgebra.2019.03.025 YJABR 17121
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Journal of Algebra
Received date:
6 February 2019
Please cite this article in press as: H.P. Tong-Viet, Orders of real elements in finite groups, J. Algebra (2019), https://doi.org/10.1016/j.jalgebra.2019.03.025
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ORDERS OF REAL ELEMENTS IN FINITE GROUPS HUNG P. TONG-VIET Dedicated to the memory of Professor Kay Magaard Abstract. Finite groups with a small number of real element orders are investigated. As an application, we determine the structure of finite groups having four real-valued irreducible characters.
1. Introduction Let G be a finite group. An element x ∈ G is real if x and x−1 are G-conjugate. A conjugacy class of G is real if it contains real elements. Moreover, an element x ∈ G is said to be strongly real if it is inverted by an involution. In this paper, we point out some connections between recent studies on the real elements and real-valued (ordinary or modular) irreducible characters (see [5, 6, 12, 13]) and some classical results in finite group theory due to Goldschmidt [9], Suzuki [14] and others. The existence or non-existence of real elements of certain orders is an important question in finite group theory. Following M. Suzuki, a finite group G is called a (C)group if the centralizer of every involution is 2-closed, that is, having a normal Sylow 2-subgroup. It turns out that a finite group G is a (C)-group if and only if G has no real element of order 2m with m > 1 being odd; equivalently, the order of every real element of G is either a power of 2 or odd (see [5, Proposition 2.7]). These groups have been studied by Suzuki himself in [14] and recently in [5] for solvable groups. In Theorem 2.3, we characterize finite (C)-groups having no real element of order 4 and we use it to deduce the following criterion for the solvability of finite groups with some restriction on the orders of real elements. Theorem A. Let G be a finite group and let p be a prime. If every real element of G is an involution or a p-element, then G is solvable. Moreover, if L = O2 (G) then (1) L is a 2-group and has no real element of order 4; or (2) O2 (L) = 1 and L has a cyclic Sylow 2-subgroup Q and a normal 2-complement P which is a p-group with CP (Q) = CP (z), where z is the unique involution in Q. Finite 2-groups satisfying (1) above have been studied by Chillag and Mann in [4]. These are exactly the finite 2-groups satisfying the condition x2 = y 2 implies (xy −1 )2 = 1 (see [4, Lemma 4.6]). Several extensions of Theorem A simply do not hold. One may Date: March 22, 2019. 2010 Mathematics Subject Classification. Primary 20E45; Secondary 20C15, 20C20. Key words and phrases. Orders of real elements; real-valued characters; strongly closed subgroups. 1
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ask whether a group G is solvable if every real element of G is a 2-element or a p-element for some odd prime p. This is not true, for a counterexample, we may take G = L2 (7). In this simple group, every real element of G has order 1, 2, 4 or 3. Moreover, if there are two odd primes involved, then G is not solvable. For example, if x ∈ A5 is real, then o(x) = 1, 2, 3 or 5. In some sense, Theorem A is best possible. Our next result describes the structure of finite groups having at most three distinct real element orders.
Theorem B. Let G be a finite group and let L = O2 (G). If G has at most three distinct real element orders, then (1) L is a 2-group and has no real element of order 8; or (2) O2 (L) = 1 and L has a cyclic Sylow 2-subgroup Q and a normal 2-complement P which is a p-group for some odd prime p with CP (Q) = CP (z), where z is the unique involution in Q. Theorem B follows easily from Theorem A together with a result in [7] stating that a finite group is 2-closed if and only if it has no nontrivial real element of odd order. Note that the proof of the aforementioned result uses only Baer-Suzuki theorem. Thus the proof of Theorem B does not depend on the classification of finite simple groups. It is well-known that the number of real-valued ordinary irreducible characters and the number of conjugacy classes of real elements of a finite group coincide. Hence if G has at most three real-valued ordinary irreducible characters, then G has at most three conjugacy classes of real elements. In particular, such groups must satisfy the hypothesis of Theorem B. Therefore, Theorem B gives a classification-free proof of the solvability of finite groups with at most three real-valued ordinary irreducible characters (see Theorem 2.5 of [12]). Let G be a finite group and let A ≤ S ≤ G. Recall that A is strongly closed in S with respect to G if whenever a ∈ A, g ∈ G, if ag ∈ S, then ag ∈ A or equivalently Ag ∩S ⊆ A for all g ∈ G. Generalizing Glauberman’s Z ∗ -theorem, Goldschmidt [9] determines the structure of finite groups possessing an abelian strongly closed 2-subgroup. It turns out that if S is a Sylow 2-subgroup of a finite group G and Ω1 (S) is abelian, then Ω1 (S) is an elementary abelian group and hence it is strongly closed in S with respect to G. Now we can see that if a finite group G has no strongly real element of order 4, then all involutions in S commute, in particular, Ω1 (S) is abelian. Hence we can apply Goldschmidt’s result to classify all finite non-abelian simple groups having no real element of order 4 (see Proposition 3.2). Using this result, we can determine the structure of finite groups having at most four real-valued irreducible characters. Theorem C. Let G be a finite non-solvable group. If G has four real-valued irreducible characters, then G ∼ = SL3 (2) × K, where K is of odd order. Finally, in the last section of the paper, we give classification-free proofs of some results in [6, 13]. 2. Three real element orders For a finite group G, we denote by Re(G) the set of real elements of G and by E(G) the set of orders of real elements of G. An irreducible (ordinary or modular) character
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of G is real-valued if it coincides with its complex conjugate. We will prove Theorems A and B in this section. Notice that if x ∈ G is a (strongly) real element of G, then any power of x is also (strongly) real. We will need the following results. Lemma 2.1. Let G be a 2-group and let M be a normal elementary abelian 2-subgroup of G such that G = M t, where t is an involution of G with [t, M ] = 1. Then G has a strongly real element of order 4. Proof. Since M is elementary abelian and t does not centralize M , there exists an involution u ∈ M such that [t, u] = 1. Let x = tu. Clearly x is a nontrivial 2-element. If x2 = 1, then (tu)(tu) = 1 which implies that tu = ut, a contradiction. Thus o(x) ≥ 4. Moreover xt = (tu)t = ut = (tu)−1 so x is a strongly real 2-element of G. Hence some power of x is a strongly real element of order 4 of G. Lemma 2.2. Let G be a finite group and let N G be of odd order. If xN ∈ G/N is a real element of G, then there exists a real element y ∈ G such that xN = yN and o(xN ) divides o(y). In particular, if E(G) consists of 2 and powers of odd primes, then E(G/N ) also consists of 2 and powers of odd primes. Proof. The first claim follows from Lemma 2.2 of [10]. Indeed, if xN is a nontrivial real element of G/N , then there exists a nontrivial real element y ∈ G such that xN = yN . Hence o(xN ) = o(yN ) divides o(y). Now assume that every real element of G has order 2 or a power of an odd prime and let xN be a real element of G/N . We can find a real element y ∈ G such that xN = yN . Since o(y) = 2 or pa for some odd prime p and an integer a ≥ 0, o(xN ) divides 2 or pa . The last claim follows. Notice that normal subgroups of a (C)-group without real element of order 4 will be (C)-groups without real element of order 4.
Theorem 2.3. Let G be a finite group and let L = O2 (G). Then G has no real element of order 2m with m = 2 or m > 1 odd if and only if one of the following holds. (1) L is a 2-group with no real element of order 4; (2) O2 (L) = 1 and L has a cyclic Sylow 2-subgroup Q and a normal 2-complement K with CK (Q) = CK (z), where z is the unique involution in Q; or (3) L is isomorphic to either SL2 (2f ) (f ≥ 2) or 2 B2 (22f +1 ) (f ≥ 1).
Proof. Let L = O2 (G). As L contains all real elements of G, G is a (C)-group without real element of order 4 if and only if L is a (C)-group without real element of order 4. Assume that G is a (C)-group having no real element of order 4. If L is a 2-group, then L has no real element of order 4, so (1) holds. Assume that L is not a 2-group. Since L has no normal Sylow 2-subgroup, it has a nontrivial real element x of odd order by Proposition 6.4 of [7]. Furthermore, by Lemma 3 in [14], xt = x−1 for some involution t ∈ L and |CL (x)| is odd. Let Q be a Sylow 2-subgroup of L. We claim that O2 (L) = 1. Assume by contradiction that O2 (L) > 1 and let M ≤ O2 (L) be a minimal normal subgroup of L. Assume first that [t, M ] = 1. Let 1 = y ∈ M . Then [y, x] = 1 as |CL (x)| is odd. Since M L, we have [y, x] ∈ M and so [y, x] = [y, x]t = [y t , xt ] = [y, x−1 ] which implies that [y, x2 ] = 1. However, as o(x)
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is odd, y ∈ CL (x2 ) = CL (x), which is a contradiction. Therefore [t, M ] > 1. By Lemma 2.1, M t has a real element of order 4, contradicting our assumption. Hence O2 (L) = 1. Assume that L is solvable. By [5, Theorem 2.4], L has a normal 2-complement K, Q is either cyclic or generalized quaternion and CK (Q) = CK (z), where z is the unique involution of Q. As generalized quaternion groups possess a real element of order 4, Q must be cyclic. This is part (2) of the theorem. Finally, assume that L is non-solvable. Let R be the solvable radical of L. We know that O2 (R) = 1. We claim that R = O2 (L). Assume by contradiction that R is not a 2 -group. Then 1 < Q0 = Q ∩ R ∈ Syl2 (R). Clearly O2 (R) is a solvable (C)-group without real element of order 4 and Q0 ≤ O2 (R). Note that O2 (R) is not a 2-group. So, part (2) implies that Q0 is cyclic and thus by [1, 39.2] R has a normal 2-complement, say R1 R. Note that 1 < R1 L and O2 (L/R1 ) = L/R1 and by Lemma 2.2, G/R1 is a (C)-group without real element of order 4; clearly L/R1 is not a 2-group and thus by the argument in the second paragraph, O2 (L/R1 ) = R/R1 ∼ = Q0 is trivial, which is a contradiction. Thus R = O2 (L) as wanted. Let L = L/O2 (L) and use the ‘bar convention’. By Lemma 2.2, L has no real element of order 4 or 2m with m > 1 odd. Moreover, L has a trivial solvable radical and has no proper normal subgroup of odd index. Let N be a minimal normal subgroup of L. Then N = S1 × S2 × · · · × Sk , where k ≥ 1 is an integer and Si ∼ = S for i = 1, 2, · · · , k and S is a nonabelian simple (C)-group. By Theorem 1 of [14], S is isomorphic to L2 (p) where p is a Fermat or a Mersenne prime; A6 or L2 (q), 2 B2 (q), U3 (q) or L3 (q) where q > 2 is a power of 2. However, as S has no real element of order 4, S ∼ = SL2 (2f ), f > 1 or 2 B2 (22f +1 ), f ≥ 1. Since S1 is not 2-closed, it has a nontrivial real element z of odd order. By [14, Lemma 3], z i = z −1 for some involution i ∈ S1 and CL (z) is of odd order. Now if k ≥ 2, then S2 ≤ CL (z) which is impossible as |S2 | is even. This shows that N is a non-abelian simple group and that CL (N ) = 1 (since CL (N ) ⊆ CL (z), CL (N ) is a normal subgroup of L of odd order). Thus L is an almost simple group with simple socle N and O2 (L) = L. Assume that N ∼ = 2 B2 (22f +1 ), f ≥ 1. In this case |Out(S)| = 2f + 1 is odd, so L = N . Assume next that N ∼ = SL2 (2f ), f ≥ 2. Then Out(N ) is a cyclic group of order f generated by a field automorphism ϕ of order f . Assume that L = N. Then L = N φ, where φ is a field automorphism of order r which is a power of 2. We have that CL (φ) ∼ = SL2 (2f1 )φ with f1 = f /r ≥ 1. Now SL2 (2f1 ) has a real element y of f1 order 2 + 1 > 1 and [y, φ] = 1 and so CL (y) is of even order, contradicting Lemma 3 of [14]. Therefore L = L/O2 (L) is isomorphic to one of the simple groups listed in part (3). Since O2 (L) = L and L/O2 (L) is non-abelian simple, we deduce that L = L O2 (L) and |L : L | = |O2 (L) : L ∩ O2 (L)| is odd, which implies that L = L . As a Sylow 2-subgroup of L/O2 (L) has at least two involutions, L has at least two involutions and so by [14, Lemma 6], we have [Q, O2 (L)] = 1. As Q ≤ CL (O2 (L)) L, we deduce that L = CL (O2 (L)). In particular, L is a quasi-simple group with Z(L) = O2 (L).
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However, the Schur multipliers of SL2 (2f ) and 2 B2 (22f +1 ) are 2-groups or trivial. This forces O2 (L) = 1 and part (3) holds. For the converse, if (1) holds, then clearly L has no real element of order 4 or 2m with m > 1 odd. If (2) holds, then L has no real element of order 2m with m > 1 odd by [5, Theorem 2.4]. Moreover, a Sylow 2-subgroup of L is cyclic, so it has no real element of order 4. Assume that (3) holds. In this case, the centralizer of every involution of L is a 2-group, so it has no real element of order 2m with m > 1 odd. If L = SL2 (2f ), then clearly L has no real element of order 4 (the Sylow 2-subgroup of L is elementary abelian). Assume now that L = 2 B2 (q) with q = 22f +1 , f ≥ 1. It remains to show that L has no real element of order 4 which follows from the argument in pages 121 − 122 of [14]. The proof is now complete. Proof of Theorem A. Suppose that the order of every real element of G is 2 or a power of a fixed prime p. Obviously, G is a (C)-group without real element of order 4. Let L = O2 (G). Then L satisfies the same hypothesis as G does. In particular, L = O2 (L) is a (C)-group without real element of order 4. We first claim that G is solvable. Indeed, it suffices to show that L is solvable. By way of contradiction, assume that L is non-solvable. By Theorem 2.3, L ∼ = SL2 (2f ) (f ≥ 2) or 2 B2 (q), where q = 22f +1 with f ≥ 1. In the former case, SL2 (2f ) contains two real elements of odd coprime orders 2f − 1 and 2f + 1, which is impossible. In the latter case, by Theorem 9 of [15], 2 B2 (q) contains at least two nontrivial real elements of odd order dividing q − 1 or q 2 + 1, respectively, so this case cannot occur. Hence L is solvable. By Theorem 2.3, L is a 2-group without real element of order 4 or O2 (L) = 1 and L has a cyclic Sylow 2-subgroup Q and a normal 2-complement K with CK (Q) = CK (z), where z is the unique involution of Q. If the first possibility holds, then conclusion (1) of Theorem A holds. Assume we are in the second possibility, so L is not a 2-group. To finish the proof of the theorem, it remains to show that K is a p-group. Notice that p must be odd as otherwise, L is a 2-group. We prove this by induction on |L|. Assume first that M := Op (K) > 1. Since |M | is odd and coprime to p, M contains no nontrivial real element of L. By Lemma 3.1(d) in [6], [Q, M ] = 1 and so Q ≤ CL (M )L which forces CL (M ) = L since L = O2 (L). By Lemma 2.2, the order of every real element of L/M is 2 or a power of p and O2 (L/M ) = L/M has a normal 2-complement K/M . By induction, K/M is a p-group. It follows that K = P M with P ∈ Sylp (K). Since M ≤ Z(K), we have K ≤ P and thus P K which implies that K = P × M . Now this implies that QP L so QP is normal subgroup of L of odd index which is impossible. Hence Op (K) = 1. Since K is solvable, we deduce that Op (K) > 1. Let L = L/Op (K). Clearly, L satisfies the hypothesis of the theorem, O2 (L) = L and |L| < |L|, by induction K = O2 (L) is a p-group. Thus K is a p-group.
Proof of Theorem B. Let L = O2 (G) and Q ∈ Syl2 (L). We know that L contains all real elements of G and so E(G) = E(L). Moreover L = O2 (L). Suppose that G has at most three real element orders. Then |E(L)| ≤ 3. Assume first that L is a 2-group. Note that if x is a real element of L then xk is also a real element of G for any integer
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k. Thus |E(L)| ≤ 3 if and only if E(L) ⊆ {1, 2, 4} or equivalently L has no real element of order 8, which proves (1). We now assume that L > 1 is not a 2-group. Clearly |L| is even and thus {1, 2} ⊆ E(L). Since L does not have a normal Sylow 2-subgroup, L has a nontrivial real element x of odd order by [7, Proposition 6.4]. Since |E(L)| ≤ 3, we see that o(x) must be an odd prime and thus E(L) = {1, 2, p} for some odd prime p. Now the result follows from Theorem A. 3. Four real-valued irreducible characters In this section, we determine the structure of finite groups with four real-valued irreducible characters. Recall that a finite group G is quasi-simple if G is perfect and G/Z(G) is a non-abelian simple group. A component of a finite group G is a quasisimple subnormal subgroup of G. The layer E(G) of G is a subgroup of G generated by all components of G. Finally, the generalized Fitting subgroup F∗ (G) is the product of the Fitting subgroup F(G) and the layer E(G). Notice that F∗ (G) is a central product of F(G) and E(G) and CG (F∗ (G)) ≤ F∗ (G) (see [1, §31]). We begin with the following result. Theorem 3.1. Let G be a finite non-solvable (C)-group. Suppose that O2 (G) = 1 and |E(G)| ≤ 4. Then G ∼ = S × K, where K is of odd order, S ∼ = A5 or SL3 (2) and E(G) = {1, 2, 3, 5} or {1, 2, 3, 4}, respectively. Proof. By Theorem B, we may assume that |E(G)| = 4. Since G is non-solvable, we know that 2 ∈ E(G) and G has a nontrivial real element of odd order ([7, Proposition 6.4]). Let L = O2 (G). We consider the following cases. (1) G has no real element of order 4. By Theorem A, it is easy to see that E(G) = {1, 2, p, q} for two distinct odd primes p and q. By Theorem 2.3, L ∼ = SL2 (2f ) (f ≥ 2) 2 2f +1 ) (f ≥ 1). Let K = CG (L). Then K is a normal subgroup of odd order of or B2 (2 G and G/K is an almost simple group with socle LK/K ∼ = L and |G/K : LK/K| = |G : LK| is odd. Assume first that L ∼ = 2 B2 (22f +1 ), f ≥ 1. By Theorem 9 and Proposition 16 of [15], L has three nontrivial real elements of odd distinct orders (which are 22f +1 − 1 and 22f +1 ± 2f +1 + 1). Thus this case cannot occur. Assume that L ∼ = SL2 (2f ) with f ≥ 2. Now L has two real elements of order 2f − 1 and 2f + 1 and both numbers must be primes. It is easy to see that f = 2 and so L = SL2 (4) ∼ = A5 . Since |Out(A5 )| = 2, we have G = L × K, where K is of odd order. In this case E(G) = {1, 2, 3, 5}. (2) G has a real element of order 4. We have E(G) = {1, 2, 4, p}, where p is an odd prime. Since O2 (G) = 1, O2 (L) = 1 and L is non-solvable with E(L) = E(G). Let Q ∈ Syl2 (L). Assume first that O2 (L) = 1. Let N be a minimal normal subgroup of G. Then N ∼ = S1 × S2 × · · · × Sk , where k ≥ 1 is an integer, S is isomorphic to a non-abelian simple (C)-group and Si ∼ = S for all i = 1, 2, · · · , k. It follows that S is isomorphic to one of the simple groups listed in Theorem 1 of [14]. As every nontrivial real element
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of odd order of S has the same order p, we deduce that S ∼ = SL3 (2) ∼ = L2 (7) and p = 3. Assume that k ≥ 2. Let x ∈ S1 be a real element of order 3 inverted by an element t ∈ S1 and y ∈ S2 be an involution. Since [S1 , S2 ] = 1, we deduce that [x, y] = 1 = [t, y] and thus (xy)t = xt y t = x−1 y = yx−1 = (xy)−1 ; so xy ∈ N is a real element of order 6, a contradiction. Thus k = 1 and N ∼ = SL3 (2). We see that K := CG (N ) is a normal subgroup of odd order of G and G/K is an almost simple group with socle N ∼ = SL3 (2). ∼ Since Aut(N ) = SL2 (7) · 2 has a real element of order 6 while G/K is also a (C)-group, we have G = N K = N × K as wanted. In this case, E(G) = {1, 2, 3, 4}. Assume that N := O2 (L) > 1. If Q has a unique involution, then Q is either cyclic or quaternion. Since L is non-solvable, Q cannot be cyclic by [1, 39.2]. Thus Q is quaternion. Now L/N is a non-solvable (C)-group with a quaternion Sylow 2-subgroup QN/N ∼ = Q. By [3, Theorem 2], L/N has a center of order 2. Since L/N is nonsolvable, it has a nontrivial real element N x of odd order inverted by an involution of L/N (by [14, Lemma 3]), which is impossible. Thus Q has at least two involutions. By [14, Lemma 6], Q centralizes N and hence CG (N ) is a normal subgroup of odd index of G. It follows that L ≤ CG (N ) and N ≤ Z(L). Let M/N = O2 (L/N ). Since |N | is odd, M = U N where U ∈ Syl2 (M ). Moreover, as N ≤ Z(L), we have U M and thus U = O2 (M ) L forcing U = 1 or O2 (L/N ) = 1 as O2 (L) = 1. By induction, we have L/N ∼ = SL3 (2). As O2 (L) = L and |N | is odd, L = L and thus L is a quasi-simple group with L/N ∼ = SL3 (2). However, the Schur multiplier of SL3 (2) is cyclic of order 2. This implies that N = 1, a contradiction. We next determine all finite non-abelian simple groups without strongly real element of order 4. Proposition 3.2. Let G be a finite non-abelian simple group. If G has no (strongly) real element of order 4, then G is isomorphic to one of the following simple groups: (1) SL2 (2f )(f ≥ 3), U3 (2f )(f ≥ 2), 2 B2 (22f +1 )(f ≥ 1); or (2) L2 (q), 5 ≤ q ≡ 3, 5 (mod 8), J1 , 2 G2 (32f +1 ), f ≥ 1. Proof. Let S be a Sylow 2-subgroup of G. We show that Ω1 (S) is abelian and hence it is elementary abelian and contains all involutions of S which implies that Ω1 (S) is strongly closed in S with respect to G and we can then invoke Corollary 1 of [9] and the result follows. Let i and j be distinct involutions in S. Then ij ∈ S is a strongly real 2-element of order m = o(ij) > 1. Since G has no strongly real element of order 4, m = 2. In particular ij = ji. Thus Ω1 (S) is abelian as it is generated by all involutions in S. The finite non-abelian simple groups in Proposition 3.2 are called Goldschmidt groups. These are exactly the finite simple groups of Lie type of Lie rank 1 in characteristic 2 and non-abelian simple groups with an abelian Sylow 2-subgroup. We are now ready to prove Theorem C which is included in the following. Theorem 3.3. Let G be a finite group with exactly four real-valued irreducible char acters. Let L = O2 (G) and Q be a Sylow 2-subgroup of L. Then one of the following holds.
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(1) G has a normal Sylow 2-subgroup. (2) O2 (L) = 1, Q is either cyclic or quaternion of order 8 and L has a normal 2-complement K with CK (Q) = CK (z), where z is the unique involution in Q. (3) G has 2-length one and Q is either homocyclic or a Suzuki 2-group. (4) G ∼ = SL3 (2) × K, where K is of odd order. Proof. Let G be a finite group with exactly four real-valued irreducible characters. Let L = O2 (G) and Q ∈ Syl2 (L). We know that G has exactly four conjugacy classes of real elements. In particular |E(G)| ≤ 4. If |E(G)| ≤ 3, then either (1) or (2) holds by Theorem B. So, we may assume that |E(G)| = 4. Also, in view of [7, Proposition 6.4], we also assume that G has a real element of order p for some odd prime p. Notice that G always has real elements of order 1 and 2. Thus {1, 2, p} ⊆ E(G). It follows that E(G) is one of the following sets {1, 2, p, p2 }, {1, 2, p, q}, {1, 2, 4, p} or {1, 2, p, 2p}, where p and q are distinct odd primes. In particular, G has only one class of involutions. (1) Assume G is solvable. If Q has at least two involutions, then Theorem IX.8.6 of [11] implies that G has 2-length one and Q is homocyclic or a Suzuki 2-group, so (3) occurs. Assume that Q has a unique involution. Then Q is either cyclic or quaternion. If Q is cyclic, then G has a normal 2-complement by [1, 39.2], so (3) holds. Assume that Q is quaternion. Since G has no real element of order 8, Q is an ordinary quaternion group of order 8. Clearly, all real elements of G have orders 1, 2, 4 and p. Thus G is a (C)-group and hence L is also a (C)-group. By [14, Lemma 3], L has a strongly real element x of order p with |CG (x)| being odd. By [3, Theorem 2], the center of L/O2 (L) is cyclic of order 2. Thus x ∈ O2 (L) and hence O2 (L) = 1. By Theorem 2.4 of [5], L has a normal 2-complement K and CK (z) = CK (Q), where z is the unique involution of Q. So (2) holds. (2) Assume that G is non-solvable. (2a) We first claim that O2 (G) = 1. By way of contradiction, assume that O2 (G) > 1 and let N ≤ O2 (G) be a minimal normal subgroup of G. Then N is an elementary abelian normal subgroup of G. Since G has only one conjugacy class of involutions, N contains all involutions of G. Clearly G/N is non-solvable and has at most four realvalued irreducible characters, by induction we have that G/N = T /N × U/N , where T, U are normal subgroups of G with T /N ∼ = SL3 (2) and U/N being of odd order. Let K be a Hall 2 -subgroup of U . Then U = KN and by Hall’s Theorem (see [1, 18.5]) U acts transitively on its Hall 2 -subgroups. Hence G = NG (K)U = NG (K)N . Assume that NG (K) < G. Let W = NG (K) ∩ N . Then W < N and W is normalized by both N (as N is abelian) and NG (K); therefore W G. As N is a minimal normal subgroup of G, we deduce that W = 1. However, this is impossible as NG (K) is nonsolvable and so it possesses an involution which does not lie in N. Thus K G. Since T ∩ K ≤ (T ∩ U ) ∩ K = N ∩ K = 1, we have G = T × K. Let T0 be the last term of the derived series of T . Since T /N ∼ = SL3 (2), T /N is perfect and so T /N = T0 N/N which implies that T /N ∼ = T0 /N0 , where N0 = T0 ∩ N G. Since N is a minimal normal subgroup of G, either N0 = 1 or N0 = N. If the former case holds, then T = T0 × N so T0 has an involution that does not lie in N . Thus N0 = N and hence T = T0 is perfect.
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Observe that T contains all real elements of G and since [K, T ] = 1, T also has four real conjugacy classes and E(T ) = E(G); so T has only one class of involutions. Now let 1 = y ∈ N . Then y T contains all involutions of T and hence |N | − 1 = |y T | = |T : CT (y)|. As N ≤ CT (y), n := |T : CT (y)| = 1 or it is the index of a proper subgroup of the simple group T /N and n + 1 = |N | is a 2-power. If the former case holds, then |N | = 2 and so T ∼ = SL2 (7). In this case, it is easy to check that T has more than 4 real conjugacy classes. Assume that the latter case holds. Using [8], we deduce that |T : CT (y)| = 7 and hence |N | = 8. By [8] again, there are exactly two perfect groups of order |T | = | SL3 (2)| · 23 = 1344. However these two perfect groups have more than one conjugacy class of involutions. This contradiction shows that O2 (G) = 1 as wanted. (2b) G is a (C)-group. Suppose by contradiction that G is not a (C)-group. Then G has a real element of order 2m for some odd integer m > 1. It follows that m = p and hence E(G) = {1, 2, p, 2p}, where p is an odd prime. Assume first that F(G) > 1. By Claim (2a), F(G) is of odd order. By induction, we have G/F(G) ∼ = SL3 (2) × U , where U is of odd order. Hence G/F(G) has a real element of order 4 and thus G also has a real element of order 4 by Lemma 2.2, which is a contradiction. We now assume that F(G) = 1. Then F∗ (G) = E(G) is a direct product of nonabelian simple groups. Let M be a minimal normal subgroup of G. Then M is a direct product of copies of a non-abelian simple group S. Since S has no real element of order 4, S is a Goldschmidt group by Proposition 3.2. However, we can check that every Goldschmidt group contains at least two nontrivial real elements of odd coprime orders. (See the proof of [13, Theorem 3.1]). (2c) The completion of the proof. By Claims (2a) and (2b), G is a non-solvable (C)-group with O2 (G) = 1 and |E(G)| = 4. Now it follows from Theorem 3.1 that G∼ = A5 × K or G ∼ = SL3 (2) × K, where K is of odd order. However, the former case cannot occur as A5 has five conjugacy classes of real elements. Thus the latter case occurs and hence (4) holds. The proof is now complete. It seems that if G is a finite group and E(G) = {1, 2, p, 2p}, where p is an odd prime, then G is solvable. Unfortunately, we are unable to prove or disprove this at the moment. 4. Some classification-free proofs In this section, we provide classification-free proofs of Theorem C of [6] on the nonexistence of real element of order 3 and Theorem 5.2 of [13] on the structure of finite groups with two real-valued irreducible p-Brauer characters, where p is an odd prime. Recall that a section of a finite group G is the quotient H/K, where K H are subgroups of G; we say that G is X-free, if no section of G is isomorphic to X, where X is a finite group. Fix an odd prime p. If G has a real element x of order p inverted by an element t ∈ G, then [t2 , x] = 1 and x, t/t2 ∼ = D2p , the dihedral group of order 2p, so D2p is a section of G. Conversely, if G has a section H/K ∼ = D2p , then H/K has
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a real element of order p which lifts to a real element of H whose order is divisible by p (see [10, Lemma 2.2]). Thus G has a real element of order p. Therefore, G has no real element of order p if and only if G is D2p -free; in other words, D2p is not involved in G. Notice that if p = 3, then D6 ∼ = S3 . By Corollary 3 of [2], if S is a finite S3 -free ∼ non-abelian simple group, then S = L2 (32f +1 ) or 2 B2 (22f +1 ) with f ≥ 1. If we assume that 3 divides the order of S, then S is isomorphic to the former group. Now by using Theorem 3.3 of [6] and the previous result instead of Theorem 2.1 of [6], we obtain a classification-free proof of the following theorem, which is Theorem C of [6].
Theorem 4.1. Let G be a finite group of order divisible by 3. Suppose that G = O2 (G). Then G has no real element of order 3 if and only if G/O3 (G) is isomorphic to a direct products of simple groups L2 (32fi +1 ), where i = 1, 2, · · · , k and fi ≥ 1. It would be interesting to classify all finite D2p -free non-abelian simple groups without invoking the full strength of the classification of finite simple groups. Fix an odd prime p. For a finite group G, we write IBrrv (G) for the set of realvalued irreducible p-Brauer characters of G. We now provide a classification-free proof of Theorem 5.2 of [13], which is stated below. Theorem 4.2. Let p be an odd prime and G be a finite group with Op (G) = 1. If | IBrrv (G)| = 2, then G has a normal Sylow 2-subgroup, which is homocyclic or a Suzuki 2-group. Proof. Suppose that | IBrrv (G)| = 2. We first show that G is solvable. By Brauer’s lemma on character tables, G has exactly one conjugacy class of nontrivial real element of p -order. It follows that |G| is even and G has a unique conjugacy class of involutions. In particular, G has no real element of order 4. If 1 < N G, then | IBrrv (G/N )| ≤ | IBrrv (G)| = 2, by induction we deduce that G/N is solvable. Hence G has a unique minimal normal subgroup N which is a direct product of copies of a non-abelian simple group S. Since G has no real element of order 4, S has no real element of order 4 and so by Proposition 3.2 S is a Goldschmidt group. However, one can check that Goldschmidt groups contain at least two nontrivial real elements of odd coprime orders. Now as G is solvable and has a unique conjugacy class of involutions, the Sylow 2-subgroup Q of G is either cyclic or quaternion (if Q has a unique involution) or Q is homocyclic or a Suzuki 2-group by [11, Theorem IX.8.6]. However, as G has no real element of order 4, Q cannot be quaternion. Now we can follow the original argument in [13] to deduce that Q G. Notice that the authors use [13, Lemma 2.1] in their proof but that lemma is only stated for p = 2 although it also holds for p odd.
Acknowledgment The author is grateful to Dan Rossi for the discussion during the preparation of this paper. He also thanks the anonymous referee for his or her numerous corrections and suggestions, which helped to improve the clarity of this paper.
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References [1] M. Aschbacher, Finite group theory, second edition, Cambridge Studies in Advanced Mathematics, 10, Cambridge University Press, Cambridge, 2000. [2] M. Aschbacher, S3 -free 2-fusion systems, Proc. Edinb. Math. Soc. (2) 56 (2013), no. 1, 27–48. [3] R. Brauer and M. Suzuki, On finite groups of even order whose 2-Sylow group is a quaternion group, Proc. Nat. Acad. Sci. U.S.A. 45 (1959), 1757–1759. [4] D. Chillag, A. Mann, Nearly odd-order and nearly real finite groups, Comm. Algebra 26 (1998), no. 7, 2041–2064. [5] S. Dolfi, D. Gluck, G. Navarro, On the orders of real elements of solvable groups. Israel J. Math. 210 (2015), no. 1, 1–21. [6] S. Dolfi, G. Malle, G. Navarro, The finite groups with no real p-elements, Israel J. Math. 192 (2012), no. 2, 831–840. [7] S. Dolfi, G. Navarro, P. H. Tiep, Primes dividing the degrees of the real characters, Math. Z. 259 (2008), no. 4, 755–774. [8] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.10.1; 2019. (https://www.gap-system.org) [9] D. M. Goldschmidt, 2-fusion in finite groups, Ann. of Math. (2) 99 (1974), 70–117. [10] R. M. Guralnick, G. Navarro, P. H. Tiep, Real class sizes and real character degrees, Math. Proc. Cambridge Philos. Soc. 150 (2011), no. 1, 47–71. [11] B. Huppert and N. Blackburn, Finite groups. II, Grundlehren der Mathematischen Wissenschaften, 242, Springer-Verlag, Berlin, 1982. [12] A. Moret´o and G. Navarro, Groups with three real valued irreducible characters, Israel J. Math. 163 (2008), 85–92. [13] G. Navarro, L. Sanus, P. H. Tiep, Groups with two real Brauer characters. J. Algebra 307 (2007), 891–898. [14] M. Suzuki, Finite groups in which the centralizer of any element of order 2 is 2-closed. Ann. of Math. (2) 82 (1965) 191–212. [15] M. Suzuki, On a class of doubly transitive groups, Ann. of Math. (2) 75 (1962), 105–145. [16] J. G. Thompson, Nonsolvable finite groups all of whose local subgroups are solvable, Bull. Amer. Math. Soc. 74 (1968), 383–437. Department of Mathematical Sciences, Binghamton University, Binghamton, NY 13902-6000, USA E-mail address: [email protected]
Hung P. Tong-Viet
PII: DOI: Reference:
S0021-8693(19)30173-5 https://doi.org/10.1016/j.jalgebra.2019.03.025 YJABR 17121
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Journal of Algebra
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ORDERS OF REAL ELEMENTS IN FINITE GROUPS HUNG P. TONG-VIET Dedicated to the memory of Professor Kay Magaard Abstract. Finite groups with a small number of real element orders are investigated. As an application, we determine the structure of finite groups having four real-valued irreducible characters.
1. Introduction Let G be a finite group. An element x ∈ G is real if x and x−1 are G-conjugate. A conjugacy class of G is real if it contains real elements. Moreover, an element x ∈ G is said to be strongly real if it is inverted by an involution. In this paper, we point out some connections between recent studies on the real elements and real-valued (ordinary or modular) irreducible characters (see [5, 6, 12, 13]) and some classical results in finite group theory due to Goldschmidt [9], Suzuki [14] and others. The existence or non-existence of real elements of certain orders is an important question in finite group theory. Following M. Suzuki, a finite group G is called a (C)group if the centralizer of every involution is 2-closed, that is, having a normal Sylow 2-subgroup. It turns out that a finite group G is a (C)-group if and only if G has no real element of order 2m with m > 1 being odd; equivalently, the order of every real element of G is either a power of 2 or odd (see [5, Proposition 2.7]). These groups have been studied by Suzuki himself in [14] and recently in [5] for solvable groups. In Theorem 2.3, we characterize finite (C)-groups having no real element of order 4 and we use it to deduce the following criterion for the solvability of finite groups with some restriction on the orders of real elements. Theorem A. Let G be a finite group and let p be a prime. If every real element of G is an involution or a p-element, then G is solvable. Moreover, if L = O2 (G) then (1) L is a 2-group and has no real element of order 4; or (2) O2 (L) = 1 and L has a cyclic Sylow 2-subgroup Q and a normal 2-complement P which is a p-group with CP (Q) = CP (z), where z is the unique involution in Q. Finite 2-groups satisfying (1) above have been studied by Chillag and Mann in [4]. These are exactly the finite 2-groups satisfying the condition x2 = y 2 implies (xy −1 )2 = 1 (see [4, Lemma 4.6]). Several extensions of Theorem A simply do not hold. One may Date: March 22, 2019. 2010 Mathematics Subject Classification. Primary 20E45; Secondary 20C15, 20C20. Key words and phrases. Orders of real elements; real-valued characters; strongly closed subgroups. 1
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ask whether a group G is solvable if every real element of G is a 2-element or a p-element for some odd prime p. This is not true, for a counterexample, we may take G = L2 (7). In this simple group, every real element of G has order 1, 2, 4 or 3. Moreover, if there are two odd primes involved, then G is not solvable. For example, if x ∈ A5 is real, then o(x) = 1, 2, 3 or 5. In some sense, Theorem A is best possible. Our next result describes the structure of finite groups having at most three distinct real element orders.
Theorem B. Let G be a finite group and let L = O2 (G). If G has at most three distinct real element orders, then (1) L is a 2-group and has no real element of order 8; or (2) O2 (L) = 1 and L has a cyclic Sylow 2-subgroup Q and a normal 2-complement P which is a p-group for some odd prime p with CP (Q) = CP (z), where z is the unique involution in Q. Theorem B follows easily from Theorem A together with a result in [7] stating that a finite group is 2-closed if and only if it has no nontrivial real element of odd order. Note that the proof of the aforementioned result uses only Baer-Suzuki theorem. Thus the proof of Theorem B does not depend on the classification of finite simple groups. It is well-known that the number of real-valued ordinary irreducible characters and the number of conjugacy classes of real elements of a finite group coincide. Hence if G has at most three real-valued ordinary irreducible characters, then G has at most three conjugacy classes of real elements. In particular, such groups must satisfy the hypothesis of Theorem B. Therefore, Theorem B gives a classification-free proof of the solvability of finite groups with at most three real-valued ordinary irreducible characters (see Theorem 2.5 of [12]). Let G be a finite group and let A ≤ S ≤ G. Recall that A is strongly closed in S with respect to G if whenever a ∈ A, g ∈ G, if ag ∈ S, then ag ∈ A or equivalently Ag ∩S ⊆ A for all g ∈ G. Generalizing Glauberman’s Z ∗ -theorem, Goldschmidt [9] determines the structure of finite groups possessing an abelian strongly closed 2-subgroup. It turns out that if S is a Sylow 2-subgroup of a finite group G and Ω1 (S) is abelian, then Ω1 (S) is an elementary abelian group and hence it is strongly closed in S with respect to G. Now we can see that if a finite group G has no strongly real element of order 4, then all involutions in S commute, in particular, Ω1 (S) is abelian. Hence we can apply Goldschmidt’s result to classify all finite non-abelian simple groups having no real element of order 4 (see Proposition 3.2). Using this result, we can determine the structure of finite groups having at most four real-valued irreducible characters. Theorem C. Let G be a finite non-solvable group. If G has four real-valued irreducible characters, then G ∼ = SL3 (2) × K, where K is of odd order. Finally, in the last section of the paper, we give classification-free proofs of some results in [6, 13]. 2. Three real element orders For a finite group G, we denote by Re(G) the set of real elements of G and by E(G) the set of orders of real elements of G. An irreducible (ordinary or modular) character
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of G is real-valued if it coincides with its complex conjugate. We will prove Theorems A and B in this section. Notice that if x ∈ G is a (strongly) real element of G, then any power of x is also (strongly) real. We will need the following results. Lemma 2.1. Let G be a 2-group and let M be a normal elementary abelian 2-subgroup of G such that G = M t, where t is an involution of G with [t, M ] = 1. Then G has a strongly real element of order 4. Proof. Since M is elementary abelian and t does not centralize M , there exists an involution u ∈ M such that [t, u] = 1. Let x = tu. Clearly x is a nontrivial 2-element. If x2 = 1, then (tu)(tu) = 1 which implies that tu = ut, a contradiction. Thus o(x) ≥ 4. Moreover xt = (tu)t = ut = (tu)−1 so x is a strongly real 2-element of G. Hence some power of x is a strongly real element of order 4 of G. Lemma 2.2. Let G be a finite group and let N G be of odd order. If xN ∈ G/N is a real element of G, then there exists a real element y ∈ G such that xN = yN and o(xN ) divides o(y). In particular, if E(G) consists of 2 and powers of odd primes, then E(G/N ) also consists of 2 and powers of odd primes. Proof. The first claim follows from Lemma 2.2 of [10]. Indeed, if xN is a nontrivial real element of G/N , then there exists a nontrivial real element y ∈ G such that xN = yN . Hence o(xN ) = o(yN ) divides o(y). Now assume that every real element of G has order 2 or a power of an odd prime and let xN be a real element of G/N . We can find a real element y ∈ G such that xN = yN . Since o(y) = 2 or pa for some odd prime p and an integer a ≥ 0, o(xN ) divides 2 or pa . The last claim follows. Notice that normal subgroups of a (C)-group without real element of order 4 will be (C)-groups without real element of order 4.
Theorem 2.3. Let G be a finite group and let L = O2 (G). Then G has no real element of order 2m with m = 2 or m > 1 odd if and only if one of the following holds. (1) L is a 2-group with no real element of order 4; (2) O2 (L) = 1 and L has a cyclic Sylow 2-subgroup Q and a normal 2-complement K with CK (Q) = CK (z), where z is the unique involution in Q; or (3) L is isomorphic to either SL2 (2f ) (f ≥ 2) or 2 B2 (22f +1 ) (f ≥ 1).
Proof. Let L = O2 (G). As L contains all real elements of G, G is a (C)-group without real element of order 4 if and only if L is a (C)-group without real element of order 4. Assume that G is a (C)-group having no real element of order 4. If L is a 2-group, then L has no real element of order 4, so (1) holds. Assume that L is not a 2-group. Since L has no normal Sylow 2-subgroup, it has a nontrivial real element x of odd order by Proposition 6.4 of [7]. Furthermore, by Lemma 3 in [14], xt = x−1 for some involution t ∈ L and |CL (x)| is odd. Let Q be a Sylow 2-subgroup of L. We claim that O2 (L) = 1. Assume by contradiction that O2 (L) > 1 and let M ≤ O2 (L) be a minimal normal subgroup of L. Assume first that [t, M ] = 1. Let 1 = y ∈ M . Then [y, x] = 1 as |CL (x)| is odd. Since M L, we have [y, x] ∈ M and so [y, x] = [y, x]t = [y t , xt ] = [y, x−1 ] which implies that [y, x2 ] = 1. However, as o(x)
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is odd, y ∈ CL (x2 ) = CL (x), which is a contradiction. Therefore [t, M ] > 1. By Lemma 2.1, M t has a real element of order 4, contradicting our assumption. Hence O2 (L) = 1. Assume that L is solvable. By [5, Theorem 2.4], L has a normal 2-complement K, Q is either cyclic or generalized quaternion and CK (Q) = CK (z), where z is the unique involution of Q. As generalized quaternion groups possess a real element of order 4, Q must be cyclic. This is part (2) of the theorem. Finally, assume that L is non-solvable. Let R be the solvable radical of L. We know that O2 (R) = 1. We claim that R = O2 (L). Assume by contradiction that R is not a 2 -group. Then 1 < Q0 = Q ∩ R ∈ Syl2 (R). Clearly O2 (R) is a solvable (C)-group without real element of order 4 and Q0 ≤ O2 (R). Note that O2 (R) is not a 2-group. So, part (2) implies that Q0 is cyclic and thus by [1, 39.2] R has a normal 2-complement, say R1 R. Note that 1 < R1 L and O2 (L/R1 ) = L/R1 and by Lemma 2.2, G/R1 is a (C)-group without real element of order 4; clearly L/R1 is not a 2-group and thus by the argument in the second paragraph, O2 (L/R1 ) = R/R1 ∼ = Q0 is trivial, which is a contradiction. Thus R = O2 (L) as wanted. Let L = L/O2 (L) and use the ‘bar convention’. By Lemma 2.2, L has no real element of order 4 or 2m with m > 1 odd. Moreover, L has a trivial solvable radical and has no proper normal subgroup of odd index. Let N be a minimal normal subgroup of L. Then N = S1 × S2 × · · · × Sk , where k ≥ 1 is an integer and Si ∼ = S for i = 1, 2, · · · , k and S is a nonabelian simple (C)-group. By Theorem 1 of [14], S is isomorphic to L2 (p) where p is a Fermat or a Mersenne prime; A6 or L2 (q), 2 B2 (q), U3 (q) or L3 (q) where q > 2 is a power of 2. However, as S has no real element of order 4, S ∼ = SL2 (2f ), f > 1 or 2 B2 (22f +1 ), f ≥ 1. Since S1 is not 2-closed, it has a nontrivial real element z of odd order. By [14, Lemma 3], z i = z −1 for some involution i ∈ S1 and CL (z) is of odd order. Now if k ≥ 2, then S2 ≤ CL (z) which is impossible as |S2 | is even. This shows that N is a non-abelian simple group and that CL (N ) = 1 (since CL (N ) ⊆ CL (z), CL (N ) is a normal subgroup of L of odd order). Thus L is an almost simple group with simple socle N and O2 (L) = L. Assume that N ∼ = 2 B2 (22f +1 ), f ≥ 1. In this case |Out(S)| = 2f + 1 is odd, so L = N . Assume next that N ∼ = SL2 (2f ), f ≥ 2. Then Out(N ) is a cyclic group of order f generated by a field automorphism ϕ of order f . Assume that L = N. Then L = N φ, where φ is a field automorphism of order r which is a power of 2. We have that CL (φ) ∼ = SL2 (2f1 )φ with f1 = f /r ≥ 1. Now SL2 (2f1 ) has a real element y of f1 order 2 + 1 > 1 and [y, φ] = 1 and so CL (y) is of even order, contradicting Lemma 3 of [14]. Therefore L = L/O2 (L) is isomorphic to one of the simple groups listed in part (3). Since O2 (L) = L and L/O2 (L) is non-abelian simple, we deduce that L = L O2 (L) and |L : L | = |O2 (L) : L ∩ O2 (L)| is odd, which implies that L = L . As a Sylow 2-subgroup of L/O2 (L) has at least two involutions, L has at least two involutions and so by [14, Lemma 6], we have [Q, O2 (L)] = 1. As Q ≤ CL (O2 (L)) L, we deduce that L = CL (O2 (L)). In particular, L is a quasi-simple group with Z(L) = O2 (L).
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However, the Schur multipliers of SL2 (2f ) and 2 B2 (22f +1 ) are 2-groups or trivial. This forces O2 (L) = 1 and part (3) holds. For the converse, if (1) holds, then clearly L has no real element of order 4 or 2m with m > 1 odd. If (2) holds, then L has no real element of order 2m with m > 1 odd by [5, Theorem 2.4]. Moreover, a Sylow 2-subgroup of L is cyclic, so it has no real element of order 4. Assume that (3) holds. In this case, the centralizer of every involution of L is a 2-group, so it has no real element of order 2m with m > 1 odd. If L = SL2 (2f ), then clearly L has no real element of order 4 (the Sylow 2-subgroup of L is elementary abelian). Assume now that L = 2 B2 (q) with q = 22f +1 , f ≥ 1. It remains to show that L has no real element of order 4 which follows from the argument in pages 121 − 122 of [14]. The proof is now complete. Proof of Theorem A. Suppose that the order of every real element of G is 2 or a power of a fixed prime p. Obviously, G is a (C)-group without real element of order 4. Let L = O2 (G). Then L satisfies the same hypothesis as G does. In particular, L = O2 (L) is a (C)-group without real element of order 4. We first claim that G is solvable. Indeed, it suffices to show that L is solvable. By way of contradiction, assume that L is non-solvable. By Theorem 2.3, L ∼ = SL2 (2f ) (f ≥ 2) or 2 B2 (q), where q = 22f +1 with f ≥ 1. In the former case, SL2 (2f ) contains two real elements of odd coprime orders 2f − 1 and 2f + 1, which is impossible. In the latter case, by Theorem 9 of [15], 2 B2 (q) contains at least two nontrivial real elements of odd order dividing q − 1 or q 2 + 1, respectively, so this case cannot occur. Hence L is solvable. By Theorem 2.3, L is a 2-group without real element of order 4 or O2 (L) = 1 and L has a cyclic Sylow 2-subgroup Q and a normal 2-complement K with CK (Q) = CK (z), where z is the unique involution of Q. If the first possibility holds, then conclusion (1) of Theorem A holds. Assume we are in the second possibility, so L is not a 2-group. To finish the proof of the theorem, it remains to show that K is a p-group. Notice that p must be odd as otherwise, L is a 2-group. We prove this by induction on |L|. Assume first that M := Op (K) > 1. Since |M | is odd and coprime to p, M contains no nontrivial real element of L. By Lemma 3.1(d) in [6], [Q, M ] = 1 and so Q ≤ CL (M )L which forces CL (M ) = L since L = O2 (L). By Lemma 2.2, the order of every real element of L/M is 2 or a power of p and O2 (L/M ) = L/M has a normal 2-complement K/M . By induction, K/M is a p-group. It follows that K = P M with P ∈ Sylp (K). Since M ≤ Z(K), we have K ≤ P and thus P K which implies that K = P × M . Now this implies that QP L so QP is normal subgroup of L of odd index which is impossible. Hence Op (K) = 1. Since K is solvable, we deduce that Op (K) > 1. Let L = L/Op (K). Clearly, L satisfies the hypothesis of the theorem, O2 (L) = L and |L| < |L|, by induction K = O2 (L) is a p-group. Thus K is a p-group.
Proof of Theorem B. Let L = O2 (G) and Q ∈ Syl2 (L). We know that L contains all real elements of G and so E(G) = E(L). Moreover L = O2 (L). Suppose that G has at most three real element orders. Then |E(L)| ≤ 3. Assume first that L is a 2-group. Note that if x is a real element of L then xk is also a real element of G for any integer
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k. Thus |E(L)| ≤ 3 if and only if E(L) ⊆ {1, 2, 4} or equivalently L has no real element of order 8, which proves (1). We now assume that L > 1 is not a 2-group. Clearly |L| is even and thus {1, 2} ⊆ E(L). Since L does not have a normal Sylow 2-subgroup, L has a nontrivial real element x of odd order by [7, Proposition 6.4]. Since |E(L)| ≤ 3, we see that o(x) must be an odd prime and thus E(L) = {1, 2, p} for some odd prime p. Now the result follows from Theorem A. 3. Four real-valued irreducible characters In this section, we determine the structure of finite groups with four real-valued irreducible characters. Recall that a finite group G is quasi-simple if G is perfect and G/Z(G) is a non-abelian simple group. A component of a finite group G is a quasisimple subnormal subgroup of G. The layer E(G) of G is a subgroup of G generated by all components of G. Finally, the generalized Fitting subgroup F∗ (G) is the product of the Fitting subgroup F(G) and the layer E(G). Notice that F∗ (G) is a central product of F(G) and E(G) and CG (F∗ (G)) ≤ F∗ (G) (see [1, §31]). We begin with the following result. Theorem 3.1. Let G be a finite non-solvable (C)-group. Suppose that O2 (G) = 1 and |E(G)| ≤ 4. Then G ∼ = S × K, where K is of odd order, S ∼ = A5 or SL3 (2) and E(G) = {1, 2, 3, 5} or {1, 2, 3, 4}, respectively. Proof. By Theorem B, we may assume that |E(G)| = 4. Since G is non-solvable, we know that 2 ∈ E(G) and G has a nontrivial real element of odd order ([7, Proposition 6.4]). Let L = O2 (G). We consider the following cases. (1) G has no real element of order 4. By Theorem A, it is easy to see that E(G) = {1, 2, p, q} for two distinct odd primes p and q. By Theorem 2.3, L ∼ = SL2 (2f ) (f ≥ 2) 2 2f +1 ) (f ≥ 1). Let K = CG (L). Then K is a normal subgroup of odd order of or B2 (2 G and G/K is an almost simple group with socle LK/K ∼ = L and |G/K : LK/K| = |G : LK| is odd. Assume first that L ∼ = 2 B2 (22f +1 ), f ≥ 1. By Theorem 9 and Proposition 16 of [15], L has three nontrivial real elements of odd distinct orders (which are 22f +1 − 1 and 22f +1 ± 2f +1 + 1). Thus this case cannot occur. Assume that L ∼ = SL2 (2f ) with f ≥ 2. Now L has two real elements of order 2f − 1 and 2f + 1 and both numbers must be primes. It is easy to see that f = 2 and so L = SL2 (4) ∼ = A5 . Since |Out(A5 )| = 2, we have G = L × K, where K is of odd order. In this case E(G) = {1, 2, 3, 5}. (2) G has a real element of order 4. We have E(G) = {1, 2, 4, p}, where p is an odd prime. Since O2 (G) = 1, O2 (L) = 1 and L is non-solvable with E(L) = E(G). Let Q ∈ Syl2 (L). Assume first that O2 (L) = 1. Let N be a minimal normal subgroup of G. Then N ∼ = S1 × S2 × · · · × Sk , where k ≥ 1 is an integer, S is isomorphic to a non-abelian simple (C)-group and Si ∼ = S for all i = 1, 2, · · · , k. It follows that S is isomorphic to one of the simple groups listed in Theorem 1 of [14]. As every nontrivial real element
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of odd order of S has the same order p, we deduce that S ∼ = SL3 (2) ∼ = L2 (7) and p = 3. Assume that k ≥ 2. Let x ∈ S1 be a real element of order 3 inverted by an element t ∈ S1 and y ∈ S2 be an involution. Since [S1 , S2 ] = 1, we deduce that [x, y] = 1 = [t, y] and thus (xy)t = xt y t = x−1 y = yx−1 = (xy)−1 ; so xy ∈ N is a real element of order 6, a contradiction. Thus k = 1 and N ∼ = SL3 (2). We see that K := CG (N ) is a normal subgroup of odd order of G and G/K is an almost simple group with socle N ∼ = SL3 (2). ∼ Since Aut(N ) = SL2 (7) · 2 has a real element of order 6 while G/K is also a (C)-group, we have G = N K = N × K as wanted. In this case, E(G) = {1, 2, 3, 4}. Assume that N := O2 (L) > 1. If Q has a unique involution, then Q is either cyclic or quaternion. Since L is non-solvable, Q cannot be cyclic by [1, 39.2]. Thus Q is quaternion. Now L/N is a non-solvable (C)-group with a quaternion Sylow 2-subgroup QN/N ∼ = Q. By [3, Theorem 2], L/N has a center of order 2. Since L/N is nonsolvable, it has a nontrivial real element N x of odd order inverted by an involution of L/N (by [14, Lemma 3]), which is impossible. Thus Q has at least two involutions. By [14, Lemma 6], Q centralizes N and hence CG (N ) is a normal subgroup of odd index of G. It follows that L ≤ CG (N ) and N ≤ Z(L). Let M/N = O2 (L/N ). Since |N | is odd, M = U N where U ∈ Syl2 (M ). Moreover, as N ≤ Z(L), we have U M and thus U = O2 (M ) L forcing U = 1 or O2 (L/N ) = 1 as O2 (L) = 1. By induction, we have L/N ∼ = SL3 (2). As O2 (L) = L and |N | is odd, L = L and thus L is a quasi-simple group with L/N ∼ = SL3 (2). However, the Schur multiplier of SL3 (2) is cyclic of order 2. This implies that N = 1, a contradiction. We next determine all finite non-abelian simple groups without strongly real element of order 4. Proposition 3.2. Let G be a finite non-abelian simple group. If G has no (strongly) real element of order 4, then G is isomorphic to one of the following simple groups: (1) SL2 (2f )(f ≥ 3), U3 (2f )(f ≥ 2), 2 B2 (22f +1 )(f ≥ 1); or (2) L2 (q), 5 ≤ q ≡ 3, 5 (mod 8), J1 , 2 G2 (32f +1 ), f ≥ 1. Proof. Let S be a Sylow 2-subgroup of G. We show that Ω1 (S) is abelian and hence it is elementary abelian and contains all involutions of S which implies that Ω1 (S) is strongly closed in S with respect to G and we can then invoke Corollary 1 of [9] and the result follows. Let i and j be distinct involutions in S. Then ij ∈ S is a strongly real 2-element of order m = o(ij) > 1. Since G has no strongly real element of order 4, m = 2. In particular ij = ji. Thus Ω1 (S) is abelian as it is generated by all involutions in S. The finite non-abelian simple groups in Proposition 3.2 are called Goldschmidt groups. These are exactly the finite simple groups of Lie type of Lie rank 1 in characteristic 2 and non-abelian simple groups with an abelian Sylow 2-subgroup. We are now ready to prove Theorem C which is included in the following. Theorem 3.3. Let G be a finite group with exactly four real-valued irreducible char acters. Let L = O2 (G) and Q be a Sylow 2-subgroup of L. Then one of the following holds.
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(1) G has a normal Sylow 2-subgroup. (2) O2 (L) = 1, Q is either cyclic or quaternion of order 8 and L has a normal 2-complement K with CK (Q) = CK (z), where z is the unique involution in Q. (3) G has 2-length one and Q is either homocyclic or a Suzuki 2-group. (4) G ∼ = SL3 (2) × K, where K is of odd order. Proof. Let G be a finite group with exactly four real-valued irreducible characters. Let L = O2 (G) and Q ∈ Syl2 (L). We know that G has exactly four conjugacy classes of real elements. In particular |E(G)| ≤ 4. If |E(G)| ≤ 3, then either (1) or (2) holds by Theorem B. So, we may assume that |E(G)| = 4. Also, in view of [7, Proposition 6.4], we also assume that G has a real element of order p for some odd prime p. Notice that G always has real elements of order 1 and 2. Thus {1, 2, p} ⊆ E(G). It follows that E(G) is one of the following sets {1, 2, p, p2 }, {1, 2, p, q}, {1, 2, 4, p} or {1, 2, p, 2p}, where p and q are distinct odd primes. In particular, G has only one class of involutions. (1) Assume G is solvable. If Q has at least two involutions, then Theorem IX.8.6 of [11] implies that G has 2-length one and Q is homocyclic or a Suzuki 2-group, so (3) occurs. Assume that Q has a unique involution. Then Q is either cyclic or quaternion. If Q is cyclic, then G has a normal 2-complement by [1, 39.2], so (3) holds. Assume that Q is quaternion. Since G has no real element of order 8, Q is an ordinary quaternion group of order 8. Clearly, all real elements of G have orders 1, 2, 4 and p. Thus G is a (C)-group and hence L is also a (C)-group. By [14, Lemma 3], L has a strongly real element x of order p with |CG (x)| being odd. By [3, Theorem 2], the center of L/O2 (L) is cyclic of order 2. Thus x ∈ O2 (L) and hence O2 (L) = 1. By Theorem 2.4 of [5], L has a normal 2-complement K and CK (z) = CK (Q), where z is the unique involution of Q. So (2) holds. (2) Assume that G is non-solvable. (2a) We first claim that O2 (G) = 1. By way of contradiction, assume that O2 (G) > 1 and let N ≤ O2 (G) be a minimal normal subgroup of G. Then N is an elementary abelian normal subgroup of G. Since G has only one conjugacy class of involutions, N contains all involutions of G. Clearly G/N is non-solvable and has at most four realvalued irreducible characters, by induction we have that G/N = T /N × U/N , where T, U are normal subgroups of G with T /N ∼ = SL3 (2) and U/N being of odd order. Let K be a Hall 2 -subgroup of U . Then U = KN and by Hall’s Theorem (see [1, 18.5]) U acts transitively on its Hall 2 -subgroups. Hence G = NG (K)U = NG (K)N . Assume that NG (K) < G. Let W = NG (K) ∩ N . Then W < N and W is normalized by both N (as N is abelian) and NG (K); therefore W G. As N is a minimal normal subgroup of G, we deduce that W = 1. However, this is impossible as NG (K) is nonsolvable and so it possesses an involution which does not lie in N. Thus K G. Since T ∩ K ≤ (T ∩ U ) ∩ K = N ∩ K = 1, we have G = T × K. Let T0 be the last term of the derived series of T . Since T /N ∼ = SL3 (2), T /N is perfect and so T /N = T0 N/N which implies that T /N ∼ = T0 /N0 , where N0 = T0 ∩ N G. Since N is a minimal normal subgroup of G, either N0 = 1 or N0 = N. If the former case holds, then T = T0 × N so T0 has an involution that does not lie in N . Thus N0 = N and hence T = T0 is perfect.
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Observe that T contains all real elements of G and since [K, T ] = 1, T also has four real conjugacy classes and E(T ) = E(G); so T has only one class of involutions. Now let 1 = y ∈ N . Then y T contains all involutions of T and hence |N | − 1 = |y T | = |T : CT (y)|. As N ≤ CT (y), n := |T : CT (y)| = 1 or it is the index of a proper subgroup of the simple group T /N and n + 1 = |N | is a 2-power. If the former case holds, then |N | = 2 and so T ∼ = SL2 (7). In this case, it is easy to check that T has more than 4 real conjugacy classes. Assume that the latter case holds. Using [8], we deduce that |T : CT (y)| = 7 and hence |N | = 8. By [8] again, there are exactly two perfect groups of order |T | = | SL3 (2)| · 23 = 1344. However these two perfect groups have more than one conjugacy class of involutions. This contradiction shows that O2 (G) = 1 as wanted. (2b) G is a (C)-group. Suppose by contradiction that G is not a (C)-group. Then G has a real element of order 2m for some odd integer m > 1. It follows that m = p and hence E(G) = {1, 2, p, 2p}, where p is an odd prime. Assume first that F(G) > 1. By Claim (2a), F(G) is of odd order. By induction, we have G/F(G) ∼ = SL3 (2) × U , where U is of odd order. Hence G/F(G) has a real element of order 4 and thus G also has a real element of order 4 by Lemma 2.2, which is a contradiction. We now assume that F(G) = 1. Then F∗ (G) = E(G) is a direct product of nonabelian simple groups. Let M be a minimal normal subgroup of G. Then M is a direct product of copies of a non-abelian simple group S. Since S has no real element of order 4, S is a Goldschmidt group by Proposition 3.2. However, we can check that every Goldschmidt group contains at least two nontrivial real elements of odd coprime orders. (See the proof of [13, Theorem 3.1]). (2c) The completion of the proof. By Claims (2a) and (2b), G is a non-solvable (C)-group with O2 (G) = 1 and |E(G)| = 4. Now it follows from Theorem 3.1 that G∼ = A5 × K or G ∼ = SL3 (2) × K, where K is of odd order. However, the former case cannot occur as A5 has five conjugacy classes of real elements. Thus the latter case occurs and hence (4) holds. The proof is now complete. It seems that if G is a finite group and E(G) = {1, 2, p, 2p}, where p is an odd prime, then G is solvable. Unfortunately, we are unable to prove or disprove this at the moment. 4. Some classification-free proofs In this section, we provide classification-free proofs of Theorem C of [6] on the nonexistence of real element of order 3 and Theorem 5.2 of [13] on the structure of finite groups with two real-valued irreducible p-Brauer characters, where p is an odd prime. Recall that a section of a finite group G is the quotient H/K, where K H are subgroups of G; we say that G is X-free, if no section of G is isomorphic to X, where X is a finite group. Fix an odd prime p. If G has a real element x of order p inverted by an element t ∈ G, then [t2 , x] = 1 and x, t/t2 ∼ = D2p , the dihedral group of order 2p, so D2p is a section of G. Conversely, if G has a section H/K ∼ = D2p , then H/K has
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a real element of order p which lifts to a real element of H whose order is divisible by p (see [10, Lemma 2.2]). Thus G has a real element of order p. Therefore, G has no real element of order p if and only if G is D2p -free; in other words, D2p is not involved in G. Notice that if p = 3, then D6 ∼ = S3 . By Corollary 3 of [2], if S is a finite S3 -free ∼ non-abelian simple group, then S = L2 (32f +1 ) or 2 B2 (22f +1 ) with f ≥ 1. If we assume that 3 divides the order of S, then S is isomorphic to the former group. Now by using Theorem 3.3 of [6] and the previous result instead of Theorem 2.1 of [6], we obtain a classification-free proof of the following theorem, which is Theorem C of [6].
Theorem 4.1. Let G be a finite group of order divisible by 3. Suppose that G = O2 (G). Then G has no real element of order 3 if and only if G/O3 (G) is isomorphic to a direct products of simple groups L2 (32fi +1 ), where i = 1, 2, · · · , k and fi ≥ 1. It would be interesting to classify all finite D2p -free non-abelian simple groups without invoking the full strength of the classification of finite simple groups. Fix an odd prime p. For a finite group G, we write IBrrv (G) for the set of realvalued irreducible p-Brauer characters of G. We now provide a classification-free proof of Theorem 5.2 of [13], which is stated below. Theorem 4.2. Let p be an odd prime and G be a finite group with Op (G) = 1. If | IBrrv (G)| = 2, then G has a normal Sylow 2-subgroup, which is homocyclic or a Suzuki 2-group. Proof. Suppose that | IBrrv (G)| = 2. We first show that G is solvable. By Brauer’s lemma on character tables, G has exactly one conjugacy class of nontrivial real element of p -order. It follows that |G| is even and G has a unique conjugacy class of involutions. In particular, G has no real element of order 4. If 1 < N G, then | IBrrv (G/N )| ≤ | IBrrv (G)| = 2, by induction we deduce that G/N is solvable. Hence G has a unique minimal normal subgroup N which is a direct product of copies of a non-abelian simple group S. Since G has no real element of order 4, S has no real element of order 4 and so by Proposition 3.2 S is a Goldschmidt group. However, one can check that Goldschmidt groups contain at least two nontrivial real elements of odd coprime orders. Now as G is solvable and has a unique conjugacy class of involutions, the Sylow 2-subgroup Q of G is either cyclic or quaternion (if Q has a unique involution) or Q is homocyclic or a Suzuki 2-group by [11, Theorem IX.8.6]. However, as G has no real element of order 4, Q cannot be quaternion. Now we can follow the original argument in [13] to deduce that Q G. Notice that the authors use [13, Lemma 2.1] in their proof but that lemma is only stated for p = 2 although it also holds for p odd.
Acknowledgment The author is grateful to Dan Rossi for the discussion during the preparation of this paper. He also thanks the anonymous referee for his or her numerous corrections and suggestions, which helped to improve the clarity of this paper.
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References [1] M. Aschbacher, Finite group theory, second edition, Cambridge Studies in Advanced Mathematics, 10, Cambridge University Press, Cambridge, 2000. [2] M. Aschbacher, S3 -free 2-fusion systems, Proc. Edinb. Math. Soc. (2) 56 (2013), no. 1, 27–48. [3] R. Brauer and M. Suzuki, On finite groups of even order whose 2-Sylow group is a quaternion group, Proc. Nat. Acad. Sci. U.S.A. 45 (1959), 1757–1759. [4] D. Chillag, A. Mann, Nearly odd-order and nearly real finite groups, Comm. Algebra 26 (1998), no. 7, 2041–2064. [5] S. Dolfi, D. Gluck, G. Navarro, On the orders of real elements of solvable groups. Israel J. Math. 210 (2015), no. 1, 1–21. [6] S. Dolfi, G. Malle, G. Navarro, The finite groups with no real p-elements, Israel J. Math. 192 (2012), no. 2, 831–840. [7] S. Dolfi, G. Navarro, P. H. Tiep, Primes dividing the degrees of the real characters, Math. Z. 259 (2008), no. 4, 755–774. [8] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.10.1; 2019. (https://www.gap-system.org) [9] D. M. Goldschmidt, 2-fusion in finite groups, Ann. of Math. (2) 99 (1974), 70–117. [10] R. M. Guralnick, G. Navarro, P. H. Tiep, Real class sizes and real character degrees, Math. Proc. Cambridge Philos. Soc. 150 (2011), no. 1, 47–71. [11] B. Huppert and N. Blackburn, Finite groups. II, Grundlehren der Mathematischen Wissenschaften, 242, Springer-Verlag, Berlin, 1982. [12] A. Moret´o and G. Navarro, Groups with three real valued irreducible characters, Israel J. Math. 163 (2008), 85–92. [13] G. Navarro, L. Sanus, P. H. Tiep, Groups with two real Brauer characters. J. Algebra 307 (2007), 891–898. [14] M. Suzuki, Finite groups in which the centralizer of any element of order 2 is 2-closed. Ann. of Math. (2) 82 (1965) 191–212. [15] M. Suzuki, On a class of doubly transitive groups, Ann. of Math. (2) 75 (1962), 105–145. [16] J. G. Thompson, Nonsolvable finite groups all of whose local subgroups are solvable, Bull. Amer. Math. Soc. 74 (1968), 383–437. Department of Mathematical Sciences, Binghamton University, Binghamton, NY 13902-6000, USA E-mail address: [email protected]