Correction to: A coprime action version of a solubility criterion of Deskins

Correction to: Monatshefte für Mathematik


In this Corrigendum we correct a missed case in the statement of Theorem 2.4 and a subsequent mistake in the proof of the main result in “A coprime action version of a solubility criterion of Deskins”, Monatsh. Math. 188, 461–466 (2019).

Keywords   Soluble groups \(\cdot \) Maximal subgroups \(\cdot \) Coprime action \(\cdot \) Group action on groups

Mathematics Subject Classification   20D20 \(\cdot \) 20D15

The main result of [1] is a coprime action version of a theorem of B. Huppert: If a finite group G has a maximal subgroup that is nilpotent with Sylow 2-subgroup of nilpotency class at most 2, then G is soluble (Satz IV.7.4 of [4]). This theorem is the completion of previous results by Huppert [5], J.G. Thompson [8], W.E. Deskins [2] and Z. Janko [6]. Professor M.D. Pérez Ramos noticed and informed us that there are some mistakes and inaccuracies in the last part of the proof of the main theorem of [1]. Thus the goal of this note is to correct them.

First, the proof of the main theorem uses two classification theorems due to Kondrat’ev [7] and to Gilman and Gorenstein [3], respectively. However, there is one simple group missed in the statement of Theorem 2.4 in [1], which joins both classifications. The correct statement is the following.

Theorem 2.4

Let G be a finite non-abelian simple group and P a Sylow 2-subgroup of G. If \(\mathbf{N}_G(P)=P\) and P has class at most 2, then \(G\cong \mathrm{PSL}(2,q)\), where \(q\equiv 7,9\) (mod 16) or \(G\cong A_7\).


This is a consequence of combining the main result of [7] and Theorems 7.1 and 7.4 of [3]. \(\square \)

As a consequence of this correction, several modifications in the proof of Step 4 of the main theorem of [1] are necessary. Furthermore, in lines 28–29, page 465 it is claimed that the subgroup K is normalised by an element of order 3 lying in S. This is not true. For the reader’s convenience we rewrite the whole proof of Step 4.

Proof of Step 4 Let N be a minimal A-invariant normal subgroup of G. We can assume that N is not soluble; otherwise by Step 1, N is not contained in M, and by maximality we obtain \(NM=G\). As a consequence, G would be soluble and the proof is finished. Therefore, we can write \(N=S_1\times \ldots \times S_n\) where \(S_i\) are isomorphic non-abelian simple groups (possibly \(n=1\)). Put \(S=S_1\). Notice that A permutes the \(S_i's\), but not necessarily this action is transitive. Let \(B=\mathbf{N}_A(S)\) and let T be a transversal of B in A. On the other hand, since M is maximal in G, we have \(\mathbf{N}_G(M\cap N)=M\), so in particular \(\mathbf{N}_N(M\cap N)=M\cap N\). Further, as \(M\cap N\) is a Sylow 2-subgroup of N, we have \(M\cap N=M\cap S \times \ldots \times M\cap S_n\), so we conclude that \(M\cap S\) is self-normalising in S. Also, it has nilpotency class exactly 2 by Lemma 2.1 and Step 3. Then by applying Theorem 2.4, we obtain \(S\cong \) PSL(2, q) with \(q\equiv 7,9\) (mod 16) or \(S\cong A_7\). We distinguish separately these cases.

Assume first that \(q\equiv 9 \) (mod 16), with \(q>9\). Then we can certainly choose an odd prime \(r\mid (q-1)/2\) and R to be a B-invariant Sylow r-subgroup of S. By Lemma 2.5(3), we know that \(|\mathbf{N}_S(R)|= q-1\), so \(\mathbf{N}_S(R)\) has odd index in S and contains properly a Sylow 2-subgroup of S. Analogously, if \(q\equiv 7 \) (mod 16), with \(q>7\), there exists an odd prime \(r\mid (q+1)/2\) and we take R to be a B-invariant Sylow r-subgroup of S. Again by Lemma 2.5(2), we know that \(|\mathbf{N}_S(R)|= (q+1)\), so \(\mathbf{N}_S(R)\) has odd index in S and hence, it contains properly a Sylow 2-subgroup of S. In both cases, we put \(R_1=\prod _{t\in T}R^t\), which is an A-invariant Sylow r-subgroup of \(\prod _{t\in T} S_1^t\). We can argue similarly to construct an A-invariant Sylow r-subgroup for each orbit of the action of A on the \(S_i's\). Hence, we can construct \(R_0=R_1 \times \ldots \times R_t\), where t denotes the number of orbits of A on the \(S_i's\), and this is certainly an A-invariant Sylow 2-subgroup of N. We conclude that \(|N:\mathbf{N}_N(R_0)|=|S:\mathbf{N}_S(R)|^n\) is odd too. Now, by the Frattini argument, \(G=N\mathbf{N}_G(R_0)\) and thus, \(|G:\mathbf{N}_G(R_0)|=|N: \mathbf{N}_N(R_0)|\). We conclude that \(\mathbf{N}_G(R_0)\) properly contains an A-invariant Sylow 2-subgroup of G, contradicting the maximality of M.

Finally, suppose that \(S\cong \mathrm{PSL}(2,9), \mathrm{PSL}(2,7)\) or \(A_7\). In all cases, the Sylow 2-subgroups of S are dihedral groups of order 8. Now, \(M\cap N\) is an A-invariant Sylow 2-subgroup of N, which is the direct product of n copies of a dihedral group, say D, of S. As M has nilpotence class two, then \([M, M\cap N]\le M'\le \mathbf{Z}(M)\), and since \(M\cap N\unlhd M\) it follows that \([M, M\cap N]\le \mathbf{Z}(M) \cap (M\cap N) \le \mathbf{Z}(M\cap N)\). This implies that every subgroup of \(M\cap N\) containing \(\mathbf{Z}(M\cap N)\) must be normal in M. We will use this property later. Now let K be one of the two subgroups of D isomorphic to the 4-Klein group, which obviously satisfies \(\mathbf{Z}(D)\le K\) and set \(K^A=\langle K^a \mid a\in A\rangle \). By the coprime action hypothesis we have that |A| is odd, and then the fact that D has exactly two subgroups isomorphic to the 4-Klein group implies that for every \(a\in A\), either \(K^a=K\), or \(K^a\) lies in some other distinct copy of S. Furthermore, \(K^A\) is a direct product of certain copies of K, each of which lies in a different copy of S. Now, if the action of A on the \(S_i's\) is transitive, we will just consider the subgroup \(K^A\), but if the action is not transitive, then we proceed as follows. For each of the orbits of the action of A on the \(S_i\), we choose j with \(S_j\) in the orbit, and choose a 4-Klein subgroup \(K_j\le D_j\), where \(D_j\) is the corresponding isomorphic copy of D appearing in \(M\cap N\). Then we define the subgroup \(K_j^A\) similarly as \(K^A\). Set \(K_0\) to be the direct product of these subgroups, one for each orbit of the action of A on the \(S_i\). We can write \(K_0=K_1\times \ldots \times K_n\), where each \(K_i\) is a 4-Klein group lying in \(S_i\). By construction, \(K_0\) is trivially A-invariant and, moreover, \(K_0\unlhd M\), because \(\mathbf{Z}(M\cap N)\le K_0\le M\cap N\). By the above proved property, we get \(M\le \mathbf{N}_G(K_0)\), which is also A-invariant. Now \(\mathbf{N}_G(K_0)=M \mathbf{N}_N(K_0)\) and \(\mathbf{N}_N(K_0)=\prod _{i=1}^n\mathbf{N}_{S_i}(K_i)\). In fact, one can easily check that when \(S\cong \mathrm{PSL}(2,9)\) or \(\mathrm{PSL}(2,7)\) then \(\mathbf{N}_{S}(K)\cong S_4\), and when \(S\cong A_7\), then \(\mathbf{N}_{S}(K)\cong (A_4 \times C_3) < imes C_2\). In all cases we get a contradiction with the maximality of M.


It is possible to give a simpler argument for the case \(S\cong A_7\) by using that \(A_7\) possesses a unique conjugacy class of \(\{2, 3\}\)-Hall subgroups. In this case, by Glauberman’s Lemma, there exists an A-invariant \(\{2,3\}\)-Hall subgroup of N, say H. Then the Frattini argument gives \(G=N\mathbf{N}_G(H)\), so \(|G:\mathbf{N}_G(H)|\) is a \(\{2, 3\}'\)-number. This implies that the A-invariant subgroup \(\mathbf{N}_G(H)\) properly contains an A-invariant Sylow 2-subgroup of G, contradicting the maximality of such Sylow 2-subgroup (Step 2).


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We would like to thank M.D. Pérez-Ramos for many helpful conversations on the subject. The first author was partially supported by Ministerio de Ciencia, Innovación y Universidades, Proyecto PGC2018-096872-B-100 and also by Proyecto UJI-B2019-03. The second author was supported by the Nature Science Fund of Shandong Province (No. ZR2019MA044) and the Opening Project of Sichuan Province University Key Laboratory of Bridge Non-destruction Detecting and Engineering Computing (No. 2018QZJ04).

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Beltrán, A., Shao, C. Correction to: A coprime action version of a solubility criterion of Deskins. Monatsh Math (2020).

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