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

## Correction to: Monatshefte für Mathematik https://doi.org/10.1007/s00605-018-1256-x

Abstract

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 $$G\cong A_7$$.

### Proof

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.

### Remark

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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## Acknowledgements

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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Correspondence to Antonio Beltrán.

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