1 Introduction

Let G be a finite group. A sequence \((g_1,\ldots ,g_n)\) of elements of G is said to be irredundant if \(\langle g_j \mid j\ne i\rangle \) is properly contained in \(\langle g_1,\ldots ,g_n\rangle \) for every \(i\in \{1,\ldots ,n\}\). Let i(G) be the maximum size of any irredundant sequence in G and let m(G) be the maximum size of any irredundant generating sequence of G [i.e. an irredundant sequence \((g_1,\ldots ,g_n)\) with the property that \(\langle g_1,\ldots ,g_n\rangle =G\)]. Clearly \(m(G)\le i(G)=\max \{m(H)\mid H\le G\}.\) The invariant m(G) has received some attention (see, e.g., [1, 2, 4, 5, 7, 9]) also because of its role in the efficiency of the product replacement algorithm [6]. In a recent paper, Fernando [3] investigates a natural connection between irredundant generating sequences of G and certain configurations of maximal subgroups of G. A family of subgroups \(H_i\le G\), indexed by a set I, is said to be in general position if for every \(i\in I\), the intersection \(\cap _{j\ne i}H_j\) properly contains \(\cap _{j\in I}H_j\). Define \({{\mathrm{MaxDim}}}(G)\) as the size of the largest family of maximal subgroups of G in general position. It can be easily seen that \(m(G)\le {{\mathrm{MaxDim}}}(G)\le i(G)\) (see, e.g., [3, Propositions 2 and 3]). However, the difference \({{\mathrm{MaxDim}}}(G)-m(G)\) can be arbitrarily large: for example if \(G={{\mathrm{Alt}}}(5)\wr C_p\) is the wreath product of the alternating group of degree 5 with a cyclic group of prime order p,  then \({{\mathrm{MaxDim}}}(G)\ge 2p\) but \(m(G)\le 5\) [3, Proposition 12]. On the other hand, Fernando proves that \({{\mathrm{MaxDim}}}(G)=m(G)\) if G is a finite supersoluble group [3, Theorem 25], but gives also an example of a finite soluble group G with \(m(G)\ne {{\mathrm{MaxDim}}}(G) \) [3, Proposition 16].

In this note we collect more information about the difference \({{\mathrm{MaxDim}}}(G)-m(G)\) when G is a finite soluble group. In this case m(G) coincides with the number of complemented factors in a chief series of G (see [4, Theorem 2]). Our first result is that the equality \({{\mathrm{MaxDim}}}(G)=m(G)\) holds for a class of finite soluble groups, properly containing the class of finite supersoluble groups (see, e.g., [8, 7.2.13]).

Theorem 1

If G is a finite group and the derived subgroup \(G^{\prime }\) of G is nilpotent, then \({{\mathrm{MaxDim}}}(G)=m(G).\)

However, already in the class of finite soluble groups with Fitting length equal to two, examples can be exhibited of groups G for which the difference \({{\mathrm{MaxDim}}}(G)-m(G)\) is arbitrarily large.

Theorem 2

For any odd prime p, there exists a finite group G with Fitting length two such that \(m(G)=3, {{\mathrm{MaxDim}}}(G)=p\) and \(i(G)=2p.\)

Notice that if G is a soluble group with \(m(G)\ne {{\mathrm{MaxDim}}}(G)\), then \(m(G)\ge 3\). Indeed, if \(m(G)\le 2\), then a chief series of G contains at most two complemented factors and it can be easily seen that this implies that \(G^{\prime }\) is nilpotent.

2 Groups whose derived subgroup is nilpotent

Definition 3

A family of subgroups \(H_i\le G\), indexed by a set I, is said to be in general position if for every \(i\in I\), the intersection \(\cap _{j\ne i}H_j\) properly contains \(\cap _{j\in I}H_j\) (equivalently, \(H_i\) does not contain \(\cap _{j\ne i}H_j\)).

Note that the subgroups \(\{ H_i \mid i \in I\}\) are in general position if and only, whenever \(I_1 \ne I_2\) are subsets of S, then \( \cap _{i \in I_1} H_i \ne \cap _{i \in I_2}H_i\) (see, e.g., Definition 1 in [3]).

Lemma 4

Let \(\mathbb F\) be a field of characteristic p. Let V a finite dimension \(\mathbb F\)-vector space, let \(H= \langle h \rangle \) where \(h \in \mathbb F^*\) such that \(\mathbb F=\mathbb F_p[h]\) and set \(G=V \rtimes H\).

If \(M_1, \ldots , M_r\) is a set of maximal subgroups of G supplementing V, then

$$\begin{aligned} M_1 \cap \ldots \cap M_r= W \rtimes K \end{aligned}$$

where W is a \(\mathbb F\)-subspace of V and K is either trivial or a conjugate \(H^v\) of H, for some \(v \in V\).

Proof

By induction on r we can assume that \(T_1=M_1 \cap \ldots \cap M_{r-1}= W_1 \rtimes K_1\), where \(W_1\) is a subspace of V and \(K_1=\{1\}\) or \(K_1 = H^v, v \in V\). The maximal subgroup \(M_r\) is a supplement of V, so we can write \(M_r=W_2 \rtimes H^{w}\), where \(W_2\) is a subspace of V and \(w \in V\). For shortness, set \(T_2=M_r\) and \(T=T_1 \cap T_2\). Since \(W_1\) and \(W_2\) are normal Sylow p-subgroups of \(T_1\) and \(T_2\), respectively, their intersection \(W=W_1 \cap W_2\) is a normal Sylow p-subgroup of T. In the case where T is not a p-group, then \(T = W \rtimes K\) where K is a non-trivial \(p'\)-subgroup of T. Then K is contained in some conjugates \(H^{v_1}\) and \(H^{v_2}\) of the \(p'\)-Hall subgroups of \(T_1\) and \(T_2\), respectively. In particular, there exists \(1\ne y \in K\) such that \(y=h_1^{v_1}=h_2^{v_2}\) for some \(h_1, h_2 \in H\). It follows that \(1\ne h_1=h_2 \in C_H(v_1-v_2)\). From \( C_H(v_1-v_2)\ne \{1\}\), we deduce that \(v_1=v_2\). Thus we have \(T_1=W_1 \rtimes H^{v_1}, T_2=W_2 \rtimes H^{v_1}\) and \(T=W \rtimes H^{v_1}\). \(\square \)

Corollary 5

In the hypotheses of Lemma 4, if \(M_1, \ldots , M_r\) are in general position, then

  1. (1)

    \(r \le \dim (V)+1\);

  2. (2)

    if \(r= \dim (V)+1\), then, for a suitable permutation of the indices, \(\bigcap _{i=1}^{r-1} M_i=H^v\) for some \(v\in V\), and \(\bigcap _{i=1}^{r} M_i=\{1\}.\)

Proof

Let \(n=\dim V\). Since the subgroups \(M_1, \ldots , M_r\) are in general position, the set of the intersections \(T_j=\cap _{i=1}^{j} M_i\), for \(j=1, \ldots , r\), is a strictly decreasing chain of subgroups. By Lemma 4, \(T_i=W_i \rtimes K_i\), where \(W_i\) is a \(\mathbb F\)-subspace of \(W_{i-1}\) and \(K_i\) is either trivial or a conjugate of H. Note that \(n-1=\dim W_1 \ge \dim W_{i} \ge \dim W_{i+1} \). Moreover, if \(\dim W_{i} = \dim W_{i+1}\) for some index i, then \(W_i=W_{i+1}\) and, since \(T_i \ne T_{i+1}\), we have that

  • \(K_1, \ldots ,K_{i}\) are non-trivial;

  • \(K_{i+1}= \cdots = K_r=\{1\}\).

In particular there exists at most one index i such that \(\dim W_{i} = \dim W_{i+1}\). As \(\dim W_1 = n-1\), it follows that we can have at most \(n+1\) subgroups \(T_i\), hence \(r \le n+1\).

In the case where \(r=n+1\), we actually have that \(\dim W_{i} = \dim W_{i+1}\) for at least one, and precisely one, index i. This implies that \(W_i=W_{i+1}\) and, setting \(J=\{1, \ldots , n+1\} {\setminus } \{i+1\}\) and \(T=\cap _{l \in J} M_{l}\), we get that \(W_{n+1}\) coincides with the Sylow p-subgroup of T. Since \(\dim W_{n+1}=0\) and \(T\ne 1\) we deduce that \(T = H^v\), for some \(v \in V\). Finally, \(T \cap M_{i+1}=\{1\}\).

\(\square \)

A proof of the following lemma is implicitly contained in Sect. 1 of [3], but, for the sake of completeness, we sketch a direct proof here.

Lemma 6

Let H be an abelian finite group. The size of a set of subgroups in general position is at most m(H).

Proof

The proof is by induction on the order of H. Let \(\Omega =\{ A_1, \ldots , A_r\}\) be a set of subgroups of H in general position. Without loss of generality we can assume that \(\cap _{i=1}^rA_i=\{1\}\). If \(m=m(H)\), then H decomposes as a direct product of m cyclic groups of prime-power order. Let B be one of these factors, and let X be the unique minimal normal subgroup of B. Since \(\cap _{i=1}^rA_i=\{1\}\), there exists at least an integer i such that X is not contained in \(A_i\). It follows that \(A_i \cap B=\{1\}\), hence \(A_i \cong A_iB/B \le H/B\) and

$$\begin{aligned} m(A_i) \le m \left( H/B \right) =m-1. \end{aligned}$$

Now, the set of subgroups of \(A_i\)

$$\begin{aligned} \Omega ^*=\{ A_j \cap A_i \mid j \ne i, \ 1 \le j \le r\} \end{aligned}$$

is in general position, hence, by inductive hypothesis, \(|\Omega ^*|=r-1 \le m(A_i)\). Therefore, \(r \le m\). \(\square \)

Proof of Theorem 1

Since

$$\begin{aligned} m \left( G\right) =m\left( G/{{\mathrm{Frat}}}\left( G\right) \right) \text{ and } {{\mathrm{MaxDim}}}\left( G\right) ={{\mathrm{MaxDim}}}\left( G/{{\mathrm{Frat}}}\left( G\right) \right) , \end{aligned}$$

without loss of generality we can assume that \({{\mathrm{Frat}}}(G)=1\). In this case the Fitting subgroup \({{\mathrm{Fit}}}(G)\) of G is a direct product of minimal normal subgroups of G, it is abelian and complemented. Let K be a complement of \({{\mathrm{Fit}}}(G)\) in G; note that, being \(G^{\prime }\) nilpotent by assumption, K is abelian. Let F be a complement of Z(G) in \({{\mathrm{Fit}}}(G)\) and let \(H=Z(G)\times K\). We have \(G=F\rtimes H\) and we can write F as a product of nontrivial H-irreducible modules

$$\begin{aligned} F = V_1^{n_1} \times \cdots \times V_r^{n_r} \end{aligned}$$

where \(V_1, \ldots , V_r\) are irreducible H-modules, pairwise not H-isomorphic.

By [4, Theorem 2] m(G) coincides with the number of complemented factors in a chief series of G, hence

$$\begin{aligned} m(G)= \sum _{i=1}^r n_i +m(H). \end{aligned}$$

Let \({\mathcal {M}}\) be a family of maximal subgroups of G in general position.

Let \(M_{0,1}, \ldots , M_{0,\nu _0}\) the elements of \({\mathcal {M}}\) containing F. We can write

$$\begin{aligned} M_{0,i} =F \rtimes Y_{i} \end{aligned}$$

where \(Y_{i}\) is a maximal subgroup of H. Note that \(Y_1, \ldots , Y_{\nu _0}\) are maximal subgroups of H in general position, hence, by Lemma 6, \( \nu _0 \le {{\mathrm{MaxDim}}}(H) \le m(H)\).

If M is a maximal subgroup supplementing F, then M contains the subgroup \(U_i = \prod _{j\ne i} V_j^{n_j}\) for some index i. In particular \(M=(U_i \times W_i)\rtimes H^v\) for some \(v\in V_i^{n_i}\) and some maximal H-submodule \(W_i\) of \(V_i^{n_i}\). Set \(C_i=C_H(V_i)\) and \(H_i=H/C_i\). Then \(\mathbb F_i={{\mathrm{End}}}_{H_i}(V_i)\) is a field and \(V_i\) is an absolutely irreducible \(\mathbb F_iH_i\)-module. Since \(H_i\) is abelian, \(\dim _{\mathbb F_i}V_i=1\), that is \(V_i \cong \mathbb F_i\), and hence \(H_i\) is isomorphic to a subgroup of \(\mathbb F_i^*\) generated by a primitive element. In particular we can apply Corollary 5 to the group \(V_i^{n_i} \rtimes H_i\). Let \(M_{i,1}, \ldots , M_{i, \nu _i}\) the maximal subgroups in \({\mathcal {M}}\) containing \(U_i\); say

$$\begin{aligned} M_{i,l}= \left( U_i \times W_{i,l}\right) \rtimes H^{v_{i,l}}, \end{aligned}$$

where \(v_{i,l} \in V_i^{n_i}\). Note that the subgroups \(\overline{M}_{i,l}=W_{i,l} \rtimes H_i^{v_{i,l}}\), for \(l \in \{1, \ldots , \nu _i\},\) are maximal subgroups of \(V_i^{n_i} \rtimes H_i\) in general position, hence, by Corollary 5,

$$\begin{aligned} \nu _i \le n_i+1. \end{aligned}$$

If \(\nu _i \le n_i\) for every \(i \ne 0\), then

$$\begin{aligned} \left| {\mathcal {M}} \right| = \sum _{i=1}^{r} \nu _i +\nu _0 \le \sum _{i=1}^{r} n_i + m (H) =m(G), \end{aligned}$$

and the result follows.

Otherwise let J be the set of the integers \(i \in \{1, \ldots ,r \}\) such that \(\nu _i = n_i+1\). By Corollary 5, we can assume that, for some \(v_i \in V_i^{n_i}\),

$$\begin{aligned} \bigcap _{l=1}^{n_i} M_{i,l}= & {} U_i \rtimes H^{v_i},\\ \bigcap _{l=1}^{n_i+1} M_{i,l}= & {} U_i \rtimes C_i. \end{aligned}$$

Recall that the \(M_{0,j} =F \rtimes Y_{j}\), for \(j=1 , \ldots , \nu _0\), are the elements of \({\mathcal {M}}\) containing F. Our next task is to prove that

$$\begin{aligned} \Omega =\{ C_i \mid i\in J\} \cup \{ Y_{j} \mid j=1, \dots , \nu _0 \} \end{aligned}$$

is a set of subgroups of H in general position.

Assume, by contradiction, that for example \(C_1 \ge \left( \cap _{i \ne 1} C_i \right) \cap \left( \cap _{ j =1}^{\nu _0} Y_{j}\right) \); then

$$\begin{aligned} M_{1, n_1+1} \ge U_1 \rtimes C_1 \ge \left( \cap _{l=1}^{n_1} M_{1,l}\right) \cap \left( \cap _{i \ne 1} \left( \cap _{ l=1}^{n_i+1} M_{i,l}\right) \right) \cap \left( \cap _{ j=1}^{\nu _0} M_{0,j}\right) \end{aligned}$$

against the fact that \({\mathcal {M}}\) is in general position. Similarly, if \(Y_{1 } \ge \left( \cap _{i \in J} C_i \right) \cap \left( \cap _{ j \ne 1} Y_{j} \right) \), then

$$\begin{aligned} M_{0, 1} = F \rtimes Y_1 \ge \left( \cap _{i \in J}\left( \cap _{l=1}^{n_i+1} M_{i,l}\right) \right) \cap \left( \cap _{ j \ne 1} M_{0,j}\right) , \end{aligned}$$

a contradiction.

Now we can apply Lemma 6 to get that \(|\Omega | \le m(H)\). Therefore, we conclude that

$$\begin{aligned} \left| {\mathcal {M}} \right| = \sum _{i=1}^{r} \nu _i +\nu _0 \le \sum _{i=1}^{r} n_i + |J| + \nu _0 = \sum _{i=1}^{r} n_i +|\Omega | \le \sum _{i=1}^{r} n_i + m (H) =m(G), \end{aligned}$$

and the proof is complete. \(\square \)

3 Finite soluble groups with \(m(G)=3\) and \({{\mathrm{MaxDim}}}(G)\ge p\)

In this section we will assume that p and q are two primes and that p divides \(q - 1\). Let \(\mathbb F\) be the field with q elements and let \(C=\langle c \rangle \) be the subgroup of order p of the multiplicative group of \(\mathbb F\). Let \(V=\mathbb F^p\) be a p-dimensional vector space over \(\mathbb F\) and let \(\sigma =(1,2,\dots ,p)\in {{\mathrm{Sym}}}(p)\). The wreath group \(H=C\wr \langle \sigma \rangle \) has an irreducible action on V defined as follows: if \(v=(f_1,\dots ,f_p)\in V\) and \(h=(c_1,\dots ,c_p)\sigma \in H\), then \(v^h=(f_{1\sigma ^{-1}}c_{1\sigma ^{-1}},\dots ,f_{p\sigma ^{-1}}c_{p\sigma ^{-1}}).\) We will concentrate our attention on the semidirect product

$$\begin{aligned} G_{q,p}=V\rtimes H. \end{aligned}$$

Proposition 7

\(m(G_{q,p})=3.\)

Proof

Since V is a complemented chief factor of \(G_{q,p},\) by [4, Theorem 2], we have \(m(G_{q,p})=1+m(H)=1+m(H/{{\mathrm{Frat}}}(H))=1+m(C_p\times C_p)=3.\) \(\square \)

Proposition 8

\(i(G_{q,p})=2p.\)

Proof

Let \(B\cong C^p\) be the base subgroup of H and consider \(K=V\rtimes B\cong (\mathbb F\rtimes C)^p\). A composition series of K has length 2p, and all its factors are indeed complemented chief factors, so \(m(K)=2p\). Now by definition \(i(G_{q,p})=\max \{m(X)\mid X\le G_{q,p}\}\ge m(K)= 2p\). On the other hand, \(m(G_{q,p})=3\) and, if \(X<G_{q,p},\) then |X| is a proper divisor of \(|G|=(pq)^p p\) and the composition length of X is at most 2p, so \(m(X)\le 2p\). Therefore, \(i(G_{q,p}) \le 2p\), and consequently \(i(G_{q,p})=m(K)=2p.\) \(\square \)

Lemma 9

\({{\mathrm{MaxDim}}}(G_{q,p})\ge p.\)

Proof

Let \(e_1=(1,0,\dots ,0), e_2=(0,1,\dots ,0),\dots , e_p=(0,0,\dots ,1) \in V\) and let \(h_1=(c,1,\dots ,1), h_2=(1,c,\dots ,1),\dots , h_p=(1,1,\dots ,c) \in C^p\le H.\) For any \(1\le i,j \le p,\) we have

$$\begin{aligned} h_i^{e_j}=h_i \text{ if } \quad i\ne j, \quad h_i^{e_i}=\left( \left( 1/c-1\right) e_i \right) h_i. \end{aligned}$$

But then, for each \(i\in \{1,\dots ,p\},\) we have

$$\begin{aligned} h_i \in \cap _{j\ne i}H^{e_j}, \quad h_i \notin H^{e_i}, \end{aligned}$$

hence \(H^{e_1},\dots ,H^{e_p}\) is a family of maximal subgroups of \(G_{q,p}\) in general position. \(\square \)

In order to compute the precise value of \({{\mathrm{MaxDim}}}(G_{q,p}),\) the following lemma is useful.

Lemma 10

Let \(v_1=(x_1,\dots ,x_p)\) and \(v_2=(y_1,\dots ,y_p)\) be two different elements of \(V=\mathbb F^p\) and let \(\Delta (v_1,v_2)=\{i\in \{1,\dots ,p\}\mid x_i=y_i\}.\) Then

  • if \(|\Delta (v_1,v_2)|=0,\) then \(|H^{v_1}\cap H^{v_2}| \le p;\)

  • if \(|\Delta (v_1,v_2)|=u\ne 0,\) then \(|H^{v_1}\cap H^{v_2}|= p^u.\)

Proof

Clearly \(|H^{v_1}\cap H^{v_2}|=|H\cap H^{v_2-v_1}|=|C_H(v_2-v_1)|.\) If \(\Delta (v_1,v_2)= \varnothing ,\) then \(C_H(v_2-v_1) \cap C^p=\{1\}\), hence \(|C_H(v_2-v_1)| \le p.\) If \(|\Delta (v_1,v_2)|=u\ne 0,\) then

$$\begin{aligned} C_H \left( v_2-v_1 \right) =\{\left( c_1,\dots ,c_p \right) \in C^p \mid c_i=1 \text{ if } \quad i \notin \Delta \left( v_1,v_2 \right) \}\cong C^u \end{aligned}$$

has order \(p^u.\) \(\square \)

Proposition 11

If \(p\ne 2,\) then \({{\mathrm{MaxDim}}}(G_{q,p})=p.\)

Proof

By Lemma 9 it suffices to prove that \({{\mathrm{MaxDim}}}(G_{q,p})\le p.\) Assume that \({\mathcal {M}}\) is a family of maximal subgroups of \(G=G_{q,p}\) in general position and let \(t=|{\mathcal {M}}|.\) Let \(M\in {\mathcal {M}}\). One of the following two possibilities occurs:

  1. (1)

    M is a complement of V in G :  hence \(M=H^v\) for some \(v\in V\).

  2. (2)

    M contains V :  hence \(M=V\rtimes X\) for some maximal subgroup X of H.

If \(M_1\) and \(M_2\) are two different maximal subgroups of type (2), then \(M_1\cap M_2= V\rtimes {{\mathrm{Frat}}}(X)\) is contained in any other maximal subgroup of type (2). Hence, \({\mathcal {M}}\) cannot contain more then two maximal subgroups of type (2). Now we prove the following claim: if \({\mathcal {M}}\) contains at least three different complements of V in G, then \(t\le p.\) In order to prove this claim, assume, by contradiction that \(t>p\). This implies in particular that in the intersection X of any two subgroups of \({\mathcal {M}}\), the subgroup lattice \({\mathcal {L}} (X)\) must contain a chain of length at least \(p-1\).

Assume that \(H^{v_1}, H^{v_2}, H^{v_3}\) are different maximal subgroups in \({\mathcal {M}}\). It is not restrictive to assume \(v_1=(0,\dots ,0)\). Let \(v_2=(x_1,\dots ,x_p)\) and \(v_3=(y_1,\dots ,y_p)\). For \(i\in \{2,3\}\), it must \(|H\cap H^{v_i}|\ge p^{p-1}\), hence, by Lemma 10, \(|\Delta (0,v_2)|=|\Delta (0,v_3)|=p-1,\) i.e. there exists \(i_1 \ne i_2\) such that \(x_{i_1}\ne 0, x_j=0\) if \(j\ne i_1, y_{i_2}\ne 0, y_j=0\) if \(j\ne i_2\). But then \(|\Delta (v_2,v_3)|=p-2\), hence \(|H^{v_2}\cap H^{v_3}|=p^{p-2}\), a contradiction. We have so proved that either \(t\le p\) or \({\mathcal {M}}\) contains at most two maximal subgroups of type (1) and at most two maximal subgroups of type (2), and consequently \(t\le 4\). It remains to exclude the possibility that \(t=4\) and \(p=3\). By the previous considerations it is not restrictive to assume \({\mathcal {M}}=\{H, H^v, V\rtimes X_1, V\rtimes X_2\}\) where \(X_1\) and \(X_2\) are maximal subgroups of H and \(|\Delta (0,v)|=2\). In particular we would have \(H\cap H^v\le C^3\): this excludes \(C^3\in \{X_1,X_2\}\) but then \(X_1\cap C^3=X_2\cap C^3={{\mathrm{Frat}}}\, H=\{(c_1,c_2,c_3)\mid c_1c_2c_3=1\},\) hence \(H\cap H^v \cap X_1= H\cap H^v\cap X_2,\) a contradiction. \(\square \)

Proposition 12

\({{\mathrm{MaxDim}}}(G_{q,2})=3.\)

Proof

By Lemma 7, \({{\mathrm{MaxDim}}}(G_{q,2})\ge m(G_{q,2})=3\). Assume now, by contradiction, that \(M_1,M_2,M_3,M_4\) are a family of maximal subgroups of \(G_{q,2}\). As in the proof of the previous proposition, at least two of these maximal subgroups, say \(M_1\) and \(M_2\), are complements of V in \(G_{q,2}\). But then, by Lemma 10, \(|M_1\cap M_2| \le 2\), hence \(M_1\cap M_2\cap M_3=1\), a contradiction. \(\square \)