FINITE GROUPS WITH INDEPENDENT GENERATING SETS OF ONLY TWO SIZES

Abstract

A generating set S for a group G is independent if the subgroup generated by $S\setminus \{s\}$ is properly contained in G for all $s \in S.$ We describe the structure of finite groups G such that there are precisely two numbers appearing as the cardinalities of independent generating sets for G.

1 Introduction

The minimal number of generators of a finite group G is denoted by $d(G).$ A generating set S for a group G is independent (sometimes called irredundant) if

$$ \begin{align*}\langle S\setminus \{s\}\rangle < G \quad\mbox{for all } s \in S.\end{align*} $$

Let $m(G)$ denote the maximal size of an independent generating set for $G.$

The finite groups with $m(G)=d(G)$ are classified by Apisa and Klopsch.

Theorem 1.1 (Apisa–Klopsch, [1, Theorem 1.6]).

If $d(G)=m(G)$ , then G is soluble. Moreover, either

• • $G/\operatorname {\mathrm {Frat}}(G)$ is an elementary abelian p-group for some prime p; or

• • $G/\operatorname {\mathrm {Frat}}(G)=PQ,$ where P is an elementary abelian p-group and Q is a nontrivial cyclic q-group for distinct primes p and $q,$ such that Q acts by conjugation faithfully on P and P (viewed as a module for Q) is a direct sum of $m(G)-1$ isomorphic copies of one simple Q-module.

In view of this result, Apisa and Klopsch suggest a natural ‘classification problem’: given a nonnegative integer c, characterise all finite groups G which satisfy $m(G) - d(G) \leq c$ . The particular case $c=1$ has been recently highlighted by Glasby (see [7, Problem 2.3]).

A nice result in universal algebra, due to Tarski and known as the Tarski irredundant basis theorem (see for example [3, Theorem 4.4]), implies that, for every positive integer k with $d(G)\leq k\leq m(G), G$ contains an independent generating set of cardinality  $k.$ So the condition $m(G) - d(G)= 1$ is equivalent to the fact that there are only two possible cardinalities for an independent generating set of $G.$

Let G be a finite group. We recall that the socle of G, denoted $\operatorname {\mathrm {soc}}(G)$ , is the subgroup generated by the minimal normal subgroups of G; moreover, G is said to be monolithic primitive if G has a unique minimal normal subgroup and the Frattini subgroup $\operatorname {\mathrm {Frat}}(G)$ of G is the identity.

In this paper, we prove the following two main results.

Theorem 1.2. Let G be a finite group with $\operatorname {\mathrm {Frat}}(G)=1$ and $m(G)=d(G)+1.$ If G is not soluble, then $d(G)=2$ , G is a monolithic primitive group and $G/\operatorname {\mathrm {soc}}(G)$ is cyclic of prime power order.

It was proved by Whiston and Saxl [15] that $m(\operatorname {\mathrm {PSL}}(2,p))=3$ for any prime p with p not congruent to $\pm 1$ modulo 8 or 10. In particular, as $d(S)=2$ for every nonabelian simple group, we deduce that there are infinitely many nonabelian simple groups G with $m(G)=d(G)+1$ . We also give examples of nonsimple groups G having ${m(G)=d(G)+1}$ in Section 4.

Theorem 1.3. Let G be a finite group with $\operatorname {\mathrm {Frat}}(G)=1$ and $m(G)=d(G)+1.$ If G is soluble, then one of the following occurs:

1. (1) $G\cong V\rtimes P$ , where P is a finite noncyclic p-group and V is an irreducible P-module, which is not a p-group; in this case, $d(G)=d(P)$ ;

2. (2) $G\cong V^t \rtimes H,$ where V is a faithful irreducible H-module, $m(H)=2$ and either $t=1$ or H is abelian; in this case, $d(G)=t+1$ ;

3. (3) there exist two normal subgroups $N_1$ , $N_2$ such that $1 \lneq N_1 \leq N_2, N_1$ is an abelian minimal normal subgroup of $G, N_2/N_1\leq \operatorname {\mathrm {Frat}}(G/N_1)$ and $G/N_2\cong V^t\rtimes H$ , where V is an irreducible H-module and H is a nontrivial cyclic group of prime power order; in this case, $d(G)=t+1.$

In Section 4, we give examples of finite soluble groups G with $m(G)=d(G)+1$ for each of the three possibilities arising in Theorem 1.3.

2 Preliminary results

Let L be a monolithic primitive group and let A be its unique minimal normal subgroup. For each positive integer k, let $L^k$ be the k-fold direct product of L. The crown-based power of L of size k is the subgroup $L_k$ of $L^k$ defined by

$$ \begin{align*}L_k:=\{(l_1, \ldots , l_k) \in L^k \mid l_1 \equiv \cdots \equiv l_k \ {\mbox{mod}}\, A \}.\end{align*} $$

In [4], it is proved that for every finite group G, there exists a monolithic group L and a homomorphic image $L_k$ of G such that

1. (1) $ d(L/\operatorname {\mathrm {soc}} L) < d(G) $ ; and

2. (2) $d(L_k) =d(G).$

A group $L_{k}$ with this property is called a generating crown-based power for G.

In [4], it is explained how $d(L_{k})$ can be explicitly computed in terms of k and the structure of L. A key ingredient (when one wants to determine $d(G)$ from the behaviour of the crown-based power homomorphic images of G) is to evaluate, for each monolithic group L, the maximal k such that $L_{k}$ is a homomorphic image of G. This integer k arises from an equivalence relation among the chief factors of G. In what follows, we give some details.

Given groups G and A, we say that A is a G-group if G acts on A via automorphisms. In addition, A is irreducible if G does not stabilise any nontrivial proper subgroups of A. Two G-groups A and B are G-isomorphic if there exists a group isomorphism $\phi : A\to B$ such that $\phi (g(a))=g(\phi (a))$ for all $a\in A$ and $g\in G.$ Following [8], we say that two irreducible G-groups A and B are G-equivalent, denoted $A \sim _G B$ , if there is an isomorphism $\Phi : A\rtimes G \rightarrow B\rtimes G$ which restricts to a G-isomorphism $\phi \colon A \to B$ and induces the identity $G \cong AG/A \to BG/B \cong G$ , in other words, such that the following diagram commutes:

Observe that two G-isomorphic G-groups are G-equivalent, and the converse holds if A and B are abelian.

Let $A=X/Y$ be a chief factor of G. A complement U of A in G is a subgroup of G such that

$$ \begin{align*} UX=G \quad\text{and}\quad U \cap X=Y. \end{align*} $$

We say that $A=X/Y$ is a Frattini chief factor if $X/Y$ is contained in the Frattini subgroup of $G/Y$ ; this is equivalent to saying that A is abelian and there is no complement to A in G. The number $\delta _G(A)$ of non-Frattini chief factors that are G-equivalent to A, in any chief series of G, does not depend on the particular choice of such a series.

Now, we denote by $L_G(A)$ the monolithic primitive group associated to A, that is,

$$ \begin{align*}L_G(A):= \begin{cases} A\rtimes (G/\mathbf{C}_G(A)) & \text{ if }A\text{ is abelian}, \\ G/\mathbf{C}_G(A)& \text{ otherwise}. \end{cases} \end{align*} $$

If A is a non-Frattini chief factor of G, then $L_G(A)$ is a homomorphic image of G. More precisely, there exists a normal subgroup N such that $G/N \cong L_G(A)$ and $\operatorname {\mathrm {soc}} (G/N) \sim _G~A$ . We identify $\operatorname {\mathrm {soc}}( L_G(A))$ with A, as G-groups.

Consider now all the normal subgroups N of G with the property that ${G/N \cong L_G(A)}$ and $\operatorname {\mathrm {soc}} (G/N) \sim _G A$ . The intersection $R_G(A)$ of all these subgroups has the property that $G/R_G(A)$ is isomorphic to the crown-based power $(L_G(A))_{\delta _G(A)}$ . The socle $I_G(A)/R_G(A)$ of $G/R_G(A)$ is called the A-crown of G and it is a direct product of $\delta _G(A)$ minimal normal subgroups G-equivalent to A.

Note that if L is monolithic primitive and $L_k$ is a homomorphic image of G for some $k\geq 1$ , then $L \cong L_G(A)$ for some non-Frattini chief factor A of G and $k \leq \delta _G(A)$ . Furthermore, if $(L_G(A))_k$ is a generating crown-based power, then so is $(L_G(A))_{\delta _G(A)}$ ; in this case, we say that A is a generating chief factor for G.

For an irreducible G-module M, set

$$ \begin{align*} r_G(M)&:=\dim_{\operatorname{\mathrm{End}}_G(M)}M,\\ s_G(M)&:=\dim_{\operatorname{\mathrm{End}}_G(M)} H^1(G,M), \\ t_G(M)&:=\dim_{\operatorname{\mathrm{End}}_G(M)} H^1(G/\mathbf{C}_G(M),M). \end{align*} $$

It can be seen that

$$ \begin{align*}s_G(M)=t_G(M)+\delta_G(M)\end{align*} $$

(see for example [10, 1.2]). Now, define

$$ \begin{align*}h_{G}(M):= \begin{cases} \delta_G(M)&\text{if }M\text{ is a trivial } G\textrm{-module}, \\ \displaystyle \bigg\lfloor\frac{s_G(M)-1}{r_G(M)}\bigg\rfloor+2=\bigg\lfloor\frac{\delta_G(M)+t_G(M)-1}{r_G(M)}\bigg\rfloor+2& \text{otherwise}. \end{cases} \end{align*} $$

By [2, Theorem A], $t_G(M)<r_G(M)$ for any irreducible G-module M, and therefore

(2.1) $$ \begin{align}h_G(M)\leq \delta_G(M)+1. \end{align} $$

The importance of $h_G(M)$ is clarified by the following proposition.

Proposition 2.1 [6, Proposition 2.1].

If there exists an abelian generating chief factor A of G, then $ d(G)=h_G(A)$ .

When G admits a nonabelian generating chief factor A, a relation between $\delta _G(A)$ and $d(G)$ is provided by the following result.

Proposition 2.2. If $d(G)\geq 3$ and there exists a nonabelian generating chief factor A of G, then

$$ \begin{align*}\delta_G(A)>\frac{|A|^{d(G)-1}}{2|\mathbf{C}_{\operatorname{\mathrm{Aut}} A}(L_G(A)/A)|}\geq \frac{|A|^{d(G)-2}}{2\log_2|A|}. \end{align*} $$

Proof. Suppose that $d(G)\ge 3$ and let A be a nonabelian generating chief factor of G.

For a finite group X, let $\phi _X(m)$ denote the number of ordered m-tuples $(x_1,\ldots ,x_m)$ of elements of X generating X. Define

$$ \begin{align*} L&:=L_G(A),\\ \gamma&:=|\mathbf{C}_{\operatorname{\mathrm{Aut}} A}(L/A)|,\\ \delta&:=\delta_G(A),\\ d&:=d(G). \end{align*} $$

In [4], it is proved that if $m\geq d(L),$ then

(2.2) $$ \begin{align}d(L_k)\leq m \quad {\text {if and only if}} \quad k\leq \frac{\phi_{L/A}(m)}{\phi_L(m)\gamma}. \end{align} $$

By the main result in [13], $d(L)=\max (2,d(L/A))$ . Since A is a generating chief factor, from the definition, we have $d(L/A) < d(L_{\delta _G(A)})=d(G)$ . As $2 < d(G),$ it follows ${d(L)<d(G).}$ Now, by applying (2.2) with $k=\delta _G(A)$ and $m=d(G)-1,$ we deduce that

(2.3) $$ \begin{align}\delta_G(A)>\frac{\phi_{L/A}(d(G)-1)}{\phi_L(d(G)-1)\gamma}.\end{align} $$

By [6, Corollary 1.2],

(2.4) $$ \begin{align}\frac{\phi_{L/A}(d(G)-1)}{\phi_L(d(G)-1)}\geq \frac{|A|^{d(G)-1}}{2}.\end{align} $$

Moreover, $A\cong S^n$ , where n is a positive integer and S is a nonabelian simple group. In the proof of Lemma 1 in [5], it is shown that

$$ \begin{align*}\gamma \leq n|S|^{n-1}|{\operatorname{\mathrm{Aut}}}(S)|.\end{align*} $$

Now, [9] shows that $|{\operatorname {\mathrm {Out}}}(S)| \leq \log _2(|S|)$ and hence

(2.5) $$ \begin{align}\gamma\leq n|S|^{n}\log_2(|S|)\leq |S|^{n}\log_2(|S|^n)=|A|\log_2(|A|).\end{align} $$

From (2.3), (2.4) and (2.5), we obtain

$$ \begin{align*} \delta_G(A)> \frac{\phi_{L/A}(d(G)-1)}{\phi_L(d(G)-1)\gamma}\geq \frac{|A|^{d(G)-1}}{2|A|\log_2|A|}=\frac{|A|^{d(G)-2}}{2\log_2|A|}.\\[-42pt] \end{align*} $$

Recall that $m(G)$ is the largest cardinality of an independent generating set of G.

Theorem 2.3 [14, Theorem 1.3].

Let G be a finite group. Then $m(G)\geq a+b,$ where a and b are, respectively, the number of non-Frattini and nonabelian factors in a chief series of G. Moreover, if G is soluble, then $m(G)=a.$

Corollary 2.4. Assume that G is a finite group with a unique minimal normal subgroup A. If A is nonabelian, then $m(G)\geq 3.$

Proof. Suppose first that G is simple. Let l be an element of G of order 2. Since $G=\langle l^x\mid x\in G\rangle $ , the set $\{l ^x\mid x \in G\}$ contains a minimal generating set of $G.$ Since G cannot be generated by two involutions, this minimal generating set has cardinality at least three. Thus, $m(G)\ge 3$ .

Suppose next that G is not simple. Let a and b be the number of non-Frattini and nonabelian factors in a chief series of G. As G is not simple, there exists a maximal normal subgroup N of G containing A and we have a chief series $1\unlhd N_1\unlhd \cdots \unlhd N_{t-1}\unlhd N_t=G$ with $N_1=A$ and $N_{t-1}=N.$ Then, $a\geq 2$ , $b\geq 1$ and $m(G)\geq a+b\geq 3$ by Theorem 2.3.

3 Proof of the main results

Let G be a finite group, let $d:=d(G)$ and let $m:=m(G)$ . Suppose that $m=d+1.$ Let A be a generating chief factor of G and let $\delta :=\delta _G(A)$ , $L:=L_G(A).$

3.1 A is nonabelian

First, suppose that $\delta \geq 2.$ By Theorem 2.3, $m\geq 2\delta $ and therefore $d\geq 2\delta -1\geq 3.$ By Proposition 2.2,

$$ \begin{align*}\delta> \frac{|A|^{d-2}}{2\log_2|A|}\geq \frac{ |A|^{2\delta-3}}{2\log_2|A|}\geq \frac{60^{2\delta-3}}{2\log_260},\end{align*} $$

but this is never true.

Suppose now that $\delta =1$ . In this case, by the main theorem in [13], $d=d(L)=\max (2,d(L/A))=2$ and therefore $m=3.$ Since L is an epimorphic image of G, we must have $m(L)\leq 3$ . However, $m(L)\geq 3$ by Corollary 2.4. Hence, $m(L)=m=3$ and therefore it follows from [11, Lemma 11] that $G/\operatorname {\mathrm {Frat}}(G)\cong L.$ Finally, by Theorem 2.3, $m(L)=3$ implies $m(L/A)\leq 1,$ and this is possible only if $L/A$ is a cyclic p-group. This concludes the proof of Theorem 1.2.

3.2 A is abelian

It follows from Proposition 2.1 and (2.1) that

$$ \begin{align*}\delta-1 \leq m-1 = d = h_G(A)\leq \delta+1.\end{align*} $$

If $d=\delta -1,$ then $m=\delta $ and this is possible if $G/\operatorname {\mathrm {Frat}}(G)\cong A^\delta .$ However, in this case, A would be a trivial G-module and therefore $d=h_G(A)=\delta =m,$ which is a contradiction.

Now suppose that $d=\delta .$ By Theorem 2.3, G is soluble and contains only one non-Frattini chief factor which is not G-isomorphic to $A.$ If A is noncentral in G, then $G/\operatorname {\mathrm {Frat}}(G)\cong L_\delta $ and $L/A$ is a cyclic p-group. However, this implies $r_G(A)\,{=}\,1,\ t_G(A)\,{=}\,0$ and $d=h_G(A)=\delta +1,$ which is a contradiction. If A is central, then $G/\operatorname {\mathrm {Frat}}(G)\cong V\rtimes P$ , where P is a finite p-group, V is an irreducible P-module and $d(P)=d.$ In particular, we obtain item (1) in Theorem 1.3.

Finally assume $d=\delta +1.$ Notice that in this case, $L=A\rtimes H,$ where A is a faithful, nontrivial, irreducible H-module, and

$$ \begin{align*}m(H)\leq m-\delta=\delta+2-\delta=2.\end{align*} $$

In particular, by Corollary 2.4, H is soluble.

If $m(H)=2,$ then $G/\operatorname {\mathrm {Frat}}(G)\cong L_\delta .$ In particular, we obtain item (2) in Theorem 1.3.

If $m(H)=1,$ then there exist two normal subgroups $N_1$ and $N_2$ of G such that $1 \lneq N_1 \leq N_2, G/N_2\cong L_\delta , N_2/N_1\leq \operatorname {\mathrm {Frat}}(G/N_1)$ and $N_1/\operatorname {\mathrm {Frat}}(G)$ is an abelian minimal normal subgroup of $G/\operatorname {\mathrm {Frat}}(G).$ As $m(H)=1$ , H is cyclic of prime power order. In particular, we obtain item (3) in Theorem 1.3.

4 Examples for Theorems 1.2 and 1.3

4.1 Monolithic groups: examples for Theorem 1.2

Let G be monolithic primitive with nonabelian socle $N= S_1\times \cdots \times S_n$ , with $S\cong S_i$ for each $1\leq i \leq n.$ The number $\mu (G)=m(G)-m(G/N)$ has been investigated in [12]. The group G acts by conjugation on the set $\{S_1,\ldots ,S_n\}$ of the simple components of $N.$ This produces a group homomorphism $G\to \operatorname {\mathrm {Sym}}(n)$ and the image K of G under this homomorphism is a transitive subgroup of $\operatorname {\mathrm {Sym}}(n).$ Moreover, the subgroup X of $\operatorname {\mathrm {Aut}} S$ induced by the conjugation action of $\textbf {N}_G(S_1)$ on the first factor $S_1$ is an almost simple group with socle $S.$

By [12, Proposition 4], $\mu (G)\geq \mu (X)=m(X)-m(X/S).$ Assume $m(G)=3.$ Observe that by Theorems 1.1 and 1.2, $G/N$ is cyclic of prime power order. If $X=S$ , then

$$ \begin{align*} 3=m(G)=m(G/N)+\mu(G)& \geq m(G/N)+\mu(X)= m(G/N)+m(S)\\ &\geq m(G/N)+3. \end{align*} $$

This implies that $G/N=1$ and $G=S$ is a simple group. If $X\neq S$ , then $G\neq N$ and

$$ \begin{align*}3=m(G)\geq m(G/N)+\mu(G)\geq 1+\mu(X).\end{align*} $$

Moreover, $X/S$ is a nontrivial cyclic group of prime power order, so

$$ \begin{align*}m(X)=m(X/S)+\mu(X)\leq 1+\mu(X)\leq 1+2=3.\end{align*} $$

By Corollary 2.4, $m(X)=3.$

The groups

$$ \begin{align*}\mathrm{P}\Sigma\mathrm{L}_2(9), M_{10},\mathrm{Aut}(\mathrm{PSL}_2(7))\end{align*} $$

are currently the only known examples (to the best knowledge of the authors) of almost simple groups X with $X\neq \operatorname {\mathrm {soc}}(X)$ and $m(X)=3.$ We believe that there are other such examples, but our current computer codes are not efficient enough to carry out a thorough investigation.

Let $S:=\mathrm {PSL}_2(7)$ and $H:=\mathrm {Aut}(\mathrm {PSL}_2(7)),$ or let $S:=\mathrm {PSL}_2(9)$ and $H\in \{\mathrm {P}\Sigma \mathrm {L}_2(9), M_{10}\}$ . Consider the wreath product $W:=H \wr \operatorname {\mathrm {Sym}}(n)$ . Any element $w\in W$ can be written as $w=\pi (a_1,\ldots ,a_n),$ with $\pi \in \operatorname {\mathrm {Sym}}(n)$ and $a_i\in H$ for $1\leq i\leq n$ . In particular, $N=\operatorname {\mathrm {soc}}(W) =S_1\times \cdots \times S_n=\{(s_1,\ldots ,s_n)\mid s_i \in S\}$ .

Proposition 4.1. Let G be the subgroup of W generated by $N=\operatorname {\mathrm {soc}}(W)$ and $\gamma =\sigma (a,1,\ldots ,1),$ where $\sigma =(1\,2\cdots n)\in \operatorname {\mathrm {Sym}}(n)$ and $a\in H\setminus S.$ If $n=2^t$ for some positive integer $t,$ then $m(G)=3.$

In particular, this gives infinitely many examples of nonsimple, nonsoluble groups G with $m(G)=d(G)+1$ in Theorem 1.2.

Proof. Suppose that $n=2^t$ for some positive integer t. Let $r:=m(G)$ ; we aim to prove that $r=3$ .

Let $\{g_1,\ldots ,g_r\}$ be an independent generating set of $G.$ Observe that

$$ \begin{align*}\gamma^n=(a,\ldots,a)\in G\setminus N\end{align*} $$

and hence $G/N$ is cyclic of order $2^{t+1}.$ Therefore, relabelling the elements of the independent generating set if necessary, we may assume $G=\langle g_1, N\rangle $ . Hence, ${g_1=\sigma (as_1,s_2,\ldots ,s_n)}$ with $s_1,\ldots ,s_n\in S.$ Moreover, for $2\leq i\leq r,$ there exists $u_i\in \mathbb Z$ such that $g_ig_1^{u_i}\in N$ . Observe that $\{g_1,g_2g_1^{u_2},\ldots ,g_rg_1^{u_r}\}$ is still an independent generating set having cardinality r.

Let

$$ \begin{align*}m=(s_2\cdots s_n, s_3\cdots s_n,\ldots,s_{n-1}s_n,s_n,1) \in N.\end{align*} $$

Then, $Y=\{g_1^m,(g_2g_1^{u_2})^m,\ldots ,(g_rg_1^{u_r})^m\}$ is another independent generating set for G having cardinality r. We have

$$ \begin{align*}y_1:=g_1^m=\sigma(b,1,\ldots,1),\end{align*} $$

with $b=as_1\cdots s_n \in \operatorname {\mathrm {Aut}} S\setminus S,$ and for $2\leq i\leq r,$ there exist $s_{i1},\ldots s_{in}\in S$ such that

$$ \begin{align*}y_i:=(g_ig_1^{u_i})^m=(s_{i1},\ldots,s_{in}).\end{align*} $$

Let $Z:=\{b, s_{ij} \mid 2\leq i\leq r, 1\leq j\leq n\}$ and $T=\langle Z\rangle .$ Since $G=\langle y_1,\ldots ,y_t\rangle \leq T \wr \langle \sigma \rangle ,$ we must have $\operatorname {\mathrm {Aut}}(S)=T.$ However, $m(\operatorname {\mathrm {Aut}}(S))=3,$ so $\operatorname {\mathrm {Aut}}(S)=\langle b, s_{iu}, s_{jv}\rangle $ for suitable $2\leq i,j\leq r$ and $2\leq u, v\leq n.$

Let $H:=\langle y_1, y_i, y_j\rangle $ and, for $1\leq k\leq n,$ consider the projection $\pi _k: N\to S$ sending $(s_1,\ldots ,s_n)$ to $s_k.$ Notice that $\pi _1(y_1^n)=b, \pi _1((y_i)^{y_1^{1-u}})=s_{iu}, \pi _1((y_j)^{y_1^{1-v}})=s_{jv}.$ In particular, $\pi _1(H\cap N)=S$ and $H\cap N$ is a subdirect product of $N= S_1\times \cdots \times S_n$ .

Recall that a subgroup D of $N=S_1\times \cdots \times S_n$ is said to lie fully diagonally in N if each projection $\pi _i: D\to S_i$ is an isomorphism. To each pair $(\Phi ,\alpha )$ , where ${\Phi =\{B_1,\ldots ,B_c\}}$ is a partition of the set $\{1,\ldots ,n\}$ and $\alpha =(\alpha _1,\ldots ,\alpha _n)\in (\operatorname {\mathrm {Aut}} S)^n$ , we associate a direct product $\Delta (\Phi ,\alpha )=D_1\times \cdots \times D_c$ , where each factor $D_j=\{(x^{\alpha _{i_1}},\ldots , x^{\alpha _{i_d}})\mid x\in S\}$ is a full diagonal subgroup of the direct product $S_{i_1} \times \cdots \times S_{i_d}$ corresponding to the block $B_j=\{i_1,\ldots ,i_d\}$ in $\Phi .$

Since $H\cap N$ is a subdirect product of $N,$ we must have $H\cap N=\Delta (\Phi ,\alpha )$ for a suitable choice of the pair $(\Phi ,\alpha ).$ As $G=\langle H,N\rangle $ , the action by conjugation of H on $\{S_1,\ldots ,S_n\}$ is transitive and hence the partition $\{B_1,\ldots ,B_c\}$ corresponds to an imprimitive system for the permutation action of $\langle \sigma \rangle $ on $\{1,\ldots ,n\}.$ So there exist $c=2^\gamma $ and $d=2^\delta $ with $c\cdot d=n$ such that

$$ \begin{align*}B_i:=\{i,i+c,i+2c,\ldots,i+(d-1)c\} \quad\mbox{for } 1\leq i \leq c.\end{align*} $$

Notice that $y_1\in H$ normalises $\Delta (\Phi ,\alpha )$ . In particular, $y_1^c$ normalises $\Delta (\Phi ,\alpha ).$ However, $y_1^c$ normalises $L=S_1\times S_{1+c}\times \cdots \times S_{1+(d-1)c}$ and acts on L as $\pi \cdot l,$ where $\pi $ is the d-cycle $(1,1+c,\ldots ,1+(d-1)c)$ and $l=(b,1,\ldots ,1)\in L.$ In particular, $\pi \cdot l$ normalises the full diagonal subgroup $D_1$ of L. Therefore, setting $\phi _i=\alpha _{1+(i-1)c}$ , for every $s\in S$ , there exists $t\in T$ such that

$$ \begin{align*}(s^{\phi_db},s^{\phi_1},s^{\phi_2},\ldots,s^{\phi_{d-1}})= (t^{\phi_1},t^{\phi_2},t^{\phi_3 },\ldots,t^{\phi_{d}}).\end{align*} $$

It follows that

$$ \begin{align*}\begin{aligned} \phi_{d}b\phi_1^{-1}\phi_2&=\phi_1,\\ \phi_{d}b\phi_1^{-1}\phi_3&=\phi_{2},\\ \cdots\\ \phi_{d}b\phi_1^{-1}\phi_d&=\phi_{d-1}. \end{aligned}\end{align*} $$

In particular, $(\phi _1\phi _d^{-1})^d\equiv b^{d-1}$ modulo $S.$ If d is even, then $b\in \langle x^2\mid x\in \operatorname {\mathrm {Aut}}(S)\rangle =S,$ against our assumption. Thus, $d=1$ and hence $c=n$ . However, this implies that ${H\cap N=N}$ and consequently $H=G.$ Thus, $m(G)=r\leq 3$ . However, $m(G)\geq 3$ by Theorem 2.3. So we conclude that $m(G)=3.$

4.2 Soluble groups: examples for Theorem 1.3

We give three elementary examples, but with the same ideas, one can construct more complicated examples. Let $S_n$ be the symmetric group of degree n and let $C_n$ be the cyclic group of order n.

The group $G:=S_3 \times C_2^t = C_3 : C_2^{t+1}$ with $t\ge 1$ satisfies $d(G)=t+1$ and ${m(G)=t+2}$ . This gives examples of groups satisfying item (1) in Theorem 1.3.

The group $G:=S_4=K:S_3$ with K the Klein subgroup of $S_4$ and the group ${G:=(C_3^t : C_2) \times C_2}$ with $C_2$ acting on $C_3^t$ by inversion also satisfy $m(G)=d(G)+1$ . These two examples yield groups satisfying item (2) in Theorem 1.3 with $m(H)=2$ in the first case and with H abelian in the second case.

As above, let K be the Klein subgroup of $S_4$ and let $G:=K:(S_3\times C_2^{t-1})$ . This gives examples of groups satisfying item (3) in Theorem 1.3.