Bounds on the Number of Maximal Subgroups of Finite Groups: Applications

(Theorem B in [4]). The number $m_n\left(G\right)$ of maximal subgroups of index n of a d-generated group G satisfies the following bounds:

(Theorem 9.5 in [9]). Let G be a d-generated group. Then, where c and $c_7$ are two absolute constants.

(Corollary 9.3 in [9]). Let G be a group. Then, $m_x\left(G\right)\ge {rk}_n\left(G\right)/n^{c_7}$ for some $x\le n^{c_7}$.

(see Theorem 6.1 in [13]). Let S and T be non-isomorphic finite simple groups. If $\left|S^a\right|=\left|T^b\right|$ for some natural numbers a and b, then $a=b$ and S and T are either ${PSL}_3\left(4\right)$ or ${PSL}_4\left(2\right)$, or are $O_{2n+1}\left(q\right)$ and ${PSp}_{2n}\left(q\right)$ for some $n\ge 3$ and some odd q.

(Theorem A in [10]). Let S be a simple group and let R be an almost simple group, such that $Soc\left(R\right)\cong S$. We can assume that $S\le R\le A=Aut\left(S\right)$.

Suppose that G is not an almost simple group and that $Soc\left(G\right)=S_1\times \cdots \times S_r$, where $\{S_1,\dots ,S_r\}$ is the set of all conjugate subgroups of a simple normal subgroup $S_1$ of $Soc\left(G\right)$, write $N=N_G\left(S_1\right)$ and $K=S_2\times \cdots \times S_r$. By a result of Gross and Kovács ([15], see also Theorem 1.1.35 in [6]), there exists a bijection between, on the one hand, the conjugacy classes in G of supplements U of $Soc\left(G\right)$ in G, such that $U\cap Soc\left(G\right)=(U\cap S_1)\times \cdots \times (U\times S_r)$ and, on the other hand, the conjugacy classes in $N/K$ of supplements $L/K$ of $Soc\left(G\right)/K$ in $N/K$. This correspondence sends conjugacy classes of maximal subgroups of G to conjugacy classes of maximal subgroups of $N/K$ and conjugacy classes of complements of M in G to conjugacy classes of complements of $N/K$ in $M/K$. From this, it follows that every primitive group of type 2 has maximal subgroups U, such that the projection ${\pi }_1\left(U\cap Soc\left(G\right)\right)$ of $U\cap Soc\left(G\right)$ onto $S_1$ is a non-trivial proper subgroup of $S_1$. In this case, by Proposition 1.1.44 and Remarks 1.1.46 in [6], G can be regarded as a subgroup of $W\cong Z\wr P_k$, where Z is an almost simple group, $P_k$ is a transitive group of degree $k>1$, and $U=G\cap (H\wr P_k)$ for a maximal subgroup H of Z.

By Remarks 1.1.46 in [6], if H is a maximal subgroup of Z and $U=G\cap (H\wr P_k)$, then $|G:U|={|Z:H|}^k$ and, by Proposition 1.1.44 in [6], U is a core-free maximal subgroup of G. Moreover, all elements in its conjugacy class, consisting of $|G:U|$ elements, are also core-free maximal subgroups of the same index.

Suppose that $S\notin \mathfrak{X}\cup \mathfrak{Y}$. Then, we can consider a core-free maximal subgroup H of Z of index $l\left(S\right)$ and construct $U=G\cap (H\wr P_k)$. Then, the conjugacy class of U in G contains at least $|G:U|={l\left(S\right)}^k$ elements.

Suppose that $S\in \mathfrak{X}$. Then, Z contains two non-conjugate core-free maximal subgroups $H_1$ and $H_2$ of index $l\left(S\right)$, and so if $U_i=G\cap (H_i\wr P_k)$, $i\in \{1,2\}$, then $|G:U_1|=|G:U_2|={l\left(S\right)}^k$, or Z contains a core-free maximal subgroup $H_3$ of index $v_S$, and so if $U_3=G\cap (H_3\wr P_k)$, then $|G:U_3|=v_S^k$. By the mentioned result of Gross and Kovács ([15], see also Theorem 1.1.35 in [6]), $U_1$ and $U_2$ are in different conjugacy classes in G. It follows that there are at least $2{l\left(S\right)}^k$ maximal subgroups of G of index ${l\left(S\right)}^k$ or at least $v_S^k$ maximal subgroups of G of index $v_S^k$.

Finally, suppose that $S\in \mathfrak{Y}$. Then, Z contain a core-free maximal subgroup H of index $v_S$, and so if $U=G\cap (H\wr P_k)$, then $|G:U|=v_S^k$. In this case, the conjugacy class of U in G contains $v_S^k$ elements. □

Suppose, first, that A is non-abelian. As in the proof of Corollary 9.3 in [7], there exists a normal section $H/N$ of G that is the direct product of $r=r_A\left(G\right)$ chief factors isomorphic to a direct product $A=A_1\times \cdots \times A_r$ of r copies of a simple group $S\cong A_i$, $1\le i\le r$, with $l\left(A\right)\le n$. Consider the groups $C_i=C_G\left(A_i\right)$, $1\le i\le r$. Then, the $C_i$ are different normal subgroups of G and the quotients $G/C_i$ are groups with a unique minimal normal subgroup isomorphic to A.

Suppose that $S\in \mathfrak{X}$. According to Lemma 2, there exists an integer $v_S\le {l\left(S\right)}^2$, such that $G/C_i$ has at least $2{l\left(S\right)}^k$ maximal subgroups of index ${l\left(S\right)}^k$ or $G/C_i$ has at least $v_S$ maximal subgroups of index $v_S$. Let $b_1$ be the number of $i\in \{1,\dots r\}$, such that $G/C_i$ has at least $2{l\left(S\right)}^k$ maximal subgroups of index ${l\left(S\right)}^k$ and let $b_2$ be the number of $i\in \{1,\dots ,r\}$, such that $G/C_i$ has at least ${l\left(S\right)}^k$ maximal subgroups of index ${l\left(S\right)}^k$. It follows that G has at least $2b_1{l\left(S\right)}^k$ maximal subgroups of index $l\left(S\right)$ and at least $b_2{v}_S^k$ maximal subgroups of index $v_S^k$. Suppose that $2b_1\ge b_2$. Then, $b_1\ge \lceil r/3\rceil$. Hence, G has at least $2\lceil r/3\rceil {l\left(S\right)}^k\ge \lceil 2r/3\rceil {l\left(S\right)}^k$ maximal subgroups M of index ${l\left(S\right)}^k$ with $Soc(G/M_G)\cong A$. Suppose, now, that $2b_1<b_2$. Then $b_2\ge \lceil 2r/3\rceil$. It follows that G has at least $\lceil 2r/3\rceil v_S^k$ maximal subgroups M of index $v_S^k$ with $Soc(G/M_G)\cong A$.

Suppose, now, that $S\in \mathfrak{Y}$. By Lemma 2, there exists an integer $v_S\le {l\left(S\right)}^2$ such that $G/C_i$ has at least $v_S^k$ maximal subgroups of index $v_S^k$ for $1\le i\le r$. For $x=v_S^k\le {l\left(A\right)}^2$, we have that $m_{x,A}\left(G\right)\ge xr$.

Finally, suppose that S is non-abelian and $S\notin \mathfrak{X}\cap \mathfrak{Y}$. By Lemma 2, $G/C_i$ has at least ${l\left(S\right)}^k$ maximal subgroups of index ${l\left(S\right)}^k$ for $1\le i\le r$. Then, for $x=l\left(A\right)={l\left(S\right)}^k$, we obtain that $m_{x,A}\left(G\right)\ge xr$.

Assume that $n=p^k$ is a power of the prime p and that there exist normal subgroups $K\le H$ of G such that $H/K=A_1\times \cdots \times A_r$ is a direct product of r non-Frattini chief factors of G isomorphic to an elementary abelian group A of order n. Let $1\le i\le r$ and let $M_i$ be a maximal subgroup supplementing $A_i$. Consider the primitive group $P_i=G/{Core}_G\left(M_i\right)$. Then, $P_i$ has a minimal normal subgroup ${\tilde{A}}_i$, namely the precrown associated with $A_i$ and M, and a maximal subgroup ${\tilde{M}}_i$ of trivial core.

Assume that ${\tilde{M}}_i\ne 1$. Then, $N_{P_i}({\tilde{M}}_i)\ne P_i$ and so $N_G\left(M_i\right)\ne G$. By the maximality of $M_i$, $N_G\left(M_i\right)=M_i$. It follows that the conjugacy class of $M_i$ in G has exactly $|G:M_i|=\left|A\right|=n$ elements.

Assume that ${\tilde{M}}_i=1$. Then, $P_i\cong A_i$ and $A_i$ is a central chief factor, in particular, $n=p$ is a prime number.

Suppose that $H/K$ has a central G-chief factors and b non-central G chief factors in a given chief series. Note that, in the case that $n\notin \mathbb{P}$, $a=0$ and $b=r$. For each of the b non-central chief factors, we can obtain with the previous construction n maximal subgroups, they have different core since the core contains the product of all other chief factors. Hence, we obtain at least $nb$ maximal subgroups. Now, suppose that $a>0$ and consider the a central chief factors, in this case, $n=p$ is prime. By the main result of [8], G has a normal subgroup with elementary abelian quotient of order $p^a$. This group has $1+p+\cdots +p^{a-1}$ subgroups of index p, and all of them have in their core the non-central chief factors. It follows that $m_{p,A}\left(G\right)\ge pb+1+p+\cdots +p^{a-1}=p(r-a)+1+p+\cdots +p^{a-1}$. If $a=0$, then $m_{p,A}\left(G\right)\ge pr$. If $a\in \{1,2\}$, then $m_{p,A}\left(G\right)\ge pr-p+1$. If $a\ge 3$, then $1+p+p^2+\cdots +p^{a-1}\ge ap$ because $1+p^2\ge 2p$, and so $m_{p,A}\left(G\right)\ge pr$. The result follows. □

Let

By Proposition 1.2 in [2], $\mathcal{M}\left(G\right)-3.5\le \mathcal{V}\left(G\right)$. Let B be a non-Frattini chief factor of G, such that ${max}_{A}f(G,A)=f(G,B)$. Let $B\cong T^k$ with T a simple group.

Suppose that $B\in \mathfrak{X}$. Then, $m_x\left(G\right)\ge m_{x,B}\left(G\right)\ge x\lceil (2/3)r_B\left(G\right)\rceil$ for $x\in \{l\left(B\right),v_T^k\}$ and $v_T^k\le {l\left(B\right)}^2$. In this case,

Suppose that $B\in \mathfrak{Y}$. Then, $m_x\left(G\right)\ge m_{x,B}\left(G\right)\ge x{r}_B\left(G\right)$ for some $x\le {l\left(B\right)}^2$. Hence,

Suppose that $B\cong C_p$ for a prime p and that G has exactly one or two central chief factors isomorphic to B in a given chief series. Then, $m_p\left(G\right)\ge m_{p,B}\left(G\right)\ge p{r}_B\left(G\right)+1-p=p\left(r_B\left(G\right)-1+(1/p)\right)$ and so

Suppose that B does not satisfy any of the previous properties. Then, for $n=l\left(B\right)$, $m_n\left(G\right)\ge m_{n,B}\left(G\right)\ge n{r}_B\left(G\right)$ and so

Consequently,

The inequality $\mathcal{V}\left(G\right)\ge d$ holds trivially. □