# Bounds on the Number of Maximal Subgroups of Finite Groups: Applications
## 2 Lower Bounds for the Number of Maximal Subgroups of a Given Index in a Group
_If, in addition,_ \(S\
ot\cong\mathrm{Alt}(6)\)_,_ \(S\) _is not of the form_ \(\mathrm{PSL}_{m}(q)\) _with_ \(q=p^{f}\)_,_ \(m\geq 3\)_, and_ \(p\in\{2,3,5,7\}\)_, or_ \(m=2\) _and_ \(q=3^{f}\)_,_ \(S\) _is not of the form_ \(\mathrm{PSU}_{m}(q)\) _with_ \(q=p^{f}\) _and_ \(p\in\{2,3,5,7\}\)_, and_ \(S\
ot\cong\mathrm{O}^{+}_{8}(q)\) _with_ \(q=p^{f}\) _and_ \(p\in\{3,5,7,11,13\}\)_, then_ \(|\mathrm{Out}\,S|\leq\log\mathrm{l}(S)\)_._  
**Lemma 2**. : _Suppose that \(G\) is a primitive group of type 2 with socle isomorphic to \(S^{k}\), where \(S\) is a non-abelian simple group._  
1. _If_ \(S\in\mathfrak{X}\)_, then_ \(G\) _has at least_ \(2\mathrm{l}(S)^{k}\) _maximal subgroups of index_ \(\mathrm{l}(S)^{k}\) _or at least_ \(v_{S}^{k}\) _maximal subgroups of index_ \(v_{S}^{k}\)_, where_ \(v_{S}\) _is defined in Theorem_ 11_;_
2. _If_ \(S\in\mathfrak{Y}\)_, then_ \(G\) _has at least_ \(v_{S}^{k}\geq\mathrm{l}(S)^{k}\) _maximal subgroups of index_ \(v_{S}^{k}\)_, where_ \(v_{S}\) _is defined in Theorem_ 11_;_
3. _If_ \(S\
otin\mathfrak{X}\cup\mathfrak{Y}\)_, then_ \(G\) _has at least_ \(\mathrm{l}(S)^{k}\) _maximal subgroups of index_ \(\mathrm{l}(S)^{k}\)_._  
Proof. : Suppose that \(G\) is not an almost simple group and that \(\mathrm{Soc}(G)=S_{1}\times\cdots\times S_{r}\), where \(\{S_{1},\ldots,S_{r}\}\) is the set of all conjugate subgroups of a simple normal subgroup \(S_{1}\) of \(\mathrm{Soc}(G)\), write \(N=\mathrm{N}_{G}(S_{1})\) and \(K=S_{2}\times\cdots\times S_{r}\). By a result of [PERSON] and [PERSON] ([15], see alsoTheorem 1.1.35 in [6]), there exists a bijection between, on the one hand, the conjugacy classes in \(G\) of supplements \(U\) of \(\operatorname{Soc}(G)\) in \(G\), such that \(U\cap\operatorname{Soc}(G)=(U\cap S_{1})\times\cdots\times(U\times S_{r})\) and, on the other hand, the conjugacy classes in \(N/K\) of supplements \(L/K\) of \(\operatorname{Soc}(G)/K\) 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 \(\pi_{1}(U\cap\operatorname{Soc}(G))\) of \(U\cap\operatorname{Soc}(G)\) 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 \(S\
otin\mathfrak{X}\cup\mathfrak{Y}\). Then, we can consider a core-free maximal subgroup \(H\) of \(Z\) of index \(\operatorname{l}(S)\) and construct \(U=G\cap(H\wr P_{k})\). Then, the conjugacy class of \(U\) in \(G\) contains at least \(|G:U|=\operatorname{l}(S)^{k}\) elements. Suppose that \(S\in\mathfrak{X}\). Then, \(Z\) contains two non-conjugate core-free maximal subgroups \(H_{1}\) and \(H_{2}\) of index \(\operatorname{l}(S)\), and so if \(U_{i}=G\cap(H_{i}\wr P_{k})\), \(i\in\{1,2\}\), then \(|G:U_{1}|=|G:U_{2}|=1(S)^{k}\), 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 [PERSON] and [PERSON] ([15], see also Theorem 1.1.35 in [6]), \(U_{1}\) and \(U_{2}\) are in different conjugacy classes in \(G\).