# 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
It follows that there are at least \(2\operatorname{l}(S)^{k}\) maximal subgroups of \(G\) of index \(\operatorname{l}(S)^{k}\) or at least \(v_{S}^{k}\) maximal subgroups of \(G\) of index \(\operatorname{v}_{S}^{k}\). 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. **Remark 1**. : _Consider the O'Nan simple group \(S\cong\operatorname{O}^{\prime}\!\operatorname{N}\), and let \(A=\operatorname{Aut}(S)\). According to [11], all the core-free maximal subgroups of \(A\) have index greater than its order. Let \(W=A\wr C_{2}\). Let \(D\) be the diagonal subgroup of the base group \(A\times A\) of \(W\), let \(H=DC_{2}\), and let \(G=(\operatorname{Soc}W)H\). In the primitive pair \((G,H)\) with simple diagonal action (see Definition 1.1.42 in [6]), we have that \(|G:H|=|A|\) is smaller than the index of any maximal subgroup of the form \(M\wr C_{2}\) (see Remarks 1.1.46 in [6]). Hence, the smallest index of a core-free maximal subgroup \(\operatorname{l}^{*}(G)\) of \(G\) is smaller than the index \(\operatorname{l}^{*}(A)^{2}\) corresponding to the product action with the core-free subgroup of smallest index \(\operatorname{l}(A)\) of \(A\). What we prove in Lemma 2 is that there are maximal subgroups of indices \(\operatorname{l}(S)^{k}\) and \(v_{S}^{k}\)._  
**Notation 3**. : _For a group \(G\) and a prime \(p\) we denote by \(\zeta_{p}(G)\) the number of central \(p\)-chi factors of \(G\) in a given chief series._  
**Notation 4**. : _Given a natural number \(n\geq 2\) and a chief factor \(A\) of a group \(G\), we denote by \(\mathfrak{m}_{n,A}(G)\) the number of maximal subgroups of \(G\) of index \(n\) for which the socle of the associated primitive group is isomorphic to \(A\)._  
Now, we are in a position to establish our lower bound. **Theorem 12**. : _Let \(G\) be a group and let \(A\) be a non-Frattini chief factor of \(G\) isomorphic to \(S^{k}\), with \(k\) a natural number and \(S\) a simple group._  
1. _If_ \(S\in\mathfrak{X}\)_, then_ \(\mathfrak{m}_{x,A}(G)\geq x\lceil(2/3)\mathfrak{r}_{A}(G)\rceil\) _for some_ \(x\in\{1(A),v_{S}^{k}\}\) _and_ \(v_{S}^{k}\leq\operatorname{l}(A)^{2}\)_;_
2. _If_ \(S\in\mathfrak{Y}\)_, then_ \(\mathfrak{m}_{x,A}(G)\geq x\mathfrak{r}_{A}(G)\) _for_ \(x=v_{S}^{k}\leq\operatorname{l}(A)^{2}\)_;_
3. _If_ \(S\) _is non-abelian and_ \(S\
otin\mathfrak{X}\cup\mathfrak{Y}\)_, then_ \(\mathfrak{m}_{n,A}(G)\geq n\mathfrak{r}_{A}(G)\) _for_ \(n=\operatorname{l}(A)\)_;_
4. _If_ \(A\cong\operatorname{C}_{p}^{k}\)_, with_ \(k\geq 2\)_, then_ \(\mathfrak{m}_{n,A}(G)\geq n\mathfrak{r}_{A}(G)\) _for_ \(n=\operatorname{l}(A)=p^{k}\)_;__._
5. _If_ \(A\cong C_{p}\)_, then_ \(\mathfrak{m}_{p,A}(G)\geq pr_{A}(G)-p+1\)_. Moreover, if_ \(\zeta_{p}(G)\
otin\{1,2\}\)_, then_ \(\mathfrak{m}_{p,A}(G)\geq pr_{A}(G)\)_._  
Proof of Theorem 12.: 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 \(r=\mathfrak{r}_{A}(G)\) chief factors isomorphic to a direct product \(A=A_{1}\times\cdots\times A_{r}\) of \(r\) copies of a simple group \(S\cong A_{i}\), \(1\leq i\leq r\), with \(\operatorname{l}(A)\leq n\). Consider the groups \(C_{i}=\operatorname{C}_{G}(A_{i})\), \(1\leq i\leq r\). 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 \(v_{S}\leq 1(S)^{2}\), such that \(G/C_{i}\) has at least \(2\operatorname{l}(S)^{k}\) maximal subgroups of index \(\operatorname{l}(S)^{k}\) or \(G/C_{i}\) has at least \(v_{S}\) maximal subgroups of index \(v_{S}\).