# 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
Let \(b_{1}\) be the number of \(i\in\{1,\ldots r\}\), such that \(G/C_{i}\) has at least \(2\operatorname{l}(S)^{k}\) maximal subgroups of index \(\operatorname{l}(S)^{k}\) and let \(b_{2}\) be the number of \(i\in\{1,\ldots,r\}\), such that \(G/C_{i}\) has at least \(\operatorname{l}(S)^{k}\) maximal subgroups of index \(\operatorname{l}(S)^{k}\). It follows that \(G\) has at least \(2b_{1}\operatorname{l}(S)^{k}\) maximal subgroups of index \(\operatorname{l}(S)\) and at least \(b_{2}v_{S}^{k}\) maximal subgroups of index \(v_{S}^{k}\). Suppose that \(2b_{1}\geq b_{2}\). Then, \(b_{1}\geq\lceil r/3\rceil\). Hence, \(G\) has at least \(2\lceil r/3\rceil\operatorname{l}(S)^{k}\geq\lceil 2r/3\rceil\operatorname{l}(S)^{k}\) maximal subgroups \(M\) of index \(\operatorname{l}(S)^{k}\) with \(\operatorname{Soc}(G/M_{G})\cong A\). Suppose, now, that \(2b_{1}<b_{2}\). Then \(b_{2}\geq\lceil 2r/3\rceil\). It follows that \(G\) has at least \(\lceil 2r/3\rceil v_{S}^{k}\) maximal subgroups \(M\) of index \(v_{S}^{k}\) with \(\operatorname{Soc}(G/M_{G})\cong A\). Suppose, now, that \(S\in\mathfrak{Y}\). By Lemma 2, there exists an integer \(v_{S}\leq\operatorname{l}(S)^{2}\) such that \(G/C_{i}\) has at least \(v_{S}^{k}\) maximal subgroups of index \(v_{S}^{k}\) for \(1\leq i\leq r\). For \(x=v_{S}^{k}\leq\operatorname{l}(A)^{2}\), we have that \(\mathfrak{m}_{x,A}(G)\geq xr\). Finally, suppose that \(S\) is non-abelian and \(S\
otin\mathfrak{X}\cap\mathfrak{Y}\). By Lemma 2, \(G/C_{i}\) has at least \(\operatorname{l}(S)^{k}\) maximal subgroups of index \(\operatorname{l}(S)^{k}\) for \(1\leq i\leq r\). Then, for \(x=\operatorname{l}(A)=\operatorname{l}(S)^{k}\), we obtain that \(\mathfrak{m}_{x,A}(G)\geq xr\). Assume that \(n=p^{k}\) is a power of the prime \(p\) and that there exist normal subgroups \(K\leq H\) 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 \(1\leq i\leq r\) and let \(M_{i}\) be a maximal subgroup supplementing \(A_{i}\). Consider the primitive group \(P_{i}=G/\operatorname{Core}_{G}(M_{i})\). 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 \(\tilde{M}_{i}\
eq 1\). Then, \(\operatorname{N}_{P_{i}}(\tilde{M}_{i})\
eq P_{i}\) and so \(\operatorname{N}_{G}(M_{i})\
eq G\). By the maximality of \(M_{i}\), \(\operatorname{N}_{G}(M_{i})=M_{i}\). It follows that the conjugacy class of \(M_{i}\) in \(G\) has exactly \(|G:M_{i}|=|A|=n\) 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 \(n\
otin\mathbb{P}\), \(a=0\) 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 \(\mathfrak{m}_{p,A}(G)\geq pb+1+p+\cdots+p^{a-1}=p(r-a)+1+p+\cdots+p^{a-1}\). If \(a=0\), then \(\mathfrak{m}_{p,A}(G)\geq pr\). If \(a\in\{1,2\}\), then \(\mathfrak{m}_{p,A}(G)\geq pr-p+1\). If \(a\geq 3\), then \(1+p+p^{2}+\cdots+p^{a-1}\geq ap\) because \(1+p^{2}\geq 2p\), and so \(\mathfrak{m}_{p,A}(G)\geq pr\). The result follows.