# Bounds on the Number of Maximal Subgroups of Finite Groups: Applications
## 3 Upper Bounds for the Number of Maximal Subgroups of a Given Index in a Group
### Bounds on the Number of Inequivalent Irreducible \(G\)-Modules
Hence, the number of completely reducible soluble subgroups of \(\mathrm{GL}_{m}(p)\) is bounded by \(p^{3.36m^{2}}p^{(1+\varepsilon_{1}(m,p))m^{2}}=p^{(4.366+\varepsilon_{1}(m,p ))m^{2}}\). In Lemma 2.3, in [21], it is shown that the number of subgroups \(X\) of \(\mathrm{GL}_{m}(p)\) such that \(X=\mathrm{F}^{*}(G)\) for some irreducible subgroup \(G\) of \(\mathrm{GL}_{m}(p)\) is at most \(p^{(15+\mathrm{o}(1))m^{2}}\), where \(\mathrm{F}^{*}(G)\) denotes the generalised Fitting subgroup of \(G\). We specify the term \(\mathrm{o}(1)\) in this expression. **Lemma 12**. : _The number of subgroups \(X\) of \(\mathrm{GL}_{m}(p)\), such that \(X=\mathrm{F}^{*}(G)\) or some irreducible subgroup \(G\) of \(\mathrm{GL}_{m}(p)\) is at most \(p^{(13.098+3\varepsilon_{1}(m,p))m^{2}}\), where \(\varepsilon_{1}(m,p)\) is defined in Lemma 11._  
Proof. : We can argue as in the proof of Lemma 2.3 in [21], where it is shown that \(\mathrm{F}^{*}(G)\) can be generated by three soluble completely reducible subgroups of \(\mathrm{GL}_{m}(p)\). If we replace the term 2.25 by 2.244 in the proof of Lemma 2.7 in [21], we obtain the following result, where the 8 is replaced by 7.78. **Lemma 13**. : _With the same notation of Lemma 2.7 in [21], if \(P\) is a \(p\)-subgroup of \(N/H\), then \(|P|\leq p^{7.784}\)._  
The following result specifies the bound \(p^{(94+\mathrm{o}(1))m^{2}}\) that appears in Theorem 2.8 in [21]. **Theorem 16**. : _The number of irreducible subgroups \(G\) of \(\operatorname{GL}_{m}(p)\) is at most \(p^{(87.7724+4\epsilon_{1}(m,p))m^{2}}\), where \(\varepsilon_{1}(m,p)\) is defined in Lemma 11._  
Proof. : We can follow the proof of Theorem 2.8 in [21] by replacing the terms \(5+\operatorname{o}(1)\) by \(4.366+\varepsilon_{1}(m,p)\), \(15+\operatorname{o}(1)\) by \(13.098+3\varepsilon_{1}(m,p)\), and 8 by 7.78 according to Lemmas 11-13, respectively. We obtain that the number of choices for \(G\) is at most  
\[p^{(4.366+\varepsilon_{1}(m,p))m^{2}+m^{2}+(13.098+3\varepsilon_{1}(m,p))m^{2 }+m^{2}+7.78\times 8.78m^{2}}=p^{(87.7724+4\varepsilon_{1}(m,p))m^{2}}.\qed\]  
The following result specifies the bound \(n^{3+\operatorname{o}(1)}\) of Lemma 3.2 in [21]. **Lemma 14**. : _The number of conjugacy classes of subgroups \(F\) of \(\operatorname{Sym}(n)\), such that \(F=\operatorname{Soc}(G)\) for some primitive subgroup \(G\) of \(\operatorname{Sym}(n)\) is at most_  
\[h(n)=2(\operatorname{ex}n)^{2}+2\operatorname{ex}n+n^{2}g_{1}(n)\log^{3}n+ \frac{(\operatorname{ex}n-1)}{8}ng_{1}(\sqrt{n})\log^{3}n=n^{3+\operatorname{ o}(1)},\]  
_where \(g_{1}\) is defined in Lemma 3._  
Proof. : We follow the proof of Lemma 3.2 in [21]. If \(F\) acts regularly on \(\Omega\), we have at most \(2\operatorname{ex}n\) choices for \(F\) up to conjugacy. If \(F\) is not regular and has diagonal action, then there are at most \(2(\operatorname{ex}n)^{2}\) choices for \(F\) up to conjugacy. Suppose that \(F\) is not regular on \(\Omega\) and has a wreath product action. Then there are at most \(\operatorname{ex}n\) choices for \(n_{0}\). By Lemma 5, \(\operatorname{Sym}(n)\) has at most \(g_{1}(n_{0})\log^{3}n_{0}\) almost simple subgroups up to isomorphism and so \(G_{0}\) can be chosen in at most \(g_{1}(n_{0})\log^{3}n_{0}\) as an abstract group. Once \(G_{0}\) is fixed up to isomorphism, \(G_{0}\) has at most \(n_{0}^{2}\) core-free maximal subgroups of index \(n_{0}\) by Theorem 1.3 in [2]. Hence there are \(n_{0}^{2}81(n_{0})\log^{3}n_{0}\) possibilities for the conjugacy class of \(G_{0}\) in \(\operatorname{Sym}(\Omega_{0})\). We can distinguish the case \(n_{0}=n\) and the rest of the cases, corresponding to \(n_{0}\leq\sqrt{n}\). Consequently, the number is bounded by  
\[2(\operatorname{ex}n)^{2}+2\operatorname{ex}n+n^{2}g_{1}(n)\log^{3}n+\frac{( \operatorname{ex}n-1)}{8}ng_{1}(\sqrt{n})\log^{3}n.\qed\]  
The following result specifies the bound \(24^{(1/6+\operatorname{o}(1))n^{2}}\) on the number of subgroups of \(\operatorname{Sym}(n)\) given in Corollary 3.3 in [22]. It is based on the proof of this last result.