# 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  
We can go further if we specify the values of the constants of this linear combination giving rise to the values of the constant \(c\). The existence of these constants follows from counting arguments in which some terms are known to be \(\mathrm{o}(1)\), but we have not found any explicit value for them. We begin by estimating the constant \(c_{1}\) of the following result. **Proposition 4** (Proposition 2.4 in [7]): _There exists an absolute constant \(c_{1}\) such that, for each \(n\), the group \(\mathrm{Sym}(n)\) has at most \(c_{1}^{n}\) conjugacy classes of primitive subgroups._  
We will prove the following result. **Proposition 5**: _The value of the constant \(c_{1}\) of Proposition 4 can be taken to be \(c_{1}=2^{42.02}\approx 4.46\times 10^{12}\)._  
The proof of Proposition 4 depends on the following result. **Theorem 14** (Theorem I in [21]): _The number of conjugacy classes of primitive subgroups of the symmetric group \(\mathrm{Sym}(n)\) is at most \(n^{c_{1}(n)}\), where \(c\) is some absolute constant and \(\mu(n)\) denotes the maximal exponent of a prime in the prime factorisation of the natural number \(n\). Consequently, \(\mathrm{Sym}(n)\) has at most \(n^{c_{2}n}\) conjugacy classes of primitive subgroups._  
We will obtain a value for this constant \(c\). **Theorem 15**: _The constant \(c\) of Theorem 14 can be taken to be \(c=1\,714.95\)._  
In our arguments, we will replace the term \(\mu(n)\) by \(\mathrm{ex}\,n\), the largest number \(r\), such that \(n=m^{r}\) for a natural number \(m\), that is, the greatest common divisor of the exponents in the decomposition of \(n\) as a product of prime powers. If \(n=p^{m}\) is a power of a prime, then \(\mu(n)=\mathrm{ex}\,n\), while, in general, \(\mathrm{ex}\,n\leq\mu(n)\). We use the following result of Palry. **Lemma 8** (see Lemma 3.4 (ii) in [22]): _The number of conjugacy classes of maximal, irreducible, soluble subgroups of \(\mathrm{GL}_{n}(p)\) (\(p\) prime) is at most_\[n^{20\log^{3}n+5}.\]  
The following two results, of [PERSON] and [PERSON], and [PERSON], respectively, concern completely reducible subgroups of \(\mathrm{GL}_{m}(p)\). **Lemma 9** (Theorem in [23]). : _If \(G\) is a completely reducible subgroup of \(\mathrm{GL}_{m}(p)\), with \(p\) a prime, then \(G\) can be generated by at most \(\lfloor(3/2)m\rfloor\) elements._  
**Lemma 10** (Theorem 3.1 in [24]). : _A soluble completely reducible subgroup of \(\mathrm{GL}_{m}(p)\) has order at most \(24^{-1/3}p^{\alpha m}\), where_  
\[\alpha=(3\log 48+\log 24)/(3\log 9)=2.24399105059531\ldots.\]  
In Lemma 1.5 in [21], it is shown that the number of completely reducible soluble subgroups of \(\mathrm{GL}_{m}(p)\) is at most \(p^{(5+\alpha(1))m^{2}}\). We specify the term \(\mathrm{o}(1)\). In the following results, we will follow the notation and the proofs of [21] and we will indicate only the differences. Hence, these are best followed with [21] at hand. **Lemma 11**. : _The number of completely reducible soluble subgroups of the linear group \(\mathrm{GL}_{m}(p)\) is at most \(p^{(4.366+\varepsilon_{1}(m,p))m^{2}}\), where_  
\[\varepsilon_{1}(m,p)=\frac{m-1+20m\log^{4}m+5m\log m}{m^{2}\log p}.\]  
Proof. : We follow the proof of Lemma 1.5 in [21]. It is shown at the end of the first paragraph there that \(M\) can be chosen in at most  
\[2^{m-1}m^{20m\log^{3}m+5m}p^{m^{2}}=p^{1+\varepsilon_{1}(m,p)}\]  
ways. By Lemma 10, once fixed a maximal soluble completely reducible subgroup \(M\) of \(\mathrm{GL}_{m}(p)\), by Lemma 10 we find that \(|M|\leq 24^{-1/3}p^{\alpha m}\). By Lemma 9, every completely reducible subgroup \(G\) of \(M\) can be generated by at most \(\lfloor(3/2)m\rfloor\) elements. Therefore, the number of completely reducible subgroups \(G\) of \(M\) is at most \((24^{-1/3}p^{2.244m})^{1.5m}\leq p^{3.366m^{2}}\).