# 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 for the Number of Socles of Primitive Groups of Type 2
By Lemma 4, we know that the number of subgroups of Out \(T\) is bounded by \(\log^{3}1(T)\leq\log^{3}n\). Therefore, \(\operatorname{Sym}(n)\) contains at most \(g(n)\log^{3}n\leq g_{1}(n)\log^{3}(n)\leq g_{2}(n)\log^{3}(n)\) almost simple subgroups. The results follow. Let us denote by \(\operatorname{ex}n\) the largest natural number \(r\), such that there exists a natural \(c\), such that \(c^{r}=n\). It is clear that \(\operatorname{ex}n\leq\log n\), and the equality holds if, and only if, \(n\) is a power of \(2\). Moreover, \(\operatorname{ex}n\) is the greatest common divisor of the exponents of the different primes appearing in the decomposition of \(n\) as a product of prime powers. Note that if \(n=n_{s}^{s}\), then \(s\) divides the exponents of the different primes appearing in the prime power decomposition of \(n\), in particular, \(s\mid\operatorname{ex}n\). The following result follows from Theorem 1.1.52 and Proposition 1.1.53 in [6] and a result of [PERSON] and [PERSON] [15] whose proof can be found in Theorem 1.1.35 in [6]. **Proposition 2**. : _Let \(P\) be a primitive group with a minimal normal subgroup \(A=S_{1}\times\cdots\times S_{r}\), where for \(1\leq i\leq r\), \(S_{i}\) is isomorphic to a given non-abelian simple group \(S\). Let \(X=\operatorname{N}_{P}(S_{1})/C_{P}(S_{1})\) and let \(U\) be a maximal subgroup of \(P\) of a given index \(n=|P:U|\)._  
_Exactly one of the following three conditions holds:_  
**Condition 1**. : \(U\cap A=(U\cap S_{1})\times\cdots\times(U\cap S_{r})\
eq 1\)_. In this case, there exists a maximal subgroup \(\bar{U}\) of \(X\), such that \(\bar{U}\cap\operatorname{Inn}X\cong U\cap S_{i}\), \(|U\cap A|=|\bar{U}\cap\operatorname{Inn}X|^{r}\), and \(n=|X:\bar{U}|^{r}=m^{r}\) for some \(m\);_  
**Condition 2**. : \(|U\cap A|=|S|^{a}\)_, where \(0<a<r\) is such that there exists an integer \(b\) with \(ab=r\). In this case, \(n=|S|^{r}/|S|^{a}=|S|^{a(b-1)}=x^{a(b-1)}\);_  
**Condition 3**. : \(|U\cap A|=1\) _and, in this case, \(n=|S|^{r}=x^{\prime}\)._  
_There exists a natural bijection between the conjugacy classes of maximal subgroups of \(X\) of trivial core and the conjugacy classes in \(P\) of core-free maximal subgroups satisfying the above Condition 1._  
**Proposition 3**. : _The number \(s(n)\) of isomorphism types of minimal normal subgroups of primitive groups of type 2 with a core-free maximal subgroup of index \(n\) satisfies the inequalities_  
\[s(n) \leq (\operatorname{ex}n-1)g(\sqrt{n})+g(n)+2(\operatorname{ex}n)^{2 }+2\operatorname{ex}n\] \[g_{2}(n)=4.8869n+1088.\]0  
Proof. : Let \(P\) be a primitive group with a unique minimal normal subgroup isomorphic to \(A=S^{m}\), where \(S\) is a non-abelian simple group, and with a core-free maximal subgroup \(U\) of index \(n\). Let \(r=\operatorname{ex}n\) and consider \(s\mid r\) (the values \(r=s=1\) are valid in this context). We first look for the possibilities of \(A\) corresponding to primitive groups \(P\) of type 2 with a core-free maximal subgroup satisfying the Condition 1 of Proposition 2. We must consider simple groups \(S\) having an almost simple group with a core-free maximal subgroup of index \(n^{1/s}\). By Lemma 3, the number of the possibilities for \(A\) is at most \(g(n^{1/s})\leq g(n)\). The Condition 1 can only hold if \(P\) is almost simple or \(\operatorname{ex}n\
eq 1\). In the former case, by Lemma 3, the number of possibilities for \(A\) is at most \(g(n)\). In the latter case, we obtain for each \(s\
eq 1\) a number of possibilities for \(A\) bounded by \(g(\sqrt{n})\). Hence, the total number of possibilities for \(A\) of this type is at most \((\operatorname{ex}(n)-1)g(\sqrt{n})+g(n)\). Now, we look for the possible socles of \(P\) satisfying the Condition 2 of Proposition 2. We must look for simple groups \(S\) with order \(n^{1/s}\) and, for each divisor \(a\) of \(s\), we can have asocie \(A=S^{ac}\) where \(c=s/a+1\).