# 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
**Theorem 17**. : _The number of subgroups of \(\operatorname{Sym}(n)\) is at most_  
\[24^{(n^{2}-1)/6}(n!)^{2}2^{17n}.\]  
Now we can prove Theorem 15. Proof of Theorem 15.: We follow the proof in Proof of Theorem I in [21]. Assume that \(G\) has abelian socle and \(n=p^{m}\), we obtain at most  
\[v_{a}(m,p)=p^{(87.7724+4\epsilon_{1}(m,p))m^{2}}=n^{(87.7724+4\epsilon_{1}(m,p ))m}\]  
Suppose now that \(G\) has a non-abelian socle \(F=L^{r}\), where \(L\) is a simple group. By Lemma 14, \(F\) can be chosen in at most \(h(n)\) ways in \(\operatorname{Sym}(n)\) up to conjugacy. The element \(\tilde{g}\) can be chosen in at most \(n^{2}r!\) ways. The total number of subgroups of \(\operatorname{Sym}(r)\) is at most \(24^{(r^{2}-1)/6+17r}(r!)^{2}\). Given \(|\tilde{O}S|\), there are at most \(n^{12r}\) choices for \(\tilde{S}\). It follows that thenumber of choices for \(\tilde{S}\) is at most \(n^{12r}\). Let \(u=\exp n\) and \(n=s^{u}\). Since \(r\leq u\), we obtain that \(G\) can be chosen in at most  
\[v_{n}(u,s) =h(n)(ut!)n^{2}24^{(u^{2}-1)/6+17u}(ut!)^{2}n^{12u}\] \[\alpha=(3\log 48+\log 24)/(3\log 9)=2.24399105059531\ldots.\]0  
ways. In order to obtain a bound for the number of conjugacy classes of primitive groups of degree \(n\), it will be enough to find a bound on \(v(u,s)=v_{a}(u,s)+v_{n}(u,s)\). Since we must obtain a bound for \(v(u,s)\) of the form \(n^{cu}=s^{cu^{2}}\), it will be enough to maximise \(\left(\log v(u,s)\right)/(u^{2}\log s)\). We can check, with the help of a computer algebra system, such as Maxima [18], that \(\left(\log v_{a}(u,s)\right)/(u^{2}\log s)\) is bounded by \(1714.94\) (the bound is attained if \(s=54\) and \(u=2\)), while the bound for \(\left(\log v_{n}(u,s)\right)/(u^{2}\log s)\) is bounded by \(47.569\) (the bound is attained for \(u=5\), \(s=1\)). Therefore, since \(1+n^{-1\,667.371u}\leq n^{0.01u}\) for \(n\geq 2\) and \(u\geq 1\), we obtain that  
\[\alpha=(3\log 48+\log 24)/(3\log 9)=2.24399105059531\ldots.\]1 \[\alpha=(3\log 48+\log 24)/(3\log 9)=2.24399105059531\ldots.\]2  
Consequently Theorem 1.1 holds with \(c=1\,714.95\). Finally, we can prove Proposition 5. Proof of Proposition 5.: To obtain the constant \(c_{1}\) of Proposition 4, in which we need a bound of the type \(c_{1}^{n}=c_{1}^{su}\), we must find a bound for \(\left(\ln v(u,s)\right)/s^{u}\), that will correspond to \(\ln c_{1}\). With the help of Maxima [18], we show that \(v_{a}(s,u)\leq c_{a}^{su}\), where \(\log c_{a}=439.662\), with the maximum attained in \((u,s)=(5,2)\), and \(v_{n}(s,u)\leq c_{n}^{su}\), where \(\log c_{n}=22.0903\), with the maximum attained in \((u,s)=(1,5)\). Since \(c_{n}/c_{a}=2^{-417.5717}\), we have that  
\[\alpha=(3\log 48+\log 24)/(3\log 9)=2.24399105059531\ldots.\]3  
where \(c_{0}=439.663\). However, we see that for \(s^{u}\geq 4096\), the value of \(\log\delta_{a}=42.019\), corresponding to \((u,s)=(12,2)\), satisfies that \(v_{a}(u,s)\leq c_{a}^{su}\), and \(\log\delta_{n}=1.402\), also corresponding to \((u,s)=(12,2)\), satisfies that \(v_{n}(u,s)\leq c_{n}^{su}\). As above, for \(s^{u}\geq 4096\), we have that  
\[\alpha=(3\log 48+\log 24)/(3\log 9)=2.24399105059531\ldots.\]4  
where \(c_{1}=42.02\). This bound also holds for all values of \(n\leq 4095\) as we can check with Magma (see [19]). Our next step will be to estimate the constant associated to the number of conjugacy classes of transitive subgroups of the symmetric group \(\operatorname{Sym}(n)\) of degree \(n\). In the next results we will follow the notation and the arguments of [7]. Hence, it will be convenient for the reader to have that paper at hand. In order to avoid confusion between the constants in [7] and constants, we will use capital letters to refer to the constants of [7] and reserve the lowercase letters for our constants when they are different.