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Suppose $G$ is topologically generated by $d$ elements. An open subgroup $H$ of index $n$ gives a continuous transitive coset actionA continuous homomorphism $G\to S_n$ is determined by the images of the $d$ topological generators, so there are at most $|S_n{|}^d$ such homomorphisms. Each has only finitely many point stabilizers. Therefore $G$ has only finitely many open subgroups of index $n$.

A Sylow subgroup of a finite group $G$ is a subgroup whose order is the largest power of a prime $p$ dividing $|G|$. The Sylow theorems say that such subgroups exist, that every $p$-subgroup lies in one, that all Sylow $p$-subgroups are conjugate, and that their number $n_p$ divides $|G|/p^a$ and obeys $n_p\equiv 1(\mathrm{mod}p)$.

For the last congruence, let one Sylow $p$-subgroup $P$ act by conjugation on the set of all Sylow $p$-subgroups. Every orbit other than a fixed point has size divisible by $p$. If $Q$ is fixed, then $P,Q\le N_G(Q)$. They are Sylow subgroups of this normalizer, so they are conjugate within $N_G(Q)$; because every element of $N_G(Q)$ fixes $Q$ under conjugation, this forces $P=Q$. Thus there is exactly one fixed point and $n_p\equiv 1(\mathrm{mod}p)$.

Now let $H<A_n$ have index $m$ with $1<m<n$. The coset action gives a homomorphismIts kernel is a normal subgroup of the simple group $A_n$. It cannot be all of $A_n$, since the action is transitive and nontrivial, so it is trivial. This would embed $A_n$ into $S_m$, contrary to $|A_n|=n!/2>m!=|S_m|$. Hence no such subgroup exists.

Suppose finally that a group $G$ of order $90$ were simple. The Sylow count satisfiesand simplicity excludes $n_5=1$, so $n_5=6$. Conjugation on these six subgroups gives a nontrivial homomorphism $G\to S_6$, which simplicity makes injective. Its image lies in $A_6$, because the composite with the sign homomorphism $S_6\to \{\pm 1\}$ must be trivial. It would therefore be a subgroup of $A_6$ of index $360/90=4$, contradicting the result just proved. Thus

Because $n$ is not a prime power, distinct primes $p<q$ divide $n$. By Cauchy's theorem, $G$ has subgroups $H,K$ of orders $p,q$. Combine the two coset actions to obtainThe kernel of the first action is contained in $H$, and that of the second is contained in $K$. Their intersection is trivial because $H\cap K=\{1\}$, so the combined action is faithful.

Since $p\ge 2$, $q\ge 3$, and $n\ge 6$,Adding fixed points embeds this symmetric group into $S_{n-1}$. Thus every group of non-prime-power order $n$ is a subgroup of $S_{n-1}$, as summarized by symmetric-group embedding at one less than the group order.

Let $k=|G/H|$. The coset action gives a nontrivial homomorphism $G\to S_k$ whose kernel is normal. Simplicity makes it injective. The sign map is trivial on the nonabelian simple image, so $G$ embeds in $A_k$. For $k\le 4$, $A_k$ is solvable, as are its subgroups, whereas a nonabelian simple group is not. Hence $k\ge 5$.

For a subgroup $H\le G$, the permutation character of the coset action on $G/H$ is the induced trivial characterand