= Solution
A <Sylow subgroup> of a <finite group> $G$ is a subgroup whose order is the largest <prime power>[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 $n_p\equiv1\pmod p$.
For the last congruence, let one Sylow $p$-subgroup $P$ act by conjugation on the set of all Sylow $p$-subgroups. Every <orbit of a group action>[orbit] other than a fixed point has size divisible by $p$. If $Q$ is fixed, then $P,Q\leq N_G(Q)$. 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 $n_p\equiv1\pmod p$.
Now let $H<A_n$ have <index of a subgroup>[index] $m$ with $1<m<n$. The <coset action> gives a homomorphism
$$
A_n\longrightarrow S_m.
$$
Its <kernel of a group homomorphism>[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 satisfies
$$
n_5\mid18,\qquad n_5\equiv1\pmod5,
$$
and 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> $S_6\to\{\pm1\}$ must be trivial. It would therefore be a subgroup of $A_6$ of index $360/90=4$, contradicting the result just proved. Thus
$$
\boxed{\text{no group of order }90\text{ is simple}.}
$$
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