= Solution
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 action>s to obtain
$$
G\longrightarrow S_{n/p}\times S_{n/q}
\longrightarrow S_{n/p+n/q}.
$$
The 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 $p\geq2$, $q\geq3$, and $n\geq6$,
$$
\frac np+\frac nq
\leq\frac{5n}{6}
\leq n-1.
$$
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>.
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