= Solution
{c}
Acting separately on disjoint sets of four and eight symbols embeds the <direct product of groups> $S_4\times S_8$ into $S_{12}$. The action is injective since fixing all twelve symbols forces both factors to be identities.
For the lower bound, $S_4$ contains a cycle of order four and $S_8$ contains disjoint cycles of lengths five and three, whose product has order fifteen. The corresponding element of $S_4\times S_8$ has <order of a group element> $\operatorname{lcm}(4,15)=60$. Every isomorphic subgroup of $S_n$ must therefore contain a <permutation> of order sixty. We can bound the <minimum number of symbols for a permutation of prescribed order> directly here.
In the <disjoint cycle decomposition> of such a <permutation>, all cycle lengths divide $60$, and their <least common multiple> must contain the factors $4$, $3$, and $5$. If these factors occur in three distinct cycles, those cycles use at least $4+3+5=12$ symbols. If two occur in the same cycle, that cycle has length at least their product: combining $4,3$ costs at least $12$, combining $4,5$ costs at least $20$, and combining $3,5$ costs at least $15$. Each possibility already requires at least twelve symbols. Thus $n\geq12$, and
$$
\boxed{n_{\min}=12.}
$$
Back to article page
