Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics I / 5 / d

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 $\mathrm{lcm}(4,15)=60$. 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 $n\ge 12$, and