Let $C_n$ be the cyclic group of order $n$ and $S_m$ be the group of permutations of $m$ elements. It's not difficult to see that every $C_n$ embeds into $S_n$, but some groups also embed on smaller symmetric groups, e.g. $C_6$ also embeds into $S_5$. By Langrange's theorem, we also know that, if $p$ is prime, the smallest $m$ such that $C_p$ embeds into $S_m$ is $p$ itself. But what about the general case? Is there any way to determine, for each $n$, the smallest $m$ such that $C_n$ embeds into $S_m$?
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See here for abelian groups. – Dietrich Burde Mar 20 '17 at 16:19
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5If $n = \prod_{i=1}^k p_i^{m_i}$ is the prime factorization of $n$, then the smallest $m$ is $\sum_{i=1}^{k}p_i^{m_i}$. – Derek Holt Mar 20 '17 at 16:22