Smallest $m$ such that $D_n$ is a subgroup of $S_m$

Question

I was wondering what was, given any $n$ the smallest $m$ such that $D_n$ is a subgroup of $S_m$.

For cycle groups $C_n$ I believe there is a relation of the form $m = \min \sum_{k=1}^{K}p_k$ with :

• $n = \prod_{k=1}^{K} p_k$
• $\forall (k,k') \in \left [ \left | 1, K \right | \right ], p_k\wedge p_{k'}=1$ where $\wedge$ denote greatest common divisor between $p_k$ and $p_{k'}$

As an example $C_{12}$ can be found in $S_7$ with the permutation of order 12 : $(1234)(567)$

So first of all is the relation between $m$ and $n$ correct for $C_n$ ? And second what would be this relation for $D_n$?

Thanks !

Answer 1

Your description is right for cyclic groups, and the minimum is obtained by taking $p_1, \dots, p_K$ to be the prime power factors of $n$. See Claim 1 here: https://mathoverflow.net/a/409831/20598.

The answer is the same for $D_n$ as for $C_n$ (for $n>2$). Indeed, take a subgroup $G \le S_m$ such that $G \cong C_n$. Then $G$ acts as a cycle on each of its orbits. Let $\tau \in S_m$ be an involution that preserves each orbit of $G$ and reverses the cycle induced by $G$. Then $\langle G, \tau\rangle \cong D_n$.

Exceptions: If $n=1$ then $\tau$ is trivial (not an involution). If $n=2$ then $\tau \in G$.

Answer 2

By definition $d(G)$ is the smallest positive integer $n$, such that $S_n$ has a subgroup isomorphic to $G$. This is called the minimal permutation degree of $G$. It is known for cyclic groups and also for semidirect products of cyclic groups, hence in particular for dihedral groups $D_n$ of order $2n$.

Theorem: The minimal permutation degree of a dihedral group $D_n$ of order $2n$ is $$ d(D_n)=\sum_{i=1}^r p_i^{e_i}, $$ where $$ n=\prod_{i=1}^r p_i^{e_i} $$ is the decomposition of $n$ in powers of distinct primes $p_1,\ldots ,p_r$. As remarked in the other answer, we have the equality $$ d(D_n)=d(C_n). $$ For a detailed proof see, for example, this thesis, section $4.3$.