The smallest symmetric group $S_m$ into which a given dihedral group $D_{2n}$ embeds

If $n > 2$, then $\mu_n$ is the sum of the prime powers appearing in the decomposition of $n$.

$D_{2n}$ is embeddable in $S_k$ if and only if the following things happen: (1) We can find an element $\sigma$ of order $n$ in $S_k$. (2) We can find an element $\tau$ of order $2$ in $S_k$ such that $\tau \sigma \tau^{-1} = \sigma^{-1}.$
(3) $\tau$ is not a power of $\sigma$. (But (3) follows from (2) if $n > 2$, since otherwise $\tau$ would commute with $\sigma$, so we would have $\sigma = \sigma^{-1}$.)

The order of an element $\sigma$ is the lcm of its cycle lengths. Say the order of $\sigma$ is $n$, and $\sigma$ moves as few elements as possible. Clearly, no prime factor can appear in the length of more than one cycle of $\sigma$. (If a factor $p^a$ appears in one cycle to a lower power than in another, simply divide the length of that cycle by $p^a$ to get a shorter permutation $\sigma$ without changing the lcm of its cycle lengths.) Also, each cycle must have prime power order, for if a cycle had length $ab$ with $a$ and $b$ relatively prime (and $> 1$), we could replace the cycle with two cycles of respective lengths $a$ and $b$ to obtain a shorter $\sigma$. Consequently the best we can do is have $\sigma$ with cycle lengths each of the prime power factors of $n$.

The only problem now is to show that $\tau$ can be chosen without increasing $k$. If one of the cycles in $\sigma$ is $(a_1, a_2, \dots, a_r)$, then let $\tau(a_1) = a_r$, $\tau(a_2) = a_{r-1}$, ..., $\tau(a_r) = a_1$. Define $\tau$ this way on each of the orbits of $\sigma$. Then $\tau$ has order $2$ and $\tau \sigma \tau^{-1} = \sigma^{-1}$.

There is also a following geometric way to explain the solution by David.

Let us recall that $D_{2n}$ is the group of symmetries of a regular $n$-gon. The idea is that one should consider the action of $D_{2n}$ on regular polygons with smaller number of vertices that are inscribed into the given regular n-gon.

To clarify the construction, let us first consider $n=6$. There are two regular 3-gons and three regular 2-gons whose vertices are the vertices of the given regular 6-gon. Every element of $D_{12}$ acts on the set of two regular 3-gons and acts on the set of three regular 2-gons, thus one has $D_{12} \rightarrow S_2 \times S_3$.

It is easy to see that this map is injective. Its composition with tautological embedding $S_2 \times S_3 \rightarrow S_{2+3}$ is the necessary map.

Now let us consider a general $n = \prod_{i=1}^s p_i^{k_i}$. A regular $n$-gon has $p_i^{k_i}$ regular $n/p_i^{k_i}$-gons, which gives us a map $D_{2n} \to S_{p_i^{k_i}}$. Now by Chinese remainder theorem and thoughtful look one proves that the map $D_{2n} \to \prod_{i=1}^s S_{p_i^{k_i}}$ is an injection.

(Equivalently, suppose that a symmetry $g \in D_{2n}$ preserves every regular $n/p_i^{k_i}$-gon in the given $n$-gon. Consider any vertex $v$ of the given $n$-gon. The vertex $v$ lies in a certain $n/p_1^{k_1}$-gon, in a certain $n/p_2^{k_2}$-gon, etc. Each of them is preserved by $g$ and their intersection consists of $v$ only, thus $v$ is preserved by $g$).

Thus $D_{2n} \hookrightarrow S_{\sum_{i=1}^s p_i^{k_i}}$ and $\mu(D_{2n}) \leqslant \sum_{i=1}^s p_i^{k_i}$. But it can not be smaller by the argument by David.