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 $\tau \sigma \tau^{-1} = \sigma^{-1}$.