If $k\mid n$, then $D_k \leq D_n$ I'm not quite sure how to show that If $k\mid n$, then $D_k \leq D_n$. I found How to show that if $k | n$, then $D_{2k} \leq D_{2n}$? but i'm not exactly sure that is a proof?
I say let $kl = n$. i.e assuming $k|n$. Then $D_k = \{1,r,r^l,r^{2l}, \cdots, r^{kl}, s, rs, \cdots, r^{kl}s \}$. I'm not really sure where to go. 
 A: Every element is of the form $r^j$ or $r^js$ with $0\le j<kl$. Now  take subset of elements with $j< kl$ and $j$ multiple of $l$. This subset is a subgroup and is isomorphic to $D_k$.
A: Here's a more geometrical answer.  Let $P_n$ be a regular $n$-gon drawn on the unit circle in the complex plane. First, note that the vertices of $P_n$ are at $e^{\frac{2\pi i}{n} j}$ for $j = 0, \ldots, n-1$.  If $n = k \ell$, then the vertices of a regular $k$-gon $P_k$ are contained within the set of vertices of $P_n$: a vertex of $P_k$ is of the form $e^{\frac{2\pi i}{k} m}$ for $m = 0, \ldots k-1$ and
$$
e^{\frac{2\pi i}{k} m} = e^{\frac{2\pi i}{k \ell} \ell m} = e^{\frac{2\pi i}{n} \ell m} \, .
$$
Thus the vertices of $P_k$ are exactly those vertices $e^{\frac{2\pi i}{n} j}$ of $P_n$ where $j$ is a multiple of $\ell = n/k$.  See the image below for $n = 6$ and $k = 3$.
$\hspace{4cm}$
Thus a rigid motion of the vertices of $P_k$ will correspond to a rigid motion of the vertices of $P_n$, giving an inclusion $D_k \hookrightarrow D_n$. 
