Let $k$ and $n$ be any integers such that $k \ge 3$ and $k$ divides $n$. Prove that $D_n$ contains exactly one cyclic subgroup of order $k$ a) Find a cyclic subgroup $H$ of order $10$ in $D_{30}$. List all generators of $H$.
b) Let $k$ and $n$ be any integers such that $k \ge 3$ and $k$ divides $n$. Prove that $D_n$ contains exactly one cyclic subgroup of order $k$.
My attempt at a), the elements of $D_{30}$ of order 10 are $r^{3n},sr^{3n},  0\le n\le 4$ so any cyclic groups of the corresponding elements would work. The generators of the elements would be of the form $\langle a^j \rangle$ where $\gcd(10, j) = 1$ so $j =1,3,7,9,11,13$.
Any ideas as to how I should attempt b)? 

Let $$D_{2n} = \langle r, s \mid r^n = 1, s^2 = 1, s r = r^{-1}s \rangle$$ be the dihedral group of order $2n$ generated by rotations ($r$) and reflections ($s$) of the regular $n$-gon.
From the presentation it is clear that every element can be put into the form $s^i r^j$ where $i$ is $0$ or $1$. So the cyclic subgroups of $D_{2n}$ are the cyclic subgroups generated by elements of the form $r^j$ and $s r^j$.


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*Since $r$ generates $C_n$ we uniquely have $C_d \le C_n \le D_{2n}$ for every $d|n$ by the lemma.

*Since $s r^i s r^i = 1$ the second form only generates $C_2$ subgroups.


This shows that there may be many different $C_2$ subgroups of $D_{2n}$, but the $C_d$ subgroups are all unique.
Lemma For $d|n$, there is a unique subgroup of $C_d$ isomorphic to $C_d$.
proof: Let $m=ab$, every cyclic group $C_m$ has exactly $a$ elements $g$ such that $g^a=1$ (in fact these elements are $b$, $2b$, ...). So $C_n$ has exactly $d$ elements such that $g^d=1$, and if $C'$ is a subgroup of $C_n$ isomorphic to $C_d$ then it too has exactly $d$ elements like this: they must be exactly the same elements then!
