If $n \geq 3$ then there is no surjective homomorphism $f: D_{2n} \to Z_n$. Claim: If $n \geq 3$ then there is no surjective homomorphism $f: D_{2n} \to Z_n$. In this case $D_{2n}$ refers to the dihedral group of order 2n.
Thoughts:
I'm thinking that the proof to this relies on orders between these groups since surjectivity is coming into play, but I don't know where to proceed from there. Any hints would be appreciated. 
Current attempt:
Suppose there is a surjective homomorphism. Then since $n \geq 3$ we have $|D_{2n}| > |Z_n|$ so by pigeonhold principle $\exists x,y \in D_{2n}$ s.t. $f(x) = f(y) = z\in Z_n$
I think from here maybe find a contradiction by working with inverses?
 A: $\;\;\;$In the dihedral group $D_n$ of size $2n,n\geq3,$ on the regular $n$-gon a rotation $R_{\theta}$ of degree $\theta, 0<\theta<360,$ never commutes with a reflection $S_j$ over the symmetry axis through vertex $j=1,2,\dots,n$. Therefore, the group center $\mathcal Z(D_n)$ is trivial ; i.e. $\mathcal Z(D_n)=\{\iota\}$. 
$\;\;\;$If such a surjective group homomorphism $\psi:D_n\to\Bbb Z_n$ exists then by the fundamental theorem of group homomorphisms we'd have $\frac{D_n}{\ker(\psi)}\approx\Bbb Z_n$ and thus $$\frac{|D_n|}{|\ker(\psi)|}=|\Bbb Z_n|\;;\;\frac{2n}{|\ker(\psi)|}=n$$ implying $\ker(\psi)=\{\iota,\alpha\}$ has size $2$. Because $\ker(\psi)$ is normal we'd have $x\alpha x^{-1}=\alpha$ for all $x\in D_n$ implying $\alpha\in\mathcal Z(D_n)$ which is impossible because the group center $\mathcal Z(D_n)$ must be trivial. Therefore, no such surjective group homomorphism exists.
A: I'll answer the question in the comments (now deleted, but the question was: is it possible that for some $y \in Z_n$ there is only 1 $x \in D_{2n}$ mapping to it, for some other $y$ there are 2 such $x$ and for yet others 3 such $x$ etc.?), since it is at the same time a hint at how to approach the original question. 
Let $f$ be the homomorphism. The claim is that for every $y \in Z_n$ there is an equal number of $x \in D_{2n}$ such that $f(x) = y$. (It then follows that this number must be 2 in this case, but for a surjection between any groups $G$ and $H$ the result also holds, just with a possibly different number).
Proof: 
Let $e_1, \ldots, e_k$ be the elements that map to the identity element $e$ of $Z_n$. Let $y$ be a general element of $Z_n$ and let $x$ be any element such that $f(x) = y$. Then, since $f$ is a homomorphism we have that $f(xe_1) = y$, $f(xe_2) = y$, ..., $f(xe_k) =y$ so there are at least $k$ elements mapping to $y$.
Conversly suppose that $x'$ maps to $y$ as well then $x^{-1}x'$ maps to $e$ so $x^{-1}x' = e_i$ for one of the $i \in 1, \ldots, k$, so $x' = xe_i$. So when we were listing the elements $xe_1, ..., xe_k$ above we had indeed found all elements that map to $y$. Hence there are exactly $k$.
EDIT: Above is a proof of the fact that, for a surjection between groups $f: G \to H$ the number of elements in $G$ mapping to a given element $y$ in $H$ is the same for all $y$. I claimed at the beginning that the proof also contained a hint for the original question, but as stated that claim is a bit vague. What I meant by that is: in the above proof there is a special role for the set of elements in $D_{2n}$ that map to the identity in $Z_n$. My hint is: look closer at this set. What properties must it have? Then can you find any set with those properties inside $D_{2n}$?
