What is the center of $D_{2n}/Z(D_{2n})$ What is the center of $D_{2n}/Z(D_{2n})$. I see that when $n=2^k$ then I have a $p$-group so the center is not trivial. But when $n$ is not power of $2$ how can i know what is the center of this group?.
Surely when $n$ is odd then the center is trivial. 
 A: Let $D_{2n}$ denote the dihedral group of order $2n$ generated by a rotation $r$ of order $n$ and a reflection $s$ of order $2$. Then $D_{2n}$ has presentation
$$D_{2n}\cong \langle r,s\mid r^n=1, s^2=1, srs=r^{-1}\rangle.$$
This presentation implies that, as a set, we have
$$D_{2n} = \{1,r,r^2,\dots, r^{n-1}, s, sr, sr^2,\dots, sr^{n-1}\}.$$
See here for the computation of the center of $D_{2n}$.
For $n=1$ or $n=2$, the dihedral groups $D_2$ and $D_4$ are abelian, and so in these case $D_{2n}/Z(D_{2n})$ is a trivial group, and thus has trivial center.
If $n\geq 3$ is odd, then the dihedral group $D_{2n}$ has trivial center. Hence $D_{2n}/Z(D_{2n})$ is isomorphic to $D_{2n}$, and thus has trivial center.
The interesting case is when $n\geq 3$ and $n$ is even. Then the dihedral group has center $Z(D_{2n})=\langle r^{n/2} \rangle$ which is the two element subgroup consisting of the identity and $r^{n/2}$.
Claim. If $n\geq 3$ and $n$ is even, then $D_{2n}/Z(D_{2n})\cong D_n$.
Proof of Claim. For clarity, suppose that $D_n$ is generated by a rotation $r_0$ of order $n/2$ and a reflection $s_0$ of order $2$ and that $D_{2n}$ is generated by a rotation $r$ of order $n$ and a reflection $s$ of order $2$. Define a homomorphism $\phi: D_{2n}\to D_{n}$ by $\phi(s^i r^j)=s_0^i r_0^j.$ I will leave it to you to show that $\phi$ is a surjective homomorphism. The kernel of $\phi$ is $\ker \phi = \langle r^{n/2}\rangle$, and so by the first isomorphism theorem
$$D_n \cong D_{2n}/\ker \phi = D_{2n}/Z(D_{2n}).$$
$$\tag*{$\blacksquare$}$$ 
The claim tells us what we need to compute the center of $D_{2n}/Z(D_{2n})$ in the case where $n\geq 3$ and $n$ is even. If $n$ is a multiple of $4$, then $D_{2n}/Z(D_{2n}) \cong D_{n}$ where $n/2$ is even. The center of $D_n$ is $\langle r^{n/4}\rangle$. Writing that in terms of the quotient group, we get that the center of $D_{2n}/Z(D_{2n})$ is $\langle r^{n/4}Z(D_{2n})\rangle$. If $n$ is even and not a multiple of $4$, then $D_{2n}/Z(D_{2n})\cong D_{n}$ where $n/2$ is odd. Hence the center of $D_{2n}/Z(D_{2n})$ is trivial.
