Find all elements of $D_4$ such that $s$ commute with them. 
Consider the Dihedral Group $D_4$.
  Find all elements of $D_4$ such that $s$ commute with them.

We know that $Z(D_4)=\{1,r^2\}$
Hence these elements commute with $s,sr$.
Now I found that $s$ commutes with  $sr^2$ because $s(sr^2)=r^2$
and $(sr^2)s=s(sr^{-2})=r^2$.
But I dont see any other elements.Where am I going wrong?
Also there is another general question.I am stuck here too:

Consider the Dihedral Group $D_n$ where  $n$ is even.
  Find all elements of $D_n$ such that $s$ commute with them.

Can someone kindly help.
 A: $r^is=sr^{-i}$. So $r^is=sr^i$ if and only if $r^i=r^{-i}$ which happens if and only if $r^{2i}=e$. Well, obviously $i=0$ satisfies this. And if $n$ is even then $i=\frac{n}{2}$ does as well.
Also, $s(sr^i)=r^i$ and $(sr^i)s=r^{-i}$. So again, $s$ commutes with $sr^i$ if $i=0$ or $i=\frac{n}{2}$ when $n$ is even. 
This means that if $n$ is odd then the centralizer of $s$ is $\langle s\rangle$ and if $n$ is even then it is $\langle s,r^\frac{n}{2}\rangle$. 
A: Mark's answer is a perfectly good one. Here is another way to recover the answer if you know a bit of geometry. I will prove a more general result.
Thm. In the euclidean plane, given a reflection $s$, the only isometries commuting with $s$ are $\pm Id,\pm s$.
We have the following well-known geometric result concerning linear isometries. 
If $s$ is a reflection of axis $L$ (a line passing through the origin), then for any isometry $\rho$ (whose center is the origin) , then $\rho s\rho^{-1}$ is a reflection of axis $\rho(L)$ (the square gives $Id$, and clearly the set of fixed points is $\rho(L)$)
Now , reflections being characterized by their axis, we need to find all isometries of the euclidean plane which stabilizes globally the axis $L$ of $s$. For rotations , we only have two choices: $Id$ or $-Id$ (look at the angles of $L$ and $\rho(L)$ with the $x$-axis). For symmetries: they are the composition of a rotation and $s$ ($\rho$ is an isometry of determinant $1$, then $\rho=(\rho s)s=\rho's,$ ands $\rho'=\rho s$ is a rotation, since it has determinant $-1$.) 
Now $\rho$ commutes with $s$ if and only if $\rho'$ does, and we are done.
To recover Mark's answer , just realize that $-Id$ lies in $D_n$ if and only if $n$ is even.
