Center of Odd Dihedral Group So I'm trying to prove the center of $D_n$ only has the identity for odd $n$ greater than or equal to 3. I know that since $D_n$ for $n$ greater than 2 is non-Abelian, the flips and the rotations do not commute in general. Additionally, I know that the only other element that can commute with every element (rotation by 180 degrees) is not going to be in an odd dihedral group- I just have a hard time saying why it holds. My question is how do I formalize this, going forward?
Edit: My question is different from the other that was pointed out, as I'm proving the center is trivial whereas the other question is an if and only if regarding an element in the center.
 A: In a dihedral group with $2n$ elements, there is a cyclic subgroup of order $n$ such that all of its elements are conjugate (within the dihedral group) to their inverses.
If an element $x$ of the cyclic group of order $n$ is central in the dihedral group, then it can only be conjugate to itself. Hence the only elements in the cyclic subgroup which are in the center of the dihedral group are equal to their inverses. So we must have $x = x^{-1}$ for all such $x$. Hence, when $n$ is odd we must have $x = 1$ for any such $x$ , as a group of odd order contains no element of order $2$.
When $n$ is odd, the only element in a cyclic group of order $n$ which is its own inverse is the identity.
A: Suppose we define the dihedral group $D_{2n} = \{ r,s | s^2 = e, r^n = e \}$. The definition of the center of a group G is $Z(G) = \{zg = gz | g\in G\}$. Also, $r^is = sr^{-i}$, for $0\leq i\leq n$. An element $z$ is in the center iff it commutes with both $r,s$. So the element $z=r^i s^k\in Z(D_{2n})$ given $zr=rz$ and $zs=sz$. This is the same as $r^i s^k r=r^{i+1} s^{k}$. This reduces to $s^kr=rs^k$ by left multiplication by $r^i$. $k$ is modulo 2, so we can have $k=0$ or $k=1$. If $k=0$ then this is fine. But now, consider $k=1$. Then $rs=sr$ but $D_{2n}$ is non-commutative for $n\geq 3$. Then we require that $k=0$ and $z=r^i\in Z(D_{2n})$. Using our second condition, $zs=sz=r^is=sr^i$. But $r^is = sr^{-i} = sr^i$, so $sr^{-i} = sr^i$. Left multiplying by $s$, and right multiplying by $r^i$, we obtain $s(sr^i)r^i = s(sr^{-i})r^i = e = r^{2i}$ by that $s^2=e$. Then $n|2i$. Since $n$ is odd and $0\leq i < n$, then $i=0$. Recall that $z=r^i\in Z(D_{2n})$. Then $z=r^0=e\in Z(D_{2n})$ for any element $z\in Z(D_{2n})$. Then $\{e\} = Z(D_{2n})$.
