I am trying to prove that the center of the dihedral group $Z(D_{10})$ is trivial.
The only way I've been able to do it so far is to draw the group table, but I am trying to find a more elegant way of proving the result. If I fix an $a \in Z(D_{10})$ and assume $ab = ba$ for all $b \in B$, I am not able to prove that $a = e$ because there isn't a way to, for example, multiply by inverses. If I broke it into cases, I could possibly use the fact that reflections invert themselves.
You could use the normal form: every element can be written uniquely as $r^is^j$ with $0\leq i\leq 4$, $0\leq i\leq 1$, subject to the rules $r^5=s^2=e$, $sr=r^4s$.
Suppose $r^is^j$ is central; then $s(r^is^j) = r^{5-i}s^{j+1}$ must equal $(r^is^j)s = r^is^{j+1}$, which tells you that $r^i=r^{-i}$, but this only holds when $i=0$ in this dihedral group. That means the element is either $e$ or $s$; but $s$ is not central, since $sr = r^4s\neq rs$. So the center is trivial.
(Alternatively, if $r^is^j$ is central, then $r^{i+1}s^j = r(r^is^j) = (r^is^j)r = r^{i-1}s^j$ if $j=1$, hence $r^{i+1}=r^{i-1}$, which is impossible. So $j=0$, but $r^i$ does not commute with $s$ if $i\neq 0$, as above.)
This follows from a general fact. Every dihedral group $D$ of order $2n$ (the group of symmetries of the regular $n$-gon), $n>2$, consists of $n$ rotations $\rho_i$ through angles $2\pi i/n$, $i=0,1,2,...,n-1$ and of $n$ reflections about lines that connect either opposite vertices or a vertex and the midpoint of the opposite side or two opposite midpoints.
Suppose $n$ is odd. Then none of the reflections commute with $\rho_1$ and none of the rotations $\rho_i, i>0$ commutes with all reflections. Hence the center of group $D$ is trivial $=\{\rho_0\}$.
If $n$ is even then the center consists of two rotations through angles $0$ and $\pi$ which are $\rho_0$ and $\rho_{\frac n2}$.
