If $$D_{2n} = \{r^i s^j, \quad 0 ≤ i ≤ n − 1, \quad 0 ≤ j ≤ 1 \}$$ is the dihedral group of order $2n$, show that the center of $D_{2n}$, $n &gt; 2$ is the trivial subgroup if $n$ is odd, and is the subgroup $\{e, r^{n/2} \}$ if $n$ is even.  I don't know how to proceed with this one. Kindly help. Thanks in advance.
I have posted a proof below, and would appreciate it if someone could review it for accuracy. Thanks!   Problem:   Let n $\in$ $\mathbb{Z}$ with $n$ $\ge$ 3. Prove the following:   (a) Z(D$_{2n}$) = 1 if $n$ is odd.   (b) Z(D$_{2n}$) = {1, r$^k$} if $n$ = $2k$.   Note that $r$ and $s$ generate D$_{2n}$ with the group presentation { $r$,$s$ | $r$$^n$ = $s$$^2$ = 1, $rs$ = $sr$$^{-1}$ }   Proof:  part (a)   Let $n$ $\ge$ 3 where $n$ $\in$ $\mathbb{Z}$ and $n$ is odd.   For any x $\in$ Z(D$_{2n}$), x must commute with both $s$ and $r$, since both are in $D_{2n}$. Note that if $x$ commutes with both $s$ an $r$, then $x$ commutes with any element of $D_{2n}$ since $r$ and $s$ generate D$_{2n}$.    Then for any x$\in$ Z(D$_{2n}$), we have $xr$ = $rx$, where x is of the form $x$ = s$^j$r$^w$ (any such element can be arranged into this form via the relation $rs$ = $sr$$^{-1}$ and the fact that $r$ and $s$ generate D$_{2n}$). Note that since r and s have finite order, the exponents are modulo n and modulo 2 for $r$ and $s$ respectively.   Then we have  s$^j$r$^w$$r$ = $r$s$^j$r$^w$.   Which implies s$^j$r$^{w+1}$ = s$^j$r$^{w +- 1}$  In the case where the power of r on the RHS is $w$ - 1, it is clear we cannot have equality unless |r| = 1, which it is not. Note that if $j$ is even then we have only the $w$ + 1 case. Hence $x$ may be of the form $1$$r$$^w$ = $r$$^w$.   Also we have that x must commute with $s$, hence $s$$x$ = $x$$s$.   Then $s$$r$$^w$ = $r$$^w$$s$.  Which implies $s$$r$$^w$ = $s$$r$$^{-w}$.   Applying s$^{-1}$ to both sides we arrive at the task of finding when r$^w$ = r$^{-w}$ i.e when $w$ = $-w$. But since r has finite order we have:   $$ \bar{w} = (n-1)*w $$   where $\bar{w}$ is the residue class of w modulo n. Hence for some a $\in$ $\mathbb{Z}$$_+$   $$ w + an = wn - w $$  $$ 2w = wn - an $$ $$ 2w = n(w-a) $$   Now since 0 $\lt$ w $\le$ n we have that the LHS is greater than 0. Also since n $\ge$ w then (w-a) $\le$ 2, since otherwise the RHS would be greater than the LHS. So since n is odd we must have that (w-a) is even and hence (w-a) = 2, implying that 2w = 2n and hence n = w.   Hence x = $r$$^w$ = $r$$^n$ = 1. So Z(D$_{2n}$) = 1.   part (b)   From the argument above any x $\in$ Z(D$_{2n}$) must be of the form $s$$^j$$r$$^w$ for j even and j $\in$ $\mathbb{Z}$. Then x is of the form $r$$^w$ for $0$ $\lt$ w $\le$ $n$. Since $n$ is even we have the equation below is satisfied when $2w$ = $n$ or $w$ = $n$, since (w-a) $\le$ 2 but also (w-a) $\ge$ 0 for (w-a) $\in$ $\mathbb{Z}$$_+$.   $$ 2w = n(w - a) $$   Hence $r$$^w$ $\in$ Z(D$_{2n}$) when $2w$ = $n$, as desired.
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