Continuing my independent journey through "Abstract Algebra" (see this previous question for context and notation), I am working on:
If $n = 2k$ is even and $n \geq 4$, show that (a) $z = r^k$ is an element of order 2 which (b) commutes with all elements of $D_{2n}$ (Dihedral group of a regular n-gon). Show also that (c) $z$ is the only non-identity element of $D_{2n}$ which commutes with all elements of $D_{2n}$.
(a) $z^2 = (r^k)^2 = r^{2k} = r^n = 1$, so the order of $z$ divides 2.
$\quad\quad$ Does the hypothesis that $n \geq 4$ come into play here?
(b) Elements in $D_{2n}$ are of the form $r^i$ or $sr^i$, $0 \leq i \leq n - 1 $
$\quad\quad$ For elements $r^i$, we have: $zr^i = r^kr^i = r^{k + i} = r^{i + k} = r^ir^k = r^iz$,
$\quad\quad$ so $z$ commutes with these elements.
$\quad\quad$ For elements $sr^i$, I am a little confused (yet still trying to remember what needs to be shown!):
$\quad\quad$ $zsr^i = r^ksr^i = (r^ks)r^i = (sr^{-k})r^i = sr^{-k + i}$
$\quad\quad$ Yet $sr^iz = sr^ir^k = sr^{k + i}$
$\quad\quad$ The two don't seem equal to me, so I must be missing something?
(c) The only thing that comes to mind is the argument that the identity in a group is unique,
$\quad\quad$ but I'm having a hard time generalizing/extending that concept to this group.
Thanks in advance for any hints!