This is an example from Dummit & Foote text.
Let $D_{2n}=<r,s:r^n=s^2=1,s^{-1}rs=r^{-1}>$.Since [$r,s$]$=$$r^{-2}$,we have $\langle r^{-2}\rangle=\langle r^2\rangle \le D'_{2n}$.Furthermore,$\langle r^2\rangle \unlhd D_{2n}$ and the images of $r$ & $s$ in $D_{2n}/\langle r^2\rangle$ generate this quotient. They are commuting elements of order $\le 2 $,so the quotient is abelian and $D'_{2n} \le \langle r^2\rangle$.Thus, $D'_{2n}=\langle r^2\rangle$. Finally,note that if $n(=\vert r \vert$) is odd,$\langle r^2\rangle=\langle r\rangle$ whereas if $n$ is even,$\langle r^2\rangle$ is of index $2$ in $\langle r\rangle$.Hence,$D'_{2n}$ is of index $2$ or $4$ in $D_{2n}$ according to whether $n$ is odd or even,respectively.
NOTIONS WHICH I AM NOT GETTING
$\color{red}{(1)}$ How $\langle r^{-2}\rangle=\langle r^2\rangle $ ?
$\color{red}{(2)}$ How $\langle r^2\rangle \le D'_{2n}$?
$\color{red}{(3)}$ How does the images of $r$ and $s$ in $D_{2n}/\langle r^2\rangle$ generate this quotient?
$\color{red}{(4)}$ How if $n(=\lvert r \rvert$) is odd,$\langle r^2\rangle=\langle r\rangle$ whereas if $n$ is even,$\langle r^2\rangle$ is of index $2$ in $\langle r\rangle$?
(1) $r^2$ is in the group generated by $r^{-2}$ (since it's $(r^{-2})^{-1}$), and $r^{-2}$ is in the group generated by $r^2$, so the two groups are equal.
(2) $D_{2n}'$ is, by definition, the group generated by the commutators; the calculation shows that $r^{-2}$ is a commutator, so the group it generates (which is the same as the group $r^2$ generates) is contained in $D_{2n}'$.
(3) $r$ and $s$ generate $D_{2n}$, so their images generate the quotient.
(4) That's how cyclic groups work. If you have a cyclic group of order $n=2s$, then $r^2$ generates the group $\{\,r^2,r^4,r^6,\dots,r^{2s}=1\,\}$, of index 2. If $n$ is odd, then you have to raise $r^2$ to the power $n$ to get the identity, so $r^2$ generates the whole group.