Let $G=D_{2n}=⟨x,y|x^2=y^n=e, $ $yx=xy^{n-1}⟩$
I need to find $G'$ [ the commutator subgroup of $G$.]
I understand that $G'$ is the subgroup generated from $ U=xyx^{-1}y^{-1} , $ $ \ \forall x,y \in G$ So, $U=\langle y^2 \rangle$
What is the strategy here?
Some ideas: if we write
$$D_{2n}=\langle\;x,y\;:\;\;x^2=y^n=1\;,\;\;xyx=y^{n-1}\;\rangle$$
then we get that
$$\forall\,k\in\Bbb N\;,\;\;[x,y^k]=x^{-1}y^{-k}xy^k=\left(xy^{-k}x\right)y^k=\left(y^{n-1}\right)^{-k}\;y^k=y^{2k}$$
and this means every element in $\;\langle\;y^2\;\rangle\;$ is a commutator.
OTOH, we have that
$$xy(yx)^{-1}=xyxy^{-1}=y^{n-1}y^{-1}=y^{-2}\in\langle\;y^2\;\rangle\implies G/\langle\;y^2\;\rangle\;\;\text{is abelian}\iff G'\le\langle\;y^2\;\rangle$$
The above yields $\;G'=[G:G]=\langle\;y^2\;\rangle\;$ .