understanding the commutator of dihedral group [duplicate]

Let $G=D_{2n}=⟨x,y|x^2=y^n=e, $ $yx=xy^{n-1}⟩$

I need to find $G'$ [ the commutator of G]

now I understand that $G'$ is the subgroup generated from $ U=xyx^{-1}y^{-1} , $ $ \ \forall x,y \in G$ So, $U=<y^2>$

whats the strategy here ?

Solution 1:

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 $\;G'=[G:G]=\langle\;y^2\;\rangle\;$ .