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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 ?


marked as duplicate by Dietrich Burde, Aditya Hase, Anastasiya-Romanova 秀, Stefan Hamcke, Cheerful Parsnip Dec 12 '14 at 15:41

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  • 1
    $\begingroup$ I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$. $\endgroup$ – GPerez Dec 12 '14 at 14:49
  • $\begingroup$ @GPerez how can i find the commutattor of group in general..i'm not getting it $\endgroup$ – Nizar Halloun Dec 12 '14 at 14:57
  • $\begingroup$ @NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators. $\endgroup$ – Cheerful Parsnip Dec 12 '14 at 15:40

Some ideas: if we write


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


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