Let $G=F( x,y)$ be a free group with two generators, assume that $H\leq G$ where $H=\langle xyxy^{-1}\rangle $, let $N$ be the normal closure of $H$. Is it true that $G/N$ has a presentation $\langle x,y\mid xyxy^{-1} \rangle $?


closed as off-topic by Matthew Towers, YuiTo Cheng, Cesareo, Parcly Taxel, Derek Holt May 25 at 8:08

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    $\begingroup$ To get the best possible answers, you should explain what your thoughts on the problem are. That way, people won't tell you stuff you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help if you show that you've tried the problem yourself. Lastly, some may consider your post rude because it is phrased as a command, not a request for help, so please consider rewriting it. $\endgroup$ – Zev Chonoles Jul 13 '13 at 5:07
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    $\begingroup$ @Danial: Do you mean $G= \langle x,y \mid \ \rangle$? In this case, just use the definition of a presentation. $\endgroup$ – Seirios Jul 13 '13 at 6:36
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    $\begingroup$ As Serios says, this is true by definition if $G$ is a free group generated by $x$ and $y$, but it is not necessarily true otherwise, because any relations between $x$ and $y$ that hold in $G$ will laso hold in $G/N$. $\endgroup$ – Derek Holt Jul 13 '13 at 8:12
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    $\begingroup$ A good references for this is the first chapter of Magnus, Karrass and Solitar's book "Combinatorial group theory". $\endgroup$ – user1729 Jul 13 '13 at 10:07

Yes, this is the definition of $\langle x, y \mid xy xy^{-1}\rangle$.

  • $\begingroup$ But shouldn't the relation $xyxy^{-1}$ be the generator of the normal group $N$ not for the subgroup $H$? $\endgroup$ – Ronald Jul 13 '13 at 16:12
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    $\begingroup$ @Danial: "the quotient of $\langle x, y \rangle$ by the normal closure of $\langle xyxy^{-1}\rangle$" is literally the definition of the expression "$\langle x,y \mid xyxy^{-1}\rangle$". I'm not sure how much clearer I can be here. $\endgroup$ – Chris Eagle Jul 13 '13 at 16:14
  • $\begingroup$ Ok, now it's convenient. I thought that $xyxy^{-1}$ should be the generator of $N$. but this is not literally true because the generated set of $xyxy^{-1}$ may not be normal.. as in our example! Thank you very much $\endgroup$ – Ronald Jul 13 '13 at 16:25

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