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Let $G=\langle x,y\mid x^3=e, y^2=e\rangle$ and otherwise unrestricted.

Let $H=\langle xyxy\rangle$.

Is $H\triangleleft G$?

And in that case, what is $$G/H$$ isomorphic to?


If it is a normal subgroup, I'd suspect $G/H\cong D_3$, but is that really true?

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    $\begingroup$ If a subgroup is normal, then it is invariant under conjugation, meaning $\forall a\in G$, $aHa^{-1}=H$ (note that individual elements may not map back to themselves, but elsewhere in $H$). Does that happen in this case (what do the $h\in H$ look like?) $\endgroup$ – Justin Benfield Mar 13 '16 at 17:51
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    $\begingroup$ On the other hand, if you define $H$ as the normal closure of $xyxy$, ie the smallest normal subgroup containing $xyxy$, then you do get $D_3$ in the quotient. $\endgroup$ – Captain Lama Mar 13 '16 at 17:55
  • $\begingroup$ It seems it can't be. There doesn't seem any way to write $x(xyxy)x^2$ in the form $xyxy\cdots xyxy$. Thanks. I'll think about the normal closure of $xyxy$. $\endgroup$ – David Molano Mar 13 '16 at 18:16
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    $\begingroup$ Here's the start of a chat about my (admittedly overthought yet rather shallow) attempt at answering this. I got fed up after about an hour and had to move on to other things. $\endgroup$ – Shaun Dec 18 '18 at 21:10
  • $\begingroup$ I've checked in GAP whether the subgroup $H$ is normal in $G$. The code IsNormal(G,H); with the standard definitions of G and H was inconclusive. $\endgroup$ – Shaun May 24 at 20:21

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