# Is the subgroup $H:=\langle xyxy\rangle$ normal in $G:=\langle x,y\mid x^3=e, y^2=e\rangle$? If so, find $G/H$.

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?

• 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?) – Justin Benfield Mar 13 '16 at 17:51
• 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. – Captain Lama Mar 13 '16 at 17:55
• 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 $xyxy$. – David Molano Mar 13 '16 at 18:16
• 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. – Shaun Dec 18 '18 at 21:10
• 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. – Shaun May 24 at 20:21