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$\newcommand{\R}{\mathbf R}$ Let $G$ be the group of homeomorphisms of $\R^2$ generated by $g$ and $h$, where $g(x, y)=(x+1, y)$ and $h(x, y)=(-x, y+1)$.

To show that $G\cong \langle a, b|\ b^{-1}aba\rangle$.

I tried the following:

Define a map $f:\langle a, b\rangle \to G$ which sends $a$ to $g$ and $b$ to $h$. Then it can be checked that $b^{-1}aba$ lies in the kernel of $f$. So $f$ factors through $\langle a, b|\ b^{-1}aba\rangle$ to give a map $\bar f: \langle a, b|\ b^{-1}aba\rangle\to G$.

What I am unable to show is that $\bar f$ is injective.

Also, here we were already given a presentation which we had to show is isomorphic to $G$. If it were not given, then is there a general way to get one?

Thank you.

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    $\begingroup$ The relation $b^{-1}ab = a^{-1}$ allows you to write every element of $G$ as $a^ib^j$ for some $i,j \in {\mathbb Z}$. So to show that $\bar{f}$ is injective, it is sufficient to show that these elements all map onto distinct elements of $G$, which you should be able to do. (Finding a normal form for the elements of a group defined by a finite presentation is a standard technique for proving that the group is isomorphic to some other more group.) $\endgroup$ – Derek Holt May 25 '16 at 18:00
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    $\begingroup$ A different approach would be to use some algebraic topology, look at the quotient space of the group acting on the plane and the fundamental group of that space will be your group (this is only because the group acts nicely), and you can use van Kampens to calculate a presentation. Check out 1.2 and 1.3 in Hatcher's algebraic topology book. $\endgroup$ – Paul Plummer May 26 '16 at 15:53
  • $\begingroup$ it would be nice that I could receive any feedback for my answer, thanks! $\endgroup$ – janmarqz Apr 9 '18 at 15:07
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    $\begingroup$ @janmarqz I apologize for not noticing. I got a notification but I thought it is an old post so I did not see the page carefully. I didn't realize that a new answer was added. Give me some time to read your answer. I am actually quite busy right now with my PhD work. Thanks. $\endgroup$ – caffeinemachine Apr 9 '18 at 15:38
  • $\begingroup$ that you said is already a feedback :) thanks again $\endgroup$ – janmarqz Apr 9 '18 at 16:17
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Direct calculations with $$\left(\begin{array}{c}x\\y\end{array}\right) \stackrel{g}\longmapsto \left(\begin{array}{c}x+1\\y\end{array}\right)\ \mbox{and} \ \left(\begin{array}{c}x\\y\end{array}\right) \stackrel{h}\longmapsto\left(\begin{array}{c}-x\\y+1\end{array}\right),$$ give you $$ \left( \begin{array}{c} x\\ y \end{array} \right) \stackrel{g^{-1}}\longmapsto \left( \begin{array}{c} x-1\\ y \end{array} \right) \ \mbox{and} \ \left( \begin{array}{c} x\\ y \end{array} \right) \stackrel{h^{-1}}\longmapsto \left( \begin{array}{c} -x\\ y-1 \end{array} \right), $$ respectively. But also $gh=hg^{-1}$ because:

$$\left(\begin{array}{c}x\\y\end{array}\right) \stackrel{h}\longmapsto\left(\begin{array}{c}-x\\y+1\end{array}\right) \stackrel{g}\longmapsto\left(\begin{array}{c}-x+1\\y+1\end{array}\right),$$ and $$\left(\begin{array}{c}x\\y\end{array}\right) \stackrel{g^{-1}}\longmapsto\left(\begin{array}{c}x-1\\y\end{array}\right) \stackrel{h}\longmapsto\left(\begin{array}{c}-x+1\\y+1\end{array}\right),$$

and this implies that $h^{-1}ghg=e$.

Take all the reduced word in the letters $g,h,g^{-1},h^{-1}$, which can brought into a canonical form $g^mh^n$ taking into account that $gh=hg^{-1}$.

So you have that the subgroup $\langle\{g,h\}\rangle$ (the subgruoup generated by $g,h$) satisfies $$\langle g,h\ |\ h^{-1}ghg=e\rangle.$$

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  • $\begingroup$ I do not see how this gives us that $G$ is isomorphic to $\langle a, b| b^{-1}aba\rangle$. It seems that you have also shown that $f$ is surjective. Am I missing something? $\endgroup$ – caffeinemachine Apr 10 '18 at 1:49
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    $\begingroup$ @caffeinemachine. Other example of how to read a presentation for a subgroup inside another group: The matrix $j= \left( \begin{array}{cc} 1&1\\ 0&1 \end{array} \right) $ is an element of the group $SL_2(\Bbb Z)$ and satisfies $ \left( \begin{array}{cc} 1&1\\ 0&1 \end{array} \right)^n = \left( \begin{array}{cc} 1&n\\ 0&1 \end{array} \right) $ then the subgroup $\langle\{ j\}\rangle$ has the presentation $$\langle j\ |\quad\rangle\cong \Bbb Z$$ of the cyclic free rank one group. $\endgroup$ – janmarqz Apr 16 '18 at 17:44

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