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I meet these three groups $$G_1=\langle a,b| (aba^{-1}b^{-1})^2=1\rangle,$$

$$G_2= \langle a, b, c| aba^{-1}b^{-1}=c,c^2=1,c\neq1,ac=ca,bc=cb\rangle$$

$$G_3=\mathbb{Z}\times\mathbb{Z}=\langle a,b|aba^{-1}b^{-1}=1 \rangle$$

I find it maybe a faithful representation of $G_2$: $$a=e^{i \alpha} \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} $$ $$b=e^{i \beta} \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} $$ $$c= \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} $$ where $\alpha$, $\beta$ and $\alpha/\beta$ are irrational number. Since $a b=-ba$.

My question is :

What's the relation between $G_1$, $G_2$ and $G_3$. To be more specific:

  1. Are $G_1$ and $G_2$ isomorphic? I think that at least $G_1$ is surjective homomorphic $\phi_1$ to $G_2$ and $G_2$ is surjective homomorphic $\phi_2$ to $G_3$. Since the faithful representation of $G_2$ is also a representation of $G_1$. If I define a faithful representation of $G_3$ such that $a= e^{i \alpha}$ $b= e^{i \beta}$ where $\alpha$, $\beta$ and $\alpha/\beta$ are irrational number, the faithful representation of $G_3$ is also a representation of $G_2$

  2. If my argument above is correct, what's kernel of $\phi_1$? I think that the kernel of $\phi_2$ should be $Z_2$, so $G_3=G_2/Z_2$.

  3. If $G_1$ is not isomorphic to $G_2$, what's the faithful linear representation of $G_1$?

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  • $\begingroup$ I think $\Bbb{Z}\times\Bbb{Z}$ should be a quotient group of $G$, no? $\endgroup$ – Jyrki Lahtonen Mar 29 '17 at 21:43
  • $\begingroup$ You get a slightly bigger quotient group of $G$ as the group of upper triangular $3\times3$ matrices with 1s on the diagonal, and where you treat the top right entry modulo two. The homomorphism is to send $a\mapsto I_3+e_{12}$ and $b\mapsto I_3+e_{23}$, when $aba^{-1}b^{-1}$ will go to $c=I_3+e_{13}$. Then $c^2=I_3+2e_{13}$, so we only view the $(1,3)$ entry modulo two. $\endgroup$ – Jyrki Lahtonen Mar 29 '17 at 22:00
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    $\begingroup$ $a\mapsto (1,2,3)(4,5)$ and $b\mapsto (1,2,3,4,5)$ gives $aba^{-1}b^{-1}\mapsto(1,4)(2,5)$ which has order 2, so we get a homomorphism onto $S_5$. We can do something similar to get a homomorphism onto $A_6$, so I think this group is quite complicated. $\endgroup$ – Josh B. Mar 29 '17 at 23:25
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    $\begingroup$ It doesn't have any simple description in terms of familiar groups. It is an infinite hyperbolic group. How did this problem arise? $\endgroup$ – Derek Holt Mar 30 '17 at 8:00
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    $\begingroup$ @346699: How many questions in total did you already ask (including the ones which are now deleted)? You should settle down on one question and do not change your questions after it is answered since it makes it impossible to write a reasonable answer. Instead, you should ask separate questions. I know answers to all three current questions but will refrain from answering since it is very likely they will morph into something else once I write an answer. $\endgroup$ – Moishe Kohan May 11 '17 at 22:04
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First of all, instead of using the set-builder notation to describe a group by its presentation, you should use the standard notation, such as: $$ \langle a, b| [a,b]^2\rangle, $$ or $$ \langle a, b| [a,b]^2 =1\rangle $$ if you prefer. Now, the group with this presentation is one of so called Fuchsian groups, you can find some discussion of these in the book by Lyndon and Schupp "Combinatorial Group Theory". Your group $G$ contains an index 4 subgroup $G_1$ isomorphic to the fundamental group of the genus 2 closed oriented surface, which has the presentation: $$ \langle a_1, b_1, a_2, b_2| [a_1,b_1] [a_2,b_2]\rangle. $$ If this is not good enough, you can also say that there exists a (non-split) short exact sequence $$ 1\to G_2\to G\to Q\to 1, $$ where $Q$ is the quaternion group. The epimorphism $f: G\to Q$ sends $a, b$ to the generators $i, j$ of $Q$ and sends their commutator to the element $k\in Q$. The subgroup $G_2$ (the kernel of $f$) is isomorphic to the fundamental group of the genus 3 closed oriented surface , which has the presentation: $$ \langle a_1, b_1, a_2, b_2, a_3, b_3| [a_1,b_1] [a_2,b_2] [a_3, b_3]\rangle. $$ In particular, every abelian subgroup of $G$ is cyclic (of infinite order or order $\le 2$).

Edit. As for your revised questions:

What's the relation between $G_1$, $G_2$ and $G_3$. To be more specific:

  1. Are $G_1$ and $G_2$ isomorphic? I feel that $G_1$ includes the cases of $G_2$ and $G_3$. Am I right?

  2. Is $G_3$ a subgroup of $G_1$? Is $G_3$ homomorphic to $G_1$?

  3. Whether there is some relation between $G_1$, $G_3$ and $Z_2$? Like $G_1/Z_2 = G_3$?

The answer are:

  1. No, they are not isomorphic. I do not understand what "includes the cases" means.

  2. No, $G_3$ is not isomorphic to a subgroup of $G_1$. No, $G_1$ is not isomorphic to the quotient group of $G_3$.

  3. I explained the relation to the quaternion group. Since $Z_2$ is not isomorphic to a normal subgroup of $G_1$, you cannot have an isomorphism $G_1/Z_2\cong G_3$. However, if you take the normal closure $N$ of $[a,b]$ in $G_1$, then indeed $G_3\cong G_1/N$.

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    $\begingroup$ To be specific about the "Fuchsian group" nature of this group, one can apply the Poincare Polygon Theorem. Let $P \subset \mathbb{H}^2$ be a square (aka regular 4-gon) in the hyperbolic plane with all angles equal to $\pi/4$. Let $f_a : \mathbb{H}^2 \to \mathbb{H}^2$ be the translation that moves the bottom of $P$ to its top, and let $f_b$ be the translation that moves the left side of $P$ to the right side. Then $f_a,f_b$ generate a discrete group of isometries of $\mathbb{H^2}$ that is isomorphic to $G$, and $P$ is a fundamental polygon for this group. $\endgroup$ – Lee Mosher Apr 5 '17 at 18:52
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    $\begingroup$ This is quite similar to an ordinary Euclidean square being a fundamental domain for $\langle a,b \,|\, aba^{-1}b^{-1}=e \rangle$, except that $f_af_bf_a^{-1} f_b^{-1}$, rather than being the identity, is a rotation by angle $\pi$ around the upper right corner of $P$, hence its square is the identity. $\endgroup$ – Lee Mosher Apr 5 '17 at 18:56
  • $\begingroup$ Thanks for your reply! I've added more details to the question. $\endgroup$ – 346699 May 11 '17 at 17:43
  • $\begingroup$ I've revise the question 1:I think that at least $G_1$ is surjective homomorphic $\phi_1$ to $G_2$ and $G_2$ is surjective homomorphic $\phi_2$ to $G_3$. Since the faithful representation of $G_2$ is also a representation of $G_1$. $\endgroup$ – 346699 May 11 '17 at 21:21

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