What's the group $G=\langle a,b| (aba^{-1}b^{-1})^2=1\rangle $? 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:


*

*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 $G_2$

*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 $G_3=G_2/Z_2$.

*If $G_1$ is not isomorphic to $G_2$, what's the faithful linear representation of $G_1$? 
 A: 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:
  
  
*
  
*Are $G_1$ and $G_2$ isomorphic? I feel that $G_1$ includes the cases of $G_2$ and $G_3$. Am I right?
  
*Is $G_3$ a subgroup of $G_1$? Is $G_3$ homomorphic to $G_1$? 
  
*Whether there is some relation between $G_1$, $G_3$ and $Z_2$? Like $G_1/Z_2 = G_3$?  

The answer are:


*

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

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

*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 $G_3\cong G_1/N$. 
