Map from group $G$ to $GL_n(\Bbb Q)$ I'm working on the following problem:

Let $G$ be the group $\langle a,b  \mid  aba^{-1} = b^2\rangle$. 
a) Show that the map $\phi: G \longrightarrow GL_n(\Bbb Q)$ defined by $\phi(a) =\begin{pmatrix} 
2 & 0 \\
0 & 1 
\end{pmatrix} = A$    and $\phi(b) = \begin{pmatrix} 
1 & 1 \\
0 & 1 
\end{pmatrix} = B$ is a group homomorphism. 
b) Prove that $A^nBA^{-n} = B^{2^n}$ for all positive integers $n$. 
c) Prove that $G$ is infinite. 

For part a), I know that such a group homomorphism $\phi: G \longrightarrow GL_n(\Bbb Q)$ would have the property that $\phi(ab) = \phi(a) \cdot \phi(b)$. But although we know $\phi(a)$ and $\phi(b)$ individually, how do we know what $\phi(ab)$ is equal to ? I can get from the relations that $ab = b^2a^{-1}$, but I'm not sure how to proceed from there. Does $\phi(ab) =\phi(b^2a^{-1}$) buy any useful relation to tell us what $\phi(ab)$ must equal? 
I was able to prove part (b). We proceed by induction. The base case $(n=1)$ checks out -- we can verify through matrix multiplication that $ABA^{-1} = B^2$. Note that this also means that $AB^jA^{-1} = (B^2)^j$, where $j$ is a positive integer.  Now, assume $A^{n-1}BA^{-n+1} = B^{2^{n-1}}$. Then $A^nBA^{-n} = AB^{2^{n-1}}A^{-1} = (B^2)^{2^{n-1}} = B^{2^n}$, and the result follows. 
For part c), my idea so far was to show that $ab$ has infinite order. Using $ab = b^2a^{-1}$, I planned on showing $a^n \neq e$ for any $n \in \{0,1,2,...\}$, but I'm not sure how to proceed. 
Thanks! 
 A: Part (a).
You must show that $\phi(a)$ and $\phi(b)$ satisfy the same relationships as $a$ and $b$. In this case there's only one relationship $$ABA^{-1}=B^2.$$.
Part (c).
If $G$ were finite then so would be the matrix group generated by $A$ and $B$. 
A: For a), as $G$ is generated by $\{a,b\}$, all elements of $G$ can be written in terms of $a,b$.
In particular, if $\phi$ is a group homomorphism and we know $\phi(a),\phi(b)$ then, using $\phi(xy)=\phi(x)\phi(y)$ for all $x,y\in G$ we can compute $\phi(x)$ for all $x\in G$. 
Now if $G=\langle S|R\rangle$ you may have been told or you could prove that $\phi:S\to H$ for some group $H$ extends to a group homomorphism $\phi:G\to H$ if and only if $\phi(r)=1$ for each $r\in R$.
In this case you need to check that $\phi(a)\phi(b)\phi(a)^{-1}=\phi(b)^2$. 
For c), both $A$ and $B$ have infinite order, I think $B$ is slightly easier to prove. Compute $B^n$ for a few small $n$, conjecture what $B^n$ is in general, prove it and you'll be done. 
A: (a) To show that the given map is a homomorphism (in the category of groups), we have to show that relations go to relations. The given relation $aba^{-1}=b^2$ in the presentation of $G$ goes to the (possibly invalid) relation $ABA^{-1}=B^2$, and we compute for this:
$$
\begin{aligned}
AB&=
\begin{bmatrix}
2&\\&1
\end{bmatrix}
\begin{bmatrix}
1&1\\&1
\end{bmatrix}
=
\begin{bmatrix}
2&2\\&1
\end{bmatrix}
\\
B^2A
&=
\begin{bmatrix}
1&1\\&1
\end{bmatrix}^2
\begin{bmatrix}
2&\\&1
\end{bmatrix}
=
\begin{bmatrix}
1&2\\&1
\end{bmatrix}
\begin{bmatrix}
2&\\&1
\end{bmatrix}
=
\begin{bmatrix}
2&2\\&1
\end{bmatrix}
\ ,
\end{aligned}
$$
so we have indeed $AB=B^2A$, equivalently $ABA^{-1}=B^2$.
(b) Induction. The relation in (a) is the beginning. Fix an $n\ge 1$. Assume the relation $A^nBA^{-n}=B^{2^n}$ is true for this $n$. Then:
$$
\begin{aligned}
A^{n+1}BA^{-(n+1)}
&=
A\ A^nBA^{-n}\ A^{-1}
\\
&=A\ B^{2^n}\ A^{-1} \text{ by assumption}
\\
&=(ABA^{-1})^{2n}
\\
&=(B^2)^{2^n}
\\
&=B^{2^{n+1}}
\ .
\end{aligned}
$$
This shows (b) inductively.
(c) The composition of the representation $\phi$ with the determinant morphism gives a map $G\to \Bbb Q^{\times}$, the image of this morphism is the infinite group generated by $2$, so $G$ is infinite. (Or observe that either $A$, or $B$ has infinite order, by letting the matrices act as usual on the rational or real plane, or on the complex plane using the homographic action $\begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot z = (az+b)/(cz+d)$. By this last actions, $B$ corresponds to the translation by one.) 
