How to get a Presentation of a Group $\newcommand{\R}{\mathbf R}$ 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.
 A: 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 
$$\langle g,h\ |\ h^{-1}ghg=e\rangle.$$
