Need help understanding how ID'ing subgroups of $\mathrm{GL}_{2}(\mathbb{C})$ works. Let $A=
\begin{bmatrix}
0&-1\\1&-1
\end{bmatrix}$ and
$B=
\begin{bmatrix}
-1&1\\0&1
\end{bmatrix}$.
Consider $H=\langle A, B \rangle \le \mathrm{GL}_{2}(\mathbb{C})$.
Which familiar subgroup is $H$ isomorphic to? The answer's $S_3$ and I want to know how to see this is so.
The explanation I was given was that it'll help to think of the elements in $S_3$ permuting the basis vectors $\{e_1, e_2, e_3\}$ of $\mathbb{R}^{3}$. I am familiar with how to do that on a surface level. For example, $(1\; 2)$ is the permutation such that
\begin{align*}
e_1 &\mapsto e_2\\
e_2&\mapsto e_1\\
e_3&\mapsto e_3
\end{align*}
Something else that was mentioned is that to get $H$ in its permutation representation we would need to focus on a specific subset/subspace of $\mathbb{R}^{3}$. Specifically, if we think about all elements of the form $x_1e_1+x_2e_2+x_3e_3$ but rather focus on the vectors where $x_1+x_2+x_3=0$, then we get $H$ if we use the basis $B=\{e_1-e_2, e_2-e_3\}$.
I understand how using the above we can see that $(1\;2)$ would then be represented (am I correctly using this word here?) by
\begin{align*}
e_1-e_2&\mapsto e_2-e_1 = -(e_1-e_2) \\
e_2-e_3& \mapsto e_1-e_3=(e_1-e_2)+(e_2-e_3)
\end{align*}
and similarly $(1\;3)$, say, would be represented by
\begin{align*}
e_1-e_2&\mapsto e_3-e_2=-(e_2-e_3)\\
e_2-e_3&\mapsto e_2-e_1=-(e_1-e_2)
\end{align*}

I'm confused as to why we focused on looking at vectors where their coefficients satisfy $x_1+x_2+x_3=0$? Also, I don't know how this still is supposed to prove $H \cong S_3$.
This isn't my homework or assignment.
Edit: I'll write down the elements in $H$ for convenience. These are: $I, A, B=B^{-1}$,
$AB=
\begin{bmatrix}
 0&-1\\-1&0
\end{bmatrix},
A^{-1}=
\begin{bmatrix}
 -1&1\\-1&0
\end{bmatrix}
A^{-1}B=
\begin{bmatrix}
 1&0\\1&-1
\end{bmatrix},
$
and then my question becomes: how do I pair each of these matrices with a permutation in $S_3$?
 A: The group $S_3$ has a natural representation on $\Bbb C^3$, namely, the group homomorphism $$\phi : S_3 \to \operatorname{GL}(\Bbb C^3), \qquad \sigma \cdot e_i := e_{\sigma(i)} ,$$ where $(e_i)$ is the standard basis of $\Bbb C$. This definition immediately implies that the matrix representations $[\phi(\sigma)]$ of $\phi(\sigma) \in \operatorname{GL}(3, \Bbb C)$ with respect to the $(e_i)$ are the corresponding permutation matrices.
Our representation $\phi$ in fact decomposes as a direct sum of two smaller representations:  For any element $v := a e_1 + b e_2 + c e_3 \in \Bbb C^3$, the sum of the coefficients of $\sigma \cdot v := a e_{\sigma(1)} + b e_{\sigma(2)} + c e_{\sigma(3)}$ is again $a + b + c$, hence the action of $S_3$ on $\Bbb C^3$ preserves the $2$-dimensional subspace $$\Bbb V := \{a e_1 + b e_2 + c e_3 : a + b + c = 0\} \subset \Bbb C^3 .$$ Put another way, $\phi$ restricts to a representation $S_3 \to \operatorname{GL}(\Bbb V)$, which by mild abuse of notation we also denote $\phi$. Fixing a basis of $\Bbb V$ then yields matrix representations $[\phi(\sigma)] \in \operatorname{GL}(2, \Bbb C)$. As you pointed out, choosing the basis $(e_1 - e_2, e_2 - e_3)$ of $\Bbb V$ yields $$[\phi(12)] = \pmatrix{-1&1\\\cdot&1} = B,$$ and similarly $$[\phi(123)] = \pmatrix{\cdot&-1\\1&-1} = A .$$
Since $A, B$ by definition generate $H$, $\phi$ restricts to an isomorphism $S_3 \to H$.
Remark The above assignment of permutations in $S_3$ to matrices in $H$ is not unique: For any $\tau \in S_3$, $\sigma \mapsto \tau \sigma \tau^{-1}$ is an automorphism of $S_3$ (in fall, all automorphisms of $S_3$ arise this way), and so $\sigma \mapsto \phi(\tau \sigma \tau^{-1})$ is another isomorphism $S_3 \to H$. Different isomorphisms correspond to different suitable choices of bases of $\Bbb V$.
A: For instance (and this is a slightly different  approach).  Note that $A$ has order three,  $B$ order two, and $ABA=B$. Thus we indeed have $S_3$, which has presentation $$\langle a,b\mid a^3,b^2, (ab)^2\rangle$$.  So
sending $A\mapsto (123)$ and $B\mapsto (12)$ determines an isomorphism (since $S_3=\langle (12),(123)\rangle $).  That's specifying that generators go to generators is sufficient to determine an isomorphism (which is in some sense what you're asking for).
As I said there are $5$ other ways to go (three possible targets for $B$, two $A$, total).
