Suppose we denote the free group on two generators as $F_2$, which is the standard one used in proving the Banach-Tarski Paradox. Now let $\Gamma(2)$ be the group of integer matrices $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ that satisfy the condition $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right) \equiv \left( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right) \pmod{2}$. Finally, let $\Gamma(2)/T$ denote the quotient group of $\Gamma(2)$ by the central order $2$ subgroup generated by the matrix $\left( \begin{smallmatrix} -1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$ which I will denote by $T$. How can we show that $F_2 \cong \Gamma(2)/T$, i.e. these two groups are isomorphic? Apparently it's known, but I haven't found a proof for this in any text. Any suggestions?
[Math] How the free group on two generators $F_2$ is isomorphic to a particular quotient group
group-theory
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If $F_3$ were a quotient of $F_2$, then $\mathbb{Z}^3$ would be, but $\mathbb{Z}^3$ cannot be generated by fewer than $3$ elements. To me it seems easier to see directly that $\mathbb{Z}^3$ needs at least $3$ generators than the corresponding statement for $F_3$, perhaps because it's easy to visualize.
The rank of a group is the smallest cardinality of a generating set. Here's a list of some facts about ranks of groups (including that the rank of $F_3$ is $3$) on Wikipedia.
For the "by induction" statement: Let $T$ be the set of matrices with entries $a_{ij}$ such that $a_{ij}\in\Bbb Z$ if $i+j$ is even and $a_{ij}\in\Bbb Z\sqrt 2$ if $i+j$ is odd, that is, of the form $$\begin{pmatrix}\Bbb Z&\Bbb Z\sqrt2&\Bbb Z\\\Bbb Z\sqrt2&\Bbb Z&\Bbb Z\sqrt2\\\Bbb Z&\Bbb Z\sqrt2&\Bbb Z\end{pmatrix}.$$ Then $T$ forms a subring of $M(3,\Bbb R)$. Clearly $I,0\in T$ and $T$ is closed under addition. To show that $T$ is closed under multiplication, suppose $A,B\in T$, with elements $(a_{ij}),(b_{ij})$. Then the product is $c_{ij}=\sum_{k=1}^3a_{ik}b_{kj}$. If $i+j$ is even, then every element in the sum is either $a_{ik}b_{kj}\in\Bbb Z\cdot \Bbb Z$ if $k$ is even or $a_{ik}b_{kj}\in\sqrt 2\Bbb Z\cdot \sqrt 2\Bbb Z=2\Bbb Z$ if $k$ is odd, so the sum is also in $\Bbb Z$, or else $i+j$ is odd, in which case one of the two terms is in $\Bbb Z$ and the other is in $\sqrt 2\Bbb Z$ hence the product is in $\sqrt 2\Bbb Z$ and the sum of these is also in $\sqrt 2\Bbb Z$. Thus $AB\in T$.
Note also that $$3a=\begin{pmatrix}3&0&0\\0&1&-2\sqrt2\\0&2\sqrt2&1\end{pmatrix}\in T\qquad3b=\begin{pmatrix}1&-2\sqrt2&0\\2\sqrt2&1&0\\0&0&3\end{pmatrix}\in T,$$ so for any product $M$ of $n$ terms selected from $\{a,b,a^{-1},b^{-1}\}$, we have $3^nM\in T$, and since applying this to $v=(1,0,0)$ extracts the top row, we have $3^nMv=(i,j\sqrt 2,k)$ for some $i,j,k\in\Bbb Z$ as desired.
"Analyzing modulo 3": Consider taking the coefficients of the matrix $\bmod 3$ (treated as elements of $\Bbb Z[\sqrt2]$). This has the effect of simply reducing the coefficient of an element of the form $\Bbb Z\sqrt2$ (there are no mixed terms $a+b\sqrt2$). Dropping the $\sqrt2$ from the $a_{ij}$, $i+j$ odd entries and denoting $-1$ as $\bar 1$, we get:
$$[3a]=\begin{pmatrix}0&0&0\\0&1&1\\0&\bar 1&1\end{pmatrix}\quad [3a^{-1}]=\begin{pmatrix}0&0&0\\0&1&\bar 1\\0&1&1\end{pmatrix}\quad [3b]=\begin{pmatrix}1&1&0\\\bar 1&1&0\\0&0&0\end{pmatrix}\quad [3b^{-1}]=\begin{pmatrix}1&\bar 1&0\\1&1&0\\0&0&0\end{pmatrix}$$
These matrices share the common pattern of having a $2\times2$ block of $1$'s, with a single $\bar 1$ in the block and $0$ outside. Now consider the following four matrices:
$$A=[3a]=\begin{pmatrix}0&0&0\\0&1&1\\0&\bar 1&1\end{pmatrix}\quad B=\begin{pmatrix}0&0&0\\0&1&1\\0&1&\bar 1\end{pmatrix}\quad C=\begin{pmatrix}0&1&\bar 1\\0&1&1\\0&0&0\end{pmatrix}\quad D=\begin{pmatrix}0&\bar 1&1\\0&1&1\\0&0&0\end{pmatrix}$$
It turns out that matrices of this form and their negatives are closed under left multiplication by the generators, which constitutes a proof of the goal because the identity matrix is not of this form. Here is the multiplication table:
\begin{array}{c|cccc}\times&A&B&C&D\\\hline [3a]=A&-A&0&A&A\\ [3a^{-1}]&0&-B&B&B\\ [3b]&C&C&-C&0\\ [3b^{-1}]&D&D&0&-D \end{array}
The $0$ elements are because $[3a^{-1}][3a]=[9I]$ is divisible by $3$ and so is equal to $0$, but can only occur if the word $M$ is not reduced (contains adjacent cancelling pairs) - for example $[3a]B=0$, but $\pm B$ only arises from a product $[3a^{-1}]x$ for some $x\in\{A,B,C,D\}$. This proves that if $M$ is any product of $n$ terms selected from $\{a,b,a^{-1},b^{-1}\}$ whose last term is $a$ and with no cancelling pairs, $[3^nM]\in\{\pm A,\pm B,\pm C,\pm D\}$, but $[3^nI]=0$. Thus $M\ne I$. Argumentation by symmetry (or with a different set of matrices) also establishes the result for words that end in $a^{-1},b,b^{-1}$, so we have that $M\ne I$ for any nontrivial word.
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For the rest of this post, let $$ A=\begin{pmatrix} 1 & 0\\ 2 & 1\end{pmatrix}, B=\begin{pmatrix} 1 & 2\\ 0 & 1\end{pmatrix}, C=\begin{pmatrix} -1 & 0\\ 0 & -1\end{pmatrix}, D=\begin{pmatrix} -1 & 0\\ 0 & 1\end{pmatrix}.$$
Note that all of $A$, $B$, $C$ and $D$ live in $\Gamma(2)$.
First, let's consider the case of $G=SL(2,\mathbb{Z}$), which is the case I think you wanted (so in your notation, we have the requirement that $ad-bc=1$). Note that in this case, $D\notin G$.
Proposition: $A$, $B$, and $C$ generate $\Gamma(2)$.
Proof: Define a mapping from $f:\ \Gamma(2)\rightarrow \mathbb{Z}^+$ by the formula $$ f:\ \begin{pmatrix} a & b\\ c & d\end{pmatrix}\mapsto |a|+|c|.$$
Let $\mathfrak{H}$ be the subgroup of $\Gamma(2)$ generated by $A$, $B$, and $C$, and let $X$ be an arbitrary element of $\Gamma(2)$. We will be done if we can show $X\in \mathfrak{H}$.
To this end, pick an element $Y\in \mathfrak{H}X$ [the right coset of $\mathfrak{H}$ containing $X$] for which $f(Y)$ is minimal.
Now letting $Y=\begin{pmatrix} a & b\\ c & d\end{pmatrix}$, consider the following cases:
Thus we see that if $Z=\langle C\rangle$, then $\Gamma(2)/Z$ is generated by $A$ and $B$. But the group generated by $A$ and $B$ is free, by the Ping-Pong Lemma.
I think this is enough for the question, but if you actually meant $G=GL(2,\mathbb{Z})$, one can still say a lot. In this case, $\Gamma(2)/Z$ is not free, but has $F_2$ as a subgroup of index 2. There are only a handful of groups which possess $F_2$ as a subgroup of index 2, and in this case one gets the isomorphism $\Gamma(2)/Z\cong F_2\rtimes C_2$, where $F_2=\langle A, B\rangle$ and $C_2=\langle D\rangle$, with $D$ acting via $A^D=A^{-1}, B^D=B^{-1}$.