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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?

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Do you know which matrices are supposed to map to the generators? Is $\Gamma(2)$ meant to contain only the invertible matrices of the given form? –  Henning Makholm Feb 7 '12 at 0:40
    
@HenningMakholm: Yes, I believe so. –  Libertron Feb 7 '12 at 0:45
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I don't have time now to work out the details, but the proof that I know starts by showing that the group ${\rm PSL}(2,{\mathbb Z})$ is isomorphic to the free product of $C_2 * C_3$ of cyclic groups of orders 2 and 3. The proof of that uses the so-called "ping-pong lemma", which you can google. Then you can identify your group as a subgroup of small index (I think it's 6) in ${\rm PSL}(2,{\mathbb Z})$ and thus show that it is isomorphic to $F_2$. –  Derek Holt Feb 7 '12 at 1:37
    
Presumably you should add the condition that $ad-bc = 1$. Otherwise, the matrix $\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}$ will be an order-two element of your group. –  Jim Belk Feb 7 '12 at 2:27
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Why is there a bounty here? Derek Holt's comment almost completely answers this question. Can you clarify what about it you don't understand? –  user641 Feb 9 '12 at 23:51
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1 Answer 1

up vote 3 down vote accepted
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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:

  • $c=0$. We know $ad-bc=1$, and so in this case $a=d=\pm 1$. But then $Y$ (or $YC$) must be a power of $B$, since $$ B^n=\begin{pmatrix} 1 & 2n\\ 0 & 1\end{pmatrix}.$$ This means $Y\in\mathfrak{H}\cap\mathfrak{H}X$, so that $\mathfrak{H}=\mathfrak{H}X$, or $X\in \mathfrak{H}$.
  • $c\neq0$, and $|a| > |c|$. Then there exists an $n\in\mathbb{Z}$ such that $-|c| < a+2nc < |c|$ [strict inequality because $a$ is odd and $c$ is even], and then $$ B^nY=\begin{pmatrix} 1 & 2n\\ 0 & 1\end{pmatrix}\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix} a+2nc & b+2nd\\ c & d\end{pmatrix},$$ so that $f(B^nY)=|a+2nc| + |c| < |c| + |c| < |a| + |c|$, contradicting the choice of $Y$. In other words, this case does not happen.
  • $c\neq 0$, and $|a| < |c|$. Then a similar argument to the one above, using $A$ instead of $B$, leads also to a contradiction. The proof is now complete.

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 $A^D=A^{-1}, B^D=B^{-1}$.

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