Is the congruence group $\Gamma(2)$ generated by the upper triangular matrix $(1, 2; 0, 1)$ and the lower triangular matrix $(1, 0; 2, 1)$ or does on need to also throw in the negation of the identity? To be specific, how do I check that the negation of the identity is not a word in the above matrices?
Yes, you need to throw in $I$. Check that the set of all matrices of the form $$\left(\begin{matrix} a&b\\\ c&d \end{matrix}\right)$$ with $b$ and $c$ even and $a\equiv d\equiv1$ (mod $4$) is a subgroup of the modular group.

$\begingroup$ It's Exercise 6 on p. 34 in M. Yoshida's "Hypergeometric functionsmy love": the group $\Gamma(2)$ is generated by the $[1,2;0,1]$, $[1,0;2,1]$ and... $[1,0;0,1]$. $\endgroup$ – Wadim Zudilin Jun 27 '10 at 12:12

1$\begingroup$ And Robin explains why the minus identity is not in the group generated by two others. $\endgroup$ – Wadim Zudilin Jun 27 '10 at 12:54

2$\begingroup$ Wadim, exercise 6 does not answer the question. $\endgroup$ – Chris Judge Jun 27 '10 at 17:01
There is already an answer posted, but I can't resist making two remarks. The first gives an alternate proof that also works for $\Gamma_n(p)$ for all $n$ and $p$ (and also gives a minimal generating set for these groups, at least when $n \geq 3$). The second says a little more about $\Gamma_2(2)$. By the way, $p$ doesn't have to be prime.
1) Let us define a surjective homomorphism $f : \Gamma_n(p) \rightarrow \mathfrak{sl}_n(\mathbb{Z}/p\mathbb{Z})$. An element $M \in \Gamma_n(p)$ is of the form $M = \mathbb{I}_n + p A$ for some matrix $A$. Define $f(M) = A$ mod $p$. Amazingly enough, this is a homomorphism! Indeed, if $N = \mathbb{I}_n + p B$, then $$f(MN) = f((\mathbb{I}_n + p A)(\mathbb{I}_n + p B)) = f(\mathbb{I}_n + p(A+B) + p^2 AB) = A+B$$ modulo $p$. This is sort of like a derivative! It is an easy exercise to check that the image of $f$ lies in $\mathfrak{sl}_n(\mathbb{Z}/p\mathbb{Z})$.
To check that $f$ is surjective, let $e_{ij}$ for $i \neq j$ be the identity matrix with a $1$ inserted into the $(i,j)$ position. Then $f(e_{ij}^p)$ is the matrix with a $1$ in the $(i,j)$ position and zeros elsewhere. To get the diagonal matrices, define $f_i$ for $1 \leq i < n$ to be the result of inserting the 2x2 matrix $(1+p,p;p,1p)$ into the identity matrix with its upper left entry at position $(i,i)$. Then $f(f_i)$ is the matrix with a $1$ at positions $(i,i)$ and $(i,i+1)$, a $1$ at positions $(i+1,1)$ and $(i+1,i+1)$, and zeros elsewhere.
The existence of $f$ implies immediately that $\Gamma_n(p)$ is not generated by the elementary matrices $e_{ij}^p$. A theorem of Lee and Szczarba says that in fact $f$ gives the abelianization of $\Gamma_n(p)$ for $n \geq 3$. Thus for $n \geq 3$ we have $[\Gamma_n(p),\Gamma_n(p)] = ker\ f = \Gamma_n(p^2)$. One can check (I've never seen this in print) that $\Gamma_n(p)$ is generated by the $e_{ij}^p$ and the $f_i$ when $n \geq 3$. For the case $n=2$, see the answers to my question here.
2) In fact, we have $\Gamma_2(2) \cong F_2 \times (\mathbb{Z}/2\mathbb{Z})$. Here $F_2$ is a rank $2$ free group generated by $e_{12}^2$ and $e_{21}^2$ and $\mathbb{Z}/2\mathbb{Z}$ is generated by the central element $(1,0;0,1)$. This can be proved in many ways : I leave it as a fun exercise!

$\begingroup$ Quite beautiful! Thank you for adding this. $\endgroup$ – Chris Judge Jun 27 '10 at 17:18
This follows from the fact that the image of $\Gamma(2)$ in $\text{PSL}_2(\mathbb{Z})$ is freely generated by the two matrices you describe. There is a geometric proof of this fact based on the fact that $\Gamma(2)$ acts properly discontinuously on the upper half plane $\mathbb{H}$ which I sketch here.