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 functions---my 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 $(1+p,p;-p,1-p)$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.