$\DeclareMathOperator\SL{SL}$For an odd prime $p$, let $$\Gamma_1(p)=\left\{\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in \SL_2(\mathbb{Z}):\begin{pmatrix}a & b \\ c & d\end{pmatrix}\equiv\begin{pmatrix}1 & * \\ 0 & 1\end{pmatrix} \mod (p)\right\}\subset \SL_2(\mathbb{Z})$$ be the congruence subgroup of level $p$. Let $e$ be the matrix $$\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}.$$

Question: What is the normal closure of $\{e\}$ in $\Gamma_1(p)$?

For $$A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in\Gamma_1(p),$$ we can see that $$AeA^{-1}=\begin{pmatrix}1-ac & a^2 \\ -c^2 & 1+ac\end{pmatrix}$$ lies in the subgroup $$\Gamma=\left\{\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in \SL_2(\mathbb{Z}):a\equiv d\equiv 1 \mod (p),\quad c\equiv 0 \mod \left(p^2\right)\right\}.$$ My guess is that the normal closure is exactly this subgroup $\Gamma$.

I have checked it for $p=3$, in which case $\Gamma$ is generated by $$\left\{\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix},\ \begin{pmatrix}4 & -1 \\ 9 & -2\end{pmatrix},\ \begin{pmatrix}7 & -4 \\ 9 & -5\end{pmatrix}\right\},$$ where $$\begin{pmatrix}4 & -1 \\ 9 & -2\end{pmatrix}=\begin{pmatrix}1 & 1 \\ 3 & 4\end{pmatrix}\begin{pmatrix}1 & -1 \\ 0 & 1\end{pmatrix}\begin{pmatrix}4 & -1 \\ -3 & 1\end{pmatrix}$$ and $$\begin{pmatrix}7 & -4 \\ 9 & -5\end{pmatrix}=\begin{pmatrix}-2 & 1 \\ -3 & 1\end{pmatrix}\begin{pmatrix}1 & -1 \\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & -1 \\ 3 & -2\end{pmatrix}$$ are conjugates of $e^{-1}$.

For $p=5$, I used Magma to obtain a set of generators of $\Gamma$, while not all of which are conjugates of $e^n$. However, it could still be possible that they are products of conjugates of $e^n$. I have no idea how it would be proven for general $p$.

  • $\begingroup$ A successive-approximation argument at least shows that the analogous normal closure with the $p$-adic integers $\mathbb Z_p$ in place of $\mathbb Z$ is what you'd like. Do you happen to know that the normal closure contains some congruence subgroup? $\endgroup$
    – LSpice
    Commented Apr 11 at 1:24
  • 4
    $\begingroup$ My guess is that the normal closure of this matrix is an infinite-index subgroup. The point here is that your subgroup of SL(2,Z) is the fundamental group of a finite cover of the modular curve, and I believe that once your prime p is large enough that cover will have positive genus. Conjugates of your matrix e will correspond to loops around cusps since e is parabolic, so at best your normal closure is the subgroup generated by the cusps. (sorry for the poor typesetting and explanation — I am writing this on a phone) $\endgroup$ Commented Apr 11 at 1:44
  • $\begingroup$ @AndyPutman You're absolutely correct! It turns out that the genus of the modular curve for $\Gamma$ is already nonzero when $p=7$ (it has genus 3). Thanks! $\endgroup$
    – Max
    Commented Apr 11 at 4:40
  • $\begingroup$ @LSpice I checked that the normal closure and $\Gamma$ do have the same image in $SL_2(\mathbb{Z}/p^n)$. It seems that the normal closure cannot be a congruence subgroup by Andy's argument when $p\ge7$. $\endgroup$
    – Max
    Commented Apr 11 at 4:45
  • $\begingroup$ @Max: Happy to help! $\endgroup$ Commented Apr 11 at 12:01

1 Answer 1


$\DeclareMathOperator{\SL}{SL} \newcommand\Z{\mathbb{Z}} \newcommand\bbH{\mathbb{H}}$ I'm at my office now, so I'll write a properly typeset version of my comment so this question will leave the "unanswered" queue.

Recall that the modular curve $Y$ is $\bbH^2/\SL_2(\Z)$. Topologically, $Y$ is a once-punctured $2$-sphere with two orbifold points. Any finite-index subgroup $\Gamma$ of $\SL_2(\Z)$ gives a finite orbifold cover $Y(\Gamma)$ of $Y$. For your subgroup $\Gamma_1(p)$, write $Y_1(p)$ for $Y(\Gamma_1(p))$.

The space $Y_1(p)$ is topologically a punctured surface of finite type. The punctures are the cusps. Your element $e$ is a parabolic element of $\SL_2(\Z)$, so considered as an element of the fundamental group of $Y_1(p)$ it corresponds to a loop that is homotopic into one of the cusps. The same is true for all conjugates of $e$. It follows that the normal closure of $e$ in $\Gamma_1(p)$ is contained in the subgroup of the fundamental group of $Y_1(p)$ generated by loops around the cusps. This is an infinite-index proper subgroup if $Y_1(p)$ has positive genus, which it does for $p \geq 7$.


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