# Normal closure of $e_{12}$ in the congruence subgroup $\Gamma_1(p)\subset \mathrm{SL}_2(\mathbb{Z})$

$$\DeclareMathOperator\SL{SL}$$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 $$p$$.

• 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? Commented Apr 11 at 1:24
• 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) Commented Apr 11 at 1:44
• @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!
– Max
Commented Apr 11 at 4:40
• @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 $p\ge7$.
– Max
Commented Apr 11 at 4:45
• @Max: Happy to help! Commented Apr 11 at 12:01

$$\DeclareMathOperator{\SL}{SL} \newcommand\Z{\mathbb{Z}} \newcommand\bbH{\mathbb{H}}$$ 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 $$p \geq 7$$.