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