$\DeclareMathOperator\SL{SL}$Let $p$ be a prime, and let $\Gamma_r$ denote the kernel of the map $\SL_2(\mathbb{Z}_p)\rightarrow \SL_2(\mathbb{Z}_p/p^r\mathbb{Z}_p)$.
Explicit formulas with formal group laws shows that for odd $p$, the derived subgroup $\Gamma_r'$ of $\Gamma_r$ is just $\Gamma_{2r}$. However, it seems that for $p = 2$, $\Gamma_r'$ is strictly smaller than $\Gamma_{2r}$. What exactly is $\Gamma_r'$ in this case?
