Does every index $p$ subgroup of $SL(2,\mathbb{Z}_p)$ contain the principal congruence subgroup $\Gamma(p)$?
Equivalently, must it be the preimage of an index $p$ subgroup of $SL(2,\mathbb{Z}/p\mathbb{Z})$?
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Sign up to join this communityDoes every index $p$ subgroup of $SL(2,\mathbb{Z}_p)$ contain the principal congruence subgroup $\Gamma(p)$?
Equivalently, must it be the preimage of an index $p$ subgroup of $SL(2,\mathbb{Z}/p\mathbb{Z})$?
Yes.
Let $H$ be the subgroup, and let $N$ be the normal closure. The index of $N$ in $\Gamma = \mathrm{SL}(2,\mathbb{Z}_p)$ has index dividing $p!$ which is not divisible by $p^2$. Hence either:
If $p > 2$, one can now apply the following observations:
The abelianization of $\Gamma(p)$ is $V = \Gamma(p)/\Gamma(p^2)$. (It's easy to write down topological generators of $\Gamma(p^2)$ using commutators when $p > 2$.)
As a module for $\Gamma/\Gamma(p) = \mathrm{SL}(2,\mathbb{F}_p)$ under conjugation, $V$ is the adjoint representation and is irreducible.
This gives a contradiction, since, if $N$ is normal, then $\Gamma(p)/(N \cap \Gamma(p))$ will be a proper quotient of $V$ under the action of $\mathrm{SL}(2,\mathbb{F}_p)$.
If $p = 2$, then you can work a little harder with explicit computations and use a similar argument (now the abelianization of $\Gamma(2)$ is something close to $\Gamma(2)/\Gamma(8)$) or instead simply make the stupid observation that $\mathrm{SL}(2,\mathbf{Z})$ is dense in $\Gamma$ and thus there is an injection from subgroups of $\Gamma$ (of any index) to subgroups of $\mathrm{SL}(2,\mathbf{Z})$ of the same index (not a bijection because of the failure of the congruence subgroup property). However, the latter is well known to have abelianization $\mathbf{Z}/12 \mathbf{Z}$ and so has a unique index $2$ subgroup.