Let $G \subseteq \mathrm{Sp}_{2g}(\mathbb{Z}_2)$ be a closed subgroup of the symplectic group over the $2$-adic integers whose image under the mod-$2$ homomorphism $\pi : \mathrm{Sp}_{2g}(\mathbb{Z}_2) \twoheadrightarrow \mathrm{Sp}_{2g}(\mathbb{F}_2)$ is isomorphic to the symmetric group $S_{2g + 1}$ (or, as a variant to this question, we could assume that the mod-$2$ image of $G$ is instead $S_{2g + 2}$). Is it known whether for $g \geq 2$ there exists such a $G$ which is *not* of finite index in $\mathrm{Sp}_{2g}(\mathbb{Z}_2)$, and if so, has an explicit example ever been given? (Note that Brumer and Kramer show as Theorem 2.1.1 in their article "Large $2$-adic Galois image and non-existence of certain abelian surfaces over $\mathbb{Q}$" that such a $G$ could not contain a transvection.)

I would be surprised if such a $G$ couldn't exist, because in that case I would expect someone to have proven this already; however, it would be nice to have a definitive answer one way or another. In the $g = 1$ case I'm able to come up with an explicit example of such a $G$, and in any case, we know it should exist as the $2$-adic Galois image associated to a CM elliptic curve given by a polynomial with full Galois group. For $g \geq 2$, I can show that there exists a $G \subseteq \mathrm{Sp}_{2g}(\mathbb{Z}_2)$ surjecting onto $S_{2g + 1}$ which does not contain $\mathrm{ker}(\pi)$ modulo $4$, but I'm finding it somewhat tricky to lift that to an infinite-index subgroup of $\mathrm{Sp}_{2g}(\mathbb{Z}_2)$.

notcontain $\mathrm{ker}(\pi)$, yet this $G$ contains a transvection and according to Theorem 2.1.1doescontain $\mathrm{ker}(\pi)$. The authors state that this case is well known and don't provide a source. $\endgroup$ – Jeff Yelton Jan 30 '18 at 16:58