What is the abelianization of $SL(2,\mathbb{Z}_p)$? What is the abelianization of the groups $SL_2(\mathbb{Z}_p)$ where $p$ is a prime number and $\mathbb{Z}_p$ denotes $p$-adic integers?
I'm guessing that it's $\mathbb{Z}/4$ for $p = 2$ and $\mathbb{Z}/3$ for $p = 3$, though I would like to see a reference or a proof.
 A: I'll assume that you are referring to the topological abelianization - that is, the quotient by the closure of the group-theoretic commutator subgroup.
The abelianization of $SL_2(\mathbb{Z})$ is $\mathbb{Z}/12$, and the commutator subgroup is a congruence subgroup of index 12 (e.g. see Theorem 3.8 of http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/SL(2,Z).pdf)
Thus, the abelianization map factors as:
$$SL_2(\mathbb{Z})\rightarrow SL_2(\widehat{\mathbb{Z}})\rightarrow\mathbb{Z}/12$$
where $\widehat{\mathbb{Z}}$ is the profinite completion. On the other hand, the (topological) commutator subgroup of $SL_2(\widehat{\mathbb{Z}})$ is a closed subgroup which certainly contains the commutator subgroup of $SL_2(\mathbb{Z})$. Since $SL_2(\mathbb{Z})$ is dense inside $SL_2(\widehat{\mathbb{Z}})$, the above factorization implies that the abelianization of $SL_2(\widehat{\mathbb{Z}})$ is also $\mathbb{Z}/12$.
Note that $SL_2(\widehat{\mathbb{Z}}) = \prod_p SL_2(\mathbb{Z}_p)$. It can be checked that there are surjective maps
$$SL_2(\mathbb{Z}_2)\rightarrow SL_2(\mathbb{Z}/4)\rightarrow\mathbb{Z}/4$$
and
$$SL_2(\mathbb{Z}_3)\rightarrow SL_2(\mathbb{Z}/3)\rightarrow\mathbb{Z}/3$$
Since the abelianization of a direct product is the direct product of abelianizations, since these two maps "generate" $\mathbb{Z}/12$, we find that $SL_2(\mathbb{Z}_p)$ must have trivial abelianization for $p\ne 2,3$, and the maps above are the abelianizations of $SL_2(\mathbb{Z}_2),SL_2(\mathbb{Z}_3)$.
