Let $p$ be a prime and denote by $\mathbb{Z}_p^{\times}$ the group of $p$-adic units. Suppose that $\chi$ is a character $\chi: \mathbb{Z}_p^{\times} \rightarrow \mathbb{C}^{\times}$. Then it is well known that
$\mathbb{Z}_p^{\times} \simeq (\mathbb{Z}/ p \mathbb{Z})^{\times} \times (1+ p \mathbb{Z}_p) $.
Do we have that the kernel of $\chi$ contains $(1+ p \mathbb{Z}_p)$ ? Any help or reference would be appreciated.
YCor's example in the comments shows that you can have continuous characters with image an arbitrarily large finite group. In fact, there are even injective homomorphisms $\mathbb{Z}_p^\times \to \mathbb{C}^\times$ (if you don't ask for continuity): Observe that $\mathbb{C}^\times \cong \mathbb{Q}/\mathbb{Z}\times \bigoplus \mathbb{Q}$ as abstract abelian group, where $\bigoplus \mathbb{Q}$ is a $\mathbb{Q}$-vector space of continuum dimension.
For odd $p$, we have $\mathbb{Z}_p^\times \cong \mathbb{F}_p^\times \times \mathbb{Z}_p$. You can map the cyclic group $\mathbb{F}_p^\times$ injectively into the $\mathbb{Q}/\mathbb{Z}$ factor, and $\mathbb{Z}_p$ injectively into the continuum-dimensional $\mathbb{Q}$-vector space, since $\mathbb{Z}_p$ is torsion free and $\mathbb{Z}_p\otimes \mathbb{Q}$ is also a continuum-dimensional $\mathbb{Q}$-vector space.
For $p=2$, we abstractly have $\mathbb{Z}_2^\times \cong \{\pm 1\} \times \mathbb{Z}_2$, so the same argument works.
