Importance of continuity of Galois representations So for a one dimensional Galois representation $\rho: G_{\Bbb Q} \to \mathbb C^{\times}$, I know that it must factor through the abelianization of $G_{\Bbb Q}$, which by the Kronecker-Weber theorem is the Galois group of the maximal cyclotomic extension of $\Bbb Q$. I want to conclude from this that $\rho$ factors through a Dirichlet character, that is, a representation of the Galois group of some finite cyclotomic extension. I've seen this question: Complex Galois Representations are Finite
This definitely gives me the answer, but my question is, can I do this without assuming $\rho$ is continuous? In general how important is the continuity assumption when talking about Galois representations? I'm wondering if this is a purely algebraic fact or if it only applies to continuous representations. I would absolutely accept a reference in lieu of a written answer, surely this is written somewhere but I haven't been able to find it. 
 A: The comment by @Mariano Suárez-Alvarez "By definition, continuity is equivalent to factoring through a finite quotient." is completely wrong. In particular, there are uncountably many surjective homomorphisms from the Galois group of $\mathbf{Q}$ to a cyclic group of order $2$, but only countably many of them are continuous, namely the ones factoring through the Galois group of the coutably many quadratic fields. So even for characters of degree $2$, it is important to consider continuity.
Added: This (naturally) provides an answer to your question: Even for characters $G_{\mathbb{Q}} \rightarrow \mu_2 \subset \mathbb{C}^{\times}$ which have finite image, some continuity is required to deduce that one has a Dirichlet character (there are only countably many such characters).
BTW: The moderator who summarily deleted the previous version of this answer as "This does not provide an answer to the question" should refrain from making mathematical judgements on questions which they apparently have no mathematical background. It is honestly embarassing that an answer both correcting important misconceptions in the comments and also answering the actual question would be deleted.
A: As an abstract group, $\Bbb Z_p$ is $q$-divisible for any prime $q \neq p$ and uncountable, while no element is infinitely divisible by $p$, so it is a direct sum of uncountably many copies of $\Bbb Z_{(p)}$ (the localisation of $\Bbb Z$ at $p$).
There are uncountably many group morphisms $\Bbb Z_{(p)} \to \Bbb C^*$ (choose the value at $1$, then for each prime $q \neq p$, you have $q$ choices for the value at $q^{-1}$, again $q$ choices for the value at $q^{-2}$, and so on), so yes, there are many group morphisms $\Bbb Z_p \to \Bbb C^*$, and many group morphisms $\hat {\Bbb Z} \to \Bbb C^*$
A: A variant on mercio's answer is that $\mathbb Z_p/\mathbb Z$ is a uniquely divisible abelian group, i.e. a $\mathbb Q$-vector space.  It has cardinality, and hence dimension, equal to the continuum.   Thus it is isomorphic to $\mathbb C$ as a $\mathbb Q$-vector space.  Any such isomorphism has infinite order (obviously) and is also incredibly far from being continuous.
