$GL(n, \mathbb{C})$ is not residually finite

Question

I saw in some text that $GL(n,\mathbb{C})$ is not residually finite since it is connected. I am not sure about this implication. I guess the proof works somehow like this: we endow $GL(n, \mathbb{C})$ with subspace topology from $\mathbb{C}^{n^2}$ and it is connected by How to show path-connectedness of $GL(n,\mathbb{C})$. For any $g\in GL(n,\mathbb{C})$, if $f:GL(n,\mathbb{C})\to H$ is a group homomorphism where $H$ is finite and $f(g)$ is nontrivial, then $f$ is automatically continuous if we endow $H$ discrete topology. Then $H$ is connected since $GL(n,\mathbb{C})$, which is absurd.

This is really my guess and I am not sure how to prove the bold part. Could anyone help me with this problem? Thank you.

Answer 1

I think that argument leads to a dead end. Instead, we use the following:

There are no non-trivial morphisms from a divisible group to a finite group.

This is because if $n$ is the order of the finite group and we write $y = x^n$ then $f(y) = f(x^n) = f(x)^n = 1$.