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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.

Ja_1941
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  • I do not agree with the other answer that this is a dead end. Indeed I believe the bold part is true, but I don't know how to prove it either. I will see if I can find a reference – Vincent Nov 22 '22 at 09:47
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    Okay, under closer scrutiny it seems that while some version of the bold statement is true even for much more general Lie groups than just $GL(n, \mathbb{C})$, any proof necessarily exploits an argument of the type given in the answer of Trevor so I do I agree with that answer that that is the best way to approach this. A general principle in dealing with paths in Lie group is that it is wise to first check how far you can get by looking at some very special paths: the paths through the origin that locally look like subsets of subgroups isomorphic to $\mathbb{R}$. Ctd in next comment – Vincent Nov 22 '22 at 10:11
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    Now in these paths, the intuitive notion of what 'continuous' means (able to draw without lifting your pencil; for every two points there is one in between etc) almost seamlessly translates to the notion of divisible group. Conversely, the definition of divisible group has the 'discreteness' we find at the other side of $f$ already baked in. So Trevor's argument is the perfect way to build the bridge between the ingredients you are dealing with here. – Vincent Nov 22 '22 at 10:13

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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 $f(y) = f(x^n) = f(x)^n = 1$.

Trevor Gunn
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