So what is the definition of a 0-dimensional manifold?

Hi Giuseppe, but $H$ is not necessarily a topological manifold. So are you saying that $H$ is a $0$-dimensional manifold so therefore discrete?

Very good point GH, this is the worst case namely when $N\sim\prod\limits_{p\leq\log(N)}p$. Then one uses the fact that $\prod_{p\leq x}(1-1/p)\sim\frac{1}{\log(x)}$. So we don't quite get a constant but $c>>\frac{1}{\log\log(x)}$ is good enough!

Hi GH, the key point is to notice that for $2p-N\leq k\leqp-1$ and $3p-N\leq k\leq N$ that $p$ divides $\binom{N+k}{k}$. Then one uses simple manipulations and congruences modulo $p$ and Wilson's theorem at one place and that's it. But the sketch of proof that I just explained does not say much about the residue class modulo $p^2$.

Yes exactly Todd

Well connected implies that every open neighboorhood which contains $1$ generates the whole group. For the converse, say that your group is not connected then the connected component of the identity, say $G^{0}$, is closed, ah ok, but may be not open. So do you have such an example at hand?

And this result is rarely applicable directly. nevertheless, assuming the connectedness you can use it to show the surjectivity of the exponential map $exp:\mathbf{C}[A]\rightarrow \mathbf{C}[A]^{\times}$ where $A$ is an $n$ by $n$ matrix in $\mathbf{C}$. Note that this implies in particular the surjectivity of the exponential map $exp: M_n(\mathbf{C})\rightarrow GL_n(\mathbf{C})$!

Hi Neil, thanks a lot for your amazingly simple proof that $GL_n(\mathbf{C})$ is connected!