Lie algebra of a connected simple algebraic group is simple and a simply connected algebraic group having the same Lie algebra

Let $G$ be a connected simple algebraic group over an algebraically closed field $C$. What I infer from this definition is that the defining polynomials of $G$ have coefficients in $C$ while $G$ may have points in $K^n$ where $K$ is a field extension of $C$. We denote the $C$-points in $G$ by $G_C$.

Now, we have a result that the Lie algebra of such a group $G$ is simple. ( I am still trying to figure out the proof using Milne's notes as I had to read the basics like what is actually meant by a Lie algebra of such a group.) Any help here will be appreciated.

Moreover, in a paper I am reading, I came across another fact that there exists a simply connected algebraic group $K$ such that $Lie(G) \cong Lie(K)$.

I have not seen the proof of this and welcome some useful suggestions so that I can atleast get an idea of the proof with some basic algebraic geometry.

I had the following doubts :

1. Is that algebraic group $K$ which we claim to exist via the above result defined over $C$ (I mean its defining equations have coefficients in $C$)?

2. Further, they say that $G$ is isomorphic to $K/H$ where $H$ is a normal algebraic subgroup of $K$. I am not able to realise this conclusion.

3. Since $K$ is simply connected, $K_C$ is also simply connected. How to realise this ?

• 1) Yes. This is not trivial, but any simply connected simple algebraic group is definable over the prime field. 2) (because of the use of Lie algebras, we should probably assume characteristic zero) ... and $H$ is also finite and central. 3) this is senseless. This makes sense if $C=\mathbf{C}$, the complex numbers, and is a nontrivial theorem (for algebraic groups, simply connected in the algebraic sense implies simply connected for complex points). – YCor Feb 12 '16 at 0:23