The maximal compact subgroup of the nilradical

Question

Let $G$ be a connected (complex) Lie group which is not abelian nor semisimple. Let $Z$ be the center of $G$ and $Z^0$ be the connected component of the identity. Suppose that $Z^0=\{e\}$.

Let $N$ be the nilradical of $G$ and $K$ be the maximal compact subgroup of $N$. Consider the adjoint representation $Ad:N\rightarrow GL(\mathfrak n)$ where $\frak n$ is the Lie algebra of $N$.

1. How to show that $Ad(K)$ is unitary? And why does that imply $K$ is contained in the center of $N$?
2. How to show that $K$ is normal subgroup of $G$
3. If $K=\{e\}$ then why this implies that $N$ is simply-connected?