Let $G$ be a connected Lie Group ($\mathrm{dim}G>1$) such that $Z(G) = \{1\}$ (center of $G$) and $\mathrm{Lie}(G) = \mathfrak g$ is a simple Lie Algebra. How do I prove that these conditions imply $G$ simple (as a group)?
Can anyone help me?
Some Ideas
If $H< G$ is a normal group of $G$, then $\overline{H}$ is a Lie Group (since $\overline{H}$ is closed), then $\mathrm{Lie}\left(\overline{H}\right)$ is a ideal of $\mathfrak{g}$, implying $\mathrm{Lie}\left(\overline{H}\right) = \{0\}$ or $\mathrm{Lie}\left(\overline{H}\right) = \mathfrak{g}$, if $\mathrm{Lie}\left(\overline{H}\right) = \{0\}$ then $H$ is discrete normal set $\Rightarrow$ $H\subset Z(G) = \{1\}$. On the other hand, if $\mathrm{Lie}\left(\overline{H}\right) = \mathfrak{g}$, $H$ is dense in $G$, and I don't know how to proceed.
Moreover, since $\mathfrak{g}$ is simple $\mathfrak g' = [\mathfrak g,\mathfrak g] =g$, which implies $G' = [G,G] = G$, then
\begin{align*}
\det \mathrm{Ad}(g) &= \det \mathrm{Ad}([g_1,h_1]\cdot \ldots \cdot[g_k,n_k])\\
&=[\det(\mathrm{Ad}(g_1),\det\mathrm{Ad}(h_1)]\cdot \ldots\cdot[\det(\mathrm{Ad}(g_k),\det \mathrm{Ad}(h_k)]\\
&= 1\cdot\ldots\cdot 1 = 1,
\end{align*}
then $\det \mathrm{Ad}(g)=1$, $\forall$ $g\in G$.
Note
$$\rho : G\to SL(n,\mathbb{R}) $$
$$g\mapsto \mathrm{Ad}(g) $$
is a homomorphism such that $\mathrm{Ker}(\mathrm{Ad})= Z(G)={1}$. Then $\rho$ is a embedding in of $G$ in $SL(n,\mathbb{R})$, however I do not know, again, how to proceed.
Finally, if $G$ is simply connected then $H$ normal and connected $\Rightarrow$ $H$ closed $\Rightarrow$ $H$ is lie group $\Rightarrow$ $\mathrm{Lie}(H)$ is a ideal of $\mathfrak{g}$, then $H=\{1\}$ or $H = G$ (using the same argument of the beginning). If $G$ is not simply connected I don't know what to do.
Best Answer
Let $\tilde{G}$ be the unique simply-connected Lie Group such that $\mathrm{Lie}(\tilde G) = \mathfrak g$. Then there exists a central discrete subgroup $\Gamma \subset \tilde{G}$, satisfying $$\tilde G /\Gamma \cong G, $$ consider the projection map $$\pi : \tilde{G} \to \tilde{G}/\Gamma \cong G, $$ it is worth mentioning that $Z(\tilde{G}) = \Gamma,$ because $\pi(Z(\tilde G)) \subset Z(G) = \{1\}$, then $Z(\tilde{G})\subset \Gamma \subset Z(\tilde{G}).$
Now, let $H$ be a normal subgroup of $G$. Note that $\tilde{H}:=\pi^{-1}(H)\triangleleft \tilde{G}$, denote $\tilde{H}_0$ the connected component of $1$ contained in $\tilde{H}$, note that $\tilde{H}_0\triangleleft \tilde{G}$, since $\tilde{G}$ is $1$-connected $\Rightarrow$ $\tilde{H}_0$ is closed in $\tilde{G}$ $\Rightarrow$ $\tilde{H}_0$ is a Lie subgroup of $\tilde{G}.$ Define $\mathfrak{h}:= \mathrm{Lie}(\tilde{H}_0)\subset \mathfrak g$, since $\tilde{H}_0 \triangleleft \tilde{G}$ $\Rightarrow$ $\mathfrak{h}$ is an ideal of $\mathfrak{g}$, since $\mathfrak{g}$ is a simple lie algebra then $\mathfrak{h} = \{0\}$ (if $\mathfrak{h} = \mathfrak g$ we would imply $\tilde{H}_0 = \tilde{G}$) and then $\tilde H$ is completely disconnected.
Let $h$ be an element of $\tilde{H}$, then the map \begin{align} F:\tilde G&\to \tilde H\\ g&\mapsto ghg^{-1} \end{align} is continous and $\tilde{G}$ is connected $\Rightarrow$ $F(G) = \{h\}$ (image of connected space by a continous function is connected, and $\tilde{H}$ is a completely disconnected space) and therefore $g h g^{-1} = h$, $\forall$ $g\in \tilde{G}$ $\Rightarrow $ $h\in \Gamma = Z(\tilde{G})$ $\Rightarrow$ $\Gamma \subset \tilde{H}\subset \Gamma $. Then $H = \pi(\tilde{H}) = \pi (\Gamma) \subset Z(G) =\{1\}$.
Q.E.D.