Why is there no closed subgroup of $\mathrm{GL}(2, \mathbb{C})$ with Lie algebra $\frak{g}$? Why is there no closed subgroup of $\mathrm{GL}(2, \mathbb{C})$ with Lie algebra 
$$\mathfrak{g} = 
\left\lbrace
\begin{pmatrix} 
it & 0 \\
0  & i\alpha t
\end{pmatrix}: t\in\mathbb{R}
\right\rbrace ?$$
Where $\alpha$ is a irrational number
 A: There is if $\alpha\in\Bbb Q$.
If $\alpha\in\Bbb R\setminus\Bbb Q$ then the closure of the
group generated by $\exp(A)$
for $A\in\frak g$ is the set of diagonal matrices $\pmatrix{a&0\\0&d}$
with $|a|=|d|=1$. The Lie algebra of this subgroup is two-dimensional.
A: Check that $\frak{g}$ is a real subalgebra of $gl(2, \Bbb C)$ and $dim_{\Bbb R} \frak{g}=1$
If $\alpha \not \in \Bbb Q$
Suppose $\exists G$ , matrix Lie group s.t $Lie(G)=\frak{g}$.
Now $H_0=\left\lbrace e^{X_t}=
\begin{pmatrix} 
e^{it} & 1 \\
1  & e^{i\alpha t}
\end{pmatrix}: t\in\mathbb{R}
\right\rbrace \subseteq G$
This implies that $\bar H_0 \subseteq \bar G=G $[As $G$ is a matrix lie group, it is closed in $GL(2, \Bbb C)$]
And $\bar H_0=\left\lbrace
\begin{pmatrix} 
e^{it} & 1 \\
1  & e^{is}
\end{pmatrix}: s,t\in\mathbb{R}\right\rbrace$; $Lie(\bar H_0) \leq Lie(G)=\mathfrak g$ and $\mathfrak h_0=Lie(\bar H_0)=\left\lbrace 
\begin{pmatrix} 
e^{it} & 0 \\
0  & e^{is}
\end{pmatrix}: s,t\in\mathbb{R}\right\rbrace \subseteq \mathfrak g$ this implies $\dim_{\Bbb R} \mathfrak h_0=2$ a contradiction.
