A Lie group that has an immersion in $\mathrm{GL}(n,\Bbb R)$ but no embedding?

Question

Question: Is there a Lie group $G$ that admits a smooth immersion $$i:G\longrightarrow\mathrm{GL}(n,\Bbb R)$$ for some $n\in\Bbb N$, but no smooth embedding $$j:G\longrightarrow\mathrm{GL}(m,\Bbb R)$$ for any $m\in\Bbb N$? (Here, $i$ and $j$ are also required to be group homomorphisms.)

The usual example of a Lie group which is immersed but not embedded is the group $\Bbb R$ with the immersion $$i:\Bbb R\longrightarrow\mathrm{GL}(2,\Bbb C),\quad t\longmapsto\begin{pmatrix}e^{it} & 0 \\ 0 & e^{i\alpha t}\end{pmatrix}$$ for $\alpha$ irrational. The so-called dense curve on the torus. However, here $\Bbb R$ do embed in $\mathrm{GL}(n,\Bbb R)$, but just in a different way: $$j:\Bbb R\longrightarrow \mathrm{GL}(2,\Bbb R),\quad t\longmapsto\begin{pmatrix}1 & t \\ 0 & 1\end{pmatrix}.$$

There are also well-known groups that have no injective homomorphism into $\mathrm{GL}(n,\Bbb R)$, but then they do not immerse either.

Answer 1

HINT:

You could consider the simply connected covering of $SL(2,\mathbb{R})$. There are others too.

Answer 2

There are countable groups (hence, Lie groups) $G$ such that there exists a faithful linear representation of $G$ but there is no faithful linear embedding of $G$. For instance, the 2-step solvable group $$ G=BS(2,1)=< a,b: aba^{-1}=b^2> $$ admits an injective homomorphism to $SL(2,R)$ but every injective homomorphism $f: G\to GL(n,R)$ will have non-closed image and, hence, will not be an embedding. The reason for the latter is that a discrete ebmedding $f: G\to GL(n,R)$ will (up to conjugation) land in the group $B$ of upper triangular matrices, but every discrete subgroup of $B$ is polycyclic (G.D. Mostow, On the fundamental group of a homogeneous space, Ann. of Math. (2) 66 (1957), 249–255), while $BS(2,1)$ is not polycyclic.