Is there a group between $SO(2,\mathbb{R})$ and $SL(2,\mathbb{R})$? Is there a non-trivial subgroup $H \subset SL(2,\mathbb{R})$ such that $H \supset SO(2,\mathbb{R})$ ?
My intuition is that, since $\dim SO(2)=1$ and $\dim SL(2)=3$, there should be some group between, but I can't point out one.
Note : in the complex case, $H:= SO(2,\mathbb{C}) \cup \left\{ \left( \matrix{ a&b \\ b&-a} \right) ~|~ a^2+b^2=-1 \right\}$ is an example.
 A: I think there's a fairly simple bare hands proof, even when you don't require $H$ to be a Lie group.
Consider the image of the unit circle under an element of $SL(2,\mathbb{R})$. This is always an ellipse, centred on the origin, where the product of the lengths of the major axis and minor axis is equal to $4$. The elements of $SO(2,\mathbb{R})$ are precisely those giving a circle (i.e., a major axis of length $2$).
If $H$ contains $SO(2,\mathbb{R})$ and also another element $A$ of $SL(2,\mathbb{R})$ which gives an ellipse with major axis $2a$, where $a>1$, then by pre-composing and post-composing $A$ with suitable rotations, we get all elements of $SL(2,\mathbb{R})$ giving an ellipse with major axis $2a$. So we can assume $A=\begin{pmatrix}a&0\\0&\frac{1}{a}\end{pmatrix}$.
Then $H$ also contains $AR_\theta A$ for every $\theta$, where $R_\theta$ is a rotation through an angle of $\theta$. For $\theta=0$ this element maps the unit circle to an ellipse with major axis $2a^2$, and for $\theta=\frac{\pi}{2}$ to a circle. By the Intermediate Value Theorem, for intermediate values of $\theta$ we can get any major axis length between $2$ and $2a^2$, and by the argument above, $H$ must then contain all elements of $SL(2,\mathbb{R})$ that give an ellipse with major axis between $2$ and $2a^2$.
Repeating, we can get any length of major axis.
A: Apparently not. There is a nontrivial 2 dimensional real Lie algebra, it can even be written as 2 by 2 matrices,
$$ 
\left(
\begin{array}{rr}
a & b \\
0 & 0
\end{array}
\right)
  $$
see http://en.wikipedia.org/wiki/Table_of_Lie_groups#Real_Lie_groups_and_their_algebras
I get that the exponential of this is
$$ 
\left(
\begin{array}{cc}
e^a & b \left( \frac{e^a - 1}{a} \right) \\
0 & 1
\end{array}
\right)
  $$
Note
$$  \frac{e^a - 1}{a} = 1 + \frac{a}{2}   + \frac{a^2}{6} + \cdots   $$
is analytic, and becomes $1$ when $a=0.$
The trouble is that you need, as a Lie algebra for something in between, matrices that are both trace-free and anti-trace free, as
$$ 
\left(
\begin{array}{rr}
a & b \\
-b & -a
\end{array}
\right)
  $$
which do not make a Lie algebra. 
Oh, the trivial Lie group of dimension 2 can be realized in $SO_4,$ maximal torus
