Is there a non-trivial subgroup H⊂SL(2,R) such that H⊃SO(2,R) ?

My intuition is that, since dimSO(2)=1 and dimSL(2)=3, there should be some group between, but I can’t point out one.

Note : in the complex case, H:=SO(2,C)∪{(abb−a) | a2+b2=−1} is an example.

**Answer**

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,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,R) are precisely those giving a circle (i.e., a major axis of length 2).

If H contains SO(2,R) and also another element A of SL(2,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,R) giving an ellipse with major axis 2a. So we can assume A=(a001a).

Then H also contains ARθA for every θ, where Rθ is a rotation through an angle of θ. For θ=0 this element maps the unit circle to an ellipse with major axis 2a2, and for θ=π2 to a circle. By the Intermediate Value Theorem, for intermediate values of θ we can get any major axis length between 2 and 2a2, and by the argument above, H must then contain all elements of SL(2,R) that give an ellipse with major axis between 2 and 2a2.

Repeating, we can get any length of major axis.

**Attribution***Source : Link , Question Author : user10676 , Answer Author : Jeremy Rickard*