We define SO(2,1) as the stabilizer of the matrix diag(1,1,-1) in the action of GL(2) on the space of symmetric matrices, given by $g.M=gMg^t$. I know that SO(2) is maximal in SL(2) and I have tried assuming there is $SO(2,1)<H<SL(3)$ and looking at the intersection of H with a copy of SL(2), but it doesn't seem to help.
Could anyone suggest how to approach this problem? Thanks.
The main step should be looking at the Lie algebras. Via the inclusion $SO(2,1)\hookrightarrow SL(3,\mathbb R)$, the Lie algebra $\mathfrak{sl}(3,\mathbb R)$ becomes a representation of $SO(2,1)$. The Lie algebra $\mathfrak{so}(2,1)$ is an $SO(2,1)$-invariant subspace, which by general results admits an invariant complement. Denoting by $\mathbb R^3$ the standard representation of $SO(2,1)$ it is well known that $\mathfrak{so}(2,1)\cong\Lambda^2\mathbb R^3$ which easily implies that $\mathfrak{sl}(3,\mathbb R)\cong\mathfrak{so}(2,1)\oplus S^2_0(\mathbb R^3)$. Here the second summand is the tracefree part in the symmetric square of $\mathbb R^3$, which is an irreducible represetation of $\mathfrak{so}(2,1)$ and not isomorphic to $\mathfrak{so}(2,1)$. Now if $\mathfrak h\subset\mathfrak{sl}(3,\mathbb R)$ is a Lie subalgebra containing $\mathfrak{so}(2,1)$, then in particular, it is an $\mathfrak{so}(2,1)$-invariant subspace. By standard results in Lie theory, this implies that either $\mathfrak h=\mathfrak{so}(2,1)$ or $\mathfrak h=\mathfrak{sl}(3,\mathbb R)$. Since there is no proper subgroup of $SL(3,\mathbb R)$ with Lie algebra $\mathfrak{sl}(3,\mathbb R)$, we must have $\mathfrak h=\mathfrak{so}(2,1)$. So it only remains to show by elementary means that $SO(2,1)$ is the maximal subgroup of $SL(3,\mathbb R)$ with Lie algebra $\mathfrak{so}(2,1)$.