$(\mathbb{R}^{3},\times)$ is a Lie algebra that does not have Lie subalgebras of dimension 2.

Question

Could anyone give me a suggestion to start solving this problem?

Proof that $(\mathbb{R}^{3},\times)$ is a Lie algebra that does not have Lie subalgebras of dimension 2.

Answer 1

Given any two linearly independent vectors in $\mathbb{R}^3$, their cross product will be perpendicular to the plane that they generate. In particular, no $2$-dimensional subspace can be closed under the Lie bracket.

Answer 2

Just find a basis $(x,y,z)$ for the cross product algebra, which is the $3$-dimensional Lie algebra $\mathfrak{so}(3)$, e.g., with brackets $$ [x,y]=z,\; [y,z]=x,\; [z,x]=y, $$ see this question. Now it is obvious that there is no $2$-dimensional subalgebra - see also this duplicate.