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I am trying to find all 1- or 2- dimensional Lie Algebras "a" up to isomorphism. This is what I have so far:

If a is 1-dimensional, then every vector (and therefore every tangent vector field) is of the form $cX$. Then , by anti-symmetry, and bilinearity:

$$[X,cX]=c[X,X]= -c[X,X]==0$$

I think this forces a unique Lie algebra because Lie algebra isomorphisms preserve the bracket. I also know Reals $\mathbb{R}$ are the only 1-dimensional Lie group, so its Lie algebra ($\mathbb{R}$ also) is also 1-dimensional. How can I show that every other 1-dimensional algebra is isomorphic to this one? Do I use preservation of bracket?

For 2 dimensions, I am trying to use the fact that the dimension of the Lie algebra g of a group $G$ is the same as the dimension of the ambient group/manifold $G$. I know that all surfaces (i.e., groups of dimension 2) can be classified as products of spheres and Tori, and I think the only 2-dimensional Lie group is $S^1\times S^1$, but I am not sure every Lie algebra can be realized as the Lie algebra of a Lie group ( I think this is true in the finite-dimensional case, but I am not sure).

I know there is a result out there that I cannot yet prove that all 1- and 2-dimensional Lie algebras are isomorphic to Lie subalgebras of $GL(2,\mathbb{R})$ (using matrix multiplication, of course); would someone suggest how to show this last? Thanks.

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    Check out the early chapters of http://www.amazon.com/Introduction-Algebras-Springer-Undergraduate-Mathematics/dp/18462804002011-03-02
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    I don't have time to answer the full question, but I don't think your assertion that R is the only Lie group of dimension 1 is true. Consider $S^1$ with its usual group operation (multiplication of complex numbers) and its standard manifold structure. In fact, you can show that when a one-dimension Lie group is compact, it must be $S^1$; when it's noncompact, it's R.2011-03-02
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    As I read the rest of your question, it's not true either that the torus is the only Lie group of dimension 2. Again, stuff like $R^2$ or $R x S^1$ works as well. The so-called classification theorem only works for compact manifolds.2011-03-03

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