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If I have two abelian Lie Algebra $L_{1} $ and $L_2$, then they are isomorphic if and only if they have the same dimension. I would a example of two Lie algebras(not abelian) that have the same dimension but they are not isomorphic.

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  • $\begingroup$ Welcome to our site! $\endgroup$ – kjetil b halvorsen Oct 4 '15 at 16:14
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If you are happy with one of them being abelian, take an abelian one, and a non-abelian one.

For instance, in dimension $2$, take $L_{1}$ to have a basis $a, b$, and $[a, b] = b$.

If you want both of them to be non-abelian, go to dimension $3$, and take $L_{1}$ to have a basis $a, b, c$ with $$ [a, b] = b, [a, c] = [b, c] = 0 $$ and $L_{2}$ to have a basis $a, b, c$ with $$ [a, b] = c, [a, c] = [b, c] = 0. $$ The two algebras are not isomorphic, because in the second one every commutator $[[x, y], z]$ is zero, which is not the case with the first one.

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An interesting example consists of the two real $3$-dimensional Lie algebras $\mathfrak{so}(3)$ and $\mathfrak{sl}(2)$. Both are of dimension $3$, and both are simple. However, the second algebra has a subalgebra of dimension $2$, which is not the case for the first one. Hence they cannot be isomorphic. For more details see also this MSE equation.

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