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I read in book written by Karin Erdmann and Mark J. Wildon's "Introduction to Lie algebras" "Let F be in any field. Up to isomorphism, there is a unique two-dimensional nonabelian Lie algebra over F. This Lie algebra has a basis {x, y} such that its Lie bracket is defined by [x, y] = x"

How to prove that Lie bracket [x,y] = x satisfies axioms of Lie algebra such that [a,a] = 0 for $a \in L$ and satisfies jacobi identity and can some one give me an example of two dimensional nonabelian Lie algebra

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    This is called a one-dimensional affine algebra. You may consider $x=d/dz$ and $y=z d/dz$. Then $\exp(ay) \exp(bx) z= \exp(a) ~z+b$.2017-04-04

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By linearity, it is enough to check the Jacobi identity on the basis elements. When there are repetitions on the Jacobi identity it is satisfied automatically. Therefore you have to check nothing!

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    For future readers, see http://math.stackexchange.com/questions/24601/classsifying-1-and-2-dimensional-algebras-up-to-isomorphism . $[x, y] = x$ is a bit misleading since you really need $[x, y] = z$ for some $z$ independent of $x, y$.2015-07-27
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In $2$ dimensional case, we have $[x,y]=0$ or $[x,y]=z=ax+by$ if $a$ is zero then by changing the variables you get what you looking for but if $a$ was not zero then divide both sides by $a$ so that it becomes $[x,y/a]=x+by/a$ now change the $x+by/a$ variable to $z$. $[ z-by/a,y/a]=[z,y/a]=z$ then change $y/a$ to $u$ so that you get $[z,u]=z$