Three dimensional Lie algebra L with dim L' = 1

[valtz]

Now suppose the derived algebra has dimension 1. Then there exits some non-zero X_{1} \in g such that L' = span{X_{1}}. Extend this to a basis {X_{1};X_{2};X_{3}} for g. Then there exist scalars$\alpha, \beta , \gamma \in R (not all zero) such that
[X_{1},X_{2}] = \alpha X_{1}
[X_{1},X_{3}] = \beta X_{1}
[X_{2},X_{3}] = \gamma X_{1}
Suppose $\alpha \neq 0$. Then construct a new basis, as follows:
e_{1} = X_{1}
e_{2} = \frac{1}{\alpha} X_{2}
e_{3} = \alpha X_{3} - \beta X_{2} + \gamma X_{1}
Since $\alpha \neq 0$, by assumption, this is a basis for the Lie algebra g. Let us
calculate the Lie brackets for this basis:
[e_{1},e_{2}] = e_{1}
[e_{1},e_{2}] = 0
[e_{1},e_{2}] = 0
This Lie algebra is seen to be the direct sum of two Lie algebras, two dimensional non abelian lie algebra \oplus 1 dimensional lie algebra

i get this from pdf in internet , i want to ask, how can that be the direct sum of two lie algebra , two dimensional non abelian lie algebra \oplus 1 dimensional lie algebra

what use from construct new basis here? what can we see from construct new basis here?
i'm sorry before, please give me a detail explanation, because my essay is about lie algebra. so I'm a little new about lie algebra

source pdf where i get : http://math.ucsd.edu/~abowers/downloads/survey/3d_Lie_alg_classify.pdf

 

[fresh_42]

We have the Heisenberg algebra spanned by $\{\,e_1,e_2,e_3\,\}$ here. It is $\mathfrak{g}= \mathfrak{Z(g)} \oplus [\mathfrak{g},\mathfrak{g}]$. Since both, center and derived Lie algebra are ideals, it is a direct sum of Lie algebras, not only vector spaces. The two dimensional part, the center $\mathfrak{Z(g)}=\langle e_2,e_3 \rangle$ is Abelian.

Another way to see it is to observe that $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ is always Abelian. Since $[\mathfrak{g},\mathfrak{g}]$ is one dimensional and $\mathfrak{g}$ three dimensional, $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ is two dimensional and Abelian.