Simple 3-dim Lie algebra 
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*If $g$ is a Lie algebra, how to prove that $Tr(ad \ a)=0$ for all $a\in [g,g]$?

*In case $dim\ g=3$ and $[g,g]=g$ how to show that $g$ is simple? 
 A: *

*$\text{ad}([X, Y]) = [\text{ad}(X), \text{ad}(Y)]$, so from here it suffices to prove that $\text{tr}([M, N]) = 0$ for any square matrices $M$ and $N$. This is equivalent to saying that $\text{tr}(MN) = \text{tr}(NM)$, which is a nice exercise. You can prove it by writing everything out explicitly but there are also nicer proofs.

*If $\mathfrak{g}$ weren't simple, it would have some nontrivial quotient, which would have dimension $\le 2$. A $2$-dimensional Lie algebra has nontrivial abelian quotients, so in either case we would conclude that $\mathfrak{g}$ has a $1$-dimensional, and in particular abelian, quotient. Hence $\mathfrak{g} \neq [\mathfrak{g}, \mathfrak{g}]$. Now take the contrapositive. 
A: 1.) We have $ad(L)=ad([L,L])=[ad(L),ad(L)]$, hence every $ad(x)$ can be written as a linear combination of elements $[ad(y),ad(z)]$, which all have trace zero. Hence $tr(ad(x))=0$ for all $x\in L$.
2.) $L$ cannot be solvable because of $[L,L]=L$. Now suppose that $I$ is a nonzero ideal, different from $L$. Hence $dim (I)=1$ or $dim(I)=2$. In the first case, $I$ is solvable and $L/I$ is of dimension $2$, hence solvable. It follows that $L$ is solvable, a contradiction. In the second case, $I$ is solvable and $L/I$ is $1$-dimensional, hence also solvable. Again it would follow that $L$ is a solvable extension of a solvable Lie algebra, hence solvable. A contradiction. It follows that $L$ has only the trivial ideals.
