A (hopefully) very simple question that has been bugging me all day! Let $L$ be a Lie algebra then the derived Lie algebra $L'$ is
$$ L' = \{ \, [u,v] : \forall u,v\in L \, \}. $$
I want to show (without introducing the notion of an ideal) that the derived algebra $L'$ is a subalgebra of $L$.
Of course, to do this I have to show it is a subspace. And here is where I fall down, as I am struggling to show it is closed under addition. If $v_{1}, v_{2} \in L' \;\;\;\;$ then I need to show $v_{1}+v_{2} \in L' \;\;\;\;$. We have
$v_{i} = [x_{i},y_{i}] \;\;$ for some $x_{i},y_{i} \in L$
then $v_{1}+v_{2} = [x_{1},y_{1}] + [x_{2},y_{2}]\;\;\;\; $, but in order to be in $L'$, this must be able to be written as a single Lie bracket. Is this logic correct? Any pointers would be much appreciated. Thanks.
The definition of $L'$ you have is wrong ; we usually take $$ L' \overset{def}= \langle \{ [u,v] \, | \, u,v \in L \} \rangle_k $$ where $k$ is the field of your Lie algebra $L$ (the brackets $\langle - \rangle_k$ means you take the $k$-span of the subset). Then your issue becomes trivial.
Hope that helps,