Classify Lie algebra with 1 dimension derived algebra Problem 3.10 from Erdmann, Wildon ask:
Find, up to isomorphism, all Lie algebras with 1-dimensional derived algebra.
(also, this book assume finite dimension Lie algebra only; I'm not sure whether it's over a general field, or just $\mathbb{C}$, but it's best to do the general field)
I am having a hard time doing this. I got apparently a lot of algebras, and it's not even easy to tell when they are isomorphic or not. Just to illustrate this, consider even this one particular case:
Let $v,a_{1},\ldots,a_{n}$ be a basis where $v$ span the derived algebra, and further assume that $[v,a_{i}]=0$; denote $m_{i,j}$ ($i<j$) to be a constant such that $[a_{i},a_{j}]=m_{i,j}v$. Well, the difficulty here is in the fact it seems like there is no restriction on what $m_{i,j}$ could be (well, aside from the fact that they cannot be all $0$). Yet not every possible set of $m_{i,j}$ produce a different Lie algebra. For example, a Lie algebra with all $m_{i,j}=1$ is isomorphic to one with all $m_{i,j}=2$ because we can scale $v$ be a constant. Other isomorphism between Lie algebras with more complicated set of $m_{i,j}$ are possible by scaling the $a_{i}$ as well. And that's not counting the fact that we can also do addition. So I can't find anyway to classify them.
Any help solving the full question, or this particularly problematic case, is appreciated. Since I only started learning Lie algebra, I don't really have any advanced tools available, and they shouldn't be needed considering this is just chapter 3. Thank you for your help.
 A: Rough outline for an elementary approach. 
Let $L$ be a finite-dimensional Lie algebra with $\dim L' = 1$.
If $L'$ is not contained in $Z(L)$, show that in this case $L$ is the direct sum of an abelian Lie algebra and an nonabelian Lie algebra of dimension $2$.
Assume then that $L'$ is contained in $Z(L)$. If $L'$ is properly contained in $Z(L)$, prove that $L$ is the direct sum $K \oplus A$, where $A$ is an abelian Lie algebra and $K$ satisfies $\dim K' = 1$ and $K' = Z(K)$. 
We are left with the case where $L' = Z(L) = \operatorname{span} \{z\}$. In this case $\dim L \geq 3$.
Prove that $L$ has a basis $\{z, f_1, g_1, \cdots, f_t, g_t\}$ such that $[f_i, g_i] = z$ for all $i$ and $[f_i, g_j] = [f_i, f_j] = [g_i, g_j] = 0$ for all $i \neq j$. ($L$ is a central product of $t$ copies of the Heisenberg Lie algebra)
To find such a basis, you could proceed by induction. Find $f_1$ and $g_1$ such that $z, f_1, g_1$ span the Heisenberg Lie algebra $H$. Apply induction to a suitable subalgebra (note that any subspace containing $z$ is a subalgebra).
Or you could also consider the bilinear form defined by $(x,y) \mapsto \lambda$, where $[x,y] = \lambda z$. This defines a symplectic bilinear form on $L/Z(L)$.
Finally, prove that for all $t \geq 1$ such an Lie algebra can be constructed.
A: Here is a proof, assuming that you know what bilinear forms are. I will consider the case when the derived subalgebra is central (this is not be the case in general). We have the following situation: $h$ is a lie algebra (over a field $k$) with the 1-dimensional center $c$ (the derived subalgebra) and the abelian quotient $g=h/c$. Then for every $a, b\in g$, the commutator $[a,b]$ belongs to $c\cong k$. By using axioms of the lie algebra, we immediately see that the map
$$
\phi: g\times g\to k, \phi(a,b)=[a,b]
$$ 
is a bilinear antisymmetric ($\phi(a,b)=-\phi(b,a)$) form on the $k$-vector space $g$. Such firms $\phi$ are known as symplectic forms. Using an analogue of Gramm/Schmidt diagonal inaction process one sees that all such forms on the given vector space are isomorphic to each other. 
Lastly, since the center of $h$ is exactly 1-dimension, you see that the form $\phi$ is nondegenerate (for every nonzero $a\in g$, $\exists b\in g$, $\phi(a,b)\ne 0$).  
Conversely, given a nondegenerate bilinear antisymmetric form $\phi$ on $g$, one defines the Lie algebra structure on the vector space $h:=g\oplus c$:
$$
[a,b]=\phi(a,b), [g,c]=0 
$$
and extend the bracket by bilinearity to the rest of $g\oplus c$. Now, its your turn to do some work: you sit down and check carefully the axioms of lie algebra for this bracket on $h$ and the fact that its center is exactly $c$. Denote the resulting Lie algebra by $Ext_\phi$. 
Suppose that we have two different (nondegenerate) antisymmetric bilinear forms $\phi, \psi$ on $g$ and $\alpha$ is an automorphism of the vector space $g$ sending the form $\phi$ to the form $\psi$:
$$
\psi(\alpha(a), \alpha(b))= \phi(a,b). 
$$
Then you again sit down and check that $\alpha$ defines an isomorphism 
$$
\alpha_*: Ext_\phi\to Ext_\psi
$$
which acts by the identity on $c$ and via $\alpha$ on $g$. 
Lastly, you need to check that if $h, h'$ are isomorphic lie algebras with 1-dimensional centers $c, c'$ and commutative quotients $h/c, h'/c'$ which are  isomorphic as vector spaces, then there exists an isomorphism
$$
\alpha: (h/c, \phi)\to (h'/c', \phi'),
$$
where $h=Ext_{\phi}$ and $h'=Ext_{\phi'}$. You prove this by observing that the isomorphism $h\to h'$ has to send center to center. 
Therefore, the set of isomorphism classes of your lie algebras is parameterized by the set of isomorphism classes of pairs $(g, \phi)$ where $g$ is a $k$-vector space and $\phi$ is a nondegenerate antisymmetric bilinear form on $g$. Since all such forms are isomorphic, there is a unique one, determined by the dimension of $g$. The corresponding central extension is called Heisenberg Lie algebra. 
