3-dim Lie algebra with two commutative elements Let $\cal g$ be a Lie algebra and let $a,b,c\in \cal g$ be such that $ab=ba$ and $[a,b]=c\not =0$. Let $\mathcal h=span\ \\{a,b,c\\}$. How to prove that $\mathcal h$ is isomorphic to the strictly upper triangular algebra $\mathcal n(3,F)$? * * * Problem: If $\mathcal h\cong n(3,F)$ then $\exists a',b',c'\in \mathcal n(3,F)$ with $a'b'=b'a'$ and $[a',b']=c'$ as in $h$ But then $c'$ must equal $0$ whereas $c\in h$ is not $0$?

The Lie algebra with $[a,b]=c$ is the $3$-dimensional Heisenberg Lie algebra $\mathfrak{h}_1$. It has a faithful linear representation given by $$ a = \begin{pmatrix} 0 & 1 & 0 \\\ 0 & 0 & 0 \\\ 0 & 0 & 0 \\\ \end{pmatrix}, \quad b = \begin{pmatrix} 0 & 0 & 0 \\\ 0 & 0 & 1 \\\ 0 & 0 & 0 \\\ \end{pmatrix}, \quad c = \begin{pmatrix} 0 & 0 & 1 \\\ 0 & 0 & 0 \\\ 0 & 0 & 0 \\\ \end{pmatrix}, $$ see here. Obviously this matrix Lie algebra is given by $\mathfrak{n_3}$, so that $\mathfrak{n_3}\cong \mathfrak{h_1}$.