# Let <mi class="MJX-tex-caligraphic" mathvariant="script">g be a Lie algebra and let a ,

Let $\mathcal{g}$ be a Lie algebra and let $a,b,c\in \mathcal{g}$ be such that $ab=ba$ and $\left[a,b\right]=c\ne 0$. Let . How to prove that $\mathcal{h}$ is isomorphic to the strictly upper triangular algebra $\mathcal{n}\left(3,F\right)$?
Problem: If $\mathcal{h}\cong n\left(3,F\right)$ then $\mathrm{\exists }{a}^{\prime },{b}^{\prime },{c}^{\prime }\in \mathcal{n}\left(3,F\right)$ with ${a}^{\prime }{b}^{\prime }={b}^{\prime }{a}^{\prime }$ and $\left[{a}^{\prime },{b}^{\prime }\right]={c}^{\prime }$ as in $h$ But then ${c}^{\prime }$ must equal $0$ whereas $c\in h$ is not $0$?
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Terrance Phillips
The Lie algebra with $\left[a,b\right]=c$ is the $3$-dimensional Heisenberg Lie algebra ${\mathfrak{h}}_{1}$. It has a faithful linear representation given by
$a=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),\phantom{\rule{1em}{0ex}}b=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right),\phantom{\rule{1em}{0ex}}c=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),$
Obviously this matrix Lie algebra is given by ${\mathfrak{n}}_{\mathfrak{3}}$, so that ${\mathfrak{n}}_{\mathfrak{3}}\cong {\mathfrak{h}}_{\mathfrak{1}}$.