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Let $\mathcal g$ be a nilpotent Lie algebra with $dim>1$. Why it is impossible that $codim \ \mathcal g'=1$?

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Let $L$ be a nilpotent Lie algebra. It is well-known that $L$ is generated by the representatives of $L/[L,L]$. If this space would be $1$-dimensional, $L$ would be generated by $1$ element, and hence would be abelian. But this is a contradiction, because then $[L,L]=0$. It follwos that $\dim L/[L,L]\ge 2$.

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Suppose that $codim g'=1$. Let $Vect\{x\}$ be a supplementary space of $g'$. Write $g'=W+[g,g']$ and $W$ is not trivial. Let $u,v\in g$ you can write $u=cx+u', u'\in g'$ and $v=dx+v', v'\in g'$. $[u,v] =[cx+u',dx+v']= c[x,v']+d[u',x]+[u',v']$. Since $u',v'\in g'$ we deduce that $[u,v]\in [g,g']$ thus the vector space generated by $[u,v], u,v\in g$ does not contain $W$. Contradiction.

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  • $\begingroup$ Do you mean the space generated by $[u,v]$ where $u,v\in \mathcal g$ or the space generated by the three elements $\{[u,v],u,v\}$? $\endgroup$ – Ronald Feb 16 '16 at 1:14
  • $\begingroup$ Yes it is what i mean, $[u,v], u.v\in g$. $\endgroup$ – Tsemo Aristide Feb 16 '16 at 1:17
  • $\begingroup$ I've a question, since $codim \ g' =1$ doesn't that mean $dim\ g/g'=1$ which implies that $g=W + g'$?. I don't understand why then $g'=W+[g,g']$? $\endgroup$ – Ronald Feb 16 '16 at 1:23
  • $\begingroup$ Yes, it what it means $\endgroup$ – Tsemo Aristide Feb 16 '16 at 1:24

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