There exists an $n_{0}\in\mathbb{N}$ such that $C_{\frak{g}}(K)\subseteq \frak{g}^{n_{0}}$ and $C_{\frak{g}}(K)\nsubseteq\frak{g}^{n_{0}+1}$

Question

If $\frak{g}$ is a nilpotent Lie algebra there exists an ideal $K$ of $\frak{g}$ of codimension 1 such that $\frak{g}$ = $K + x\mathbb{F}$.

How can I prove that there exists an $n_{0}\in\mathbb{N}$ such that

$C_{\frak{g}}(K)\subseteq \frak{g}^{n_{0}}$ and $C_{\frak{g}}(K)\nsubseteq\frak{g}^{n_{0}+1}$ ?

score 1 · Accepted Answer · answered Sep 28 '17 at 14:47

1

$\newcommand{\g}{\mathfrak g}$ $C_{\g}(K) \subset \g$ so it's enough to find $n$ with $C_{\g}(K) \nsubseteq \g^n$. On the other hand $K$ is not abelian (else $\g$ would be abelian) so $C_{\g}(K)$ is not trival but $\g$ is nilpotent so there is $m$ with $\g^m = 0$.

answered Sep 28 '17 at 14:47

Nicolas Hemelsoet

8,509

Yes, you're right, thank you very much. – fer2017 Sep 28 '17 at 15:19
@fer2017 : You are welcome ! – Nicolas Hemelsoet Sep 28 '17 at 15:22

There exists an $n_{0}\in\mathbb{N}$ such that $C_{\frak{g}}(K)\subseteq \frak{g}^{n_{0}}$ and $C_{\frak{g}}(K)\nsubseteq\frak{g}^{n_{0}+1}$

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