Let $V=(x)$ be the 1-dimensional vector subspace of a Lie algebra $\mathfrak{g}$ generated by $x$. Let $\mathfrak{h}$ is the smallest Lie ideal containing $V$. Is $[\mathfrak{h}^n, \mathfrak{h}^m]= \mathfrak{h}^{n+m}$ ??
Recall that, the Lie ideal $\mathfrak{a}^n$ is defined inductively:
\begin{align} \mathfrak{a}^n &= \begin{cases} \,\, \mathfrak{a} & \text{if $\quad$ } n=1,\\ \,\,[\mathfrak{a}, \mathfrak{a}^{n-1}] & \text{if $\quad$ } n>1, \end{cases} \end{align}
No. Let L be the 5-dimensional Lie algebra over a field F with basis e1,e2,e3,e4,e5 and non-zero products [e2, e3] = e1, [e1, e4] = e1, [e2, e4] = e2, [e3, e5] = e3, [e2, e5] = −e2 and let x = e5. Then h = Fe1 + Fe2 + Fe3 + Fe5, h^2 = Fe1 + Fe2 + Fe3, [h^2,h^2] = Fe1, but h^n = h^2 for all n > 1.