Let $\mathfrak{r}$ be the radical of $\mathfrak{g}$. If $\mathfrak{r}$ is abelian, then $\mathfrak{g}$ is semisimple?

What if $\mathfrak{g}$ is the Lie algebra of an algebraic complex linear group?

ntaraxq
2022-07-02
Answered

Let $\mathfrak{g}$ be a complex linear Lie algebra. Assume that the center $\mathfrak{z}$ of $\mathfrak{g}$ is trivial

Let $\mathfrak{r}$ be the radical of $\mathfrak{g}$. If $\mathfrak{r}$ is abelian, then $\mathfrak{g}$ is semisimple?

What if $\mathfrak{g}$ is the Lie algebra of an algebraic complex linear group?

Let $\mathfrak{r}$ be the radical of $\mathfrak{g}$. If $\mathfrak{r}$ is abelian, then $\mathfrak{g}$ is semisimple?

What if $\mathfrak{g}$ is the Lie algebra of an algebraic complex linear group?

You can still ask an expert for help

Mekjulleymg

Answered 2022-07-03
Author has **14** answers

Your question whether $\mathfrak{g}$ is semisimple is equivalent to whether necessarily $\mathfrak{r}=0$. The answer is no. A counterexample is given by

$\mathfrak{g}=\{\left(\begin{array}{ccc}a& b& d\\ c& -a& e\\ 0& 0& 0\end{array}\right):a,b,c,d,e\in \mathbb{C}\}$

where the radical

$\mathfrak{r}=\{\left(\begin{array}{ccc}0& 0& d\\ 0& 0& e\\ 0& 0& 0\end{array}\right):d,e\in \mathbb{C}\}$

is two-dimensional.

$\mathfrak{g}=\{\left(\begin{array}{ccc}a& b& d\\ c& -a& e\\ 0& 0& 0\end{array}\right):a,b,c,d,e\in \mathbb{C}\}$

where the radical

$\mathfrak{r}=\{\left(\begin{array}{ccc}0& 0& d\\ 0& 0& e\\ 0& 0& 0\end{array}\right):d,e\in \mathbb{C}\}$

is two-dimensional.

Cierra Castillo

Answered 2022-07-04
Author has **6** answers

Another explantion?

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$k=?$

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$\left[\begin{array}{ccc}1& 0& 1\\ 0& 0& 1\\ 0& 0& 1\end{array}\right]$

do this matrix transformation options:

1) rotate and scaling

2) translate and scaling

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$\left[\begin{array}{ccc}1& 0& 1\\ 0& 0& 1\\ 0& 0& 1\end{array}\right]$

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$\left(\begin{array}{ccc}1& 3& 3\\ 2& 6& -3.5+k\end{array}\right)$

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The equation of the parabola is y = ___.

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Let G be a finite group with $card\left(G\right)={p}^{2}q$ with $p<q$ two ' numbers. We denote $s}_{q$ the number of q-Sylow subgroups of G and similarly for p. I have just shown that ${s}_{q}\in \{1,{p}^{2}\}$. Now I want to show that

$\underset{S\in Sy{l}_{q}\left(G\right)}{\cup}S\setminus \left\{1\right\}={\dot{\cup}}_{S\in Sy{l}_{q}\left(G\right)}S\setminus \left\{1\right\}$

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