On the solvability of the Affine Group

Question

Let $n\in\mathbb{N}$. For each $A\in \mathcal{M}_{n\times n}(\mathbb{R})$ and $b\in\mathbb{R}^{n}$, define $T_{A,b}:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ the affine transformation

$$T_{A,b}(x)=Ax+b$$

Set $G=\{T_{A,b}: det(A)\neq 0\}$. It is easy to see that $G$ is a group under the composition of functions. Is this group a solvable and not nilpotent group?

I'm reading the book "Affine Maps, Euclidean Motions and Quadrics" but there is no mention of that on it.

score 0 · Answer 1 · answered May 27 '21 at 22:54

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If $n>1$, this group is not solvable since it contains $Sl(n,\mathbb{R})=\{ A\in Gl(n,\mathbb{R}), det(A)=1\}$ which is semi-simple.

answered May 27 '21 at 22:54

Tsemo Aristide

87,475

The elements of $G$ are the affine transformations $T_{A,b}$ with $det(A)\neq 0$, not the matrix. – José Luis Camarillo Nava May 27 '21 at 23:30
What about for the case n=1? – José Luis Camarillo Nava May 27 '21 at 23:31
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If $n=1$, it is solvable since $[G,G]$ is the group of translations. – Tsemo Aristide May 27 '21 at 23:33
Thank you so much... – José Luis Camarillo Nava May 27 '21 at 23:34

On the solvability of the Affine Group

1 Answers1