I proved that $\mathrm{Aff}(n) \cong O(n) \rtimes \mathbb{R}^n$ and $\mathrm{Aff}(n) \cong \mathbb{R}^n \rtimes O(n)$, where $\mathrm{Aff}(n)$ is the affine group, $O(n)$ the orthogonal group, $\rtimes$ denotes the semi-direct product of two groups.

Does it make sense that $\mathrm{Aff}(n) \cong \mathbb{R}^n \rtimes O(n)$ holds? I saw that most sources tend write down $\mathrm{Aff}(n) \cong O(n)\rtimes \mathbb{R}^n$ and as a consequence I am not sure whether my proof of showing $\mathrm{Aff}(n) \cong \mathbb{R}^n \rtimes O(n)$ makes sense. I could give the proof if there is interest in that.

  • $\begingroup$ Where did you see $O(n)\rtimes \mathbb{R}^n$? Are you sure it wasn't $O(n)\ltimes \mathbb{R}^n$? $\endgroup$ – FredH Mar 31 at 15:29
  • $\begingroup$ What is the difference between those expressions? $\endgroup$ – Dani Mar 31 at 16:25
  • $\begingroup$ With $\rtimes$, the normal subgroup is on the left; with $\ltimes$, it is on the right. $\endgroup$ – FredH Mar 31 at 16:26
  • $\begingroup$ I think it was $\ltimes$. But then why is $O(n) \rtimes \mathbb{R}^n \cong \mathrm{Aff}(n)$ not true? Both $O(n)$ and $\mathbb{R}^n$ are normal subgroups right? $\endgroup$ – Dani Mar 31 at 16:30
  • $\begingroup$ Because $O(n)$ is not a normal subgroup of $\mathrm{Aff}(n)$. $\endgroup$ – FredH Mar 31 at 16:33

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