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How do I prove that subgroups $O'$ of the group of motions on the plane fixing a point $p$ (say) is the conjugate of the group of all orthogonal transformations by translations i.e. $O' = t_{p}\ O\ {t_{p}}^{-1}$ where $O$ is the group of all orthogonal transformations?

It is easy to see that $t_{p}\ O\ {t_{p}}^{-1} \subseteq O'$. How do I prove the other way round? Please help me in this regard.

Thank you very much.

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  • $\begingroup$ Show that the group of translations $\mathbb R^2$ is normal. Can you show that $\mathrm{Isom}(\mathbb R^2)/\mathbb R^2) \cong O(2)$? And in fact, there is a section $O(2) \to \mathbb R^2$ so we have a semidirect product. This might help in getting some of the structure down. $\endgroup$ – Andres Mejia Dec 24 '18 at 16:42
  • $\begingroup$ I don't know semidirect product. Actually I am following Artin's algebra where these things are discussed without using the concept of semidirect product. $\endgroup$ – Dbchatto67 Dec 24 '18 at 16:57
  • $\begingroup$ As far as I know $G/\Bbb R^2 \simeq O(2)$ where $G$ is the group of motoins in the plane. $\endgroup$ – Dbchatto67 Dec 24 '18 at 17:15
  • $\begingroup$ the semidirect product bit is inessential (although helpful) $\endgroup$ – Andres Mejia Dec 24 '18 at 17:15
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We take an element $m' \in O'.$ Then $m$ is a rigid motion fixing the point $p.$ Consider $m = {t_p}^{-1} m' t_{p}.$ Then $m$ is a rigid motoin fixing the origin. We know that the subgroup $G_0$ of the group of motions in $\Bbb R^n$ fixing the origin is same as the orthogonal group $O$ in dimension $n.$ So $m \in O.$ Therefore $m' \in t_{p}\ O\ {t_{p}}^{-1}.$ So $O' \subseteq t_{p}\ O\ {t_{p}}^{-1}.$ Also It is easy to see that $t_{p}\ O\ {t_{p}}^{-1} \subseteq O'.$ Hence $t_{p}\ O\ {t_{p}}^{-1} = O'.$ This completes the proof.

QED

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