# Subgroups of the group of motions on the plane fixing a point is the conjugate of the group of all orthogonal transformations by translations.

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.

• 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. – Andres Mejia Dec 24 '18 at 16:42
• 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. – Dbchatto67 Dec 24 '18 at 16:57
• As far as I know $G/\Bbb R^2 \simeq O(2)$ where $G$ is the group of motoins in the plane. – Dbchatto67 Dec 24 '18 at 17:15
• the semidirect product bit is inessential (although helpful) – Andres Mejia Dec 24 '18 at 17:15

## 1 Answer

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