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.
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