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Let $ \ G \ $ be a non-abelian subgroup of isometries in the plane $ \ \mathbb{R}^2 \ $ whose elements are all orientation-preserving . Prove that $ \ G \ $ contains a non-trivial translation .

Answer

Since the group $ \ G \ $ contains isometries with orientation -preserving , then

$ t_v \rho_\theta \in G $ , where $ \ t_v \ $ is translation and $ \ \rho_\theta \ $ is rotation about origin .

So if the rotation $ \ \rho_\theta \ $ be identity rotation , then

$ \rho_\theta= identity=I_\theta \ $ .

Then ,

$ t_v \circ I_\theta \in G \ \Rightarrow t_v \in G \ $.

Thus $ \ G \ $ contains non-trivial translation $ \ t_v \ $ .

Am I right ?

I need confirmation of my work .

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    $\begingroup$ Unfortunately, there is no proof in what you have written. The fact that you haven't used the non-abelian assumption (which is important here) is one indication of that. $\endgroup$
    – user357151
    Nov 11 '17 at 22:07
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It seems the following.

For each $\varphi\in\Bbb R$ put $A(\varphi)=\begin{pmatrix} \cos\varphi & \sin\varphi \\ -\sin\varphi & \cos\varphi \end{pmatrix}$. It is easy to check that $A(\varphi)^{-1}= A(-\varphi)$ and $A(\varphi)A(\psi)= A(\varphi+\psi)$ for each $\varphi, \psi\in\Bbb R$. For each orientation-preserving isometry $f$ of the plane $\Bbb R^2$ there exist an angle $\varphi(f)\in\Bbb R$ and a vector $a(f)\in\Bbb R^2$ such that $f(x)=A(\varphi(f)) x+a(f)$ for each $x\in\Bbb R^2$. It is easy to check that $ A(\varphi(fg))=A(\varphi(f))A(\varphi(g)) $ for each $f,g\in G$. Since the group $G$ is non-abelian, there exist elements $f,g\in G$ such that $fg\ne gf$. Put $h=f^{-1}g^{-1}fg\ne e$. Then $ A(\varphi(h))$ is the identity matrix, so $a(h)\ne 0$ and thus $h$ is the required translation.

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