# How to prove $\text{Sym}^+(H)$ is an abelian group.

Let $$H$$ be a shape, i.e., a non-empty compact set in $$\mathbb{R}^2$$. Denote by $$\text{Sym}(H)$$ the set of isometries preserving $$H$$ and by $$\text{Sym}^+(H)$$ the set of positive orientation-preserving isometries preserving $$H$$. Prove that $$\text{Sym}^+(H)$$ is an abelian group.

If $$f\in\text{Sym}(H)$$ then $$f(H)=H, A_f$$ is an orthogonal matrix, i.e $$A\in O(2)$$.
If $$f\in\text{Sym}^+(H)$$ then $$f(H)=H, A_f=\begin{pmatrix} \cos x & -\sin x\\ \sin x & \cos x\end{pmatrix}$$.
To prove $$\text{Sym}^+(H)$$ is an abelian group, we need to prove it's indeed a group and $$fg=gf\quad\forall f,g\in\text{Sym}^+(H)$$. But how can I prove this? Could someone show me?

• Essentially, such an isometry has to fix the center of mass of $H,$ but that is messy to define for sets with no interior points. Commented Jul 3 at 22:37
• You can restrict tot the case when $H$ is convex, I think. I might be wrong. Commented Jul 3 at 22:39
• Can you not compute $f \circ g = g \circ f$ ? Commented Jul 3 at 23:19
• First prove that $\text{Sym}^{+}$ is a group. Next, it's a subgroup of $SO(2)$, which is the group of rotations. Do you see why this group (and so its subgroups) are abelian? Commented Jul 3 at 23:19
• Yeah, once you have a fixed point $O$, each symmetry $f$ gives you the linear transformation $x \mapsto f(x + O) - O$ - and this makes an injective homomorphism $\operatorname{Sym}(H) \to O(2)$. Commented Jul 3 at 23:56