# Is there exists such figure $A$ on plane $E^2$ which isometry group is isomorphic to $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$

I suppose that there are no such $$A \subset E^2$$ which satisfy $$\text{Iso}(A) \simeq \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$$

But I'm stuck on showing this in formal way. Can we use here Classification of Euclidean plane isometries theorem to prove it ?

One way to prove this is to prove that the isometry group of $$E^2$$ does not even contain a subgroup isomorphic to $$\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$$. Yes, the classification of Euclidean isometries will help, but you also need to know about some special subgroups of the group of isometries. In particular, you can use the following facts:

• For every finite group of isometries $$G$$ of $$E^2$$, there exists a point $$x \in E^2$$ fixed by each element of $$G$$.

So we can assume that your subgroup $$G = \text{Iso}(A) \approx \mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_2$$ fixes some point $$x$$.

• The subgroup $$\Gamma_x < E^2$$ of all isometries that fixes $$x$$ can be decomposed as a split short exact sequence $$1 \to S^1_x \to \Gamma_x \to \mathbb Z_2 \to 1$$ where $$S^1_x$$ is the circle group of rotations around $$x$$, and a splitting $$\mathbb Z_2 \to \Gamma_x$$ is given by any reflection across a line through $$x$$.

So, $$G$$ is contained in $$\Gamma_x$$.

• The normal subgroup $$S^1_x$$ contains a unique order $$2$$ element, namely the $$180^\circ$$ rotation around $$x$$. That element generates an order 2 subgroup $$R_x$$, which is a normal subgroup of $$\Gamma_x$$.

To finish the proof we consider two cases.

If $$G$$ contains $$R_x$$ then $$G \cap S^1_x = R_x$$, because of uniqueness of the order $$2$$ element in the group $$S^1_x$$. It follows that the homomorphic image of $$G$$ in $$\mathbb Z_2$$ is isomorphic to $$G/R_x$$ which has order $$4$$.

Whereas if $$G$$ does not contain $$R_x$$ then $$G \cap S^1_x$$ is trivial, by the same uniqueness argument. It follows the homomorphic image of $$G$$ in $$\mathbb Z_2$$ is isomorphic to $$G$$ which has order $$8$$.

In either case one gets a contradiction: $$\mathbb Z_2$$ only has order 2.

• you mean, fixed by each element of G? Oct 1 '19 at 22:05
• Yes, I''ll fix that. Oct 1 '19 at 22:06
• By the way, why $Ker\{\phi : G \rightarrow \mathbb{Z}_2\} = R_x$ in the first case ? Oct 2 '19 at 19:05
• I expanded on that point, it is a consequence of the uniqueness of the $180^\circ$ rotation. Oct 3 '19 at 12:39