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.

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

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