gr.group theory - Number of subgroups in a Bieberbach group.

That is not an answer.
I want to give an example where the argument of Erdős does not work directly.

Consider an action of group $Gamma$ on $mathbb R^3$ generated by the reflections $r_1, r_2$ and $r_3$ correspondingly in the lines $x=z=0$ and $x+1=z=0$ and $x-y=z-1=0$.

Each of the reflections $r_i$ generate a maxiamal $mathbb Z_2$-subgroups, all of them are nonconjugate.
These groups corespond to three singular circles, say $Sigma_i$ in the factor $X=mathbb R^3/Gamma$.
($X$ is homeomorphic to $S^3$ and $Sigma_1$, $Sigma_2$, $Sigma_3$ form Borromean rings, but all this is not important.)

Let us try to mimic argument of Erdős.
Take subsets $X_i$ of $X$ of midpoints $m$ between $xin X$ and a closest $x_0inSigma_i$ to $x$.
As in the argument of Erdős we have $mathrm{vol}, X_i>tfrac{1}{2^3}cdotmathrm{vol}, X$.
BUT $X_1cap X_3$ has interior points and here argument brakes into parts.

Comments

• Since fixed point sets are 1-dimensional, it would be enough to take $min [xx_0]$ such that $tfrac{|mx_0|}{|xx_0|}=tfrac1{2sqrt[3]{2}}$.
  But even in this case one has interior points in $X_1cap X_2$ (the borderline in this example seems to be $tfrac13$).




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