Isometry group of an integer as of the corresponding $\Omega(n)$-parallelotope

Question

Let $\prod_{i}p_{i}^{a_{i}}$ be the prime factorization of a positive integer $n$ and let's consider the $\Omega(n)$-parallelotope built with, for all $i\in I$, $a_{i}$ pairwise orthogonal vectors of norm $p_i$, so that there are $\omega(n)$ different lengths for the sides of this parallelotope. Let's consider the isometry group of this parallelotope, as a part of the affine space $\mathbb{R}^{\Omega(n)}$, denoted by $G(n)$. Let's write $\pi_{\mathcal{G}}(x):=\vert\{n\leqslant x, G(n)\cong\mathcal{G}\}\vert$. Can we obtain an asymptotics for $\pi_{\mathcal{G}}(x)$ from the structure of $\mathcal{G}$?
Thanks in advance.

Answer 1

Well, I'm fond of this question, and suggest the following shadow of beginning of a possible answer...

For the sake of simplicity, let's write : $~~~~\forall~n\in \mathbb{N^*}~~~~\dfrac{\mathbb{Z}}{n\mathbb{Z}}=\mathbb{Z}/n~.$

$\bullet~$ When $~n~$ is a prime $~p,~$ then the parallelotope is degenerated and become a segment of lenght $~p,~$ so, if we note $~Ht~$ the half-turn around the middle of this segment, $$~G(p)=\{Id,~Ht\}=\mathbb{Z}/2~$$

$\bullet~$ For any $~n\geq2,~$ the diagonals of any $n$-parallelotope intersect at one point and are bisected by this point. Inversion $~\mathfrak{I}~$ in this point leaves the $n$-parallelotope unchanged, thus $~G(n)~$ contains $~\mathbb{Z}/2~$ as a subgroup, so $~~Card(G(n))=\#(G)~~$ is even.

Thus, for any $~n\geq2,~$ and any group $~\mathfrak{G}~$ of odd order, $~~~G(n)\neq \mathfrak{G}~~~$ and $$~~~\forall x\geq 1~~~~\pi_{\mathfrak{G}}(x)=0~$$

In particular, for all integers $~n\geq 2,~m\geq 0,~~~G(n)\neq \mathbb{Z}/ (2m+1)~~~$ and $$~~~\forall x\geq 1~~~~\pi_{\mathbb{Z}/ (2m+1)}(x)=0.~$$

$\bullet~$ As soon as $~n~$ possess two prime divisors $~p,~q~$, $~G(n)~$ contains at least four different isometries which naturally act on the ``rectangle" defined by $~(p,q)~$ and leave invariant all the other orthogonal directions, so $~\#(G(n))\geq 4.~$ Thus we have $$n~prime~~\Longleftrightarrow~~G(n)=\mathbb{Z}/2$$

and the function $~\pi_{\mathbb{Z}/2}~$ is the ``usual" $~\pi(x)=Card(\{p\leq x~;~p~prime\})~$ and, by the PNT, $$~~\pi_{\mathbb{Z}/2}(x)~\sim_{\infty}~\dfrac{x}{\ln(x)}$$

$\bullet~$ If $~n=pq~$ with $~p,~q~$ distinct primes, $~G(n)~$ is the only non-cyclic group of order $4,$ i.e. the dihedral group $~D_2~$ or $~\mathbb{Z}/2 \times \mathbb{Z}/ 2.~$

Now, if $~n~$ has at least $~3~$ prime factors, $~G(n)~$ contains at least $~8~$ distinct isometries and $~\#(G)\geq 8,~$ so we have :

$$n=pq,~~~~p,~q~~distinct~primes~~\Longleftrightarrow~~G(n)=\mathbb{Z}/2 \times \mathbb{Z}/ 2$$

$\bullet~$ In the same way, we'll have, if $~D_4~$ is the dihedral group of order $~8,~$ :

$$n=p^2,~~p~~prime~~\Longleftrightarrow~~G(n)=D_4$$

so that

$$~~\pi_{D_4}(x)~\sim_{\infty}~\dfrac{2\sqrt{x}}{\ln(x)}$$

I hope all this is not awfully an stupidly wrong. If not, it's certainly the beginning of a longer story, and I must refresh my memory on the symmetry groups of the n-dimensional Hypercube and Hyperrectangle, but, first of all, I'm hungry and smelling some imprudent Durin's folks approaching from the town-lake of Dale, and, as Dragon-King under the Mountain, it's my duty to roast and eat them fast.

Harsh criticism welcome. Yours truly, Smaug, from the Lonely Mountain of Erebor.