Let $n$ be a natural number and $D_n$ be the set of divisors. We can make this set to a ring by observing that each divisor $d$ has

$$0 \le v_p(d) \le v_p(n)$$

Hence we can add two divisors $d,e$ by setting:

$$d \oplus e := \prod_{p | n} p^{v_p(d)+v_p(e) \mod (v_p(n)+1)}$$

and similarily we can multiply them by setting: $$d \otimes e := \prod_{p | n} p^{v_p(d) \cdot v_p(e) \mod (v_p(n)+1)}$$

Then, if $n = p_1^{a_1} \cdots p_r^{a_r}$, this ring will be isomorphic to the ring

$$\mathbb{Z}/(a_1+1) \times \cdots \times \mathbb{Z}/(a_r+1)$$

If $n$ is squarefree, than this reduces to :

$$d\oplus e = \frac{de}{\gcd(d,e)^2}$$

$$d\otimes e = \gcd(d,e)$$

and the ring of divisors is a boolean ring as defined here Boolean ring of unitary divisors / Structure of unitary divisors? and here https://math.stackexchange.com/questions/3799607/does-this-characteristic-polynomial-factor-into-linear-factors-over-the-integers/3799759

If we consider the addition table ($\oplus$) of this ring as a matrix, than it is clear that the sum of divisors $\sigma(n)$ is an eigenvalue to the eigenvector:

$$(1,\cdots,1)$$

Here is as an example the addition ($\oplus$) table for $n=12$:

$$\left(\begin{array}{rrrrrr} 1 & 2 & 3 & 4 & 6 & 12 \\ 2 & 4 & 6 & 1 & 12 & 3 \\ 3 & 6 & 1 & 12 & 2 & 4 \\ 4 & 1 & 12 & 2 & 3 & 6 \\ 6 & 12 & 2 & 3 & 4 & 1 \\ 12 & 3 & 4 & 6 & 1 & 2 \end{array}\right) $$

I have checked numerically ($n=1,\cdots,60$) that

$$\sigma(n) = |A_n^k|_2^{1/k}, \forall k \ge 1$$

where $A_n$ is the addition matrix of this ring.

*Is there a proof for this last equality (where $|.|_2$ denotes the spectral norm)?*(This question is proved here: https://math.stackexchange.com/questions/3800389/ring-of-divisors-of-a-natural-number-and-the-sum-of-divisors-as-an-eigenvalue-an )

Similarliy we can make the set $U_n$ of unitary divisors into a boolean ring by setting:

$$a\oplus b = \frac{ab}{\gcd(a,b)^2}$$

$$a \otimes b = \gcd(a,b)$$

I have checked numerically similarliy to the above ($\sigma^*(n) = $ sum of unitary divisors):

$$\sigma^*(n) = |B_n^k|_2^{1/k}, \forall k \ge 1$$

where $B_n$ is the addition matrix of $U_n$.

- Can this be proven?

To each eigenvalue $\lambda$ with eigenvector $v_{\lambda}$ of $B_n$ we can associate a "stabilizer group" $V_{\lambda} \le U_n$:

$$V_{\lambda} = \{u \in U_n| \left < (u\oplus u_1,\cdots,u \oplus u_r)^T ,v_{\lambda}\right >=\lambda \}$$

Then it seems that:

$$\lambda = \sum_{v \in V_{\lambda}} v - \sum_{u \in V_{\lambda}^C} u$$

Is $V_n$ a subgroup of $U_n$?

Is $\lambda$ equal to the right hand side of the last equality?

Thanks for your help!

Related questions:

Boolean ring of unitary divisors / Structure of unitary divisors?