2
$\begingroup$

Fix a positive integer $r$. Describe the solutions to the system of equations given by:

$$\begin{equation}\sum_{1\leq i\leq r}X_i^2\equiv0\pmod{X_k}(1\leq k\leq r)\end{equation}$$

Example: In the case that $r=3$, $(X_1,X_2,X_3)=(1,5,13)$ is a solution, because $1^2+5^2+13^2=195=3\cdot 5\cdot 13$ is divisible by $1$, $5$ and $13$.

Motivation: The degrees of the irreducible characters $\text{Irr}(G)$ of a finite group $G$ satisfy the following relation:

$$\begin{equation}\sum_{\chi\in\text{Irr}(G)}\chi(1)^2=|G|\end{equation}$$

Furthermore, for every $\chi\in\text{Irr}(G)$, $\chi(1)$ divides $|G|$. So the broader question is: is there some way to arithmetically characterize sequences of positive integers which are the irreducible degrees of some finite group?

The example I gave above does not qualify. No group has irreducible degrees $(1,5,13)$. Such a group would have $195$ elements, but since $195$ is a squarefree number, our hypothetical group with $195$ elements would have to be solvable. But nontrivial solvable groups must have nonprincipal linear characters, so $(1,5,13)$ cannot be its set of irreducible degrees.

(27th Aug) Example:

In the case that $r=4$, $(X_1,X_2,X_3,X_4)=(1,1,2,4)$ is a solution, because $1^2+1^2+2^2+4^2=22=2\cdot 11$. But this does not correspond to the irreducible degrees of some finite group $G$. Suppose it did, so $|G|=22$. Then $G$ must be non-abelian, so $G\cong C_{11}\rtimes C_2$. However, every irreducible character of $C_{11}\rtimes C_2$ has either degree $1$ or $2$, which contradicts the assumption that $G$ has an irreducible character of degree $4$.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Both of your corrections are correct, I have edited the post. $\endgroup$
    – semisimpleton
    Jun 16 at 11:17
  • $\begingroup$ For a fixed $|G|$, the problem reduces to representing $\frac1{|G|}$ as an Egyptian fraction with squared denominators (each of which also divides $|G|$). A somewhat similar system appears in our research on integral modular fusion categories in arxiv.org/abs/2302.01613 $\endgroup$
    – Max Alekseyev
    Aug 27 at 14:31

1 Answer 1

Reset to default
2
$\begingroup$

Even for $r=3$ the question isn't easy. For instance a special case is Markov's diophantine equation $a^2+b^2+c^2=3abc$, for which one infinite family of solutions is $(1, F_{2n-1}, F_{2n+1})$, where $F_i$ is the $i$th Fibonacci number.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ I think the general question of finding all positive integer solutions to the system of congruences might be beyond current technology, but the restricted question of looking for solutions corresponding to irreducible degrees of a finite group might be tractable. Thus I believe this is an interesting question, because it seems to lie on the boundary of what’s currently approachable $\endgroup$
    – Stanley Yao Xiao
    Jun 16 at 22:39
  • $\begingroup$ @Stanley, for, say, $n=3$, isn't it known what finite groups have exactly three conjugacy classes (and, thus, exactly three irreducible degrees)? cefns.nau.edu/~falk/old_classes/511/extras/threeclass.html says the only ones are the group of order three, and the symmetric group on three letters. See also math.stackexchange.com/questions/1581996/… $\endgroup$
    – Gerry Myerson
    Aug 27 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.