3
$\begingroup$

Burnside's lemma (for finite group) tells us that the square sum of dimensions over all inequivalent irreducible representations is equal to the order of the group, i.e.

$$ d_1^2 + d_2^2 + \cdots + d_r^2 = |G| $$

where $d_i$ is the dimension of representation $D^{(i)}(g)$. It seems that for all finite groups, this equation has only one solution (for $d_i$). Is this true and how can I prove it or find a counter-example?

$\endgroup$
  • $\begingroup$ Sounds rather unlikely to me. $\endgroup$ – Lord Shark the Unknown Jun 15 '18 at 4:24
  • $\begingroup$ Even if you know $r$ (and that is also not always the case) the $d_i$ are not uniquely determined. Thus, this equation is very helpful, but is not the final word on the $d_i$. (it so turns out that $d_i$ is a multiple of $|G|$ for each $i$, but even then uniqueness cannot be guaranteed). $\endgroup$ – астон вілла олоф мэллбэрг Jun 15 '18 at 4:39
  • $\begingroup$ @астонвіллаолофмэллбэрг: Other way around: each $d_i$ divides $|G|$. $\endgroup$ – anomaly Jun 15 '18 at 5:01
  • $\begingroup$ Yes, sorry : $d_i$ is a divisor of $|G|$. While this reduces the choices of $d_i$ greatly, it still does not guarantee uniqueness, but makes the above equation supremely helpful. $\endgroup$ – астон вілла олоф мэллбэрг Jun 15 '18 at 5:31
6
$\begingroup$

$$8 \times 1^2 + 2 \times 2^2 + 4^2 = 4\times 1^2 + 7 \times 2^2 = 32.$$ There are groups of order $32$ (numbers $6$ and $18$ in the database) with these character degrees.

$\endgroup$
3
$\begingroup$

If a finite group $G$ has a representation with $d_i=2$ and $d_j=9$ then one can replace these by $6$ and $7$ without affecting the sum of the $d_i^2$s. So the equation $d_1^2 + d_2^2 + \cdots + d_r^2 = |G|$ may have multiple solutions in positive integers.

$\endgroup$
  • $\begingroup$ But does this actually correspond to some pair of non-isomorphic groups? $\endgroup$ – Tobias Kildetoft Jun 15 '18 at 6:11
  • 2
    $\begingroup$ Probably not, but it makes the point that to determine the degrees of a groups characters, one has to do some group theory, Arithmetic is not sufficient... @TobiasKildetoft $\endgroup$ – Lord Shark the Unknown Jun 15 '18 at 6:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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