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Assume that $G = \langle a, b \rangle$ is a finite non-abelian group which is generated by an involution $a$ and an element $b$ of order $n$ ($n\geq 3$) such that for every (complex) representation $\varphi$ of $G$ the matrix $\varphi(a) + \varphi(b) + \varphi(b^{-1})$ has only rational eigenvalues.

Question: Is there an upper bound on the order of $G$?

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  • $\begingroup$ Where do you get the number 25 from? $\endgroup$
    – Yemon Choi
    Commented Jun 6, 2013 at 4:11
  • $\begingroup$ @Yemon: 25 probably came from some incompletely reproduced homework problem, as one can take $G$ to be an infinite finitely generated simple group, with $a$ and $b$ generating a subgroup isomorphic to, say $Z_2 *Z_3$. $\endgroup$
    – Misha
    Commented Jun 6, 2013 at 4:29
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    $\begingroup$ Do you mean $\tau(a) + \tau(b) + \tau(b^{-1})$? I don't know what you mean by addition in a non-abelian group (although addition in the group ring would yield the above formula). $\endgroup$
    – S. Carnahan ♦
    Commented Jun 6, 2013 at 8:43
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    $\begingroup$ I don't know what you mean by background and motivation. I was came across this problem when I was studying Q-groups in the book "Structure and Representations of Q-Groups" by Dennis Kletzing. I needed to know, what can happen if we put more restriction. Is that produce an infinite class of groups like Q-groups, or not. This was the most obvious extension I could have think of. Characters turning away many other information about representations. Problem get a lot challenging when we look representation. I know the answer for this, if I was just taking characters rather than representations. $\endgroup$
    – katie
    Commented Jun 7, 2013 at 6:34
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    $\begingroup$ @katie, thank you. That is the sort of information that can be helpful when inserted into the text of the question. $\endgroup$
    – S. Carnahan ♦
    Commented Jun 7, 2013 at 9:10

2 Answers 2

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If you are looking for a more abstract proof based on representation and group theory, I suggest you to work on a more general question as follows:

Assume that $G=\langle S\rangle$ is a finite group which is generated by $S=S^{-1}$ ($1\not\in S$) such that for every complex representation $\phi$ of G the matrix $\sum_{s\in S} \phi(s)$ has only rational eigenvalues. Then is $|G|$ bounded above by a function of $|S|$?

The answer to the above question is positive and a crude known bound is $\frac{|S|(|S| − 1)^{2|S|} − 2}{|S| − 2}$.

Note that, what you need by the representation theoretic assumption is that the eigenvalues of the linear transformation $T=\sum_{s\in S}s$ on the vector space $\mathbb{C}(G)$ are all rational (and so integer). Note that $T$ is an element of the group ring $\mathbb{C}(G)$.

You may find some related graph theoretic results in the following paper:

Alireza Abdollahi and E. Vatandoost, Which Cayley graphs are integral?, The Electronic Journal of Combinatorics 16 (2009), #R122.

Sorry for the self-promotion!

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Finally, I found an answer! A friend of mine, brought this to my attention. Apparently, there is a connection between this and eigenvalues of Cayley graphs. This problem is equivalent with finding bound on the number of nodes of a cubic Cayley graphs which all its eigenvalues are integers. A. J. Schwenk ("Exactly thirteen connected cubic graphs have integral spectra". Proceedings of the International Graph Theory Conference at Kalamazoo, (Y. Alavi and D. Lick, eds.) Springer-Verlag. May 1976.) has characterized all cubic graphs. He show that the number of nodes is maximum $30$. His result is not about Cayley graphs, but consequently it proves bound of $24$ with my numeric calculations. I read his paper, but was painfully difficult to call it a proof. I still prefer to find a a more abstract proof based on representation and group theory!

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