Most questions that I can find on here (or anywhere else on the internet) deal with constructing a character table given a description of the group. I'm trying to answer a question which goes the other way though:

Given the following character table, where $\alpha=(-1+i\sqrt{7})/2$, what is the order of each $g_i$?

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I've had a look in Representations and Characters of Groups by James & Liebeck, but I can't find anything that (at a first glance) seems particularly geared towards answering this kind of question.

Either references or just an explanation of the answer here would be great!


1 Answer 1


There is a book around 1968-71 which is a conference proceedings of a Conference in Oxford (maybe "Oxford Symposium on Finite Group Theory", edited by M.B. Powell and G. Higman), in which an article by Higman himself outlines how to characterize the alternating group $A_{n}$ by its character table. In it, a result implicit in Brauer's development of modular representation theory, they note the following result: let $R$ be a ring of algebraic integers containing all entries of the character table of a finite group $G$. Let $p$ be a rational prime divisor of $|G|$, and let $\pi$ be a prime ideal of $R$ containing $p$. Then two elements $x,y \in G$ have $G$- conjugate $p^{\prime}$-parts if and only if $\chi(x) \equiv \chi(y)$ (mod $\pi$) for each irreducible character in $G$. I won't give here the orthogonality relations from modular character theory which imply this result. Recall that we may write uniquely $x = uv = vu$ where the order of $u$ is a power of $p$ and the order of $v$ is prime to $p$. The element $v$ is known as the $p^{\prime}$ (or sometimes $p$-regular) part of $x$.

Applying this with $p =2$, for example, we see that $g_{2}$ has the same $2^{\prime}$-part as $g_{1}= e$, so that the order of $g_{2}$ is a power of $2$. Similarly, the order of $g_{3}$ is a power of $2$. Likewise, the order of $g_{4}$ is a power of $3$. As for $g_{5}$ and $g_{6}$, this depends of course what $\alpha$ is. You would normally need to be told this. However, I think you can see from the other character values that $g_{5}$ and $g_{6}$ have to have order a power of $7$. With a little more thought, $g_{4}$ has order $3$. Looking at centralizer orders (there must be an involution somewhere!), we see that $g_{2}$ has order $2$. Looking at centralizer orders again, the Sylow $2$-subgroup is non-Abelian of order $8$ and $g_{3}$ must have order $4$. In fact, $g_{5}$ and $g_{6}$ each have order $7$.

(I should also have noted above that continuing the above argument, Higman notes that it is possible to determine the prime divisors of the order of each element of $G$ from its character table).

Actually, as noted in the comments below, this argument is "overkill"for this particular problem, though this more general argument has its own interest.

  • $\begingroup$ This is an amazing answer, thank you! It will take me some time to read through it though, and the language is not overly familiar to me. This question actually came up on a third-year undergraduate character theory exam, and so I feel that there should be a (possibly more clunky) way of answering this specific example that uses slightly less slick ideas, since there isn't usually too much crossover between module content at this point in my degree, and the idea of rational prime divisors hasn't been mentioned at all. Any ideas on what the intended answer might have been along the lines of? $\endgroup$
    – Tim
    Commented Jun 6, 2015 at 19:39
  • 1
    $\begingroup$ Well, I suppose that we can deduce that $g_{2}$ has centralizer of order $8$ and $g_{3}$ has centraizer of order $4$. From that, I suppose it already follows that $g_{2}$ and $g_{3}$ each have order a power of $2$. Arguing as I indicated above, it follows from this that $g_{2}$ has order $2$ and $g_{3}$ has order $4$. From the centralizer orders, you can also deduce that $g_{4}$ has order $3$ and $g_{5}$ and $g_{6}$ each have order $7$ ( I had missed that $\alpha$ was actually given). $\endgroup$ Commented Jun 6, 2015 at 20:21
  • $\begingroup$ Thanks, I'll have a work through of all that to see if it sinks in ok. And no, you didn't missed it, I just realised that I'd forgotten to include it so edited it after you said so! $\endgroup$
    – Tim
    Commented Jun 6, 2015 at 20:30

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