Various questions have been posted on MO (some answered, some not) involving the character table of a finite group $G$ over a splitting field such as $\mathbb{C}$ of characteristic 0. My basic question is more about references:

1) Are there textbook or other convenient sources summarizing properties of $G$ which can or can't be deduced from a knowledge of its character table?

By itself this is a rather artificial question, since one doesn't usually know a character table without already knowing a lot about the group, but it provides good exercise material. Recently I was going over some standard theory with a graduate student and came across old notes from a course I taught decades ago, but I can't recall which books I consulted at the time.

Two sorts of information are typically deduced from a character table using the orthogonality relations: (a) numerical data, such as the order of $G$ or more generally all orders of centralizers and hence classes; (b) normal subgroup data, starting with the fact that normal subgroups are intersections of the kernels of characters and then deducing orders and inclusions of such subgroups. In particular, one can pinpoint the center and derived group, as well as determine whether or not $G$ is simple. On the other hand, it's well known that nonisomorphic groups can have the same character table (e.g., the two nonabelian groups of order 8); in particular, the character table can fail to predict the orders of the class representatives labelling columns.

One substantive question of this type which I'm unclear about is this:

2) To what extent does the character table determine properties of $G$ ranging from solvability to nilpotency?


4 Answers 4


For nilpotency, you can deduce the character table of $G/Z$ from the character table of $G$. First, determine $Z$. Second, throw out all the representations where $Z$ is not in the kernel. Third, merge the conjugacy classes which have the same trace in every representation. (This works because the irreducible representations of $G/Z$ are exactly the irreducible representations of $G$ with kernel containing $Z$, and because the inverse images of the conjugacy classes of $G/Z$ in $G$ are unions of conjugacy classes of $G$.) Then iterate.

  • 2
    $\begingroup$ You can deduce solvability as well; you can easily find series of normal subgroups, then the order of their quotients needs to be prime. A good source for a lot of questions like this is Isaacs's Character Theory book; he doesn't discuss all these things in one place, but a lot of them are scattered throughout the text. $\endgroup$
    – Steve D
    May 21, 2012 at 19:15
  • $\begingroup$ Sorry, I meant "prime power" order above; if you can find all prime order quotients, the group is supersolvable! $\endgroup$
    – Steve D
    May 21, 2012 at 19:16
  • $\begingroup$ I think there is a typo in the first line- you mean "from the character table of $G$" I think. Solvability is similar. You can find all the minimal normal subgroups and their orders from the character table, and if $M$ ia any one of them, you can find the character table of $G/M$ from the character table of $G$. If $G$ is solvable, then all such $M$ should have prime power order. On the othe hand if all such $M$ have prime power order, youcan work inductively- if $G$ is not solvbale, you must find a normal subgroup $N$ so that $G/N$ has a minimal normal subgroup not of prime power order. $\endgroup$ May 21, 2012 at 19:17
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    $\begingroup$ @Geoff: Meanwhile I hope you are taking notes for your forthcoming definitive (but concise) text on finite group characters. Ideally all of this material would be set up as an extended exercise, but at the moment it's complicated to locate all details in the literature. $\endgroup$ May 21, 2012 at 22:16
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    $\begingroup$ @NathanielMayer Yes, because the commutator subgroup of the $p$-group, the commutator of the commutator, etc. are all normal subgroups, and the quotient of each one by the next is abelian. $\endgroup$
    – Will Sawin
    Oct 21, 2017 at 20:40

Lots of properties related to solvability can be deduced from the character table of a group, but perhaps it is worth mentioning one property that definitely cannot be so determined: the derived length of a solvable group. Sandro Mattarei constructed such examples, including examples of $p$-groups with identical character tables but different derived lengths. I think that no examples are known where the difference in derived lengths exceeds 1, however.


Although it is not possible in general to determine from the character table the order of the class representative labeling a column, it is nevertheless possible to determine all the prime divisors of this order. A proof of this fact can be found in [Chapter 22, Theorem 1.1 (vi)] of

Karpilovsky, Gregory Group representations. Vol. 1. Part B. Introduction to group representations and characters. North-Holland Mathematics Studies, 175. North-Holland Publishing Co., Amsterdam, 1992. pp. MR1183468

Another interesting fact is that given a class representative $g$ it is possible to know whether or not $g$ is a commutator. I have no reference for this at hand, but this is well known to my knowledge.

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    $\begingroup$ This is slightly off-topic but, to clarify Anvita's comment: In a finite group $G$, an element $g$ is a commutator if and only if $$\sum\limits_{\chi \in Irr(G)} \frac{\chi(g)}{\chi(1)}\neq 0.$$ This result is stated in the paper by Liebeck, O'Brien, Shalev and Tiep which proves the Ore conjecture; they remark that it follows from a result of Frobenius (see Lemma 2.5 of that paper). $\endgroup$
    – Nick Gill
    May 23, 2012 at 12:50
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    $\begingroup$ @Anvita: Do you know where the result quoted by Karpilovsky comes from? His books mostly consisted of compilations from research papers by other people, and I suspect one of these would be the original source. $\endgroup$ May 24, 2012 at 15:41
  • $\begingroup$ @Jim: In this book, Karpilovsky gives a proof following "an argument due to Isaacs (1976)" referring to the book M.I.Isaacs, Character Theory of Finite Groups, Academic Press, New York-San Fransisco- London. 1976. There is ineed such a proof in Isaacs' book (see chapter on Brauer's theorem there, Theorem (8.21) ) attributed to G.Higman, but no explicit reference given. $\endgroup$
    – Anvita
    Jun 3, 2012 at 22:33

For your reference requests, Marty Isaacs' book:

Character Theory of Finite Groups (Dover Books on Mathematics) [Paperback] I. Martin Isaacs (Author)

Is a classic.

  • $\begingroup$ See also comment thread on the accepted answer. $\endgroup$
    – Alex B.
    May 22, 2012 at 0:14
  • $\begingroup$ @Alex: yes, I should have looked at it first :( $\endgroup$
    – Igor Rivin
    May 22, 2012 at 0:30

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