Looking at finite groups $G$. Of course the character table is (up to permutation of rows and columns) determined by $G$ up to isomorphism. I thought about why the converse is not true (question 1)?

Question 2

Given a complete set of characters of a finite group $G$, but not the group table (or the generators). What is exactly the minimum amount of information that is missing, necessary to determine the group $G$ (i.e. the group table) up to isomorphism, uniquely?

Own efforts

I have been looking at the famous example of the quaternion group $Q$ and the dihedral group of order $8$, $D_4$. They have up to permutation of rows and elements the same character table. However, they disagree in the order. I understand that a large part of the information necessary to determine the group table up to isomorphism must be contained in the character table, but I fail to pinpoint what is exactly the missing information in the general case.

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    $\begingroup$ Knowing that orders of elements correspond still does not determine the group up to isomorphism. But I would have to hunt around for an example. $\endgroup$
    – Derek Holt
    Commented Apr 5, 2018 at 15:05
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    $\begingroup$ For examples of non-isomorphic groups where orders of elements correspond see here. For some information, see this MO-question. $\endgroup$ Commented Apr 5, 2018 at 15:20
  • $\begingroup$ @DietrichBurde: Yes, that hits the nail on the head, thank you! (Except even the down to earth answers there on MO are quite involved (maybe could someone explain it in simpler terms?)). And then it seems to me that the issue is not completely settled - right? $\endgroup$ Commented Apr 5, 2018 at 15:32
  • $\begingroup$ Moreover, it seems that for Abelian groups indeed $G^\vee \simeq G$ holds and the trouble only occurs for non-Abelian $G$. $\endgroup$ Commented Apr 5, 2018 at 16:54

2 Answers 2


One possible answer is given by the following paper.

Hoehnke, H.-J.; Johnson, K. W. The 1-, 2-, and 3-characters determine a group. Bull. Amer. Math. Soc. (N.S.) 27 (1992), no. 2, 243–245.

Here is a brief explanation. For a character $\chi$ of a finite group $G$, define the corresponding $2$-character $\chi^{(2)} : G \times G \rightarrow \mathbb{C}$ by $$\chi^{(2)}(x,y) = \chi(x)\chi(y) - \chi(xy)$$ for all $x, y \in G$. We also define the corresponding $3$-character $\chi^{(3)}: G \times G \times G \rightarrow \mathbb{C}$ by $$\chi^{(3)}(x,y,z)=\chi(x)\chi(y)\chi(z)−\chi(x)\chi(yz)−\chi(y)\chi(xz)−\chi(z)\chi(xy)+\chi(xyz)+\chi(xzy)$$ for all $x, y, z \in G$.

There are similar definitions of $k$-characters $\chi^{(k)}$ and these go back to Frobenius. A paper of Formanek and Sibley from 1991 shows that the group determinant determines $G$. One consequence of this is that $G$ is determined by its irreducible characters $\chi$ and their $k$-characters $\chi^{(k)}$.

In their paper, Hoehnke and Johnson improved this by showing that the irreducible characters $\chi$ along with $\chi^{(2)}$ and $\chi^{(3)}$ suffice to determine $G$:

Theorem. Let $G$ be a finite group with complex irreducible characters $\chi_1$, $\ldots$, $\chi_t$. Then $G$ is determined up to isomorphism by the $\chi_i$, $\chi_i^{(2)}$, $\chi_i^{(3)}$, $1 \leq i \leq t$.

In a sense this result is optimal, since in the following paper Johnson and Sehgal show that the knowledge of $\chi$ and $\chi^{(2)}$ does not determine $G$ in general.

Johnson, Kenneth W.; Sehgal, Surinder K. The 2-character table does not determine a group. Proc. Amer. Math. Soc. 119 (1993), no. 4, 1021–1027.


The following papers might be of interest

  1. Sandro Mattarei. Character tables and metabelian groups. J. London Math. Soc. (2) 46 (1992), no. 1, 92-100.
  2. Sandro Mattarei. An example of $p$-groups with identical character tables and different derived lengths. Arch. Math. (Basel) 62 (1994), no. 1, 12-20.
  3. Sandro Mattarei. On character tables of wreath products. J. Algebra 175 (1995), no. 1, 157-178.

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