Why do the characters not determine the group?

Question

Motivation/Context

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.

Answer 1

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.

Answer 2

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.