Why do the characters not determine the group?

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.

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.