Validating a Character Table for a Given Finite Group

Question

Suppose we have some finite group $G$, and we have computed the first few rows and columns of the table. I was wondering how one would verify that one have computed the correct character table of a given group (up to some permutation of the rows and columns of the entries). Presumably even after all the orthogonality conditions, a priori there might be multiple ways to complete the character table, at least without looking at the group/representations themselves.

From my understanding, it seems that in textbooks, one would exhibit the representation to a given character. In general how does one 'check' if one has the correct character table of a given group? Are there any way to avoid exhibiting these irreducible representations?

Daniel Robert-Nicoud gave an answer here mentioning Burnside's Algorithm, so in principle one could execute the algorithm on the given group $G$ and check if it matches your table but I don't think people would do this in real life, using pen and paper.

I've looked around for answer and it seems that most question are either about constructing a group from a given character table, or questions about how to compute a character table of a particular group. This is my first question on the site so I apologize if this question was asked previously.

Answer 1

In general, it is not easy to decide if a given matrix is a character table. For instance, the matrix $$\begin{pmatrix} 1&1&1&1\\ 1&1&-1&1\\ 2&2&0&-1\\ 6&-1&0&0 \end{pmatrix}$$ looks like the character table of the non-existent Frobenius group $C_7\rtimes S_3$ (I learned this example from Gunter Malle).

On the other hand, let me list some ad hoc strategies to construct a character table of a finite group $G$ by hand:

• Whenever you know a quotient $G/N$ you can inflate $\mathrm{Irr}(G/N)$ to $G$ (this is a no-brainer). In particular, you get all linear characters (i.e. degree 1) as inflations from $G/G'$. More generally, if $\chi\in\mathrm{Irr}(G)$ and $\chi_N\in\mathrm{Irr}(N)$, then $\chi\psi\in\mathrm{Irr}(G)$ for all $\psi\in\mathrm{Irr}(G/N)$ (Clifford theory).
• An irreducible character multiplied with a linear character is irreducible.
• Use the orthogonality relations. In particular, $|G|=\sum_{\chi\in\mathrm{Irr}(G)}\chi(1)^2$.
• Use $\chi(1)\bigm||G:Z(G)|\bigm|G|$ for $\chi\in\mathrm{Irr}(G)$. If $A\le G$ is abelian, then $\chi(1)\le|G:A|$ (a theorem of Ito).
• For $\chi\in\mathrm{Irr}(G)$, $g\in G$ and $z\in Z(G)$ we have $\chi(gz)=\chi(g)\chi(z)/\chi(1)$.
• If $\chi\in\mathrm{Irr}(G)$ and $g\in G$ has order $n$, then $\chi(g)$ is a sum of $\chi(1)$ many $n$-th roots of unity. In particular, $\chi(g)$ is an algebraic integer and $\chi(g)\in\mathbb{Z}$ if $g^2=1$. Moreover, $|\chi(g)|\le\chi(1)$ with equality if and only if $g\in Z(\chi)$ (a normal subgroup of $G$).
• For $H\le G$, one gets a permutation character $\pi$ from the (transitive) action of $G$ on the set of (left) cosets by (left) multiplication. The multiplicity of the trivial character $1_G$ as a constituent of $\pi$ is $1$. Also, $\pi-1_G$ is irreducible if and only if the action is $2$-transitive, i.e. $G=H\cup HgH$ for $g\in G\setminus H$.
• Every non-linear character has a zero (a theorem of Burnside).
• Every Galois automorphism of the cyclotomic field $\mathbb{Q}_{|G|}$ has the same cycle type on the rows as on the columns of the character table. In particular, if a character $\chi$ has a non-integer value, then there is a Galois conjugate character (with the same degree). Moreover, $|N_G(\langle g\rangle)/C_G(g)|=|\mathbb{Q}_{|\langle g\rangle|}:\mathbb{Q}(g)|$ for every $g\in G$. Hence, computing $|N_G(\langle g\rangle)/C_G(g)|$ hints which irrationalities to expect.
• If $\frac{|G|}{\chi(1)}$ is not divisible by a prime $p$, then $\chi$ vanishes on all elements whose order is divisible by $p$ (a theorem of Brauer).
• Multiply known characters and subtract known constituents. If you are lucky, you end up with a new irreducible character (irreducibility can be checked with the inner product).
• If $|G|$ is divisible by a prime $p$ only once, then the characters of $p'$-degree are known to a large extend (Brauer's theory of blocks of defect $1$, see Navarro's book "Characters and blocks of finite groups" for details).

Answer 2

Adding to the excellent answer of @BrauerSuzuki I wanted to comment on the point:

For $H≤G$, one gets a permutation character $π$ from the (transitive) action of $G$ on the set of (left) cosets by (left) multiplication. The multiplicity of the trivial character $1_G$ as a constituent of $π$ is 1. Also, $π−1_G$ is irreducible if and only if H is a maximal subgroup.

This is not quite correct (a counterexample is e.g. any group $G$ of odd order). A correct statement is: $π−1_G$ is irreducible if and only if $G$ acts 2-transitively on $G/H$ (or equivalently, $G=H\cup HgH$ for some $g\not\in H$). This implies that $H$ is maximal, but not conversely.