# Validating a Character Table for a Given Finite Group

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.

• I am not certain what you are asking. Are you asking about how to complete a (possible) partial character table to a complete character table, or are you asking how to check whether a candidate for the complete character table is correct? In any case, it is (in general) much easier to compute the character table than the corresponding irreducible representations, so computing the representation is not a useful way of checking its accuracy. Commented Sep 4, 2021 at 7:42
• I was trying to compute a character table for a particular group, and just by using row/col orthogonality I noticed I had a free parameter on how to finish the table. I'm not sure if I just missed a relation, or perhaps some statement about some table that would have eliminated this choice. I guess I wasn't sure if you are supposed to be able to determine the entirety of the table if you only start with the first two rows and columns. Commented Sep 4, 2021 at 8:02
• For context, I had an assignment involving computing the character table for $D_n$ (of order $2n$). I'm reasonably sure that I have the right table and it wasn't too difficult to compute the irreducible representations. I'm wondering how one would know in general if they had the right table. Commented Sep 4, 2021 at 8:04
• For a specific group (as $D_n$) one starts with the linear characters (inflations from $G/G'$). Now for $D_n$ it is perhaps the easiest to notice that there is a $2$-dimensional real representation (after all $D_n$ is the group of symmetries of the regular $n$-gon). Then you can multiply the $2$-dimensional character with the linear characters. Alternatively you could induce linear characters from the normal subgroup $C_n$ (rotations) to $D_n$. Commented Sep 4, 2021 at 18:11
• This is really just an instance of a more general question, how do you check that the results of a mathematical calculation are correct? There is often a collection of sanity checks that you apply to the solution (such as orthogonality relations of a character table), but in the end, the only way is to re-check your calculations, or if possible recalculate using a different method. In the case of a computer calculation, if possible check again using a completely independent implementation (and of course check that the algorithm is theoretically correct). Commented Sep 6, 2021 at 7:54

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).

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.

• Thanks for pointing this out! I confused primitivity with 2-transitivity. I will correct my answer accordingly. Commented Jan 22, 2023 at 6:53
• Good observation. In principle this sort of observation should be made as a comment or possibly an edit on the answer, but I see that this is the first answer from this account. Sometimes moderators convert this sort of answer into a comment, but I'm choosing to leave this as is. Commented Jan 25, 2023 at 16:52