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The character table is square and invertible

Statement

Let $G$ be a finite group. The character table of $G$ has as many rows as columns, and the table matrix is invertible.

Facts & Assumptions

Given: A finite group $G$ with irreducible characters $χ_{1}, \dots, χ_{r}$ and conjugacy-class representatives $g_{1}, \dots, g_{s}$ .

[F1]

The irreducible characters form an orthonormal basis of the class functions, so there are as many of them as there are conjugacy classes (The irreducible complex characters form an orthonormal basis of $c f (G)$ ).

[F2]

The columns of the character table satisfy the second orthogonality relation: $\sum_{i} χ_{i} (g_{a}) \overline{χ_{i} (g_{b})} = ∣ C_{G} (g_{a}) ∣ δ_{a b}$ (The second orthogonality relation for irreducible complex characters).

Proof

technique · direct

1.1

By [F1], $r$ equals the number of conjugacy classes of $G$ , which is the number $s$ of columns; hence the table is square.

F1given

1.2

Rescale the $a$ -th column of the table by the positive number $1 / \sqrt{∣ C_{G} (g_{a}) ∣}$ , forming the matrix $U_{i a} = χ_{i} (g_{a}) / \sqrt{∣ C_{G} (g_{a}) ∣}$ . By [F2], distinct columns of $U$ are orthogonal and each column has squared norm $1$ , so $U$ has orthonormal columns.

F2givenalgebra

2.1

A square matrix with orthonormal columns has linearly independent columns, hence is invertible. Since $U$ is obtained from the table by multiplying columns by nonzero scalars, the table matrix is invertible too.

step 1.1step 1.2algebra∎

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Sources

Peter Webb, A Course in Finite Group Representation Theory, Theorem 3.4.3 (standard reference, not scraped)
Pavel Etingof et al., Introduction to Representation Theory, Section 3.5 (standard reference, not scraped)