## row sums of character tables

If you ever worked with finite groups, chances are you stared at group character tables, e.g. ones in Atlas of Finite Groups, or ones produced by a CA system, such as GAP. Each row stores a character $\chi$ of an irreducible representation of a group $G,$ columns correspond to the values $\chi(C_i)$ of $\chi$ on representatives of the conjugacy classes $C_1,\dots C_k$ of $G.$ The following must be known, but so far I haven’t found any references.

Proposition. $S_\chi = \sum_{i=1}^k \chi(C_i)$ is a nonnegative integer, namely
it is the multiplicity of $\chi$ in the character $\theta$ of $G$ acting on itself by conjugation.

Indeed, note that $\theta(C_i)=\frac{|G|}{|C_i|},$ as it counts the order of the centraliser in $G$ of an element in $C_i$. Then using the First Orthogonality Relation of characters on $\chi$ and $\theta$ completes the proof: $S_\chi=\frac{1}{|G|}\sum_j |C_j|\chi(C_j)\theta(C_j),$ as claimed.

Update (21.04.09) It turns out to be well-known: see e.g. Solomon, Louis, “On the sum of the elements in the character table of a finite group”. Proc. AMS 12(1961) 962–963.

The below conjecture was actually made by Richard L. Roth in “On the conjugating representation of a finite group”. Pacific J. Math. 36(1971), 515-521.

And it was disproved by Edward Formanek in “The conjugation representation and fusionless extensions”. Proc. AMS 30(1971), 73–74.

Question. When does $S_\chi=0 ?$ It does happen, e.g. take a nontrivial 1-dimensional representation of an abelian group (as an abelian group acts trivially on itself, only trivial character will occur in $\theta$).
More generally, if $\chi$ is an irreducible character such that $\chi(1_G)\neq \chi(z)$ for $1_G\neq z\in Z(G),$ then $\chi$ does not occur in $\theta,$ and so $S_\chi=0.$ Is the latter also only if? (Here $Z(G)$ denotes the centre of $G.$)

Update (21.04.09) See the 3rd reference above for a negative answer to this.

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