Skip to main content

All Questions

Filter by
Sorted by
Tagged with
17 votes
1 answer
1k views

Explicit character tables of non-existent finite simple groups

In connection with the historical development of the classification of finite simple groups, I am interested in a particular aspect that seems to be less well-documented than the main narrative of ...
5 votes
1 answer
342 views

Product of all conjugacy classes

Related to this post of my coauthor Sebastien, I should also mention that one can also prove the following dual result: For any finite group G, the following identity holds: $$ \left(\prod_{j=0}^m \...
5 votes
0 answers
351 views

Adjoint identity on finite nilpotent groups

Let $G$ be a finite nilpotent group. Consider the following (adjoint) identity from [BuPa, Theorem 9.6]: $$\left(\prod_{\chi \in \mathrm{Irr}(G)} \frac{\chi}{\chi(1)} \right)^2 =\frac{|Z(G)|}{|G|}\...
12 votes
2 answers
926 views

Finite groups with integral character table

The character table of a finite group will be called integral if all its entries are integers. There are $11$ such groups up to order $16$, namely $C_1$, $C_2$, $C_2^2$, $S_3$, $D_8$, $Q_8$, $C_2^3$, $...
6 votes
1 answer
276 views

Reference request: an elementary result on characters of finite abelian groups

The referee of a paper I submitted to a journal asked me to include a reference of the following elementary result on characters of finite abelian groups: Let $A$ be a finite abelian group of order $...
3 votes
2 answers
315 views

Character kernels in the lattice of subgroups of a finite abelian group

I am looking for any efforts that have been made to characterize the character kernels (equivalently, the subgroups yielding cyclic quotients) inside the lattice of subgroups of a finite abelian group....
5 votes
1 answer
216 views

To whom is the internal characterization of $Q$-groups due?

A group is said to be a $Q$-group if the character of any complex representation is rational valued. A well-known internal characterization of $Q$-groups is the following: $G$ is a $Q$-group if and ...