3
$\begingroup$

Let $G$ be a finite group. Its length $\ell(G)$ is the length $n$ of any Jordan-Hölder filtration $1=G_n\subset\cdots\subset G_0=G$ ($G_i$ is normal in $G_{i-1}$ and $G_{i-1}/G_i$ is simple).

Can $\ell(G)$ be read (easily) in the table of characters of $G$ ?

For instance $\ell(G)=1$ (that is $G\ne1$ is simple) if and only if there does not exist a character $\chi\ne\bf1$ such that $\chi(g)=\chi(e)$ for some $g\ne e$.

A suggestion for starting: consider the case of a $p$-group of order $p^m$. Because its J-H factors are $p$-groups and simple, they are all isomorphic to $C_p$ and therefore $\ell(G)=m$. Is there any general fact concerning the table of $p$-groups, that is directly related to $m$ ?

Improve this question
$\endgroup$
9
  • $\begingroup$ If I recall correctly, the character table does allow one to locate the normal subgroups (as unions of some conjugacy classes). From this you can probably recover the length, but I'd have to check the details. $\endgroup$
    – Jim Humphreys
    Commented Feb 2, 2017 at 15:01
  • $\begingroup$ @Jim. I agree with your first sentence (this is basically what I meant when I considered the case $\ell(G)=1$). However, once you know the list of normal subgroups, how do you reconstruct the Joradn-Hölder factors ? $\endgroup$
    – Denis Serre
    Commented Feb 2, 2017 at 15:17
  • $\begingroup$ Do you mean that $\ell(G) = 1$ if and only if there does not exist a character $\chi\neq 1$ such that...? $\endgroup$
    – benblumsmith
    Commented Feb 2, 2017 at 15:59
  • 1
    $\begingroup$ Note that the length of any solvable group is the number of prime factors in its order (counted with multiplicity). Moreover, the order of a group is recoverable from its character table, as $\sum \chi(1)^2$. Thus, to have a hope of building a counter-example, we need to use non-solvable groups. $\endgroup$
    – David E Speyer
    Commented Feb 2, 2017 at 17:35
  • 2
    $\begingroup$ David's point is that to answer your question in the negative you need two groups with the same character table but different lengths; this and the observation about the order shows that one of these groups must be non-solvable. $\endgroup$
    – Kevin Buzzard
    Commented Feb 2, 2017 at 21:56

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .