3
$\begingroup$

I am working on a problem in character theory where I try to bound the derived length of a solvable group using information about its characters. In my specific case, it will be extremely helpful for me if I knew that the center was non-trivial. Here is what I know about the group:

  • $G$ is solvable.
  • the derived subgroup $G'$ is a $p$-group.
  • $G''$ is the unique minimal normal subgroup in G.
  • $G$ has a faithful irreducible character.

Of course, a lot of information can be deduced from the above. But in particular, I want to know if $G$ has a non-trivial center. This may not be deduced from the above information, but maybe if some additional condition is satisfied?

$\endgroup$
3
  • 2
    $\begingroup$ Any finite group with a unique minimal normal subgroup must have a faithful irreducible character, so your final condition is redundant. $\endgroup$
    – Derek Holt
    Oct 25, 2018 at 8:27
  • $\begingroup$ Your group has derived length $3,$ so I am curious to know how this fits with the context you state for the question ( in other words, how would it help to have a non-trivial center)? $\endgroup$ Oct 25, 2018 at 9:17
  • $\begingroup$ I am trying to prover that my group is supersolvable. Because it has a faithful irreducible character the center is cyclic and then so is G''. This (I think) will help me trying to find a normal series in which every quotient is cyclic. $\endgroup$ Oct 25, 2018 at 14:16

1 Answer 1

4
$\begingroup$

There are such groups with trivial centre. One such (possibly the smallest) is a group $G$ of order $448$ with the shape $2^{3+3}:7$. It has derived group $G'$ of order $64$, and $G''$ has order $8$ and is the unique minimal normal subgroup of $G$.

This is $\tt{SmallGroup}(448,179)$ in the databases in GAP and Magma. You can compute its character table, and it has two faithful irreducible characters of degree $14$.

Edit: In fact $\tt{SmallGroup}(108,17)$ with structure $3^{1+2}:4$ is a smaller example.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.