8
$\begingroup$

During my research, I encountered the class of finite groups $G$ satisfying the following conditions:

  1. $G=\langle x_1,\dots, x_n\rangle$
  2. All the $x_i$ belong to the same conjugacy class of $G$ (hence they have the same order, say $m$)
  3. $\langle x_i \rangle \cap \langle x_j \rangle\neq 1$ for all $i,j$.

My experiments using GAP indicate that a group $G$ with the properties described above has a non-trivial center. Therefore, I am currently working on proving this result which would be enough to prove what I need.

Remark: The statement is trivially true when $n=2$.

Approaches: I have verified the statement using brute-force computations and GAP's SmallGroups library for groups up to order 250. An inductive argument on the number of generators appears to be a promising approach, given the strength of the inductive hypothesis. However, I am having difficulty extending this argument to include the additional generator in the case where $n>2$.

I would appreciate any comments or suggestions on how to proceed with the proof. Thanks in advance.

$\endgroup$
2
  • $\begingroup$ Why is the statement true for $n=2$? Given $x,y \in G$ and positive integers $m$ and $n$ with $G=\langle x, yxy^{-1} \rangle$ and $x^m=(yxy^{-1})^n \ne e$, it is not clear that $x=yxy^{-1}$. $\endgroup$
    – Geoffrey Trang
    Commented Aug 7 at 17:26
  • 2
    $\begingroup$ @GeoffreyTrang: The subgroup $\langle x_1\rangle\cap\langle x_2\rangle$ is centralized by both $x_1$ and $x_2$. $\endgroup$
    – Steve D
    Commented Aug 7 at 17:27

1 Answer 1

Reset to default
10
$\begingroup$

Here is a counterexample.

$G = \langle x,y,z \rangle = S_{11}$ with

$x=(1,2)(3,4,5)(6,7,8,9,10)$,

$y = (1,2)(3,4,6)(7,8,9,10,11)$,

$z=(1,6)(3,4,5)(7,8,9,10,11)$.

$\endgroup$
3
  • $\begingroup$ +1. Just came to post (essentially) the same answer after playing around in GAP, but you beat me to it :) $\endgroup$
    – Steve D
    Commented Aug 7 at 17:24
  • $\begingroup$ Thanks for your fast and correct counterexample. Unfortunately there is one. However, what if the group is solvable, in addition to the assumptions above? Does it have non-trivial center this time? $\endgroup$
    – fspa
    Commented Aug 7 at 17:50
  • $\begingroup$ @fspa That should probably be a separate question. $\endgroup$
    – Derek Holt
    Commented Aug 8 at 7:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .