2
$\begingroup$

Let $S$ be a non-trivial simple group and suppose $S \trianglelefteq G$ if $C_G(S)=1$ then $S$ is characteristic in $G$. To prove this let $\phi$ be an automorphism of $G$ and note that the intersection $S\cap \phi(S)$ can't be trivial since otherwise $S$ commutes with $\phi(S)$ in $G$. Therefore since both $S$ and $\phi(S)$ are simple $S =S \cap \phi(S) = \phi(S)$.

Is there a non-trivial perfect centerless group $P \trianglelefteq G$ such that $C_G(P)=1$ which is not a characteristic subgroup of $G$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Let $S$ be a finite simple group and $V$ a faithful absolutely irreducible module for $S$. Then $W = V \otimes V$ is a faithful irreducible module for $S \times S$.

Let $G = W \rtimes (S \times S)$ be the corresponding semidirect product of $W$ with $S \times S$. Then $G$ has two normal subgroups of the form $W \rtimes S$, and they are both perfect with trivial centralizer in $G$, but they are not characteristic in $G$, because there is an automorphism that interchanges them.

For example, we could take $S = A_5 \cong {\rm SL}(2,4)$ and $V$ the natural module of order $4^2$, giving a group $G$ of order $2^8 \times 60^2=921600$. (I checked that one on the computer.)

Improve this answer
$\endgroup$
5
  • $\begingroup$ Why don't you use $\rtimes$ (rtimes)? $\endgroup$
    – YCor
    Commented Oct 9, 2015 at 13:26
  • 1
    $\begingroup$ @YCor It's the ATLAS notation for a split extension. $\endgroup$
    – Jay Taylor
    Commented Oct 9, 2015 at 13:35
  • $\begingroup$ I've changed it to $\rtimes$, which is probably preferable because it is more standard. The ATLAS notation is useful for conciseness. $\endgroup$
    – Derek Holt
    Commented Oct 9, 2015 at 13:42
  • $\begingroup$ Thank you for your answer. I hope you don't mind if I ask a few questions. By a faithful module $V$ for $S$ do you mean an abelian group $V$ together with a monomorphism from $S$ to $Aut(V)$? By an irreducible module do you mean one which has no non-trivial submodules? What is an absolutely irreducible module? Finally what is the underlying abelian group for the natural module of order $4^2$? $\endgroup$
    – Nex
    Commented Oct 13, 2015 at 7:13
  • $\begingroup$ This is all standard terminology, and you could find the answers on wikipedia or planetmath. The underlying abelian group of any module over a finite field is elementary abelian. $\endgroup$
    – Derek Holt
    Commented Oct 13, 2015 at 8:30

You must log in to answer this question.

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