0
$\begingroup$

Prove that if $G$ is a finite nilpotent group and $x \in \omega(G)$, then $x$ acts by conjugation as a power automorphism on G.

$\omega(G)=\cap N_G(H)$ where each $H$ is subnormal in $G$.

A power automorphism is one which leaves every subgroup of $G$ invariant, i.e. $\forall H \le G, f(H)=H$.

I know every subgroup is subnormal and clearly $\omega(G)$ normalizes each, so $H^{x}=H$. However, I don't understand what the question is asking.

The wording "$x$ acts by conjugation as a power automorphism on $G$" aren't clear to me. Can someone please explain what the question means and how to answer it?

There is also another question that asks to show $\omega(G) \subseteq Paut(G)$, and I don't understand how can these be subsets. (Here $Paut(G)$ is the collection of power automorphisms on $G$.)

  • 0
    "$x$ acts by conjugation as a power automorphism on $G$" simply means that the homomorphism $f_x:g\mapsto g^x$ is a power automorphism. I think $\omega(G)\subseteq Paut(G)$ means that the map $\omega(G)\to aut(G)$ mapping $x$ to $f_x$ restricts to an injective map $\omega(G)\to Paut(G)$ – 2017-01-12
  • 0
    @RobertChamberlain Thanks so much. Leave your comment as an answer. – 2017-01-12

1 Answers 1

1

As requested, I leave my comment as an answer:

"$x$ acts by conjugation as a power automorphism on $G$" simply means that the homomorphism $f_x:g\mapsto g^x$ is a power automorphism.

I think $\omega(G)\subseteq Paut(G)$ means that the map $\omega(G)\to aut(G)$ mapping $x$ to $f_x$ restricts to an injective map $\omega(G)\to Paut(G)$.

  • 1
    Something seems to be wrong here: the center is in every normalizer, so it is in the Wielandt subgroup. Unless the group has a trivial center, your map is not injective. In an abelian group the Wielandt subgroup is the entire group, but every conjugation is trivial. You need to find a non-conjugation way to embed the Wielandt subgroup in the group of power automorphisms. – 2018-02-13