2
$\begingroup$

Let $G$ be a group and $H, K\unlhd\ G$ (normal subgroups), with $K\leq H$ such that $\left[H, Aut(G)\right]\leq K$. Why does this imply that $H/K\leq Z(G/K)$? $Z(G)$ denotes the center of $G$.

$\endgroup$
  • $\begingroup$ What do you mean by $[H,Aut(G)]$? Where is this commutator taken? $\endgroup$ – Tobias Kildetoft May 8 '13 at 16:45
  • $\begingroup$ An element in $[H, Aut(G)]$ is described by $h^{-1}h^\alpha$ where $h\in H$, $\alpha\in Aut(G)$ and $h^\alpha$ is the image of $h$ by the automorfism $\alpha$. $\endgroup$ – graozovsky May 8 '13 at 17:14
  • 1
    $\begingroup$ Hint: Consider the automorphism $\varphi_g$ of $G$ that conjugates by some fixed $g\in G$. The fact that $[H,\varphi_g]\leq K$ then tells you what? $\endgroup$ – Tobias Kildetoft May 8 '13 at 17:23
  • $\begingroup$ I've seen it. Thanks! $\endgroup$ – graozovsky May 8 '13 at 17:26
3
$\begingroup$

Here is my comment as an answer with the details:

We know that for all $\varphi\in\rm{Aut}(G)$ and all $h\in H$ we have $h^{-1}\varphi(h)\in K$. In particular, this holds for the automorphisms $\varphi_g$ given by $\varphi_g(x) = g^{-1}xg$ (for $g\in G$).

Thus, we have that for all $h\in H$ and $g\in G$, $h^{-1}\varphi_g(h) = h^{-1}g^{-1}hg = [h,g]\in K$ which precisely means that $H/K\leq Z(G/K)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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