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 $G$.

  • $\begingroup$ 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
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    $\begingroup$ 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

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 $H/K\leq Z(G/K)$.

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