Timeline for p-subgroups of automorphisms of $p$ groups
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 11, 2011 at 14:23 | answer | added | Maurizio Monge | timeline score: 2 | |
| Feb 10, 2011 at 15:01 | comment | added | Someone | @Maurizio: You are right. I hadn't expected it to be true and didn't want to look for a counterexample. As you solved Martin's question completely, I wouldn't mind if you post your solution as answer. | |
| Feb 10, 2011 at 14:13 | comment | added | Maurizio Monge | Actually the construction provided by Someone does provide a composition series made by groups that are normal in $G$: in fact, $H_{i+1}=[H_i,H]$ is the biggest quotient of $H_i$ on which $H$ acts trivially by conjugation, and in particular $G$ acts trivially and all the intermediate subgroups $H_{i+1} \subseteq K \subseteq H_i$ are normal in $G$. | |
| Feb 10, 2011 at 13:11 | comment | added | Someone | @Martin: You might call them $P$-characteristic. They don't have to be invariant under automorphisms not in $P$. Yes, the $H_i$ are a normal series which you can refine. | |
| Feb 10, 2011 at 13:08 | comment | added | Martin David | The subgroups $[H_i,H]$ are characteristic subgroups of $G$, isn't it? Then the series $1\subseteq H_r\subseteq \cdots \subseteq H_0=G$ will be a normal series (i.e. all subgroups in the series will be normal in $G$), and then we can refine the series. | |
| Feb 10, 2011 at 13:03 | history | edited | Martin David | CC BY-SA 2.5 |
added 90 characters in body
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| Feb 10, 2011 at 13:01 | comment | added | Someone | I should have mentioned that $u(x) x^{-1}$ is simply the commutator $[u, x]$ as elements of $G \rtimes P$. | |
| Feb 10, 2011 at 12:59 | comment | added | Someone | $G$ is normal in the semi-direct product $G\rtimes P =: H$. Define $H_0 := G$ and $H_{i+1} := [H_i, H]$ for $i\ge 0$. All $H_i$ are normal in $H$ and $H_{i+1} < H_i$ as long as $H_i \ne 1$. Refining the chain $H_0 > H_1 > H_2 >...> 1$ such that all factors are cyclic instead of abelian, you are done. But I doubt that you can choose the refinement to contain only normal subgroups. | |
| Feb 10, 2011 at 12:40 | history | asked | Martin David | CC BY-SA 2.5 |