Timeline for Do isomorphic semi-direct products correspond to conjugate automorphisms?
Current License: CC BY-SA 4.0
10 events
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| S Apr 22, 2021 at 20:45 | history | suggested | Mike Pierce | CC BY-SA 4.0 |
Improved the title and clean the body
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| Apr 22, 2021 at 17:56 | review | Suggested edits | |||
| S Apr 22, 2021 at 20:45 | |||||
| Jan 22, 2011 at 9:18 | vote | accept | Soluble | ||
| Jan 22, 2011 at 7:25 | comment | added | Soluble | Many books of algebra/group theory, and many articles on Semi-direct product give the following theorem: "If H is cyclic and f,g:H --> Aut(N) are homomorphisms, such that f(H) and g(H) are conjugate subgroups of Aut(N), then these two homomorphisms give isomorphic semi-direct product of N by H. But I didn't see, whether its converse is true, in any book, article, even not as an exercise, or any counterexample. I worked for small groups, but couldn't get. | |
| Jan 22, 2011 at 6:07 | comment | added | T.B. | @Rahul: I look back at the exercise and Zev was right. Sorry, although I do think you perhaps can get faster and more detailed response to this type of question at math.stackexchange.com as I did when I had tried it. | |
| Jan 22, 2011 at 6:04 | comment | added | David Roberts♦ | If you consider $f$ and $g$ to be functors from the one-object groupoids determined by $H$ to $Grp$ (with the object part picking out $H$), then there is a natural transformation $f\Rightarrow g$ precisely when for each $h\in H$ there is an $a\in Aut(N)$ such that $af(h)a^{-1} = g(h)$. I'm pretty sure that a natural transformation from $f$ to $g$ induces an isomorphism between the extensions determined by $f$ and $g$. You may need your isomorphism to commute with the maps to $H$. in order to reverse this argument. | |
| Jan 22, 2011 at 4:34 | comment | added | Soluble | Dear Tin Bui, I didn't see that this question in Dummit-Foote. I am asking it for following reason. If $G$ has a normal subgroup $N$, and a subgroup $H$, with $N\cap H=1$, then $G$ is a semidirect product of $N$ by $H$. Consider simple case when $H$ is cyclic. Then To determine possible dofferent semidirect products of $N$ by $H$, we can look for only those homomorphisms from $H$ to $Aut(N)$, such that the images of $H$ under homomorphisms are not conjugate. But is this sufficient to get non-isomorphic semidirect products? I worked for small groups, but I couldn't find counterexample. | |
| Jan 22, 2011 at 4:13 | answer | added | Alex B. | timeline score: 7 | |
| Jan 22, 2011 at 4:07 | comment | added | T.B. | This is a question from Dummit and Foote and I do not think it's suitable for mathoverflow. | |
| Jan 22, 2011 at 3:31 | history | asked | Soluble | CC BY-SA 2.5 |