Timeline for Non split extension isomorphic (as a group) to a split extension
Current License: CC BY-SA 3.0
14 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Apr 13, 2014 at 16:36 | comment | added | Yassine Guerboussa | Perhaps this approach is useful. Assume that $B$ is a subgroup of $G$ which is isomorphic to $A$ as a $G$-module, such an isomorphism can be extended to an isomorphism $\phi : A \rightarrow B$ of $E$-modules. This induces an isomorphism between the abelian groups $Der(E,A)$ and $Der(E,B)$ in the obvious way. I wonder if $\phi$ induce a ring isomorphism or more generally if there is a ring isomorphism between $Der(E,A)$, $Der(E,B)$. If so Then $E$ splits over $B$. (The multiplication in $Der(G,A)$ is the composition of map) | |
| Apr 12, 2014 at 0:37 | comment | added | Gro-Tsen | I did some experimental testing with Gap, and (barring errors in my code) I've ruled out the existence of examples with the order of $E$ being, or, of course, dividing, one of: 128, 192, 288, 160, 240, 360, 400 or 336. | |
| Apr 11, 2014 at 13:41 | comment | added | Damien Robert | @Yves: thanks for the reference! There is another infinite example here mathoverflow.net/a/42184/26737 | |
| Apr 11, 2014 at 13:05 | comment | added | Derek Holt | @Geoff Robinson We want a finite group $E$ with two isomorphic abelian normal subgroups $N_1$ and $N_2$ such that: (i) $E/N_1$ and $E/N_2$ are isomorphic groups and, using this isomorphism, $N_1$ and $N_2$ are equivalent modules under the actions induced by conjugation; and (ii) $N_1$ has a complement in $E$ but $N_2$ does not. | |
| Apr 11, 2014 at 11:39 | comment | added | Derek Holt | @Yves yes, that was (almost) exactly the example I had in mind! | |
| Apr 11, 2014 at 10:59 | comment | added | YCor | About an easy infinite example: Let $A$ be the product (of order 8) of cyclic groups of order 2 and 4. Let $G$ be the (restricted or unrestricted) direct product of countably many copies of $A$. Then $G$ is isomorphic to $G\times G$ and can also obviously be decomposed as a non-split extension of $G$ by $G$. | |
| Apr 11, 2014 at 10:55 | comment | added | YCor | user.math.uzh.ch/ayoub/PDF-Files/DIRECT.PDF (J. Ayoub, The direct extension theorem. J. Group Theory 9 (2006), 307-316.) | |
| Apr 11, 2014 at 10:44 | comment | added | Gro-Tsen | @YvesCornulier: Where did Joseph prove this? | |
| Apr 11, 2014 at 8:08 | comment | added | Derek Holt | I can think of examples in which $A$ and $G$ are infinite. At the moment I would not like to bet either way on whether there exist finite examples. | |
| Apr 11, 2014 at 1:02 | comment | added | Damien Robert | @Robinson: but $(\Z/2\Z) \times A_5$ is not isomorphic as a group to $SL(2,5)$, no? | |
| Apr 10, 2014 at 21:36 | comment | added | YCor | More generally than the abelian case, J. Ayoub proved that a finite group that is a direct product $A\times B$ cannot be written as a non-split extension of $A$ by $B$. | |
| Apr 10, 2014 at 16:07 | history | asked | Damien Robert | CC BY-SA 3.0 |