Timeline for Can any reductive $k$-group be written as a semidirect product of $k$-linear groups?
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9 events
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| Oct 26, 2016 at 18:22 | comment | added | LSpice | What does "in a non-trivial way" mean? If $G$ is connected and semisimple, then your exact sequence is $1 \to G \to G \to 1 \to 1$. | |
| Oct 12, 2014 at 22:43 | comment | added | Jim Humphreys | @user: "I've read .... "? Do you recall where this is written? | |
| Oct 4, 2014 at 11:08 | comment | added | Paul Levy | Perhaps a better question is whether such a splitting exists if we replace $(G,G)$ by its connected component and similarly for $G/(G,G)$. I think the answer is still no: suppose $G^\circ=\GL_n$ and $G/G^\circ$ is cyclic of order 2, generated by $\theta$ satisfying $\theta g= ({^{t}}g^{-1})\theta$ and $\theta^2=-I$. If $n$ is even then there is a splitting (since $-I$ is the square of an element of $\SO_n$) but if $n$ is odd, then there is no element of $\theta G^\circ=G\setminus G^\circ$ of order 2. However, I think the corresponding statement might be true if we assume $G^\circ$ is simple. | |
| Sep 26, 2014 at 0:47 | comment | added | Venkataramana | As to the original question, take $G$ to be a finite group (e.g the group of upper triangular unipotent matrices matrices with entries in $Z/3Z$. Then the commutator is the centre and the extension above does not split. If it did, the group would be abelian. | |
| Sep 26, 2014 at 0:43 | comment | added | Venkataramana | @anon: what you say is not right: take the torus inside the diagonals in $GL_n$ all of whose entries except the first one are $1$. This maps iso to $GL_n/SL_n=G_m$. | |
| Sep 25, 2014 at 22:05 | review | Close votes | |||
| Sep 26, 2014 at 15:32 | |||||
| Sep 25, 2014 at 18:02 | history | edited | user58639 | CC BY-SA 3.0 |
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| Sep 25, 2014 at 17:55 | review | First posts | |||
| Sep 25, 2014 at 17:58 | |||||
| Sep 25, 2014 at 17:51 | history | asked | user58639 | CC BY-SA 3.0 |