Timeline for Can group solvability be detected from identities among the generators?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22, 2013 at 17:01 | comment | added | Matt Brin | oops, see my comment to Alireza's answer. | |
| May 22, 2013 at 5:59 | comment | added | Matt Brin | one commutator is the product of (12) and (23) and the other is the product of (12) and (14). That is (12) multiplied (12) conjugated by the rotation (1234) "one way" and (12) multiplied by (12) conjugated by the rotation (1234) "the other way." I still get (123) and (124). | |
| May 21, 2013 at 6:25 | comment | added | Gerry Myerson | Write $a=(12)$, $b=(1234)$. Then $X_1$ just has the two elements $a^{-1}b^{-1}ab$ and $b^{-1}a^{-1}ba$ (and the identity), right? And I think these come to $(132)$ and $(123)$, respectively. | |
| May 21, 2013 at 1:32 | comment | added | Matt Brin | Your example is right if you go to a bigger group. For S_n, the set X_1 is {(123), (12n)}. Then X_2 still generates a Klein-4 group. Then X_3 is trivial, but S_n is not solvable. | |
| May 21, 2013 at 0:22 | comment | added | Matt Brin | I get X_1={(123), (142)}, and X_2={(14)(23), (12)(34)} which seems to generate Klein-4. I should recheck. | |
| May 21, 2013 at 0:08 | history | answered | Gerry Myerson | CC BY-SA 3.0 |