Timeline for A finite group that has no decomposition of given cardinality
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 26, 2018 at 14:57 | comment | added | Taras Banakh | Unfortunately, my inductive "proof" of decomposability $A_n$ also failed. | |
| Nov 26, 2018 at 14:40 | comment | added | Ilya Bogdanov | No,sorry, I was wrong, as not any finite group is a product of Sylow subgroups (one per prime)... | |
| Nov 26, 2018 at 14:30 | comment | added | Taras Banakh | Very interesting! | |
| Nov 26, 2018 at 14:20 | comment | added | Taras Banakh | It seems that the decomposability of the alternating groups $A_n$ can be proved by induction on n. Now I am trying to write down the proof (not very complicated). | |
| Nov 26, 2018 at 14:15 | comment | added | Ilya Bogdanov | At least, not directly. Check what happens with $A_{35}$ and $a$ being the product of maximal powers of 3 and 5 in $|A_{35}$. | |
| Nov 26, 2018 at 13:25 | comment | added | Taras Banakh | It seems that your argument allows to prove that any alternating group $A_n$ is $a{\times}b$-decomposable for any $a,b$ with $ab=|A_n|$. Right? | |
| Nov 25, 2018 at 20:16 | history | answered | Ilya Bogdanov | CC BY-SA 4.0 |