Timeline for Why do we define quotient groups for normal subgroups only?
Current License: CC BY-SA 4.0
27 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jun 22, 2022 at 11:01 | history | bounty ended | Martin Sleziak | ||
| S Jun 22, 2022 at 11:01 | history | notice removed | Martin Sleziak | ||
| S Jun 16, 2022 at 7:14 | history | bounty started | Martin Sleziak | ||
| S Jun 16, 2022 at 7:14 | history | notice added | Martin Sleziak | Reward existing answer | |
| S Jun 16, 2022 at 7:14 | history | bounty ended | Martin Sleziak | ||
| S Jun 16, 2022 at 7:14 | history | notice removed | Martin Sleziak | ||
| Jun 14, 2022 at 18:00 | history | tweeted | twitter.com/StackMath/status/1536770452730920963 | ||
| S Jun 14, 2022 at 7:10 | history | bounty started | Martin Sleziak | ||
| S Jun 14, 2022 at 7:10 | history | notice added | Martin Sleziak | Reward existing answer | |
| S Mar 11, 2021 at 1:08 | history | suggested | Jollywatt | CC BY-SA 4.0 | Consistently use `*` for group operation. Don’t need revision history in question. |
| Mar 10, 2021 at 23:22 | review | Suggested edits | |||
| S Mar 11, 2021 at 1:08 | |||||
| Jan 22, 2019 at 4:34 | history | edited | Martin Sleziak | edited tags | |
| May 13, 2018 at 21:55 | comment | added | lhf | See math.stackexchange.com/a/1014784/589 | |
| May 13, 2018 at 20:45 | answer | added | user1551 | timeline score: 4 | |
| Sep 24, 2015 at 13:51 | history | edited | Martin Sleziak | edited tags | |
| Dec 16, 2010 at 19:50 | comment | added | user1119 | An alternative definition of normal subgroup might of interest. The algebra book of your choice would probably contain a proposition that says "kernels of homormorphisms are precisely normal subgroups". That might help you. | |
| Dec 14, 2010 at 17:26 | answer | added | Arturo Magidin | timeline score: 196 | |
| Dec 14, 2010 at 14:57 | vote | accept | Aleksei Averchenko | ||
| Dec 14, 2010 at 14:09 | comment | added | Matt E | Yes (as is easily checked). | |
| Dec 14, 2010 at 13:33 | answer | added | Tobias Kildetoft | timeline score: 70 | |
| Dec 14, 2010 at 13:07 | history | edited | Aleksei Averchenko | CC BY-SA 2.5 | added 67 characters in body |
| Dec 14, 2010 at 13:02 | comment | added | Aleksei Averchenko | Do you mean that if $xH = zH$ then it is necessary for $H$ to be a normal subgroup in order for the equality $(xy)H = (zy)H$ to hold for all $y \in G$? | |
| Dec 14, 2010 at 12:34 | answer | added | Alf | timeline score: 8 | |
| Dec 14, 2010 at 12:30 | comment | added | Alex B. | To define the group operation, you have to fix coset representatives. You need to check that the coset you get as a result of the group operation is independent of the coset representatives you chose. But that is only true if the subgroup is normal. So before you "trivially show" that your operation gives a group, you should check that you have a well-defined operation on cosets in the first place. | |
| Dec 14, 2010 at 12:22 | comment | added | Aleksei Averchenko | What do you mean by well-defined? It is a group, which is trivial to show. | |
| Dec 14, 2010 at 12:20 | comment | added | KCd | That operation on cosets is well-defined if and only if H is a normal subgroup. If H is just a subgroup, what you call "left quotient group" has the more standard name "set with a left group action". More precisely, the coset spaces G/H describe essentially all the examples of sets with transitive left G-actions. It is part of the general theory of groups acting on sets. | |
| Dec 14, 2010 at 11:54 | history | asked | Aleksei Averchenko | CC BY-SA 2.5 |