Timeline for Finite indexed simultaneously normal subgroup in a normal union of two groups
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 21, 2011 at 20:40 | comment | added | HJRW | @Todd, @Igor - Actually '-ize' and '-ise' are both acceptable in British English; '-ize' is often referred to as the 'Oxford spelling' (see wikipedia). I think the key lesson here is not to try to correct others' spelling unless you actually have some idea about spelling yourself (and probably best not even then). | |
| Nov 21, 2011 at 18:50 | history | edited | user6976 | CC BY-SA 3.0 |
added a part of proof that was missing as was pointed out by Ashot Minasyan.
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| Nov 21, 2011 at 18:17 | comment | added | user6976 | @Ashot: Indeed, I missed the case when $A^g$ is inside $B$ (so $A^g=A^g\cap B$). But that case is even easier. | |
| Nov 21, 2011 at 17:08 | comment | added | Todd Leason | I think Drike (deleted comment) and the others are right: "normalize" is American English and "normalise" British English. More examples can be found on spelling.org/free/instructional/…. An example in mathematical context is the announcement talks.cam.ac.uk/talk/index/34171 where "generalise" appears. Sorry, if my comment led to confusion. | |
| Nov 21, 2011 at 16:03 | comment | added | user6976 | I am not a native English speaker, so my spelling cannot be trusted. Google gives 10 mil links for "normalize" and 3 mil links for "normalise". I guess this makes both versions possible. | |
| Nov 21, 2011 at 15:44 | comment | added | Igor Rivin | @Todd: perhaps @Drike is a native English speaker (as opposed to a native, or otherwise, American speaker)? Google "normalize" and you will be enlightened. | |
| Nov 21, 2011 at 13:25 | comment | added | Todd Leason | @Drike: Reading Mark's answer carefully also shows that the correct spelling is "normalize" (not "normalise"). | |
| Nov 21, 2011 at 12:44 | comment | added | Drike | @Mark Sapir : Thanks ! You are proving even more : if some group K normalises $A\cup B$, then it normalises $A\cap B$, and your key point is that a union of two distinct subgroups is not a group. | |
| Nov 21, 2011 at 12:17 | vote | accept | Drike | ||
| Nov 21, 2011 at 12:00 | history | edited | user6976 | CC BY-SA 3.0 |
added 105 characters in body
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| Nov 21, 2011 at 11:44 | history | answered | user6976 | CC BY-SA 3.0 |