Timeline for On order of subgroups in abelian groups
Current License: CC BY-SA 4.0
16 events
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| Aug 1, 2018 at 13:08 | comment | added | Gerry Myerson | I thought it was a reference to the painter, Homonymous Bosch. | |
| Aug 1, 2018 at 4:59 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
Removed the deprecated (abstract-algebra) tag - see the tag info: https://mathoverflow.net/tags/abstract-algebra/info (if there are some other suitable tags, choose them instead.)
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| Aug 7, 2012 at 4:01 | history | edited | José Hdz. Stgo. | CC BY-SA 3.0 |
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| Aug 7, 2012 at 3:53 | history | edited | José Hdz. Stgo. | CC BY-SA 3.0 |
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| Feb 8, 2011 at 22:24 | history | edited | José Hdz. Stgo. | CC BY-SA 2.5 |
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| May 28, 2010 at 22:26 | vote | accept | José Hdz. Stgo. | ||
| Mar 24, 2010 at 1:58 | comment | added | José Hdz. Stgo. | Actually, I meant same name as the present MO discussion. | |
| Mar 24, 2010 at 0:47 | comment | added | Pete L. Clark | @JHS: Yes, that's what homonymous means. Is there another paper called "On Orders of Subgroups in Abelian Groups: An Elementary Solution of an Exercise of Herstein"? (If you mean that the paper has the same name as itself, then you're using the term in a way I haven't seen before. I honestly didn't understand until now.) | |
| Mar 23, 2010 at 21:45 | comment | added | José Hdz. Stgo. | The only (big) theorem proved up to that point is Lagrange's one. Of course, he's already developed the standard criteria for subgroups and the formula $|HK| = |H||K|/|H \cap K|$ (when both $H$ and $K$ are finite subgroups of $G$). The notions of normality, quotient group, and homomorphism won't be featured until subsequent sections. @Pete: Isn't "homonymous" supposed to mean having the same name? | |
| Mar 23, 2010 at 14:01 | comment | added | Pete L. Clark | I like Tobias's solution: it seems simple, direct and powerful. Since the Monthly article exists, I gather the use of quotient groups must be out of bounds? | |
| Mar 23, 2010 at 13:52 | comment | added | Pete L. Clark | I don't have immediate access to the article, but: why "homonymous"? | |
| Mar 23, 2010 at 11:53 | comment | added | Tobias Kildetoft | I am not completely familiar with what theorems are allowed for this, but proving that an abelian group has a subgroup of any order dividing the order of the group requires only very little. Namely the existence of a subgroup of order $p$ for a prime $p$ dividing the order of the group (a very simple proof for abelian groups) and the correspondence of subgroups containing $H$ and subgroups of $G/H$. Now the result follows by induction. | |
| Mar 23, 2010 at 7:44 | answer | added | Robin Chapman | timeline score: 9 | |
| Mar 23, 2010 at 7:14 | history | edited | José Hdz. Stgo. | CC BY-SA 2.5 |
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| Mar 23, 2010 at 6:56 | history | edited | José Hdz. Stgo. | CC BY-SA 2.5 |
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| Mar 23, 2010 at 6:49 | history | asked | José Hdz. Stgo. | CC BY-SA 2.5 |