Timeline for Nice examples of groups which are not obviously groups
Current License: CC BY-SA 4.0
68 events
| when toggle format | what | by | license | comment | |
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| Oct 20, 2024 at 9:10 | comment | added | suckling pig | Again, we need more big lists. So inane. | |
| Oct 20, 2024 at 9:03 | answer | added | P Vanchinathan | timeline score: 0 | |
| Mar 6, 2021 at 22:59 | history | protected | user26857 | ||
| Dec 29, 2019 at 20:38 | history | edited | Simon Fraser | CC BY-SA 4.0 | deleted 1 character in body |
| Apr 7, 2017 at 13:07 | answer | added | tattvamasi | timeline score: 3 | |
| Mar 19, 2017 at 16:05 | answer | added | Andres Mejia | timeline score: 3 | |
| Sep 27, 2016 at 19:33 | review | Close votes | |||
| Sep 28, 2016 at 0:41 | |||||
| Mar 12, 2016 at 11:38 | answer | added | J.-E. Pin | timeline score: 3 | |
| Feb 12, 2016 at 14:20 | history | edited | johnny09 | CC BY-SA 3.0 | corrected spelling |
| Apr 19, 2015 at 1:38 | answer | added | jop | timeline score: 14 | |
| Feb 15, 2015 at 10:54 | answer | added | MattAllegro | timeline score: 3 | |
| Feb 1, 2015 at 22:15 | answer | added | MattAllegro | timeline score: 3 | |
| Jan 29, 2015 at 1:12 | answer | added | Joseph Zambrano | timeline score: 9 | |
| Jan 29, 2015 at 0:44 | answer | added | Timotej | timeline score: 1 | |
| Jan 29, 2015 at 0:08 | answer | added | MattAllegro | timeline score: 1 | |
| Jan 28, 2015 at 13:17 | answer | added | KCd | timeline score: 20 | |
| Apr 10, 2014 at 22:07 | answer | added | Man | timeline score: 5 | |
| Jan 8, 2014 at 4:48 | answer | added | janmarqz | timeline score: 8 | |
| Sep 3, 2013 at 15:14 | answer | added | Seirios | timeline score: 26 | |
| May 8, 2013 at 18:30 | comment | added | Giuseppe Negro | @ DominicMichaelis: @julien: This must be related to the fact that this question is on top of the "Hottest Questions this Month" page, which is what you get when you access the site without previous login. So it is receiving a lot of views from unregistered users. | |
| May 3, 2013 at 8:54 | history | edited | Dominic Michaelis | CC BY-SA 3.0 | added 3 characters in body |
| Apr 23, 2013 at 16:59 | history | edited | Dominic Michaelis | CC BY-SA 3.0 | added 84 characters in body |
| Apr 22, 2013 at 4:54 | history | tweeted | twitter.com/#!/StackMath/status/326197204194848768 | ||
| Apr 18, 2013 at 1:29 | comment | added | Nick Matteo | The axiom of choice is equivalent to the statement that for any nonempty set there is a binary operation which makes that set into a group. | |
| Apr 17, 2013 at 17:10 | answer | added | Josh | timeline score: 8 | |
| Apr 17, 2013 at 17:03 | comment | added | Orat | You may also interested in a surprising isomorphism $(\mathbb{Z}[x], +) \cong (\mathbb{Q}_\text{pos}, \times)$. (Use the fundamental theorem of arithmetic.) | |
| Apr 17, 2013 at 14:50 | comment | added | Julien | Just curious: how does the number of views keep increasing exponentially when the question is no longer on the front page? | |
| Apr 17, 2013 at 3:04 | answer | added | user17945 | timeline score: 8 | |
| Apr 16, 2013 at 21:44 | answer | added | 2'5 9'2 | timeline score: 55 | |
| Apr 16, 2013 at 21:30 | answer | added | 2'5 9'2 | timeline score: 44 | |
| Apr 16, 2013 at 21:23 | comment | added | MyUserIsThis | Oh, I just saw you wrote my example yourself in the question... I feel stupid. | |
| Apr 16, 2013 at 21:22 | comment | added | MyUserIsThis | Symmetrical difference induces a group structure on a universe of sets. Not REALLY surprising, but it looked a bit weird the first time I saw it. | |
| Apr 16, 2013 at 21:05 | comment | added | Yong Hao Ng | I just learned that the set of Cauchy sequences can form groups or rings, does that count? This is obviously coming from the idea of completion. i.e. if $(x_k)$ and $(y_k)$ are 2 Cauchy sequences it turns out $(x_k)+(y_k)$ and $(x_k)\times (y_k)$ are also Cauchy sequences and they form a commutative ring with identity. However, if this is valid perhaps someone else should write it as an answer since I am still learning. | |
| Apr 16, 2013 at 20:05 | comment | added | Dominic Michaelis | @YACP mh maybe I gonna ask them when I am more in the theory of those. Well as Martin said, it is more the question: "what remains when one removes the uninteressting stuff?" | |
| Apr 16, 2013 at 19:58 | comment | added | user26857 | Just wondering: what's next? Nice examples of rings/fields/modules which are not obviously rings/fields/modules? | |
| Apr 16, 2013 at 15:08 | answer | added | Rasmus | timeline score: 10 | |
| Apr 16, 2013 at 13:42 | answer | added | MJD | timeline score: 137 | |
| Apr 16, 2013 at 13:22 | answer | added | Tobias Kildetoft | timeline score: 25 | |
| Apr 16, 2013 at 4:22 | answer | added | user641 | timeline score: 14 | |
| Apr 16, 2013 at 4:19 | answer | added | user641 | timeline score: 27 | |
| Apr 15, 2013 at 21:14 | comment | added | Aaron Mazel-Gee | @MartinBrandenburg: You can see it that way, or you can also just pinch the last suspension coordinate. (Like you say, these are essentially the same observation.) Of course, either argument gives that any $[\Sigma A,-]$ admits a group structure. | |
| Apr 15, 2013 at 20:33 | answer | added | user44441 | timeline score: 151 | |
| Apr 15, 2013 at 20:11 | comment | added | Martin Brandenburg | @Aaron: Yes but this is just an equivalent problem, to find a cogroup structure on $S^n$. Of course this is also trivial when one already knows something about loop spaces ... seems to depend on the perspective. | |
| Apr 15, 2013 at 18:58 | comment | added | Aaron Mazel-Gee | @HSN: For $n\geq 1$, the group structure on $\pi_n(X)$ just comes from the homotopy co-group structure on $S^n$, which is commutative for $n \geq 2$. Does this count as "not obvious"? From the functorial perspective, it's clear that this is actually the only possible way to have a natural group structure on a functor of the form $[A,-]$. | |
| Apr 15, 2013 at 18:37 | answer | added | Qiaochu Yuan | timeline score: 72 | |
| Apr 15, 2013 at 18:28 | comment | added | Qiaochu Yuan | Sometimes a group will be isomorphic to a group which is obviously a group, but the isomorphism itself isn't obvious. Let's keep this in mind. ("Obvious" is a property of a description of a group, not a property of a group.) | |
| Apr 15, 2013 at 18:28 | history | made wiki | Post Made Community Wiki by Qiaochu Yuan | ||
| Apr 15, 2013 at 18:24 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 | edited title |
| Apr 15, 2013 at 17:22 | answer | added | Julien | timeline score: 55 | |
| Apr 15, 2013 at 17:10 | answer | added | Jyrki Lahtonen | timeline score: 36 | |
| Apr 15, 2013 at 16:32 | answer | added | user54358 | timeline score: 5 | |
| Apr 15, 2013 at 16:12 | comment | added | Dominic Michaelis | @MartinBrandenburg No it was thought as out of the blue, as my knowledge about general constructions of groups is very limited, and I am searching for some nice example when one learns what a group is. | |
| Apr 15, 2013 at 16:09 | comment | added | user1729 | @MartinBrandenburg: I presume the author is merely trying to find realisations of groups where it is not obvious that the realisation is a group. So, for example, an HNN-extension of a group is always a group, and this is a pretty much trivial assertation... | |
| Apr 15, 2013 at 16:02 | comment | added | Martin Brandenburg | And your question also doesn't refer to general constructions of groups, right? For example groups defined by generators and relations, stabilizer groups of group actions, automorphism groups of objects of categories, semidirect products (in particular wreath products, holomorphs), free products, amalgamated sums, HNN extensions, etc. I have to ask since this already produces millions of interesting examples and even more unsolved questions about them, but your question suggests that a group has to come "out of the blue" in order to be non-trivial. | |
| Apr 15, 2013 at 15:40 | comment | added | Dominic Michaelis | @MartinBrandenburg yeah I would like to have some examples for those. | |
| Apr 15, 2013 at 15:36 | comment | added | Martin Brandenburg | @Dominic: Do monoids (which are obviously monoids) also count, who surprisingly turn out to be groups? | |
| Apr 15, 2013 at 15:34 | comment | added | HSN | @BenjaLim: I think I wrote homotopy groups, not homology groups. I agree that these are pretty obviously groups, indeed. I'm sorry if I haven't been clear there. | |
| Apr 15, 2013 at 15:23 | answer | added | Martin Brandenburg | timeline score: 183 | |
| Apr 15, 2013 at 15:18 | comment | added | user38268 | @HSN Homology groups are obviously groups. You are taking a quotient of a subgroup by another subgroup. How is that not a group? | |
| Apr 15, 2013 at 15:14 | history | edited | Alexander Gruber♦ | CC BY-SA 3.0 | edited body |
| Apr 15, 2013 at 15:11 | answer | added | Stano | timeline score: 21 | |
| Apr 15, 2013 at 15:10 | answer | added | user38268 | timeline score: 34 | |
| Apr 15, 2013 at 15:09 | comment | added | Jim | Homotopy groups of sphere's. Pretty easy to give an intuitive definition but it's very unintuitive that they have a group law at all, let alone that they are generally abelian. | |
| Apr 15, 2013 at 15:09 | answer | added | Alexander Gruber♦ | timeline score: 95 | |
| Apr 15, 2013 at 15:07 | comment | added | HSN | Maybe the fundamental group is one you're looking for - or homotopy groups, more generally? Since it is obvious that you can compose based maps, while it still isn't immediate that they form groups - you need to do some work for that. Apart from that, many subgroups of symmetric groups $S_n$ arise in examples and those can be well-hidden groups, I think. This may not be what you're looking for at all, though. | |
| Apr 15, 2013 at 15:07 | comment | added | user1729 | @Shahab: Well, it is obviously a group. The clue is in the name... (also, hand-waving, as groups are symmetries and all symmetries are groups, then this is obvious...) | |
| Apr 15, 2013 at 15:03 | comment | added | user10575 | Automorphism group of a graph? | |
| Apr 15, 2013 at 15:02 | history | asked | Dominic Michaelis | CC BY-SA 3.0 |