Timeline for Examples of common false beliefs in mathematics
Current License: CC BY-SA 2.5
24 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 8 at 18:14 | comment | added | მამუკა ჯიბლაძე | Before seeing this, I was convinced that I've been taught homological algebra too early :D | |
| Jan 21 at 0:31 | comment | added | printf | I must say that I have also wondered about this, but I decided against this based on the following "meta" reason: if this were true, it would be a very important result that would be discussed (or at least mentioned) in every textbook and course on group theory. Since, however, no textbooks mentioned this, I concluded it was not true... | |
| Nov 17, 2022 at 1:12 | comment | added | Sidharth Ghoshal | basically you can attack it like this, we define our product: $(n_1, q_1) * (n_2, q_2) = (f(n_1,q_1,n_2,q_2), q_1q_2)$ Now if you insist this is associative you can start adding constraints on f $((n_1, q_1) * (n_2, q_2)) * (n_3, q_3) = (n_1, q_1) * ((n_2 , q_2) * (n_3, q_3))$ $(f(n_1,q_1,n_2,q_2), q_1q_2, n_3, q_3) = f(n_1, q_1, f(n_2,q_2,n_3,q_3), q_2q_3)$ and then with a lot of pain and suffering this becomes nicer but still has some strange cocycle action. I'm 99.9% sure this is how it was found historically since it only depends on first principles and wish they taught us this. | |
| Nov 17, 2022 at 1:03 | comment | added | Sidharth Ghoshal | Yea in our abstract algebra class I always found the definition of the semidirect product strange (like its amazing it works but where did this come from?). Much later I TRIED to just classify all the ways that a normal subgroup and quotient group could be "combined" (and of course failed) but realized when you try to express their multiplication you end up with rather odd looking formula that depends on an automorphism, a cocycle, and something else, and by setting that cocycle parameter to the identity you can recover the semidirect product. | |
| Jan 17, 2022 at 20:46 | comment | added | Qiaochu Yuan | @Whelp: yes, take $G = C_4, N = C_2$, so that $N \times G/N = C_2 \times C_2$. | |
| Jan 14, 2022 at 13:42 | comment | added | Whelp | Is there a counterexample to N x G/N being homomorphic to G? | |
| Apr 3, 2021 at 3:45 | comment | added | Cayley-Hamilton | @DuchampGérardH.E. $G$-grp is actions of a group on a group by group homomorphisms, and the other category is groups with maps from $G$. | |
| Apr 3, 2021 at 3:33 | comment | added | Duchamp Gérard H. E. | @DeanYoung Could you clarify the definition of your categories $G$-Grp and $G\uparrow$ Grp ? (thanks) | |
| Feb 10, 2021 at 8:31 | comment | added | Simon1729 | I just got that one wrong even after having read this comment a few months ago. I thought that every solvable group is a semidirect product of cyclic groups. (Quaternion group is a counterexample) | |
| Aug 12, 2020 at 1:48 | comment | added | Ali Taghavi | What is a counter example on the context of CONNECTED lie groups.That is all things are connected G,N, G/N? | |
| Aug 12, 2020 at 1:44 | comment | added | Ali Taghavi | @MarianoSuárez-Álvarez yes I totally agree! | |
| Jul 23, 2020 at 0:01 | comment | added | Student | Group cohomology <3 | |
| Feb 4, 2018 at 9:53 | comment | added | Cayley-Hamilton | Or another condition for an exact sequence $0 \rightarrow N \rightarrow G \rightarrow H \rightarrow 0$ to be isomorphic to one of the form $0 \rightarrow N \rightarrow N \rtimes H \rightarrow H \rightarrow 0$ is if $G \rightarrow H$ has a section. | |
| Feb 4, 2018 at 4:33 | comment | added | Cayley-Hamilton | This is false, but something very nice is true and related to this: $F : G \text{-Grp} \rightarrow G \uparrow \text{Grp}$ where $\phi : G \rightarrow \text{Aut}(H)$ is sent to $G \rightarrow H \rtimes_{\phi} G$ where $g \mapsto (1, g)$ is a left adjoint. So the exact sequences $0 \rightarrow N \rightarrow G \rightarrow G / N \rightarrow 0$ which come from semi-direct products are the "free" ones in a sense. | |
| May 4, 2011 at 0:35 | comment | added | Selene Routley | I tripped up on this one for a VERY long time too! Given experiences cited here, I would strengthen Kevin's comment and say it proves :) we do a terrible job explaining semidirect products. And I second the comment about short exact sequences. | |
| Apr 14, 2011 at 17:57 | comment | added | roy smith | maybe the easiest way to see this, as suggested by Kevin's example, is to think of abelian groups and ask whether every subgroup is a direct factor. I.e. this has little to do with true semi direct products, and more to do, as Fabrizio observed, with splitting maps. | |
| Jun 9, 2010 at 1:42 | comment | added | Cory Knapp | It took me a long time to realize that was false as well... Still being an undergrad, I often catch myself trying to use that "theorem". | |
| May 10, 2010 at 11:14 | comment | added | Fabrizio Polo | Remember being confused by this too. It became much clearer when I formally was taught about short exact sequences. Then you can see exactly the obstruction to such a decomposition. | |
| May 5, 2010 at 1:49 | comment | added | Emerton | Schur--Zassenhaus says that this is true if $N$ and $G/N$ have coprime orders, so there is some intrinsic pressure in the subject towards this. Coupled with the fact that it is true for the first non-trivial non-abelian example ($A_3$ inside $S_3$), it's easy to see how this misconception arises. | |
| May 5, 2010 at 0:27 | comment | added | Kevin McGerty | This suggests that we do a terrible job of talking about semi-direct products no? | |
| May 4, 2010 at 23:33 | comment | added | Mariano Suárez-Álvarez | It is a sad state of things, but my impression is that most people coming out of the standard introductory course to groups have more or less the sam belief :( | |
| May 4, 2010 at 21:30 | comment | added | Kevin Buzzard | umm Z/4Z contains Z/2Z? | |
| May 4, 2010 at 21:28 | comment | added | Nate Eldredge | Argh! Me too! What is a good example? | |
| May 4, 2010 at 21:22 | history | answered | Qiaochu Yuan | CC BY-SA 2.5 |