Timeline for Can we ascertain that there exist an epimorphism $G\rightarrow H?$
Current License: CC BY-SA 3.0
21 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 28, 2013 at 4:10 | answer | added | Alexander Gruber | timeline score: 35 | |
| Aug 27, 2013 at 23:14 | comment | added | fedja | Looks like a counterexample was just posted on MSE: math.stackexchange.com/questions/221152/… | |
| Nov 22, 2012 at 19:32 | comment | added | user9072 | @Peter Mueller: Since the merging happened it is not surprsing that the link no longer works. For context see tea.mathoverflow.net/discussion/1467/… (in a nutshell, OP wished to have no accepted answer and thus gave-up the upvotes, as technically there was no other solution) | |
| Nov 22, 2012 at 17:56 | comment | added | Peter Mueller | The MO link in the revised question is broken. But more seriously: Where are all the many upvotes for this great question gone? | |
| Nov 22, 2012 at 14:57 | history | edited | j.c. |
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| Nov 22, 2012 at 8:41 | history | 18 | |||
| Nov 22, 2012 at 5:32 | history | asked | Kerry | CC BY-SA 3.0 | |
| Nov 9, 2012 at 16:15 | comment | added | Peter Mueller | My question in the comment above has an easy negative answer: The alternating group $A_6$ contains $C_2\times C_2$, and also $C_3\times C_3$. So $C_6\times C_6$ is a subgroup of $A_6\times A_6$. However, $C_6$ is not a subgroup of $A_6$. | |
| Nov 9, 2012 at 6:22 | vote | accept | CommunityBot | ||
| Nov 22, 2012 at 8:41 | |||||
| Nov 9, 2012 at 6:01 | comment | added | Gerhard Paseman | This is suggestive of Lovasz's result on uniqueness of nth roots in certain finite structures. You might see if his results extend to this problem. Gerhard "Ask Me About System Design" Paseman, 2012.11.08 | |
| Nov 9, 2012 at 1:12 | comment | added | Kerry | I met him today and he said one of the professor in my department might have found something related to your comment. Nothing is concrete yet, though. | |
| Nov 7, 2012 at 13:34 | comment | added | Peter Mueller | I'm wondering about the following kind of dual question: Suppose that $G\times G$ is isomorphic to a subgroup of $H\times H$. Does this imply that $G$ is isomorphic to a subgroup of $H$? (Here again, only finite groups are considered.) Note that we cannot replace subgroup by normal subgroup here: Let $H=A_4$ be the alternating group of order $12$. Then $(C_2\times C_2)\times 1$ is normal in $A_4\times A_4$, but $A_4$ has no normal subgroup of order $2$. | |
| Nov 5, 2012 at 14:33 | answer | added | Someone | timeline score: 1 | |
| Nov 1, 2012 at 7:10 | answer | added | François Brunault | timeline score: 15 | |
| Oct 29, 2012 at 14:35 | comment | added | François Brunault | Is it true that the kernel of $G \times G \twoheadrightarrow H \times H$ must be decomposable? This is true (I think) when $G$ is abelian. | |
| Oct 28, 2012 at 20:26 | comment | added | François Brunault | The comments on Math SE actually proved that it's true whenever $H$ is abelian, and also proved that $G \times G \cong H \times H$ implies $G \cong H$ (by Krull-Remak-Schmidt). | |
| Oct 28, 2012 at 19:52 | answer | added | Ian Agol | timeline score: 30 | |
| Oct 28, 2012 at 16:20 | comment | added | Kerry | Thanks. Perhaps I should be clearer; he proved this holds for all abelian groups and simple groups. Similarly he also tried to make use of Krull-Schmidt. So the comments are helpful but did not offer anything new. | |
| Oct 28, 2012 at 9:56 | comment | added | YCor | With a software, I'd suggest to try with $G$ a 2-group (of order 16,32,64...), and mod out $G\times G$ by central subgroups of order 4. | |
| Oct 28, 2012 at 9:36 | comment | added | YCor | The comments in MathSE included the remarks that it's true when $G$ is abelian (by direct checking) and when the epimorphism is an isomorphism (by the Krull-Schmidt theorem). Certainly it's far from the general picture but I wouldn't deny them as "constructive answers". |