Timeline for does this set of permutations form a group? And more
Current License: CC BY-SA 3.0
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Apr 7, 2019 at 4:00 | vote | accept | T. Amdeberhan | ||
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Mar 5, 2017 at 9:14 | comment | added | Martin Rubey | I think it's very confusing if you edit your question changing just two words. Wouldn't it be more appropriate to set a bounty? | |
| Mar 5, 2017 at 3:22 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 4, 2017 at 12:56 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 3, 2017 at 23:18 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 3, 2017 at 22:50 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 3, 2017 at 21:57 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 3, 2017 at 20:38 | answer | added | user44191 | timeline score: 10 | |
| Jan 3, 2017 at 20:28 | comment | added | user44191 | Do you have any examples for $m, n \neq 1$? It's not a group for $m = n = 2$. | |
| Jan 3, 2017 at 19:40 | comment | added | John Shareshian | It seems to me that $U_{mn}$ contains the full stabilizer $G$ in $S_{mn}$ of a partition of $[mn]$ into $n$ parts of size $m$. As $G$ is a maximal proper subgroup of $S_{mn}$ when $m,n>1$, we see that if $U_{mn}$ is a subgroup, then it is $G$ or $S_{mn}$. It should not be hard to show that neither $U_{mn}=G$ nor $U_{mn}=S_{mn}$ holds when $m,n>1$. | |
| Jan 3, 2017 at 19:37 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 3, 2017 at 18:33 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 3, 2017 at 18:21 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 3, 2017 at 16:35 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 3, 2017 at 16:06 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 3, 2017 at 15:51 | history | asked | T. Amdeberhan | CC BY-SA 3.0 |