Timeline for Existence of n-axial elements in groups with at least 2 ends
Current License: CC BY-SA 4.0
6 events
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| Sep 19, 2019 at 10:33 | comment | added | Ville Salo | I prove this in two parts above because the first lemma was what I needed myself and the second is just to show you can get an axial from it. But I think the first lemma could be strengthened so axial elements drop more directly, by instead for any $A$ finding $k$ so $h_1$ is connected to $kh_2$ in $G \setminus (A \cup kA)$. The proof should be exactly the same, and then $k$ seems clearly axial if $h_1$ and $h_2$ are in distinct components of $G \setminus A$. | |
| Sep 19, 2019 at 8:24 | history | edited | Ville Salo | CC BY-SA 4.0 |
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| Sep 19, 2019 at 8:08 | history | edited | Ville Salo | CC BY-SA 4.0 |
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| Sep 19, 2019 at 7:24 | history | edited | Ville Salo | CC BY-SA 4.0 |
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| Sep 19, 2019 at 7:17 | history | edited | Ville Salo | CC BY-SA 4.0 |
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| Sep 19, 2019 at 6:56 | history | answered | Ville Salo | CC BY-SA 4.0 |