Timeline for An application of ping-pong lemma
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| May 16, 2023 at 15:34 | comment | added | Carl-Fredrik Nyberg Brodda | I really like how many different (yet, under the hood, the same!) answers this question attracts. It showcases just how beautiful even the most basic objects of combinatorial/geometric group theory are; so imagine how beautiful the rest of it is! :-) | |
| May 16, 2023 at 15:00 | answer | added | Peter LeFanu Lumsdaine | timeline score: 3 | |
| May 16, 2023 at 14:16 | answer | added | Peter LeFanu Lumsdaine | timeline score: 7 | |
| May 16, 2023 at 9:56 | answer | added | YCor | timeline score: 7 | |
| May 16, 2023 at 8:22 | history | edited | HJRW |
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| May 15, 2023 at 21:29 | answer | added | Luc Guyot | timeline score: 4 | |
| May 15, 2023 at 14:43 | history | became hot network question | |||
| May 15, 2023 at 13:10 | answer | added | Benjamin Steinberg | timeline score: 5 | |
| May 15, 2023 at 8:31 | answer | added | Sam Nead | timeline score: 7 | |
| May 15, 2023 at 8:02 | comment | added | HJRW | The easiest and most illuminating way to prove this is to draw the Stallings core graph for the subgroup $H$. If you're determined to use ping pong, you should think about the complementary components of the geodesic $[1,b^{d-1}]$ in the Cayley tree. | |
| May 15, 2023 at 7:08 | comment | added | YCor | In any case it is true. If $A,B$ are groups, then in $A\ast B$, the subgroups $aBa^{-1}$ for $a\in A$ generate their free product. The proof is quite immediate: consider the semidirect product $B^{\ast A}\rtimes A$ ($A$ acting by shift) and map $A\ast B$ to it in the obvious way. | |
| May 15, 2023 at 7:06 | comment | added | YCor | Why don't you simply write $H=\langle b^{-k}ab^k,\; k=0,\dots,d-1\rangle$? | |
| May 15, 2023 at 6:38 | history | asked | Shiv Parsad | CC BY-SA 4.0 |