Timeline for Probability that a random element of a group is trivial
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 18, 2022 at 6:11 | comment | added | HJRW | you missed my point. Never mind. | |
| Oct 17, 2022 at 22:32 | comment | added | R W | @HJRW - To quote Bertrand Russell, "Occam's razor entia non multiplicanda praeter necessitatem is [...] the supreme methodological maxim in philosophizing." | |
| Oct 17, 2022 at 7:38 | comment | added | HJRW | The proof can be phrased without mentioning these things, but it's not correct to say that it has "nothing to do with them". Your argument happens on a Cayley graph, a kind of Schreier graph. The generating set can be equivalently thought of as an epimorphism $q:F\to G$ where $F$ is a free group, and the question equivalently asks for the probability that a random element of $F$ is in $\ker q$. The cogrowth measures the density of $\ker q$ in $F$, so is intimately related to this. (Indeed, the OP found the answer in a paper on cogrowth before you posted your answer.) | |
| Oct 16, 2022 at 12:23 | history | edited | R W | CC BY-SA 4.0 |
added 303 characters in body
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| Oct 16, 2022 at 12:16 | comment | added | R W | @HJRW - I honestly fail to see this: the argument is much more general. I will add a comment about it. | |
| Oct 16, 2022 at 6:11 | comment | added | HJRW | Very nice! Your first sentence is not correct, though: the question has something to do with all those things. | |
| Oct 14, 2022 at 20:47 | history | edited | R W | CC BY-SA 4.0 |
EDIT: typos
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| Oct 14, 2022 at 18:58 | vote | accept | Xiyan | ||
| Oct 14, 2022 at 18:56 | history | answered | R W | CC BY-SA 4.0 |