Timeline for Subgroups generated by two random elements
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 12, 2021 at 23:00 | comment | added | YCor | @FusRoDah you've edited writing "I understand that the question is too general in the present setting" but this is a post by another user. I'm not sure you can say this on the other user's behalf. Also, uniformly taking a random subgroup among those with 2 generators, is not the same as uniformly taking the subgroup generated by a random pair. I tend to believe the new question is so personal that it should rather be posted separately. | |
| Nov 12, 2021 at 22:01 | history | edited | FusRoDah | CC BY-SA 4.0 |
Split the question up into chunks, improved notation.
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| Oct 30, 2021 at 6:48 | comment | added | HJRW | There’s masses of work on this topic, especially in the case where the finite group is simple. I’d check out the papers of Martin Liebeck and his collaborators. | |
| Oct 28, 2021 at 2:48 | comment | added | Joseph Van Name | In fact, by taking into consideration two random elements $f,g$ in $S_{n}$ or $A_{n}$ such that there is some $A\subseteq S_{n}$ where $|A|=k$ and $f[A]=g[A]=A$, then we obtain a much larger lower bound for the arithmetic mean of the index. | |
| Oct 28, 2021 at 1:57 | comment | added | Joseph Van Name | For the permutation group, if we are looking at the arithmetical mean of the order (corresponding to the harmonic mean of index), then we get about $\frac{7}{8}n!$. If we are looking at the harmonic mean of the order (corresponding to the arithmetic mean of the index), then there is about a $\frac{1}{n}$ that two random elements would have the same fixed point. In this case, the arithmetic mean of the index will be at least $\approx\frac{9}{4}$, so the harmonic mean of the order is at most $\approx\frac{4}{9}n!$. | |
| Oct 28, 2021 at 1:35 | comment | added | Joseph Van Name | "mean order or index"-The arithmetical mean of the order is the harmonic mean of the index and vice versa. Is there any reason we should restrict our attention to the arithmetical and harmonic mean instead of looking at a more general class of means? | |
| Oct 27, 2021 at 23:54 | comment | added | Benjamin Steinberg | Two random elements of S_n generate A_n or S_n with the obvious probability of each. This is due to John Dixon. | |
| Oct 27, 2021 at 23:21 | review | Close votes | |||
| Nov 12, 2021 at 22:01 | |||||
| Oct 27, 2021 at 22:45 | comment | added | markvs | Depends on the finite group. For example if $G$ is cyclic of prime order $p$, these numbers can be easily computed. If $G$ is $S_n$ for large $n$, the numbers are much harder to compute but the problem seems manageable (look at the maximal subgroups). For other groups it is too hard to even try. | |
| Oct 27, 2021 at 22:21 | history | asked | Daniel Sebald | CC BY-SA 4.0 |