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Suppose that we have a finite group $G$ and choose elements $a, b \in G$ at random. What can be said about the order of the subgroup generated by $a$ and $b$? Mainly, what is the expected order, $\epsilon_1(G)$, and index of such a subgroup?

EDIT: I understand that the question is too general in the present setting. Therefore, splitting the question up might be a good idea.

  1. For which groups $G$ is the mean order easily computable? The comments suggest that $\mathbb Z_n$ and $S_n$ belong to that category.

  2. Are there any asymptotic results available, depending on $|G|$? While $\epsilon_1(G)$ or $\sum_{|G|=n} \epsilon_1(G)$ might fluctuate wildly, $\sum_{|G|\leq n} \epsilon_1(G)$ could be more tractable.

  3. The mean order is the arithmetic mean of the orders of all 2-generator (+ the trivial and 1-generator) subgroups. The mean index is the harmonic mean of orders of all 2-generator subgroups, so that we may suggestively denote it by $\epsilon_{-1}(G)$. And similarly, we can look at other expected values, $\epsilon_\alpha(G)=\sum_{a,b\in G} |\langle a,b \rangle|^\alpha$. Are there any results, in the above senses, available for these values?

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    $\begingroup$ 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. $\endgroup$
    – markvs
    Commented Oct 27, 2021 at 22:45
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    $\begingroup$ Two random elements of S_n generate A_n or S_n with the obvious probability of each. This is due to John Dixon. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 27, 2021 at 23:54
  • $\begingroup$ "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? $\endgroup$
    – Joseph Van Name
    Commented Oct 28, 2021 at 1:35
  • $\begingroup$ 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 $\approx\frac{4}{9}n!$. $\endgroup$
    – Joseph Van Name
    Commented Oct 28, 2021 at 1:57
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    $\begingroup$ @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. $\endgroup$
    – YCor
    Commented Nov 12, 2021 at 23:00

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