Conversation started Feb 4, 2019 at 21:13.
Mike Miller
Mike Miller
Feb 4, 2019 21:13
An element of S_n may be written by partitioning {1, 2, ..., n} into smaller sets, and applying a cyclic permutation on each of them. This is usually written like (134)(256)(7), say.
1010011010
1010011010
The content compiles just fine in my tex editor:(
Mike Miller
Mike Miller
A cyclic permutation on k letters has order k, so we should understand what that says about products of cyclic permutations (where the sets they act on are disjoint).
@1010011010 see LaTeX in chat on the sidebar.
1010011010
1010011010
\begin{equation}
\int \frac{\operatorname{Li}_s(x)}{x+c_1} = \frac1{x+c_1}(-1)^s \left[x\sum_{i=2}^{s} (-1)^{i}\operatorname{Li}_{i}(x) - (1-x)\log(1-x) -x\right] + (-1)^s\int dx \frac1{\left[x+c_1\right]^2} \left[x\sum_{i=2}^{s} (-1)^{i}\operatorname{Li}_{i}(x) - (1-x)\log(1-x) -x\right]
\end{equation}
@MikeMiller Yeah thanks I know about that, there was just a bug in the code somewhere, I guess the other comment can be deleted
Mike Miller
Mike Miller
Gotcha
@Shaun Let me write some notation to simplify what I want ro say.
1010011010
1010011010
The above appears to have the same integral on both the RHS and LHS (just take the $i=s$ term of the summation under the integral sign on the RHS)
Does this automatically mean I cannot take the polylogarithm as the function to be integrated?
Mike Miller
Mike Miller
Feb 4, 2019 21:19
@Shaun Let $\sigma_1, \sigma_2, \cdots, \sigma_k$ be cyclic permutations on some subset of $\{1, \cdots, n\}$. We say that $\text{Supp}(\sigma_i)$ is the set of elements so that $\sigma_i(x) \neq x$. We demanded above that $\text{Supp}(\sigma_i) \cap \text{Supp}(\sigma_j) = \varnothing$ for all $i \neq j$.
(I am introducing notation instead of specifying the subsets because the latter would make the notation way more hellish.)
Now, let's say $|\text{Supp}(\sigma_i)| = s_i$. Then $\sigma_i$ has order $s_i$, and $\sum_{i=1}^k s_i \leq n$. We have equality iff none of the $s_i$ are zero.
Now, taking $(\sigma_1 \sigma_2 \cdots \sigma_k)^j = \sigma_1^j \sigma_2^j \cdots \sigma_k^j$. This is because the domains of these permutations are disjoint (so they all commute with one another).
Therefore, the order of $(\sigma_1 \cdots \sigma_k)$ is the least $j$ so that all of the $\sigma_i^j = 1$.
That is, the order of $\sigma_1 \cdots \sigma_k$ is the greatest common denominator of the orders of the $\sigma_i$.
The question "what are the orders of elements of $S_n$", then, is the same as "if $p = \{s_1, \cdots, s_k\}$ is some sequence of numbers with $1 < s_i$ and $\sum s_i \leq n$, what is $\gcd(s_1, \cdots, s_k)$? If we allow $p$ to vary over all partitions of numbers at most $n$, what are the possible numbers this produces?"
Then in those questions, people carried out this number theoretic calculation.
Mostly it is a matter of brute computation. The fact that $1 < s_i$ reduces the amount of computation somewhat.
In particular, if you want an element of order exactly 2n, the discussion in one of those posts shows that this cannot be any prime power n. This reduces us to 6, 10, 12 as our first possible examples.

It is not true of n=6: there the possible partitions as above are {2,2}, {2,3}, {2,4}, {2,2,2}. The gcds in each case are 2, 6, 4, 2: never 12.

For n=10, take the partition {5, 4}.
 
Conversation ended Feb 4, 2019 at 21:29.