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Let $S_n$ be the symmetric group on $\{1, \ldots, n\}$. Let \begin{align} T=\sum_{g\in S_n} g. \end{align}

Are there some references about the factorization of $T$?

In the case of $n=3$, we have \begin{align} & T=1 + (12) + (23) + (12)(23) + (23)(12) + (12)(23)(12) \\ & = 1 + (12) + (23) + (12)(23) + (23)(12) + (23)(12)(23) \\ & = (1 + (12))((12) + (23) + (23)(12)) \\ & = (1 + (23))((12) + (23) + (12)(23)). \end{align} Has this problem been studied in some references?

Thank you very much.

Edit: the group algebra I consider is $\mathbb{C} S_n$.

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    $\begingroup$ What do you call "the factorization of $T$"? All the ways to write it as a product of two elements? more than two? some ways only? Note that since $T$ is conjugation-invariant, whenever you have a factorization, all conjugates yield a conjugation (in your example, you conjugate by the transposition $(13)$). $\endgroup$ – YCor Sep 16 '16 at 8:22
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    $\begingroup$ And could you specify you consider the group algebra over which ring? the ring of integers? $\endgroup$ – YCor Sep 16 '16 at 8:50
  • $\begingroup$ Take a set of generators. Then all information about factorings (in your sense, which may need to be made more precise) involving one of those generators is encoded in the corresponding Cayley graph, by looking at its symmetries/automorphsms. Question: how easy is it to retrieve that info from the Cayley graph? And are all the Cayley graphs for, say, non redundant sets of generators of $S_n$ known? $\endgroup$ – Wolfgang Sep 16 '16 at 8:57
  • $\begingroup$ You'll get a load of factorizations for every subgroup (all elements of the subgroup multiplied by a bunch of coset reps). The set of all such factorizations is conjugation invariant, to reference the comment of @YCor... But, now, I guess it'd be interesting to know what other reps can occur. $\endgroup$ – Nick Gill Sep 16 '16 at 8:57
  • $\begingroup$ Put another way, you're trying to write the group as a product: $S_n=A.B$, where $|A| \cdot |B|=n!$ (so there are no repetitions). This seems kind of tricky to me -- it's like a Zappa-Szep product for subsets, rather than subgroups... en.wikipedia.org/wiki/Zappa%E2%80%93Sz%C3%A9p_product $\endgroup$ – Nick Gill Sep 16 '16 at 9:00
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A famous factorization is $(1+X_1) (1+X_2) \cdots (1+X_n)$ where $X_1=0$, $X_k= (1,k)+(2,k)+\cdots +(k-1,k)$ for $2\leq k\leq n$. $X_k$ is called a Jucys-Murphy element, though this factorization is due to Alfred Young in 1902. Jucys gave the $q$-analogue $$ (q+X_1)(q+X_2)\cdots(q+X_n) =\sum_{\pi\in S_n} q^{c(\pi)}\pi, $$ where $c(\pi)$ denotes the number of cycles of $\pi$. See for instance https://en.wikipedia.org/wiki/Jucys-Murphy_element.

Another useful factorization is $T_2T_3\cdots T_n$, where $$ T_k = \sum_{j=1}^k (k,k-1,\dots,k-j+1). $$ This also has a $q$-analogue: $$ T_2(q)T_3(q)\cdots T_n(q) = \sum_{\pi\in S_n} q^{\mathrm{inv}(\pi)} \pi, $$ where $\mathrm{inv}(\pi)$ is the number of inversions of $\pi$ and $$ T_k(q) = \sum_{j=1}^k q^{j-1}(k,k-1,\dots,k-j+1). $$ Zagier in 1992 found a deep factorization of $T_k(q)$ which was used by Philip Hanlon and me in http://math.mit.edu/~rstan/papers/qdef.pdf (see Theorem 2.1).

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If you view T as an element of the group algebra $\mathbb{C}[S_n]$ and the latter as a product of matrix algebras one for each inequivalent irrep of $S_n$, then $T$ is simply the scalar element $n!$ on the trivial representation and is $0$ on all others.

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Here is one such expression for $T$ ( there being endless possibilities): for each $n,$ let $U_{n} = \sum_{x \in \langle (12 \ldots n) \rangle} x$. Then $T = U_{2}U_{3}\ldots U_{n-1}U_{n}$ for each $n$. The proof is an easy induction, the result being clear when $n = 2$. If $n > 2,$ then by induction we have $U_{2} \ldots U_{n-2}U_{n-1} = \sum_{y \in S_{n-1}} y.$ Since, (as noted in comments), we have $S_{n-1} \langle (12 \ldots n) \rangle$ ( where this time we consider $S_{n-1}$ as the stabilizer of $n$), we see that $T = U_{2}U_{2} \ldots U_{n-1}U_{n}.$

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For a different factorization due to Diaconis see e.g. here, Remark 2.1.4. See also further references therein.

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  • $\begingroup$ It seems to be a very interesting formula but I have trouble figuring out what is $e$ there. Is it just the identity permutation? Identity is denoted by $id$ before that... $\endgroup$ – მამუკა ჯიბლაძე Sep 24 '16 at 13:59
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    $\begingroup$ The identity permutation. $\endgroup$ – Igor Pak Sep 24 '16 at 14:10

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