Factorization in the group algebra of symmetric groups

Question

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 $\mathbb{C} S_n$.

Answer 1

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).

Answer 2

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.

Answer 3

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 $T = U_{2}U_{2} \ldots U_{n-1}U_{n}.$

Answer 4

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