Jucys-murphy elements commute with each other. In the group algebra $\mathbb{C}[S_n]$, for $1<i<j\le n$, $X_i=(1\ i)+(2\ i)+...+(i-1\ i)$ and $X_j=(1\ j)+(2\ j)+...+(j-1\ j)$ commute with each other.
I have been trying to do it computationally by multiplying out each term but I am getting lost. Since there are transpositions that don't commute, I am finding it difficult to keep track. Please help me.
 A: The standard way to see this goes as follows: the $i$th Jucys-Murphy-Young element is the orbit sum
$$X_i=\sum_{1 \leq j < i} (ij)=\frac{1}{|S_{i-2}|}\sum_{w \in S_{i-1}} w(1i)w^{-1} $$ which commutes with $S_{i-1}$, and in particular with $X_1,\dots,X_{i-1}$.
A: Well, this is one occasion, it might not be such a bad idea to do it computationally after all (that is, I am not aware of a better method). It is true that some transpositions don't commute, but we can pinpoint exactly which..:
The arbitrary elements in the expansions $X_iX_j$ and $X_jX_i$ are $(ki)(mj)$ and $(mj)(ki)$. These are definitely equal when the set $\{k,m,i,j\}$ has exactly $4$ elements. The only cases that are left are $k=m<i<j$ and $k<i=m<j$.
Therefore we have to show that $$\sum_{k=1}^{i-1}(ki)(kj)+\sum_{k=1}^{i-1}(ki)(ij)=\sum_{k=1}^{i-1}(kj)(ki)+\sum_{k=1}^{i-1}(ij)(ki)$$
Notice however that $(ki)(kj)=(kji)=(ij)(ki)$ and $(kj)(ki)=kij=(ki)(ij)$, that is both expressions are equal to:$$\sum_{k=1}^{i-1}(kij)+(kji)$$
Anyway, I would like to see a better proof but I think at some point you'd have to do a similar computation. By the way, if you are studying Jucys-Murphy elements, it's really worth it to see how Okounkov and Vershik use them to build the representation theory of the Symmetric group. These notes are helpful if you do decide to study the paper..
