[proofplan] We prove pairwise commutativity by expanding the commutator $[X_k,X_l] := X_kX_l-X_lX_k$ for $k<l$. The terms where the two transpositions have disjoint supports commute and therefore contribute zero. The only remaining terms occur inside the three-letter subgroup on $\{i,k,l\}$, where two explicit transposition product identities give cancellation for each $i<k$. Summing these cancellations proves $[X_k,X_l]=0$, and the cases involving $X_1$ or $k=l$ are immediate. [/proofplan]

[step:Reduce pairwise commutativity to commutators with $1<k<l$] For elements $A,B \in \mathbb{C}[S_n]$, define their commutator by \begin{align*} [A,B] := AB-BA. \end{align*} It is enough to show $[X_k,X_l]=0$ whenever $1<k<l\le n$. Indeed, if $k=l$, then $[X_k,X_k]=0$ by definition. If one of the indices is $1$, then $X_1=0$, so $[X_1,X_l]=0$ and $[X_l,X_1]=0$ for every $l$. Finally, if $l<k$, then \begin{align*} [X_k,X_l] = -[X_l,X_k], \end{align*} so the case $l<k$ follows from the already-treated increasing-order case applied to the pair $(l,k)$. [/step]

[step:Expand the commutator and separate the potentially noncommuting terms]Fix integers $k,l$ with $1<k<l\le n$. Since multiplication in $\mathbb{C}[S_n]$ is extended bilinearly from multiplication in $S_n$, we have \begin{align*}[X_k,X_l]=\sum_{i=1}^{k-1}\sum_{j=1}^{l-1}[(i\ k),(j\ l)].\end{align*} For fixed $i$ with $1\le i<k$, the transpositions $(i\ k)$ and $(j\ l)$ have disjoint supports whenever $j\notin\{i,k\}$. In that case the two permutations act on disjoint subsets of $\{1,\dots,n\}$, hence their compositions agree in both orders: \begin{align*} (i\ k)(j\ l)=(j\ l)(i\ k). \end{align*} Thus \begin{align*} [(i\ k),(j\ l)] = 0 \end{align*} for every $j\notin\{i,k\}$. Therefore only the terms $j=i$ and $j=k$ remain, and \begin{align*}[X_k,X_l]=\sum_{i=1}^{k-1}\left([(i\ k),(i\ l)]+[(i\ k),(k\ l)]\right).\end{align*}[/step]

[guided]Fix $k,l$ with $1<k<l\le n$. We want to prove that $X_kX_l-X_lX_k=0$, so we expand using the definition of the group algebra. The product in $\mathbb{C}[S_n]$ is obtained by multiplying permutations in $S_n$ and then extending bilinearly, hence \begin{align*} [X_k,X_l] = \left[\sum_{i=1}^{k-1}(i\ k),\sum_{j=1}^{l-1}(j\ l)\right] = \sum_{i=1}^{k-1}\sum_{j=1}^{l-1}[(i\ k),(j\ l)]. \end{align*} Now fix one index $i$ with $1\le i<k$. Which terms in the inner sum can fail to commute with $(i\ k)$? The transposition $(i\ k)$ only moves the two letters $i$ and $k$. The transposition $(j\ l)$ only moves the two letters $j$ and $l$. Since $l>k$, the letter $l$ is distinct from both $i$ and $k$. Therefore the supports $\{i,k\}$ and $\{j,l\}$ are disjoint exactly when $j$ is neither $i$ nor $k$. When the supports are disjoint, the permutations commute because each transposition leaves the two letters moved by the other transposition fixed. Thus, for $j\notin\{i,k\}$, \begin{align*} (i\ k)(j\ l)=(j\ l)(i\ k), \end{align*} and consequently \begin{align*} [(i\ k),(j\ l)] = (i\ k)(j\ l)-(j\ l)(i\ k)=0. \end{align*} Hence the only possible contributions for this fixed $i$ are the two terms with $j=i$ and $j=k$. Summing over all $i=1,\dots,k-1$ gives \begin{align*} [X_k,X_l] = \sum_{i=1}^{k-1} \left( [(i\ k),(i\ l)] + [(i\ k),(k\ l)] \right). \end{align*} This reduction isolates the whole problem inside the three letters $i,k,l$.[/guided]

[step:Verify the three-letter cancellation for each index $i<k$] Fix $i$ with $1\le i<k$. We use the standard convention that products of permutations are composed from right to left. In the subgroup of $S_n$ acting on the three letters $\{i,k,l\}$, direct evaluation on $i,k,l$ gives \begin{align*} (i\ k)(i\ l)=(k\ l)(i\ k) \end{align*} and \begin{align*} (i\ l)(i\ k)=(i\ k)(k\ l). \end{align*} Using these identities, compute \begin{align*}[(i\ k),(i\ l)] + [(i\ k),(k\ l)] = (i\ k)(i\ l) - (i\ l)(i\ k) + (i\ k)(k\ l) - (k\ l)(i\ k).\end{align*} Substituting the two product identities gives \begin{align*}[(i\ k),(i\ l)] + [(i\ k),(k\ l)] = (k\ l)(i\ k) - (i\ k)(k\ l) + (i\ k)(k\ l) - (k\ l)(i\ k) = 0.\end{align*} [/step]

[step:Sum the cancellations to obtain pairwise commutativity] From the commutator expansion and the three-letter cancellation, we obtain \begin{align*}[X_k,X_l]=\sum_{i=1}^{k-1}\left([(i\ k),(i\ l)]+[(i\ k),(k\ l)]\right)=\sum_{i=1}^{k-1}0=0.\end{align*} Thus $X_kX_l=X_lX_k$ for every $1<k<l\le n$. By the reduction step, the same equality holds for all $1\le k,l\le n$. Therefore the Jucys-Murphy elements $X_1,\dots,X_n$ commute pairwise in $\mathbb{C}[S_n]$. [/step]