Generate $S\_n$ by transpositions $s\_i$ of (i) and (i+1). Both $S\_3$ and $S\_4$ have single elements of maximal word norm associated with this presentation. In fact, the Cayley graph of $S\_3$ can be seen as a tiling of $S^1$, and the Cayley graph of $S\_4$ a tiling of $S^2$. The element of maximal length is then antipodal to e.


Does every symmetric group $S\_n$ have a single element of maximal word norm? If so, is there a formula for its length l(n)?



Does every symmetric group S_n have a single element of maximal word norm?
It is amazing how a fact that I was taught in a middle school can be proved using big theories where I don't understand half of the words. Let me add a straightforward proof (for $S\_n$ and only $S\_n$).


For a permutation $\sigma:\{1,\dots,n\}\to\{1,\dots,n\}$, let $\lambda(\sigma)$ denote the number of inversions in $\sigma$, that is the number of pairs $(i,j)$ such that $i<j$ and $\sigma(i)>\sigma(j)$. Then $\lambda(\sigma)$ equals the length of $\sigma$ with respect to the generating set $\{s\_i\}$.


Indeed, left-multiplying $\sigma$ by $s\_i$ only interchanges $\sigma(i)$ and $\sigma(i+1)$, and hence changes $\lambda(\sigma)$ by at most 1. Therefore the length is bounded below by $\lambda$. On the other hand, if $\sigma$ is not the identity, there exists $i$ such that $\sigma(i+1)<\sigma(i)$, then left-multiplying by $s\_i$ decreases $\lambda(\sigma)$ by 1. Repeating this procedure, one reaches the identity from $\sigma$ by exactly $\lambda(\sigma)$ multiplications by generators.


Now it is clear that the maximum length equals $n(n-1)/2$ and is attained only at the order reversing permutation (the one given by $\sigma(i)=n+1-i$ for all $i$).


