Case of 𝔖3:
s2∘s1=s4≠s1∘s2=s3.
s3∘s2=s1≠s2∘s3=s5.
s4∘s5=s1≠s5∘s4=s2.
In fact, look at 𝔖3’s table, there is no commuting elements in 𝔖3 excepted its identity.
Case of 𝔖n (n>2):
Let u∈Z(𝔖n), k∈ℕn, s∈𝔖n such that s is a n−1-cycle and s(k)=k.
Then s∘u(k)=u∘s(k)=u(k). Thus, u(k)=k because of s; and it’s true for any k. Thus u is 𝔖n’s identity.
What’s happening if n=2? s is already the identity and we already showed that the two-element group is abelian.