Let α ∈ S n α∈Sn for n ≥ 3 . n≥3. If α β = β α αβ=βα for all β ∈ S n , β∈Sn, prove that α α must be the identity permutation; hence, the center of S n Sn is the trivial subgroup.

Sn is the trivial subgroup

Step-by-step explanation:

To prove that α must be the identity permutation and also Sn is the trivial subgroup we have to make assumptions :

when we assume   α  is not an identity permutation

α(i) = -j   but since n ≥ 3 we will make  another assumption of a number k

hence we will choose : β = ( i k )

from these assumptions we can deduce that

        αβ (i) = α(k) ≠ -j   since α(i) = -j

also   βα (i) = β(-j) = -j

from the equations above we can conclude that

 αβ ≠  βα   which is in contradiction to the expression : αβ=βα . this simply shows that  α must be the identity permutation, and the number of permutations of Sn is true for every number of permutation of Sn provided it is other than  the identity

hence the center Sn is the trivial subgroup

question is incomplete.