Question

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.

Answer 1

Sn is the trivial subgroup

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