Show that a permutation with odd order is an even permutatio

Show that a permutation with odd order is an even permutation. Let H be a subgroup of S_n. If H has odd order, prove that H is a subgroup of A_n.

Solution

(a) We know that a permutation Sn is called even if it can be written as a product of an even number of transpositions (ie, cycles of the form (ij)). A permutation Sn is called odd if it isn’t even.

Suppose that ²ⁿ⁺¹ = e for some integer n. Writing as a product of m transpositions, and plugging this into ⁿ , we see that a product of m(2n + 1) transpositions is equal to e. However, e can only be written as a product of an even number of transpositions (The identity permutation is an even permutation.). Thus m(2n + 1) is even and thus m is also even, which proves that is an even permutation as desired.

(b) If H is a subgroup of Sn, and if H contains an odd permutation, then exactly half the elements of H are odd permutations. If H has an odd order, this is impossible for exactly half of its elements to be odd permutations. Hence H contains no odd permutations at all, i.e., H is a subgroup of An.