How to prove that the only subgroup of the symmetric group Sn of order n!/2 is An?

Why isn’t there other possibility?

Thanks 🙂

**Answer**

As mentioned by yoyo: if H⊂Sn is of index 2 then it is normal and Sn/H is isomorphic to C2={1,−1}. We thus have a surjective homomorphism f:Sn→C2 with kernel H. All transpositions in Sn are conjugate, hence f(t)∈C2 is the same element for every transposition t∈Sn (this uses the fact that C2 is commutative). Sn is generated by transpositions, therefore C2 is generated by f(t) (for any transposition t∈Sn), therefore f(t)=−1, therefore ker f=An.

**Attribution***Source : Link , Question Author : ShinyaSakai , Answer Author : user8268*