AnA_n is the only subgroup of SnS_n of index 22.

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 HSn is of index 2 then it is normal and Sn/H is isomorphic to C2={1,1}. We thus have a surjective homomorphism f:SnC2 with kernel H. All transpositions in Sn are conjugate, hence f(t)C2 is the same element for every transposition tSn (this uses the fact that C2 is commutative). Sn is generated by transpositions, therefore C2 is generated by f(t) (for any transposition tSn), therefore f(t)=1, therefore ker f=An.

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

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