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Is $A\_{n}$ the only subgroup of $S\_{n}$ which satisfies that $|A\_{n}|=\\frac{|S\_{n}|}{2} = \\frac{n!}{2}$

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Yes. Suppose N is a subgroup of S_n of index 2. Then we have a surjective homomorphism f:S_n->Z_2 (the cyclic group of order 2) such that ker f= H.
Observe now that the image of any transposition is 1 (why?)
This implies that the image of every even permutation is 0 hence Ker f=A_n. So we must have H=A_n
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