Let An (the alternating group on n elements) consist of the set of all even permutations in Sn. Prove that An is indeed a subgroup of Sn and that it has index two in Sn and has order n!/2. First of all, I need clarification on the definition of an alternating group. My book wasn't really good in explaining it. My attempt at a solution is limited but I was thinking I can prove that An is a subgroup by showing that its elements are closed under composition and it is closed under inverses, but I need to better understand what an alternating group is before I start a proof. Also the order n!/2, I believe has to do with calculating permutations nPn-2 = n!/2! = n!/2. Thanks in advance.