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Adv Alegbra HW Solutions 79

# Adv Alegbra HW Solutions 79 - 79 p | m By Euclids lemma we...

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79 p | m ! . By Euclid s lemma, we must have p | k for some k m , and this contradicts m < p . Therefore, G is not simple. 2.135 If n 3, prove that A n is the only subgroup of S n of order 1 2 n ! . Solution. The case n = 3 is very easy, and n = 4 has been done in Exercise 2.89; therefore, we may assume that n 5. If H is a second such subgroup, then H S n because it has index 2. By the second isomorphism theorem, H A n A n . Since A n is simple, either H A n = A n or H A n = { ( 1 ) } . In the fi rst case, A n H , so that A n = H because they have the same order. In the second case, we contradict the product formula, which now says that 1 4 ( n ! ) 2 | A n || H | = | A n H | ≤ | S n | = n ! . Here is a second proof. If H S n has index 2, then Proposition 2.97(i) says that α 2 H for every α S n . Since α 2 is always an even permutation, we have α 2 H A n for all α S n . However, if α is a 3-cycle, then it has order 3, and so α = α 4 = 2 ) 2 . Therefore, H contains every 3-cycle, and so A n H , by Exercise 2.127(i). Since both of these subgroups have
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