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Unformatted text preview: Math 4124 Monday, February 28 February 28, Ungraded Homework Exercise 3.5.11 on page 111 Prove that S 4 has no subgroup isomorphic to Q 8 . In Q 8 we have the relation i j = k , and of course i , j , k all have order 4; in other words we have the product of two elements of order 4 giving an element of order 4. Therefore if Q 8 was isomorphic to a subgroup of S 4 , we would have in S 4 the product of two elements of order 4 giving an element of order 4. Since an element of order 4 in S 4 must be a 4cycle, which is an odd permutation, we would now have the product of two odd permutations giving an odd permutation. But the product of two odd permutations is always an even permutation, so we have a contradiction. Therefore Q 8 cannot be isomorphic to a subgroup of S 4 . Let G = D 8 , the dihedral group of order 8. Prove that G has exactly one normal subgroup of order 2, and that there are exactly 5 normal subgroups containing this subgroup....
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 Spring '08
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 Math, Algebra

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