Adv Alegbra HW Solutions 53

Adv Alegbra HW Solutions 53 - Every proper subgroup of S 4...

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53 Solution. True. (ii) G is a subgroup of itself. Solution. True. (iii) The empty set is a subgroup of G . Solution. False. (iv) If G is a fi nite group and m is a divisor of | G | , then G contains an element of order m . Solution. False. (v) Every subgroup of S n has order dividing n ! . Solution. True. (vi) If H is a subgroup of G , then the intersection of two (left) cosets of H is a (left) coset of H . Solution. False. (vii) The intersection of two cyclic subgroups of G is a cyclic sub- group. Solution. True. (viii) If X is a fi nite subset of G , then X is a fi nite subgroup. Solution. False. (ix) If X is an in fi nite set, then F = { σ S X : σ moves only fi nitely many elements of X } is a subgroup of S X . Solution. True. (x) Every proper subgroup of S 3 is cyclic. Solution. True. (xi)
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Unformatted text preview: Every proper subgroup of S 4 is cyclic. Solution. False. 2.53 Let H be a subgroup of a f nite group G , and let a 1 H , . . . , a t H be a list of all the distinct cosets of H in G . Prove the following statements without using the equivalence relation on G de f ned by a ≡ b if b − 1 a ∈ H . (i) Prove that each g ∈ G lies in the coset gH , and that gH = a i H for some i . Conclude that G = a 1 H ∪ ··· ∪ a t H . Solution. Absent. (ii) If a , b ∈ G and aH ∩ bH ±= ∅ , prove that aH = bH . Conclude that if i ±= j , then a i H ∩ a j H = ∅ . Solution. Absent....
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