Adv Alegbra HW Solutions 72

Adv Alegbra HW Solutions 72 - G ≤ ker ϕ Conversely if G...

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72 the last congruence by Wilson s theorem. If the statement of the exercise is false, then each k between 1 and p 1 is congruent mod p to some ia i , and so Y k = ( p 1 ) !≡ Y i ia i mod p . But Y ia i = Y i Y a i ≡[ ( p 1 ) !] 2 mod p . It follows that Q ia i ( 1 ) 2 = 1mod p . Since p is odd, 1 ±≡ 1mod p , and this is a contradiction. 2.113 If G is a group and x , y G ,de f ne their commutator to be xyx 1 y 1 , and de f ne the commutator subgroup G 0 to be the subgroup generated by all the commutators. (The product of two commutators need not be a commutator, but the smallest instance of this occurs in a group of order 96.) (i) Prove that G 0 C G . Solution. Absent. (ii) Prove that G / G 0 is abelian. Solution. Absent. (iii) If ϕ : G A is a homomorphism, where A is an abelian group, prove that
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Unformatted text preview: G ≤ ker ϕ . Conversely, if G ≤ ker ϕ , prove that im ϕ is abelian. Solution. Absent. (iv) If G ≤ H ≤ G , prove that H C G . Solution. Absent. 2.114 True or false with reasons. (i) Every group G is isomorphic to the symmetric group S G . Solution. False. (ii) Every group of order 4 is abelian. Solution. True. (iii) Every group of order 6 is abelian. Solution. False. (iv) If a group G acts on a set X , then X is a group. Solution. False. (v) If a group G acts on a set X , and if g , h ∈ G satisfy gx = hx for some x ∈ X , then g = h . Solution. False....
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