Adv Alegbra HW Solutions 73

Adv Alegbra HW Solutions 73 - last two because the f rst...

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73 (vi) Ifag roup G acts on a set X , and if x , y X , then there exists g G with y = gx . Solution. False. (vii) If g G , where G is a f nite group, then the number of conjugates of g is a divisor of | G | . Solution. True. (viii) Every group of order 100 contains an element of order 5. Solution. True. (ix) Every group of order 100 contains an element of order 4. Solution. False. (x) Every group of order 5 8 contains a normal subgroup of order 5 6 . Solution. True. (xi) If G is a simple group of order p n , where p is a prime, then n = 1. Solution. True. (xii) The alternating group A 4 is simple. Solution. False. (xiii) The alternating group A 5 is simple. Solution. True. (xiv) The symmetric group S 6 is simple. Solution. False. 2.115 Prove that every translation τ a S G , where τ a : g 7→ ag , is a regular permutation. Solution. Absent. 2.116 Prove that no pair of the following groups of order 8, I 8 ; I 4 × I 2 ; I 2 × I 2 × I 2 ; D 8 ; Q , are isomorphic. Solution. None of the f
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Unformatted text preview: last two because the f rst three are abelian while D 8 and Q are not abelian. Now I 8 ± ∼ = I 4 × I 2 and I 8 ± ∼ = I 2 × I 2 × I 2 because I 8 has an element of order 8 and the other two groups do not. Similarly, I 4 × I 2 ± ∼ = I 2 × I 2 × I 2 because I 4 × I 2 has an element of order 4 and I 2 × I 2 × I 2 does not. Finally, D 8 ± ∼ = Q because Q has only 1 element of order 2 while D 8 has more than one element of order 2. 2.117 If p is a prime and G is a f nite group in which every element has order a power of p , prove that G is a p-group. Solution. If q is a prime divisor of | G | with q ±= p , then Cauchy ’ s theorem gives an element in G of order q , contrary to the hypothesis....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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