Adv Alegbra HW Solutions 63

Adv Alegbra HW Solutions 63 - 63 Solution We use Exercise...

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63 Solution. We use Exercise 2.69: Q has exactly one element of order 2, while D 8 has 5 elements of order 2. 2.88 If G is a f nite group generated by two elements of order 2, prove that G = D 2 n for some n 2. Solution. Absent. 2.89 (i) Prove that A 3 is the only subgroup of S 3 of order 3. Solution. Absent. (ii) Prove that A 4 is the only subgroup of S 4 of order 12. (In Exer- cise 2.135, this will be generalized from S 4 and A 4 to S n and A n for all n 3.) Solution. If H is a subgroup of order 12, then H is normal (it has index 2), and so it contains all the conjugates of each of its elements. It must contain 1. We count the number of conjugates of the various types of permutations (each count uses Exercise 2.24: ( 12 ) has 6 conjugates; ( 123 ) has 8 conjugates; ( 1234 ) has 6 conjugates; ( )( 34 ) has 3 conjugates. The only way to get 12 elements is 1 + 3 + 8; but this is A 4 . 2.90 (i) Let A be the set of all 2 × 2 matrices of the form A = £ ab 01 ± , where a ±= 0. Prove that A is a subgroup of GL ( 2 , R ) . Solution. It is a routine calculation to show that if
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