Adv Alegbra HW Solutions 74

Adv Alegbra HW Solutions 74 - 74 2.118 Prove that a finite...

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Unformatted text preview: 74 2.118 Prove that a finite p -group G is simple if and only if |G | = p . Solution. Absent. 2.119 Show that S4 has a subgroup isomorphic to D8 . Solution. In Example 2.139, we have seen that D8 can be viewed as certain permutations of the 4 vertices of a square. 2.120 Prove that S4 /V ∼ S3 . = Solution. S4 /V is a group of order 24/4=6, hence Proposition 2.135 shows that it is isomorphic to either S3 or I6 . But S4 /V is not abelian: for example, (1 2)V(1 3)V = (1 3)V(1 2)V because (1 2)(1 3)[(1 3)(1 2)]−1 = (1 2)(1 3)(1 2)(1 3) = (2 3) ∈ V. / Therefore, S4 /V ∼ S3 . = 2.121 (i) Prove that A4 ∼ D12 . = Solution. A4 has no element of order 6, while D12 does have such an element. (ii) Prove that D12 ∼ S3 × I2 . = Solution. We may suppose that D12 = a , b , where a 6 = 1 = b2 and bab = a −1 . We know that the subgroup a has order 6, hence index 2, and so there are two cosets: D12 = a ∪ b a . Thus, every element x ∈ D12 has a unique factorization x = bi a j , where i = 0, 1 and 0 ≤ j < 6. Define H = a 2 , b ; now H ∼ S3 , = for it is a nonabelian group of order 6; note that H D12 because it has index 2. If we define K = a 3 , then | K | = 2 and K D12 : it suffices to prove that aa 3 a −1 ∈ K (which is, of course, obvious) and ba 3 b ∈ K ; but ba 3 b = a −3 = a 3 ∈ K . It is plain that H ∩ K = {1} and H K = D12 , and so D12 ∼ H × K ∼ S3 × I2 , = = by Proposition 2.127. 2.122 (i) If H is a subgroup of G and if x ∈ H , prove that C H (x ) = H ∩ C G (x ). Solution. C G (x ) = {g ∈ G : gx = xg } while C H (x ) = {g ∈ H : gx = xg }. It follows easily that each of C H (x ) and H ∩ C G (x ) contains the other. (ii) If H is a subgroup of index 2 in a finite group G and if x ∈ H , prove that either |x H | = |x G | or |x H | = 1 |x G |, where x H is the 2 conjugacy class of x in H . ...
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