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# Read e-book online On Anharmonic Co-ordinates PDF

By Hamilton W.R.

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This action is nontrivial, so the corresponding homomorphism φ : G → Sn is nontrivial. Therefore ker(φ) is trivial, since G is simple. Thus G can be embedded in Sn . Then An ∩φ(G) is a normal subgroup of φ(G), so since G is simple, either φ(G) ⊆ An , or An ∩φ(G) = e . The second case implies |G| = 2, since the square of any odd permutation is even, and this cannot happen since n > 2. 3. Prove that if G contains a nontrivial subgroup of index 3, then G is not simple. Solution: If G is simple and contains a subgroup of index 3, then G can be embedded in A3 .

2 Conjugacy 1. Prove that if the center of the group G has index n, then every conjugacy class of G has at most n elements. Solution: The conjugacy class of a ∈ G has [G : C(a)] elements. Since the center Z(G) is contained in C(a), we have [G : C(a)] ≤ [G : Z(G)] = n. ) 2. Let G be a group with center Z(G). Prove that G/Z(G) is abelian iff for each element x ∈ Z(G) the conjugacy class of x is contained in the coset Z(G)x. Solution: First suppose that G/Z(G) is abelian and x ∈ G but x ∈ Z(G). For any a ∈ G we have ax = zxa for some z ∈ Z(G), since Z(G)ax = Z(G)xa in the factor group G/Z(G).

We conclude that N and G have the same number of Sylow p-subgroups. 12. Prove that if G is a group of order 105, then G has a normal Sylow 5-subgroup and a normal Sylow 7-subgroup. Solution: Use the previous problem. Since 105 = 3 · 5 · 7, we have n3 = 1 or 7, n5 = 1 or 21, and n7 = 1 or 15 for the numbers of Sylow subgroups. Let P be a Sylow 5-subgroup and let Q be a Sylow 7-subgroup. At least one of these subgroups must be normal, since otherwise we would have 21 · 4 elements of order 5 and 15·6 elements of order 7.