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By J. Milne

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27. For 1 < k ≤ n, there are needed so that we don’t count n(n−1)···(n−k+1) k distinct k-cycles in Sn . The 1 k is (i1i2 . . ik ) = (ik i1 . . ik−1 ) = . . k times. Similarly, it is possible to compute the number of elements in any conjugacy class in Sn , but a little care is needed when the partition of n has several terms equal. For example, the number of permutations in S4 of type (ab)(cd) is 1 4×3 2×1 = 3. × 2 2 2 The 12 is needed so that we don’t count (ab)(cd) = (cd)(ab) twice. For S4 we have the following table: Partition Element No.

Proof. Any α ∈ An is the product of an even number of transpositions, α = t1 t1 · · · tm tm, but the product of two transpositions can always be written as a product of 3-cycles:    (ij)(jl) = (ijl) case j = k, (ij)(kl) = (ij)(jk)(jk)(kl) = (ijk)(jkl) case i, j, k, l distinct,    1 case (ij) = (kl). GROUP THEORY 33 Recall that two elements a and b of a group G are said to be conjugate a ∼ b if there exists an element g ∈ G such that b = gag −1 , and that conjugacy is an equivalence relation.

Because each of G1 and H1 is normal in G, G1 H1 is a normal subgroup of G, and it properly contains both G1 and H1 . But they are maximal normal subgroups of G, and so G1 H1 = G. 2). Similarly G/H1 ≈ G1 /G1 ∩ H1 . Hence K2 =df G1 ∩ H1 is a maximal normal subgroup in both G1 and H1 , and G/G1 ≈ H1 /K2 , G/H1 ≈ G1 /K2 . Choose a composition series K2 K3 · · · Ku . We have the picture: G1 G H1 13 G2 ··· Gs K2 ··· Ku H2 ··· Ht . Jordan showed that corresponding quotients had the same order, and H¨ older that they were isomorphic.

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