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With 1 = 1H, as necessary. Rule (iii) tells us simply that c1 = c. We now apply the ﬁrst two rules. Since we don’t what a1 is, let’s denote it 2: a1 = 2. Similarly, let a2 = 3. Now a3 = a31, which according to (ii) must be 1. Thus, we have introduce three (potential) cosets 1,2,3, permuted by a as follows: a a a 1 → 2 → 3 → 1. What is b1? We don’t know, and so it is prudent to introduce another coset 4 = b1. Now b4 = 1, and so we have b b 1 → 4 → 1. We still have the relation cba. We know a1 = 2, but we don’t know what b2 is, and so set b2 = 5.

Let G be a ﬁnite group, and let (G : 1) = pr m with m not divisible by p. (a) Any two Sylow p-subgroups are conjugate. (b) Let sp be the number of Sylow p-subgroups in G; then sp|m, and sp ≡ 1 mod p. (c) Any p-subgroup of G is contained in a Sylow p-subgroup. GROUP THEORY 41 Let H be a subgroup of G. Recall (p27) that the normalizer of H in G is NG (H) = {g ∈ G | gHg −1 = H}, and that the number of conjugates of H in G is (G : NG (H)). For a Sylow p-subgroup P , the number of conjugates of P is (G : NG (P )) = (G : 1) (G : 1) m = = .

24) that conjugating an element preserves the type of its disjoint cycle decomposition. 1234... 26. (ijk) = (1234... ijk4... )(123)(ijk4... ) . S. 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.