Math 210 A : Algebra , Homework 4

Solution. Henceforth, for the sake of notation when we must use variables, let i denote the element i+ k in the symmetric group on n elements, where i+ k ≤ n. S1 is the trivial group, so has one conjugacy class. S2 is abelian, so it has two conjugacy classes of one element each. We proved in class that cycle type is invariant under conjugation, so every conjugacy class may contain only one cycle type. We use this for the next two cases. The conjugacy classes for S3 are {e}, {(1 2), (1 3), (2 3)}, and {(1 2 3), (1 3 2)}. We see this since (1 2)(1 3)(1 2) = (2 3) and (2 3)(1 3)(2 3) = (1 2), and (2 3)(1 2 3)(2 3) = (1 3 2). Therefore each conjugacy class contains all elements of a given cycle type. The conjugacy classes for S4 are {e}, the transpositions, the 3-cycles, the 4-cycles, and the 2-2-cycles, i.e. {(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. We can see the 2-2-cycles are all conjugate by conjugating the first listed element by (2 3) and (2 4). The transpositions are all conjugate since, for any (i j) and (k l), we have (i k j l)(i j)(i j k l)−1 = (k l). The 3-cycles are all conjugate since, for any (i j k) and (l mn), we must have at most one element can differ between the two. Without loss of generality, we have (i j k) and (i j l) (the other cases being analogous). Then (k l)(i j k)(k l) = (i j l), so all 3-cycles are conjugate. Finally, given two 4cycles (i j k l) and (i k l j) (the other cases being analogous), we have (j k l)(i j k l)(j k l)−1 = (i k l j) as required. This completes the classification.