Compute the centralizer of each member of A4SolutionSolution

Compute the centralizer of each member of A4

Solution: Since any power of an element a commutes with a, the centralizer C(a)

always contains the cyclic subgroup hai generated by a. Thus the centralizer of (1, 2, 3)

always contains the subgroup {(1),(1, 2, 3),(1, 3, 2)}.

take a 4-cycle, we have a(x, y, z, 4)a1 = (u, v, w, 4),

since a just permutes the numbers x, y, and z. This means that w 6= z, so (u, v, w, 4)

is not equal to (x, y, z, 4). Without doing all of the calculations, we can conclude that

no 4-cycle belongs to C(a). This accounts for an additional 6 elements. A similar

argument shows that no 3-cycle that includes the number 4 as one of its entries can

belong to C(a). Since there are 6 elements of this form, we now have a total of 21

elements (out of 24) that are not in C(a), and therefore C(a) = <a>.

Finally, since A4 contains the three products of transpositions and the six 3-cycles

that include 4, we have nine elements (out of 12 in A4) that do not commute with

(1, 2, 3). Thus in A4 we get the same answer: C(a) = <a>