Group and its elements

Question

Let A4 be the alternating group on 4 elements.

1) Find all of its elements and put them in the form of product of disjoint cycles

2)Find all subgroups of A4 isomorphic to K4 -Klein 4 group

3)Find all of the left costs of the subgroups in 2

Answer 1

(1) The elements are e = (1)(2)(3)(4) the identity element, as well as all the 3-cycles and the products of disjoint 2-cycles. A 3-cycle is even because (ac)(ab) = (abc) [here composing from right to left as in function composition.] Specifically (123),(132),(124),(142),(134), (143), (234),(243) are all the 3-cycles (4 choices of which element to omit and 2 choices of how to orient the remaining elements in the cycle). The products of disjoint 2-cycles are (12)(34), (13)(24), and (14)(23). The total number of elements is 1 (identity) + 8 (3-cycles) + 3 (products of disjoint 2-cycles), for a total of 12, which agrees with |A_4|=4!/2

(2) There is only one of these. The Klein 4-group has order 4, so a group isomorphic to K4 does not contain any 3-cycle (such elements have order 3, and 3 does not divide 4). There are only four other elements, the products of disjoint two-cycles. Let's compute to make sure these form a subgroup: one example suffices

(12)(34)(13)(24) = (14)(23)

So the set H = {e, (12)(34),(13)(24),(14)(23)} is a subgroup (finite subgroup test). The mapping (12)(34) -> (1,0), (13)(24)-> (0,1), and (14)(23) -> (1,1), with e -> (0,0) is an isomorphism from H onto the Klein 4-group.

(3) The trivial coset is H, the subgroup itself. If we multiply H by any element of H that does not belong to H, say (123), we obtain another coset:

(123)H = {(123), (123)(12)(34), (123)(13)(24), (123)(14)(23) } = {(123), (134), (243), (142)}

The other coset must contain the remaining 3-cycles. Pick any of these cosets to serve as representative, say (124).Then the coset is (124)H = {(132),(143),(234),(124)}

Thus the cosets are H, (123)H, (124)H