Hi,

Set H=<(12)(34),(123)>. Then H=A_{4}: 2 and 3 both divide the order of H but since A_{4}has no subgroup of order 6 (if it did a Sylow 3 subgroup of A_{4}would be normal in A_{4}), H=A_{4}.

Here's a Cayley diagram for A_{4}and the given generators. The products of cycles are formed "left to right" and the elements are post multiplied by the generators. If you prefer products "right to left", just think of the elements being pre multiplied by the generators.