Prove that A ₄ has no subgroups of order 6. (prove by contradiction broken up into 3 cases which each lead to a different kind of contradiction) An outline is given below:

Show that the converse of Lagrange's Theorem is false: That is, show that if k divides the order of G, G may not have a subroup of order k. In fact, A ₄ which has 12 elements does not have any of order 6.

The 12 elements of A4 are: 3-Cycles: (123), (132), (124), (142), (134), (143), (234), (243) Four other permutations which aren’t 3-cycles: (1), (12), (34), (13),(24), (14), (23)

Suppose A4 had a subgroup H of order 6. Then H must contain at least one 3-cycle. (Why?) If H contains a 3-cycle, then it must also contain the inverse of that 3-cycle, so the number of 3-cycles in H must be even— therefore 2, 4, or 6. If H contains six 3-cycles, then we have a contradiction (what is it?). If H contains just two 3-cycles, then H must contain all four of the non-3-cycle permutations, but those four permutations form a subgroup, and 4 does not divide 6, which is a contradiction (show this!) Therefore H contains four 3-cycles—call them a, a ⁻¹ , b, and b ⁻¹ . H must also contain ab, ab ⁻¹ , and (1)—which adds up to seven elements unless two of these elements are the same. Show that’s not the case.