Prove that A4 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:
1) 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, A4 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.
1-I have a difficulty in finding the contradiction in the case when H contains six 3-cycles, could anyone explain for me that contradiction?
2-Also I have a question in that sentence "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." why $a^{-1}b$ & $ba^{-1}$ must not also be in H, could anyone clarify this for me please?Also How can I show that this is not the case?
Many thanks!
For the first question, a group has to contain an identity element, which is not a 3-cycle.
For the second, it's not necessary to find $a^{-1}b$ and $ba^{-1}$ -- it's enough to show that $ab$ and $ab^{-1}$ are too many. We're looking for a contradiction. To show that it's not a contradiction, what would it mean if $ab=ab^{-1}$, or $ab=1$?