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Prove that A_4 (a group of order 4!//2 = 24/2 = 12) has no subgroup H...

Prove that A_4 (a group of order 4!//2 = 24/2 = 12) has no subgroup H of order 6 by showing:

1) all squares of elements A_4 must be H since (A_4:H) =2

2) All eight 3 -cycles are squares and hence must be in H.

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