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Permutation 3 (brace yourself)

  1. Dec 1, 2004 #1
    Prove that A4 has no subgroup of order 6 in this way (and this way only):

    Suppose that A4 has a subgroup H of order 6. Explain (in one sentence) why H must contain a 3-cycle. WLOG(without loss of generality) let this be (1,2,3).
    Then H must have iota,(1,2,3) and (1,2,3)^-1=(3,2,1).
    Now apply the result from my permutation 2 thread( Prove: a group of even order must have an even number of elements of order 2) to H.
    As a result all that's left to do is take three cases, according to which element of order 2 in A4 belongs to H.
    In each case compute products and take inverses in H to show H always has more than 6 elements (contradiction)
  2. jcsd
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