Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted



Similar Discussions: Permutation 3 (brace yourself)
  1. Permutation mapping (Replies: 4)

  2. Multiplying permutations (Replies: 10)

  3. Permutation Group (Replies: 29)

Loading...