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Is HK a Subgroup

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Let H, K be subgroups of S_5, where H is generated by (1 2 3) and K is generated by (1 2 3 4 5). Is HK a subgroup of S_5?


    2. Relevant equations
    HK is a subgroup iff HK = KH.

    3. The attempt at a solution
    Is there an easy way of answering this question without computing HK and KH?
     
  2. jcsd
  3. Nov 11, 2008 #2
    Maybe the simplest thing to try is to show that HK is a subgroup directly from the definition of subgroup.

    HK is obviously non-empty. Let x = (1 2 3), let y = (1 2 3 4 5) and let a = xiyj and b = xkym be elements of HK.

    Then ab-1 = xiyj-mx-k. Let n be such that k + n = i. Then xiyj-mx-k = xk(xnyj-m)x-k = cd where c is in H and d is in K, since cojugating preserves cycle structure. Right? Thus, ab-1 is in HK and HK is a subgroup.
     
  4. Nov 11, 2008 #3

    Dick

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    I don't believe that proof for a minute. The product of the two groups contains all kinds of cycle structure. Like HK contains (3,4,5). That's a 3-cycle and it's not in H. And I'll tell you another thing. (3,4,5) is not in KH. I haven't figured out any special tricks to know that yet. I did it the hard way.
     
  5. Nov 11, 2008 #4
    Coming from you, that's not a good sign. Can you tell me the flaw in my proof?
     
  6. Nov 11, 2008 #5

    Dick

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    I can't really tell you because I don't see how it proves anything. You seem to be thinking everything in HK has the cycle structure of a disjoint three cycle and a five cycle. And the three cycle must belong to H and the five cycle must belong to K. Is that what you are thinking? I can only guess. It's not true. (1,2,3)*(1,3,5,2,4)=(2,4)*(3,5). So?
     
  7. Nov 11, 2008 #6
    That's exactly what I was thinking. I realize now that it is wrong. Oh well.
     
  8. Nov 11, 2008 #7

    Dick

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    That doesn't mean there isn't some trick you can use to show HK is not equal to KH without evaluating all 15 products in both sets (which really isn't that bad, only 8 of them are nontrivial in each set). It just means I haven't thought of one.
     
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