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Abstract Algebra help needed

  1. Oct 15, 2006 #1
    1) prove that H is a subgroup of S5 (the permutation group of 5 elements). every element x in H is of the form x(1)=1 and x(3)=3, meaning x moves 1 to 1 and moves 3 to 3. does your argument work hen 5 is replaced by a number greater than or equal to 3?

    2) Let G be a group. prove or disprove that H={g^2:g is an element of H} is a subgroup.
     
  2. jcsd
  3. Oct 15, 2006 #2

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    What exactly is H in 1?

    For 2, note that it is a subgroup if G is abelian, but the proof of this should show there's no reason to expect it to be true in general, and it shouldn't be hard to find a counterexample.
     
  4. Oct 15, 2006 #3
    i actually fouund that out also but im having a very hard time finding an example. maybe im just not seeing it?
     
  5. Oct 15, 2006 #4

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    Just look at some non-abelian groups. You only need to find some elements a and b of G such that aabb is not the square of any element of b. Note a must not commute with b. S_3 is the smallest non-abelian group, but the associated H is A_3, a subgroup, so this won't work. Try A_4, S_4, etc.
     
  6. Oct 15, 2006 #5
    edit:

    so ridiculously lost
     
    Last edited: Oct 16, 2006
  7. Oct 16, 2006 #6

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    Well, I said the smallest non-abelian group, S_3, won't work. The next smallest is A_4, and it does work. One way to prove this is to show the size of the set of elements which are squares does not divide the order of the group. EDIT: Not that it really matters, but, I realized A_4 isn't the next smallest. There are the dihedral groups, etc.
     
    Last edited: Oct 16, 2006
  8. Oct 16, 2006 #7
    i get that |G| = 12 and |H| = 12

    i have that A4={e, (123), (234), (341), (412), (132), (243), (314), (421), (12)(34), (13)(24), (14)(23)}

    then i get that H= {e, (132), (243), (314), (421), (123), (234), (341), (12)(34), (13)(24), (14)(23)}

    or i get that G= G^2 = H

    what am i doing wrong?
     
  9. Oct 16, 2006 #8

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    What element a has a^2=(12)(34)?
     
  10. Oct 16, 2006 #9
    oooh thanks!
     
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