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Groups again

  1. Mar 26, 2005 #1
    if a group of order 2p ( p prime) is abelian....then does it have exactly one element of order 2 ?? if a group is non abelian...i could figure out that there are p elements of order 2. but the abelian case is a bit confusing...
    also..is it like...any group of order 2p has an element of order p?

    if a group has orer p^a , a>=1 where p is prime....then i've gotta show that G has an element of order p.
    can i say that any non-identity element in G can have order p or p^2 or p^3......or p^a. then if x in G is of the form x^(p^i) =e ....we can say, (x^(p^(i-1))^p= e and we've found an element x^(p^(i-1)) that is of order p....it seems too simple to be correct.
     
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  3. Mar 27, 2005 #2
    suppose G is abelian and has 2 elements of order 2, say a and b...then {e,a,b,ab} becomes a subgroup of G. but 4 does not divide 2p unless p=2. if p=2 then anyway {e,a} is a subgroup of G of order 2( p in this case).
     
  4. Mar 27, 2005 #3
    thanks for the help!
    here's another question on the same lines: if G is a group where a^2=e for each a in G. show that order of G is 2^n for some n >=0. it's clear that the group is abelian and also clear that 2 divides O(G). how do you proceed further? by any chance, is it induction that we're supposed to use?
     
  5. Mar 27, 2005 #4

    matt grime

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    The first answer also follows by the observation that any abelian group of order 2p for p an odd prime is cyclic.

    The second follows from the fact that if p is a prime and p divides |G| then there is an element of order p in G.
     
  6. Mar 27, 2005 #5
    if G is a group such that a^2=e for each a in G. show that order of G is 2^n for some n>=0.
    please help...
    is the group of 3 non-singular upper triangular matrices a normal subgroup of
    GL(3,R), the group of 3 cross 3 non singular matrices over R.
     
  7. Mar 27, 2005 #6

    matt grime

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    I've answered the a^2=e for all a in G one.

    As for the p^r one, all elements have order dividing p^r say the order of g is p^s, then p^{s-1] has order p.
     
  8. Mar 27, 2005 #7

    matt grime

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    And why don't you just do the matrix one? it simple boils down to multiplying matrices together.

    Hint: it's probably easier to find a counter example.
     
  9. Mar 27, 2005 #8

    matt grime

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    Just to put here another useful result: if G is any group and all of its elements have order 2 then it is abelian. proof an exercise (hint what is (xy)^{-1})
     
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