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Group theory, isomorphism

  1. Oct 16, 2009 #1
    Let G be a finite group. For all elements of G (the following holds: g^2=e(the idendity.) So , all except the idendity have order two.

    Proof that G is isomorphic to a finite number of copies of Z_2 ( the group of adittion mod 2, Z_2 has only two elements (zero and one).)

    I can try to tell you what I have already tried, but please, can someone give a hint in the right direction..? I basically need a bijection from G to (Z_2)^m, but no idea how I can do it( In particular: they should have the same size as they are finite.)
     
  2. jcsd
  3. Oct 16, 2009 #2
    Define [tex] \phi : G \rightarrow Z_2 [/tex], so that it is forced to be surjective; i.e. tell the identity where to go. Show [tex] \phi [/tex] is a hom. Then look at the kernel. There is only one possibility.
     
  4. Oct 16, 2009 #3
    I am sorry, I am not able to see it. I keep on thinking about the fact that m products of Z_2 has order 2^m, which is supposed to be equal to n( the order of the group G..)

    I dont even get your advice: How can we tell the identity where to go!? The identity(of G) should always go to idendity(of Z_2). ..

    Sorry, I am confused...
     
  5. Oct 18, 2009 #4
    No one who can provide some insights..?
     
  6. Oct 21, 2009 #5
    Show the group is isomorphic to [tex] Z_2 \times Z_2 \times \ldots [/tex], so

    G must be commutative since it is the case that forall g, [tex] g = g^{-1} [/tex] (since gg=1)
    In particular, [tex] ab = (ab)^{-1} = b^{-1}a^{-1} = ba (by assumption) [/tex] So that G is commutative.

    Next, you could use the fundamental theorem of finite abelian groups, says that any abelian finite group is isomorphic to [tex] Z_{m_1} \times Z_{m_2} \times \ldots [/tex]

    Next, look to intuitionize what is happening, look at subgroups generated by a single non identity element. There will be the |G|- 1 of them and they will all have order 2. There is only one group of order 2 up to isomorphism. What is the next step?
     
  7. Oct 21, 2009 #6
    We know by cauchy that the order of G is in the form 2^k. Applying this combined with the fundamental theorem is the proof.
     
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