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Isomorphism between G and Z x Z_2 if G has a normal subgroup isomorphic to Z_2

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data
    If G contains a normal subgroup H which is isomorphic to [tex]\mathbb{Z}_2[/tex], and if the corresponding quotient group is infinite cyclic, prove that G is isomorphic to [tex]\mathbb{Z}\times\mathbb{Z}_2[/tex]

    3. The attempt at a solution
    [tex]G/H[/tex] is infinite cyclic, this means that any [tex]g\{h1,h2\}[/tex] is generated by some [tex]\gamma\{h1,h2\}[/tex] with [tex]\gamma\in G[/tex]. [tex]\gamma=g^n[/tex] because H is normal. But now?
     
  2. jcsd
  3. Jan 14, 2010 #2
    Do you know any general property of a group [tex]G[/tex] such that, when [tex]G[/tex] has this property and [tex]N[/tex] is a normal subgroup of [tex]G[/tex], you can conclude that [tex]G \cong N \times G/N[/tex]?

    (Hint: in the direct product [tex]H \times K[/tex] of two groups, what is the relationship between the subgroups [tex]H \times 1[/tex] and [tex]1 \times K[/tex]?)
     
  4. Jan 14, 2010 #3
    I do not know any general property of this kind... the subgroups H x 1 and 1 x K only have the identity in common and (H x 1)(1 x K)=H x K, but I do not see how this helps...
     
  5. Jan 14, 2010 #4
    The "general property of [tex]G[/tex]" I was referring to is "[tex]G[/tex] is abelian". One way to understand the thing that makes direct products special is that the factors commute with each other: in the product above, [tex](h, 1)(1, k) = (1, k)(h, 1) = (h, k)[/tex].

    In your original problem, what happens if [tex]G[/tex] is abelian? What happens if it's not?
     
  6. Jan 16, 2010 #5
    I really do not understand... what is the use of the fact that the factors of the direct products commute with each other?
     
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