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Math Help - Exact sequence 'counterexample'

  1. #1
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    Exact sequence 'counterexample'

    I would like to show that if the sequence:

    0 \to \mathbb{Z} \to G \to \mathbb{Z}/2\mathbb{Z} \to 0

    is exact, then G need not necessarily be isomorphic to (\mathbb{Z}/2\mathbb{Z}) \oplus \mathbb{Z}. Just one example of G that is not isomorphic to this (where the above sequence is still exact) would be enough. I've tried many and I can't seem to find a single valid one.. any hints?
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  2. #2
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    This is not my area of expertise, so excuse me if I make a dumb error, but can't we just let G be the integers with the first map multiplication by 2, and the second map sending n to n(mod 2)?

    Edit: I didn't check if \mathbb{Z} is isomorphic to the direct sum. The obvious map is probably an isomorphism.
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  3. #3
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    If I'm not mistaken, you can take G = \mathbb{Z} and define \Phi : \mathbb{Z} \to G, \Psi : G \to \mathbb{Z}_2 by
    \Phi (1) = 2, \ \Psi(k) = k (mod 2)

    then \Phi is injective, \Psi is surjective and Im \Phi = Ker \Psi, so the sequence is exact.

    Edit: Ah, Steve beat me to it. Woops.
    Last edited by Defunkt; February 15th 2011 at 08:25 AM.
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  4. #4
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    Quote Originally Posted by DrSteve View Post
    This is not my area of expertise, so excuse me if I make a dumb error, but can't we just let G be the integers with the first map multiplication by 2, and the second map sending n to n(mod 2)?

    Edit: I didn't check if \mathbb{Z} is isomorphic to the direct sum. The obvious map is probably an isomorphism.
    Quote Originally Posted by Defunkt View Post
    If I'm not mistaken, you can take G = \mathbb{Z} and define \Phi : \mathbb{Z} \to G, \Psi : G \to \mathbb{Z}_2 by
    \Phi (1) = 2, \ \Psi(k) = k (mod 2)

    then \Phi is injective, \Psi is surjective and Im \Phi = Ker \Psi, so the sequence is exact.

    Edit: Ah, Steve beat me to it. Woops.
    Yes, that is exactly the problem! Virtually any decent map you can come up with gives G=\mathbb{Z} and this is isomorphic to \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}. So really you need to be able to find something quite strange, like an exact sequence where G =\mathbb{Z}^2. But I've tried many functions and it seems all but impossible.
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  5. #5
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    Wait a second, \mathbb{Z} is not isomorphic to \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}. The latter group has torsion elements, like (1,0). The former group is torsion-free. (You could also invoke the Fundamental Theorem of Finitely Generated Abelian Groups.)
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  6. #6
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    Quote Originally Posted by roninpro View Post
    Wait a second, \mathbb{Z} is not isomorphic to \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}. The latter group has torsion elements, like (1,0). The former group is torsion-free. (You could also invoke the Fundamental Theorem of Finitely Generated Abelian Groups.)
    Great! Don't know how I missed that.. the 'obvious' homomorphism between them looked like an isomorphism to me.
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  7. #7
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    Quote Originally Posted by Capillarian View Post
    Great! Don't know how I missed that.. the 'obvious' homomorphism between them looked like an isomorphism to me.
    You might need to check the difference between a short exact sequence and a split exact sequence. Being a short exact sequence is a necessary condition for being a split exact sequence. Given a short exact sequence, there are some additional conditions to be satisfied for being a split exact sequence.
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