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Thread: Exact sequence 'counterexample'

  1. #1
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    Exact sequence 'counterexample'

    I would like to show that if the sequence:

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

    is exact, then G need not necessarily be isomorphic to $\displaystyle (\mathbb{Z}/2\mathbb{Z}) \oplus \mathbb{Z}$. 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 $\displaystyle \mathbb{Z}$ 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 $\displaystyle G = \mathbb{Z}$ and define $\displaystyle \Phi : \mathbb{Z} \to G$, $\displaystyle \Psi : G \to \mathbb{Z}_2$ by
    $\displaystyle \Phi (1) = 2, \ \Psi(k) = k (mod 2)$

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

    Edit: Ah, Steve beat me to it. Woops.
    Last edited by Defunkt; Feb 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 $\displaystyle \mathbb{Z}$ 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 $\displaystyle G = \mathbb{Z}$ and define $\displaystyle \Phi : \mathbb{Z} \to G$, $\displaystyle \Psi : G \to \mathbb{Z}_2$ by
    $\displaystyle \Phi (1) = 2, \ \Psi(k) = k (mod 2)$

    then $\displaystyle \Phi$ is injective, $\displaystyle \Psi$ is surjective and $\displaystyle Im \Phi = Ker \Psi$, 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 $\displaystyle G=\mathbb{Z}$ and this is isomorphic to $\displaystyle \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}$. So really you need to be able to find something quite strange, like an exact sequence where $\displaystyle G =\mathbb{Z}^2$. But I've tried many functions and it seems all but impossible.
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  5. #5
    Senior Member roninpro's Avatar
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    Wait a second, $\displaystyle \mathbb{Z}$ is not isomorphic to $\displaystyle \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}$. The latter group has torsion elements, like $\displaystyle (1,0)$. 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, $\displaystyle \mathbb{Z}$ is not isomorphic to $\displaystyle \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}$. The latter group has torsion elements, like $\displaystyle (1,0)$. 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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