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Math Help - [SOLVED] Proper Subgroups of Q

  1. #1
    Member Haven's Avatar
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    [SOLVED] Proper Subgroups of Q

    Prove that \mathbb{Q} is not isomorphic to a proper subgroup of itself.

    So, far i have:
    Let G \leq (\mathbb{Q},+)
    Let \phi : G \rightarrow (\mathbb{Q},+)
    Let \frac{p}{q} not in G
    Since \phi is bijective \phi(x) = \frac{p}{q} for some x \in G

    I know there's a contradiction here, but I can't see it
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  2. #2
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    Quote Originally Posted by Haven View Post
    Prove that \mathbb{Q} is not isomorphic to a proper subgroup of itself.

    So, far i have:
    Let G \leq (\mathbb{Q},+)
    Let \phi : G \rightarrow (\mathbb{Q},+)
    Let \frac{p}{q} not in G
    Since \phi is bijective \phi(x) = \frac{p}{q} for some x \in G

    I know there's a contradiction here, but I can't see it
    Assume G be a proper subgroup of (\mathbb{Q},+) such that \phi : \mathbb{Q} \rightarrow G is a bijection.
    Let \frac{p}{q} \in (\mathbb{Q} \setminus G). Since \phi is a bijection, if x = \frac{p/q}{\phi(1)}, then \phi(x)=\frac{p}{q}, where \phi(x)=x\phi(1) and \phi(1) \neq 0. It contradicts that \frac{p}{q} is not a member of G. Thus (\mathbb{Q}, +) is not isomorphic to a proper subgroup of itself.
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by aliceinwonderland View Post
    Assume G be a proper subgroup of (\mathbb{Q},+) such that \phi : \mathbb{Q} \rightarrow G is a bijection.
    Let \frac{p}{q} \in (\mathbb{Q} \setminus G). Since \phi is a bijection, if x = \frac{p/q}{\phi(1)}, then \phi(x)=\frac{p}{q}, where \phi(x)=x\phi(1) and \phi(1) \neq 0. It contradicts that \frac{p}{q} is not a member of G. Thus (\mathbb{Q}, +) is not isomorphic to a proper subgroup of itself.
    I do not believe this is a proof. You assume \phi(x) = x\phi(1), but this is not necessarily so. Essentially, you have produced a generic proof which says that a proper subgroup of a group cannot be isomorphic to the group itself. This is not so. An obvious counter-example is the infinite cartesian product of the integers. \mathbb{Z} \times \mathbb{Z} \times \ldots \cong <1> \times \mathbb{Z} \times \ldots.

    However, the result you used is true, but needs to be proven.

    Notice that for x \in \mathbb{Z}, x\phi = x. So the result now just needs to be shown for the denominators.

    x\phi = 1/p \Rightarrow (px)\phi = 1 \Rightarrow x = 1/p as the kernel is trivial.

    Thus, \frac{p}{q} \phi = \frac{p\phi}{q\phi} = \frac{p}{q} as required.

    Or something along those lines...
    Last edited by Swlabr; February 6th 2010 at 12:20 AM.
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  4. #4
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    Quote Originally Posted by Swlabr View Post
    I do not believe this is a proof. You assume \phi(x) = x\phi(1), but this is not necessarily so. Essentially, you have produced a generic proof which says that a proper subgroup of a group cannot be isomorphic to the group itself. This is not so. An obvious counter-example is the infinite cartesian product of the integers. \mathbb{Z} \times \mathbb{Z} \times \ldots \cong <1> \times \mathbb{Z} \times \ldots.

    However, the result you used is true, but needs to be proven.

    Notice that for x \in \mathbb{Z}, x\phi = x. So the result now just needs to be shown for the denominators.

    x\phi = 1/p \Rightarrow (px)\phi = 1 \Rightarrow x = 1/p as the kernel is trivial.

    Thus, \frac{p}{q} \phi = \frac{p\phi}{q\phi} = \frac{p}{q} as required.

    Or something along those lines...
    I do not believe this is false. If \phi is an isomorphism between (Q, +) and its subgroup, then \phi(x)=x\phi(1) must hold. Here is why.

    Let x be an arbitrary member in Q, let's say
    x= p/q (if x=0, it is trivial).
    \phi (\frac{p}{q})=\phi(\frac{1}{q} + ... +\frac{1}{q})=p \phi(\frac{1}{q}).
    \phi (1)=\phi(\frac{q}{q})=q\phi(\frac{1}{q}). It follows that \phi(\frac{1}{q})=\frac{\phi(1)}{q}.
    Then, \phi(x)=\phi(\frac{p}{q})=\frac{p}{q}(\phi(1))=x\p  hi(1).
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by aliceinwonderland View Post
    I do not believe this is false. If \phi is an isomorphism between (Q, +) and its subgroup, then \phi(x)=x\phi(1) must hold. Here is why.

    Let x be an arbitrary member in Q, let's say
    x= p/q (if x=0, it is trivial).
    \phi (\frac{p}{q})=\phi(\frac{1}{q} + ... +\frac{1}{q})=p \phi(\frac{1}{q}).
    \phi (1)=\phi(\frac{q}{q})=q\phi(\frac{1}{q}). It follows that \phi(\frac{1}{q})=\frac{\phi(1)}{q}.
    Then, \phi(x)=\phi(\frac{p}{q})=\frac{p}{q}(\phi(1))=x\p  hi(1).
    I apologise for any ambiguity in my previous post. I was trying to say the claim required proof, not that it was wrong.
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