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  1. #1
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    group theory question

    Let G be the group of complex numbers with modulus 1 under multiplication? Is G \cong G \times G. Justify?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Chandru1 View Post
    Let G be the group of complex numbers with modulus 1 under multiplication? Is G \cong G \times G. Justify?
    Noting that your group is Abelian, I shall give you a somewhat cryptic hint: What two comditions must be fulfilled for a vector space to be a direct sum of two vector spaces?
    Last edited by Swlabr; June 7th 2009 at 11:39 PM. Reason: Ridding myself of some scepticism over infinite-dimensional vector spaces and direct sums...
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  3. #3
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    Answer

    IF V is the vector space then

    v \in V may be written as v=w1+w2 where W1 and W2 are subspaces and W1 \cap W2=(0).

    Ok.

    cant this problem be done by viewing something upon the Kernel. Like if it was isomorphic then K=e..so on
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Chandru1 View Post
    IF V is the vector space then

    v \in V may be written as v=w1+w2 where W1 and W2 are subspaces and W1 \cap W2=(0).

    Ok.

    cant this problem be done by viewing something upon the Kernel. Like if it was isomorphic then K=e..so on
    Do you understand how my hint gives you the result?...

    Another approach would be to note that G \lhd G \times G and so we can quotient out G in both G and G \times G.

    I'm not sure where looking at the Kernel of the isomorphism would really get you. It made me conjour up the above paragraph, but I'm not sure what it would really do. However, in both of my solutions the actual group is irrelevent (as long as it is non-trivial), so I would suspect that there exists a solution that actually uses the group given...

    Also, put the text [math-] and [/math-] (without the "-") around your maths to make it LeTeX-ey.
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  5. #5
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Chandru1 View Post
    Let G be the group of complex numbers with modulus 1 under multiplication? Is G \cong G \times G. Justify?
    Itís not too hard to see that G\times G\not\cong G as follows.

    Suppose \phi:G\times G\to G is a homomorphism. Then \phi(1,1)=1 since homomorphisms map identities to identities. Hence 1=\phi(1,1)=\phi((-1,-1)(-1,-1))=(\phi(-1,-1))^2 and so \phi(-1,-1)=\pm1. Similarly \phi(-1,1)=\phi(1,-1)=\pm1. It follows that \phi maps the four-element set \{(1,1),(-1,1),(1,-1),(-1,-1)\} to the two-element set \{1,-1\}. So, what can you conclude here? Can \phi be injective? Can it be an isomorphism?
    Last edited by TheAbstractionist; June 8th 2009 at 04:26 AM.
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