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Math Help - Which group modulo the additive reals is isomorphic to the multiplicative reals.

  1. #1
    Kep
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    Which group modulo the additive reals is isomorphic to the multiplicative reals.

    Hi There.

    Simple question to pose, maybe not too easy to answer.

    Let G be a group such that it's quotient over the additive reals is isomorphic to the multiplicative reals.

    Which familiar group is G isomorphic to?

    Equivalently, what is the product group of the additive reals with the multiplicative reals isomorphic to?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Kep View Post
    Hi There.

    Simple question to pose, maybe not too easy to answer.

    Let G be a group such that it's quotient over the additive reals is isomorphic to the multiplicative reals.

    Which familiar group is G isomorphic to?

    Equivalently, what is the product group of the additive reals with the multiplicative reals isomorphic to?
    I would guess the complex numbers under addition. You are quotienting out a copy of the reals (under +) to get a copy of the reals (under *). Apparently the group you start with is one you are `familiar' with. Well, you only know a couple of groups which are uncountable...

    I hope that helps for the moment!

    EDIT: Although (\mathbb{C}, +) \cong (\mathbb{R} \times \mathbb{R}, +) so you will have to look quite hard for this copy of the reals, it certainly isn't obvious...

    EDIT2: Powers seems to work. You are wanting to turn addition into multiplication, so powers seem to be a sensible choice.

    a, b) \mapsto e^{a+b}" alt="\phia, b) \mapsto e^{a+b}" />. Clearly e^{0+0} = e^0 = 1 and (a,b)\phi * (c,d)\phi = e^{a+b}e^{c+d} = e^{a+b+c+d} = (a+c, b+d)\phi = ((a, b)+(c, d))\phi.

    You now just need to prove that the kernel is isomorphic to the reals. However, the kernel is the set (a, -a) which is isomorphic to the reals under the isomorphism (a, -a) \mapsto a.
    Last edited by Swlabr; May 14th 2010 at 02:15 AM.
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