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

#### Kep

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?

#### Swlabr

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 $$\displaystyle (\mathbb{C}, +) \cong (\mathbb{R} \times \mathbb{R}, +)$$ 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.

$$\displaystyle \phi a, b) \mapsto e^{a+b}$$. Clearly $$\displaystyle e^{0+0} = e^0 = 1$$ and $$\displaystyle (a,b)\phi * (c,d)\phi = e^{a+b}e^{c+d} = e^{a+b+c+d} = (a+c, b+d)\phi = ((a, b)+(c, d))\phi$$.

You now just need to prove that the kernel is isomorphic to the reals. However, the kernel is the set $$\displaystyle (a, -a)$$ which is isomorphic to the reals under the isomorphism $$\displaystyle (a, -a) \mapsto a$$.

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