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

• May 14th 2010, 01:40 AM
Kep
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?
• May 14th 2010, 01:53 AM
Swlabr
Quote:

Originally Posted by 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?

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.

$\phi:(a, b) \mapsto e^{a+b}$. 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 $(a, -a) \mapsto a$.