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

Kep

Jan 2010
21
0
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 2009
1,176
412
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 \(\displaystyle (a, -a) \mapsto a\).
 
Last edited: