Show that R^x/<-1> is isomorphic to the group of positive real numbers under multipli

Printable View

• Dec 9th 2010, 07:03 AM
kathrynmath
Show that R^x/<-1> is isomorphic to the group of positive real numbers under multipli
Show that R^x/<-1> is isomorphic to the group of positive real numbers under multiplication.

I know I need to show we have a homomorphism, and is one - to one and onto in order to be isomorphic. I know all that, but I don't know what function mapping to use.
• Dec 9th 2010, 09:19 AM
tonio
Quote:

Originally Posted by kathrynmath
Show that R^x/<-1> is isomorphic to the group of positive real numbers under multiplication.

I know I need to show we have a homomorphism, and is one - to one and onto in order to be isomorphic. I know all that, but I don't know what function mapping to use.

An idea: check the function $\displaystyle \phi :\mathbb{R}^{*}\rightarrow \mathbb{R}^+ \,,\,\,\phi(x):=|x|$ ...

Tonio
• Dec 9th 2010, 02:21 PM
kathrynmath
I'm curious as to how you chose that function?
• Dec 9th 2010, 03:28 PM
tonio
Quote:

Originally Posted by kathrynmath
I'm curious as to how you chose that function?

Well, you need a multiplicative function, i.e. $\displaystyle \phi \,\,s.t.\,\,\phi(xy)=\phi(x)\phi(y)$ and I don't know that many, so....

Tonio
• Dec 9th 2010, 06:53 PM
kathrynmath
So you can just choose any multiplicative function?
so assume f(x1)=f(x2)
|x1|=|x2|
x1=x2
1-1
onto because a maps to |a|
Now f(x)f(y)=|x||y|=|xy|=f(xy) homomorphism
Thus isomorphic.
• Dec 9th 2010, 06:59 PM
tonio
Quote:

Originally Posted by kathrynmath
So you can just choose any multiplicative function?

No, not any. Say, $\displaystyle f(x):=1$ doesn't work. Do you know any other multiplicative functions for our problem?

Tonio

so assume f(x1)=f(x2)
|x1|=|x2|
x1=x2
1-1
onto because a maps to |a|
Now f(x)f(y)=|x||y|=|xy|=f(xy) homomorphism
Thus isomorphic.

.
• Dec 9th 2010, 07:00 PM
tonio
Quote:

Originally Posted by kathrynmath
So you can just choose any multiplicative function?
so assume f(x1)=f(x2)
|x1|=|x2|
x1=x2
1-1
onto because a maps to |a|
Now f(x)f(y)=|x||y|=|xy|=f(xy) homomorphism
Thus isomorphic.

No. My function is NOT an isomorphism since it is not 1-1, but you can apply the first isomorphism theorem for groups...

Tonio
• Dec 9th 2010, 08:22 PM
kathrynmath
If G is a group, N is a normal subgroup of G, and H is any subgroup of G. Then HN is a subgroup of G, H intersect N is a normal subgroup of H and HN/N isomorphic to H/(H intersectN) by the first isomorphism theorem.

I'm not quite sure how to use that.
• Dec 10th 2010, 03:42 AM
tonio
Quote:

Originally Posted by kathrynmath
If G is a group, N is a normal subgroup of G, and H is any subgroup of G. Then HN is a subgroup of G, H intersect N is a normal subgroup of H and HN/N isomorphic to H/(H intersectN) by the first isomorphism theorem.

I'm not quite sure how to use that.

That is not the standard first isomorphism theorem for groups, which you can read in one of the following sites:

Isomorphism theorem - Wikipedia, the free encyclopedia

http://www.tjsullivan.org.uk/math/morphthm.pdf (pages 5-6)

http://marauder.millersville.edu/~bi...o/firstiso.pdf

If you look at the above carefully then you'll understand why the absolute value function solves your problem.

Tonio