Prove these groups are not isomorphic

[gtfitzpatrick]

Homework Statement

prove that R under addition is not isomorphic toR^*, the group of non zero real numbers under multiplication.

Homework Equations

The Attempt at a Solution

\varphi:(R,+) \rightarrow (R^* , .)
let \varphi(x) = -x
then \varphi(x+y) = -(x+y) = -x-y
\neq \varphi(x)\varphi(y) = (-x)(-y) = xy \neq -x-y
since these are not equal it proves they are not isomorphic?
im a little confused about this, am i doing this right?
Thanks all

 

[lanedance]

how about considering multiplication by zero...

and for clarity, with
<br /> \varphi:(R,+) \rightarrow (R^* , .)<br />

your notation is not very clear, to clean it up a bit how about writing
<br /> X = \varphi(x)<br />

try re-writing the argument you made and see if it makes sense

 

[micromass]

\varphi:(R,+) \rightarrow (R^* , .)
let \varphi(x) = -x
then \varphi(x+y) = -(x+y) = -x-y
\neq \varphi(x)\varphi(y) = (-x)(-y) = xy \neq -x-y
since these are not equal it proves they are not isomorphic?
im a little confused about this, am i doing this right?
Thanks all

This does not suffice to prove that they are not isomorphic. What you have done now, is prove that \varphi is not an isomorphism. However, there could be another function, that does yield an isomorphism. It's not because one function is not an isomorphism, that there doesn't exist a function that is!

You'll have to proceed by contradiction. Suppose that there does exist an isomorphism \varphi:(R,+)\rightarrow (R^*,.) (and you know nothing else of this function, only that it's an isomorphism!), then try to derive a contradiction.

 

[gtfitzpatrick]

thanks for the replys people,
so i know nothing else about this function only that it is isomophic, So the 2 groups are really the same only "labelled" differently. I am not sure how to use this...

 

[micromass]

So, since you know it is an isomorphism, then there must be an element x which gets sent to -1. But where does x+x gets sent to?

 

[HallsofIvy]

Homework Statement

prove that R under addition is not isomorphic toR^*, the group of non zero real numbers under multiplication.

Is this really what the problem says? I can't help but wonder about \phi(x)= e^x where \phi is from R to R^*. Why is that not an isomorphism?

 

[Dick]

Is this really what the problem says? I can't help but wonder about \phi(x)= e^x where \phi is from R to R^*. Why is that not an isomorphism?

R* contains negative numbers. It's not onto.

 

[gtfitzpatrick]

So, since you know it is an isomorphism, then there must be an element x which gets sent to -1. But where does x+x gets sent to?

im not sure if I am thinking along the right lines but x+x get sent to x^2 where as -1-1 = -2 gets sent to +1?

 

[Dick]

im not sure if I am thinking along the right lines but x+x get sent to x^2 where as -1-1 = -2 gets sent to +1?

Partially right. But confused. If phi(x)=(-1) then phi(x+x)=phi(x)*phi(x)=1. Now what does phi(0) have to be? 0 is the identity in R+, yes?

 

[gtfitzpatrick]

Partially right. But confused. If phi(x)=(-1) then phi(x+x)=phi(x)*phi(x)=1. Now what does phi(0) have to be? 0 is the identity in R+, yes?

phi(0) = (0)
phi(0+0) = 0 in (R^*, .) but its supposed to be the group of non zero real numbers so we have a contradiction?

 

[micromass]

Why would phi(0)=0??

 

[gtfitzpatrick]

If phi(x)=(-1) then phi(x+x)=phi(x)*phi(x)=1

so phi(0)=(-1) then phi(0+0)=phi(0)*phi(0)=1

so phi(0) = (1)?

 

[Dick]

If phi(x)=(-1) then phi(x+x)=phi(x)*phi(x)=1

so phi(0)=(-1) then phi(0+0)=phi(0)*phi(0)=1

so phi(0) = (1)?

phi(0)=1 all right. But your argument is hardly convincing since it relies on the statement phi(0)=(-1)??! Where did that come from? Can't you really think of something a little more persuasive??