Prove these groups are not isomorphic

[gtfitzpatrick]

Homework Statement

prove that R under addition is not isomorphic to$$R^*$$, 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
$$\varphi:(R,+) \rightarrow (R^* , .)$$

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

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 to$$R^*$$, 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??