MHB Group R^× isomorphic to the group R?

[mathjam0990]

Question: Is the group R^{x} isomorphic to the group R? Why?
R^{x} = {x ∈ R | x not equal to 0} is a group with usual multiplication as group composition. R is a group with addition as group composition.

Is there any subgroup of R^{x} isomorphic to R?

What I Know: Sorry, I would have liked to show some steps I took, but Not sure where to begin. I tried, but couldn't get too far. Or well, I can say I know for something to be isomorphic the function should be,
1)Injective
2)Surjective
3)Homomorphism f(ab)=f(a)f(b) for all a,b in group

Do I just show that all elements in real numbers with multiplication defined maps to real numbers has those 3 properties above?

Thank you!

 

[Euge]

Hi mathjam0990,

To show that $R^\times$ is isomorphic to $R$, you need to construct a function $f : R^\times \to R$ (or a function $f : R \to R^\times$) that satisfies all three conditions you labeled. However, be careful with the homomoprhism part. To show that a function $f : R^\times \to R$ is a homomorphism, you would need to prove that for all $a,b\in R^\times$, $f(ab) = f(a) + f(b)$.

It turns out though that $R^\times$ is not isomorphic to $R$. A way to show that two groups $(G_1, *)$ and $(G_2,\cdot)$ are not isomorphic is to prove that the number of solutions to $x * x = e_{G_1}$ is not equal to the number of solutions of $x\cdot x = e_{G_2}$. Now, there is only one solution to $x + x = 0$ in $R$, namely $x = 0$. However, there are two solutions to $x\cdot x = 1$ in $R^\times$ -- they are $1$ and $-1$. Therefore, $R^\times$ is not isomorphic to $R$.

Consider the set $R^{>0}$ of all positive real numbers. It is a subgroup of $R^\times$ by the subgroup test: it contains the identity $1$ of $R^\times$, the product of positive numbers is positive, and the inverse of any positive number is positive. I claim that $R^{>0}$ is isomorphic to $R$. To start you off, consider the function $f : R \to R^{>0}$ given by $f(x) = e^x$. Show that $f$ is an isomorphism.

 

[mathjam0990]

Okay, that all makes sense. So basically a way of proving two groups are not isomorphic is to take two elements of each group and multiply (or add) them based on whatever their respective group calls for. Then, if the number of solutions for getting the identity (for each group respectively) are not equal, then they are not isomorphic? Is that correct?

If so, do I still have to show anything for the injective, surjective and homomorphic properties or does the above suffice alone to prove they are not isomorphic?

Thank you!

 

[Deveno]

Just to be clear, suppose we have an isomorphism: $f:G \to G'$.

Suppose $a \in G$ has order $2$. This means $a\ast a = e$, but $a \neq e$.

Since an isomorphism is bijective, and $f(e) = e'$ (the identity of $G'$), we have $f(a) \neq e'$.

Since $f$ is a homomorphism, $f(a) \ast f(a) = f(a\ast a) = f(e) = e'$.

Therefore, if $a$ has order $2$, so does $f(a)$.

So if one group has an element of order $2$, and another does not, they cannot be isomorphic.

It does not help to suppose that $G'$ may have elements of order $2$ that lie OUTSIDE of the image $f(G)$, because isomorphisms are surjective, also.

So I would review Euge's comments in this light.

 

[mathjam0990]

Okay, that makes a bit more sense to combine your answer and his together. Thank you!

 

[Euge]

Given isomorphic groups $(G,*)$, and $(G',\cdot)$, let $f : G \to G'$ be an isomoprhism. Let $X(G)$ represent the set of solutions to $x*x = e_G$ and let $X(G')$ represent the set of solutions to $x\cdot x = e_{G'}$. There is a mapping $\Phi : X(G) \to X(G')$ given by $\Phi(x) = f(x)$, for all $x\in X(G)$. This mapping is well-defined because group isomorphisms send elements of order two to elements of order two (as Deveno has shown). Now $\Phi$ is a bijection (i.e., an injection and surjection), for it has an inverse $\Psi : X(G') \to X(G)$ given by $\Psi(x) = f^{-1}(x)$. Therefore, $X(G)$ and $X(G')$ have the same cardinality (recall, if there is a bijection between two sets $A$ and $B$, then they have the same cardinality). In particular, if $G$ has infinitely many solutions to $x*x = e_G$, then $G'$ has infinitely many solutions to $x\cdot x = e_{G'}$. If $G$ has finitely many solutions to $x*x = e_G$, then $G'$ has the same number of solutions to $x\cdot x = e_{G'}$.