# Group R^× isomorphic to the group R?

• MHB
• mathjam0990
In summary: This is consistent with what has been established by Euge and Deveno, and more or less answers the questions you have asked. However, you should keep in mind that this does not give you the full toolset to determine whether two groups are isomorphic. You should also consider the subgroup test, which basically says that if you have a subset $H$ of a group $(G,*)$, then $H$ is a subgroup of $(G,*)$ if it contains the identity of $(G,*)$ and is closed under the operation of $(G,*)$.
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!

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.

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!

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.

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

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 $x\cdot x = e_{G'}$.

## What is the definition of "Group R^× isomorphic to the group R"?

The term "Group R^× isomorphic to the group R" refers to the concept of two groups, R^× and R, having the same structure or pattern of elements. This means that there is a one-to-one correspondence between the elements of the two groups and the group operations are preserved. In simpler terms, the two groups are essentially the same, just written differently.

## Why is it important to study the isomorphism between these two groups?

Studying the isomorphism between R^× and R can help us better understand the structure and properties of these groups. It also allows us to make connections between different mathematical concepts and to apply techniques and theorems from one group to the other. Additionally, understanding isomorphism can help us identify and classify different groups, which can have important implications in various branches of mathematics.

## How can we determine if two groups are isomorphic?

To determine if two groups are isomorphic, we need to find a function, called an isomorphism, that maps the elements of one group to the elements of the other group while preserving the group structure. This means that the function must be one-to-one, onto, and preserve the group operations. If such a function exists, then the two groups are isomorphic. However, proving that two groups are not isomorphic can be more challenging and often requires a deeper understanding of the groups' properties.

## What is the difference between isomorphism and homomorphism?

Isomorphism and homomorphism are both types of functions between groups. However, isomorphism is a stronger condition than homomorphism. While isomorphism requires the function to be one-to-one and onto, homomorphism only requires the function to preserve the group operations. In other words, homomorphism only requires that the structure of the groups is preserved, while isomorphism also requires that the elements of the groups are matched in a one-to-one manner.

## Can two groups have the same elements but not be isomorphic?

Yes, it is possible for two groups to have the same elements but not be isomorphic. This is because isomorphism not only considers the elements of the groups but also their group operations. Even if the elements of two groups are the same, if their group operations are different, then the two groups are not isomorphic. For example, the groups of integers under addition and multiplication have the same elements, but they are not isomorphic because their group operations are different.

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