# Showing a composition is isomorphic to additive group

Consider the law of composition $(x,y) \rightarrow \sqrt{x^3+y^3}$ on the set of real numbers. Show that this law of composition defines a group isomorphic to the additive group ($G$) of $\mathbb{R}$.

My attempt: So I know the additive group on $\mathbb{R}$ is just the set of all real numbers that 1) have an identity such that : $x+e = e+x$ for all $x \in G$ i.e. ($e=0$) 2) have an additive inverse 3) are associative

My confusion arises because Im not sure what composition of two elements for instance $(x_1,y_1) \circ (x_2,y_2)$ looks like. Do I add $\sqrt{x_1^3+y_1^3} + \sqrt{x_2^3+y_2^3}$ which then has two values both in $\mathbb{R}$ and then I can see that we are just adding real numbers, which makes sense for the isomorphism, or is the law of composition $\sqrt{(x_1+x_2)^3+(y_1+y_2)^3}$, in which case im a bit lost.

How do I know which is the proper way to compose in the original group? Im heavily leaning towards the first option, because then we can have that $(0,0)$ be our identity element, and $-(x,y)$ be our inverse and we have a subgroup, but I dont understand why this is the law of composition to choose.

In terms of showing that it is an isomorphism, unfortunatley, I am completely lost.

Any help is appreciated, thanks!

• The law of composition given in the problem does not give a group structure on $\mathbb{R}^2$, but on $\mathbb{R}$. That is, it defines the group $(\mathbb{R}, *)$, where $a * b = \sqrt{x^3+y^3}$. Nov 19, 2012 at 22:36

I approach this question by figuring out the most fundamental question: What is the isomorphism here, if there is one?

First of all, the isomorphism has to be from $(\mathbb{R},\circ)$ to $(\mathbb{R},+)$ with $x\circ y:=\sqrt{x^3+y^3}.$ Thus, we know that it is $f: (\mathbb{R},\circ) \to (\mathbb{R}, +)$.

We have, of course, the primary condition of an isomorphism (using our $f$ and aforementioned notation): $$f(a \circ b)=f(a)+f(b) \quad \forall a,b \in (\mathbb{R}, \circ).$$

Our choice of $f$, then, would be $f(z)=z^{3}$. We see that this satisfies the primary condition of an isomorphism:

\begin{align} f(a\circ b)&=\left(\sqrt{a^3+b^3}\right)^3\\ &=a^3+b^3.\\ f(a)&=a^3.\\ f(b)&=b^3.\\ \therefore f(a\circ b)&=f(a)+f(b). \end{align}

Thus, we have an isomorphism and we can state $(\mathbb{R}, \circ)\cong (\mathbb{R}, +)$.

N.B. The identity and inverses are trivially preserved. If necessary, show that.

• Thanks for the edit, MSEoris.
– 000
Nov 20, 2012 at 21:41

First of all, both of your proposed composition laws are incorrect. Recall that a composition law is a way of taking two elements of your group and getting a third, i.e., taking $x$ and $y$ to get $z$. Now if we compose the result with $w$, what we should get is $(x*y)*w=z*w.$ One thing this means is that applying composition twice is a function that takes three elements and gives us a fourth. Your mistake was treating elements of the group as pairs of numbers, when actually each element is a single real number. So let's see what that means in the context of the group here. Let $z=x*y=\sqrt{x^3+y^3}.$ Then $$(x*y)*w=z*w=\sqrt{z^3+w^3}=\sqrt{(\sqrt{x^3+y^3})^3+w^3}=\sqrt{x^3+y^3+w^3}.$$ That's pretty neat! It means that in this group, it's easy to compute things: $x*y*z$ is just $\sqrt{x^3+y^3+z^3}.$ So now let's look for the desired isomorphism with $\mathbb{R}.$ Call the isomorphism $\phi:\mathbb{R}:\to$ your group. We know because $\phi$ is a homomorphism that $$\phi(x+y)=\phi(x)*\phi(y)=\sqrt{\phi(x)^3+\phi(y)^3}$$ and that $\phi$ must have an inverse $\phi^{-1}$ so that $$\phi^{-1}(\sqrt{x^3+y^3})=\phi^{-1}(x*y)=\phi^{-1}(x)+\phi^{-1}(y).$$ How can we find a $\phi$ satisfying these properties? A good way to begin is to look at the way in which multiplication in your group resembles addition in $\mathbb{R},$ and then to figure out what precise manipulation you'd have to make to switch from one operation to the other.

Your group is not on the set $\mathbb{R}^2$, but on $\mathbb{R}$. The composition here is given by $x \oplus y = \sqrt{x^3 + y^3}$.

An isomorphism from $G = (\mathbb{R},+)$ to this group is given by $x \mapsto x^{1/3}$. I leave it to you to verify that this works.

• I'm curious. Do you think that it would be preferred that the given mapping is from $(\mathbb{Z}, \circ)$ to $(\mathbb{Z},+)$ rather than the other way around? I'm aware this is a very minor point, but people often make a big deal about the minor points. In my opinion, it matters very little since isomorphism is an equivalence relation in this context.
– 000
Nov 19, 2012 at 22:52
• I appreciate this insight and between this and some of the other points brought up I understand the correct way to interpret this. What im now wondering now is what would be the difference in the notation if my set were on $\mathbb{R}^2$. Nov 19, 2012 at 22:56
• I am not certain, but I don't think you have an implied extension to $\mathbb{R}^2$. You could define it yourself in a certain way, but there is no implication of what it means given the definition $x\circ y:=\sqrt{x^3+y^3}$. I may be wrong, but I have no clue how one would interpret $(x_1,y_1)\circ(x_2,y_2):=\sqrt{(x_1,y_1)^3+(x_2,y_2)^3}$, which is the only natural implication of the notation.
– 000
Nov 19, 2012 at 23:01