Showing a composition is isomorphic to additive group Consider the law of composition $(x,y) \rightarrow  \sqrt[3]{x^3+y^3}$ 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 $\sqrt[3]{x_1^3+y_1^3} + \sqrt[3]{x_2^3+y_2^3}$ 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 $\sqrt[3]{(x_1+x_2)^3+(y_1+y_2)^3}$, 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!
 A: 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 $x\circ y:=\sqrt[3]{x^3+y^3}.$ 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:
$$
\begin{align}
f(a\circ b)&=\left(\sqrt[3]{a^3+b^3}\right)^3\\
&=a^3+b^3.\\
f(a)&=a^3.\\
f(b)&=b^3.\\
\therefore f(a\circ b)&=f(a)+f(b).
\end{align}
$$
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.
A: Your group is not on the set $\mathbb{R}^2$, but on $\mathbb{R}$. The composition here is given by $x \oplus y = \sqrt[3]{x^3 + y^3}$.
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.
A: 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 $z=x*y=\sqrt[3]{x^3+y^3}.$ Then $$(x*y)*w=z*w=\sqrt[3]{z^3+w^3}=\sqrt[3]{(\sqrt[3]{x^3+y^3})^3+w^3}=\sqrt[3]{x^3+y^3+w^3}.$$
That's pretty neat! It means that in this group, it's easy to compute things: $x*y*z$ is just $\sqrt[3]{x^3+y^3+z^3}.$ 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 $$\phi(x+y)=\phi(x)*\phi(y)=\sqrt[3]{\phi(x)^3+\phi(y)^3}$$ and that $\phi$ must have an inverse $\phi^{-1}$ so that $$\phi^{-1}(\sqrt[3]{x^3+y^3})=\phi^{-1}(x*y)=\phi^{-1}(x)+\phi^{-1}(y).$$ 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. 
