A unique operation on a set that makes it a group From Rotman's "Introduction to Group Theory":

Let $G$ be a group and let $X$ be a set having the same number of elements as $G$. If $f:G\rightarrow X$ is a bijection, there is a unique binary operation that can be defined on $X$ so that $X$ is a group and $f$ is an isomorphism.

Let $|G|=|X|=n$. Let the operation for $G$ be given by "$+$". Let us denote the operation for the set $X$ as "$\ast$" (we don't know if it's a group yet, though).
We can construct a bijection $f:G\rightarrow X$. Thus, $x=f(a)$ for every $x\in X$ and $a\in G$. We can see that $f(a+b)\in X\;\;\forall a,b\in G$.
My question is this: What should be my first step in approaching this problem? I was thinking of analyzing the equation:
$$f(a+b)=f(a)\ast f(b)$$
but can't think of anything useful to do with this.
I'm looking for hints on how to get started, not full answers please.
Thanks

this is really just an exercise in abstraction. in essence we use the bijection to pull back the group structure on $G$ into the hitherto unstructured set $X$. we may represent the image of $x$ in $G$ as $x_G$ and if the inverse of this map is $\phi$ we may define an operation $\ast$ on $X \times X$ by:
$$
x \ast y = \phi(x_G y_G) \\
x^{-1} = \phi((x_G)^{-1})
$$
