An Introduction to Theory of Groups by Joseph J. Rotman - Exercise 1.44 Question:

Let $G$ be a group, let $X$ be a set and let $f: G \longrightarrow X$ be a bijection. Show that there is an unique operation on $X$ so that $X$ is a group and $f$ is an isomorphism.

My attempt:

Define an operation $*$ on $X$ such that $f(a \ \cdot \ b) = f(a) * f(b)$ for an arbitrary $a, b \in G$, so $f(a), \ f(b), \ f(a) * f(b) \in X$, because $a, b \in G$ are arbitrary and $f$ is a bijection, so $*$ is an binary operation on $X$. We show now that $(X, *)$ is a group. In fact,
$\bullet \ f(e)$ is the neutral element of $(X,*)$, because $f(e) * f(a) = f(e \ \cdot \ a) = f(a) = f(a \ \cdot \ e) = f(a) * f(e)$, where $e$ is the neutral element of $(G, \cdot)$ and $a$ is an arbitrary element of $(G,\cdot)$.
$\bullet$ $f(a^{-1})$ is the inverse element of $f(a)$ in $(X,*)$, because $f(a^{-1}) * f(a) = f(a^{-1} \ \cdot \ a) = f(e) = f(a \ \cdot \ a^{-1}) = f(a) * f(a^{-1})$
Therefore $(X,*)$ is a group and $f: G \longrightarrow X$ is an isomorphism by the way it was defined. We show now the oneness of operation $*$. Suppose there are two operations $*$ and $\circledast$ on $X$ such that $(X,*)$, $(X,\circledast)$ are groups and $f: (G, \cdot) \longrightarrow (X,*)$, $f: (G, \cdot) \longrightarrow (X,\circledast)$ are isomorphisms, so $id_X: (X,*) \longrightarrow (X,\circledast)$ is an isomorphism.

I think this last isomorphism should help me tho prove what I want, but I don't know how this isomorphism ensure the oneness of operation $*$. I would like to a hint. Thanks in advance!
 A: $f$, which is a specific (fixed) bijection, is an isomorphism of both $(X,*) \longleftrightarrow (G,\cdot)$ and $(X,*) \longleftrightarrow (G,\circledast)$.
So $(1): \forall x_1,x_2 \in X : f^{-1}(x_1* x_2) = f^{-1}(x_1)\cdot  f^{-1}(x_2)$.
And $(2): \forall x_1,x_2 \in X : f^{-1}(x_1\circledast x_2) = f^{-1}(x_1)\cdot  f^{-1}(x_2)$.
$(1)$ and $(2)$ are just the composition parts of the definition of a isomorphism between two groups.  But note that the RHS of both of these are identical. 
Now take $f$ of both sides in $(1)$ and in $(2)$; this is valid since $f$ is a bijection:
$$(1'): \forall x_1,x_2 \in X : x_1* x_2 = f(f^{-1}(x_1)\cdot  f^{-1}(x_2))$$
$$(2'): \forall x_1,x_2 \in X : x_1\circledast x_2 = f(f^{-1}(x_1)\cdot  f^{-1}(x_2))$$
Again the RHS are identical in $(1')$ and $(2')$. 
$$\forall x_1,x_2 \in X : x_1* x_2 = x_1\circledast x_2$$
Therefore $*$ and $\circledast$ have identical "multiplication tables" and are the same operation.
A: Pick $x,y\in X$, suppose $f(a)=x,f(b)=y$, we must have $x*y=f(a)*f(b)=f(ab)$.
So $*$ must be defined in this way, we must have $x*y=f(f^{-1}(x)*f^{-1}(y))$.
