Proving an Isomorphism regarding Group law for Elliptic Curves (independent of neutral element) - Fulton Exercise 5.36 This is Fulton's (Algebraic Curves) Exercise 5.36 for reference.
If we consider $C$ an irreducible cubic, and $O,O'$ simple (smooth) points on $C$ we have that each gives rise to a group operation $+,+'$ on the set of simple points $C^{\circ}$. Let $ Q= \phi(O,O')$. We define $A+B = \phi(O,\phi(A,B))$ and similar for $+'$. We assume that $+,+' $ are commutative and associative.
Now we need to show that the map $\alpha :(C^{\circ},+,O) \to (C^{\circ},+',O')$ by $\alpha(P) = \phi(Q,P)$ is a  group isomorphism. I'm not getting the homomorphism part at all, bijectivity is fine.
This question is similar : Proving a group isomorphism from $(S,+)$ to $(S,+')$ but it doesn't define the isomorphism the same way, here $F(P)= \phi(O,\phi(P,O'))$ where $\alpha(P)=\phi(P,\phi(O,O'))$.
What I have is $\alpha(A)+'\alpha(B) = \phi(O',\phi(\phi(Q,A),\phi(Q,B))$ and $\alpha(A+B) = \phi(Q,\phi(O,\phi(A,B))$ But I do not see equality at all.
Even just hints are appreciated. Thanks!
 A: I will denote by $\mathcal C$ the given curve. Lines through two points $A,B$ are denoted by $\mathcal L_{A,B}$.
In order to have a more compositional notation, it is useful to denote as in loc. cit the point $\phi(A,B)$ by $A*B$. Then the composition star has the following properties for simple points $A,B,C,D$:

*

*commutativity, $A*B=B*A$,

*idempotency (for $A$), $A*(A*B)=B$,

*$3\times 3$-match, $(A*B)*(C*D)=(A*C)*(B*D)$.

The last property is best seen in a picture:

This is Proposition 3, page 124 in Fulton's book, applied for the two intersections $\mathcal C\cdot\mathcal C_1$ and $\mathcal C\cdot\mathcal C_2$,
where $\mathcal C_1,\mathcal C_2$ are the reducible cubics (each product/union of three lines):
$$
\begin{aligned}
\mathcal C_1 &=\mathcal L_{AB}\ \mathcal L_{CD}\ \mathcal L_{A*C\ B*D}\ ,\\
\mathcal C_2 &=\mathcal L_{AC}\ \mathcal L_{BD}\ \mathcal L_{A*B\ C*D}\ .
\end{aligned}
$$
Now we can work:
$$
\begin{aligned}
\alpha(A)+'\alpha(B)
&=O'*(\alpha(A)*\alpha(B)) &&\text{definition of $+'$}
\\
&=O'*((Q*A)*(Q*B)) &&\text{definition of $\alpha$}
\\
&=O'*((Q*Q)*(A*B)) && 3\times 3 \text{ inside the right operator}
\\
&=(Q*O)*((Q*Q)*(A*B)) &&\text{definition of $Q=O*O'$, idempotency for $O$}
\\
&=(Q*(Q*Q))*(O*(A*B)) && 3\times 3 
\\
&=Q*(O*(A*B)) &&\text{by idempotency for $Q$, $Q*(Q*Q)=Q$}
\\
&=Q*(A+B) &&\text{by definition of $+$}
\\
&=\alpha(A+B)&&\text{by definition of $\alpha$ .}
\end{aligned}
$$
