$\newcommand\O{\mathcal O}$I was reading Silverman and Tate's Rational Points on Elliptical Curves. In page 21 of the same book it was written
We also want to mention that there is nothing special about our choice of $\O$; if we choose a different point $\O'$ to be the zero element of our group, then we get a group with exactly the same structure. In fact, the map $$P\longmapsto P+(\O'-\O)$$ is an isomorphism from the group "$C$ with zero element $\O$" to the group "$C$ with zero element $\O'$."
My question is how is this an isomorphism between the points? Let $\phi$ be the aforesaid map. To prove it's an isomorpshim we have to show it's a bijection first. That is easily done. Now I have to prove that $$\phi(P)+_{\O'}\phi(Q)=\phi(P+_\O Q) $$
$\newcommand\O{\mathcal O}$Let $C$ be an elliptic curve and $\ast:C\times C\to C$ be the operation that takes $P,Q\in C$ to the third point $P\ast Q$ on the line through $P$ and $Q$. If we choose a zero element $\O$, we obtain a group $(C,+)$ with $$ P+Q = \O \ast (P\ast Q). $$ Picking a different zero $\O'$ gives a group $(C,\oplus)$ with \begin{align} P\oplus Q = \O' \ast (P\ast Q). \end{align}
From these definitions, we have $$ (P\oplus Q)\ast\O' = P\ast Q = (P+Q)\ast\O. $$ Thus, by applying $\ast\,\O$ to this equation, we get $$ (P\oplus Q)+\O' = P+Q. $$ We conclude $$ P\oplus Q = P+Q-\O'. $$
Now let's check if \begin{align} \Phi : (C,+) &\longrightarrow (C,\oplus) \\ P &\longmapsto P+(\O'-\O) \end{align} is a homorphism. First note that $\O'-\O=\O'$ since $\O$ is the zero element in $(C,+)$, so \begin{align} \Phi(P+Q) &= P+Q+\O' = (P+\O')+(Q+\O')-\O' \\&= \Phi(P)+\Phi(Q)-\O' = \Phi(P)\oplus\Phi(Q). \end{align}