What is the proof of - $E[2]$ torsion group is isomophic to $Z_2 \oplus Z_2$? This is from the book "Elliptic Curves" by Lawrence Washington.
$E$ is an elliptic curve over $K$
$E[n] = \{P \in E(\bar K) \mid nP = \infty\}$
where $\bar K$ is the algebraic closure of $K$
The points of $E[2]$ are ${\infty, (e_1, 0), (e_2, 0), (e_3, 0)}$
I understood how they get up to the above.
But after this the book says
As an abstract group, this is isomorphic to $Z_2 \oplus Z_2$
They haven't expanded further on this - so it looks like the above is something which should be obvious but I don't understand how the above is true.
For it to be isomorphic to $Z_2 \oplus Z_2$, I think the following needs to be shown
$(e_1, 0) + (e_2, 0) = (e_3, 0)$ and so on
How do we prove the above?
 A: I assume that your elliptic curve is given in Weierstrass form $$E:y^2=x^3+c x^2+ax+b$$ with $c=0$. You factorize $$x^3+cx^2+ax+b=(x-e_1)(x-e_2)(x-e_3)$$
So that $e_1+e_2+e_3=c=0$.
For $E$ to be an elliptic curve (not a singular cubic) you need that the $e_j$ are distinct (and $char(K)\ne 2$).
In the group law of $E$ you have that $-(u,v)=(u,-v)$ so the 2-torsion is the points with either $v=0$ or $v=\infty$, which gives $$E[2]=\{ (\infty,\infty),(e_1,0),(e_2,0),(e_3,0)\}$$
$E$ is an abelian group. Whence $E[2]$ is a subgroup.

*

*$(e_1,0)+(e_2,0)$ can't be $(\infty,\infty)$ as otherwise you'd have $(e_1,0)=-(e_2,0)=(e_2,0)$ so $e_1=e_2$.


*$(e_1,0)+(e_2,0)$ can't be $(e_1,0)=(e_1,0)+(\infty,\infty)$ as otherwise you'd have $(e_2,0)=(\infty,\infty)$.
Similarly $(e_1,0)+(e_2,0)$ can't be $(e_2,0)$.


*Whence $(e_1,0)+(e_2,0)=(e_3,0)$.
Similarly $(e_i,0)+(e_j,0)=(e_k,0)$ for $i\ne j\ne k$.


*Otherwise, in general, $E[2]$ is a $\Bbb{F}_2$-vector space so it is automatically $\cong (\Bbb{F}_2)^n$ with $2^n=|E[2]|$ ie. $n=2$.
A: From @reuns' answer, we get
$e_1 + e_2 + e_3 = 0$
Let $P = (\infty,\infty), Q = (e_1,0), R = (e_2,0),S = (e_3,0)$
Let $N = \{x, y\} = Q + R$
The formula for point addition for elliptic curves is
$m = \frac {y_2 - y_1}{x_2 - x_1}$
$m = \frac {0 - 0} {e_2 - e_1} = 0 $
$x = m^2 - e_1 - e_2 \bmod p$
$x = -e_1 - e_2$
$e_1 + e_2 + e_3 = 0$, so $e_3 = -e_1 - e_2$
So $x = e_3$
$y = m(x_1 - x) - y_1 \bmod p$
$y = m(e_1 - e_3) - 0 \bmod p$
$y = 0$
So $N = Q + R = \{e_3, 0\} = S$
Likewise we can prove that
$R + S = Q$
$S + Q = R$
