In the book Elliptic curves number theory and cryptography (Lawrence C. Washington 2nd eddition), page 79, there is a theorem (Theorem 3.2): Let $E$ be an elliptic curve over a field $K$ and let $n$ be a positive integer. If the characteristic of $K$ does not divide $n$, or is 0, then $E[n] ≃ Z_n ⊕ Z_n$.
It then mentions "choose a basis $\{\beta_1, \beta_2\}$ for $E[n] ≃ Z_n⊕Z_n$. This means that every element of $E[n]$ is expressible in the form $m_1\beta_1+m_2\beta_2$ with integers $m_1, m_2$. Note that $m_1, m_2$ are uniquely determined mod $n$. Let $α : E(K) → E(K)$ be a homomorphism. Then $α$ maps $E[n]$ into $E[n]$. Therefore, there are $a, b, c, d ∈ Z_n$ such that $α(\beta_1) = a\beta_1 + c\beta_2$, $α(\beta_2) = b\beta_1 + d\beta_2$. Therefore each homomorphism $α : E(K) → E(K)$ is represented by a $2 × 2$ matrix
$ α_n= \begin{pmatrix} a & b\\ c & d\\ \end{pmatrix}$
Composition of homomorphisms corresponds to multiplication of the corresponding matrices".
I understand that every element of $E[n]$ is expressible in the form $m_1\beta_1+m_2\beta_2$ but I don't understand how we get to the fact that $α(\beta_1) = a\beta_1 + c\beta_2$. Surely if alpha is the $2×2$ matrix stated above then $α(\beta_1) = a\beta_1 + c\beta_1$?
In the text they say: Let $E$ be an elliptic curve over a field $K$ ... then $$ E[n]\cong Z_n + Z_n $$ The text gives no proof yet - they say it will come in section 3.2.
Let $n$ be such that the above theorem holds. Choose a basis $\{\beta_1, \beta_2\}$ of $Z_n + Z_n$. Since they are isomorphic, we identify a point of $E[n]$ with a point of $Z_n + Z_n$. Let $(m_1, m_2)$ be a point in $Z_n + Z_n$, so $(m_1, m_2) := m_1\beta_1 + m_2\beta_2$. Let $\alpha : E[\bar{K}] \rightarrow E[\bar{K}]$ be a homomorphism.
This is omitted from the text, but if we restrict the domain of $\alpha$ to $E[n]$, we get $\alpha_n : E[n] \rightarrow E[\bar{K}]$. But $n\alpha_n(m_1, m_2) = \alpha_n(n(m_1, m_2)) =\alpha_n(1) = 1$ (because $\alpha$ is a homomorphism) so $\alpha_n : E[n] \rightarrow E[n]$.
Since $\alpha_n(m_1, m_2)\in Z_n + Z_n$ we can represent it as a linear combination of the basis elements. (this is what it means for the result of $\alpha_n$ to be in $Z_n + Z_n$) i.e. There are some $a, b, c, d\in Z_n$ (the order of which is chosen so that their order in the following matrix is nice) such that: $$ \alpha_n(\beta_1) = a\beta_1 + c\beta_2 \\ \alpha_n(\beta_2) = b\beta_1 + d\beta_2 $$ Therefore if we calculate $\alpha_n$ on our arbitrary point: $$ \alpha_n(m_1, m_2) = \alpha_n(m_1\beta_1 + m_2\beta_2) $$ This is what it means for $(m_1, m_2)$ to be in $Z_n + Z_n$. $$ \alpha_n(m_1\beta_1 + m_2\beta_2) = m_1\alpha_n(\beta_1) + m_2\alpha_n(\beta_2) $$ This is because $\alpha_n$ is a homomorphism. $$ m_1\alpha_n(\beta_1) + m_2\alpha_n(\beta_2) = am_1\beta_1 + cm_1\beta_2 + bm_2\beta_1 + dm_2\beta_2 $$ This comes from the way we defined $a, b, c, d$ in terms of the result of $\alpha_n$ on the basis elements. $$ am_1\beta_1 + cm_1\beta_2 + bm_2\beta_1 + dm_2\beta_2= (am_1 + bm_2)\beta_1 + (cm_1 + dm_2)\beta_2 $$ Collecting like terms. $$ (am_1 + bm_2)\beta_1 + (cm_1 + dm_2)\beta_2= (am_1 + bm_2, cm_1 + dm_2)_{\in Z_n + Z_n} $$
Instead of doing this whole process every time, we can notice that the result of $\alpha_n$ is a matrix multiplication. $$ \alpha_n(m_1, m_2) =\begin{pmatrix}a &b\\c &d\end{pmatrix}\begin{pmatrix}m_1\\ m_2\end{pmatrix} $$