How does the Galois group act on $E[m]$?

Question

Let $K$ be a perfect field and $\bar{K}$ its algebraic closure and say the characteristic of $K$ is $0$. Suppose I have an elliptic curve over $K$. It (the book I am reading) says that the Galois group $G(\bar{K}/K)$ acts on $E[m]$ because: if $[m]P = O$ then $$ [m](P^{\sigma}) = ([m]P)^{\sigma} = O^{\sigma} = O. $$ I don't understand why the first equality is true. Any explanation is appreciated. Thank you.

Answer 1

$m\sigma(p)=\sigma(p)+...+\sigma(p)=\sigma(p+...+p)=\sigma(mp)=\sigma(\mathcal O)=\mathcal O $

we can explain how the Galois group $G(\bar{K},K)$ acts on $E[m]$ by form: We have $$ 0\to E(\bar{K})[m]\to E(\bar{K}) \to E(\bar{K})\to 0 $$ then $$ 0\to E({K})[m]\to E({K}) \to E({K}) \to Hom(G(\bar{K},K),E(\bar{K})[m])\to Hom(G(\bar{K},K),E(\bar{K}))\to ... $$