 Solution

In this setting a formal group law means a one-dimensional commutative series $F(X,Y)\in {\mathcal{O}}_K[[X,Y]]$ withThese identities force $F(X,Y)=X+Y+$ terms of total degree at least two. There is a unique formal inverse $i_F(T)=-T+O(T^2)$ with $F(T,i_F(T))=0$.

A morphism to another formal group law $G$ is a series $h(T)=cT+O(T^2)$ satisfying $h(F(X,Y))=G(h(X),h(Y))$. The invertible morphism criterion for formal group laws isNecessity follows by comparing linear coefficients in $h^{-1}\circ h=T$. For sufficiency, solve recursively for a compositional inverse $j(T)=c^{-1}T+\cdots$: at every stage the new coefficient is obtained by dividing by the unit $c$, so all coefficients stay in ${\mathcal{O}}_K$. Applying $j$ to the morphism identity shows that $j$ is itself a group homomorphism of formal group laws.

In particular, the multiplication series satisfies $[n{]}_F(T)=nT+O(T^2)$. Since $p\nmid n$, the multiplication isomorphism of a formal group law makes $[n{]}_F$ invertible over ${\mathcal{O}}_K$. Its series and inverse converge on ${\mathfrak{m}}_K$, so multiplication by $n$ is bijective on $E_1(K)$, and on $E_1(L)$ for every finite extension $L/K$.