I have started learning about formal groups defined over a R1NG, $R$. Then, I studied multiplication by $m$ map on a formal group. The definition is as follows:
Def: Let $F$ be a formal group over $R$. Define the multiplication by $m$ map $[m](T)$ inductively as follows $[0](T)=0$ and $[m+1](T)=F([m](T), T).$ (So that each $[m](T)$ is a power series in $T$.)
Then the notes I am reading claims that such map defines a homomorphism from $F$ to $F$ given that $F$ is a commutative formal group but I cannot seem to figure why?
I suppose that by homomorphism, I should show the following equation holds:
$$[m](F(X,Y))=F([m](X), [m](Y))$$ for each $m$.
I can see this equation holds for $m=0,1.$ But in general situation, I tried to do an induction on $m$ and really got nowhere. I also thought about maybe writting out the terms but it was very very messy.
Wondering if there is a slick way of showing this homomorphism? Thank you so much in advance!
Edit 1: There was a bad notation that I used, which I should have noticed earlier in stating the equation I want to prove.
Edit 2: $F$ is meant to be a commutative formal group.