= Solution
{c}
Let $\omega_G=h_G(T)dT$. The pullback through the <formal group> homomorphism is
$$
f^*\omega_G=h_G(f(T))f'(T)dT.
$$
The identity $f(F(X,T))=G(f(X),f(T))$ and the <chain rule> show that this pullback is invariant under $F$: translation by $T$ before $f$ becomes translation by $f(T)$ afterward. By part (i) it is a scalar multiple of $\omega_F$. Its coefficient at zero is $h_G(0)f'(0)=f'(0)$, so
$$
\boxed{f^*\omega_G=f'(0)\omega_F.}
$$
Apply this to the multiplication-by-$p$ endomorphism $q(T)=[p](T)$. Since the linear coefficient of $F$ is $X+Y$, repeated addition gives $q'(0)=p$. If $\omega_F=h(T)dT$ with $h(0)=1$, the pullback formula gives
$$
h(q(T))q'(T)=p\,h(T),\qquad
q'(T)\in pR[[T]],
$$
because $h(q(T))$ is a <unit in a ring>. Write $q(T)=\sum_{m\geq1}a_mT^m$. Then $ma_m\in pR$ for every $m$. If $p\nmid m$, choose <integers> $u,v$ with $um+vp=1$. It follows that
$$
a_m=u(ma_m)+vp a_m\in pR.
$$
Choose $b_m\in R$ with $a_m=pb_m$ for those $m$ not divisible by $p$, and define
$$
f(T)=\sum_{p\nmid m}b_mT^m,\qquad
g(T)=\sum_{j\geq1}a_{pj}T^j.
$$
Coefficient comparison now proves
$$
\boxed{[p](T)=pf(T)+g(T^p).}
$$
No cancellation of $p$ was used, so this <prime-multiplication coefficient splitting for formal groups> remains valid when $p$ is a <zero divisor> in $R$.
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