Let $G$ be a 1-dimensional, commutative formal group over a ring $R$. Give $G$ a coordinate $x$ and let $A\subset R$ be the subring generated by the coefficients of the corresponding formal group law $F(x,y)= \sum\_{ij}a\_{ij}x^iy^j$. So $G$ is really defined over $A$.


Call a finite subgroup $K\subset G$ *special* if it is the kernel of a homomorphism $T:G\rightarrow \phi^\*G$ for some ring map $\phi:A\rightarrow R$. ($\phi^\*G$ is the formal group over $R$ with formal group law $\phi^\*F(x,y)=\sum\_{ij}\phi(a\_{ij})x^iy^j$.)


What's an example of a finite subgroup $K\subset G$ that is not (isomorphic to) a special subgroup? (I guess that would be the same as asking that $G/K$ does not inject into any $\phi^\*G$.)


In all the cases I've tried (including the additive, multiplicative, and universal formal group laws) I seem to have convinced myself that all subgroups are special. Which leads me to suspect that maybe special subgroups are not so special.


$\textbf{Edit}$: To be explicit, a subgroup $K$ corresponds to a monic polynomial $f\_K(x)\in R[x]$ with nilpotent lower order coefficients and such that $f(F(x,y))\equiv 0\ \text{mod}\ (f(x),f(y))$. That subgroup $K$ is special if furthermore there is some invertible power series $u(x)\in R[[x]]$ and a ring map $\phi:A\rightarrow R$ as above such that $u(F(x,y))f(F(x,y))=\phi^\*F(u(x)f(x),u(y)f(y)).$



Some special subgroups of formal groups
Let me specialize heavily to the case of formal groups (group laws) of dimension one over a $p$-adic ring $\mathfrak o$, i.e. the ring of integers of a finite extension $k$ of $\Bbb Q\_p$.


I still am uncertain about what category you’re thinking of. If we restrict further to formal groups of finite height (the endomorphism $[p]$ being of finite degree $p^h$), then these things become $p$-divisible groups, or, if you like, ind-finite objects. For instance the kernel of $[p^n]$ will be a finite $\mathfrak o$-group-scheme, $K\_n=\ker([p^n])=\mathrm{Spec}(\mathfrak o[[x]]/([p^n](x))\,)$, and you have natural maps $K\_n\hookrightarrow K\_{n+1}$, and you see that $\projlim\mathfrak o[[x]]/([p^n](x))\cong\mathfrak o[[x]]$. In this sense, your $G$, if indeed a formal group of finite height over $\mathfrak o$, is the union of its finite subgroups. This is the viewpoint that I tend to work with.


Now, let’s consider just one fairly simple case, where the formal group law has all its coefficients in an unramified extension $A$ of $\Bbb Z\_p$, even in $\Bbb Z\_p$ itself, and suppose the height is $h=2$ for simplicity. This means that $[p](x)\equiv px+ux^{p^2}\pmod{x^{p^2+1}}$, where $u$ is a unit of $A$, and the congruence ignores all terms in the power series of degree $>p^2$. Look at the Newton polygon and see that all the $z\in\overline k$ with $v\_p(z)>0$ and $[p](z)=0$ have $v\_p(z)=\frac1{p^2-1}$, plus of course $0$. So $p^2$ in all, and thus they form an elementary $p$-group of order $p^2$.


Now take any of the cyclic subgroups of $\ker[p]$, call it $\Gamma$. One proves that
$$
\pi\_\Gamma(x)=\prod\_{\gamma\in\Gamma}F(x,\gamma)\,,
$$
which is defined over a totally ramified extension $A'$ of $A$ (actually of degree $p+1$), is a morphism into another formal group, which I will abuse language in calling $G/\Gamma$.


I ask you to believe that I have shown you a formal group $G/\Gamma$ that, as far as I can see, will prove to you that $\Gamma$ is not a special subgroup of $G$, once you see that the formal group law of $G/\Gamma$ is not isomorphic to that of $G$, not even with a morphism $\varphi^\*$ of the type you allow. (I think, because I’m not sure what properties you allow $\varphi^\*$ to have.)


How do I know that $G/\Gamma$ is nothing like $G$? By Newtonian magic, you see that the Newton polygon of $[p]\_{G/\Gamma}$ has vertices at $(1,1)$, $(p,\frac1{p+1})$, and $(p^2,0)$. The important fact is that this polygon is not the same as that of $[p]\_G$; and since the shape of the Newton polygon of $[p]$ is an invariant, it follows that there is no way for $G/\Gamma$ to be isomorphic to $G$.


In the appropriate category, the map from $G$ to $G/\Gamma$ is onto. You can show, for instance, that if $v\_p(\eta)>0$, there is $\xi$ in a finite extension of $k(\eta)$ such that $v\_p(\xi)>0$ and $\pi\_\Gamma(\xi)=\eta$.


(All of this is in an old and poorly-written paper of mine, [Finite subgroups and isogenies…](https://www.jstor.org/stable/1970443). **EDIT**:The “Newtonian magic” involves the “Newton copolygon”, also called the valuation function. I’ll bet a nickel that somebody else has explained it better than I can, but it’s in a later paper of mine, [Canonical subgroups of formal groups](https://www.jstor.org/stable/1998685), and I fear that it’s at most a little better-written than the other. Copolygon talk begins on p. 109.)


