Is a formal subscheme of a formal group scheme also a formal group scheme? I'm fairly sure the answer to this is no (since https://www.math.upenn.edu/~chai/papers_pdf/local_rigidity_corrected_2016.pdf is a tricky proof of 'yes' under certain conditions).  But I'm really confused, since the conditions required by a power series to express a formal group law seem like they all hold (obviously) modulo quotienting by an ideal.  I think I'm missing something really obvious - but I'm not sure what...
 A: Disclaimer: I'm only like 60% sure that I know the definitions, so take what I'm saying with a grain of salt and/or edit it to make it correct.
$\DeclareMathOperator{\Spf}{Spf}
\require{AMScd}$
Take the additive group law $F(x,y)=x+y$. This is giving the multiplication for a group structure on $G:=\Spf k[[x]]$, namely the map $G\times G\to G$ given by the ring map $k[[x]]\to k[[x_1,x_2]] \cong k[[x_1]]\widehat\otimes k[[x_2]]$ sending $x$ to $x_1+x_2$.
Closed subschemes of $G$ corresponds to ideals of $k[[x]]$, which is a DVR, so all proper subschemes are cut out by $x^n$ for some $n$. Say $H$ is cut out by $x^n$. For $H$ to be a subgroup we would need to be able to fill the last arrow in this diagram:
$$
\begin{CD}
G @<<< G\times G\\
@AAA @AAA \\
H @<\exists?<< H\times H.
\end{CD}
$$
Which is the same as being able to fill this diagram of rings:
$$
\begin{CD}
k[[x]] @>x\mapsto x_1+x_2>> k[[x_1,x_2]] \\
@VVV @VVV\\
\frac{k[[x]]}{x^n} @>\exists?>> \frac{k[[x_1,x_2]]}{(x_1^n,x_2^n)}.
\end{CD}
$$
Note that if $H$ is cut out by $x^n=0$, then $H\times G$ is cut out in $G\times G$ by $x_1^n$, and similarly for $G\times H$, so their intersection is cut out by the ideal sum $(x_1^n,x_2^n)$.
The missing arrow above will exist whenever $x^n$ maps to $0$. Since $x\mapsto x_1+x_2$ and it's a ring homomorphism, we have
$$
x^n\mapsto (x_1+x_2)^n = \sum_{i+j=n} \binom{n}{i} x_1^ix_2^j.
$$
Which I believe is only in the ideal $(x_1^n,x_2^n)$ if $n=1$ (so we get the trivial group) or the characteristic of $k$ divides $n$.
The most intuitive explanation that I have for this is that if you had two $R$-points of $H\subseteq G$, i.e. two maps $\alpha_1,\alpha_2:k[[x]]\to R$ such that $\alpha_i(x)^n=0$, their sum in the group law is by definition the composition
$$
k[[x]] \xrightarrow{x\mapsto x_1+x_2} k[[x_1]]\otimes k[[x_2]] \xrightarrow{\alpha_1\otimes \alpha_2} R,
$$
so it sends $x$ to $\alpha_1(x)+\alpha_2(x)$. Now the fact that $\alpha_i(x^n)=0$ doesn't imply that $\left(\alpha_1(x)+\alpha_2(x)\right)^n=0$, which is why $H$ is not necessarily closed under the group law.
