A smooth group of multiplicative type is the only closed subscheme containing all its finite subgroups

By [F1] write $G=D_k(M)$ with $M$ finitely generated, so $\mathcal{O}(G)=k[M]$. By [F3] fix the decomposition $M={\mathbb{Z}}^r\oplus \mathbb{Z}\mathrm{/}(n_1)\oplus \cdots \oplus \mathbb{Z}\mathrm{/}(n_t)$ with $1<n_1\mid \cdots \mid n_t$; write $F=\mathbb{Z}\mathrm{/}(n_1)\oplus \cdots \oplus \mathbb{Z}\mathrm{/}(n_t)$ and $N=n_t$ when $t\ge 1$, and $N=1$, $F=0$ when $t=0$. By [F4], applied repeatedly, $k[M]\cong k[{\mathbb{Z}}^r]{\otimes }_{k}k[F]$ and $G\cong {\mathbf{G}}_m^r{\times }_k{D}_k(F)$, while by [F2] the group algebra $k[\mathbb{Z}\mathrm{/}(n)]\cong k[u]\mathrm{/}(u^n-1)$ and $O({\mathbf{G}}_m)=k[\mathbb{Z}]$.

I claim that $G$ is reduced. By [F5] every local ring ${\mathcal{O}}_{G,x}$ is regular, hence a domain by [F6], and therefore reduced; a scheme whose local rings are all reduced is reduced.

I claim that no $n_i$ is divisible by $p=\mathrm{char}k$; in characteristic zero this is vacuous, so suppose $p>0$ and suppose $p\mid n_i$ for some $i$; write $n_i=p^{v}m$ with $v\ge 1$ and $p\nmid m$. Then in $k[u]$ one has $u^{n_i}-1=(u^m-1{)}^{p^v}$, and $u^m-1\mathrm{\ne }0$ is a nonzero nilpotent in $k[u]\mathrm{/}(u^{n_i}-1)$ because $(u^m-1{)}^{p^v}=u^{n_i}-1=0$; hence $k[\mathbb{Z}\mathrm{/}(n_i)]$ is not reduced. By [step 1.1], $k[M]\cong k[{\mathbb{Z}}^r]{\otimes }_{k}k[F]$ and $k[F]$ has $k[\mathbb{Z}\mathrm{/}(n_i)]$ as a tensor factor, so a nonzero nilpotent of $k[\mathbb{Z}\mathrm{/}(n_i)]$ produces a nonzero nilpotent $1\otimes z$ of $k[M]$; this contradicts [step 1.2], since $\mathcal{O}(G)=k[M]$ reduced means $k[M]$ is reduced. Hence $p\nmid \mathrm{\mid }F\mathrm{\mid }$.

I claim that for every integer $j$ with $N\mid j$ and $j\in {\mathcal{N}}_k$ one has $G_j(k)={\mu }_j(k{)}^r\times D_k(F)(k)$, where $D_k(F)(k)={\prod }_i{\mu }_{n_i}(k)$. Indeed $G(k)=\mathrm{Hom}(M,k^{\times })\cong (k^{\times }{)}^r\times D_k(F)(k)$ by [F2], and $G_j(k)$ is the set of characters $\chi$ of $M$ with ${\chi }^j=1$, i.e. $\mathrm{Hom}(M\mathrm{/}jM,{\mu }_j(k))$. Since $M\mathrm{/}jM\cong (\mathbb{Z}\mathrm{/}j{)}^r\oplus {⨁}_i\mathbb{Z}\mathrm{/}\gcd (j,n_i)\cong (\mathbb{Z}\mathrm{/}j{)}^r\oplus F$, and ${\mu }_j(k)$ contains all $n_i$-th roots of unity because $n_i\mid j$ and $j\in {\mathcal{N}}_k$ with $k$ algebraically closed, this set is ${\mu }_j(k{)}^r\times {\prod }_i{\mu }_{n_i}(k)$.

I claim that $T:={\bigcup }_{j\in {\mathcal{N}}_k}{G}_j(k)$ is dense in $\mathrm{\mid }G\mathrm{\mid }$. By [step 3.1], $T$ contains $({\bigcup }_j{\mu }_j(k){)}^r\times D_k(F)(k)$, where $j$ runs over the multiples of $N$ in ${\mathcal{N}}_k$. The inner union is the set of all roots of unity whose order lies in ${\mathcal{N}}_k$: it is infinite, and an infinite subset of ${\mathbf{G}}_m(k)=k^{\times }$ is dense in ${\mathbf{G}}_m$ because a nonzero polynomial has finitely many roots; its $r$-fold Cartesian power is dense in ${\mathbf{G}}_m^r$: a Laurent polynomial vanishing on that grid is zero, by induction on $r$ and comparison of coefficients after fixing the other variables in the infinite set. Thus $({\bigcup }_j{\mu }_j(k){)}^r$ is dense in ${\mathbf{G}}_m^r$. By [step 2.1] the $n_i$ lie in ${\mathcal{N}}_k$, so $k[\mathbb{Z}\mathrm{/}(n_i)]\cong k^{n_i}$ and every point of the finite group $D_k(F)\cong {\prod }_i{\mu }_{n_i}$ is a $k$-point, while $G\cong {\mathbf{G}}_m^r{\times }_k{D}_k(F)$ by [step 1.1]; a product of a dense subset with the full point set of the second factor is dense. Hence $T$ is dense in $\mathrm{\mid }G\mathrm{\mid }$.

By [step 4.1] the set $T\subseteq Z(k)$ is dense in the reduced finite-type $k$-scheme $G$, and $k$ is algebraically closed, so [F7] applies with $S=T$ and gives $Z=G$.