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3 added 860 characters in body

If we have a central extension of group schemes $1\rightarrow B \rightarrow C\rightarrow A\rightarrow1$ with $A$ abelian, then we get a commutator mapping $\Lambda^2A\rightarrow B$ (of sheaves as $\Lambda^2A$ in general is not a group scheme) and the extension is abelian precisely when this map is zero. Hence for an non-abelian extension to exist there must be a non-zero map $\Lambda^2A\rightarrow B$. Let us now assume that $A=\mu_n$ and consider first the case when $n=p$, the characteristic of the field $k$ (which we may assume is algebraically closed). A non-zero map $\Lambda^2A\rightarrow B$ would give a non-zero map $A\rightarrow\mathrm{Hom}(A,B)$, where the right hand side is the sheaf of group homomorphisms. As the Frobenius map is zero on $\mu_p$ we may replace $B$ by its Frobenius kernel so we may assume that $B$ is either $\mu_p$ or the Cartier dual of $\alpha_{p^m}$. Now, as sheaves $\mathrm{Hom}(A,B)$ is isomorphic to $\mathrm{Hom}(D(B),D(A))$, where $D(-)$ denotes the Cartier dual. However $D(\mu_p)=\mathbb Z/p$ so when $A=\mu_p$ we get that $\mathrm{Hom}(D(A),D(B))=\mathbb \mathrm{Hom}(D(B),D(A))=\mathbb Z/p$ and there is only the zero map from $A=\mu_p$ into it. In the other case $D(A)=\alpha_{p^m}$ and $\mathrm{Hom}(\alpha_{p^m},\mathbb Z/p)$ is zero. If instead $n=p^k$, the argument is the same. The case when $n=\ell^k$ is even simpler so in all cases all possible commutator maps are zero and the extension is commutative.

(When $A=B=\mathbb G_a$ then there are candidates for commutator maps and in fact $(a,b)(a',b')=(a+a',b+b'+a^pa')$ gives a non-commutative central extension which I imagine is the fake Heisenberg group.)

Addendum: A general comment is that it is more convenient to work with sheaves (in the fppf topology say) as that means that we essentially can pretend that we work with set-theoretic groups. It is however also necessary if we want to see the commutator map as a map $\Lambda^2A\to B$ as the sheaf $\Lambda^2A$ (of $A$ considered as an abelian sheaf) is in general not reprsentable. The $\langle-,-\rangle\colon\Lambda^2A\to B$ view point is convenient as it allows us to do what one usually does when having a pairing: We get for instance a map $A\to\mathrm{Hom}(A,B)$ given by $a\mapsto (a'\mapsto \langle a,a'\rangle)$ just from the fact that $\langle-,-\rangle$ is biadditive.

I have implicitly assumed that $B$ is of finite type (as I claim that its Frobenius kernel is finite) even though it may not be necessary (a limit argument anyone?).

2 Fixed missing transposition

If we have a central extension of group schemes $1\rightarrow B \rightarrow C\rightarrow A\rightarrow1$ with $A$ abelian, then we get a commutator mapping $\Lambda^2A\rightarrow B$ (of sheaves as $\Lambda^2A$ in general is not a group scheme) and the extension is abelian precisely when this map is zero. Hence for an non-abelian extension to exist there must be a non-zero map $\Lambda^2A\rightarrow B$. Let us now assume that $A=\mu_n$ and consider first the case when $n=p$, the characteristic of the field $k$ (which we may assume is algebraically closed). A non-zero map $\Lambda^2A\rightarrow B$ would give a non-zero map $A\rightarrow\mathrm{Hom}(A,B)$, where the right hand side is the sheaf of group homomorphisms. As the Frobenius map is zero on $\mu_p$ we may replace $B$ by its Frobenius kernel so we may assume that $B$ is either $\mu_p$ or the Cartier dual of $\alpha_{p^m}$. Now, as sheaves $\mathrm{Hom}(A,B)$ is isomorphic to $\mathrm{Hom}(D(A),D(B))$, \mathrm{Hom}(D(B),D(A))$, where$D(-)$denotes the Cartier dual. However$D(\mu_p)=\mathbb Z/p$so when$A=\mu_p$we get that$\mathrm{Hom}(D(A),D(B))=\mathbb Z/p$and there is only the zero map from$A=\mu_p$into it. In the other case$D(A)=\alpha_{p^m}$and$\mathrm{Hom}(\alpha_{p^m},\mathbb Z/p)$is zero. If instead$n=p^k$, the argument is the same. The case when$n=\ell^k$is even simpler so in all cases all possible commutator maps are zero and the extension is commutative. (When$A=B=\mathbb G_a$then there are candidates for commutator maps and in fact$(a,b)(a',b')=(a+a',b+b'+a^pa')$gives a non-commutative central extension which I imagine is the fake Heisenberg group.) 1 If we have a central extension of group schemes$1\rightarrow B \rightarrow C\rightarrow A\rightarrow1$with$A$abelian, then we get a commutator mapping$\Lambda^2A\rightarrow B$(of sheaves as$\Lambda^2A$in general is not a group scheme) and the extension is abelian precisely when this map is zero. Hence for an non-abelian extension to exist there must be a non-zero map$\Lambda^2A\rightarrow B$. Let us now assume that$A=\mu_n$and consider first the case when$n=p$, the characteristic of the field$k$(which we may assume is algebraically closed). A non-zero map$\Lambda^2A\rightarrow B$would give a non-zero map$A\rightarrow\mathrm{Hom}(A,B)$, where the right hand side is the sheaf of group homomorphisms. As the Frobenius map is zero on$\mu_p$we may replace$B$by its Frobenius kernel so we may assume that$B$is either$\mu_p$or the Cartier dual of$\alpha_{p^m}$. Now, as sheaves$\mathrm{Hom}(A,B)$is isomorphic to$\mathrm{Hom}(D(A),D(B))$, where$D(-)$denotes the Cartier dual. However$D(\mu_p)=\mathbb Z/p$so when$A=\mu_p$we get that$\mathrm{Hom}(D(A),D(B))=\mathbb Z/p$and there is only the zero map from$A=\mu_p$into it. In the other case$D(A)=\alpha_{p^m}$and$\mathrm{Hom}(\alpha_{p^m},\mathbb Z/p)$is zero. If instead$n=p^k$, the argument is the same. The case when$n=\ell^k$is even simpler so in all cases all possible commutator maps are zero and the extension is commutative. (When$A=B=\mathbb G_a$then there are candidates for commutator maps and in fact$(a,b)(a',b')=(a+a',b+b'+a^pa')\$ gives a non-commutative central extension which I imagine is the fake Heisenberg group.)