Let $m \ge 1$ be an integer, let $k$ be a field of characteristic $0$, and let $$ 1 \rightarrow \mathrm{GL}_n \rightarrow E \rightarrow \mathbb{Z}/m\mathbb{Z} \rightarrow 1 $$ be an extension of $k$group schemes. Since $E$ acts on $\mathrm{GL}_n$ by conjugation, there is an induced $k$group scheme homomorphism $\mathbb{Z}/m\mathbb{Z} \rightarrow \mathrm{Out}(\mathrm{GL}_n) = \mathrm{Aut}(\mathrm{GL}_n)/\mathrm{Inn}(\mathrm{GL}_n)$. Suppose that this homomorphism is trivial, i.e., that the conjugation action of $E$ on $\mathrm{GL}_n$ is by inner automorphisms. Does this imply that the extension is split, i.e., that $E \cong \mathrm{GL}_n \times \mathbb{Z}/m\mathbb{Z}$ compatibly with the extension structure?

1$\begingroup$ I know nothing about group schemes, but if we were just talking about group extensions, then such an extension would not necessarily split as a direct product. For example if $k$ does not contain $4$th roots of $1$ (e.g. $k = {\mathbb Q}$) and $m=2$, then an element outside of ${\rm GL}_n$ could centralize ${\rm GL}_n$ and square into $I_n$. $\endgroup$ – Derek Holt Nov 1 '14 at 4:24

1$\begingroup$ Could you elaborate? I think this (in purely group theoretic terms with $k = \mathbb{Q}$) would give a desired counterexample, so I would be happy to accept it if you post an answer with more details. $\endgroup$ – Kestutis Cesnavicius Nov 1 '14 at 5:17
To make my comment more precise, suppose that $k$ contains no primitive $2m$th roots of $1$, where $m=2^j$ for some $j \ge 1$. Then we can form the central product of $G={\rm GL}_n(k)$ with a cyclic group $\langle x \rangle$ of order $2^{j+1}$, where we amalgamate $I_n$ with the element of order $2$ in the cyclic group.
Formally, this is the quotient $E = (G \times \langle x \rangle)/\langle (I_n,x^{2^j})\rangle$ of the direct product by a central subgroup of order $2$. With the natural embedding $G \hookrightarrow E$, we have $E/G \cong {\mathbb Z}/m{\mathbb Z}$, but we do not have $E \cong G \times {\mathbb Z}/m{\mathbb Z}$, because there is no suitable element of order $m$ centralizing $G$.
In general, with your hypothesis, the centralizer of $G$ in $E$ is equal to $\langle Z(G),x \rangle$, where $Z(G) \cong k^\times$ consists of the scalar matrices, and $x^m \in Z(G)$. So the full extension splits as a direct product depends if and only if the extension $1 \to Z(G) \to C_E(G) \to {\mathbb Z}/m{\mathbb Z}$ splits, and this depends on $k^\times$. In particular, if $k$ is algebraically closed or, more generally, if all nonzero elements of $k$ have $m$th roots in $k$, then all such extensions split as direct prodcuts.
In the category of commutative group schemes over a field $k$, $Ext^1(\mathbb{Z}/m\mathbb{Z},\mathbb{G}_m)=k^\times/k^{\times m}$ (write the $Ext$ sequence for $0\to \mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/m\mathbb{Z}\to 0$). This takes care of the case $n=1$, and the argument in Holt's answer suggests that the general case is essentially the same.

2$\begingroup$ For a simple explicit example, you can take $k=\mathbb{R}$ and $n=1$, and take for $E$ the subgroup of $\mathrm{GL}_2$ generated by the center and the rotation $\begin{pmatrix}0&1\\1&0\end{pmatrix}$. $\endgroup$ – Laurent MoretBailly Nov 1 '14 at 14:50

1$\begingroup$ Thanks. I should've asked this question with $k$ algebraically closed, as I've originally intended. $\endgroup$ – Kestutis Cesnavicius Nov 1 '14 at 15:40