Is the category of affine fppf groups closed under normal quotients? Let $S$ be a scheme and let $N$, $G$ be affine flat group schemes of finite presentation over $S$.
If we assume that $N$ is a closed normal subgroup of $G$, we may form the fppf quotient sheaf $G/N$, which
is a sheaf of groups. By using descent and Artin's representability result for algebraic spaces, it follows that $G/N$ is an algebraic space which is
flat, separated and of finite presentation over $S$. By a classic result in the case where $S$ is the
spectrum of a field, it also follows formally that $G/N$ has affine fibres. My question is:
Is $G/N$ always affine over $S$?
Phrased in another way: Is the category of affine fppf group schemes over $S$
a semi-abelian subcategory of the semi-abelian category of fppf group sheaves over $S$?
Another related question is if a flat, affine, finitely presented group scheme always may be embedded in
the automorphism group of a locally free coherent sheaf on the base. There is a remark about this in
SGAIII_1 EXPOSE VI_B 11.11.1, where it is stated without proof that this is supposed to be true for some
base schemes under certain regularity conditions.
EDIT:
A third related question is if someone knows of an example of a separated, flat, finitely presented group algebraic space with affine fibres which is not affine.
 A: In general, the quotient $G/N$ is not representable. Lemma X.14 of Raynaud's book "Faisceaux amples sur les schémas en groupes et les espaces homogènes" gives a counter-example with $S=\mathbb{A}^2_k$ (which is regular 2-dimensional), $G=(\mathbb{G}_{a,S})^2$ and $N\subset G$ étale over $S$.
However if $S$ is locally noetherian of dimension at most 1, then $G/N$ is representable by a scheme : this is theorem 4.C of Anantharaman's thesis "Schémas en groupes, espaces homogènes et espaces algébriques sur une base de dimension 1". If you assume moreover that $S$ has finite normalization (e.g. $S$ excellent), then $G/N$ is affine if $G$ is. Indeed by Serre's cohomological criterion, affineness is not affected by nilpotents hence you may base-change by $S_{red}\to S$ and hence assume that $S$ is reduced (note that the formation of $G/N$ commutes with base change). Then let $S'\to S$ be the normalization, a finite morphism, and let $G',N'$ be the pullbacks to $S'$. The restriction of $(G/N)'=G'/N'$ to the generic points of $S'$ is affine (since we are then over a field) and because $S'$ is Dedekind, it follows that $(G/N)'$ is affine (Anantharaman prop. 2.3.1). Finally $(G/N)'\to G/N$ is finite surjective hence $G/N$ is affine by Chevalley's theorem.
A: Yes, $G/N$ is always an affine group scheme. This is explained, for example, in one of the last chapters of Waterhouse's book on affine group schemes (an excellent reference for these questions).
[Edit] As Matthieu points out, this is only correct over a field.
As to your second question, consider the product $(\mathbb Z/2 \mathbb Z) \times \mathbb A^2_k$ (here $k$ is a field), which is an affine group scheme over $\mathbb A^2_k$, and delete the closed point $(1, 0)$.
