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After some thought my pessimism (as expressed in my concurrence with the answer of Milne) has abated somewhat. If I were bold enough I would conjecture the following (assuming that the characteristic …

I think that an example of non-equivalent permutation sets is given by $G=(\mathbb Z/p\mathbb Z)^n$ for $n>2$ (and $p$ a prime). Then the automorphism group is $\mathrm{GL}_n(\mathbb Z/p\mathbb Z)$, t …

It is not really a question of inner forms. What happens is that the algebraic group $G_2$ has an extra endomorphism $\varphi$ whose square is the Frobenius map (over the appropriate finite field). Ju …

This group is best understood in terms of the universal coefficient formula, i.e., in terms of the homology of the involved group. Hence, if $A$ is any abelian group we have $H_1(A)=A$ and the additio …

It seems clear to me that blocks should have something to do with the decomposition of the category as a direct product of subcategories. A decomposition into a product of two factors corresponds exac …

Yes, this is true. The derivative of $n_G$ at the identity element $e$ is multiplication by $n$ on the Lie algebra (tangent space at $e$) which gives that the kernel is finite and the image of $n_G$ h …

A simple yet useful way of getting examples is to look at step $2$ nilpotent groups. Hence we consider central extensions $1\to A\to H\to B\to1$, where $A$ and $B$ are abelian. Taking commutators indu …

Being systematic a $2$-special group is specified completely by two $\mathbb Z/2$-vector spaces $U=G/Z(G)$ and $V=Z(G)$ together with the induced square map $\Gamma^2U\to V$ which is non-degenerate in …

In Groupes algébriques et corps de classes Serre classifies the $2$-dimensional commutative unipotent connected algebraic groups $G$ (VII:11). With the exception of the product of the additive group w …

The $\sqrt{|G|}$ bound is true [for algebraically closed fields -- PLC] in any characteristic. In fact if $R$ is the Jacobson radical of $k[G]$, then $k[G]/R\cong\prod_iM_{n_i}(k)$ where $i$ runs over …

For $r=2$ a $k$-subset can be thought of as a graph with vertices $[n]$ and $k$ edges. Hence the number of orbits is equal to the number of isomorphism classes of graphs on $n$ vertices and $k$ edges. …

As Simon points out, the answer is no in a simple case and if you think about his argument the answer should probably be no as soon as $G^p$ is infinite. However, there is a statement that is very clo …

I interpret your statement as being concerned with conjugation in $\mathrm{SL}_2(\mathbb Z)$. In that case I think that the arguments given only give that the groups are cyclic of order $1$, $2$, $3$, …

This looks like a (slightly) non-additive version of Grothendieck's theory of "extensions panachées" (SGA 7/I, IX.9.3). There he considers objects (in some abelian category) $X$ together with a filtat …

Such automorphisms appear naturally when one tries to analyse the group of automorphisms of $G$ preserving $H$ ($H$ may for instance be a characteristic subgroup so that it is preserved by all automor …