16

The result you quoted appears in this reference: G. Szasz, Die Unabhängigkeit der Assoziativitätsbedingungen, Acta. Sci. Math. Szeged 15 (1953), 20-28.
The Szasz theorem requires that the set $S$ have at least four elements, though it is also true for sets of size $3$.
Szasz' proof is constructive and goes as follows. Assuming $a$, $u$, $v$ and $...

7

No. $C_1$ is a retract of $C_2$, so $M(C_2,C_1)\simeq C_2$ would have to be a retract of $M(C_2,C_2)\simeq C_4$, which it isn't.

6

The answer is no. Notice that $A(-,C_1)$ is a functor $F : \mathsf{Grp} \to \mathsf{Grp}$ with $F(C_n) \cong C_{n+1}$. But there is no such $F$. There is a split monomorphism $C_1 \to C_2$, hence $F$ would induce a split monomorphism $C_2 \to C_3$, contradiction.

answered Jan 30 at 3:30

Martin Brandenburg

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6

It is clear from standard properties on commutators that $H(A)$ is nilpotent of class at most $2$ with derived group and Frattini subgroup contained in $\langle z \rangle$
(just consider $H(A)/\langle z \rangle$, which is elementary Abelian of order $p^{2n})$.
The general property of commutators that you need is that we always have $[a,bc] = [a,b]^{c}[a,c]$....

6

It seems to me that the structure of ${\rm Aut}^{m}(C_{n})$ also depends heavily on the prime factorization of $n$, and I don't really see any reason to expect the answer to Q1 to be any more tractable than determining the structure of ${\rm Aut}(C_{n})$.
For example (just to illustrate) , if we choose a prime $p$, greater than $3$, and then we take a pair ...

5

It can happen that the whole group has no automorphism of order 2 although the Fitting subgroup does:
Let
$N$ be a non-abelian 2-generated $p$-group for some prime $p$, such that $N/Z(N)$ has no automorphism of order $2$;
$L$ a nontrivial group of order coprime to $2p$ with no automorphism of order $2$.
I claim that the wreath product $$...

4

If I understand correctly, you are counting the conjugacy classes of elements of order $p$ in $\operatorname{GL}(n,\mathbb{F}_p) $ — or equivalently, of matrices $N$ in $\operatorname{M}_n(\mathbb{F}_p) $ with $N^p=0$, but $N\neq 0$. If $p\geq n$ you get all nilpotent nonzero matrices, hence indeed $P(n)-1$ conjugacy classes using Jordan normal form, but ...

4

New version (existence hinted in previous version): If $G$ is a non-trivial finite (solvable) group of odd order with $\Phi(G) = 1$, then $G$ has an automorphism of order $2$.
It is well-known and easy to check that $F = F(G)$ is a product of minimal normal subgroups of $G$, each an elementary Abelian $p_{i}$-group for some prime $p_{i}$.
Also, $F$ is well-...

3

I don't think that there is a different natural construction of a group scheme associated to a finite group. You can assume that authors mean this construction unless otherwise stated.
Notice, however, that this construction can be also done in the functorial setup, and a little bit more general and elegant as well:
Let $S$ be any base scheme. If $G$ is ...

answered Feb 15 at 13:31

Martin Brandenburg

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2

By Venkov's proof of the Evens-Venkov theorem, if you take a faithful representation $G\rightarrow SO(m)$, then the cohomology $H^*(BG)$ is a finitely generated module for the image of $H^*(BSO(m))$. But the situation is quite different if you only allow yourself Euler classes, and allowing Euler classes of all representations doesn't help much.
There ...

2

I have come across a fairly recent result which pertains to this question. It is in this paper:
Guralnick, Robert M.; Maróti, Attila; Pyber, László, Normalizers of primitive permutation groups, Adv. Math. 310, 1017-1063 (2017). ZBL1414.20002.
An arXiv version is here. The paper examines the situation when $U$ is primitive. They show that, in all but a ...

1

I think it's time to write an answer.
Let $G = G(A)$, so $|G| = p^{|V|+|E|}$ with $|G'| = p^{|E|}$ and $G'$ and $G/G'$ are both elementary abelian ($p$-groups with that property are called special $p$-groups, and it is conjectured that almost all finite groups of order up to some bound are special $2$-groups, but that's not relevant to this question.) Note ...

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