11
votes
Accepted
Does every sequence of group epimorphisms (between finitely generated groups) contain a stable subsequence?
Yes. This is just (metrizable) compactness in the space of normal subgroups of $G$. It is enough to assume that $G$ is countable (finitely generated plays no role).
Namely, let $N(G)\subset 2^G$ be ...
11
votes
Are large powers of polynomials linearly independent?
We have used this problem for our
Student Olympiad in Algebra at Moscow State University
(in Russian, Пятнадцатая олимпиада, задача 8).
So, here is a completely elementary solution.
Exercise 1.
Show ...
11
votes
Accepted
Analogous results in geometric group theory and Riemannian geometry?
I think Cheeger's inequality is a good example.
Riemannian geometry version
Let $M$ be a closed Riemannian $n$-manifold. Say that a $n-1$ dimensional submanifold $N$ separates $M$ if the complement of ...
11
votes
Accepted
When are two semidirect products of two cyclic groups isomorphic
The paper of Basmaji, "On the isomorphisms of two metacyclic groups" (Proc AMS 1969) gives a complete answer to the question of when two finite metacyclic groups with the same $m$ and $n$ ...
10
votes
Analogous results in geometric group theory and Riemannian geometry?
Here is a very classical example. As stated in the comments, Gromov was an early proponent of importing ideas from geometry to group theory, but already thirty years earlier there was work in this ...
8
votes
Accepted
Morita equivalences and centers of some algebras
The answer is that in your matrix $\left(\begin{smallmatrix} 0&x_0\\0&0\end{smallmatrix}\right)$, the $x_0$ denotes the isomorphism of modules given by left multiplication by $x_0$, so it ...
8
votes
Accepted
Where has this structure been observed?
This is an infinite commutative diagram on $M$ (viewed as a category with a single object $\bullet$).
$\require{AMScd}$
\begin{CD}
\vdots @. \vdots @. \vdots\\
@VVR_y(0,2)V @VVR_y(1,2)V @VVR_y(2,2)V\...
8
votes
Accepted
Question to limit groups (over free groups)
You need to prove the following folklore lemma, which is well known to researchers in the field but perhaps not written down anywhere. The proof is a nice exercise.
Folklore lemma: Let $S$ be a ...
7
votes
Concept associated to the Eudoxus reals
This method can be used to construct the fields $\mathbb{Q}_p$ and the ring $\mathbb{A}_{\mathbb{Q}}$ of adeles over $\mathbb{Q}$. See T.D.J. Hermans' Bachelor's thesis: https://www....
7
votes
Accepted
Subalgebras of quadratic algebras that are not quadratic
Here is an example that shows that you can expect things to get as bad as it goes (I learned about this algebra from the wonderful article The Non-Commutative Gröbner Freaks by Green, Mora, and ...
6
votes
Accepted
Algebra with three anti-commutator relations
I ran a quick test on magma, asking it the dimension of the algebra defined by your relations with (chosen pretty much at random) $p=5$, $u=2$, $v=3$, $w=4$, and with the extra relations $a^3=0$, $b^3=...
6
votes
The discriminant of the Okada algebra
Using some results in progress with Jeanne Scott, I can compute the few next one. Note that these computation rely on thing that are not yet fully proved. So it might differ from what you are actually ...
6
votes
When are two semidirect products of two cyclic groups isomorphic
This is too long for a comment and solely deals with the case of coprime $m,n$.
Suppose we have an isomorphism $f:C_m\rtimes_k C_n \to C_m\rtimes_k' C_n$. I would like to name the generators $x,y$ on ...
5
votes
Accepted
Equivalences of categories of complexes of modules
The answer is yes by the same type of Morita theory, namely $Z(Ch(R))\cong R$, where $Z(A)$ is $End(id_A)$, the ring of endomorphisms of an abelian category
EDIT : sorry, I hadn't seen that you were ...
5
votes
Accepted
Markov property for groups?
Yes the Markov property is what is described in the text further down.
The terminology Markov is not directly related to Markov chains in probability theory. A group is called Markov if there exists a ...
5
votes
Accepted
Classification of real Clifford algebras
The classification is described in quite some detail in the Wiki on the topic.
In particular, the case $(p,q)=(1,3)$ is isomorphic to $M_2(\mathbb H)$ and the case $(p,q)=(3,1)$ is $M_4(\mathbb R)$.
...
5
votes
Accepted
Gluing data for modules over a ring with idempotents
If $m=2$, your data gives no information about the relationship between $e_1M$ as an $e_1Ae_1$-module and $e_2M$ as an $e_2Ae_2$-module. So this is false. For example, take a quiver with two vertices, ...
4
votes
Accepted
Subfields of division rings of degree $2$ which are not invariant
(This is basically a more detailed version of Eoin's comment.)
I assume that you are considering division algebras over a field $k$, i.e., $Z(A) = k$. If $B$ is a subalgebra of dimension $2$ of $A$, ...
4
votes
The discriminant of the Okada algebra
Define $a_0:=1,a_1:=x_1$ and recursively for $k\ge2$ $$a_{k}:=x_{k}a_{k-1}-y_{k-1}a_{k-2}.$$ So $a_k$ is the factor starting with $x_1\cdots x_k$. (Note that most of them have their signs flipped ...
3
votes
A non-example of a graded Frobenius algebra
A pretty trivial example is an algebra with the zero product. Less trivial is $k[x,y]/(x,y)^2$ where $x$ has degree $1$ and $y$ has degree $2$. One can make up various similar examples, e.g., $k[x,y,z]...
3
votes
Are large powers of polynomials linearly independent?
$\require{AMScd}
\require{enclose}$EDIT : As noted by Zach Teitler, the argument below only proves that for $m\gg0$, the family $\left\{P_1^{\otimes m}, \dotsc, P_k^{\otimes m} \right\}$ is a free ...
2
votes
An algebra map between Hopf algebras that does not commute with the counit
Let $H$ be the Hopf algebra of functions on an algebraic group $G$. The map $\phi$ defines a map of algebraic varieties $\hat{\phi}:G\rightarrow G$. The counit condition you try to impose is ...
2
votes
Accepted
Polynomial identities satisfied by the Weyl algebra in prime characteristic
This is a result of Lopatin and Rodriguez-Palma.
You can find it (in Portuguese) in the PhD thesis of Rodriguez Palma: https://repositorio.unicamp.br/Busca/Download?codigoArquivo=502605
I will ...
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