abstract algebra – Completed tensor product discrete topology

Hi I have a question it should be easy to answer but I dont see how to do it Let us suppose that $k$ is a field and $A$ is pre-adic $k$-algebra, that is a $k$-algebra linearly topologized such that there exist an ideal of definition I such that the collection ${I^n}_{ngeq 1}$ is a neighborhood basis for $0$. And let us suppose that we have another $k$-algebra $B$ without any specified topology on it. We can attach to $B$ the discrete topology. My question is if in this situation it holds that
$$Awidehat{otimes}_{k}Bsimeq varprojlim (A/I^notimes_k B)$$

abstract algebra – How to prove that $varphi$ is an injective homomorphism and why it is surjective?

Here is the question I want to solve:

Prove that the subgroup of $SL_2(mathbb F_3)$ generated by $begin{pmatrix}
0 & -1 \
1 & 0
end{pmatrix}$
and $begin{pmatrix}
1 & 1 \
1 & -1
end{pmatrix}$
is isomorphic to the quaternion group of order $8.$(use a presentation for $mathcal{Q}_8$)

Here is a solution I found online:

A presentation for $mathcal{Q}_8,$ is $$langle i,j | i^2 = j^2, i^4 = 1, ij = -ji rangle$$
Which means that $i,j$ generates $mathcal{Q}_8$ and they satisfy the relations $i^2 = j^2, i^4 = 1, ji = -ij = i^3j.$

Let $A = begin{pmatrix}
0 & -1 \
1 & 0
end{pmatrix}$
and $B = begin{pmatrix}
1 & 1 \
1 & -1
end{pmatrix}.$
Define $varphi: {i,j} rightarrow langle A, B rangle $
as $varphi(i) = A$ and $varphi(j) = B.$

Now, note that by direct calculation , the subgroup generated by $A$ is $$langle A rangle = { begin{pmatrix}
0 & -1 \
1 & 0
end{pmatrix}, begin{pmatrix}
-1 & 0 \
0 & -1
end{pmatrix}, begin{pmatrix}
0 & 1 \
-1 & 0
end{pmatrix}, begin{pmatrix}
1 & 0 \
0 & 1
end{pmatrix} }$$

And also, by direct calculation , the subgroup generated by $B$ is $$langle B rangle = { begin{pmatrix}
1 & 1 \
1 & -1
end{pmatrix}, begin{pmatrix}
-1 & 0 \
0 & -1
end{pmatrix}, begin{pmatrix}
-1 & -1 \
-1 & 1
end{pmatrix}, begin{pmatrix}
1 & 0 \
0 & 1
end{pmatrix} }$$

And it is clear that $|A|=|B| = 4$ and $A^2 = B^2.$

Also, by direct calculation, we have that $BA = begin{pmatrix}
1 & -1 \
-1 & -1
end{pmatrix} = A^3B.$

Here by the lemma $varphi$ extends to an injective homomorphism $$bar{varphi} : mathcal{Q}_8 rightarrow langle A, B rangle $$
Also, by construction $bar{varphi}$ is surjective. Hence the required is proved.

My questions are:

1-I do not know what lemma says that $varphi$ is an injective homomorphism, tthis problem is #10 in section 2.4 of Dummit and Foote (3rd edition) and I could not find any lemma that said this. Could anyone help me in proving that $varphi$ is an injective homomorphism please?

2- Why by construction is $bar{varphi}$ is surjective? what confuses me is that $langle A, B rangle $ is the subgroup generated by $A$ and $B$ and not the set that contains $A$ and $B,$ Could anyone explain to me this please?

abstract algebra – How do I find the isomorphisms between the quotient groups in D12?

I am given $𝐷_{12}={1,𝑥,𝑥^2,𝑥^3,𝑥^4,𝑥^5,𝑦,𝑥𝑦,𝑥^2𝑦,𝑥^3𝑦,𝑥^4𝑦,𝑥^5𝑦}$

I know that the normal subgroups are as follows: $⟨𝑥⟩,⟨𝑥^2⟩,⟨𝑥^3⟩,⟨𝑥^2,𝑦⟩,⟨𝑥^2,𝑥𝑦⟩$.
First, I just need some help understanding why these are the normal subgroups. After this, I would appreciate help identifying how to find the elements within these normal subgroups. I know that $langle x^3rangle={1,x^3}$, but some help identifying the others, especially the last two would help a bunch.

After I have this information, I am to identify the quotient groups within $D_{12} and then identify which groups the quotient groups are isomorphic to. How would I go about doing this? It can be as simple as using Cayley tables. I am just lost on how to start.

abstract algebra – Second Isomorphism Theorem and Jordan-Holder

In another posting, there was a question about the following:

Let $G$ be a finite non-trivial group with the following two composition series:

${e} = M_0 triangleleft M_1 triangleleft M_2 = G$

${e} = N_0 triangleleft N_1 triangleleft cdots triangleleft N_r = G.$

Prove that $r=2$ and that $G/M_1 cong G/N_1$ and $N_1/N_0 cong M_1/M_0$.

The person posting the question went on to state that “By the second isomorphism theorem I know that $M_1N_2/N_2 cong M_1/(N_2cap M_1)$

Here is my question. In order to use the second isomorphism theorem with $M_1$ and $N_2$ don’t we need to know that $M_1 leq N_G(N_2)$? And if so, then how do we know that $M_1$ actually is in the normalizer of $N_2$ in $G$?

If I can get over this hurdle, I understand the remainder of the original posting. Perhaps this is something obvious, but please help me see whatever it is that I am missing.

Thanks in advance.

c# – Refactoring a class to an abstract base class – keep default implementation in place or move it to an an other class?

I have a class which aggregates some temporal data from a database and provides a bunch of methods to query said data. It looks like this:

public class InfoProvider
{
    public IEnumerable<Contract> GetContracts() => GetContracts(DateTime.Today);
    public IEnumerable<Contract> GetContracts(DateTime at) => GetContracts(at, at);
    public IEnumerable<Contract> GetContracts(DateTime from, DateTime to) { /* query contracts */ }

    public IEnumerable<Job> GetJobs() => GetJobs(DateTime.Today);
    public IEnumerable<Job> GetJobs(DateTime at) => GetJobs(at, at);
    public IEnumerable<Job> GetJobs(DateTime from, DateTime to) { /* query jobs */ }

    /* more methods like above ... */
}

Now I need to extend my application to support different representations of the data in the database. Those representations are identical for the most part but have some subtle differences. For example GetContracts in representation 1 should return contracts of type A, B or C while in representation 2 it should return contracts only of type A or B. On the contrary, GetJobs would be the same in both representations.

I want to do this extension by refactoring InfoProvider into an abstract class and derive concrete classes, InfoProviderRepresentation1 and InfoProviderRepresentation2 for each representation. But I waver where to put the existing code. I see those two options:

  1. Leave it in the original class
public abstract class InfoProvider
{
    public IEnumerable<Contract> GetContracts() => GetContracts(DateTime.Today);
    public IEnumerable<Contract> GetContracts(DateTime at) => GetContracts(at, at);
    public virtual IEnumerable<Contract> GetContracts(DateTime from, DateTime to) { /* query contracts */ }

    public IEnumerable<Job> GetJobs() => GetJobs(DateTime.Today);
    public IEnumerable<Job> GetJobs(DateTime at) => GetJobs(at, at);
    public virtual IEnumerable<Job> GetJobs(DateTime from, DateTime to) { /* query jobs */ }
}

public class InfoProviderRepresentation1 : InfoProvider
{
}

public class InfoProviderRepresentation2 : InfoProvider
{
    public override IEnumerable<Contract> GetContracts(DateTime from, DateTime to) { /* ... */ }
}
  1. Move the code to an other abstract class and derive my concrete classes from there:
public abstract class InfoProvider
{
    public IEnumerable<Contract> GetContracts() => GetContracts(DateTime.Today);
    public IEnumerable<Contract> GetContracts(DateTime at) => GetContracts(at, at);
    public abstract IEnumerable<Contract> GetContracts(DateTime from, DateTime to);

    public IEnumerable<Job> GetJobs() => GetJobs(DateTime.Today);
    public IEnumerable<Job> GetJobs(DateTime at) => GetJobs(at, at);
    public abstract IEnumerable<Job> GetJobs(DateTime from, DateTime to);
}

public abstract class DefaultInfoProvider
{
    public virtual IEnumerable<Contract> GetContracts(DateTime from, DateTime to) { /* query contracts */ }
    public virtual IEnumerable<Job> GetJobs(DateTime from, DateTime to) { /* query jobs */ }
}

public class InfoProviderRepresentation1 : DefaultInfoProvider
{
}

public class InfoProviderRepresentation2 : DefaultInfoProvider
{
    public override IEnumerable<Contract> GetContracts(DateTime from, DateTime to) { /* ... */ }
}

I find option 2 a bit more readable. On the other hand it needs more boilerplate code than option 1. Does option 2 have any more advantages over option 1?

abstract algebra – Centralizer of subalgebras in tensor product

Let $k$ be a field, $E,F$ two non-zero $k$-algebras and $C,D$ their respective sub-$k$-algebras. Write $C’$ for the centralizer of $C$ in $E$ and $D’$ for the centralizer of $D$ in $F$. I want to show that
$$C’otimes_kD’=(C’otimes_kF)cap(Eotimes_k D’).$$
The author asserts that this is an application of the following result:

Let $k$ be a field, $E$ a right $k$-vector space, $F$ a left
$k$-vector space, $(M_alpha)_{alphain A}$ a family of subspaces of
$E$ and $(N_beta)_{betain B}$ a family of subspaces of $F$. Then
$$(bigcap_{alphain A}M_alpha)otimes_k(bigcap_{betain
B}N_beta)=bigcap_{(alpha,beta)in Atimes B}(M_alphaotimes_k
N_beta).$$

Let $M_0:=E$, $M_1:=C’$, $N_0:=F$ and $N_1:=D’$. Then
$$C’otimes_kD’=(C’cap E)otimes_k(D’cap F)=(M_0cap M_1)otimes_k(N_0cap N_1).$$
The last part is equal to $(Eotimes_kF)cap(Eotimes_k D’)cap(C’otimes_kF)cap(C’otimes_kD’)$. This is not what I needed. Am I applying the quoted result incorrectly?

Logic of defining a function on an abstract set

Obviously in practice it is common to define functions given a set $S$, however I have never thought about the concept behind this and why “I am allowed to do that”. Please keep in mind that I have never read any logic lecture notes or visited any lecture on those topics, besides the standard material. If one is in the setting of an algebraic structure, e.g. a group $G$, and one wants to prove a statement, e.g. $$varphi:G to G, g mapsto 2g$$ is a function, what is the logic behind this being true for an arbitrary group? It is standard to prove this by assuming a group $G$ is given and then showing the statement.

Q$1$: Is this to be interpreted to be given an explicit but arbitrary/unknown group? What I mean by that is that this is to be interpreted as $G$ being any explicit group, such as a set consisting of a multiplication map satisfying the usual properties of the definition?

In this setting one does not know what set one is working with, however one would know that an “explicit” set $G$ is underlying, which means that it is known which elements belong to $G$ and which do not. Therefore it is possible to speak of “$g in G$“, is this correct? As far as I can tell, this is necessary in order to do this. This allows to even make sense of a statement for all $g in G : P(g)$. Since one knows which elements belong to $G$ and which do not, one can then define an assignment for those elemnts, an example given above.

One could sum up that I am confused whether it needs justification to talk about “for all elements of the set the property $P$ holds” without knowing the set and why one is able to do so. Since defining a function is sort of a statement of that kind, I found this to be fitting. This is then solved since one knows that $G$ is some explicit but unknown/arbitrary set. If anything of what I said is wrong I would appreciate corrections, since those are only my thoughts. Thank you in advance for any comments!

abstract algebra – Co-isosimple modules are co-cyclic.

First I want to give the definitions of co-isosimple and co-cyclic modules.

If a non-zero module $M$ is isomorphic to all its non-zero quotients,then it is called $textbf{co-isosimple}$.

A module $M$ is called $textbf{co-cyclic}$ if there is a simple module which is essential in $M$.

In the paper which this lemma stated, they just said that since a co-isosimple module is both Artinian and uniserial, then it is co-cyclic. If I can prove this last sentence, then I complete the proof.

I know that every Artinian module has a simple submodule and every submodule of a uniserial module is essential. Therefore if a module is both Artinian and uniserial, then it has a simple essential submodule, that is, it is co-cyclic. However, I cannot prove a co-isosimple module is artinian and uniserial. I need some hints.

abstract algebra – why we take $p ge n+m-1$ ? why not $p ge n+m$?

I have some confusion in this statement

Theorem :The subset of nilpotent elements of $R$ form an ideal of $R.$

It is written that

Let $N $ be the subset of nilpotent element
let $x,y in N$ .let $x^n=0$ and $y^m=0$
By the Binomial theorem $(x-y)^p=0$ for $p ge n+m-1$
Thus $x-y in N$

My question : Im not getting why we take $p ge n+m-1$ ? why not $p ge n+m$ ?

My thinking : Here $x^n=y^m=0 implies x=y$.Now take $p = n+m-1$ then $$(x-y)^{n+m-1}=
frac{(x-y)^{n+m}}{(x-y)}=frac{(x-y)^{n+m}}{0}=infty$$

This leads to a contradiction to $(x-y)^p=0$