groups-rings-fields

[dusty raft]

Okay this might get downvoted but uhh here it is if you know the seifert van kampen this can be reduced to a problem in finding two loops in S^1 \wedge S^1

And showing that they are not homotopic

Anyways this is not very helpful if you don't know algebraic topology so I'll just stop

[crystal vale]

i am doing alg top and that question is in my notes

[dusty raft]

Oh cool. Look at the wedge of two circles then

I'll give you a headaway

You can show pi_1(S^1 wedge S^1) is non abelian without computing the fundamental group

[crystal vale]

i have to read about what is exactly wedge, is it co-product of pointed spaces S^1?

[dusty raft]

It's just an "8"

Think geometrically

Very vague but yk

[crystal vale]

i don't get how this is related to my question?

[dusty raft]

What is the fundamental group of the "8"

[crystal vale]

sorry no idea

[dusty raft]

So you haven't seen seifert van kampen theorem yet

That's okay

You'll see it later on

[crystal vale]

i saw this version

[dusty raft]

It's basically this pushout

Yes

What is the fundamental group of a circoe

[crystal vale]

Z

[dusty raft]

.

[crystal vale]

but in S1 wedge S1, what are my X0, X1, X2?

[dusty raft]

Let me just draw it for you and I gtg soon

Gimme a minute

[crystal vale]

okay

[crystal vale]

okay

okay so fundamental group of S1 wedge S1 is exactly that my F( isomorphic)

[dusty raft]

Yeah the entire point is to use less Algebra now, but it's too much hassle sometimes, but also it's a fun thinking geometrically excercise

[crystal vale]

but how do i find its fundamental group?

[dusty raft]

It's okay if you can't, don't waste much time one it

What

[dusty raft]

Seifert van kampen

[crystal vale]

Seifert and van Kampen gives that fundamental group is pushout, does it gives full construction of fundamental group?

[dusty raft]

I'm saying you had to prove that it's non abelian right

You can do that geometrically too

[crystal vale]

how?

[dusty raft]

How about you ping me or dm me later i gtg, basically this

[dusty raft]

Sorry this

[crystal vale]

okay i will back to you

thank you

[limber tapir]

It does, because we know what pushouts of groups are

[dusty raft]

So I'm back now and another hint is you can easily draw the universal cover of "8"

[noble nexus]

Why are you guys crazy overcomplicating this lmao

You shouldn't have to think about algebraic topology to answer this question

(though of course you can if you want)

[rocky cloak]

Yeah, its basically just, there exists a group with two elements that don't commute, so by universal property it's not abelian

[quiet pelican]

I feel like this is way more complicated than “figure out why finding a group with two elements that don’t commute solves your problem”

[dusty raft]

Yeah I think I said it's a fun geometric excercise

Uh i have no intent in saying it's easier or better

In any aspect

[sick schooner]

If I have non-trivial groups G and H is there an easy argument that G x H cannot be isomorphic to a free product of possibly infinitely many other non-trivial groups? If either G or H has nontrivial center, then also G x H has non trivial center and then its clear to me, but I dont see or found an easy argument for the general case.

[noble nexus]

In a free product distinct factors do not commute, whereas in G x H every element in G commutes with every element of H

I believe this forces G and H to be entirely contained in one factor

but you'd have to work out the details

[sick schooner]

Ill need some time to think about this (im not a group theorist) thank you

I dont quite see how the commutativity quite gives you that G and H have to be in the same factor, for example you have that (g,1) and (1,h) commute, but that could also be because they are both a power of some word w in the free product (ill think about it a bit more though)

[balmy python]

Hi, I'm looking at a question that is about proving that for a field F with q elements if d | q-1 then x^d - 1 has d distinct roots in F.

I went to consider F^x which of course has q-1 elements but came to the point where I want to find a subgroup of order d but cannot really prove its existence. I just wanted to know if it is possible? the ideas I had in my head were cauchy's theorem or the classification of finitely generated abelian groups. I haven't really learnt about sylow's theorems so don't know much about that

[rapid cave]

F^x is cyclic

[terse crystal]

And any finite subgroup of F^x for any field F is cyclic

[rapid cave]

Which you can prove by noting that F^x has at most one subgroup of every order, and showing that there must be a subgroup of order n = |F^x| by counting the the amount of elements of each ordet in F^x.

[balmy python]

what if i wasn't to use that fact?

Could i perhaps use the classification of finitely generated abelian groups to say that we can have a subgroup of order d?

Ah

I mean say if theres a finite abelian group G you can write it as an isomorphism of cyclic groups of prime power order, these prime power orders must divide the order of the group. if you have d | n where n is the size of the group, writing d as a product of prime factors can you not extract subgroups from their individual p torsions subgroups?

[velvet hull]

Sure, but at this current state you don’t know anything about what that prime decomposition looks like, nor what the prime factorisation of d looks like

[balmy python]

ah so the primes dont have to be primes that are part of the decomposition of the size of the group?

[velvet hull]

It’s possible that d has “too high” of a prime power that makes it impossible, for instance you just can’t find an element of order 8 in Z/2Z x Z/4Z

Of course that’s impossible in actuality but a priori you don’t know that

[rapid cave]

Note that pentinum is looking for a subgroup of order d

Not a cyclic subgroup

[balmy python]

oh yeah i accidentally said this 😭

[velvet hull]

In that case, I think you could actually brute force it with this proof idea then

[balmy python]

ok yeah

just wanted to ask one more thing just to confirm

a cyclic group with prime factor higher than what appesrs in the torsion group of finitely generated abelian group wouldn't be possible because of lagranges theorem right

like if A x B is isomorphic to C then |A| must divide |C|

[velvet hull]

Sure yeah

[balmy python]

ok 😭

[velvet hull]

Also I will say that the proof of the fundamental theorem is basically using the sylow theorems

[balmy python]

stress makes me rethink the simple stuff

[balmy python]

Ah really?

in lectures we used smith normal forms

is that to do with sylow theorems?

[velvet hull]

Ah, that’s interesting, so you did the PID module method

Lmao that’s fun

Yeah that’s an alternate way of doing it

[balmy python]

ahhh

yeah we covered no sylow theorems

we did do some stuff which builds up to that

should sylow theorems usually be covered in a second module of group theory?

[velvet hull]

I can’t speak on what your school does but it should be

[balmy python]

damn 😭

[visual swift]

Is there anyone here expert in Lorentz group representations?

[balmy python]

Ok so formally, is this a good way to say that for a finite field F and d dividing the order of the multiplicative group, F^x contains a subgroup of order d?

By the classification of finitely generated abelian groups:

F is isomorphic to F{p_1} x F{p_2} x F{p_k}

where p_i are the distinct primes in the prime factorisation of |F^x| and F{p_i} is the respective p primary torsion subgroup

We then have each p-primary torsion subgroup isomorphic to a product of cyclic groups of order powers of p

From d's prime factorisation, it contains a power of the each of primes above. By taking a subgroup of order of this prime power from each of the individual groups and taking their products, we have a subgroup of order d.

[white oxide]

Well, by induction we can reduce to the case where $m = 1$, $n = 2$, i.e.
[
\begin{tikzcd}
{f(A)} && B \
\
{\mathfrak{p}_2} && {?} \
\
{\mathfrak{p}_1} && {\mathfrak{q}_1}
\arrow[from=1-1, to=1-3]
\arrow[from=3-1, to=1-1]
\arrow[from=3-1, to=3-3]
\arrow[from=3-3, to=1-3]
\arrow[from=5-1, to=3-1]
\arrow[from=5-1, to=5-3]
\arrow[from=5-3, to=3-3]
\end{tikzcd}
]
where all arrows are inclusions. It makes sense to consider $V(\mathfrak{q}_1)$, which contains $\mathfrak{q}_1$. Since $f^$ is closed, $f^\bigl(V(\mathfrak{q})\bigl) = V(E)$ for some $E \subseteq A$. I would like this $E$ to be $\mathfrak{p}_2$ but the problem is I don't have a choice of $E$.

[cloud walrus]

okeyokay

[rocky cloak]

Well, does V(E) contain p1?

[white oxide]

Yeah right? Since p_1 = f^{-1}(q_1) and q_1 certainly contains q_1

Ah

so E is a subset of p_1 which is a subset of p_2, so that p_2 is in V(E)

aka p_2 = f^{-1}(q) for some q which contains q_1

and this is q is the desired prime ideal

Thanks!

[white oxide]

If f: A -> B is a ring homomorphism and p is a prime ideal of A, is it always true that f(p) is a prime ideal of f(A)?

[supple ice]

no

[agile burrow]

No. For instance take f: Z -> Q to be the inclusion. The image of a non-zero prime ideal (p) generates all of Q

[white oxide]

Hm okay

The reason I'm asking is that I am trying to show b $\implies$ c. So far, the proof has seemed natural to me:

b $\implies$ c. Let any $\overline{\mathfrak{a}} \in \text{Spec}(A/\mathfrak{p})$. Then $\mathfrak{a} \coloneqq \pi^{-1}(\overline{\mathfrak{a}})$ is a prime ideal of $A$ which contains $\mathfrak{p}$. Since $f$ has the going up property, there is $\mathfrak{b} \in \text{Spec } B$ such that
[
\begin{tikzcd}
{f(A)} && B \
\
{\mathfrak{a}} && {\mathfrak{b}} \
\
{\mathfrak{p}} && {\mathfrak{q}}
\arrow[from=1-1, to=1-3]
\arrow[from=3-1, to=1-1]
\arrow[from=3-1, to=3-3]
\arrow[from=3-3, to=1-3]
\arrow[from=5-1, to=3-1]
\arrow[from=5-1, to=5-3]
\arrow[from=5-3, to=3-3]
\end{tikzcd}
]
commutes, where all arrows are inclusions. Let $\overline{\mathfrak{b}}$ be $\mathfrak{b}$ mod $\mathfrak{q}$. We claim that $f^*(\overline{\mathfrak{b}}) = \overline{\mathfrak{a}}$.

However this won't work unless $\mathfrak{a}$ is a prime ideal of $f(A)$, and similarly $\mathfrak{p}$ is a prime ideal of $f(A)$

[cloud walrus]

okeyokay

[supple ice]

do u think u can add a condition so that this statement becomes true ?

what do u think if ker(f) is contained in p ??

[rocky cloak]

The correspondence theorem is your friend

[tardy hedge]

In k[x1,…xn] with standard Z^n-grading, hilberts basis theorem says every ideal is finitely generated, but is every homogeneous/graded ideal finitely generated by homogeneous elements?

[supple ice]

yes

[tardy hedge]

Ok hm i mean i guess thats not trivial right but also doesnt seem hard to show

[tardy hedge]

Yeah i guess its just cause the homogeneous components of any element in a graded ideal are also in the ideal

[supple ice]

exactly

[tardy hedge]

f(p) is a prime ideal of f(Z) no? But not of Q

[agile burrow]

Oh good point, I misread the original question

[stone totem]

guys i need help, can't show that the action is free(aka) intersection of stab(Hi)={1}

1. and 3) are easy but 2) seems quite complex

i have showed that card(stab(Hi)) is 6 by lagrange

and so 6=3*2 so a 3 sylow group and 1 our 3 (2 sylow groups)

but then wtf am i supposed to do

[rocky cloak]

So I don't know french, so there might be some details I'm missing, but it's not true that a group of order 24 with 4 3-sylow subgroups acts faithfully on them. For example A4xC2. So I guess there must be some more info given

Okay, I'm guessing the second sentence is that the Sylow subgroups are not unique, while A4xC2 has a unique 2-sylow subgroup

[stone totem]

yeah the second sentence is that none of it's sylow groups are normal

[rocky cloak]

So I guess one thing to notice is that the intersection of the stabilizers either has order 1 or 2, so you just need to deal with the order 2 case

[stone totem]

why?

because the stabiliser of Hi is Hi with other elements

and the intersection of all Hi is the neutral

[rocky cloak]

Notice that Hi is normal in Stab(Hi) and has order 3

[stone totem]

yeah that's the 3 sylow group

and then there can be 1 2-sylow or 3-2sylow

for each stab(hi)

what i supposed is that if i take g in the intersection of the stab(Hi) g has to necessarily belong to 1 of thses 2 sylow

[rocky cloak]

It might also be useful to use that the intersection of the stabilizers is a normal subgroup

[stone totem]

well yeah but you wanna show that if it is different from {1} than it has to be a sylow group

so it can't normal and then absurd

[rocky cloak]

Well, that's a bit more than you need. The Sylow subgroup will never have size 2 after all

[stone totem]

why?

[rocky cloak]

Because it has size 8

[stone totem]

ah in this case yes

[rocky cloak]

Anyway, if you mod out a group of size 2 you'll get one of size 12. How many sylow subgroups can a group of size 12 have?

[stone totem]

what do you mean mod out?

[rocky cloak]

G/H

[stone totem]

ah ok

well 3*2^2

either 1 or 4 3 sylows and either 1 or 3 2 sylows

[rocky cloak]

Can you have 4 3-sylow and 3 2-sylow simultaneously?

[stone totem]

nope cos you lack elements prolly

i mean you can

4*2+3=12

[rocky cloak]

3 elements don't get you very many 2-sylow subgroups though

[stone totem]

wdym?

wait no you can't

because the 2 sylow groups are of card 4 not 2

so 4x2+3x3>12

but does this have to do with my exercise??

[rocky cloak]

Your exercise required that there be more than one of each sylow subgroup

[fading basalt]

I remember seeing somewhere that subgroups of P\GammaL(V) containing PGL(V) are semidirect products of PGL(V) with a cyclic group. Is it true? If so, anyone has a reference?

[rocky cloak]

P\GammaL(V) is the semidirect product of PGL(V) being acted on by Aut(k) where k is your field.

So this would be true if Aut(k) is cyclic (for example if k is a finite field)

[fading basalt]

Yeah I'm working over a finite field. Something doesn't click for me though. Does it just follow from intersecting the short exact sequence with the subgroup H?

[rocky cloak]

Yes, so certainly H/PGLV is cyclic, then you can consider the image of H/PGLV in PGammaLV under the splitting to form a semidirect product.

[fading basalt]

Great, thank you!

[spark veldt]

So apparently we first form $H={(m,\theta(m)^{-1}:m\in M}$ where $\theta:M\to N$ is the isomorphism and construct $G=(U\times V)/H$. I'm wondering why they constructed it like this? Like what's the motivation behind this construction just by looking at the problem.

[cloud walrus]

bluepianist

[marble hinge]

I am trying to prove something simple, but it looks like something is going wrong... maybe I am missing some simple trick

when I just try to derive xax-1 = yay-1 from xy-1a = axy-1 -- I get them in the wrong order

i.e. I get x-1 a x = y-1 a y

and for the life of me, I can't swap the order of those 🙂 I tried to do the usual tricks with inverting both sides, etc.

(sorry for bad handwriting, that was not originally intended to be shared 🙂

[marble hinge]

So, any suggestions please? Or is it just a mistake in the book? Now I am getting a feeling that the author just made a mistake and confused two notations (left and right) in one set of exercises: he is using right cosets of centralizer subgroup later, but seems to use this xax-1 form of conjugates (which corresponds to left actions/cosets) instead of x-1 a x form...

full context:

[white oxide]

Ah, I think it's just because if p is a prime ideal then A/p is isomorphic to f(A)/f(p), and since the former is an integral domain so is f(A)/f(p), hence f(p) is prime

wait never mind the kernel is not always equal to p

[tulip otter]

i think that there is a mistake in the question, tho i might be wrong lol

[rocky cloak]

Consider the free group on x, y and let a=y^- x

Then
xax^- = xy^- = y a y^-
but
xy- a = x y^-2 x =/= a xy^- = y- x^2 y^-

So this is probably a typo

[marble hinge]

Mm, I’m not yet very familiar with free groups…

[marble hinge]

A typo? It looks more like he needs to restate this whole set of exercises with x-1 a x definition of conjugates…

To make it work and be consistent with each other, no?

[rocky cloak]

The free group on x, y is just the set of all sequences of symbols from {x, y, x^-, y^-} such that a symbol never appears directly before it's inverse, and the operation is concatenation (cancelling inverses)

[rocky cloak]

I mean, that's what a typo is. Something that needs to be rewritten

Swapping x for x^- isn't exactly a major rewriting effort

[marble hinge]

Ok, for me typo is like when you type a instead of b 🙂

Or “dnoe” instead of “done”

[rocky cloak]

Well typing x instead of x^-

[marble hinge]

And also typo happens just once 🙂

[tulip otter]

well here he wrote x and y instead of x^{-1} and y^{-1}

[marble hinge]

But here he did it several times in a row

In all those exercises

So this is more like a mistake

[vapid vale]

not to jump on the pedant train but i do think typo implies a manual mistake which this isn't so i agree with sphynx lol

[marble hinge]

But anyway, doesn’t matter how I call it

Thanks for confirming that it’s a mistake/typo in the book

[white oxide]

Wait but this might not be the case here?

Is it an implicit condition? otherwise I don't think it's true

[supple ice]

this statement is false and if you add my condition it's true

[white oxide]

Ah so they should have stated it in the problem statement?

[supple ice]

I m talking in general I m not talking about your exercice

[white oxide]

Yeah well for my exercise it looks like it's necessary

b $\implies$ c. We first show that if $f: A \to B$ is a surjective ring homomorphism and $\mathfrak{p}$ is a prime ideal with $\ker f \subseteq \mathfrak{p}$, then $f(\mathfrak{p})$ is a prime ideal of $B$. Indeed, let $b_1 b_2 \in f(\mathfrak{p})$ so that $f(p) = b_1 b_2$ for some $p \in \mathfrak{p}$. Since $f$ is surjective, $f(a_1) = b_1$ and $f(a_2) = b_2$ for some $a_1$, $a_2 \in A$. Thus $p - a_1 a_2 \in \ker f \subseteq \mathfrak{p}$, so that $p - a_1 a_2 = p'$ for some $p' \in \mathfrak{p}$. Therefore, $a_1a_2 = p - p' \in \mathfrak{p}$, so that $a_1$ or $a_2 \in \mathfrak{p}$, which is to say that $b_1$ or $b_2 \in f(\mathfrak{p})$. Thus, $f(\mathfrak{p})$ is prime.

\
\newline
Let any $\overline{\mathfrak{a}} \in \text{Spec}(A/\mathfrak{p})$. Then $\mathfrak{a} \coloneqq \pi^{-1}(\overline{\mathfrak{a}})$ is a prime ideal of $A$ which contains $\mathfrak{p}$. Since $f:A \to f(A)$ is surjective, the above result implies that $f(\mathfrak{a})$ and $f(\mathfrak{p})$ are prime ideals of $f(A)$, so that by hypothesis there exists $\mathfrak{b} \in \text{Spec } B$ with
[
\begin{tikzcd}
{f(A)} && B \
\
{f(\mathfrak{a})} && {\mathfrak{b}} \
\
{f(\mathfrak{p})} && {\mathfrak{q}}
\arrow[from=1-1, to=1-3]
\arrow[from=3-1, to=1-1]
\arrow[from=3-1, to=3-3]
\arrow[from=3-3, to=1-3]
\arrow[from=5-1, to=3-1]
\arrow[from=5-1, to=5-3]
\arrow[from=5-3, to=3-3]
\end{tikzcd}
]
commutative, where all arrows are inclusions. Let $\overline{\mathfrak{b}}$ be $\mathfrak{b}$ mod $\mathfrak{q}$. We claim that $f^*(\overline{\mathfrak{b}}) = \overline{\mathfrak{a}}$. Indeed, if $a + \mathfrak{p} \in \overline{\mathfrak{a}}$, then $f(a + \mathfrak{p}) = f(a) + \mathfrak{q} \in \mathfrak{b} + \mathfrak{q} = \overline{\mathfrak{b}}$, for $f(\mathfrak{a}) \subseteq \mathfrak{b}$. Conversely, suppose that $f(a) + \mathfrak{q} = b + \mathfrak{q}$ for some $b \in \mathfrak{b}$. Then $f(a) - b \in \mathfrak{q} \subseteq \mathfrak{b}$, so that $f(a) \in f(\mathfrak{a})$. Thus $f(a) = f(a')$ for some $a' \in \mathfrak{a}$, so that $a - a' \in \ker f \subseteq \mathfrak{a}$. Therefore $a \in \mathfrak{a}$, so that $a + \mathfrak{p} \in \overline{\mathfrak{a}}$, as desired.

[cloud walrus]

okeyokay

okeyokay
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rocky cloak]

I don't think there's anything missing from the problem statement of your exercise. Just from the lemma you've formulated

For the purposes of the exercise you may as well assume A=f(A) because A/p = f(A)/f(p) since p contains the kernel of f

[tardy hedge]

The “irrelevant” ideal is the ideal generated by homogeneous elements of nonzero degree?

[quiet pelican]

Yes

[tardy hedge]

Irrelevant

Lol

[quiet pelican]

It’s irrelevant because it doesn’t correspond to a (non-empty) closed set in projective space

[tardy hedge]

Oh hm

[quiet pelican]

It “should” correspond to the origin, but that’s not in projective space

[tardy hedge]

Does any subring inherit the same grading? I know not all submodules do , like (x+y) in k[x,y] with Z^2 grade

[rocky cloak]

If the subring isn't generated by homogeneous elements you'll get some trouble.

Like the subring generated by x+x^2 in k[x, y] can't really inherit a grading

[tardy hedge]

You know what i need to do. Review direct sum vs direct product and countable vs uncountable stuff for that etc

I still dont have all the differences in my head

[tardy hedge]

Same as the ideal generated by (x+x^2)? Whats the difference between that and the subring

I learned recently defn of finite generation for algebras and how a finitely generated ring means finitely generated as a Z-algebra so how does that definition work into that

[rocky cloak]

I mean, an ideal is closed under multiplication from the ring, while a subring is just closed under multiplying things in the subring

So generally the subring is smaller except it contains 1

[marble hinge]

What are the ways to prove that a group of order 9 is abelian?

I remember a theorem that says that if |G| = p^2 for prime p, then G is either cyclic or isomorphic to Z_p x Z_p -- both abelian. So that works. But that theorem is not introduced yet in the book (Herstein) where this exercise comes from.

I followed steps in the proof of that theorem, and now know how to prove it more directly: we need to show that the size of the center is 9. Center is a subgroup, so its order must be 1, 3 or 9 (by Lagrange thm). We need to show that 1 and 3 are impossible and when order(C) is 9, we have G abelian, which is what we want to prove.

order(C) = 1 is impossible due to consequence of the class equation (and the fact that only center has order 1 in there, the rest of conjugacy classes must divide 9 and be greather than 1), and if order(C) is 3, then G/C is cyclic and using another exercise we have G abelian.

But all this machinery is still not relevant to this chapter, which is a chapter about homomorphisms, isomorphism theorems (FHT and so on), and some applications (Cauchy theorem for abelian groups proven by induction on group order, and Sylow theorem for abelian groups).

[south patrol]

This is the usual proof I have seen to show that groups of p^2 are abelian

This isn't really any machinery though right? It seems a bit simpler than the things you mention in the chapter, but just needs a bit of ingenuity

[marble hinge]

yeah, agreed that it's possible to come up with this independently, but not easy 🙂 I only proved the individual steps (that were laid out by exercise sets in another book - in Pinter)

there was one simple problem set that culminated in proving that if G/C is cyclic then G is abelian. And another set used that to prove the whole thing about p^2 via class equation

but I thought maybe there is some easier way that I am missing...

based on some clever homomorphism, or isomorphism, or something

[thorn jay]

p-groups have nontrivial centers

then Z(G) = G as G/Z(G) cannot be nontrivially cyclic

ah you mentioned that yes

[tulip otter]

well G/Z(G) is cyclic if Z(G)=G

[elfin wraith]

I guess if you want a really elementary way, 9 isn’t too big of a number and you can just work out the possibilities

But yeah I think the things mentioned are the sensible approaches

[thorn jay]

Suppose G is not cyclic. Then every nonidentity element is of order 3. I.e. the number of order 3 subgroups is (|G| - 1)/2 = 4. Let G act on the set of subgroups of order 3 by conjugation. Then one has that 4 is some sum of divisors of 9, i.e. some sum of 1, 3 or 9. Since 9 isn't possible, 1 must appear somewhere and there is a normal subgroup N of order 3. Take any other subgroup H. This defines a homomorphism f : H -> Aut(N) = C2 by conjugation. This is trivial, so the join of N and H is abelian. But the join of N and H is all of G so we are done.

not complicated at all

[delicate orchid]

write out all possible 9x9 cayley tables and observe they are symmetric along the diagonal

[thorn jay]

Write out at all reduced 9x9 latin squares and observe that every square whose transformation group has order 9 is symmetric along the diagonal

[tall igloo]

write down all 9^81 functions G^2 --> G for G a set with 9 elements, cross out the ones that don't satisfy the group axioms, and observe the remaining ones are all symmetric along the diagonal

[karmic moat]

<@&268886789983436800>

it's in other channels too

forgot to mention

[manic cairn]

yes yes im checking everywhere

[karmic moat]

tyty

[thorn jay]

wwhat happened lol

[karmic moat]

some scamming thing

[thorn jay]

yum

[south patrol]

Maths

[balmy python]

how do you get that 4 is some sum of divisors of 9

[delicate orchid]

it's 1+3 boss

[balmy python]

yeah i get that but

ah nvm

I was thinking he was trying to imply that or smth

[south patrol]

G acts on the order 3 subgroups, of which there are 4. This partitions the 4 elements into orbits, and each orbit has cardinality dividing that of G

[elfin wraith]

Any other method is woke nonsense. Work from the axioms like god intended.

[south patrol]

I think I can probably give another proof of this result, lemme try

[balmy python]

do you not mean 9 elements?

[south patrol]

No, I mean that like G acts on a set of cardinality 4

[delicate orchid]

cheeky H^2(Z/3Z, Z/3Z)

[balmy python]

ah okay

so his method was to prove theres a normal subgroup by considering G acting on the groups of order 3

and then define a homomorphism onto Aut(N) by conjugation?

[balmy python]

this is the confusing bit

[south patrol]

Sps G has order 9. If any element have order 9 then gg. Otherwise every nontrivial element has order 3. Pick any nontrivial elements a and b which aren't the same or inverse, so we get distinct subgroups H,K of cardinality 3. Observe that both are normal and they have trivial intersection (for cardinality reasons), so G = H x K and we are done

[thorn jay]

Yes

If N is a normal subgroup of G then any subgroup of G acts on N by conjugation

[delicate orchid]

a normal subgroup is fixed setwise under conjugation so any map in the larger group descends to an automorphism of the normal subgroup

[balmy python]

so basically right

i had an exam where instead of 9 it was 49

[thorn jay]

But my proof is by design kind of needlessly complicated lol

[south patrol]

My proof works for any p instead of 3

[balmy python]

and i did this exact same proof but didnt show they were normal

so then ended up getting 2/7 💔

[south patrol]

Oof sorry

[balmy python]

they wanted us to notice non trivial centre of a p group

[south patrol]

2/7 sounds harsh tho oof

Sorry to hear

[balmy python]

yeahhh

[elfin wraith]

Rough, my group theory exam was graded similarly lol, I got tripped up on like 3 small sub parts and still dropped 25 marks guy was brutal about the marking. Not unfair, but like, basically no one got any partial marks for working

[balmy python]

still did better than average tbh

[south patrol]

What would be funnier is getting 6/7

[elfin wraith]

Still, 75 was pretty good

[balmy python]

the test was outta 20

i got 14

average was 7

[elfin wraith]

Average was 7

[thorn jay]

Grown ass person

[balmy python]

cant believe i messed up one question 😭

[thorn jay]

That is crazy

[balmy python]

yeah its a pretty top university as well 💀

[thorn jay]

How

[balmy python]

ngl the exam was quite easy

[south patrol]

Wonder if I know where you are from mutual friends lol

[balmy python]

stuff like construct an abelian group of size 1000

[south patrol]

Was it imperial

[proud vigil]

mm as in like Z/1000Z?

[delicate orchid]

the cyclic group of order 1000 lol

[balmy python]

yessirr

[south patrol]

I think u said

Nice

[balmy python]

non abelian*

[elfin wraith]

There was some question about show G is abelian iff “sub object” and “quotient” are and despite seeing a million proofs of that form I couldn’t work it out, I remember trying some absolutely schizo looking exactly sequence nonsense

[proud vigil]

that makes way more sense LOL

[elfin wraith]

Turns out it basically followed by a random ass lemma I forgot

[delicate orchid]

Z/125Z x D_8

[balmy python]

there was also one about orbits of a symmetric group acted on by conjugations

[south patrol]

S_6 x C_(500/3)

[elfin wraith]

Imperial does tests and shit? You don’t just have a big fuck off exam at the end of the year? Is there a UK uni that’s actually sensible?

[balmy python]

why all that hassle when theres D_500

[south patrol]

Well these are two different questions

No way imperial is sensible

[balmy python]

we have midterms 😭

its so annoying

[south patrol]

Bro is american

[balmy python]

because it ruins the flow of work through term

[elfin wraith]

Mid terms are better than just one big exam icl

[thorn jay]

Couldnt be just any subobject and quotient (c.f. Dihedral groups)

[south patrol]

Icl = imperial college London?

[south patrol]

Bro said c.f.

[balmy python]

tbh itd have been better if they were end of term exams or smth

[elfin wraith]

Yeah I was vague because I don’t remember the exact details lol, hence the quotation marks

[south patrol]

https://tenor.com/view/flower-wilting-gif-4884848566557404043

[thorn jay]

The book im writing called dihedral groups that is just a list of all dihedral groups

Its uh, not finished yet

[balmy python]

ngl imperials 2nd year probability course is quite nice though

[elfin wraith]

It was something with derived series or something iirc

[balmy python]

ive also heard good stuff about measure theory

[thorn jay]

Rahhh

[south patrol]

Woah derived

Actually this terminology is lowkey getting annoying

[thorn jay]

Different derived 😔

[delicate orchid]

can't wait for the second edition containing all generalised dihedral groups

[south patrol]

You look up derived lie algebra or derived algebraic group and it comes up with ts 🥀

[delicate orchid]

it's actually slightly concerning I can't immediately do this. One direction is trivial at least

[thorn jay]

Deluxe edition labelling all coxeter groups

[balmy python]

can i have an example?

[elfin wraith]

I’ve actually not done any maths in ages and it’s stressing me out

[delicate orchid]

No because it's trivial to prove in complete generality

[thorn jay]

Its not true in general

[elfin wraith]

I don’t think it’s generically true if that helps

[thorn jay]

Abelian groups arent closed under extensions

[elfin wraith]

It was a more specific question but I don’t remember the details

[south patrol]

Idk if this immediately implies it

[balmy python]

its more like the words arent getting to my head 😭

[thorn jay]

Again, c.f. My upcoming book Dihedral groups

[south patrol]

Maybe it does

Lol

Oh wait there's an iff

[thorn jay]

If H and G/H being abelian would imply G is abelian then abelian groups are closed under extensions, and vice versa

[south patrol]

Okay then I agree lol

[delicate orchid]

N normal in G, all g in G, n in N gng^-1 in N, gnhg^-1 = gng^-1ghg^-1 so conjugation is a group homomorphism mapping N to N, it's invertible thus an automorphism

[south patrol]

I don't see how this would be implied in general

Well I do see for specific example of dihedral though

[delicate orchid]

are we doing contrapositive

[south patrol]

There the hypothesis obviously holds but not abelian

If that's what u mean

[delicate orchid]

ok what hypothesis because it is not true that any subobject of D_2^n is a quotient

[thorn jay]

Y'all what i meant was that abelian groups not being closed under extensions is a consequence of the backwards direction not holding

[balmy python]

ah this is just an inner automorphism right

[south patrol]

Oh lol

So it's not even a proof I guess as stated lol

[delicate orchid]

no because you're conjugating by an element of G, if it was an element of N it would be

[south patrol]

But ye dihedral is good example

S3 is the only nonabelian group I ever think of for counterexamples and ig works here lol

[delicate orchid]

holy shit I completely misread the question

[thorn jay]

Any solvable group, i believe

[delicate orchid]

I think I should go to bed I genuinely hallucinated like, 4 words that just are not in nope's post LMFAO

[thorn jay]

Cuz those are precisely the groups which are made out of abelian groups

By Jordann Holder ig

[delicate orchid]

Oh my science

[south patrol]

Or by definition

[elfin wraith]

I think the actual question was something about derived series if that conjures the correct result up in your brain

[thorn jay]

Idk im tired

[south patrol]

Depending on how they r defined

[south patrol]

Derived

[balmy python]

ahhh

[thorn jay]

Chat dont go to sleep at 5 for a week straight its not good for you

[south patrol]

Ok I'll do it for a month straight

[thorn jay]

Yes the more the better

[elfin wraith]

I also wouldn’t recommend sleeping all day every day for about a month

[thorn jay]

This is society

[elfin wraith]

IVT says there’s a sweet spot in the middle

[south patrol]

Ok I'll do it for two months

[elfin wraith]

I haven’t yet found it but it must be there

[delicate orchid]

Ts aint continuous broklahoma

[south patrol]

There was a point when I didnt know the IVT, and now I know the IVT, so

[thorn jay]

Use newton method or wtv the fuck numerical mfs are dreaming up

[elfin wraith]

Bro said time isn’t continuous, get a load of Zeno

[south patrol]

Woah I am teaching around the IVT this term

[thorn jay]

If youre teaching before it and after it then you must teach it somewhere along the way

[delicate orchid]

It warms my heart to see that you’re not completely jaded. Oh to be a 2nd year again

[balmy python]

how did you think of this before D_1000 btw 😭

[delicate orchid]

It’s got D_8 in it

[south patrol]

What is the best proof

[elfin wraith]

I think I need to do supervisions for like Reimann integration next term, I’m sure as shit hoping I remember that nonsense when I see it again

[delicate orchid]

The smallest non-abelian group that both works and isn’t Q_8

[elfin wraith]

It’s just kinda nebulously called analysis 2, no clue what it really covers beyond integration

[thorn jay]

Boy howdy am i excited for probability and statistics next quarter

[south patrol]

Ig just like what

[delicate orchid]

Is this guy trolling? Mods??

[south patrol]

0th step is knowing how to spell his name

Sorry

[thorn jay]

See it may be hard to notice but i used an advanced social technique called sarcasm

[south patrol]

If f(a) < 0 < f(b) and a < b then consider sup {x : f(x) < 0} n do a lil conditioning lol

[balmy python]

I just want to undersrand this, can you rxplain the last bit more detailed if you dont mind?

[thorn jay]

I can try lol

[elfin wraith]

I’m on my phone leave me be, and nothings getting spelled correctly at this hour

[delicate orchid]

“Join” you aint a carpenter brotato chip ✌️

[south patrol]

Probably cutest way to phrase it is composing f with its sgn if it never hits 0 and then u say f is locally constant n argument a lil cleaner

Eh

Might introduce connectedness tbh briefly

Actually I should not discuss pedagogy on this server

[balmy python]

also like what was the motivation behind considering the action on the subgroups of order 3

[south patrol]

Sabrina?

[delicate orchid]

Ts aint tuff bromula one 🖖

[south patrol]

Wynaut

[delicate orchid]

If it’s more natural you can also prove this by considering the conjugation action on elements rather than subgroups

[elfin wraith]

You can point them to the nlab page for it, and too simple to be simple. First years appreciate category theory

[south patrol]

Pointing anyone to the nlab is a mistake

[thorn jay]

Well im a first year and i appreciate category theory so..

Must be true

[delicate orchid]

I just checked and it’s actually readable until it brings out the classic “0-truncated cohesive oo-groupoid” with no warning

[elfin wraith]

IIRC it’s just like extremely passive aggressive around the “is the empty set connected” part lol

But yeah everything gets a bit nlaby eventually

[south patrol]

Tbh the page now seems to have nothing like this

[tall igloo]

and people learn from mistakes no

[twilit wraith]

Mods ban this guy his chudjak is too relatable

[south patrol]

Tbh I like this stuff in negative thinking

That's a fun article imo

[delicate orchid]

That’s a direct quote from the page on connected topological spaces

[south patrol]

Ah

Okay ye that is funny

[thorn jay]

So first assume G is noncyclic (of course if it is cyclic its abelian be default). Then we want a normal subgroup of order 3. We can do this by considering all subgroups of order 3, and trying to deduce that one of them must be fixed under the action of conjugation (i.e. be normal).

So, there are 8 nonidentity elements, none of which have order 9, so they must be order 3. But each order 3 subgroup has two nonidentity elements, so we get 8/2 is 4 order 3 subgroups.

So we have a group of order 9 acting on a set of order 4. Then by the orbit stabilizer theorem, the size of each orbit divides 9. It of course cant be 9, so it must be either 1 or 3. There are no ways of writing 4 using only 3, so there must be an orbit of size 1: normal subgroup, say, N, of order 3.

Now take any other subgroup H of order 3. Its fairly clear that N \cap H = 1, and NH = G. Both are cyclic, and so to prove that G is abelian it is therefore enough to show that elements in N commute with those in H.

To do this consider the action of conjugation of H on N. This produces a homomorphism f : H -> Aut(N) given by h |-> c_h. But Aut(N) ≈ Aut(C3) ≈ C2. There is however no nontrivial homomorphism from C3 to C2 (by langrange), so H has a trivial conjugation action on N. This is equivalent to saying that elements of N and H commute, so G must be abelian as desired.

@balmy python

As i said its silly

[delicate orchid]

9 = 1+1+1+3+3 or 1+1+1+1+1+1+1+1+1 so G/Z(G) cyclic so G abelian. Smoking gun 🚬 🔫

[thorn jay]

Not silly enough

[balmy python]

how do you get Aut(N) = Aut(C3)?

[thorn jay]

N is of order 3 hence isomorphic to C3

Because 3 is prime

[sonic coral]

this lattice is wrong, right? H_5 is the center of D_8 which is also the frattini subgroup

[rocky cloak]

Yeah H5 is contained in H7 and H8 aswell.

[sonic coral]

sweet thanks

[hexed bloom]

Given a group G which acts on a non empty set X, is it possible that the set of orbits is not countable ?

Nevermind

[thorn jay]

Consider a trivial action on a noncountable set

[marble hinge]

I've also found this proof, which is similar to yours:

(from https://lovekrand.github.io -- which has an ad-hoc solution manual to Herstein's "Topics in algebra")

I suspect though that Herstein's intended solution used non-trivial center idea, because the previous problem (no. 9) asked to prove that the center is a normal subgroup

I didn't sleep well last night, so decided to sample a new algebra book -- Isaacs "Algebra: A Graduate Course". Just glimpsed over all I've already learnt up to isomorphism theorems and simple groups. An interesting thing he is suggesting is to work on a project that finds orders of all simple groups up to 1,000 (or maybe just taking the list of orders and proving that any group of order not in the list can't be a simple group), this is somewhat similar to finding prime numbers, but there are surprisingly few. He says that proving that there are no simple subgroups of order 720 is an order of magnitude more difficult than sifting out the rest.

anyone done something like that? (that probably is not the best project for me personally, but sounds remotely interesting)

[rapid cave]

I remember some messages about this

in here I think

[quiet pelican]

I’d need to do some sieve stuff to see how much you can eliminate easily
Like I know every simple group has order divisible by 12, and 3 different primes (unless it’s prime-order cyclic)
But I also know those aren’t easy things to prove

You can sieve out ||prime powers||, ||4k+2||, ||pq|| quite easily
There’s probably also some other small two-primes stuff you can sieve out easily (as well as ||pqr||)
How much is left?

[marble hinge]

searched for "isaacs" but didn't find anything, searched for "1000", searched for "simple group", found this one 🙂

a recipe to produce an infinite number of exercises of this kind 😄

[marble hinge]

ah, it also has to be non-abelian, now I am re-reading that

[quiet pelican]

Non-abelian is a trivial restriction tbh

[marble hinge]

he claims that there are only 5 orders below 1,000 that are simple non-abelian groups

so one needs to sieve 994 numbers out to complete the project 🙂

[quiet pelican]

||pq^2|| is also easy

[marble hinge]

"How much is left?" -- ah, this is too advanced for me now, I'm just starting to study this for the first time, so no way I can prove all of that myself, I am just curious

[quiet pelican]

By the time you finish an average first group theory course, everything I’ve spoilered should be within reach

[marble hinge]

cool. Yeah, I am planning to get to the Sylow theorems and group actions in the next section, so that should unlock a lot 🙂

just need to wrap up those isomorphism theorems

but I don't quite see the point (or use) of the Diamond theorem yet

I've proved it as an exercise, but it felt somewhat artificial

[quiet pelican]

219 left, including this

The problem is that I don’t see anything nice that isn’t diminishing returns

pq^3 (which is messy but doable with elementary tools) only takes out 40ish

[rapid cave]

probably some require non-elementary tools.

do you know what tools are needed @quiet pelican?

[quiet pelican]

I can probably do it if you give me Feit-Thompson /hj
lemme see what I get with just Burnside’s theorem + pqr + prime powers

[rapid cave]

how does burnside help?

[quiet pelican]

Burnside’s theorem, not lemma

The p^a q^b order is never simple one

[rapid cave]

isn't it that its solveable?

[quiet pelican]

Which implies not simple

Look at the first term in the derived series
It’s either trivial (in which case the group is abelian), or it’s a non-trivial proper normal subgroup (by solubility)

And the answer is that gets you down to 113

No point with Feit-Thompson if you have Burnside
That only does another 12 (which is probably not bad by hand)

[quiet pelican]

(Also the actual proof of Burnside that I know proves it by “not simple” + induction on the order)

[torn sonnet]

Hi, guys, can someone help me with a video on groups fields rings, I am struggling to understand

[quiet pelican]

Next filter: ||largest prime factor is larger than the square root (the sylow of order p is normal)||
Result: 82

Next filter: ||largest prime factor is not [product of rest - 1], and is larger than half that (similar sylow stuff)||
Result: 69

The problem is not only finding filters, but finding ones that aren’t a pain to program, and will eliminate a bunch

[sonic coral]

i’ve done this up to 500ish

these are basically the results i used to wipe out a lot of them initially

then just case by case and spam lemma 8

[quiet pelican]

Bit the bullet
Did it properly (eliminating the ones that can’t work for trivial Sylow reasons)
41 left

Feit Thompson would eliminate another 5 at this point if I wanted

[sonic coral]

there’s only 21 by hand up to 500

[quiet pelican]

I’ve got 16 I think
But I used Burnside as a hammer

[sonic coral]

but if you use hardcore results then just abuse burnsides p^a q^b theorem and it’s easy

[quiet pelican]

18

[sonic coral]

it takes care of the hardest ones with all the powers of 2 and 3

[quiet pelican]

Burnside doesn’t help as much as you’d think

[quiet pelican]

The worst ones I’ve found are generally 2^a 3^b m, where m is a product of small primes

[sonic coral]

yeah looking at the ones i did by hand those are long

also the ones where m is 1

[quiet pelican]

These are the ones I have left
||[60, 120, 132, 168, 180, 240, 252, 264, 280, 300, 315, 336, 360, 380, 396, 420, 480, 495, 504, 520, 525, 528, 540, 552, 560, 600, 612, 616, 660, 672, 720, 728, 735, 756, 760, 792, 840, 900, 924, 945, 960]||

[sonic coral]

get rid of 60 168 360 504 and 660

[quiet pelican]

Yeah I know those ones work 🙃
I was just gonna keep them, and aim to get the list down to 5

[sonic coral]

i’d be surprised if you can without having to tackle them individually

[quiet pelican]

I'm aiming to get it more to like 20 before having to say "yeah I could do these one at a time, but bleh"

The reason I’m happy to use Burnside is like
I could manually prove all the useful cases of burnside for my purposes (there’s not that many)

[sonic coral]

looking at the ones i did by hand these are all of them

plus the ones on the power of two primes

it’s kinda silly after a little bc you just do the same thing over and over

240 was kinda rough i think

i think i didn’t do it right

but using burnsides transfer theorem it’s easy

[quiet pelican]

Yeah
I’m surprised 4k+2 + “doesn’t obviously fail sylow” + pqr gets you so far

[sonic coral]

my professor did this exercise as a grad student

one day i asked him ab 1008 since it’s double the order of a simple group. he wrote it up for me but id have to find it somewhere

[quiet pelican]

One of my supervisors did the character table of every group up to like 200 lol
Which is just
Insane

[elfin wraith]

Thats horrific surely not lol

I dont think even wew would put himself through that

[sonic coral]

that’s mental

[elfin wraith]

How many groups even are there up to order 200?

Like thats got to be a ridiculous number in and of its self, like a couple thousand or so right? Im guessing someone with GAP installed and check that pretty easily

[low quiver]

How do I start it?

[sonic coral]

definitely at least 200

[elfin wraith]

Proof???

[sonic coral]

Z/n for n<= 200

[twilit wraith]

Consider an element x in G and consider the subgroup generated by x

[low quiver]

subgroup generated by x?

[twilit wraith]

What can we say about the left cosets of the subgroup

[low quiver]

ax: where a is belongs to group G

Am I right?

[twilit wraith]

Hmm close, though I moreso mean general properties of cosets

[quiet pelican]

You don’t need to write them all “explicitly”
Like you can write down the character table for all dihedrals quite easily

[twilit wraith]

Actually let's start with the subgroup <x>

We know x must have finite order since G does

Whats the order of the subgroup <x>?

[elfin wraith]

Ok yeah true but like, still a crazy number no? I also am just curious how many groups there are up to order 200 now

Is there a known approximation of the number of groups of order n up to isomorphism?

[low quiver]

group of x is same as group G?

I have no idea

[quiet pelican]

The number of groups of order 2^{floor(log_2(n))}

[low quiver]

Can I take the discussion to a help channel?

[twilit wraith]

Sure

[quiet pelican]

Looking at OEIS 1, it’s probably in the ca 5k range for “number of groups up to order 200”

The vast majority of those (> 4k) are of order either 128 or 192 (= 3 * 64)

The spikes are at every multiple of 32 (except 32 itself) and at 144

[rocky cloak]

gap> Sum(List([1..200]), {x} -> Length(AllSmallGroups(x)));
6065

[rocky cloak]

gap> Length(AllSmallGroups(128))+Length(AllSmallGroups(192));
3871

Seems your estimate is off by a hair

[elfin wraith]

Yeah even with like the ones you can group up and knock out quickly, working out the character table for 6000 groups sounds like insanity lol

If only computers didn’t exist and I could be revered for publishing the big book o character tables instead of doing hard things

[swift prawn]

I was wondering if I could get some help part c of this. I followed the hint but I don't see the isomorphism

a is easy and b follows immediately from the first isomorphism theorem

Intuitively it's clear why c is true, since you are crushing all the p-powers of the elements of G and "setting them to the identity" so you should get something isomorphic to K I just maybe don't see where it comes from

[modest dune]

I would probably just construct a morphism to K and look at its kernel

[swift prawn]

sure

but like what map, K is a kinda tricky subgroup

probably the fact that G is finite abelian is important, so you could try and decompose it as G ~= \oplus Z_{p_i}^{a_i} (or something.. idk)

[modest dune]

g -> g^(|G|/p) maybe?

[swift prawn]

hmm

i'll try it out

[quiet pelican]

That doesn’t work for G = C_p x C_p

[swift prawn]

i found some insane solution which essentially says that G/H and K are F_p vector spaces of the same cardinality and so they are isomorphic

but i feel like that is overkill

there is an action of F_p on K for example which is just a * x = x^a and the action is well defined since all of the elements have order p

same with G/H (since all elements have order p)

[quiet pelican]

I don’t think there should be a “natural” way to do it
Because think about what your representatives look like for C_{p^n}

[swift prawn]

yea they're gross

no they have literally the same cardinality

but finite vector spaces of the same cardinality are isomorphic

"overkill" is a misnomer I guess. It just doesn't seem elementary I guess. I felt like there should be some way to solve this without appealing to that fact

[rocky cloak]

You can also appeal to the classification of finite abelian groups I guess.

[swift prawn]

can you maybe explain how? I was trying to figure this out but I couldn't see how it worked out

[elfin wraith]

That feels significantly more overkill to me but two proofs never hurts I guess lol

[swift prawn]

which feels more overkill aha

[rocky cloak]

By the classification any finite abelian group where all elements has order p is isomorphic to (Cp)^n

[swift prawn]

ah sure

fair enough

I was doing too much

[rocky cloak]

There's no canonical isomorphism so any proof must boil down to "any two groups with such and such property are isomorphic"

[swift prawn]

yea makes sense

tricky problem

[elfin wraith]

Not if you just write "Clear"

[swift prawn]

i'm sure the graders of the qualifying exams will be very content with this

[rocky cloak]

This one small trick solves any problem!
Mathematicians hate him

[mint seal]

nice problem

I wonder if it's possible to count how many such isomorphisms there are

[rocky cloak]

Should be
(p^n - 1)(p^n - p)(p^n - p^2)...(p^n - p^n-1)

[mint seal]

remarkable! how do you figure?

and what's n?

oh from the classification result

[rocky cloak]

https://groupprops.subwiki.org/wiki/Order_formulas_for_linear_groups

[mint seal]

so |K| = p^n

[thorn jay]

Size of GL(F_p, n), can be calculated using some fucky combinatorial argument and a little number theory iirc

Saw it on some stack exchange once

[rocky cloak]

Or just linear algebra:

The first column is any nonzero vector, the next any vector not in the span of the first, next any vector not in the span of the first two, etc

[mint seal]

very very nice

[thorn jay]

I dont know lol i could be wrong too

Anytime i see finite fields my brain screams number theory and stops working

[vocal pebble]

So you'd need to divide by p-1 right?

Nvm we are counting the number of elements and they give you different elements

[sacred ginkgo]

Hi! What is the notation here?

What is $S_X$? It isn't explicitly mentioned in the textbook (Rotman's Introduction to Theory of Groups), but does it just mean the family of subgroups of the set $X$?

[cloud walrus]

Redfern Station

[rapid cave]

Symmetric group over X

I.e. all bijections with composition

[sacred ginkgo]

Thank you!

[thin stream]

hello, i have maybe a silly question :)
since $\mathrm{End}_A(V)$, the group of A-module endomorphisms $V\to V$, can be given a ring structure by $(T+S)v:=Tv+Sv$ and $(T\times S)v:= (T\circ S)v$, can i give its group of units $\mathrm{End}_A(V)$ the structure of a division ring? when can i upgrade this to a field - if any two endomorphisms commute with each other? isn't this a crazy strong condition, which puts mad limits on our choice of vector space V?

[cloud walrus]

blutac

[thin stream]

my hypothesis is that this means V has to be 1-dimensional, and thus already a field.... to prove this i'm trying to construct two maps V->V which don't commute...

but i'm not sure how!

[velvet hull]

is V a vector space or an A-module?

also, you need another strong condition other than commutativity for it to be made into a field - you need the units (plus the 0 map) to be closed under addition

[thin stream]

a-module, sorry :)

[thin stream]

this is true! i just proved that not even commutativity holds tho...

$T(x_1,...,x_n):=(x_1,0,...,0)$ and $S(x_1,...,x_n):=(x_2,x_1,0,...,0)$ has $ST(x_1,...,x_n)=(0,x_1,0,...,0), \ TS(x_1,...,x_n)=(x_2,0,...,0)$

would be curious about counterexamples for units not being closed under addition though!

[cloud walrus]

blutac

[thin stream]

yk what i think the units are closed under addition only if it's a vector space...

[velvet hull]

take R^2, the identity plus the swap map because the matrix with only 1s entries

[thin stream]

those definitely commute

ohhh but they dont add

my bad hehe

[velvet hull]

actually with this explicit construction you immediately get that if V is a fd vector space it has to be dimension 1

[thin stream]

okay so i guess the only times End_A(V) is a field is when A is a field and V is one-dimensional! which means the only fields are... well, fields :)

yeah thats awesome tysm

[velvet hull]

I was looking to invoke some higher theorems to prove that but this is sufficient

[thin stream]

thanks for ur help! this is a pretty neat result

like a trash version of a structure theorem hehe

all fields that are also endomorphism rings are trivial

[velvet hull]

you can do a similar proof for infinite dimensional vector spaces using a hamel basis

[thin stream]

havent seen any infinite dimensional vector space stuff / functional analysis stuff, this idea came up when i was doing some representation theory... i might send this to a buddy who would know what to do, but for now i shall return to my revision

thanks a lot friend :D

[velvet hull]

the idea is straightforward - assuming axiom of choice every vector space admits a basis. Then pick two vectors in the basis, and consider the swap map that swaps those two vectors and fixes everything else in the basis

that plus the identity is also noninvertible

[thin stream]

makes sense

appreciate chu

[velvet hull]

if V is just a module over an arbitrary ring A, then this question becomes... quite a bit harder to answer

if A is an integral domain I suspect you can probably make similar arguments by taking the field of fractinos

[thin stream]

yeah that makes sense

[velvet hull]

if V is finitely generated over A, then you might be able to do the usual technique of localizing + nakayama's lemma to get some result

[naive lance]

i am looking for a proof of the following statement or source

every real orthononal finite dim rep of the homogeneous lorentzgroup (which is non compact) is trivial

[south patrol]

You can have other things like with some simple modules

[rocky cloak]

Modules for which End(M) is a division ring are called bricks. Famously simple modules are bricks, but there are non-simple examples.

For example let A be the ring of 2x2 upper triangular matrices (over R let's say) and let M = R^2 with the usual action. Then End_A(M) = R, but M is not simple

[near tapir]

I’m trying to prove that if R[x] is a PID, then R is a field.

I’m assuming r is a nonzero non-unit in R and considering the ideal (r, x) in R[x].
Can I say that (r, x) = (1) because x is irreducible and is of degree 1 and cannot be a factor of r because R is an integral domain. Therefore since R[x] is a PID that means (r, x) = (1) because 1 is a gcd of r and x?

[tardy hedge]

Why does that show R is a field

[near tapir]

It doesn’t but I wanted to confirm that part.

But if (r, x) = (1) then 1 = rf(x) + xg(x) and then g(x) must be zero.

[tardy hedge]

Ah because of degrees?

[near tapir]

Yes the degree of xg(x) is bigger than or equal 1 if g(x) is not zero

[tardy hedge]

Yeah

x cant be a factor of r also because degree argument no?

Why did u say because R is a domain

[vapid vale]

bricks :D

[near tapir]

Because deg f(x) + deg g(x) = deg f(x)g(x) holds in integral domains?

[tardy hedge]

Ok yeah i guess u still used a degree argument then

[near tapir]

Then the factors of r must have degree 0

[wooden rover]

Hi! Would the product or coproduct of artinian rings be as well artinian?

I’ll try to think about it in about a few minutes but for now I don’t have paper on me xd

I think a finite product could work because of the number of maximal ideals being finite match

[rocky cloak]

Finite product of artinian rings is artinian.

Coproduct need not be, not even coproduct in the category of commutative rings

[thorn jay]

A finite product of posets satisfying the descending chain condition satisfies the descending chain condition, and it so happens that the poset of ideals of R1 x ... x Rn is the product of the corresponding posets of ideals

[wooden rover]

Okok thank you

I think I could prove it in my head by induction

But I thought coprod could’ve worked

Ah ok no I think I understand why

Thanks a lot

[wooden rover]

Oh another thought: even countable coproducts don’t work ?

Ah nvm

[rocky cloak]

The coproduct of just two things doesn't work

[thorn jay]

Coproducts of algebraic structures are generally unfun

In large because construction of it doesnt exactly give a nice closed form, unlike products

[rocky cloak]

Forgetful functor and coproduct no get along

[thorn jay]

If only...

The funny thing is some varieties have coproduct naturally isomorphic to the product, some have products distributing over coproducts, and some have coproducts distributing over products

So the intermediate value theorem implies a spectrum of distributivity

I propose this to be the new millenium problem

[rocky cloak]

Just have to wait 974 years

[thorn jay]

Ah ive got time

[knotty badger]

I came across the construction of colimits using finite coproducts, coequalisers and filtered colimits recently

It’s pretty cool

[thorn jay]

Oh i see

You take all finite "subcolimits" and then the filtered colimit induced by inclusion of sets?

[knotty badger]

Yeah you can make small coproducts by using the filtered category of finite subsets of your indexing set

[thorn jay]

Thats so peak

[knotty badger]

It came up from me reading through maclane finally

I can actually say that I’ve read all of “categories for the working mathematician” now

[thorn jay]

I dont think ive ever read all of a math textbook

In part because i keep buying new ones

[knotty badger]

I had a plane ride so I had a lot of time to kill

Read all of Cheng’s “the joy of abstraction” and Awodey’s “Category Theory”

Now I’m planning to move onto Borcreaux

[thorn jay]

Checking off every category theory text lol

[knotty badger]

I guess I wanted to shore up my existing knowledge

I’ve already read through Leinster’s “basic category theory” and Riehl’s “category theory in context”

[thorn jay]

Ive bought a book that i now realize needs algebraic number theory lol

[knotty badger]

Idk what that is!

[thorn jay]

Field extensions of Q

Or smt

[knotty badger]

Oh ok

[thorn jay]

And then algebraic geometry is there too

[knotty badger]

I’ve also gotta live up to the title of “cat theory #1 fan”

[south patrol]

I.e. every characteristic 0 field

[quiet pelican]

What’s the book?

[thorn jay]

Knots and primes

Yk the primes in the name shouldve clued me in

However i was too infatuated with the knots part

[thorn jay]

Meant algebraic extensions

But that doesnt include padics

[quiet pelican]

Become algebraic number theorist 💜

[south patrol]

Ye

[terse crystal]

Can you give me an example?

[rocky cloak]

Say K = Q(sq(2), sq(3), sq(5), ...) adjoining square roots of all primes. Then K is a field so artinian, but the coproduct of K with itself in the category of commutative rings is
K(x)K = K[x2, x3, ...]/(xp^2 - p) which has an infinitely generated ideal (xp - sq(p) : p prime)

I think K(x)K should be the subring of K^N of eventually constant sequences

[terse crystal]

Oh. Thank you

What do you mean, I thought K coprod K is K[x_2,x_3,…]/(x_p ^2 -p: p)

How does it become subring of K^N… as you said

[rocky cloak]

Well notice that K[x2]/(x2^2 - 2) = K^2 and
K[x2, x3]/(...) = K^4 etc

So the colimit of these is definitely some subring of K^N. But the inclusions are a little complicated, so maybe it's not as simple as eventually constant things

[terse crystal]

Oh

[terse crystal]

Oh, sizes are different, Hom_K (K[x_p]/(x_p ^2-p), K) and Hom_K (eventually constant sequences in K^N, K). Former a homomorphism need to select sqrt(p) or -sqrt(p) for each p, cardinality of reals, latter just mapping (a_1,… constantly a,…) to a, cardinality of N

[rocky cloak]

So thinking about it properly you can see K(x)K as the colimit of
K -> K^2 -> (K^2)^2 -> ...
where each map is the diagonal map.

So the resulting colimit should be periodic sequences with period a power of 2.

[terse crystal]

Oh. Thank you. That makes sense

[terse crystal]

Wait why diagonal? a->(a,a). I mean K->K[x_2]/(x_2 ^2-2)->K^2, this must be the diagonal?

[thorn jay]

Does this hold for arbitrary rings?

[rocky cloak]

No, it's very specific to the constructed K

[thorn jay]

Alright, i wouldve been surprised lol

[rocky cloak]

I mean, I guess there is a choice of isomorphism K[x2]/... -> K^2, but the most natural choice sends K diagonally and x2 to (sq2, -sq2)

This makes it a homomorphism of K-algebras

[terse crystal]

I see. K[x]/(x^2-2) \cong K[x]/(x-sqrt(2)) prod K[x]/(x+sqrt(2)) ?

No wonder diagonal

[lethal cipher]

Okay, so I need to prove that there is no finite group G s.t. Aut(G)=Z/11Z.

What I've figured out so far:
If such a G exists, G is abelian.

I also know that |Aut(G)|>= phi(n) (numbers less than n relatively prime to n) for an abelian group

However, there's way too many cases to try to brute force it from here. So, I could use a bit of help figuring out my next steps

[velvet hull]

since G is finite abelian, the thing you should be thinking about in figuring out what approach to take is the classification theorem
hint - ||if G is the direct product of two subgroups, is there anything that can be said?||

[lethal cipher]

Are you referring to it being a direct product of cyclic groups?

I'm also with you on that front.

My biggest issue is figuring out how to deal with the automorphism group when the order of the cyclic groups is not relatively prime

[velvet hull]

hmmm, you dont need to worry about that case in this situation

[lethal cipher]

really? Why is that?

[velvet hull]

so the automorphism group might be bigger, yes, but Aut(AxB) certainly contains a copy of Aut(A) x Aut(B)

[lethal cipher]

Absolutely, and there are quite a few where |Aut(A)||Aut(B)|<11

[velvet hull]

what else do we know about 11

[lethal cipher]

For instance, if n=16

There is no way that a product of phi(n)'s could ever be 11.

That is fine

The issue I'm having is dealing with the remianing term for the order of the automorphism group

[velvet hull]

11 is prime

[lethal cipher]

I agree

[velvet hull]

but the only finite group with a trivial automorphism group is the trivial group and Z/2Z

so if Aut(A) and Aut(B) are both nontrivial, what can be said about Aut(AxB)

[lethal cipher]

Aut(AxB) is also non-trivial

[velvet hull]

more than that, |Aut(A)| divides |Aut(AxB)|

[lethal cipher]

Aut(A) is iso to a subgroup of Aut(AxB) of course

Ahhhh!

[velvet hull]

you got it

[lethal cipher]

Also, doesn't Z/2Z also have a trivial automorphism group?

[velvet hull]

oh yeah you're right that's the single edge case that needs to be dealt with

[lethal cipher]

Okay, so now we just gotta look at powers of 2

true, but what if we have mixing products? LIke (Z/2Z)^4

Oh, again....

[velvet hull]

the size of Aut(Z/2Z^n) is easy to bound below (computing its actual size is a little more subtle but thats not needed)

maybe bound isnt the right word but there is an obvious subgroup of Aut(Z/2Z^n) that you can pinpoint

[lethal cipher]

Well, the only thing an automorphism would really do is "swap" the coordinates around (i.e. phi(1,0)=(0,1), that kinda thing)

[velvet hull]

okay, so what is the size of this subgroup?

[lethal cipher]

nC2

Unless I'm crazy....because these just relate to transpositions

[velvet hull]

there might be more, there is no guarantee that you have to send a "standard basis vector" to another one - this is essentially trying to identify all linear isomorphisms of Z/2Z^n as a Z/2Z vector space

[lethal cipher]

Ah, I see how that could potentially be a problem

[velvet hull]

but no matter that is not needed

what can be said about nC2

remember that it has to divide |Aut(G)| = 11

[lethal cipher]

Unless n=2, 2 has to divide nC2

[velvet hull]

unless n=1 you mean

wait why is it nC2

it's n!

[lethal cipher]

...permutations...duh -_-

[velvet hull]

okay now finish the proof lol

[lethal cipher]

I did just say we are swapping the coordinates around

Well, hold on one sec, there are still some extra edge cases I don't think have been tracked yet: like G= Z/2Z x Z/4Z

Oh wait. disregard!

The only option we had at this point was (Z/2Z)^n since we already established |Aut(Z/4Z)| would have to divide |Aut(G)|

Great, so this is the last case!

Since for n>1, 2 divides n! (which is the order of the 'permutation group' of Aut(Z/2Z)^n), n must be 1

Thus, G=Z/2Z....but this automorphism group is trivial. So the proof is complete 😄

[velvet hull]

well I would phrase it as G = Z/nZ x (Z/2Z)^k for some n!=2

from the results we have derived

but the reasoning is loosely the same

[lethal cipher]

Wouldn't this fact be true, irrelevant to whether the Automorphism group of B is non-trivial?

[velvet hull]

yes!

[lethal cipher]

Maybe I jumped a step, but then that means that n has to be exactly 2, since that will give us the only trivial automorphism group

*trivial

[velvet hull]

since the size of aut Z/nZ can never be 11, yes

[lethal cipher]

precisely, which is very easy to check

This was incredibly helpful. Thank you so much HChan!

[velvet hull]

actually I just realised that there is a little more straightforward proof than the approach we took - any non-boolean abelian group has to have a nontrivial order 2 automorphism by taking the inverse

[amber burrow]

Question is to show for $P_i \in Syl_{p_i}(G)$ where $G$ is a finite group that if we have $H < G \implies H < N_G(H)$, we have $P_i \unlhd G$ for for every Sylow subgroup

[cloud walrus]

donut123

[amber burrow]

They say $P \unlhd N$ and so $P$ is characteristic in $N$ (because sylow), and since $N \unlhd N_G(N)$ we have $P \unlhd N_G(N)$. They say that this means $N_G(N) \leq N$. Why?

[cloud walrus]

donut123

[velvet hull]

what is N?

[amber burrow]

oh yeah, $N = N_G(P)$

[cloud walrus]

donut123

[velvet hull]

ah, yeah in that case this follows from the lemma that N_G(N_G(P)) = N_G(P) if P is a sylow p group

[amber burrow]

shouldnt $N_G(N_G(P)) = G$?

[cloud walrus]

donut123

[velvet hull]

well that is what it happens to be in this case

but that result holds in general

[amber burrow]

well im saying, like if P is anything, shouldnt taking the normalizer of a normalizer be the whole group?

[velvet hull]

you will get to the whole group eventually yes but that's not what we are looking to prove here

[amber burrow]

oh for some reason i thought the normalizer of a subgroup is normal in the whole group

[velvet hull]

because normality is not transitive

yes

[amber burrow]

but its actually the subgroup that is normal in the normalizer

wait how do we know N_G(N_G(P)) = N_G(P) is P is a sylow p group

[velvet hull]

P is characteristic in N_G(P)

if something normalizes the normaliser, then it has to induce an automorphism of N_G(P)

so it has to fix P as well

its a simple proof

useful result

[amber burrow]

this applies even if P isnt a sylow p group right?

oh wait no

[topaz solar]

A handy bit is all Sylow p-subgroups are conjugate

So if it’s normal there’s only one

[amber burrow]

this feels like a weird thing for dummit and footeto gloss over

[topaz solar]

Well, I’m not sure it does

I think that part of Sylow is stated explicitly?

[amber burrow]

oh im talking about the characteristic lemma thing

they def state the sylow conjugate thing

N_G(N_G(P)) = N_G(P) is P is a sylow p group

[topaz solar]

Well that’s kinda what this exercise is

[amber burrow]

oh its not exercise its like the proof they give of some theorem

[velvet hull]

for some extra info, groups where the normalizer of any subgroup strictly increases are called nilpotent

such groups enjoy some very nice properties

[amber burrow]

hmmm the definition they give is with upper/lower central series

[velvet hull]

in particular what you are proving now can be used to show that a group is nilpotent iff it is a direct product of p groups

[topaz solar]

Interesting characterization tho

[velvet hull]

it is an equivalent definition, which is what you are showing in part now

[amber burrow]

what is the relevancy of P being normal in N_G(N)?

[velvet hull]

N_G(N_G(P)) = N_G(P) but the normalizer of every proper subgroup in G increases

so what do you conclude

[topaz solar]

It’s a normal Sylow p group

[amber burrow]

so that tells us its the only sylow p subgroup

[topaz solar]

Yes

So that must be characteristic fr

[amber burrow]

yes

but how does that give N_G(N) <= N

[topaz solar]

If you normalize N_G, then conjugation is an automorphism of N_G

This fixes P by characteristic

So it normalizes P

[amber burrow]

N_G of what?

[topaz solar]

N_G(P) with P characteristic in N_G(P)

Normalizing N_G(P) is exactly being in N_G(N_G(P))

So N_G N_G P \leq N_G P

Sometimes you just gotta chase elements

For a nice proof anyway

[amber burrow]

im still confused 😢 conjugating by something in G \N_G(P) does not necessarily land in N_G(P), so how do we know it is automoprhism?

[topaz solar]

Well, it’s in the normalizer iff that conjugation is an automorphism of N_G(P), no?

[amber burrow]

oh ok yes makes sense

[topaz solar]

Kind of “what’s the definition”

A common algebra step

[amber burrow]

ok so what I have figured is:

by definition $P \unlhd N_G(P)$. Since $P$ is sylow, $P$ is also characteristic in $N_G(P)$. Now for any $g \in N_G(N_G(P))$, $g$ gives an automorphism on $N_G(P)$ by conjugation. Since $P$ is characteristic, this automorphism fixes $P$, so in particular $gPG^{-1} = P$ and $gN_G(P)g^{-1} = N_G(P)$. Therefore, $g \in N_G(P)$, so we have $N_G(N_G(P)) \subset N_G(P)$.

[cloud walrus]

donut123

[topaz solar]

Yeah I think that works

[desert quarry]

Hello anybody here preparing for cuet pg for mathematics

[golden quail]

Guys this is from abstract algebra course for the proof of one step subgroup test

Basically if has a^-1 for all a then its a sub group

He assumes inverse elements exists in the sub group.
And to prove that they exists the previously established fact of existancr of idenitiy elements depedent on the assumption that inverse element exists to prove that inverse elements exist

I mean isnt inverse element the condition we are looking for since its a 'test' to determine if something is a sub group

This bit assumes existance of inverse elements then uses that assumption to "prove" that inverse elements exists

This might have been a mistake while making the course

Probably a mistaken in phrasing

[quiet pelican]

It assumes inverses exist in G (which they do), and uses that (combined with the subgroup test) to deduce that they exist in H

[golden quail]

I was watching a youtube course and they didnt mentioned that they did it differently by assuming inverses exists (which is okay since thats the condition if the test) and in next state used the fact that we have identity element ' to show they exist is sub group' which dosent make much sense

[rocky cloak]

I'm not sure I'm following what you're asking, but in the picture they're showing that if H is a nonempty subset of a group such that
ab^-1 is in H whenever a and b are, then H contains the identity and is closed under inverses. And the point would be to give an alternative characterization of subgroups.

But assuming ab^-1 is in H is not the same as assuming inverses are in H. Only once we have the identity in H can we set a equal to the identity to prove that H has inverses

[golden quail]

Ya that was the source of confusion my bad

[golden quail]

If you are assuming ab^-1 exists then you are assuming b^-1 exists in sub set isnt that like assuming inverse element exists in sub set

[rocky cloak]

You're not assuming b^-1 is in H. You're just assuming ab^-1 is in H and then proving that b^-1 will also be in H

[golden quail]

But isnt ab^-1 contain b^-1 under multiplicative operation are we taking ab^-1 as a single element

Dosent it describe operation between a and b^-1 how that is possible if b^-1 in H.
Oh wait i get it okay we are assuming result of a (operation) b^-1 exists in sub group

Okay i get it now

Lol that was silly to overlook

@rocky cloak thanks

[amber burrow]

Im trying to follow this argument regarding groups of order $30$. Let $|G| = 30$, $P \in Syl_5(G)$, $Q \in Syl_3(G)$. Suppose either is normal, WLOG $P \unlhd G$. Then $Q \leq N_G(P) = G \implies PQ = QP \leq G$. What confuses me is they say "Note also that if eitehr $P$ or $Q$ is normal, then both $P$ and $Q$ are characteristic subgroups of $PQ$". $P$ is a sylow subgroup of $PQ$ so normal is equivalent to characteristic, but why is $Q$ characteristic?

[cloud walrus]

donut123

[amber burrow]

actually nvm i guess this is fairly obvious, since if $P$ is characteristic then under any automorphism $\phi \colon PH \to PH$ and $p\in P, h\in H$, we have $\phi(ph) = \phi(p)\phi(h) \in PH$. We can't have $\phi(H) \in P$ because $\phi$ has to be bijective.

[cloud walrus]

donut123

[rocky cloak]

I don't know what whoever wrote this argument had in mind, but I can't really think of a way to prove they're characteristic without first showing they are normal.

How I would do it is notice the automorphism group for P has order 4, so the only way for Q to act on P by conjugation is trivially, meaning P and Q commute and the product is direct.

In any event you'll have to use something about 3 and 5, since the same statement isn't true for 2 and 3 for example.

Could it perhaps be that they already have a classification of groups of order 15 that they are (implicitly) referencing?

[rocky cloak]

An element of PH isn't necessarily in P or H, it can be the product of two things

[amber burrow]

Yeah they already had a classification for groups of order 15

So we already know that if we have a subgroup of order 15 it is cyclic

So we are trying to make one

[rocky cloak]

Okay, well then I think they're just saying, if either P or Q is normal, then PQ is a group of order 15, so badabing badaboom

[amber burrow]

Yeah, but for some reason they want to show that if either P or Q is normal in G, both must be normal

[rocky cloak]

If either P or Q is normal, then PQ is a subgroup of order 15, so P and Q are characteristic and it has index 2 so PQ is normal.

Then characteristic subgroup of normal subgroup is normal

[amber burrow]

Yes, but how do u get that P and Q are characteristic in PQ

[rocky cloak]

Because PQ = Z/15

[amber burrow]

theres a dummit and foote execise which says "Let $G$ be a nontrivial finite abelian group of rank $t$. Prove that the rank of $G$ equals the maximum of the ranks of its Sylow subgroups". But, isn't non-zero rank only for infinite groups?

[cloud walrus]

donut123

[quiet pelican]

Rank = size of minimal generating set (which I believe is equal to the size of any “independent” generating set for finite abelian groups, but I’m not certain)

[amber burrow]

ah ok they define free rank as the exponent in the invariant factor decomposition

[copper kestrel]

abstract algebra 2 starts next monday

terrified 💔

[modest dune]

why should you be terrified of a good time

[copper kestrel]

bc im bad at proofs

but oh well

the learning is interesting

the solving is scary

[tardy hedge]

I think rings and modules are easier than groups

Assuming thats what its covering

[copper kestrel]

we are talking about rings yea :D

[tardy hedge]

Nice

[tardy hedge]

Ya cuz then u realize o shit maybe i dont know anything ☺️

😢

[copper kestrel]

exactly

[delicate orchid]

Absolute insane take btw

[rocky cloak]

Just depends on the adjectives you like.

Artinian rings and finite length modules I'd argue are easier than finite groups.

Without any restriction I guess both are just infinitely difficult

[rapid cave]

Hi, I am trying to get this isomorphism here from the universal property of tensor products, so I need to show every bilinear map
$\varphi: S^{-1}A \times M \rightarrow P$ factors through the map $\frac{a}{s}, m \mapsto \frac{am}{s}$. So I need to show the map $\psi(\frac{am}{s}) = \varphi(\frac{a}{s}, m)$ is well defined. Here I am stuck.

[cloud walrus]

ExpertEsquieESQUIE

[velvet hull]

that's hard, since you are kind of trying to brute force the isomorphism

use the UP a little differently: obtain a bilinear map from S-1A x M to S-1M, which factors uniquely through the tensor product via the UP

[rapid cave]

so a/s, m --> am/s is bilinear. and by the universal property it factors through \otimes.

[velvet hull]

yes

[rapid cave]

how does this give an isomorphism

[velvet hull]

well thats what you have to show

but this is the easy part

all we really needed the UP for was to guarantee that this map is well-defined

[rapid cave]

I think showing this map is bijective is not that hard

but is there a way to show this by giving an inverse map?

so it would be a categorical isomorphism

[velvet hull]

you can do it but it sounds a lot harder

[restive idol]

The category of R-modules is balanced, so it would be right?

[rocky cloak]

You can show that the map
S-A x M -> S-M satisfies the universal property of the tensor product (hence is the tensor product)

[rapid cave]

that is what I tried to do originally

[rapid cave]

.

[rocky cloak]

I guess write down the condition for
am/s = bn/t

[rapid cave]

ramt = rbns for some r in S

[rocky cloak]

And then you want to prove
phi(a/s, m) = phi(b/t, n)

[rapid cave]

yes

[rocky cloak]

So manipulate to get something using this equation

I don't know if I have good hints, but just step by step:

• ||st is a unit, so phi(a/s, m) = phi(b/t) iff the same is true after multiplying by st||
• ||stphi(a/s, m) = phi(at, m)||
• ||= phi(1, atm)||
• ||= phi(1/r, ratm)||
• ||= phi(1/r, rbns)||
• ||= st phi(b/t, n)||

[rapid cave]

st is a unit?????

[rocky cloak]

In S^-A yes

That's kinda the point of S^-A

[rapid cave]

I see

so all of these equalities are in S-1A

[rocky cloak]

Well they're in P which is an S^-A module

I guess you can avoid it by just introducing
st/st on the inside

[rapid cave]

ok

[rocky cloak]

phi(a/s, m) = phi(ta/st, m)

[rapid cave]

thanks @rocky cloak @velvet hull @restive idol

[little ocean]

Hi, I hope I'm not interrupting you guys, but I have a question. In my abstract algebra syllabus it is stated that a quotient of a noetherian ring is also noetherian, but with no proof. They instead gave it as an exercise. Any tips on how to handle this?

On the same note, how do I prove that a quotient of an artenian ring is also artenian?

[rocky cloak]

Do you know the correspondence theorem? I.e. what are the ideals of a quotient ring

[little ocean]

No

This one??

Sorry for the Dutch, it's literally straight out of my syllabus

[rocky cloak]

Looks like it could be, but my dutch is weak

[little ocean]

But my English math vocabulary is really bad

[rocky cloak]

Though the analogous one for rings, this seems to be about groups

[little ocean]

Let G be a group, N a normal group of G and let H = G/N. If K is a subgroup of H, then there exists a unique A (subgroup of G) that contains N such that K = A/N

I see nothing about this in my chapter about tings

[rocky cloak]

Yes, so for rings if R is a ring, I an ideal and J an ideal of R/I then it is equal to K/I for some ideal K in R

Anyway, I would try to use this to prove your exercise, and then you can maybe think about proving this / figuring out if you were already taught it afterwards

[little ocean]

I'm pretty sure I was not taught this theorem for rings

And how exactly would I use this to prove the exercise?

Ok how about this

[thorn jay]

Oh cool another dutchie

[little ocean]

Belgian, ectually

[twilit wraith]

This is third isomorphism theorem right

[thorn jay]

Fourth

[twilit wraith]

Ah close

[thorn jay]

Third for me is (A/R)/(S/R) ≈ A/S naturally for R < S

[little ocean]

So if R is noetherian, then if there exists a chain of ideals $I_1 \subseteq I_2 \subseteq \dots $, then there exists a $k \in N$ such that $I_l = I_k$ for all $l > k$

[cloud walrus]

Nico

[little ocean]

How about I just look at all these ideals, but then in R/I?

[thorn jay]

That is the idea

Correspondence theorem says that the ideal poset of R/I is nothing more than an "upper chunk" of the ideal poset of R

[little ocean]

What

[thorn jay]

Poset of ideals under inclusion?