groups-rings-fields

[prime sigil]

I tried to do it like this, but got stuck so I went another way

[tribal moss]

Unfortunately that is not a homomorphism.

Since we're trying to make i=2, a natural choice to try would be a+bi -> (a+2b) mod 5, though.

[velvet hull]

(it doesn't work, but might there be a way to make it work?)

[coral spindle]

To define a ring homomorphism out of Z[i], say a map f : Z[i] → R for some ring R we need f(i)^2 = f(i^2) = f(-1) = -1_R. So you always want to send f(i) to something which looks like a square root of -1 in R.

This is why we know that whatever homomorphism you choose needs to have f(i) = 2 or 3

This is a quick way to greatly reduce the number of options you're looking at when you're looking for homomorphisms that might work

[prime sigil]

Forgot to check it :(

But I think I understand it now

Thanks for the help

[tribal moss]

Of course, then you do need to show that the kernel of your morphism is indeed (2-i).

[unreal sedge]

Does anyone know of a good youtube channel that provides algebra lectures?

I've seen that MiT has analysis lectures but not so much algebra I think

[stark mist]

There's Nathan Kaplan from UCI for Dummit/Foote and Benedict Gross from Harvard for Artin.

[fervent solstice]

This is probably dumb but how do you prove that if P(a_1,...,a_n)=0 then P is in (X_1-a_1,...,X_n-a_n)? Sources mention the Nullstellensatz but I need this for proving the Nullstellensatz

[velvet hull]

consider the evaluation homomorphism in k[x1,...,xn] that sends f(x1, ..., xn) to f(a1,...,an). its kernel is exactly x1-a1, ..., xn-an.

[narrow temple]

I think this boils down to showing that the kernel of the "evaluation at (a1,...,an) homomorphism" k[x1,...,xn] -> k given by x_i -> a_i is given by (x1-a1,...xn-an)

[fervent solstice]

yeah i was boiling it down the other way around so uh

[narrow temple]

In the spoiler is a small hint on how to approach it. I have done the same exercise a few days ago and that hint was given in the book.

[velvet hull]

oh yeah, that's another way of doing it

[fervent solstice]

ok, got it, thanks.

[fervent solstice]

how would this work?

like how do you show that's the kernel without using the result we're trying to prove

[velvet hull]

well, notice that (x1-a1, ..., xn-an) is a maximal ideal as the quotient by it is k, a field

but then it must be contained inside the kernel, therefore ????

[narrow temple]

lol, i totally missed that

[fervent solstice]

oh nice

thank you both

[humble dirge]

I know it's a dumb question to ask. But I am not able to access the website for now..
I just want a book for ring theory from where I can review , pid,ufd, euclidean , dedekind ...

[elfin wraith]

Reid’s undergrad commutative algebra

I don’t remember how much Dummit and Foote covers but I imagine at least some of that’s in there

It’s also possibly in Atiyah Macdonald but I don’t remember exactly how much ring theory they assume off the bat because I know it’s at least some

[humble dirge]

Can you share the pdf or any source where can I get. As I know only libgen and it's not working properly

[elfin wraith]

I am not allowed to endorse piracy here

[humble dirge]

Okay thanks!!!

[swift tundra]

Personally I am a big Richard Borcherds fan. I think he has some lectures in group theory, but it is at the graduate level.

[velvet hull]

I get a feeling that's not what they meant by "algebra lectures"

[tardy hedge]

I never liked his videos

[swift tundra]

Yeah fair they are a bit niche. I just like the old school paper vibe. And I like how he tries to give historical background for major theorems.

[tardy hedge]

yea, i just think theyre hard to learn from

but i appreciate the vibe

[swift tundra]

Yeah you’re right, I kinda just wanted to plug him because I have found his channel helpful sometimes

[velvet hull]

fair enough

[swift tundra]

Yeah totally. I never rely just on the videos, I like to follow a textbook as well. But idk if there is any channel (that I know of) where the channel itself is sufficient to learn higher math.

[tardy hedge]

ya true

DanielChenMaths is pretty nice

he has lots of super advanced topics presented in a very organized and clear way

[south patrol]

I love this guy

[tardy hedge]

ikr

another adventure in pure mathematics

🔥

[south patrol]

Do you mean Chan tho

[tardy hedge]

Probably

[south patrol]

Then yeay

I appreciate a youtuber doing SERIOUS algebraic geometry

Rather than idk random integrals

[velvet hull]

did you just have a stroke

[tardy hedge]

Lol

[white oxide]

Could I please have a hint on the first part of exercise 15? I've been considering the finite case, say (m_1, 0, 0, m_4, 0, 0, m_7, 0) \in M but I'm struggling

My issue is that I'm having trouble seeing what elements $\mu_1(m_1)$ can relate to, for instance. In the quotient we have $\mu_1(m_1) = \mu_{1j}(m_1)$ for all $j \geq 1$ but this doesn't really tell me anything, since I can't relate it to $m_4, m_7$, say (I hope that makes sense)

[cloud walrus]

okeyokay

[velvet hull]

you can't relate x_i to x_j inside M, sure, but mu_ij (x_i) is inside M_j when you consider x_i as an element of M_i

[white oxide]

Yeah, but all we get from that is that x_i = mu_ij(x_i) in the quotient right which also doesn't relate to x_j

unless we can show that mu_ij(x_i) = x_j or smt

or maybe I'm confused by what you said

[velvet hull]

no no no, x_i = mu_ij(x_i) inside M

but forget about M, x_i projects onto some element x_i inside M_i

and mu_ij does send x_i (inside M_i) to some x_j (inside M_j)

[white oxide]

ye I was talking about in M

[white oxide]

how though? this isn't clear to me, maybe I need to think more

[velvet hull]

it's just because mu_ij is a homomorphism from M_i to M_j

by definition

[white oxide]

oh no I meant to the given x_j

what I'm doing is I'm considering an arbitrary element of M, it has a finite number of nonzero coordinates x_j

I'm trying to find a x_i such that u_i(x_i) relates to those coordinates

I understand that mu_ij is a map from M_i to M_j (really the images of M_i and M_j under the injection, but whatever)

wait really?

I was thinking about representatives of the quotient groups, since those are by definition elements of C which have finite nonzero coordinates that was what I concluded

or maybe I'm confused

sorry I don't follow

I know that we can sort of pass through the m_i

what I mean by that is say we start with m_1 \in M_1, then m_1 = mu_{1, 2}(m_1) = some element of M_2 for instance, like you said

but we can't choose this element of M_2

similarly we can go from that element of M_2 to any element of M_i, i >= 2, but again we have no choice

wait...

[velvet hull]

ignore what I said, it is true, I was thinking about a different construction

[white oxide]

all good

anyways yeah I'm just stuck on what I wrote above

[velvet hull]

is I countable here

or is it some arbitrary indexing set

[white oxide]

arbitrary directed set

[velvet hull]

what I meant to say is, it it totally ordered

not countable

[white oxide]

No, all they say is that I is partially ordered and is a directed set

I assume that doesn't imply totally ordered?

maybe the second assertion is easier to prove

[velvet hull]

so here's my intuition about the direct limit:
Think about the directed system as a diagram. Then, if I pick out any element x_i inside some arbitrary M_i, one thing I can do with x_i is I can "chase" x_i downstream into the M_j's where I have a morphism from M_i to M_j.
Then when we impose that specific relation on C to get M, what we are essentially saying is that we declare two elements x_i and x_j (which may or may not be in the same module) to be the same, if their streams eventually join up and become the same

in other words, M is the module consisting of all distinct streams you can trace inside the directed system, where addition and multiplication is defined pointwise

[white oxide]

what we are essentially saying is that we declare two elements x_i and x_j to be the same
is this the correct picture to have in mind?

[velvet hull]

I don't understand what the picture is doing

[white oxide]

well we have some element say x_1 in M_1, and we can keep on passing to different M_i by the mu_ij

say we stop at some M_k, then we can take this as our x_k (sorry it should be x_k instead of x_j)

then by the definition of the equivalence relation, x_j = x_i in M

[velvet hull]

by the way, this gives a really intuitve answer for the second part:
under this interpretation, what it means for an element (stream) to be the same as 0 (the 0 stream), is that that element, if you chase it downstream far enough, eventually meets up with the 0 stream (i.e. exists j >= i such that mu_ij(x_i) = 0)

[white oxide]

yes I guess that makes sense, but it involves showing that the element actually gets to zero in the first place which I feel like is most of the hard work

[velvet hull]

the description for what happens for the first part is similar, but you are chasing upstream instead of downstream

[white oxide]

and anyways all we do is chase a single element down, and it's mapped to random elements, as I've said

we need to get it to map to given elements

[velvet hull]

no, so let's take our I to be N to make things simple:

if I say that x1 in M1 is the same as, say, x3 in M3, that does not mean that mu_13(x1) has to be equal to x3

[white oxide]

oh I was just giving mu_13(x_1) a name

i decided to name mu_1k(x_1) x_k

i guess it was a bad choice of name lol

but that was my interpretation of what you first said

[velvet hull]

the point is that we aren't just looking at a single element

if I say that x1 in M1 is the same as, say, x3 in M3, that does not mean that mu_13(x1) has to be equal to x3, but they eventually meet up at some M_k

if we just look at what happens to a single element then that's really boring

[white oxide]

I see, so we have mu_13(x_1) = x_1, and eventually mu_13(x1) becomes x3 after more applications of mu is what you're saying?

[velvet hull]

yes, exactly

that is what the direct limit consists of

[white oxide]

Okay now I need to understand why that's true lol

so this is for any arbitrary x3 right

[velvet hull]

well, let's say they are the same in M_k

then in M, x_3 == mu_{3k}(x_3) == mu_{1k}(x_1) == x_1

and that's exactly what the ideal generated by x_i - mu_ij(x_i) is trying to do

it unifies the streams

[white oxide]

but why is mu_{3k}(x_3) == mu_{1k}(x_1)?

like how do we know they become the same in some M_k

[velvet hull]

that's the assumption

right now I'm showing you that if two streams meet up, then they are the same modulo the submodule D

in other words I'm showing you why if x1 in M1 and x3 in M3 eventually meet up at some M_k, then x1 == x3 in M

that make sense?

[white oxide]

sorry I have to go right now but thanks for the help, I'll read over this when I hav etime

[nimble folio]

I'm trying to prove that if $G$ is a group, $a \in G$, and $|a| = n$, then $|a^k| = \frac{n}{\gcd(k,n)}$. I was able to prove that $\left( a^k \right)^\frac{n}{\gcd(k,n)} = e$, and now I'm trying to show that $\frac{n}{\gcd(k,n)}$ is the smallest positive integer which this is true. \nl
        
        \noindent Most arguments online prove this by showing if $m > 0$ and $a^m = e$, then $\frac{n}{\gcd(k,n)} \mid m$. However, is there a way to prove this by showing if $0 < m < \frac{n}{\gcd(k,n)}$, then $a^m \neq e$? All my attempts at this have failed.

[cloud walrus]

clubsoda14

[nimble folio]

I thought I figured it out here

[tall igloo]

(note that this is just a modular arithmetic question: what is the smallest m such that mk is a multiple of n)

[nimble folio]

Uhh

I’m very confused ngl

[tall igloo]

well, what does it mean for a^k to have order m? it means (a^k)^n = e. that is, a^{kn} = e. but we know precisely which powers of a are e -- they are multiples of the order of a

[swift tundra]

I’ll have to check this channel out!!

[tardy hedge]

Ya he’s sweet

[unreal sedge]

FOR b: Let $e_r$ denote the right identity element that satisfies $m \ast e_r = m$. Then, by substitution, for the binary operation in (b), we have $m^2 e_r = m$. Rearranging gives us $e_r = 1/m \notin \mathbb{Z}$. Thus, there is no right identity element for the binary operation $m \ast n = m^2n$

[cloud walrus]

whyhello

[unreal sedge]

Can someone check my work?

Idk if this is the right channel for binary operations but this is in my algebra book im studying lol

[coral spindle]

This is an appropriate place to post this question, I would say, yes.

[coral spindle]

This is not really a sufficient justification.

You have just said "Thus, there is not right identity..." but you haven't really explained why you've been able to conclude that

Besides, you say that 1/m is not in Z, but this isn't always true, for example for m=1.

To actually prove what you want, you should set m to a specific number, and conclude something.

Let me demonstrate:

Suppose $e \in \mathbb N$ is some right identity for the operation. Then by definition of a right identity we would require $2 \ast e = 2$. By the definition of the binary operation, this would mean that $2^2e = 2$, and dividing by $4$ we conclude $e = 1/2$, but since $1/2 \notin \mathbb N$, this is absurd, and we conclude that there is no right identity for this operation.

[cloud walrus]

Boytjie

[unreal sedge]

Wouldn't doing this not be a sufficient justification? I mean it only provides justification for a a single case.

Would instead specifying that $e_r = 1/m \notin \mathbb{Z}$ if and only if $m \neq 0, 1$ be more general?

[cloud walrus]

whyhello

[unreal sedge]

Then we would include that a right identity doesnt exist under those conditions

I'm not sure though, I'm only an amateur lol

[coral spindle]

No, it is sufficient justification

[unreal sedge]

sorry i meant to say single case

[coral spindle]

The property of an identity is of the form "for all x, x.e = x"

When we want to disprove a proposition of the form "for all x, something" we need to find a counterexample

That is, we need to show that there exists an x such that the inner thing does not hold.

Finding a single counterexample, in particular setting m=2, was sufficient for this reason.

Your proof actually didn't mention a single counterexample, which is why it was insufficient.

[unreal sedge]

ahhhh I see

I think my issue was that I was trying to determine if there was an identity or not without any sort of mind with proof and logic lol (an issue that I just recognized a while back that im trying to fix)

The question just said to "determine" so I think I just threw that out the roof for that part of the question

Thanks for the help

[cedar oracle]

who can tell me about ECC on ring field?

[flat tree]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[vast verge]

Since I use this channel semi-regularly, would it be worth making a thread where I can ask questions and not clog up the channel?

[tardy hedge]

I mean the point of this channel is to ask questions and discuss algebra so by posting questions you are just using this channel as intended lol

[white oxide]

yeah I guess I'm trying to understand how to prove this assumption

[vast verge]

How do I know that this is the case?

[unborn terrace]

Hint: what can you say about normal subgroups in an abelian group?

[vast verge]

I forgot, sorry 😅

I really should get to making more anki cards

[crystal vale]

so i am proving that quotient of solvable group is solvable say G is solvable group and N is normal subgroup of G.

So i have to show G/N is solvable.
say, 1 = H_0 <= H1 <= ...<= H_n = G is given chain such that H_i normal in H_{i+1} and H_{i+1}/ H_{i} is abelian,

Now for G/N, what choice of K_i we have such that {N} = K_0/N <= K1/N<=....<= K_n/N= G/N.

so i am thinking of taking K_i = subgroup generated by {H_i u N }. is that correct way?

[ivory grotto]

hello, i wasnt sure where to ask this question but this is prolly a fine place for it. Can someone give me an example of an isomorphism? And whether the properties are the same, or like, just the shape.

[fervent solstice]

turns out that this was dumb, since any ideal that's not principal in k[X,Y] is just going to generate a point and the intersection between a point and anything is just dumb

[fervent solstice]

(Z/nZ,+) being isomorphic to ({e^(2i pi k/n), k in {0,...,n-1}},×) through the isomorphism f: k |-> e^(2 i pi k/n)

[ivory grotto]

this looks like alien language to me lol. could you write it in layman's terms if possible? sorry

[fervent solstice]

Z is the set of integers
nZ is the set of integers multiple of n
You can define an equivalence relation between any two integers i and j, say we write it i<->j whenever i-j is in nZ, i.e. whenever i and j are congruent mod n.
Z/nZ is the resulting quotient set, i.e. the set of equivalence classes. It can be represented as the set {0,...,n-1} because everything is congruent to one of these integers mod n.
(Z/nZ,+) is a way to write "the group over the set Z/nZ imbued with the operation +"

Is there anything still unclear in my previous answer or in this one?

[ivory grotto]

what is a quotient set? and the set of equivalence classes?

[fervent solstice]

when you have an equivalence relation such as the one I defined, you get equivalence classes: two integers are in the same equivalence class if and only if they are equivalent through that relation. In my example, whenever two integers are congruent to each other mod n, they are in the same equivalence class. If they're not congruent mod n then they're not in the same equivalence class.

A "quotient set" is just the set of equivalence classes of a given equivalence relation.

[ivory grotto]

ahhh I see. that actually made sense

[fervent solstice]

If I'm to find an example that you might get more easily, what have you seen already? Vector spaces? Rings, fields?

[ivory grotto]

I just dont understand, where are the two isomorphic sets?

[ivory grotto]

vector spaces

[fervent solstice]

ok so as vector spaces over the field of reals R, the vector spaces C and R^2 are isomorphic, through the isomorphism f : (x,y) |-> x+iy from R^2 to C

In my former example, the isomorphic structures were groups, not vector spaces. They were the groups (Z/nZ,+) and another group, that one over the set {e^(2 i pi k/n), k in {0,...,n-1}} imbued with multiplication

[ivory grotto]

I see. In this case, why can't we say that C and R^2 are isomorphic, through the isomorphism; f: x+iy |-> (x,y) from C to R^2?

[fervent solstice]

You can say that too

I just consider it a bit less rigorous to write it like that, as a function of not a complex number z but of its "unpacked" algebraic form x+iy, since writing it like that already assumes that you can uniquely decompose any complex number into its real and imaginary parts, which is kind of the very point of writing the isomorphism you're already writing (maybe what I'm saying here is a bit confusing but it's not that important)

but f: z |-> (Re(z), Im(z)) is perfectly fine

Moreover it is the inverse isomorphism to the one I told you about

[ivory grotto]

its not that confusing, i think i get it. i think you mean that the complex numbers were defined as an isomorphism of R^2, rather than the inverse

[fervent solstice]

"a vector space isomorphic to R^2" would be the correct terminology, instead of "an isomorphism of R^2". The "isomorphism" here is the function that transforms elements of R^2 into elements of C in a one-to-one fashion (or the other way around, since by definition any isomorphism has an inverse isomorphism)

[ivory grotto]

ohhhhh i didnt know isomorphism is a function

[ivory grotto]

when you say one-to-one, is it tantamount to ordered pairs? or am I far off?

[ivory grotto]

i thought it was like a comparison to show certain similarities between sets and stuff

[fervent solstice]

It's not related to anything about order or pairs, it's just that every element of R^2 is sent to exactly one element of C, and every element of C is the image of exactly one element of R^2. That's what I meant by "one-to-one"

[fervent solstice]

Strictly speaking isomorphisms are functions, but more loosely speaking one can show evidence of there being an isomorphism between two sets by showing how things from one set would be transformed into stuff from the other set through that function

[ivory grotto]

alright. thank you for taking your time, now it's become wayyyy clearer because i kept seeing the term and had no clue as to what it meant lol

[fervent solstice]

well when used loosely it means "these two structures are actually the same thing, here's how one thing in the first one corresponds to a thing in the second one"

[ivory grotto]

hmm, gonna jump the gun here or something. Does that mean that e^iθ and the matrix:
[cosθ -sinθ]
[sinθ cosθ]
are connected through an isomorphism, or is that too crazy?

[fervent solstice]

that is perfectly correct actually, the group of the rotation matrices imbued with multiplication and the group of unit complexes (the elements of C of the form e^(iθ)), also imbued with multiplication, are isomorphic

[white oxide]

are you reading fraleigh? it might be useful to read the introduction, where he displays tables for binary operations of small groups, and then you will get a concrete grasp of what it means for two structures to be equal. all an isomorphism is doing is replacing the names of the elements in one table with the names of elements in the other table

[ivory grotto]

I see. so we basically connected two different subjects of maths in this way. thats cool

[ivory grotto]

I'm not really reading anything, I just search stuff abt different subjects. And the terms that I don't understand, I come here to ask about

[tardy hedge]

Ya true i kind of thought of it in that way too at first. It is cool

A lot of more advanced math is like doing that kind of thing

[ivory grotto]

I mean if we could create bridges along every math subject, then a lot of problems could be thought over in a wayyyyy different point of view. Idk how advanced maths works, but if I'm right abt that I'll seem smart

[tardy hedge]

Ya thats how it is basically

There are always like so many different points of view of the same thing ur studying

[crystal vale]

is there any group such that [G:G] = G, [G:G], denotes character subgroup?

so non commutative simple group

A5

is there any group such that it is not simple but [G:G] = G?

[tidal schooner]

I'm guessing from the notation that "character subgroup" means the commutator/derived subgroup? If so, then A5 × A5 works. More generally, the class of groups with this property is called perfect groups

[crystal vale]

so to prove every finite group of order >1 has composition series, we will do it by induction.

For | G | = 2, it is clear, assume it is hold for all k< n.

Now Let |G | = n, if G is simple we are done.

Now, assume G is not implies there exists normal subgroup of G such that 1 < N < G. Now the family of proper normal subgroup of G is non-empty, and we can now use Zorn's lemma, also since G is finite group, so if we take any increasing chain from that family then at some point it will terminate.

Now, it implies there is maximal normal subgroup N so |N| < |G|.

Now we can use composition series of N and since G/N is simple we will have composition series of G.

is it correct?

[worthy solar]

can someone tell me the correct phrasing of this question.

If A is a field and B is ring and you construct a ring isomorphism between A and B then B must also be a field.
context for the question: https://en.wikipedia.org/wiki/Complex_number#Abstract_algebraic_aspects

Basically I just want to prove what wiki page seems to be implying

[tidal schooner]

Yes, if two rings are isomorphic and one of them is a field, then they're both fields

[crystal vale]

thank you, got it

[worthy solar]

I guess I need to try that as an exercise. Thanks. There isnt anything special about the type of "ring". Like unital or non unital, correct?

(tbh my definitions ae kinda flacky so I might be derping and saying something wrong)

[tidal schooner]

I mean, if a ring is isomorphic to a field, then it'll necessarily be unital

[worthy solar]

Oh because both have to :behave" the same under "multiplication"

because of the ring iso

identity mapping to identity kind of thing I suppose? (feel like this was some property I have forgotten))

[south patrol]

At least in algebra when one says ring they usually mean it has identity anyway, and ring maps send 1 to 1 by definitiom

And indeed this example fits into that anyway

[worthy solar]

Yes I knew the example did, I just didnt know if the question in "complete generality" would work or if there was some dumb edge case.

But yes thanks 🙇‍♂️

[vast verge]

Can someone explain where this comes from?

[crystal vale]

anyone?

[south patrol]

Looks good, but Zorn's lemma is very unnecessary. The sets are finite, so a maximal subgroup must exist (pick the biggest proper subgroup)

[tidal schooner]

It's a general fact that if $H$ and $K$ are subgroups of a group $G$, then $HK$ is also a subgroup of $G$ if (and only if) $HK=KH$. There isn't enough context in this screenshot for me to tell what groups you're working with, but it appears that $H_1 H_2=H_2 H_1$ in this case, so the general fact applies

[cloud walrus]

harmacist

[crystal vale]

Yes, i mentioned that increasing chain will terminate

Thank you for verifying

[south patrol]

Eh it will be introduced where needed, like some topics commutative algebra or Galois theory or even graph theory lol may need it, or you may just be taught it in set theory

[tardy hedge]

i think first time i saw it was the proof of every unital ring has some maximal ideal

i think i also saw it for proof that every module can be embedded into an injective module

i dont remember the proof line by line lol

i just remember it relied on zorn

but also that proof wasnt that involved i dont think

[south patrol]

Ye it is quite a standard application

[south patrol]

O hm

It is used for proving Baer's criterion

And then yeah ig one can prove the theorem for modules by using injectivity of Q and Q/Z

[cerulean quest]

What is field modulo an element

Like if you have a field and you modulo what is it

Like

F is a field

x is element of F

what is F modulo x?

[south patrol]

I mean usually modding out by an element means you consider the ideal generated by that element and form the quotient

But in a field you get F/xF = 0 unless x = 0

[cerulean quest]

what about like

R[x] modulo x^2 + 1

[south patrol]

R[x] is not a field tho

But try using first iso theorem

[cerulean quest]

Oh

It's ahlfor problem I dont know algebra but I think I dont need to know it very well

[south patrol]

I mean as theorems go this is a really shott proof

[tribal moss]

Good thing too, because fields have too few ideals for quotients to be interesting.

[south patrol]

But yeah ig peoole have had hundreds of years

[south patrol]

See previous message aha

[tribal moss]

Ah.

[south patrol]

Well it is first iso theorem but if you dont know about the algebra i would not worry

Should not be relevant to any complex analysis in Alfohrs

[cerulean quest]

Oh ok thank u

[south patrol]

Only iso theorem i remember the name of

Lol

Ye

[tribal moss]

Do I need to drop https://en.wikipedia.org/wiki/Isomorphism_theorems#Note_on_numbers_and_names again?

[south patrol]

Yeah it is a mess

[minor fulcrum]

I hate this so much

[south patrol]

Makes me feel better about this lol

At least they are so well-known it is almost ok

But yeah

Also weird to have numbering for theorems like this lol

[crystal vale]

\begin{document}

\begin{Theorem}
Let $G$ be a finite group and $H \trianglelefteq G$. Prove that there is a composition series of $G$, one of whose terms is $H$.

\vspace{0.5cm}
\textbf{Proof}: Since we proved that every finite group $G$ has a composition series, there is a composition series for $H$:
[
{1} = H_1 \trianglelefteq H_2 \trianglelefteq \ldots \trianglelefteq H_m = H.
]
such that $H_{i+1}/H_i$ is simple group.
\vspace{0.5cm}
Similarly, for $G/H$, there is composition series:
[
{H} \trianglelefteq \overline{K_1}\ldots \trianglelefteq \overline{K_m} = G/H.
]
such that $\overline{K_{i+1}}/\overline{K_i}$ is simple.
\vspace{0.5cm}
Now define $K_i = \phi^{-1}(\overline{K_i}$), where $\phi:G\rightarrow G/H$ natural mapping, then $H\leq K_i$.
\vspace{0.5cm}
Also it is easy to check, $K_i\trianglelefteq K_{i+1}$, and $\overline{K_i} = \phi^{-1}(\overline{K_i})/H $.
\vspace{0.5cm}
So, $K_{i+1}/K_i$ is simple group.
\vspace{0.5cm}
Hence,
[
{1} = H_1\trianglelefteq \ldots H = K_1 \trianglelefteq \dots \trianglelefteq K_m = G
]
is compostion series of $G$, contains $H$ as term.
\end{Theorem}

\end{document}

[cloud walrus]

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

[crystal vale]

is it correct?

[untold torrent]

My answer to question number 3 is 3

Is it wrong?

[crystal vale]

No, take H = {1,0} and K = {0,2} of Z

[untold torrent]

{0} is a subgroup no?

[crystal vale]

It is

[untold torrent]

Here intersection would be 0 no?

[crystal vale]

According to your answer, if H intersection K is a subgroup then HK is a subgroup, which is not true

[untold torrent]

HK is subgroup if it is commutative

HK=KH@crystal vale

[crystal vale]

but i think you are confused with if word

the answer should be d

[untold torrent]

No no i am saying HK=KH then i can say HK is subgroup of G

[crystal vale]

but for this you need H and K to be subgroup

HK is subgroup iff H and K are subgroup and HK=KH

[untold torrent]

If G is abelian then H is abelian?

[crystal vale]

yes

[crystal vale]

but here H and K are only non empty subset not subgroup

[untold torrent]

Oppps

[untold torrent]

I haven't read such terms yet

[untold torrent]

Could you explain it a little more

[crystal vale]

no no, its not for you

[untold torrent]

So you took two subsets of group z+

[crystal vale]

here HK = H + K, because operation is addition

[crystal vale]

yes

[velvet hull]

by the previous question, it's enough to prove that you can extend G/ G' into a compo series

but G is a finite group, so G/G' is a finite abelian group

[untold torrent]

H+K would be {0,1,2}?

[crystal vale]

no

[crystal vale]

actually in Dummit they not use derived series for solvable group

[velvet hull]

huh, whats your definition

[crystal vale]

mentioned here, a subnormal series such that factors are abelian

[velvet hull]

oh, it's still the same proof

[crystal vale]

yes

but i used different way

can you verify it?

[crystal vale]

also here i proved 7

[velvet hull]

it looks correct, yes

just an application of correspondence theorem

[crystal vale]

which one 7?

[velvet hull]

7

[crystal vale]

it is an idea

[velvet hull]

right, so it suffices to show that a finite abelian group has a compo series

that's not too hard to prove

[crystal vale]

yeah

[velvet hull]

and what can the compo factors of an abelian group be?

[crystal vale]

i proved that every finite group has comp series

[crystal vale]

abelian and simple so cyclic

[velvet hull]

you don't need it for those questions, but you should also prove that any two compo series of a group is unique up to rearranging of the factors when you have time

[crystal vale]

yes it will comes in exercise later

it is follow up exercise

[crystal vale]

so here in ii), i also proved that H_{i+1}/H_i is simple so it will has prime order, so to prove iii), it will follow from Jordan Holder theorem, but can it be done without Jordan Holder Theorem

[velvet hull]

you mean is simple and abelian?

[crystal vale]

yes

[crystal vale]

so it cannot be prove without Jordan Holder theorem?

[velvet hull]

but you haven't proven the jordan holder theorem yet, right?

[crystal vale]

yes

[rocky cloak]

I mean if you have some composition factor K1/K0
1 < K0 < K1 < G
you can just intersect it with this H-sequence to show it is solvable

[crystal vale]

can you elaborate it?

[rocky cloak]

Intersect K1 with the H sequence.

Then you get a sequence of normal subgroups in K1, such that K1 is solvable.

So [K1: K1] is not K1 -> K1/K0 abelian

[crystal vale]

actually i am not working with derived series

in dummit they define G is solvable if there is subnormal chain such that factors are abelian

[rocky cloak]

Yeah, that would be the H-sequence from (ii)

[crystal vale]

yes, all subgroup of G is solvable

[rocky cloak]

And also all quotients, so K1/K0 is solvable, but also simple

[crystal vale]

yes k1/K0 is solvable but how simple?

oh yeah it simple

we assumed

so K1/K0 simple and solvable imply K1/K0 abelian?

yes

got it

so there is based on fact that every subgroup and every quotient of group is solvable

thank you Jagr, Hchan

[untold torrent]

Question number 3

How can I discard option (3)

I took z6 which is abelian group

H={0,2}

K={0,3,4}

H+K={0,2,3,4,5} which is not a subgroup of z_6

H intersection K is {0} which is subgroup

Is this the correct example here

[crystal vale]

any hint? please don't use derived series idea

[candid patrol]

Let $k$ be the minimal integer such that $G^{(k)} \cap H \neq 1$ and define $A = G^{(k)} \cap H$, it should works

[cloud walrus]

UGOBEL

[crystal vale]

Dummit didn't introduce the G^(k) notion yet

[candid patrol]

Of course you did lol

G^(k) = D(D^(k-1)(G))

[crystal vale]

Is there no other way?

[rocky cloak]

Just replace G^(k) with whichever sequence you have in your definition of solvable

[crystal vale]

okay

[crystal vale]

say 1 <= K1 <= K2 <=...<= Kn = G be a solvable series such that each K_i is normal in G, then do intersection with H we get solvable series of H, and for minimum i, such that K_i intersection H \ = {1} is abelian because series is solvable and it is easy that K_i intersection H is normal , right?

[weary frost]

The only thing that I hang up on is that this abelian subgroup is normal

[crystal vale]

Because H is normal and K is normal therefore H intersection K normal

[white oxide]

Is there any reason why we’d care to compute the fundamental group of Spec R where R is a commutative ring

[south patrol]

What do you mean by fundamental group?

[velvet hull]

if I would guess, zariski topology on specR

[south patrol]

What i mean is there is the étale fundamental group

If you do the Zariski topology then Spec R is very often contractible

[white oxide]

Fascinating

[south patrol]

Les points génériques

[velvet hull]

les misérables --> me studying hartshrone

[thorn jay]

i am interested as well, why would we care about the étale fundamental group?

what does it tell us

[south patrol]

A nice example is that if K is a field then the étale fundamental group of Spec K is the (absolute) Galois group of K. So this is super cool

And also the definition is very analogous to that of the usual fundamental group of a topological space, and in fact there are nice comparison results between the étale fundamental group of a scheme and the fundamental group of the corresponding complex manifold (when such a thing makes sense)

[thorn jay]

fundamental group of the one point space is nontrivial

yes yes i know we are taking the affine scheme into consideration

[thorn jay]

that is awesome!

[tribal moss]

Can a definition of this "etale fundamental group" fit in a chat post?

[thorn jay]

maybe if you have nitro

lol

[south patrol]

It is a bit complicated, but to get a feel for it, do you know about how covering spaces relate to the fundamental group?

[tribal moss]

Um.

I'd guess the fundamental group must be the fiber of a universal covering space, but that’s mostly vibe.

[vast verge]

Can someone explain why this isn't working?

[south patrol]

Yes this is one thing. Like the point is that (pointed) coverings of a space correpond to subgroups of the fundamental group, which is basically encoded by the fact that if E -> X is a cover then you get an action of the fundamental group on the fibres

And yeah the group acts freely and transitively on the fibres of the universal cover

[tidal schooner]

You've got the wrong formula there. In your notation, try defining $\phi(a):=aga^{-1}$ instead

[cloud walrus]

harmacist

[south patrol]

Anyway, the idea is to replace the idea of a covering space with something more appropriate to algebraic geometry

[tribal moss]

Just to check my understanding, should that be normal subgroups?

[south patrol]

These are the étale maps (or finite étale maps are more important here, which should be like finite covers)

[south patrol]

There is a correspondence for all subgroups, and the normal subgroups correspond to so-called normal covering spaces

Which are when the action on fibres is transitive

[tribal moss]

Ok, will have to read up on that, I suppose.

[tidal schooner]

Section 1.3 of Hatcher's Algebraic Topology has all the relevant results

[tribal moss]

"Etale" is a scary word. Like a warning sign: "This is such a weird unintuitive concept that it cannot even be spoken about in English".

[south patrol]

But yeah here the picture is probably simplest in the case of fields. If k is a field, then the "finite connected coverings" of Spec k are of the form Spec K where K/k is a finite separable extension

And indeed the map k -> K (dual to Spec K -> Spec k) has a sort of covering vibe in that there is an action of Gal(K/k) on K which fixes k

(Indeed this is basically the definition of Galois group)

[tribal moss]

(And if I've understood correctly, it means completely unrelated things depending on where you put the diacritics).

[south patrol]

Yeah, étale and étalé i think.

Though the former is the more commonly used one (by far, at least for me)

[vast verge]

Can someone explain this part?

[long geyser]

what are you struggling with?

that aHa^{-1} is a subgroup of order 7 or that it equals H?

[vast verge]

Hmm okay

Maybe it's another property of cosets that I forgot

[delicate orchid]

nothing to do with cosets

if you conjugate a subgroup you end up with another subgroup because conjugation is a group homomorphism

[lavish ferry]

H is isomorphic to any of its conjugates in G. To see this consider any $a \in G$ and define $f: H \to H^a$ by $f(h) = aha^{-1}$. Then $f$ is an isomorphism

[cloud walrus]

Ellie

[lavish ferry]

This immediately gives that H has equal cardinality to any of its conjugates in G

[vast verge]

Here's a cool diagram that was included in this chapter btw

[tulip otter]

I am not sure how to start

also is this true if we dont exclude the trivial group ?

ah wait, by every element it is implicitly implied that the unit is excluded right?

[tidal schooner]

Hint: ||Pick any g ∈ G. Then conjugation by g is an automorphism of G. If the only automorphism is the identity map, then what does that say about g?||

[tidal schooner]

Yes

[tulip otter]

I see

[tulip otter]

this means that g commutes with all elements of G, but the conjugation by any g is the identity here so that all elements of G commute with each other, ie G is abelian

[tidal schooner]

Yeah!

[tulip otter]

so it remains to check the order of the elements

let me think about this for a bit

ok, so consider the map f:G->G defined by f(x)=x^3, this is an automorphism of G so it must be the identity so that x^3=x which means x^2=e. So if x is not the unit, it must be of order 2

wait

this map need not be an automorphism

[tidal schooner]

Careful, your map f might not be bijective (for example if G contains an element of order 3, which it doesn't, but we don't know that yet)

[tribal moss]

OK, I think I get that after thinking more. Say, the figure-eight has the free group {a,b} as fundamental group and a universal cover that looks like the diagram on the left. The (non-normal) cyclic subgroup <a> would then correspond to the cover shown on the right? And the fiber at a point can be identified with the set of left cosets of <a>.
And the action that is not transitive is not directly the fundamental group, but the group of ... hmm, "deck transformations" of the cover, says Wikipedia. Which in this case would be trivial, I think.

[tulip otter]

yes exactly

alright so i will think about something else

ohhh

maybe the map defined by f(x)=-x?

[tidal schooner]

Bingo

[tulip otter]

I see

[candid patrol]

-x 😵
x^(-1) 🤓

[tulip otter]

yea idk why i just thought about -x instead of x^(-1)

[candid patrol]

Just kidding you are right

[tulip otter]

this exercise was nice

[tulip otter]

yea ik

[tulip otter]

tysm, your idea of using other automorphisms was very nice

[tidal schooner]

Np, nice work!

[candid patrol]

And what do you think about G ?

Nvm, it is not that easy

[tulip otter]

is it something related to what the author said about viewing G as a vector space over Z_2?

or is it something else

[candid patrol]

Yes 👍

[lavish ferry]

Checking here, the idea is that G is equal to its centre?

[tidal schooner]

Yes

[lavish ferry]

Thanks

[tulip otter]

ohhh wait, maybe i got why this is the case

so if G is the trivial group then it indeed consists of only one element as a vector space over Z_2, since Z_2={0,1} so that the result of multiplication elements of G by scalars of Z_2 is either 1e=e or 0e=e

[tribal moss]

I think I see now. If I get you right, the plan is first re-characterize the fundamental group as the automorphism group of the universal cover (or, I suppose, if we don't know there is an universal cover then an inverse (?) limit of the automorphism groups of all possible covers), and then retain that characterization while replacing "cover" by a different concept that still have some "morally similar" symmetries?

[tulip otter]

ah nvm what i was thinking about next doesnt work

[tribal moss]

Consider that an invertible linear transformation of the vector space is in particular an automorphism of its additive group.

[south patrol]

Yes, exactly. And also like i should say this is not just ad hoc, like there is a rich theory of étale cohomology alongside this étale fundamental group and it essentially recovers the usual cohomology of complex manifolds

(Specifically, the point is you can view (nice) varieties over C as complex analytic things)

[tribal moss]

I despaired slighty after finding https://en.wikipedia.org/wiki/Étale_morphism (listing nine equivalent definitions, each of which is completely opaque to me), whereas the https://ncatlab.org/nlab/show/étale+space looks more accessible (which kind of bizarro world have I ended up in when Nlab is more readable than Wikipedia?) so I can at least see it's analogous to coverings.

[south patrol]

They are quite different things to one another

I think yeah the way there are tons of definitions is quite daunting (it was to me anyway) and wikipedia is being a bit encyclopedic here lol

The most intuitive to me here is that f should be "smooth and of dimension 0". This basically captures the idea that a covering map should be something like a submersion and the source should be of the same dimension as the target

[tribal moss]

Wait, which of them are you saying is the concept you were talking about?

[south patrol]

The former (étale morphisms)

[tribal moss]

😭

[south patrol]

lol

[tulip otter]

so you probably told me this to let me keep in mind that any invertible linear transformation of G would be the identity map (?), but i couldnt come up with anything that leads to the solution even after thinking about it for a while

[tribal moss]

If the vector space has two linearly independent vectors v1 and v2, then you can extend {v1,v2} to a basis and consider the transformation that interchanges the coefficients of v1 and v2.

[south patrol]

I think so yeah

[tribal moss]

This would contradict the assumption that G has no nontrivial automorphisms.
^ @tulip otter

[tulip otter]

can it also be seen as a contradiction to the fact that v_1 and v_2 are linearly independent by choosing an x in G with the coefficient of v_1 being 0 and that of v_2 being 1 when x is written as a linear combination of the basis containing v_1 and v_2?

because by interchanging the coefficients of v_1 and v_2 in this case you would get v_1=v_2 right?

i am trying to use your argument in some other way just to check if i am understanding things correctly

[tulip otter]

also i think now i understand this

[tribal moss]

I suppose -- since the assumptions at that point are contradictory, there can be various ways to reach obvious nonsense from them.

The main point is: a vector space of dimension >= 2 always has nontrivial automorphisms.

[velvet goblet]

Ohh, I see where your idea came from

I was thinking about this for a while since the question was asked (and after your hint) but for some reason I totally disregarded the fact that now that we are thinking about vector spaces, facts about dimension,linear independence etc can be used too

[tulip otter]

Ah now your point is clear to me. Tbh I would've never reached the answer anytime soon if not for your guidance. Tysm and have a great day

[tribal moss]

So now we've learned: the only groups that have no nontrivial automorphisms are the trivial group and C_2.

[velvet hull]

jumping into this conversation, but won't a dimension 1 vector space also have nontrivial automorphisms?

[tulip otter]

What is C_2?

[velvet hull]

v -> cv

[tribal moss]

In this context the vector space was over F_2, so 1 is the only nonzero scalar :-)

[velvet hull]

:/

[tribal moss]

The cyclic group of order 2.

[tulip otter]

Ohh ok

[tulip otter]

But there is something weird, the order of C_2 is 2, so the number of automorphisms of C_2 should be the order of the permutation group of elements of C_2 which has order 2 right ?

[tribal moss]

No, because not all of those permutations are group homomorphisms.

[tulip otter]

Ah I see yes

Tysm for the clarification

[dull ginkgo]

Can the polynomial ring over an R-module generally be described as the Symmetric algebra over the "dual" of the module, Hom(M,R)

[barren sierra]

doing some qual studying

what does gcd(a, b) mean in a general domain E?

hmmm

[velvet hull]

it's the common divisor that is divisible by every other common divisor

(which is unique up to a unit, of course)

[barren sierra]

domain = integral domain

I guess I don't see how this is well defined necessarily

like Z[sqrt(-5)] is a classic example of a domain which has no well defined GCD

$6 = 2*3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$ but what is $gcd(5, 2 \cdot (1 + \sqrt{-5}))$?

[cloud walrus]

Spamakin🎷

[barren sierra]

oh I see

[barren sierra]

this holds if D is a GCD domain right?

not just PID?

cool

[velvet hull]

you mean ufd right

[tardy hedge]

What was the point of this example

[barren sierra]

gcd domain is more general than UFD

[barren sierra]

I mean gcd(6, 2 * (1 + sqrt(-5)) whoops

[velvet hull]

but then.... the gcd always exists in a gcd domain by assumption?

[barren sierra]

yes that is definition of gcd domain

[velvet hull]

isn't the defintion of a gcd domain a ring where the gcd of two elemenets always exists

so like

[barren sierra]

D, not E

[velvet hull]

oh, you mean in your problem

idk sounds true, too tired to check right now

[tardy hedge]

Oh, so this does not have a well-defined gcd? Is it like both 2 and 1+sqrt(-5) are gcds?

[barren sierra]

and both of those numbers don't divide each other

but both 2 and 1 + sqrt(-5) are irreducible in Z[sqrt(-5)]

hence gcd is not well defined

[tardy hedge]

Idk why im confused abt something

Oh is the fact that 2 and 1+sqrt(-5) are the only common divisors relevant here

So if gcd exists there then one of them has to divide the other

But theyre both irreducible so that cant happen

[tidal schooner]

No, being a GCD domain without being a PID isn't strong enough. Let $D=\bZ[x]$, which is a UFD (hence a GCD domain), and let $E=\bQ[x]$, which is a Euclidean domain (hence a domain). Then $\gcd(2,x)$ equals $1$ in $D$ but $2$ in $E$

[cloud walrus]

harmacist

[barren sierra]

Ah I see

Hmmm I'll check my proof tomorrow

It seemed like I could use only a GCD domain in my proof so there's gotta be a mistake

[tardy hedge]

Thats neat i should study this stuff more

Havent really studied those properties too much tbh

[tidal schooner]

Addendum: In this example, $2$ and $1$ differ by a unit, but we can replace $\bQ[x]$ with $\bZ[x/2]$ to get an example where they don't

[cloud walrus]

harmacist

[tardy hedge]

Differ in what sense?

[tidal schooner]

You can multiply one by a unit to get the other

[vast verge]

I'm currently on my final chapter of group theory before moving onto ring theory

Using the fundamental theorem of finite abelian groups to decompose abelian groups and analyse their subgroups is pretty fun

Can't wait until the fundamental theorem of finite non-abelian groups gets released

[delicate orchid]

https://en.wikipedia.org/wiki/Classification_of_finite_simple_groups

funny!

[tribal moss]

Only the simple ones, unfortunately.

[tulip otter]

Proof: $H$ is solvable, so it suffices to find a sequence $$G=G_0\supset G_1\dots\supset G_n=K$$ of subgroups of $G$ such that $G_{i+1}$ is normal in $G_i$ and $G_i/G_{i+1}$ is abelian. Now $G/K$ is solvable, so there exists a sequence $$G/K=G_0/K\supset G_1/K\supset\dots\supset G_n/K={e'}$$ of subgroups of $G/K$ such that $G_{i+1}/K$ is normal in $G_i/K$ and $(G_i/K)/(G_{i+1}/K)$ is abelian. But $(G_i/K)/(G_{i+1}/K)\cong G_i/G_{i+1}$, So $G_i/G_{i+1}$ is abelian, $G_{i+1}$ is normal in $G_i$ for all $i$ and $K=f^{-1}({e'})$ where $f:G\to G/K$ is the canonical map. Hence the required sequence of subgroups of $G$ mentioned at the beginning exists.

[cloud walrus]

pirateking0723

[tulip otter]

is this correct? or is it missing anything and/or has some flaws?

[tribal moss]

The original idea was that we perhaps understand all the finite groups by (a) figuring out what the finite simple groups are, and then (b) understanding how non-simple groups are "built up" from simple groups.
Part (a) of this program was eventually completed, but my impression is that no real progress has been made on (b).
For example the quaternion group Q_8 is not simple, but has a normal subgroup isomorphic to C_2. The quotient is the Klein 4-group C_2 × C_2, but it's not clear how to start from C_2 and C_2×C_2 and end up with Q_8 other than already knowing that answer.

[tulip otter]

maybe i should've mentioned that {e'} refers to K/K (or simply {K} )

[brazen ravine]

In Image 2, are we assuming the shorthand $nx$ as $\sum_{i=1}^n x$? We are in (G, +) meaning that there is no notion of "multiplication"

[cloud walrus]

Hurricane

[tulip otter]

yes, nx is summing x n times just like the summation your wrote

just a notation to avoid writing \sum... or x+x+...+x n times whenever you want to talk about this

just like how x^n is used instead of xx..x n times

[vivid hemlock]

you tend to use whatever notation is most helpful in my experience, nx makes more sense than x^n under addition

[rocky cloak]

There is group cohomology, which can be used to classify extensions in a way not too dissimilar to semidirect products.

Not that group cohomology is that easy to compute though, but it gives you some tools.

And there's also a generalization that works when the normal subgroup is non-abelian, but I'm not sure that one's actually useful for computing anything.

[unreal sedge]

hello beautiful people

May someone check my work?

[cedar vault]

* must be defined on S, so it must be a map S x S -> S. In this case if you do so, T U T' = Z is closed under *

[unreal sedge]

So since S=R it's undefined

i see

[tulip otter]

The binary operation you defined maps from N^2 to N, so -5 cant be used as an input of this operation, also assuming that you extend that to become an operation Z^2->Z, your example doesnt show that TUT' is not closed

because -5 in Z=TUT'

hint concerning the counterexample: ||Let T to the set of all negative integers and T' be the set of all positive integers (note that none of these contains 0||

[unreal sedge]

It's either late and I'm slow or I can't read english

[unreal sedge]

so instead, I should use * : R x R -> R?

[rocky cloak]

R, Z and N are all closed under addition. So these won't work as counter examples

You probably want to start with identifying some sets that are not closed under addition

Then see how you can write them at unions of things

[unreal sedge]

Alright thanks

Would you suggest I do a complete rewrite or tweak some things?

Like restart

actually I'll restart

clean slate

[unreal sedge]

I've redone the question

Let $S=\mathbb{R}$. We will define $T=\mathbb{Z}+={n\in\mathbb{Z}:n>0}$ and $T'=\mathbb{Z}-={n\in\mathbb{Z}:n<0}$. Define $\star$ to also be a binary operation such that $\star :\mathbb{R \times R \rightarrow R}$, $(a,b) \mapsto a+b$. It then follows that $T\subset S$ and $T'\subset S$. If $a,b \in T$, then $a+b>0 \in T$. Thus, $T$ is closed under $\star$. If $a,b \in T'$, then $a+b<0 \in T'$ which would imply that $T'$ is closed under $\star$. The intersection $T \cap T'=\emptyset$ is also trivially closed under $\star$ because there are no elements that can break closure under $\star$. The union of $T$ and $T'$ provides us $T \cup T'={...,-3,-2,-1,1,2,3,...}$. If we apply the binary operation $\star$ on two elements of $T \cup T'$, say $-3$ and $3$, we are given $(-3)+3=0 \notin T\cup T'$. And thus, $T\cup T'$ is not closed under $\star$.

[cloud walrus]

whyhello

[unreal sedge]

I used the hint by the way

Now that I'm thinking about it, I could've used the set of all prime numbers as a counterexample

[cedar vault]

Many such sets

[unreal sedge]

I'll have to look online for more examples

Maybe in the morning

[south patrol]

I think like essentially any two (nice) choices of T and T' as subsets of R will work:

Suppose T, T' and T u T' are closed under addition and additive inverses. Then T contains T' or vice versa

[tulip otter]

Proof: $H$ is solvable, so it suffices to find a sequence $$G=G_0\supset G_1\dots\supset G_n=K$$ of subgroups of $G$ such that $G_{i+1}$ is normal in $G_i$ and $G_i/G_{i+1}$ is abelian. Now $G/K$ is solvable, so there exists a sequence $$G/K=G_0/K\supset G_1/K\supset\dots\supset G_n/K={e'}$$ of subgroups of $G/K$ such that $G_{i+1}/K$ is normal in $G_i/K$ and $(G_i/K)/(G_{i+1}/K)$ is abelian. But $(G_i/K)/(G_{i+1}/K)\cong G_i/G_{i+1}$, So $G_i/G_{i+1}$ is abelian, $G_{i+1}$ is normal in $G_i$ for all $i$ and $K=f^{-1}({e'})$ where $f:G\to G/K$ is the canonical map. Hence the required sequence of subgroups of $G$ mentioned at the beginning exists.

[cloud walrus]

pirateking0723

[tulip otter]

is this proof correct

[crystal vale]

yes if you done all proofs of your claims, and i think it is typo, K is solvable not H

[crystal vale]

have you showed that subgroup of G/K, say G' then G' = K1/K, such that K1 is subgroup of G and K \subset K1?

[tulip otter]

ah yes mb it should be K

[tulip otter]

no, i havent done this (neither did the author)

but it is clear that the elements of G' should be of the form xK for some x in G, so now let K_1 be the set of all these x. Then for x,y in K_1, (xy^(-1))K=xKy^(-1)K\in G' (since G' is a subgroup of G/K) which means that xy^(-1)\in K_1 so that K_1 is a subgroup of G

and it follows that K\subset K_1

[crystal vale]

I should say take pre image of G' under natural mapping

[tulip otter]

ah yes, thats definitely faster

I need to always keep this natural maps in mind because they seem to facilitate the work and they are pretty useful in proofs

[tulip otter]

as this shows for example

[tulip otter]

tysm for your time, have a great day/night

[barren sierra]

Is there a way to do this that's not Smith Normal Form and not guessing and playing around with stuff until you get a plausible answer?

neither seem viable for a qualifying exam setting

[elfin wraith]

I believe it may just be SNF, I’m racking my brain trying to remember how we did it in my comalg or possibly LA class because I remember that being a pretty standard “easy” problem

I think you just kinda work under the assumption that the setter has been nice and given you something not too awful to work with

[barren sierra]

bleh computing the matrices for SNF is so annoying

[elfin wraith]

Or possibly it was Gröbner basis stuff, I’ll come back if I remember how to actually do it (it’s late and I’m not at home with any of my notes), but I think it is an affirmative to the question of is there a way to do this which isn’t awful

[barren sierra]

Grobner by hand would also be awful

[elfin wraith]

I mean if they’ve been nice to you it should only be like maximum 3 rounds of computing S polynomials lol

[barren sierra]

Grobner bases should be computed by hand once or twice when first learning and then by computer every time after

S polynomials?

[elfin wraith]

I do not disagree

[barren sierra]

oh oh S polynomials for Grobner bases

yea

I thought S polynomials were something about SNF and I was confused

[elfin wraith]

Yeah no I’m sure it was just do SNF and pray it wasn’t too ass when we had this in my comalg class (or LA class whichever it was)

[exotic bone]

According to an AI, the best way to do this sort of thing is with Hermite Normal Form calculations — essentially Gaussian elimination but using Bézout coefficients from extended Euclidean algorithm to handle when you need to do division.

[barren sierra]

"according to an AI" no one asked you then

[exotic bone]

Don't worry, I've thought it through and the method should work

I just figured that AI should at least know names of algorithms like these even if it can't do the details

Anyway, I'll do the computation with your example to demonstrate.

[coral spindle]

Please do not respond to people with AI responses. If people wanted AI responses, they would go to AI, not discord. This is also against server rules.

[tall igloo]

if you want to find misinformation about mathematics online, you can just go to vixra.org instead of an LLM. it's even environmentally cleaner!

[south patrol]

Or sometimes even arxiv

[thorn jay]

thats how they get you

[dull ginkgo]

Kinda stuck trying to show that these valuations of polynomials are linearly independent. The issue is that it jumps over even derivatives so I can’t just use the Newton technique

[proud vigil]

could you maybe construct the polynomial and show it's unique?

i'm kind of guessing you can use the 0 and 1 to create indicator expressions x and (1-x) that you tack onto respective power series-looking things

idk if proving uniqueness would be straightforward from there though

[proud vigil]

i have an idea

we know the map from 2n+1 -degree polynomials to these coefficient vectors is linear since evaluations and derivatives are linear, so we just need to show that the kernel of the map is the 0 polynomial. then our map is invertible and we know there will be an invertible map from coefficient vectors to polynomials

from the 0 evaluations, we know the constant and all odd-power coefs of our polynomial must be 0, leaving us with just the even-power coefs. I think you can construct a map specifically from the 1 evaluations to the even power coefs and get a matrix that is necessarily invertible (this kind of looks like any generic derivatives -> polynomial map since the derivatives are smaller than the exponents), and so the only way for the evaluations to be 0 is if the even coefs are all 0 as well

i think this works but im not sure if im making any logical ambiguities or errors

[tulip otter]

for a, let G be a simple finite abelian group and let x\in G with x neq e (e being the identity of G). Consider the cyclic subgroup H=<x> of G generated by x, then H is normal, and (H:1)|(G:1) implies H=G (the only normal subgroups of G are G and {e} since it is simple) so that G is cyclic.

Now suppose that G has composite order k, and let p be a prime divisor of k. Then H_1=G where H_1=<p> (again since G is simple), but H_1\subsetneq G, which gives a contradiction. Hence G is cyclic of prime order

is this good enough for a?

and any hint about b?

[proud vigil]

your a looks fine

[tulip otter]

for b i thought that maybe i can somehow find a tower in which H_i/H_{i+1} is simple and then the result would follow from a

[proud vigil]

i think for b you can use the fact that if p is a prime factor of a group size, there exists an element of order p in the group

[thorn jay]

yes, and as the group is abelian, you can quotient by that generated subgroup of order p.

[tulip otter]

but if i do that, then the quotient may have a composite order right? maybe i should do a quotient by a subgroup of order (G:1)/p so that the resulting quotient group has prime order?

[thorn jay]

continue that process

[tulip otter]

which?

[tulip otter]

this?

[tulip otter]

or this

[thorn jay]

this

[tulip otter]

but what i mean is that assume the order of G is n, if i consider an element a of order p and the subgroup G_1=<p>, then the quotient G/G_1 would have order n/p which may not be a prime right?

but G/G_1 must be of prime order

or is it somehow forced to have prime order?

[proud vigil]

the Hs are meant to basically have a prime factor removed each step

you wouldn't make H1 a prime cyclic, you would want to use a prime cyclic to get to H1

enpeace can probably fix whatever im saying sorry i havent done algebra in ages ;-;

[tulip otter]

nw

[tulip otter]

I am not sure that i follow

Now i have 2 interpretations which are completely opposite

so what i understood so far is that i take the element of order p in G and let G_1=<p> then i do the quotient G/G_1

but from the point of taking the quotient my problem starts, you lose me there

[proud vigil]

like, first let's take an existing prime factor p, and make a prime cyclic subgroup of size p, call it Cp

then take G/Cp

this is where im forgetting but i think you can do a canonical projection or something to get a subgroup that is of size |G|/p such that taking the quotient gives you a cyclic group of order p

but G/Cp (or whatever the corresponding actual subgroup is in G) would be H1

[tulip otter]

so this would be the second group in the sequence ?

[proud vigil]

yes

[tulip otter]

oh wait i see

[proud vigil]

but G/Cp isn't in the right form 😭

[tulip otter]

i was flipping

[tulip otter]

yea i understand your point

G/C_p isnt even a subgroup of G

but there is a subgroup of G with the same order of G/C_p

then take that as H_1 (or whatever you want to call it)

so now the next step is to take a subgroup of H_1 generated by a prime divisor of its order and then repeating the same process to obtain H_2, then H_3, then ...

this process will definitely continue until reaching {e} since the order of G can be written as a product of prime

did i understand it correctly? or did i miss something along the way

[proud vigil]

no yeah that's exactly it

[proud vigil]

the only thing that for some reason is foggy in my brain is the construction of this but it sounds like u know what it is so thats fine

[tulip otter]

by the construction you mean the way to get into actual subgroups of G instead of quotient groups?

[proud vigil]

yeahh

its not a hard thing but for some reason my googling is failing

[tulip otter]

well, i think you can restrict the canonical projection f:G->G/H_1 to (G-ker f)U{e} here e is the identity of G to obtain an isomorphism between this and G/H_1

[tidal schooner]

Here's a faster way of doing it. ||First, show by induction on |G| that every finite group has a Jordan-Hölder series. Then if G is finite and abelian, apply part (a) to its JH series||

[tulip otter]

what is a jordan holder series

let me look it up rq

[tidal schooner]

Different name for a composition series (every quotient is simple)

[tulip otter]

ohh i see

[proud vigil]

whoa thats smart

[tulip otter]

for |G|=1, the result follows immediately. Assume that the result holds for |G|<n, now let |G|=n and let p be a prime such that p|n, consider the subgroup X=<p> of G generated by p. Then |G/X|=n/p<n so that there exists a subgroup G_1 of G with |G_1|=n/p for which there exists a jordan holder series by induction. Furthermore, |G/G_1|=p which means that G/G_1 is simple and this completes the proof (proof of the hint)

Hence, by a, b follows

[tidal schooner]

The induction step as written might not work since X might not be a normal subgroup of G. But instead you can say that if G is simple, then 1 < G is a JH series, otherwise take a maximal normal subgroup N. Show that G/N is simple (e.g., with the fourth isomorphism thm), then the inductive hypothesis finishes the proof

[tulip otter]

why would X not be normal- wait what did i even write, i meant to write X=<a> where a in G has order p

X=<p> doesnt even make sense

[tulip otter]

so X is cyclic and hence normal right?

[tidal schooner]

Not all cyclic subgroups are normal, even if they have prime order. For example <s> isn't normal in the dihedral group

[tulip otter]

ah yes you are right

[tulip otter]

i am not familiar with the fourth isomorphism theorem

i am thinking about some other way to prove this

[thorn jay]

its alr
also haii, welcome to the server

[proud vigil]

hii

omg i have the new person symbol

[tardy hedge]

yes u do u do u are new u are new

And when you're with the crew, you knew that this was something great to do 🎶

[tulip otter]

wait, isnt G/N simple by induction

[tidal schooner]

N has a composition series by induction, but I don't think simplicity of G/N follows from induction

[tulip otter]

but isnt the induction being done about simplicity?

[tidal schooner]

The induction is on the statement "a group of order n has a composition series"

[tulip otter]

so we are supposing that if |G|<n then it has a simple series no?

[tulip otter]

not a simple one?

[tidal schooner]

Yeah

[tulip otter]

ah yes it is not on the simplicity of G/N itself

mb rn i am not concentrating much since i am sleepy

[tidal schooner]

All good

[tulip otter]

so is there a way to prove that G/N is simple without the fourth isomorphism theorem so that i continue thinking about it? or do i look up the 4 iso thm and then try using it?

[tidal schooner]

You can prove it directly, which entails proving one the parts of the 4th iso thm anyway. ||Let π: G → G/N, and let H be a nontrivial normal subgroup of G/N. Note that H contains N (as an element of G/N). Deduce that the preimage π^-1(H) is a normal subgroup of G that contains N, and apply the maximality of N||

[torpid veldt]

😠

[proud vigil]

i like magmas

such a funky word

i feel like monoids and semigroups should swap definitions

semi group feels like it should be closer to actual groups :(

[thorn jay]

algebraic structures have funny names

"quandle"
"spindle"
"shelf"

[tidal schooner]

Singquandle, psyquandle, stuquandle, generalized Legendrian quandle

[tall igloo]

why? I agree semigroup is a bad definition, but monoid feels pretty intuitive when you realize such things can be modeled by categories with one element

[thorn jay]

saw a paper of someone studying topological biquandles

💔

[thorn jay]

categories, or their more popular name, monoidoids

[rocky cloak]

I mean mono=1, and monoids are things with 1, so would be a pretty bad name for semigroups

Which whole deal is they don't have 1

[south patrol]

Just to like amplify this, this is an instance of the "correspondence theorem" for groups, which I highly recommend using / knowing about when learnimg about solvable groups @tulip otter

[tidal schooner]

Yeah and for reference the 4th iso thm, the correspondence thm, and the lattice thm are all the same result

[south patrol]

Oh lol

[thorn jay]

the best isomorphism theorem 🙂‍↕️

purely because its of the form of an isomorphism of lattices

[south patrol]

Honestly I had not rly heard of it being called the 4th iso theorem (feels funny cause to me it does not feel like an isomorphism theorem at all lol)

[south patrol]

It is probably my favourite too

[south patrol]

In particular yeah as it is about lattices / internal structure rather than about one algebraic thing being iso to another

[tidal schooner]

My favorite is the 1st (normie opinion)

[thorn jay]

and it is especially powerful in conjunction with the third ( (G/N)/(M/N) ≈ G/M )

[south patrol]

It is cool

[thorn jay]

the rest cannot exist without this one, completely valid

the one I perhaps like the least is the second one? which is probably only because the generalisation to universal algebra is a bit ugly

[south patrol]

Though one thing i find amusing is these theorems sort of evaporate in uh "higher algebra" ig

Unless I am being silly

[thorn jay]

define higher algebra

as in, which higher algebra

[tall igloo]

like ∞-categories and such

[thorn jay]

scary...

i will stick to my pleb math

[south patrol]

I guess easiest example is working in the derived category of an abelian cat

[south patrol]

Lol it is funny how higher algebra can either mean like this stuff or like

Intro abstract algebra or smth

Like Hall's book

[thorn jay]

i would argue universal algebra is a form of higher algebra but you all would laugh at me and stone me

[south patrol]

Actually this would be a funny joke gift for my advisor

[south patrol]

Nah this makes sense

I won't stone you

[thorn jay]

a great honor

honour*

either im too tired to spell

[thorn jay]

do it do it do it do jt

[thorn jay]

the same way knot theorists laugh at people working with quandles but honestly just for the name alone thats justified

[south patrol]

Dw we can laugh at knot theorists

[thorn jay]

laugh triangle

[south patrol]

Woah correcting to honour. Americans destroyed

[thorn jay]

i am a mann't of class

[south patrol]

I like mann't

[thorn jay]

i think it fits

im not a man but not a woman, but also not really androgynous im just mann't

[tall igloo]

Hmm where's your tea again? Oh that's right, in the Boston harbor 🦅

Anyways this is late since my phone was bugging out but

[thorn jay]

at the far right should be "I study quandale dingles" it is the pinnacle

oh my god my brain is so rotted atm

[tidal schooner]

Me when my previous institution renamed the Modern Algebra class to Algebra:

[south patrol]

Especially turkish ones

[south patrol]

I like this

[thorn jay]

this is a fucking actual title:
"bridging colorings of virtual links from virtual biquandles to biquandles"
"quandle coloring quivers of general torus links by dihedral quandles"
fee foo fum ass

[thorn jay]

troll people asking for algebra books by sending them algebra chapter 0

[tidal schooner]

Also this

[thorn jay]

honestly surprised it wasnt Mohamed fucking Elhamdadi

although I believe he posted a paper about internal self distributive objects and their cohomology?

[tall igloo]

i thought of another version that was "all math is algebra" "nooo math is more than algebra" "all math is algebra"

[thorn jay]

https://www.academia.edu/8605608/COHOMOLOGY_OF_CATEGORICAL_SELF_DISTRIBUTIVITY?source=swp_share

lol

[knotty badger]

-# damn im glad I don’t like algebra if people who like it are this elitist/supremacist

[thorn jay]

its cool though, apparently lie algebra cohomology is a special case of this

[thorn jay]

hey im not elitist or supremacist qwq

[tall igloo]

who said i liked algebra 💀

[south patrol]

Lol

Tell me you are an algebraist type person without telling me

[tall igloo]

damn i was going to say im not but i spend all my time thinking about E_\infty-rings so

[thorn jay]

mental illness

[south patrol]

Is this referring to hatfuel' meme or more generally

I spend all my time thinking about uh idk

[thorn jay]

as opposed to what i do, which is decidedly not

currently i am busying myself with quandles

[tall igloo]

i am also wondering this and also can't tell how serious this comment in. to be clear, the meme and comment later are overly exaggerated, and if this was not clear then my apologies

[knotty badger]

sorry I took a nap

yeah it was hatfuel’s stuff

[wispy patio]

im kinda new to groups and rings, so i apologize for my ignorance but is this like a correct def for groups?

havent include the 4 properties, just what are groups and binary operations

ill consider the rest later (associativity, identity, inverses and abelian groups)

[knotty badger]

so far this is just a set with a binary operation

[wispy patio]

yeah

[velvet hull]

so it's not the correct definition

[wispy patio]

well its not complete

actually ur right

wait lemme just add the 4 thingies of groups

[knotty badger]

the "closure" axiom for groups is a little contested

[wispy patio]

yeahh needs a little working

[knotty badger]

some people argue that it's just part of the definition of a binary operation - you have a map $X \times X \to X$, so by definition it lands in $X$

[wispy patio]

checking as we speak

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

however, in practice what often occurs is the following

you have a "parent" set $Y$ and a binary operation $Y \times Y \to Y$, which may not make $Y$ a group

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and then you look at a subset $X \subseteq Y$, and the restriction of the binary operation, so $X \times X \to Y$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and then you have to check that this really lands in $X$, which is what the "closure" axiom is for

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[wispy patio]

ohh wait this is better than the axiom i have now

[knotty badger]

for example, $Y$ could be all square matrices, and $X$ could be invertible square matrices

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[wispy patio]

ok ok, ill replace it for my notes

[knotty badger]

so in practice, "closure" is a useful sanity check, because often the binary operation for a group comes from restricting some larger binary operation, and then it really is a nontrivial condition that it still lands in the group

[tall igloo]

yes, I take responsibility for this and I am sorry. Just to be clear about my actual opinions on this and not the meme-ified versions: algebra just happens to be very useful for many (almost all? I think?) fields of math, but it is certainly incorrect to say all of math can be reduced to studying algebra. And of course studying algebra does not make you smarter by any metric. I'm not even an algebraist, I just use algebra a lot since it's one of the main tools in my area

[wispy patio]

so essentially what youre saying is that if we have a set Y, the binary operation Y*Y->Y

[knotty badger]

yep, such as multiplication of square matrices

[wispy patio]

meaning for any Y Y*Y lands on Y

[wispy patio]

that makes sense thx

[knotty badger]

yep, since multiplying two square matrices gives a square matrix

you could restrict to the subset X of invertible square matrices

but then it's a nontrivial check to show that the product of two invertible square matrices is an invertible square matrix

and not just any ol' square matrix

[wispy patio]

enlighten me on invertible square matrices? i skipped linear alg

[knotty badger]

they're simply square matrices $A$ for which there exists another square matrix $B$ such that $AB = I$ and $BA = I$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[wispy patio]

ahh you meant that

if exist A exist B where B is the inverse of A

make sense

[knotty badger]

you can use the matrix multiplication operation on invertible square matrices to get a map $X \times X \to Y$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

but you have to check that this map actually lands in $X$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

that's where "closure" comes in

[wispy patio]

i see, btw the first sentence is more like the intro to binary operations of G can you maybe suggest me a better opening line? so that i dont miss the entire point of groups to be associative, and other 3 properties

really need my notes to be nice and consice

[knotty badger]

so, a $\textit{group } (G, \cdot)$ is defined to be a set $G$ with a binary operation $\cdot : G \times G \to G$ satisfying the following three properties:

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

1. There exists an element $e \in G$ such that $\forall g \in G, e \cdot g = g \cdot e = g$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[wispy patio]

identity

[knotty badger]

2. For every $g \in G$ there exists some $h \in G$ such that $g \cdot h = h \cdot g = e$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[wispy patio]

inverse

[knotty badger]

3. For every $g, h, k \in G$ we have that $(g \cdot h) \cdot k = g \cdot (h \cdot k)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[wispy patio]

associativity?

[knotty badger]

yep!

[wispy patio]

okiee, thx a lot

[knotty badger]

-# equivalently, a group is a one-object groupoid

[wispy patio]

the identity is unique, therefore the identity is unique. the proof is left as an exercise for the reader

[knotty badger]

ofc this is just one way to axiomatise groups, there are several others

e.g. it's possible to axiomatise groups in a way where you never need to use $\exists$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[velvet hull]

yeah, I can do it with only ∀!

it's a joke, it's just by duality of the quantifiers

$\exists x P(x) \iff \neg \forall x \neg P(x)$

[cloud walrus]

HChan

[velvet hull]

once you're mathematically matrue enough it the english statements should make sense anyway

"there exists a human that is a girl" <=> "not all humans are boys"

and the duality is just the general version of that

I mean, you've done enough group theory, I say you're pretty mature

enough to know quite a bit, but also enough to know that there is so much more

[elfin wraith]

You should, there’s very few situations in which you should be using logical quantifies when you could write it in plain English

You should also understand quantifiers, but like, you shouldn’t be writing with them if you can help it, it’s just less clear

[unreal sedge]

hello again my beautiful people,

May someone check my work?

I have to move on from binary operations at some point but I'm doing all the exercises in this one section

[velvet hull]

0 doesn't have to be in S

for instance, what if your set consists of only positive integers?

the detail is that S is closed under addition, and not necessarily subtraction

or, it is what's implied and I'm misunderstanding the question

[unreal sedge]

The positive integers is not a finite subset though

Neither are the negative integers

[velvet hull]

sure, I'm just pointing out that 0 doesn't have to be in S if you're strictly only considering addition

[unreal sedge]

enlighten me, I'm studying on my own lol

oh

i didnt see that part

oops

🤦‍♂️

[glad osprey]

You start with defining S = Z, but I'm guessing you mean S should be a subset of Z? Also, you're defining * to be equal to + for no reason, just use + directly

I know the question asks for subsets of Z, but then he shouldn't write S = Z

[unreal sedge]

noted

[ashen heron]

Proving it works for a specific class of sets is insufficient, youre asked to find all subsets that work

[unreal sedge]

alright

[unreal sedge]

This may seem like a dumb question but how would I introduce + directly

[glad osprey]

Wdym introduce it? Everyone knows what addition is, you don't have to define it

Btw, {0, n} for non-zero n is not closed under addition

[unreal sedge]

I see

[unreal sedge]

wait so how would I phrase it though? I was thinking "Let Z have the binary operation +"? Would that work?

I apologize if I seem a little slow 🌚

Or just carry out + as you would normally

[glad osprey]

I probably wouldn't write that line at all, I would just start arguing: these sets are closed under addition because blah blah blah. Any other sets is either not closed under addition or not finite, because of reasons A, B and C

[unreal sedge]

alright thank you.

[glad osprey]

Btw, it doesn't seem like you need to define S = Z either. It seems like you're defining stuff just for the sake of defining. At some point you write that 0 is in S, but of course we know that already. Or did you mean that 0 has to be in every subset closed under addition?

[unreal sedge]

I think I defined S=Z because of my laziness to write \mathbb{Z} all the time LOL

[ashen heron]

$\bZ$

[cloud walrus]

pika

[unreal sedge]

I was trying to be to some level "rigorous" and introduce everything, including stuff that might not be introduced. There's a student at my local university that I look up to and ask for help often times (i'm in high school) and the feedback I get when he reads my proofs would be "why" or "you didnt introduce this blah blah but this is here"

$\bZ$

[cloud walrus]

whyhello

[unreal sedge]

bruh

I couldve written that instead of mathbb{Z}

[ashen heron]

youd need to set things up in your preamble for your local set up

\bZ is available by default with the tex bot here

for more info #latex-help

[unreal sedge]

oh

[glad osprey]

I see, but introducing S and * is not being rigorous, it's just confusing. Z and + already exist and are well known, you don't have to reinvent stuff

[unreal sedge]

I see

I guess it just takes practice lol

Thanks for the help guys. I'll come back as soon as I get a new proof up

[south patrol]

Or write $\mathbf Z$

[cloud walrus]

Prismatic Potato

[south patrol]

😏

[unreal sedge]

ooh

[tulip otter]

Ohhh I see, I need much more practice to begin using such results naturally lol

[tulip otter]

I will definitely check this and keep it mind for future results concerning solvable groups and other results in which it would be useful

Tysm everyone, have a great day/night

[unreal sedge]

We love algebra

[fading acorn]

me when a group is just a groupoid with one object and group homomorphisms are just functors between groups

[proud vigil]

That’s true :0

I just don’t like semi groups then

I want them to have a more fun name

and sound less similar to groups 😭

[rocky cloak]

Assoccioids?

Nulloid (as a pun on monoid)

[frigid shard]

this is unironically the most enlightening point of view

group elements are automorphisms

[swift tundra]

Cayley’s theorem moment

[tardy hedge]

bruh ive been grading multivariable calc midterms ts shit so lame lmao

[frigid shard]

my condolences

[thorn jay]

anything egregious?

[delicate orchid]

this one takes the cake for me

[thorn jay]

i propose the term vocabulary

[tardy hedge]

its just like if someone doesnt know what theyre doing they just write down some random formulas sort of thing and hope for the best ig

[delicate orchid]

my thesis

[thorn jay]

horrible

i wish you luck

[vast verge]

How do you define multiplication between the integers and generic ring elements?

Is it just repeated addition?

(I finished the group theory section, yay)

[south patrol]

Given any abelian group you can define n.x as x + ... + x n times yes

There is another way to view this that makes the exercise almost trivial though

Like the element 1 makes sense in any (unital) ring and you can just add that

[vast verge]

This isn't necessarily a unitary ring so idk

I'll try by showing that their difference is 0

[cedar vault]

For positive integers this makes sense. And im sure you would have suspected that for negative integers its just the additive inverse of the positive integer counterpart

[south patrol]

Okay then this

[thorn jay]

you can also expand
(∑_i=1^n a)(∑_j=1^m b)
using distributivity, might not be the most fun though lol

[south patrol]

Lol

This is what I would do too

[cedar vault]

Most probably what was intended as well

[vast verge]

Also stuck on this

Any hints?

[thorn jay]

if the additive group is generated by some element a, then every element must be of the form n*a, where n is positive. then use the above exercise

[vast verge]

Hmm ok, I'll take a look now

[coral spindle]

Correct!

[vast verge]

Cool, I'll try looking at this one now #groups-rings-fields message

[thorn jay]

very clean

[vast verge]

Thank you

[thorn jay]

the mathematician's old trick of using a lemma youll "prove later"

[vast verge]

I've always been wondering whether people are doing that already with the riemann hypothesis

[thorn jay]

people do it all the time, its also a good way to up the importance of a conjecture, if you can show that many cool theorems can be proven from it

[cedar vault]

https://share.google/BtcvmNBtG75qAAWp1

[vast verge]

Cool, thanks

[cedar vault]

Not to mention that this is how dis proof by contradiction is done as well

[thorn jay]

i was thinking abt that yeah

[thorn jay]

this is really cool, especially the first answer

though many of it is going over my head

i do not know much number theory

[south patrol]

That is cute

[thorn jay]

instance of the fact that the rng generated by a single element is commutative

[vast verge]

I don't know what can be the inverse

[knotty badger]

Ah I think this is the geometric series trick

If you’ve seen that before