groups-rings-fields

[long geyser]

suppose its distance to two of the sides is >= 1

then form the triangle using those sides

you now have a triangle that is bigger than the original triangle, yet contained in it

ig by area

[south patrol]

sure

geometrically you can do this sort of construction

well

[long geyser]

ye that is what I mean

[south patrol]

basically divide it up into rigth angled triangles

[rotund aurora]

I suppose writing stuff down is not too bad either. Given $x+\omega y$, $x,y\in\Q$ you want to find $a,b\in\Z$ with $|a+b\omega+x+y\omega|=(a+x)^2-(a+x)(b+y)+(b+y)^2<1$, you can choose a,b such that $|a+x|, |b+y|<=1/2$ and then that expression is $\leq 3/4<1$

[south patrol]

i guess you have to know the like apothem

but you can do that too lol

[cloud walrus]

croqueta3385

[long geyser]

that is what I've been too lazy to do for the past 20 minutes

ty for doing it for me lmao

[south patrol]

i guess there is also another meme way lol

well it's what you sid ig

the area is like uh sqrt(3)/4 or smth

idk

well eh irrelevant now lol

[long geyser]

thanks for the help, tho I still don't think I figured out how the hint is supposed to be used

like

this is part 2 of the question

so surely, rotman didn't expect the reader to use the complex norm

for part i

I'm guessing

[lilac mango]

@long geyser The complex norm doesn't work though, no? Since a²+b² isn't an integer in general

Let me rephrase that

|x|² for x in Z[w] isn't an integer in general

In Z[i] it is, but here it is not

[kind temple]

N(x) = x conj(x)

is what it means in this context i believe

[lilac mango]

Yes but that isn't euclidean distance

Which they used because of the triangle argument if I am reading correctly

[kind temple]

no, like, the norm of a + bX is (a + b)(a - b) where X is an indeterminant

i think this is how artin defines it

[lilac mango]

Yes I know my point is that they are trying to prove it is a euclidean domain using a different norm

[kind temple]

ah, i missed that. wasn't reading closely enough

[long geyser]

um

unsure what you mean, but when I substituted omega = -1/2 + sqrt(3)/2 i

it did turn out to be just complex norm squared

a^2 - ab + b^2 I mean

a + b omega = (a - b/2) + (sqrt(3)b/2) i

(a - b/2)^2 + (sqrt(3)b/2)^2 = a^2 - ab + b^2

unless I fked up a calculation

[lilac mango]

Oh really

In that case sure

[lilac mango]

So this is false

[long geyser]

If a n b are integers, then a^2 - ab + b^2 is an integer

I hope that is not controversial

[kind temple]

is norm defined anywhere else in the text? @long geyser

[lilac mango]

He yeah lol

[long geyser]

how would it ask questions about it otherwise

[lilac mango]

For some reason I thought a²-ab+b² wasn't the euclidean norm squared

[long geyser]

@kind temple

[lilac mango]

Good to know I guess

[kind temple]

then all you need to show is that delta is multiplicative lol

[long geyser]

yes?

I never asked how to show it

[kind temple]

?

[long geyser]

my question was related to the hint that was given for the first part of the question

if you show the degree function is the complex norm squared, (ii) is trivial, so I'm assuming the hint in (i) wasn't related to the complex norm

this is part (i)

[kind temple]

oh, misunderstood🫡

[long geyser]

[untold basalt]

Let $G$ be a group and ${H_i}_{i=1}^n$ normal subgroups of $G$ such that $H_1 \dots H_n=G$. Assuming that each $g \in G$ has a unique representation as a product $h_1 \dots h_n$ with $h_i \in H_i$, how can I show that $\phi: G \to H_1 \times \dots \times H_n$ such that $\phi(h_1 \dots h_n)=(h_1, \dots ,h_n)$ is a homomorphism?

[cloud walrus]

Heywood Jablome

[wraith cargo]

Use induction on n, and use the fact that all the H's are normal and that the decomposition is unique
Here's the solution for n=2
||Write h1h2h1'h2'. Now because H1 and H2 are normal we can write h2h1' = h1'h2'', so h1h2h1'h2' = h1h1'h2''h2'. We can also write that h2h1' = h1''h2, so h1h2h1'h2' = h1h1''h2h2'. But the decomposition of h1h2h1'h2' into h1'''h2''' is unique, so we have that h2''h2' = h2h2' and h1h1' = h1h1''. Thus we have from our first equality that h1h2h1'h2' = h1h1'h2h2' and hence the map is a homomorphism.||
I maybe didn't use the best notation so it'll take a bit to parse it

[next obsidian]

I think you have the composition mixed up

Once you apply b after doing a

It sjould be

1 2
3 4

Yeah?

Wait

Sorry

1 3
2 4

The 90 degree counter clockwise should be

aba

Since aba = b^-1

[dusty ivy]

U mean c?

Ooh god I found my mistake

Lemme redraw it now

[tribal moss]

You probably don't want to use both b and c as generators.

[dusty ivy]

So I shouldn't define a, b and c like that?

[tribal moss]

I'm not sure what your goal is, so I'm a little wary of saying "shouldn't". But usually the dihedral group is considered to be generated by just one reflection and one rotation.

[dusty ivy]

Thanks 🙏

[frank cosmos]

In the third line of the accepted answer, why is Im f = Lhttps://math.stackexchange.com/questions/1350582/equivalent-condition-for-split-exact-sequence?rq=1

[noble lynx]

this is in the context of adjoining elements of a superring to some base ring, the inclusion doesnt really mean inclusion of sets but the existence of an injective ring homomorphism?

[tribal moss]

Depends a bit on your underlying formalism. Usually these things are written under the assumption that the base ring is literally a subset of the superring, but if there's a reason you can't keep up that pretense, you'll need to work in terms of injective homomorphisms, yes.

[noble lynx]

if you use injective homomorphisms formalism, then does inclusion both ways imply equality(isomorphism)?

[tribal moss]

You'll need some extra footwork to show that in that case, I think.

[rocky cloak]

If psi f = id, then f is injective, so gives an isomorphism between it's domain and image.

I think they got N and L switched up though.

[noble lynx]

hmm, inclusion notation really makes it confusing I feel like then

[tribal moss]

There's a reason why the convention is to pretend we're talking about honest subsets of the larger ring.

[frank cosmos]

thats what i was thinking

i am trying to show that condition (1) implies that there is a homomorphism phi:L to M such that g \circ phi=id

do you have any hint?

[noble lynx]

but doesn't this pretending fall apart, when you try to use two way inclusion to prove equality or something?

[frank cosmos]

I know that M is isomorphic to Im f \oplus Ker psi

which is Ker f \oplus Ker psi since this is an exact sequence

but i dont see how to construct a right inverse for g out of this

[tribal moss]

If the smaller ring isn't literally a subset of the larger ring, you can start by injecting it; the smaller ring is then isomorphic to its image in the larger ring, so you can do the rest of the argument actually inside the larger ring and once the dust has settled transfer the result back along the original inclusion.

[frank cosmos]

(a right inverse thats also a homo)

[rocky cloak]

The proof in the accepted answer works just fine. They've just swapped the names of L and N

[noble lynx]

yeah nm actually, the book literally defines R[u] to be relative to a superring R' that contains R as sets anyway

[frank cosmos]

what is the phi?

[tribal moss]

Yeah. It's possible to do it the other way if you invest the necessary work in keeping the inclusions straight, though -- that can be necessary if you want to phrase the argument in category-theoretic language, e.g. because you hope to transfer it to a different category.

[frank cosmos]

maybe that phi(l)=(0,l)?

[rocky cloak]

You have M = ker (+) Im ~= N (+) L. So phi is just given by this isomorphism

[frank cosmos]

wdym

[vapid vale]

M = ker psi + im f

(check this)

[feral agate]

Yooo, how do I introduce the space that irreducible representation $\rho'$ is a part of when we consider $\rho$ as the representation we decomposed into irreducible representations?

[cloud walrus]

FrankF

[frank cosmos]

yes

[rocky cloak]

Well, just like I said
M is equal to a direct sum, each summand is isomorphic to L and N respectively, so that gives you an isomorphism with the direct sum

[frank cosmos]

im trying to find a right inverse for g

[rocky cloak]

Are you trying to find it or proving that it exists?

[frank cosmos]

prove that it exists

[rocky cloak]

Well then your done

[frank cosmos]

i have an isomorphism M with N + L

so then is the inverse map for g given by phi(x)=(0,x)

[vapid vale]

do you know why skowronski calls 0 -> L -> M -> N -> 0 an extension of L by N? im assuming i just haven't read enough to understand what motivates this terminology in this context

[rocky cloak]

Like the answer on MSE says, if you restrict g to ker(f psi), you get an isomorphism. So a right inverse for g is just an inverse to this isomorphism

[frank cosmos]

hmm i see

LMAO i completely missed that line at the end

sorry!

[barren sierra]

https://math.stackexchange.com/questions/3174538/what-is-the-intuition-behind-short-exact-sequences-of-groups-in-particular-wha

I like the answer here

[vapid vale]

i meant specifically as opposed to an extension of N by L

which appears to be more conventional in homological algebra - i can accept rationalizations of both

[rocky cloak]

Well, L embeds into M, so L is the thing being extended. And that matches for example "extension of rings" or "field extension".

The other way (extension of N by L) probably comes from N being first in the notation Ext(N, L) (which comes from the Hom-functor)

I'm not sure that one is particularly more motivated than another in any given context. You just have to pick a convention.

[vapid vale]

right

alright thanks

[barren comet]

I tried to extend GF(2) the finite field of two elements into
$GF(2^5)$ by adjoining $\alpha$ root of $x^5 + x + 1$ or $\alpha^5 + \alpha + 1 = 0$

[cloud walrus]

Jack_The_Wizard

[barren comet]

But if I compute $\alpha^{40}$ by using $\alpha^{31} = 1$ then I compute $\alpha^9 = \alpha^4 + \alpha + 1$

But I can also look at $(\alpha + 1)^8$ and get the final result as $\alpha^4 + \alpha^3 + 1$.

[cloud walrus]

Jack_The_Wizard

Jack_The_Wizard

[barren comet]

This is bad as then alpha cubed = alpha

And I can't figure out what went wrong at all

It's been bothering me

[rocky cloak]

Maybe you can step through what you're doing a little closer?

Like a^9 = a^40 = a^4 + a + 1

I'm with you there

Then you look at (a+1)^8 and what happens exactly?

[barren comet]

I consider the binomial expansion

[rocky cloak]

The binomial expansion should get you a^8 + 1

[barren comet]

Mhm then I do $\alpha^3\alpha^5 + 1$

[cloud walrus]

Jack_The_Wizard

[rocky cloak]

Sure, so (a+1)^8 = a^4 + a^3 + 1....

[barren comet]

Yes

[rocky cloak]

And then what

[barren comet]

And $\alpha^{40} = (\alpha + 1)^8$

[cloud walrus]

Jack_The_Wizard

[barren comet]

So I have a^4 + a^3 + 1 = a^4 + a + 1

Sorry for being messy I'm on the phone

[rocky cloak]

I see, well

(x^2 + x + 1)(x^3 + x^2 + 1) = x^5 + x +1

So your polynomial isn't irreducible.

That should be the problem

[barren comet]

Ahh

So the field extension only works if the polynomial is irreducible

I kind of assumed it would be since it has no roots

Like neither 0 nor 1

[rocky cloak]

Yes, the ring
F2[x]/(x^5 + x + 1) will actually be GF(2^2) x GF(2^3), so not a field

And you won't necessarily have a^31 = 1

[barren comet]

Thanks a ton

Really i appreciate it, my writeup was a bit rough starting out, thanks again.

[glad osprey]

Finding a root only tells you whether a polynomial has a linear factor, but polynomials of degree > 3 can be reducible without having a linear factor

[chilly ocean]

If I have a ring homomorphism from a group ring RG does it hold linearity all the way down? i.e

F(Σ r_g . g) = Σ r_g . F(g) ?

since each g is itself embedded in RG?

[coral spindle]

You need to write F(r_g) unless you’re talking about R-algebras, but apart from that you are right.

[chilly ocean]

say F: RG -> S, then S needs to be a R-algebra?

[coral spindle]

No

You just need to decide where the coefficients in RG land

[chilly ocean]

sorry then I don't see what you mean. aren't coefficients by definition elements of the ring R?

[coral spindle]

Not of S

[graceful elm]

how do rings of matrices work? for example, M_3(Z/6Z)? i've been asked to count how many ideals it would have and i have no idea how to even begin to conceptualize this ring

I believe i've just read that there is a bijection between the ideals of M_n(R) and the ideals of R (for any ring R). Is this true?

[lusty marlin]

Yes, J is an ideal of M_n(R) if and only if J=M_n(I) for some ideal I of R

[graceful elm]

wonderful, thank you

[limpid ferry]

This is how my teacher write down the lattice for Z_18, could anyone help me explain why <1> is at the bottom? Also, he said something about 1 being the identity but I thought the identity was 0

[vast quiver]

your teacher should have put <0> (where 0 is the identity), I think they slipped up and were thinking of Z_18 as a group with the operation being written as multiplication, maybe with a generator x, so that the elements are
{1=x^0, x^1, x^2, ..., x^17},
and 1 is the identity. however, above they were writing <2> for example, not <x^2>, so again you're right that they should have written <0>

[sullen pumice]

I've just read a bit on fields in my abstract alg textbook, but it didn't explicitly state that it must be closed under multiplication and addition, but are these closure properties not apart of a field?

or is it just that these are so obvious that it don't need mentioning

[vast quiver]

when the book stated the definition of a field, maybe it stated something like addition is a function
+: F x F —> F
where the notation itself is stating that addition is “closed”, ie addition is a function that takes in two elements of F, and it must give you another element of F (not something in a different or larger set)

[sullen pumice]

i see thank you. In my book it did not state this, so I was a little bit confused.

[vast quiver]

saying that addition is a function F x F —> F or in words saying that addition is a binary operation (same thing by definition) is I guess implicitly saying that addition is “closed”

[sullen pumice]

right, so I guess in a way it follows quickly from the definition, so the author didn't mention it

and the same applies to multiplication as well?

[vast quiver]

yep, whenever an author says a binary operation, they mean a function F x F —> F, so it by definition has to land inside F

[sullen pumice]

i see i see

ok great, thank you so much!

[vast quiver]

yeah no problem!

[old zenith]

hi i dont understand why the definition of direct sum and product differ for infinite vector spaces

so going by product and co-product

the direct sum is the co-product, so we've got this kind of diagram with the coproduct, with f o i_n = f_n

i just really dont quite get why the direct product isnt also just the coproduct

[old zenith]

nvm i understand now

the coproduct is defined st f is uniquely defined by each f_n and i_n

if you have an element e with infinite support, f(e) can be defined arbitrarily

[dusty ivy]

Is this way correct

[kind temple]

what axis are you reflecting over for a?

[dusty ivy]

Vertical

[kind temple]

looks good

[dusty ivy]

But why 90° rotations doesn't work

[kind temple]

wdym by don't work

[dull marsh]

I think they mean why don't they commute with reflections

[dusty ivy]

I tried clockwise 90 degree rotation as b but I can't draw it correctly

[dull marsh]

You've answered your question then

[dusty ivy]

Yea but why don't they commute

[dull marsh]

Are you looking for answer more meaningful than just a counterexample?

[dusty ivy]

Meaningful

I'm not sure if it's only I can't draw it or if it's something that can't be drawn

[dull marsh]

Interestingly enough I think this has to do with the fact that rotation by 180 degrees is equivalent to reflection about the origin when you consider its action on a single line passing through the origin

[crystal vale]

What if I try to make a group by generators x and y such that x^(p^2) =1 and y^p =1 and xy = x^(-1)y and p is a prime number.

Then is it non-Abelian group of order p^3?

I just try to make a non-abelian group of order p^3.

[lusty marlin]

xy = x^-1 y implies x²=1

So x has order 2

[crystal vale]

Wait

xy = yx^(-1)

I am trying to find a relation but first I am using this relation which is used in dihedral

[lusty marlin]

Why do you want to make a non-abelian group of order p³?

[crystal vale]

Because it is a question

[tribal moss]

That gives you x^n·y = y·x^(-n), in particular x^-1·y = yx.
Now x·y^m = y^m·x^((-1)^m), and setting m=p here gives you x·1 = 1·x^-1 when p is an odd prime.

If you also have x^(p²)=1, that kills x completely.

[tribal moss]

Hint: The standard example I know of is a matrix group over F_p.

[crystal vale]

Group of invertible matrix, right? But it has order (p^(n) -1)( p^n - p).....(p^(n) -p^(n-1))

[tribal moss]

Yeah, you you'll need to restrict to a subgroup of that.

[crystal vale]

Yes I need to think of it

[tribal moss]

"A matrix group" generally means a subgroup of GL(n,F), not necessarily the entire GL(n,F).

[crystal vale]

Okay thank you

[tribal moss]

Further hint: ||If you consider triangular matrices, there are some entries you can choose freely without risking a change in whether the matrix is invertible.||

[crystal vale]

I am not sure but what if we take 3×3 matrices with upper triangular matrices with diagonal 1, but its inverse also has this form, right?

Yes inverse also has this form

[crystal vale]

Is it correct?

[tribal moss]

Yes :-)

[crystal vale]

Okay, thanks a lot

[untold basalt]

Prove that the group of invertible n x n matrices GL(n,F) with entries from the field F is not solvable for n \geq 5.

I thought that maybe this group has A_5 as a subgroup?

[dim widget]

prove that the commutator subgroup of Sl_n(F) is Sl_n(F)

[dim widget]

That’s another way

[untold basalt]

Ah yes that was my second guess, thanks!

[dim widget]

Actually it’s much easier to show that the group contains S_5

I think your way is better

[long geyser]

if I show that xH cap yK is closed under multiplication by H cap K from the right, is that sufficient to show it is a left coset of H cap K?

uh, no

could be a union of cosets

nvm, figured out the solution, just struggled with coming up with a candidate for the coset representative

but it was staring at me

[long geyser]

need help with this
if we call Y the closure of X under all conjugates, it is easy to see that the subgroup generated by Y is contained in N (since Y itself is)
I'm struggling with the converse direction, and I'm pretty sure the way to do it is showing that the subgroup generated by Y is itself normal

I tried to use this construct, but unable to show that the set of words on Y is closed under conjugation

[tribal moss]

Conjugation is a homomorphism, so when you conjugate a word you can conjugate each letter in it separately instead.

[long geyser]

ahhh that's right

why didn't I think of that

thanks

[tribal moss]

How is "generated by" defined here? Intersection of all containing subgroups?

[long geyser]

yes

[tribal moss]

Ok, then it makes sense.

[cloud walrus]

Hello1

Hello1

[frank cosmos]

can i use this lemma here?

[cloud walrus]

Hello1

[frank cosmos]

(When showing Hom(P,M) is a functor for P projective defined as above)

[cobalt heath]

Indeed

You can check exactness piecewise, between each 2 arrows.

[weary spade]

what do you call a group with no identity element

[rose prism]

useless

[weary spade]

ok but listen

[rose prism]

no

[weary spade]

pls

consider the set {i, -i, j, -j, k, -k} in R3 with the cross product

it forms some sort of algebraic structure similar to a group

but there's no identity

let me draw you a picture maybe

[rose prism]

its called a semigroup

[weary spade]

so it's a semigroup?

[rose prism]

no

the thing you originally asked

i took to mean “group axioms minus identity”

your thing is not even associative

[kind temple]

holy shite

i TA'd for a class

and the teacher called monoids semi-groups

[weary spade]

wdym? the cross product is associative

[rose prism]

no

[kind temple]

so i had to get used to calling monoids semi-groups

my fault memorylessfunctor

[weary spade]

yes it is

[rose prism]

are you sure about that

[weary spade]

oh shit

nvm

[rose prism]

incidentally R^3 equipped with cross product is an example of an algebraic structure which is more commonly studied than semigroups/monoids/whatever

namely a lie algebra

but the underlying vector space structure is quite relevant there

[weary spade]

ok this is cool

this is what i was looking for kinda

[rose prism]

is it

[weary spade]

i mean it doesn't really describe the i j k thing i was talking about but you gave me something to read

so thanks i guess

[rose prism]

enjoy

[tribal moss]

That's not total either; you'll need a 0 for i×i.

[kind temple]

by zero, you mean identity?

[tribal moss]

I mean the cross product of any vector by itself it the zero vector, which is not listed in {i, -i, j, -j, k, -k}.

[weary spade]

ok let me try drawing the cayley graph

it should be generated by i and j

[tribal moss]

A further problem is that if you say "like a group but without identity" instead of "like a monoid but without identity", then you're implying that inverses exist, and it's difficult to define what that even means while still leaving it possible not to have an identity.

The table above suggest that such a thing would be called an "associative quasigroup", but that is just a group.

[weary spade]

nvm there aren't even generators

or it can't be generated by i and j alone i think

[tribal moss]

The trouble there is again lack of associativity, which means that not every combination of generators can be expressed as a linear word.

[weary spade]

word?

[tribal moss]

A product of a finite sequence of generators.

[weary spade]

ah i see

that makes sense actually

so i guess this is why quaternions are useful

their multiplication actually forms a group and such

[tribal moss]

Yeah.

[weary spade]

it's almost like you need 4 dimensions

3d doesn't work

i'm sure that's some theorem we talked about in class

[tribal moss]

That's the Frobenius Theorem about real division algebras.

[weary spade]

hmm. i wonder what would happen if you tried 3d. like just 1, i, j with no k. but then you run into the issue of defining the products

like what should ij be

[south patrol]

I forget about what is due to whom

Cause there is also Radon-Hurwitz stuff

Ah yeah I think what I have in mind is Hurwitz which is for normed division algebras

But normed division algebras needn't be associative, so they are not even necessarily division algebras

Lol

e.g. octonions

[fresh gate]

how are alternating groups simple?

oh

normal subgroup

I see.....

nevermind

ignore me

now that I think about it

if it was what I think it was it would have only been prime cyclic groups lol

[tribal moss]

"the three infinite classes of such, namely" (then goes on to list four classes)

[rocky cloak]

I guess only the first 3 are infinite. And that the last class only contains the Tits group (at least if we removed the overlap)

[south patrol]

I guess it is still kinda poorly written imo

[rocky cloak]

It is, but I guess it's fine. They also explain it a little more in [note 1]

[tribal moss]

I read note 1 and it wasn't even obvious to me what it had to do with the item it's attached to.

[rocky cloak]

It's saying that if you take the derived subgroups of this family of Ree groups you get an infinite family of simple groups. And the first group in this family is the Tits group (and all others are groups of lie type).

[tribal moss]

It's not saying what "Ree groups" have to do with the text the note is attached to.

[rocky cloak]

The text is talking about the derived subgroups of groups of lie type, then it gives a family of groups of lie type and their derived subgroups

[tribal moss]

So it's talking about one of the confusingly many families of "groups of Lie type" and saying that the Tits group is the derived group of one of those?

[rocky cloak]

Yeah, it's basically just trying to argue that the Tits group is not sporadic, since it fits into an infinite family

[tribal moss]

But wait, if the groups of Lie type are simple (they are the family listed just above), how can they have distinct derived groups at all?

[rocky cloak]

Not all groups of lie type are simple

[tribal moss]

That's pretty confusingly written, then.

[rocky cloak]

I guess. It also just is somewhat complicated and confusing I guess

[tribal moss]

Especially since they go to the length of specifying "cyclic groups of prime order" and "alternating groups of degree at least 5".

[rocky cloak]

I guess it would feel silly to say that a group is simple if it is a "simple group of lie type"

It also doesn't really help that "group of lie type" isn't really that precisely defined anyway

[dim widget]

You mean in the article?

[ornate tiger]

i proved that it is associative but i have a problem proving that it has e

[dim widget]

the identity is just 0

[tribal moss]

(A shortcut is to prove that it is isomorphic as a set-with-binary-operation to R\{0} under multiplication).

[ornate tiger]

how did u know that ? is it the same as normal filed R ?

[rocky cloak]

Yeah, Wikipedia doesn't commit to any specific definition, but just talks about specific families of groups of lie type

[dim widget]

that’s annoying. I guess it is kind of elaborate to define compared to the other families

someone should edit

[tribal moss]

The article (especially the "List of finite simple groups" one) also seems to presuppose that the reader already knows how to read its ^number Letter _number notation for specifying each of the groups it lists.

[dim widget]

Just try it out

it is the same identity as in R with addition yes

[dire turret]

i guess we want some b such that a + b + ab = a, so b = 0 is a pretty natural choice

[dim widget]

indeed it is the only choice

[dire turret]

so true

[ornate tiger]

ok thx guys

[sudden condor]

I was doing an exercise that's like "show that fibre products and coproducts exist in the category Ab"

but i don't think i really understand fibre products

can't you just equip any product with two maps to a final object to make it a fiber product

(pls don't spoil the solution btw)

[fresh gate]

when is G ≅ (G / N) X N?

[tribal moss]

Yes, every product can be made as a fiber product, but you cannot make every fiber product as a product.

"Show that fiber product exist" means to show that for every two morphisms with the same codomain you can complete them into a pullback square.

[sudden condor]

ohhh

thank you

[tribal moss]

You need to guess a "complement" of N inside G, which satisfies the conditions are given at https://en.wikipedia.org/wiki/Direct_product_of_groups#Algebraic_structure

[next obsidian]

There’s also a cohomological interpretation (let’s just say stuff is abelian), which is when G corresponds to the trivial class in Ext^1(G/N,N)

[fresh gate]

what does Ext mean?

[next obsidian]

Don’t worry about it, it’s more trouble to explain everything. It’s the derived functor of Hom but also classifies extensions. You’ll see it at some point, so just take this answer as a little taste of something to come later

[untold basalt]

hey a question regarding nomenclature: does 'splitting field of a polynomial' usually mean the smallest field that contains all its roots or just a fileld that contains all its roots?

[fading field]

smallest

[untold basalt]

My professor seems to be using them interchangably

[fading field]

for example C is not the splitting field of x^2 + 1 over Q

[untold basalt]

I see, thank you.

so one would just say that x^2+1 splits into linear factors over C because C contains it's splitting field?

[south patrol]

Well, saying "because" seems a bit wrong

[tribal moss]

[whatever polynomial] splits into linear factors over C because C is algebraically closed.

[delicate bloom]

the egg got a crack in it because it was ran over by a dump truck

[sly crescent]

Up to isomorphism, how many commutative unital rings with 8 elements are there?

[rocky cloak]

Z/8,
Z/4[x]/(2x, x^2),
F8,
Z/2[x]/(x^3),
Z/2[x, y]/(x, y)^2

Should be all the connected ones

Then there's
(Z/2)^3
(Z/2)x(Z/4)
(Z/2)x(Z/2[x]/(x^2))

And I guess that's it

Finite rings are artinian and commutative artinian rings are products of local rings.

To classify local artinian rings you think about what the residue field can be and what the radical can be.

[sly crescent]

You forgot (Z/2)xF4

[long geyser]

I'm struggling with the last sentence here: how do we know that the minimal poly of omega has m distinct roots? to be clear, the entire discussion doesn't depend on this number, just knowing they are distinct is enough

I'm confused about the m part, not the distinct part

[rocky cloak]

A degree m polynomial has m roots

[sly crescent]

What does the second half of Z/2[x,y]/(x,y)^2 mean?

[long geyser]

but why is irr(omega, k) degree m, I guess something fundamental is getting missed by me

[sly crescent]

Is it a typo and supposed to be (x,y^2)?

[wraith cargo]

I think it's supposed to be product of (x,y) with itself

[sly crescent]

What is (x,y)?

[wraith cargo]

The ideal generated by x and y

[sly crescent]

Oh, I see now

[rocky cloak]

I think that might be a typo actually. The minimal polynomial must have degree strictly less than m

[long geyser]

I see, thanks
btw this book has its fair share of mistakes so a type wouldn't be out of the ordinary

[rocky cloak]

omega being seperable is the main point anyway

[long geyser]

yeah

[limpid ferry]

If H is a subgroup of G, then (aH)(bH) =abH, where a and b are in G, is it correct?

[chilly ocean]

H has to be normal in G

then yes

[ornate tiger]

if i'm not mistaken this is what they mean right

[sonic coral]

yes

[ornate tiger]

ok thx

Does det(AB) = det(A) * det(B) ?

[dull marsh]

Yes

[ornate tiger]

thx

oh quick question for all the matrices or just in SL_2(R) ?

[dull marsh]

The question clearly says "in SL_2(R)"

[ornate tiger]

no i mean does this det(AB) = det(A) * det(B) for all matrices

[dull marsh]

Ah

That holds for all matrices A, B, yes

As long as they are square matrices of equal size

[ornate tiger]

oh ok thx

[untold basalt]

Suppose that $|G|=p^nm$ with $p$ prime and $gcd(p,m)=1$. How do I show that $\binom{|G|}{p^n} \not \equiv 0 \mod p$ ?

[cloud walrus]

Heywood Jablome

[untold basalt]

this comes up in sylow's theorems

[rocky cloak]

Consider the set X of all subset of G of cardinality p^n.

Let P be the sylow subgroup of G, and let P act on X by left multiplication. Since every orbit has order p^k for some k
|X| = # fixed points (mod p)

The fixed points are exactly the cosets of P, which there are m of.

So (|G| choose p^n) = m (mod p)

This uses that G has a sylow subgroup, in case you haven't proven that yet, just replace G by for example the cyclic group of order |G|, which you know has a sylow subgroup

[ornate tiger]

what is S_n ?

is it like a special group like GL_n ?

[rocky cloak]

It's the nth symmetric group. I.e. the group of bijections from {1, ..., n} to itself

[rocky cloak]

Field with one element has entered the chat (please ignore this message)

[ornate tiger]

so if they asked us to prove that some subsets are sub groups for S_n and if yes show that they abelian or not how to do that

this for example

[dim widget]

it's a subgroup

[ornate tiger]

do we need to check the normal subgroubs properties ? like non empty identity etc

[dim widget]

Yes

but they are easy to check

[rocky cloak]

In general, you would have to check whether it's closed under the operation, whether ab = ba and so on.

But in this case, if you just try to simplify the equation
w^2 = w
a little, you might realize what's happening

[ornate tiger]

can u show me an example that it is closed under operation

[dim widget]

okay $\mathbb{Z}$ is closed under addition because if x, y are integers so is $x+y$.

[cloud walrus]

Math_Discord_Final_Girl

[dim widget]

It's closed under inversion because $-x$ is an integer if $x$ is an integer.

[cloud walrus]

Math_Discord_Final_Girl

[ornate tiger]

why we didn't use w^2 = w

[dim widget]

because i showed you an example (as you requested), I didn't solve your homework for you.

[dim widget]

@ornate tiger you should follow this hint of jagr2808, it will lead you to happiness

[ornate tiger]

no i'm depressed

[dim widget]

that's because you're not following the hint

[ornate tiger]

ok i'll try thx for both of you

what if i failed

[dim widget]

there are other more important things in life than solving abstract algebra exercises

so it'll be okay either way

[ornate tiger]

i need a degree

[dim widget]

there are lots of things to study which are related to math but don't depend on your ability to solve abstract algebra questions

[low wyvern]

Guys, Supposing r in a ring R, how do I prove -(-r) = r ...I feel like I am being stupid but I am in a hurry tbh and cannot think clearly

[dim widget]

0 = (1 + -1)r = r + -r

[low wyvern]

Oh yes, thank you I am a tool

[ornate tiger]

What are the possible orders of elements in S_4 and why ?

what i should do to know them

[lilac mango]

Do you know how to write permutations as product of disjoint cycles?

[ornate tiger]

show an example

[lilac mango]

Lets say you have the permutation 1->2, 2->4, 3->3, 4->1

That is (124)(3)

[ornate tiger]

right

[lilac mango]

Ok ok so in general (probably you already know this), any permutation can be written as a product of disjoint cycles

And uniquely

Well "uniquely"

You could have written (124)(3) as (241)(3) for example

But (241) is the same cycle as (124)

[ornate tiger]

ok wait |S4| = 24 so that's a lot like if we considered (124)(3) = (241)(3) then now we have 23 and we still have a lot to check

is this right ?

[lilac mango]

No that's not right

[lilac mango]

These permutations are the same

Oh wait I see what you are saying

Thought you were saying something else

Well the point is you don't have to check every permutation separately

Say if you have a permutation of the form (124)(3)

Or in general

(abc)(d)

What is its order?

[ornate tiger]

to know the order what do we need to reach ?

[lilac mango]

Well you could just compute it

What is ((abc)(d))²

It's (acb)(d) which isn't the identity

[ornate tiger]

the identity here is (abc)(d) ?

[lilac mango]

But ((abc)(d))³=
((abc)(d))²((abc)(d))=
(acb)(d)(abc)(d)=
(a)(b)(c)(d)

[lilac mango]

No it's the identity in S4

The permutation that sends 1 to 1, 2 to 2, 3 to 3 and 4 to 4

[ornate tiger]

so it's this (a)(b)(c)(d) ?

[lilac mango]

Yep

[ornate tiger]

waht is the order of this

[lilac mango]

3

[ornate tiger]

because it has 3 elements ?

[lilac mango]

I mean in a sense but no

Think about what happens to a if you apply the permutation three times

[ornate tiger]

oh we reached the identity when we raised to the third power

[lilac mango]

The first time it goes to b

[ornate tiger]

so it has the order of 3

[lilac mango]

The next time to c

The next to a

It takes 3 "turns" to reach a back

Same happens to b and c

And d just stays put all the time

[ornate tiger]

this is fales ?

[lilac mango]

That's true

So in general when you have a permutation of the form (abc)(d) it will have order 3

What other "type" of permytation can we have?

[ornate tiger]

(ab)(cd) ?

[lilac mango]

Yes

There are two other types

[ornate tiger]

so the order is what 2+2 ?

[lilac mango]

No

[ornate tiger]

so how to calculate it like the prev one

[tribal moss]

The order is how many times you need to repeat the permutation until you get back to the identity.

[ornate tiger]

so it's 2 ?

[tribal moss]

Yes.

[ornate tiger]

what are they

[lilac mango]

Well theres of course (a)(b)(c)(d)

And theres (abcd)

[ornate tiger]

the order of this is 1

[lilac mango]

Yes that's the identity

[ornate tiger]

and this is 4 ?

[lilac mango]

Yep

[ornate tiger]

ok so we do 1+4+3+2 ?

[lilac mango]

No no we now know what's the order of each possible type

[tame carbon]

Hello, sorry to butt in - Are there any good resources for group theory problems with worked solutions? Trying to expand my knowledge

[lilac mango]

We now need to count how many permutations there are for every possible type

Say for (a)(b)(c)(d) the only option is (1)(2)(3)(4)

So we have one element of order 1

Now how many permutations of the type (ab)(cd) are there?

[ornate tiger]

yeah it's true but how did u calculate it is there a formula for this ?

[lilac mango]

You have to count them

There isn't a formula

I mean the options were (1)(2)(3)(4), (2)(3)(4)(1), etc.

But here the order doesn't matter

They still represent the same permutation

So it's just one

Lets imagine for a sec we are in S3 and want to count the elements of cycle type (abc)

Let's say we start with a=1

Then there are two options for b and c is forced

[ornate tiger]

so there's 6 ? at the end

[lilac mango]

And we notice that we already counted every possible permutation of type (abc) since (231) is just (123)

[lilac mango]

No

[ornate tiger]

there's 2

[lilac mango]

Yep

You have to eliminate redundancy

[ornate tiger]

yeah but that does not help like S3 for (a)(b)(c) is 1 and for (abc) is 2 for (ab)(c) is also 2 ig right

[lilac mango]

(ab)(c) is 3

[ornate tiger]

it goes for a to b to c ? why not a to b to a ?

[lilac mango]

Oh yes the order is 2

But there are 3 permutations of that type is what I'm saying

[ornate tiger]

yeah

so in S3 what can we get u said we can't say it's 2+2+1

[lilac mango]

What we can say is well

Theres one of type (a)(b)(c)

So theres 1 of order 1

Theres 2 of type (abc)

So we have 2 of order 3

And theres 3 of type (ab)(c)

So 3 of order 2

And we are done

So:
Order 1: 1
Order 2: 3
Order 3: 2

Notice how 1+2+3=6=3! so we counted all permutations

[ornate tiger]

what if we had S12 this is hard

[lilac mango]

Yes it is lmao

But it's easier than counting permutations one by one

I wouldn't say it is hard though

It is long

[ornate tiger]

yeah i want to see if i understood what u said in S4

[lilac mango]

Ok lets see

[ornate tiger]

order 1: 1

this is the easy one

[lilac mango]

Oh wait theres a type we didnt count I just realized

(ab)(c)(d)

[ornate tiger]

what order is this 2 ?

[lilac mango]

Yes

We have two ways of getting order 2

[ornate tiger]

i hear voices in my head like randy orton i'm going to rko this topic

[ornate tiger]

the other one is (ab)(cd)

so do we sum the 2 ways

[lilac mango]

Yes

[ornate tiger]

so the order 2 : 4 ? or what like i know (ab)(cd) has the order 2 and (ab)(c)(d) has the order 2 like does that mean we have 4 ?

[lilac mango]

No we have to count how many (ab)(cd) there are first

How would you do that

[ornate tiger]

well we have (ab)(cd) we can have (ac)(bd) , (ad)(bc)

right ?

[lilac mango]

Yes

[ornate tiger]

so there's 3 here

[lilac mango]

Yes

One way to think of it is well a has to have some partner

It is either b, c or d

So 3 choices

And the rest is forced

[ornate tiger]

and (ab)(c)(d) we can get (ac)(b)(d) and (ad)(b)(c)

[lilac mango]

No theres more

[ornate tiger]

did a lost his partner

[lilac mango]

Yes a can have no partber lmao

[ornate tiger]

and now b and c partners

[lilac mango]

Or

[ornate tiger]

(ab)(c)(d) , (ac)(b)(d) , (ad)(b)(c) , (bc)(a)(d) , (bd)(a)(c) ?

5 ?

[lilac mango]

theres one left

[ornate tiger]

(dc)(a)(b)

[lilac mango]

Yep

[ornate tiger]

ok here we have 6 and (ab)(cd) has 3

so 9

[lilac mango]

Yes

Lets go for (abc)(d)

[ornate tiger]

wait i think there's a mistake (ab)(cd) is 2 cycles but (ab)(c)(d) is 3 cycles

do we only care about the order

like the order of 2

[lilac mango]

Yes

[ornate tiger]

ok cool

for (abc)(d) we can get (abc)(d), (adc)(b) , (adb)(c) (bdc)(a)

[lilac mango]

Theres more

Like first you choose who misses out in the fun

4 options

And then it's just counting the possibilities for (abc)

[ornate tiger]

how did u know there's 4

there's 7 overall right

[lilac mango]

a, b, c and d

Thats why

[lilac mango]

No

Like how do you create a permutation of this kind. Well, you chose one out of the four options you have to leave him out

And then the other 3 can be ordered in two ways

(abc) or (acb)

[ornate tiger]

but they are the same

[lilac mango]

Just like we did in S3

[lilac mango]

No they are not

The first one sends a to b and the sexond one sends a to c

They are the same "type" but they are different as permutations

(123)(4) and (132)(4) are not the same

[lilac mango]

Its what we did here

[ornate tiger]

ok let me try find all of them
(abc)(d) ,(abd)(c) , (acb)(d) , (adb)(c) , (adc)(b) ,(bca)(d) , (bcd)(a) , (bdc)(a) , (bad)(c) , (bda)(c)

i feel there's more

[lilac mango]

There should be 8

Yes (bca)(d) and (abc)(d) are the same

You are xounting it twice

[ornate tiger]

yeah true

[lilac mango]

Same with (adb)(c) and (bad)(c)

[ornate tiger]

really i don't see that

[lilac mango]

Here a->d, b->a, c->c and d->b

Thats true for both

One way to count them is just saying "ill put the smallest one first always"

Like (abc) you can write it as (bca) and (cab)

But following the convention that the smallest one (in this case a) goes first always, you dont overcount stuff

[tribal moss]

Alternatively you accept the overcounting and correct for it afterwards: There are 4! ways to write something of the form (_ _ _) (_), but each permutation of that type has exactly 3 ways to write it, so divide by 3 to undo the overcounting.

[ornate tiger]

ok i see that they are the same
order 1: 1
order 2: 9
order 3: 8
order 4: ?

i think we will get 6

so we get 24

[lilac mango]

Yes

[ornate tiger]

ok i will go back to everything u said to understand better and keep it in my head thanks for your time

[vivid birch]

Ler G be a free abelian group with generators U, V and H be another free abelian group with generators a,b,c. Let f: G -> H be the homomorphism defined as

f(pU + qV) = (q-p)a + (p-q)b + (p + q)c

How do I determine Im(f)?

[long obsidian]

If I have a polynomial ring in several variables like for instance Z[x,y,z] I'm pretty sure the set of all monomials like XYZ, zzzzz, XYX form a basis of the space. But I'm not really sure how to say this with the language of modules.

Do you say the monomials form a basis of Z[X,Y,Z] as a Z module?

[dull ginkgo]

Yes. (It’s an algebra)

Every element in R[X,Y,Z] (R a ring in general) is a finite sum over scaled monomials in 3 variables

[rocky cloak]

The image will be things of the form
(q-p)a + (p-q)b + (p+q)c

or what do you mean determine?

[dull ginkgo]

In fact this basis can be ordered. Can you think of a way to order them?

[vivid birch]

Oh, I want to know what elements generate the image

[rocky cloak]

The image of U and V will, or the image of any other choice of generators for G

2c and a - b + c for example

[long obsidian]

Well X,Y,Z has an order. I can't remember what the set X,Y,Z is in relation to R[X,Y,Z] I guess it's a subset of the generators as a ring.

But probably if you use the alphabetical order of X, Y, Z then if you have in multi index notation a=(a1,a2,a3) say W^a=X^a1 Y^a2 Z^a3 then I guess if you had another monomial W^b with b=(b1,b2,b3) it feels like you can do some type of lexographical order where W^a<W^b (say i in {1,2,3} is the smallest such that ai not bi) whenever ai<bi

I think I've heard of others but I'm not sure. This makes me think of grobner bases

[dull ginkgo]

There’s plenty of them ye

I remember reverse lexicographic order I think because it is way fucking easier to construct inductively

There’s this really silly way of describing it

It’s like a “symmetric polynomial” in the lower ones where addition is instead >

Or at least how I found it when doing some silly combinatorics stuff before I did algebra stuff

[coarse stag]

Is it possible to define a DFT over an arbitrary ring as long as you choose a unit with order N? And if so, are there any ways to actually compute the corresponding prime?

[dull ginkgo]

You need a principal root of unity

That’s literally the whole page lmao

[coarse stag]

yeah, thats what i figured, ty

[livid flare]

Hello, Is there any example of an non connected solvable subgroup of SL2(C) ?

[rocky cloak]

Z as [1, n; 0, 1]

[dim widget]

the center is ${\pm 1}$

[cloud walrus]

Math_Discord_Final_Girl

[grizzled mango]

About the 49 exercise in the Gallian's book, 5° ed.

I used the Fundamental Theorem of Cyclic Groups to find all the subgroups of <a> and <b>, which I assumed to be the cyclic subgroups of G with orders 4 and 5, respectively, and I found only <a²> of order two and the trivial {e} of order 1. Is there any other way that can help me derive more cyclic groups of G with orders other than these? I'm racking my brain here and so far haven't seen another way.

[terse crystal]

4,5 are coprime

Consider the kernel of Z->(Z/mZ sum Z/nZ)

[grizzled mango]

Thank you for your help! However, I am not yet familiar with the kernel concept, as I am still in Chapter 4 of Gallian's book. So far, it has only covered definitions and theories about groups, subgroups, finite groups, and now, in Chapter 4, cyclic groups. Could you suggest an approach that is limited to the elementary theorems about cyclic groups?

It seems that Gallian will only start discussing kernel in Chapter 10.

[terse crystal]

My god…

I would read a different one…

[rocky cloak]

If x has order 4 and y has order 5, what is the order of xy?

[terse crystal]

You sure you are saying chapter 10 not section 10…

[grizzled mango]

We might be talking about the same concept but using different names. So far, I haven't come across the word "kernel" or any notation similar to what you showed me. Could it be that "kernel" is what Gallian calls the "generator of the cyclic group"?

And yes, I found the word "kernel" appearing only in Chapter 10.

[terse crystal]

Just follow sir jagr

[grizzled mango]

Great! 😃

[grizzled mango]

I will check out this one, thanks! 🙂

[slim kayak]

It only has meaning once you have interacted with the concept of a homomorphism. Once you have, the kernel is just the set of all elements getting send to the identity element and will turn out to be part of a very important class of subgroups having a special property and is generally extremely useful really anywhere in math

(200 pages until group homomorphisms is cursed, jesus)

[south patrol]

Should be page 2 lol

[slim kayak]

grothi is turning in his grave, no relative viewpoint in miles

[limpid ferry]

I just started learning Normal group recently and the definition of an operation being well-defined, that is, if aH = cH and bH=dH, then (aH)(bH)=(cH)(dH).
This brought me to the question related to set theory: Given the sets A,B,C,D. Is it true that, if A=C and B=D, then the Cartesian product AxB=CxD?

[rocky cloak]

Yes, this is true.

Most constructing you'll encounter are well defined, simply because things that aren't well defined don't make much sense

[velvet goblet]

i have a question is it true that for all commutative groups G the centeralizer and the normalizer of any subset A of G are the same ?

[limpid ferry]

Wait, so, given a group G with its subgroup H and a,b \in G. Can aHbH be well-defined, like I described, without H being G's normal subgroup? I doubt it is not

[rocky cloak]

So you can define aHbH simply as {xy| x in aH, y in bH}. But then the problem is that the result might not be another coset.

So this does not define a function G/H x G/H -> G/H

(unless H is normal)

[rocky cloak]

But not that aHbH is completely different from the cartesian product, so not really related to your previous question

[limpid ferry]

It might not be another coset of H. But what I mean is if (aH)(bH) = (cH)(dH)= some subset G' of G then? (if aH = cH, bH=dH, and H is not normal)

[tribal moss]

This is at best a very confusing description of "well-defined" and at worst directly wrong.

[rocky cloak]

Yes, the definition
{xy | x in aH and y in bH}
only depends on the sets aH and bH. Not by what we chose to call them or anything like that

However, some people might define
aHbH = abH
this does give you a "mapping" of cosets to cosets, but now it might not be well defined

[tribal moss]

It would be better to write the problematic definition as something like

aH # bH = (ab)H
with an explicit symbol on the LHS to make it clear that it's not (or at least not a priori) trying to define simply the usual product of subsets of a group.

The question is then whether this defines any function from pairs-of-cosets to cosets at all.

If it does (and that turns out to be exactly when H is normal) then there is a function.
If it doesn't then sayign that "# is not well-defined" is common but strictly speaking nonsense because the point is that there is no # which satisfies the above equation for all a and b. So there's no concrete thing that we can look at and say, "that is a thing, but the thing is not well defined".

[limpid ferry]

I think my words here was quite not correct, the full definition should be: "(aH)(bH)= abH is a well-defined operation on left cosets of H in G if aH=cH and bH=dH, then (aH)(bH) = (cH)(dH).
What I really wanted to see was just if "(aH)(bH) = (cH)(dH) without H being normal"

[tribal moss]

That's still confusing. What you should say is:

aH # bH = (ab)H defines an operation on left cosets of H in G if aH=cH and bH=dH together imply (ab)H = (cd)H.

[long geyser]

can we not in general describe k(z_1, ... , z_n) as rational functions on z_1, ... , z_n? that would avoid the need for induction

[south patrol]

Yeah lol

More generally given a group G and field K, the fixed points K^G form a subfield

Here take G = <sigma> and K^G contains z_1,...,z_n hence K

[long geyser]

the K^G thing comes later on in the book

[south patrol]

Lol sure, I'm just mentioning how it fits into smth larger

[long geyser]

for this guy, I'm getting confused with regards to the application of second iso thm

for this, you need G_{i+1}N to be normal in whatever ambient group, right? I don't see what ambient group G_{i+1}N is normal in

oh

ok I

'm so dumb

I missed that the entire point of the calculations in between is to show G_i+1N is normal in G_iN

nvm

sometimes I'm so focused on fact checking the individual claims that I miss the big picture

[long geyser]

does my proof look good?

well, sec

[cloud walrus]

lems

[long geyser]

ig I should mention that I know [G : H] is finite as G is

and forgot my final sentence: Now, repeat this process with $ K $ instead of $ H $ until $ \left| G / K \right| $ is of prime order.

[long obsidian]

What is a basis of the single variable rational functions Q(x) as a Q module?

Is it something like the set of all ab^-1 where b is in Q[x] and is irreducible?

[cobalt heath]

A lot, I bet

Would at least include 1/(q(x))^n for irreducible q.

I think homogenization would be great to simplify this

[long obsidian]

What is homogenization? Are you talking about the graded structure of Q[x]?

[cobalt heath]

I mean making a polynomial homogeneous.

You introduce additional variable to match the degree.

[long obsidian]

Oh okay. Hmm let me think about it I'm not sure why homogenization of a polynomial would help

[cobalt heath]

Q(x) sadly does not have a grading structure. Homogenizing is to give it a grading.

[long obsidian]

Actually I'm looking online and this seems like a more complicated problem that could even depend on the axiom of choice. Hmm..

[cobalt heath]

Well, anything requires axiom of choice tbh

Like IIRC, existence of maximal ideal generally require AoC

[spare isle]

Is there a proper term for functions of the form f(x,x) = x akin to idempotence? (Where f does not necessarily give rise to a binary operation)

[delicate bloom]

can you elaborate a bit, do you mean to say f(x,y)=(x+y)/2 would be a special case when f(x,x)=x or something else

[spare isle]

What I mean is that $f: A \cross A \to B$ with $A \subseteq B$ and for all $a \in A \subseteq B$, $f(a,a) = a$ (with the rest of $f$ purposely not left defined for generality)

[cloud walrus]

JJCUBER

[delicate bloom]

what's the interest in this

this doesn't feel like idempotence to me, kind of feels like a random thing

[spare isle]

well I'm trying to ask if there is a more general class of "functions"/relations which are considered idempotent (more general than that of an idempotent binary operation)

(I'm just curious)

[delicate bloom]

I see I think now, f restricted to A is idempotent

[spare isle]

yes

however, is it valid to say this for a function which isn't a binary operation? IIRC, an idempotent function is typically one where f^2 = f (f^2(x) = f(x) for all x in the domain)

[delicate bloom]

well I'm reading it as f is your binary operation, and A is the subset of idempotent elements

[spare isle]

As I said, f is not (generally) a binary operation; that would require $f: A \cross A \to A$, but I have $f: A \cross A \to B$ with $A \subseteq B$, which means it could be the case that $A \neq B$

[cloud walrus]

JJCUBER

[delicate bloom]

sure, so what

[long obsidian]

SayR is a ring and consider the fraction field R(x). I think as R vector spaces R(x)=R[x]+R_p where R_p is a set of rational functions of the form f(x)/g(x)^j with j>0 and deg f<deg g. I think this is basically partially fraction decomposition.

What could be the generalization of this to several variables? Say R(x,y). I'm pretty sure R[x,y] is a subspace with basis the set of monomials. Then R(x,y)=R[x,y]+S where S is some other subspace analogous to R_p but it's not clear to me how I could generalized partial fractions...

I found some description in terms of rep theory but I'm really not following it

[spare isle]

You said you are reading it as f is my binary operation, but it’s not…?

[delicate bloom]

the restriction to elements x that satisfy f(x,x)=x is clearly from AxA to A

[spare isle]

Ah I see what you mean

Although, wouldn’t it be possible to have $f(a_1, a_1) = a_1$ and $f(a_2, a_2) = a_2$ with $f(a_1, a_2) = b$ for some $a_1,a_2 \in A, b \in B$?

[cloud walrus]

JJCUBER

[spare isle]

For example, f(x,y) = x/(y-1) gives f(0,0) = 0 and f(2,2) = 2, but f(2,0) = -2. So let A={0,2} and B={-2,0,2}; in this case, f is not a binary operation when restricted to A x A.

[delicate bloom]

if you call the set A the set of idempotent elements of f, I don't see what being a binary operator or not matters

[spare isle]

I guess I just don’t understand from where the terminology “idempotent elements of f” is coming from. Are we just defining it however we want, or does this come from some branch of study relating to functions/relations/etc?

[delicate bloom]

feels obvious enough to me

[long obsidian]

Say R is a ring and consider the field of fractions R(x) and say I is an ideal of R. I know that R[x]/I=(R/I)[x] . Is it true that similarly R(x)/I=(R/I)(x)?

[untold cloud]

For a field, the only ideals are the field itself and 0, so the ideal I in R may not be an ideal in R(x)

So R(x)/<I> is either R(x) or 0, <I> means the ideal in R(x) generated by I

R(x)/<I> may not even be a field

[rapid junco]

is the following true

if (x+1)^n = (x + f(x))^n then f(x) = 1?

Trying to formalize this using polynomial equality

[lusty marlin]

What is x and what is f?

[rapid junco]

f is a function of x

[lusty marlin]

What set are we working inside

[rapid junco]

f: R -> R

rational function

[lusty marlin]

and n is one fixed natural number?

[rapid junco]

yeah

[lusty marlin]

If n is odd then we can take the nth root to directly obtain f=1 for all x

If n is even then either f=1 for all x or x+1 = -x-f(x) for all x

[rapid junco]

what are you formally using here

viewing polynomials over which field

field of rational functions?

[lusty marlin]

By R you didn't mean the real numbers?

[rapid junco]

right but this works in generality over field of rational functions

[lusty marlin]

My bad then

[rapid junco]

no no R is reals

[lusty marlin]

Ok

[lusty marlin]

What works?

[rapid junco]

nvm nvm

like for example I am trying to find the roots of this function over R:

but like theres a bunch of garbage that one can get rid of,

and I noticed one can probably get (x+1)^n = (x + f)^n for some f, right

but then f is a function of x, so we just solve for f = 1 to solve for the original equation right?

[lusty marlin]

How?

[rapid junco]

well I havent worked it out yet

[lusty marlin]

Seems like a wild guess to me

Oh wait for some f

How does that even help?

[rapid junco]

well (x+1) and x + f are polynomials, so their coefficients agree iff they agree

[kind temple]

the left hand side is a degree n poly, so the right hand side must also be degree n, i.e., f(x)^n has degree n. if f is of degree m, then you have m + n = n, so m = 0 and f is constant

[rapid junco]

implying f = 1, but f is a function of x

[kind temple]

f(-1) = 1 so f = 1

[rapid junco]

ah right sorry, I am trying to find x for some function f such that f(x) = 1

[kind temple]

what does this mean then? is this equality for all x?

[rapid junco]

no

i am just trying to find x such that this holds for a specific function f

[kind temple]

and f is a polynomial?

[rapid junco]

rational function

some sort of rational function

[kind temple]

okay

that changes things

[rapid junco]

but its denominator never vanishes

[ornate tiger]

how to prove that (Q,+) is not Cyclic

[kind temple]

what do you know about infinite cyclic groups (up to isomorphism, perhaps)?

[ornate tiger]

i know nothing i'm trying to read some stuff about them but i don't know where and what to focus on i don't need to waste a lot of time on it

[kind temple]

if you know nothing than how r u going to understand any explanation given for your problem?

[ornate tiger]

give me some fast resource to understand a little about them

[kind temple]

wikipedia cyclic groups. or read through the theorems in artin’s algebra or dummit and foote

[severe tide]

What is the "lifting lemma" for commutative diagrams?

[ornate tiger]

oh ok does it take a lot of time ? like i have a homework and there's alot of question left

can i fnish this question in 1 hour ?

[kind temple]

this question seems easily google-able

[ornate tiger]

oh cool less time

[untold basalt]

I have replaced G with the cyclic group of order p^nm but now it's not clear to me why Au=A for all u in U \iff A is a coset of U.

[rocky cloak]

Well, do you know the definition of a coset?

Or perhaps better, when are a and b part of the same coset?

[ornate tiger]

Prove that (Q,+) is not cyclic .
is this a good proof ?

[mighty kiln]

But Z is not a subset of Q

Have you even read the output

[barren sierra]

Any ideas for the isomorphism? I don't even have a candidate map in mind 💀

[lusty marlin]

This isn't a proof, this is an answer by chatgpt

Good luck learning maths from chatgpt lmao

[mighty kiln]

What's the kernel of G → π1(G)

[barren sierra]

N2

But that gets me a map G/N2 -> H1 which doesn't really help?

[mighty kiln]

The map is induced by the identity map G → G

You just have to show that it's the same quotient on both sides

[barren sierra]

Huh?

[mighty kiln]

H1/N1 and H2/N2 are quotients of G

I claim that the map described by "for any g ∈ G, the coset of g is sent to the coset of g" is well-defined and an isomorphism

[barren sierra]

Hmmm

Ok I'll think about that

[noble lynx]

not sure how to approach this one, the chapter proves the theorem that any sub module of a free module (over a PID) is free of lower rank, and the proof sort of constructs a basis by quotienting some things. Is that the way to go from here or is there another way I am not seeing?

[rocky cloak]

Composing with H1 -> H1/N1 you get that
H1/N1 = G/N1N2

[ashen heron]

!nogpt

[flat tree]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[ornate tiger]

oh i didn't know

[ashen heron]

[spare isle]

Would you at least be able to point me in the right direction (resources, relevant topic names, etc)? I'm just trying to learn about this category of functions, but saying it feels obvious doesn't really help. You have already made a false claim too under the guise of "clearly" being true, which hasn't really helped with my confusion.

For example, all of these refer to a binary operation, but I am asking about whether there exist resources/materials/actual definitions on generalizing idempotency to functions which aren't (generally) binary operations: https://proofwiki.org/wiki/Definition:Idempotence

[south patrol]

Idk, sounds good to me

[coral shale]

wait what?

[coral spindle]

Read the 'proof' that ChatGPT provided

[coral shale]

im going to have to unlearn math now

[spare isle]

Can't wait for classes in two decades called, "MathGPT" where all the material is just AI generated

[south patrol]

Clearly the best proof that Q is not isomorphic to Z as Z-modules is that Aut_Z(Z) is finite whilst Aut_Z(Q) isn't

😼

[rocky cloak]

I'm not sure what kind of resources you're after.

Like you can extend the definition f(x, x) = x, to any setting where that makes sense if you want. I'm not sure there is much more interesting to say. Why are you wondering?

[spare isle]

Well I was originally merely asking the other person if there was a properly defined/studied class of functions which extend the idempotent characteristic in this way (from beyond binary operations). This was all just out of pure curiosity/in the pursuit of learning something new.

[ornate tiger]

Ok i know a little bit now about cyclic groups how to prove (Q,+) is not cyclic

Give me tips

[spare isle]

You can suppose some value is a generator and derive a contradiction

[rocky cloak]

I see. Not that I'm aware of. The important of idempotents really is about how they behave like projections for associative operations.

I don't think functions with f(x, x) = x are any more significant than functions satisfying any other equation.

[spare isle]

By projection, are you referring to this: https://en.wikipedia.org/wiki/Projection_(linear_algebra)
Ah I found a different definition for projections which is more related to what you are saying; I understand now. Thanks for your insight!

[coral spindle]

Draw out a picture of what cyclic subgroups of Q look like. Why can this never be the whole of Q? You should be able to spot the issue visually.

[ornate tiger]

And also give me tips for this question
If G in non abelian does G have no sub cyclic group ?

[coral spindle]

what is a sub-cyclic group?

Do you mean a cyclic subgroup? No, all groups have cyclic subgroups.

[spare isle]

(the trivial group)

[ornate tiger]

G is not abelian

[coral spindle]

So?

Still, like I said, all groups have cyclic subgroups. You need to be clearer on what you are asking.

[ornate tiger]

Wait

[ornate tiger]

first about this question does it not work because like we can send 1 to 1/1 and 2 to 1/2 so we can send n to 1/n but the problem is in 1/n+1

[ornate tiger]

can u give me an example ?

such that H < G

[south patrol]

Every element generates a cyclic subgroup

[ornate tiger]

so we can just take H = {e} ?

[south patrol]

Sure but I mean this also shows that every non-trivial group has a non-trivial cyclic subgroup

by taking any non-identity element and considering the subgroup it generates

[ornate tiger]

oh ok

what about the first question what do u think about what i said ?

prove (Q,+) is not cyclic

first about this question does it not work because like we can send 1 to 1/1 and 2 to 1/2 so we can send n to 1/n but the problem is in 1/n+1

[spare isle]

what do you mean by "send?"

[ornate tiger]

like it's a function

[spare isle]

do you mean for any value you claim to be a generator you find a "smaller one" through said function?

[ornate tiger]

um

like wait we need to take from Q some numbers and what they generate right ?

like if we take 1/2

it will generate 1/2 , 1/4 etc

[spare isle]

You are trying to prove that it isn't cyclic, not what elements within it generate per se

[ornate tiger]

right ?

[coral spindle]

No.

[ornate tiger]

so what

[spare isle]

(Q, +) is an additive group, so this is not correct

[ornate tiger]

ok so it goes 1/2 , 1 , 3/2 ?

[south patrol]

Though this is still false if you take (Q^x, x)

[south patrol]

yes

is that all of Q

[ornate tiger]

no

[south patrol]

Note there is also 0 and -1/2

(for example)

[ornate tiger]

there's 3/4 missing

[ornate tiger]

Q doesn't have a zero this is what i know right

[spare isle]

This is a good observation. Now take x to be some supposed generator (note that it isn't a generator; we are assuming it is to derive a contradiction)

[south patrol]

It does

Otherwise it wouldn't be a group

[spare isle]

It does...

[south patrol]

I think you are thinking of Q^x

like non-zero rationals under multiplication

[spare isle]

By this, are you referring to Q^*? If so, I've never seen the x notation before.

[ornate tiger]

no just the normal Q

[south patrol]

Both are commonplace notations

Well, $\mathbb Q^\times$ and $\mathbb Q^*$

[cloud walrus]

potato

[spare isle]

Lets start with this: is 0 in Q?

[south patrol]

Some people differentiate them unfortunately lol

[ornate tiger]

yeah

after what u said

[south patrol]

Like I've met people who use * to mean "non-zero elements" and x to mean units

though ofc here those coincide

[ornate tiger]

for us when we use * that means it has no 0

[spare isle]

Interesting. That probably stems from people using S^* before learning abstract, hence leading to them assuming it just removes zero elements.

[south patrol]

what is S^*?

never seen that

[south patrol]

But no by "them" I mean professors lol

[spare isle]

Sorry I was just trying to generalize the set, though it's not necessarily right to just say any set S (I should have written R^*)

[spare isle]

oh

[ornate tiger]

so now let's take x to be some generator

how to show this with x it's easier with numbers

like can we say x = p/q

[spare isle]

yes! (just don't forget to properly define your variables in your proof)

[ornate tiger]

i'm bad with proofs 😦

[long geyser]

first simplification you can make is that

if p/q generates Q, so does 1/q

so can take p = 1 here

[ornate tiger]

yeah true

[south patrol]

Chad

[long geyser]

now what is the obvious problem, what number can't you reach by 1/q and its integer multiples

[ornate tiger]

1/q+1 ?

[south patrol]

that does work for 1/(q+1) yes

[long geyser]

well, that would be true if you mean 1/(q+1) but there is a "better" choice

I'd go with 1/(2q)

[ornate tiger]

mmmm

[south patrol]

there's also like

if Q were cyclic there'd be an isomorphism Q->Z

so f(x) = 1 for some x

so f(x/2) uh

Jk, that is the same proof

But yeah

[ornate tiger]

ok so in proofs first i define x

i can start with suppose x in Q such that x = p/q

[tribal moss]

But what if it was torsion?

Better to start with a surjective homomorphism Z-> Q, since that does exist for all cyclic groups.

[south patrol]

Well I assumed that lol but yeah fair in isolation that isn't right

[long geyser]

pinging this

[south patrol]

yes this is great

[ornate tiger]

is this good now ?

[long geyser]

that is not a proof

[ornate tiger]

😦

[coral spindle]

You have not justified why x does not generate it.

[ornate tiger]

[coral spindle]

You have asserted it but not explained why.

I might as well say: it is cyclic because clearly x generates 1/(q+1).

[ornate tiger]

what's missing

a lot ?

i tried

[long geyser]

I recommend book of proof by hammack

[noble lynx]

still stuck on this

[ornate tiger]

can u help me what to do now because this homework is for today

[long geyser]

oh boy

you just have to justify your steps, that's all I can say

you say we can take p=1 but don't explain why, and you also don't explain why 1/q can't generate 1/(q+1)

[ornate tiger]

because it will not be isomorphic ?