groups-rings-fields

[vapid vale]

yeah local rings are incredibly important and fundamental

[frigid epoch]

do you know an example of a 2d or higher-dim local ring

[vapid vale]

but i also came across them before i saw one good use for them and i also thought they were kind of arbitrary for a while so i relate lol

[vapid vale]

the dimension can really be anything depending on the dimension of your scheme

[rocky cloak]

k[[x1, ..., xn]]

[vapid vale]

if the dimension of a variety is n then the local ring at a closed point will have dimension n

[frigid epoch]

oh it's that simple huh

[vapid vale]

the dimension of a local ring is just the codimension of the point

[frigid epoch]

okay yeah because functions can vanish at just that point, or a curve passing through that point, or a surface, or a hypersurface etc.?

makes sense

[vapid vale]

yeah

[raw delta]

also thanks for your answer but it seems like formal definitions as i understand. i am not talking about this

can we aslo say that these concepts focus operations?

more than numbers

(numbers are just example)

they focus structure of operations

[frigid epoch]

wait does that mean rings like C⁰ or C^k on an n-dimensional manifold are n-dimensional

[vapid vale]

those are infinite dimensional

[frigid epoch]

ah why

[vapid vale]

i dont remember a good explanation, i guess intuitively there are just way too many choices

[vapid vale]

i think thats fair to say

[raw delta]

what do you mean by ''symmetry'' in this context

[swift root]

yes, essentially its the philosophy "if it behaves like X then we may as well treat it like X"

so in the future if you come across something that behaves like X (i.e. satisfies group axioms, ring axioms or wtv) then you immediately have a big toolkit of stuff to use

for groups, it started with permutations of roots of polynomials, but it later ended being incredibly important in many other fields, like topology

this is only possible because we abstracted to concept of "symmetry"

[copper kestrel]

im a bit confused on this, i imagine we need gauss' lemma, product of primitive polynomials is primitive, but other than that idk

maybe we can set up a euclidean algorithm?

wait no euclidean algorithm doesnt work here

this lemma might work

and since every nonconstant irreducible in D[x] is primitive

i think i have to prove that though even if it says it in my book

[copper kestrel]

okay update

i really hope this works :')

[azure cairn]

Is conjugation specific to the field of complex numbers

No right

Id assume you'd generalize complex conjugate to some form of automorphism over a general field?

[rocky cloak]

The term "complex conjugation" refers specifically to the automorphism of the complex numbers.

More generally for a field extension F/K one says two elements x, y in F are conjugate (over K) if there is an automorphism taking one to the other.

In other settings, like group theory, conjugation (by g) refers to the automorphism
x |-> g x g^-1

[karmic moat]

Complex conjugation is an automorphism of C that fixes elements of R. More generally, if you have a field extension F/K then you can look at automorphisms of F which fix K, and the group of such automorphisms is called the Galois group of K/F

C/R is a Galois extension and the Galois group is just {identity, complex conjugation}

[azure cairn]

Interesting ty both

[ripe harbor]

Why is $(G_i \cap H) G_{i - 1} /G_{i - 1} \unlhd G_i/G_{i - 1}$? It's clear that it's a subgroup, but how do we justify quickly that it's a normal one?

[cloud walrus]

ILikeMathematics

[quiet pelican]

Is H a normal subgroup of G?

[ripe harbor]

No, just H <= G

1 = G_0 \unlhd G_1 \unlhd ... \unlhd G_n = G

[ripe harbor]

And we know that (even though its probably not helpful here) $(G_{i - 1} \cap H) \unlhd (G_i \cap H)$

[cloud walrus]

ILikeMathematics

[quiet pelican]

Then isn’t this trivially false
If G_0 = e and G = G_1 = S_3, and H = <(1 2)>, then it doesn’t hold for i = 1

[ripe harbor]

Sorry I shouldve mentioned this, here is the entire context: Let $G$ be solvable and $$1 = G_0 \unlhd G_1 \unlhd \cdots \unlhd G_n = G$$ with $G_i/G_{i - 1}$ abelian, and $H \leq G$. Then [...] and $(G_i \cap H) G_{i - 1} /G_{i - 1} \unlhd G_i/G_{i - 1}$

[cloud walrus]

ILikeMathematics

[ripe harbor]

S_3 isnt abelian

[quiet pelican]

In which case G_{i+1}/G_i is abelian, so any subgroup of it is normal

[ripe harbor]

Oh

Thanks!

[ripe harbor]

@rocky cloak kind of unrelated haha but since as if I remember like there was some blackboard/whiteboard/... discussion here a while ago and I didn’t ask: Do you use a chalkboard at home or just paper?

[ripe harbor]

Let $G$ be a solvable group and $N \unlhd G$. Then $G/H$ is also solvable. Can we justify it like this? We know that for $f: G \to H$ a group homomorphism, if $G$ is solvable, then so is $\im(f)$. Now take $$p_H: G \to G/H, \quad g \mapsto g + H.$$ This is surjective and so $\im(f) = G/H$ is solvable.

[velvet hull]

But why is im(f) solvable? That is true exactly because of this statement

[ripe harbor]

Thats a theorem I have from my book; the proof went solving G and applying f on the chain

Do you have a more elementary proof?

[velvet hull]

Okay sure

That works

[ripe harbor]

Alright, thanks

Let $N_1 \unlhd N_2$. Then apparently we can find $U_i$ so that $$N_1 = U_1 \unlhd U_2 \unlhd \cdots \unlhd U_n = N_2$$ with $|U_i/U_{i- 1}|$ is prime. Apparently, we can justify this by the first Sylow theorem. But why? Wouldnt that require $U_i$ and $U_{i+ 1}$ to differ only by a prime power

[cloud walrus]

ILikeMathematics

[rocky cloak]

I don't have that kind of space at home, so paper

[crystal vale]

sorry unrelated but does ipad worth it for study?

[kind temple]

i love mine

[rocky cloak]

I don't know about worth it for study, but for doing presentations online / creating lecture notes it's pretty good

[crystal vale]

So do you guys have ipad or ipad Air?

[rocky cloak]

Probably ipad, because I don't know what ipad air is

[crystal vale]

It is a different model with better specs

[fading acorn]

i have ipad pro

I usually use it outside of my desk where i can’t tex up my notes

[crystal vale]

does every subgroup of order p in group of order p^n is normal?

[rocky cloak]

Note that this would imply every subgroup is normal (so no)

[crystal vale]

I have to show in S4 there is only one subgroup of order 12 which is A4.

So if there is any other subgroup H of order 12 which is not A4 then H contains exactly 6 elements of A4 and 6 elements of odd permutations which implies H intersection A4 is a subgroup of A4 of order 6 which is not possible.

Is it correct?

[candid patrol]

Why H contains 6 éléments of A6 ?

[candid patrol]

Not sure, take Q16

[rocky cloak]

Q16 has just one subgroup of order 2, but Q16/C2 has a few that are not normal (giving rise to subgroups of Q16 that are not normal)

So in general the argument would be:

Let G be a p-group and H a subgroup. Pick N a maximal normal subgroup in H. Then H/N cannot have any elements of order p, so N=H

[crystal vale]

I proved that in S_n if H contains any odd permutation then H contains exactly the same number of even and odd permutations

[marble hinge]

If a group G has order pq (p, q - prime), |Z(G)| = 1 and all non-id elems have order p, the book claims that indices of centralizers of all non-id elems must be q then.

Not quite sure why it is true. I reasoned as follows so far:

centralizer is a subgroup, so its order must be 1, p, q, pq. It can't be pq (because then this elem will be in the center and we know that |Z(G)| = 1. It can't be 1, because every element of <x> commutes with all p other elements of <x>

It can still be p or q

The book claims that it is always p, but why can't it be q?

(here I assume that q > p, of course)

[candid patrol]

p and q odd *

[marble hinge]

yeah, there is additional assumption that p doesn't divide q-1, so p can't be 2

I additionally thought about this: if |C(x)| > p for some element, that it must commute with elements of other cyclic subgroup, not only with its own

[candid patrol]

wdym by all non-id elems have order p ? Like we got an element of order q by Cauchy

[marble hinge]

well, that was an assumption in a proof by contradiction

and it eventually led to contradiction, so right, it can't be true (by Cauchy or otherwise)

but the thing is that I don't understand that particular step in the proof

let me show you the whole thing

"then the centralizer of every nonidentity element has index q" <-- I don't see how this follows

(and how it follows that easily that it doesn't even require any justification)

but right, the proof might be done more easily just saying that we have element of order q by Cauchy

[jade mason]

Shouldn't the centralizer be of the form {e} U disjoint union of p-cycles and hence have order 1+k(p-1)?

[candid patrol]

Stab is a subgroup of G

[marble hinge]

yes

[marble hinge]

this is probably related to what I started writing here: "I additionally thought about this: if |C(x)| > p for some element, that it must commute with elements of other cyclic subgroup, not only with its own"

so if it commutes with other cyclic subgroup, then |C(x)| >= p + (p-1). And in general |C(x)| = p + k(p-1) for some k

[candid patrol]

So the centralizer cannot be of order q, otherwise, there would be an element of order q

(That's not Cauchy)

[marble hinge]

why?

[candid patrol]

By applying the first isomorphism theorem to the projection restricted to H yes, but why should H contain a non-trivial element of A_4

[candid patrol]

G contains a cyclic subgroup H of index p prime => G has an element of order p

H = <h>

[crystal vale]

?

[jade mason]

Yeah it feels like it should give a contradiction but I dont think the formula I gave really helps

[marble hinge]

mmm, still not sure why it is the case, how do we get that element? Is it some well known result that doesn't even require justfication?

say, we have a group of 14 elements, and we have a subgroup of 2 elements somewhere, how do we build an element of order 7?

[candid patrol]

I mean, how can you assume that H cap A4 is non trivial ?

[candid patrol]

sorry not index but order *

[marble hinge]

I think my obseration doesn't give any contradiction. Let's say p = 3, q = 5, k = 1, then we have q = p + k(p-1) and everything seems to be good...

[crystal vale]

So H contains 6 elements of A4 right?

[candid patrol]

Yes but why

[marble hinge]

aaah, right, then it becomes obvious, thanks

[delicate orchid]

they previously stated that we may assume that if a permutation group contains an odd element then it has an equal number of odd and even elements

[marble hinge]

OK, so indeed, if all elems are of order p, then there are no subgroups of order q, centralizers or not (because q is prime and then subgroup of order q is cyclic and all its elements have order q, which is a contradiction if q != p)

[cloud walrus]

UGOAT

[delicate orchid]

it's order 12 so contains an involution, all involutions are odd. I didn't feel like that really needed to be stated

[candid patrol]

(1 2) isn't odd

We can use a neat result on the union of two subgroups to conclude that H contains a non-trivial odd permutation

[delicate orchid]

are we dead ass?

[candid patrol]

(1 2)(3 4) isn't odd *

[delicate orchid]

ok sure that's fair

Syl_2(S_4) is D_8, all elements of order 4 in S_4 are odd, hence either H \cap Syl_2(S_4) is C_4, in which case we are done, or C_2^2. If this C_2^2 contains a transposition we are done, hence this C_2^2 is O_2(S_4) and the full group is this with a C_3. It isn't abelian as O_2(S_4) is self centralising. As O_2(S_4) \cong C_2^2, H^2(O_2(S_4), C_3) = 0 so any ext. is split. |Hom(C_3, Aut(O_2(S_4)))| = 1 and so this action is uniquely determined and is easily seen to be isomorphic to A_4.

ah actually easier way of doing it, any such subgroup is index 2 and thus normal, therefore is a disjoint union of conjugacy classes. There is a unique class of elements of order 3 and by the above argument this would need to be closed under left multiplication by O_2(S_4), but the union of O_2(S_4) with this class of order 3 elements constitutes all even permutations in S_4, and thus is A_4

[candid patrol]

O_2 is the 2-core ?

[delicate orchid]

yur

[candid patrol]

it's been a while !

[delicate orchid]

I felt it strange I had to consider the cohomology of the C_2^2 so just read my second solution it's cleaner

[crystal vale]

Is my argument wrong?

[candid patrol]

No but you need to explain why H cap A4 isn't trivial

[tough crow]

Hey people, I need some help in structuring how I should learn algebra over the next 2 months (I won't be doing algebra just for 2 months ofc but I have other things planned after the next 2 months which I can't do now so I wanna utilise the time I have rn)

I will be studying for about 1-2 hours a day consistently (and more if time permits)

I have no prior algebra knowledge other than having done LADR 4th ed, and in the past I have done baby rudin ch 1-7, spivak CoM ch 1-3, and folland upto ch 2

I have a book I wanna do afterwards which states the following for its algebra pre requisites: "some basic definitions and results from linear algebra and the theory of rings and ideals are needed" so if I can learn algebra in a way to cover this within the next 2 months that would be nice as well

[crystal vale]

H contains exactly 6 elements of A4?

[delicate orchid]

yeah but why

[candid patrol]

ok but why

[delicate orchid]

ah

[crystal vale]

I proved that any subgroup of S_n, if it contains odd permutations then they have same number of odd and even permutations otherwise subgroup of A_n

[candid patrol]

but why would H contains at least one non trivial odd permutation XD

[crystal vale]

No it is my assumption it contains odd permutations otherwise I am done

Because my original question was if H is subgroup of order 12 then H = A4

So if it is not A4 then H contains some odd permutation

[candid patrol]

ok ok

[delicate orchid]

The answer I was thinking of was "S_4 = A_4 u (12)A_4, so if H \cap A_4 = 1 all non-trivial elements of H would have to be odd, contradiction unless H = C_2, which isn't the case as |H| = 12" but I suppose this is essentially what notknow is saying

I think I'm just confused as to what exactly we're assuming and what we have to prove

[crystal vale]

Leave it

Any hint for part b? Also I am thinking can I write any two degree separable extension in terms of F(√D)?

[candid patrol]

For a field F with characteristic not equal to 2, every extension of degree 2 is generated by the root of a polynomial of the form x^2 - D yes

[crystal vale]

how?

let F(a) is extension of degree 2 over F

[delicate orchid]

I think this works: assume F(a) = F[x]/(x^2+bx+c), then by the quadratic formula a = -b/2+sqrt(b^2-4c)/2 so F(a) = F(\sqrt(b^2-4c)/2) = F[x]/(x^2-(b^2-4c)/4)

[crystal vale]

Yes

[crystal vale]

For c part, I have to show K1 intersection K2 = Q(√3), is there any other way without computation?

[quiet pelican]

You just have to show they’re not equal, and both contain sqrt(3)
There’s a very quick way to show they’re not equal (hint: ||one of them is a subset of the reals||)

[crystal vale]

i showed they are not equal

i got it

[chilly radish]

Did you end up using degree considerations? The degree of K1 cap K2 has to be 1, 2, or 4 over Q. Can't be 1 since it has sqrt(3) and can't be 4 since the fields are not equal

[crystal vale]

Yes

And the degree of Q(√3) over Q is 2

[chilly radish]

Yea

Just wanted to make sure you didn't end up brute-forcing it 🙂

[crystal vale]

No

[ripe harbor]

How do we prove that for $\sigma = (i_1, \dots, i_k)$, $$\tau \sigma \tau^{-1} = (\tau(i_1), \dots, \tau(i_k))?$$ Let $\tau = (a_1, \dots, a_m)$, then $\tau^{-1} = (a_m, \dots, a_1)$ and $$a_m \overset{\tau^{-1}}\to a_{m - 1} \overset{\sigma}\to \sigma(a_{m - 1}) \overset{\tau} \to \tau(\sigma(a_{m - 1})).$$ Thats not quite what we wanted yet

[cloud walrus]

ILikeMathematics

[ripe harbor]

and re-posting this one from yesterday:

[chilly radish]

Your presentation tells you that
\tau\sigma\tau^-1 should take \tau(i_j) to \tau(i_j+1), so start by writing a_m as \tau(i_j) for some j.

[chilly radish]

If U_i, U_i+1 differ by a prime power then U_i+1/U_i is a p-group.

It can be shown that a p-group contains a normal subgroup of order p. This follows from the center being nontrivial for p groups (a fact I hope you've seen) and Cauchy's theorem , so we can find some U' such that U'/U_i has order p and U' is normal in U_i+1.

You continue inductively by now considering U_i+1/U'

[azure cairn]

The intuition behind the correspondence theorem is that subgroups of the domain and codomain of a homomorphism can be pushed forward and pulled back right?

I'm interpreting it as like, the domain of the homomorphism gets sent to a "smaller" version of itself, with elements collapsing to 1 and the other elements being sent injectively to distinct elements

Similar intuition I had when thinking about the psuedoinverse in ljnalg ig

Then thr kernel is the difference between subgroups in domain and codomain

[chilly radish]

You can always pull back, but you only get the correspondence when the Hom is surjective

It also preserves normality (again only in this situation)

[ripe harbor]

Ah, ok, thanks

[chilly radish]

That's not really how this works. Many elements get sent to 1, but the others don't get mapped injectively, since multiplying by something in the kernel doesn't change the image

[ripe harbor]

Oh, so we basically build up the chain by going through all the prime powers that differ and always increasing by one, right?

[chilly radish]

Yea, exactly

[ripe harbor]

Thanks!

[chilly radish]

Of course!

Maybe decrease by 1 is more accurate

Every step you do the order of the quotient becomes smaller by one factor of p

But I think that's what you meant

[ripe harbor]

right

[ripe harbor]

Oh and also, this is kind of unrelated haha but do you use a chalkboard at home, or just paper?

[chilly radish]

I have a whiteboard. I use it mainly when doing computational stuff, to keep track of a lot of information, or just when I'm brainstorming, otherwise I'm usually typing. I will sometimes do scratch work on paper but I'm less of a fan

[ripe harbor]

Ah, alright. I was thinking of getting some kind of chalkboard but my room isn't very large. How big is your whiteboard/do you not run out of space?

[chilly radish]

Mine is 48 inches by 36 inches, double sided.

I don't usually write out full proofs or anything. I do still run out of room but not too often to worry about it, but it depends what exactly your workflow is like. A lot of times I will leave things up on the whiteboard for a while

Anyways we should move this convo to one of the general channels if you'd like to continue discussing since it's gotten off-topic

[ripe harbor]

Alright, thanks, that was all

[crystal vale]

For a), so norm of alpha is just the product of its conjugates so if p(x) is the minimal polynomial of alpha then norm will be (-1)^nf(0) where n depends on degree of f so it is in F, right?

[rocky cloak]

Yes, that's right. Alternatively might be better to show that it's fixed by the Galois group

[crystal vale]

yes, i thought of showing by this way

[rocky cloak]

And I guess it should be (-1)^n f(0)^m where m depends on the size of the orbit of alpha

[crystal vale]

oh yes my bad

oh then i don't think my argument is correct

[chilly radish]

You can also directly prove that the element is fixed by all the elements of Gal(L/F)

[crystal vale]

yes let me do

[proper barn]

is anything but 0_R in a ring R idempotent and nilpotent?

[crystal vale]

i am confused here, so this product is taken over all the embeddings of K into alg closure of F so is there possible that alpha^2 can be one product

[rocky cloak]

The same factor can appear several times yes, if that's what you're asking

[rocky cloak]

If x is idempotent, what is x^n equal to?

[crystal vale]

yes

is it exactly the product of its conjugates with same multiplicity?

[rocky cloak]

It will be the product of its conjugates to the power of [K : F(alpha)]

[crystal vale]

i think it is best to show fixed by G

[proper barn]

Ooh its x !
so x^n = x = 0

if nilpotent aswell

[crystal vale]

so this product runs over g\in G, right?

[proper barn]

yay thank you so much !

[crystal vale]

g just restricted to K

[azure cairn]

i mean cosets of the kernel*** (beisdes the trivial one)

like umm under a homomorphism, cosets of the kernel besides the identity coset map injectively

i will think about this for a bit longer

[chilly radish]

Why are you excluding the identity coset

[chilly radish]

Yea you're correct then

[azure cairn]

right nvm

[chilly radish]

It's the identity element, it gets mapped to the identity

[azure cairn]

i don't think this the correspondence theorem though

cuz it discusses subgroups not cosets of them

might be referencing something else idk ill figure it out

[chilly radish]

The last thing you talked about is just how to think of a surjective homomorphism as a quotient map, but before that we were discussing the correspondence theorem

Surjective homomorphisms are equivalent to quotients, so thinking about it in terms of surjective homomorphisms is equivalent

[ripe harbor]

Other than 1, what's the normal subgroup of S_3?

[stone sky]

Hint it’s a very familiar group

And it’s the kernel of the sign homomorphism

So we know it’s normal

[ripe harbor]

Ah, so just A_3?

And we also know that [S_3 : A_3] = 2, so there can't be any more, so S_3 is simple. Thanks! So S_3 has the normal subgroups 1, A_3 and S_3

[delicate orchid]

I presume you mean A_3 is simple, which is true

[ripe harbor]

Wait, I asked the wrong question haha

[stone sky]

Bingo

A_3 is simple

[ripe harbor]

Right, so it's just 1 and A_3 itself

How can we be sure though?

[crystal vale]

for S3, why don't you write all the subgroups?

[ripe harbor]

Ah

1, A_3 and S_3

So A_3 can only have 1 and A_3

Thanks

[crystal vale]

what are the subgroups of S3?

[stone sky]

Well A_3 has prime order, so it must be cyclic and thus isomorphic to Z_3, but Z_3 only has two subgroups, itself and {e}

So A_3 is simple

[azure cairn]

i see now, ty for ur help

[ripe harbor]

1, <(1, 2)>, <(1, 3)>, <(2, 3)>, <(1, 2, 3)>, S_3 right?

Are there any more

[delicate orchid]

there's only 3 conjugacy classes so you only need to check if 1 and the set of all 3-cycles or 1 and the set of all 2-cycles form subgroups to classify the normal subgroups

[crystal vale]

yes

[crystal vale]

i got it, thank you

[ripe harbor]

1 is a normal subgroup and all the other ones except (1, 2, 3) have sign -1, so only those two come into question, right?

Thanks

[crystal vale]

sorry what do you mean by sign?

[stone sky]

Correct. Because if the sign = 1 then they are in the kernel of the homomorphism. Thus, they are normal subgroups

[stone sky]

However you missed (1,3,2)

[lusty marlin]

Odd and even permutations

[crystal vale]

For part a), do I need Galois extension?

I don't think

Oh to make a group, i think i need that

I mean in the definition of N, the maps from K to alg closure of F, the set of such maps not need to be group

[velvet hull]

norm is only well-defined for galois extensions

[rocky cloak]

You don't really need it to be a galois extension with this definition, but it makes it a little easier to prove

[velvet hull]

oh you're replace it with the aut group instead

yeah it doesn't strictly require galois then

[ripe harbor]

When we say something like $G \times H \cong J$ with $G, H, J$ groups, then how is the group homomorphism defined? Usually, we want $$f(ab) = f(a)f(b),$$ but here, is it instead $$f(au, bv) = f(a, b)f(a, v)f(u, b)f(u, v)?$$

[cloud walrus]

ILikeMathematics

[swift root]

why would it be different

it wouldnt be a group isomorphism if it was the bottom one

G × H is still a group

[ripe harbor]

Ah, we can still treat it the same, as if f took only one argument, and then just let a in G x H, b in G x H

[swift root]

f does only take one argument

an element of G × H

[marble hinge]

Yeah, the argument will be a pair

[ripe harbor]

Well but that's like two arguments

[swift root]

informally

but thats abuse of notation

[marble hinge]

It will be f((x,y))

[swift root]

you can think of it as a concept with an attitude

sometimes we like to think of it as a function with two arguments

sometimes we dont

depends on the context

[marble hinge]

Also somewhat related: currying/uncurrying of functions

[swift root]

Set is a cartesian closed category

[marble hinge]

Monad is a burrito

[swift root]

what burrito would be the free group monad

[marble hinge]

Not sure… tasty burrito? 😋

[ripe harbor]

It's a shame that we don't have a homomorphism $$U \times V \to UV, \quad (u, v) \mapsto uv.$$ The kernel of that should be $U \cap V$ and it's obviously surjective, we would get $$(U \times V)/(U \cap V) \cong UV \implies |UV| = \frac{|U| |V|}{|U \cap V|}$$ which is true but that's just not a homomorphism..

[cloud walrus]

ILikeMathematics

[grim iron]

Hey !

Here is my problem (Come from my professor Book). So I have done question 1,2,3 but 4 I have spend a lot of time without going anywhere. Currently, I have a surjection from M to Coker(f). However, this application is not injective (This application come from the following diagramm). Moreover, I have check that (1, x) ~ (2, y) iff f(x) = y
(1, x) ~ (1, x') iff f(x) = f(x') and for 2 just y = y'.
So I see that M1 is actually in M by M2 but I don't know where to go after this...

If someone has any idea I take it !
Thanks you for your help

[supple ice]

for I = {1,2} with transition map f : M1 -> M2 the colimit is obtained by taking M1 disjoint union M2 and identifying each (1,x) with (2,f(x)) hence every class has a representative (2,y) and the map M2 -> M, y -> [(2,y)] is surjective it’s also injective because (2,y) ~ (2,y’) forces y=y' so M is iso to M2 canonically therefore the induced map M -> Coker(f) is just the usual quotient M2 -> M2/im(f) which is surjective but not injective unless im(f)=0; so "colim is iso to coker(f)" is not true in general

[rocky cloak]

Not with the aut group, but with the set of embeddings of K into the algebraic closure of F.

Or a definition that also works for non-seperable extensions: the determinant of left multiplication

[azure cairn]

is sigma a conjugation automorphism out of Curiosity

[rocky cloak]

Sometimes you do. You do exactly when u and v commute for any u in U, v in V

[azure cairn]

takes F/K to F/K and leaves K elements intact

[marble hinge]

@ripe harbor Here is my proof of this, without using Sylow theorems and partially based on an example in D&F and an exercise from there. Probably much more complicated than using Sylow, but still fun to do:

[candid patrol]

If you don't want to use Sylow's Theorems, you can simply use the fact that a subgroup of minimal prime index is normal

[marble hinge]

I used it there ^^

in the proof

to show that H = <x> is normal

[candid patrol]

ok mb

[marble hinge]

or do you mean that there is some quicker way?

while using this fact?

[marble hinge]

I mean this part:

[candid patrol]

I was thinking about semi-direct product

[marble hinge]

Hm, fixed some typos in the proof, but now I don't know how to edit the image in the message :) So pasting it one more time

[marble hinge]

Ah, I don't know much about that yet

[azure cairn]

is $\phi^{-1}(a)$ usual notation for the fiber of a homomorphism? i don't like that $\phi^{-1}$ is being used when it's not well defined

[cloud walrus]

Altanis

[knotty badger]

Yes i think so

You’re applying the preimage to a singleton

[azure cairn]

aa

any alternative notations lmao or do i have to get used to it

[azure cairn]

it is well defined, i meant that it makes me think phi is an isomorphism when it's not

[knotty badger]

I guess you could do $\varphi^{-1}({a})$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[azure cairn]

lol

[knotty badger]

Idk if there’s other standard notation

[azure cairn]

yeah is hould just get used to it

[stone sky]

It’s the standard notation

I’ve never seen anything else

[ripe harbor]

Hm, nice, definitely longer than with Sylow. How long did finding that take?

[ripe harbor]

But well, if I want to prove the result at the end for general subgroups U, V, that isnt necessarily the case

[azure cairn]

In the Correspondence Theorem for a group homomorphism $\phi: G \to \mathcal{G}$, suppose you have two corresponding subgroups $H \le G$ and $\mathcal{H} \le \mathcal{G}$. I want to show $|H| = |\ker \phi| \times |\mathcal{H}|$ is true. Is this proof OK?

\begin{proof}
Consider the restricted homomorphism $\phi|_H: H \to \mathcal{H}$, which is well-defined and surjective. Then we can consider the cosets of $\ker \phi|_H$ and use the Counting Formula, yielding $|H| = |\ker \phi|_H| \times [H:\ker \phi|_H]$. Note that since $\ker \phi \subseteq H$, $|\ker \phi|_H| = |\ker \phi|$. Moreover, note $[H: \ker \phi] = |\im \phi|_H| = |\mathcal{H}|$. Thus $|H| = |\ker \phi| \times |\mathcal{H}|$ as deisred. 
\end{proof}

[cloud walrus]

Altanis

[crystal vale]

For part b), is there any other way to show this, do I need to find the minimal polynomial?

[glass sinew]

yeah lgtm. never seen this called the counting formula before lol

[glass sinew]

god looking at canonical decomposition really cleared up a lot of brain fog about homomorphisms. 1st iso thm seems like such a natural consequence of the group/ring structure now. also just a really pretty diagram

[crystal vale]

To show the existence of theta, i have to show these all characters are distinct, how do I show?

[velvet hull]

yes you should find the minimal polynomial, but it is not as hard as you think

[feral slate]

Where is this from?

[velvet hull]

notice that you have
\begin{align*}
\alpha \sigma (\theta) &= \alpha \sigma (\theta) \
(\alpha \sigma \alpha) \sigma ^2(\theta) &= \alpha \sigma (\alpha \sigma (\theta)) \
(\alpha \sigma \alpha \sigma^2 \alpha ) \sigma ^3(\theta) &= \alpha \sigma ((\alpha \sigma \alpha )\sigma (\theta)) = \alpha\sigma (\alpha\sigma (\alpha\sigma (\theta))) \
&\vdots
\end{align*}

So what your expression is really equal to is $$\theta + (\alpha \sigma )(\theta) + (\alpha \sigma )^2(\theta) + (\alpha \sigma )^3(\theta) + \dots + (\alpha \sigma)^{n-1}(\theta)$$

[cloud walrus]

HChan

[glass sinew]

me!

[swift root]

it's like the fundamental theorem of arithmetic for algebra

[crystal vale]

I see thank you

[feral slate]

Oh cool

[marble hinge]

Hmm, now I realise that this is basically identical to the canonical decomposition of any function into surjective/injective components, that we were discussing here in details a couple of days ago

I.e. to this:

[velvet hull]

it is not just identical, it is in fact the same thing

it just so happens that the injective surjective components of a group homomorphism are themselves group homomorphisms

[marble hinge]

That’s basically my retelling of a proof from D&F (for the abelian part) plus my solution to the exercise for the cyclic part, so I haven’t done it all from scratch.

[knotty badger]

mfw regular category

[marble hinge]

A regular category is a finitely complete category which admits a good notion of image factorization. A primary raison d’être behind regular categories C is to have a decently behaved calculus of relations in C (see Rel).

[glass sinew]

exactly what got me looking into it lolol

when you realize each fiber is just a coset of the kernel it follows directly

[marble hinge]

Yeah, but what is interesting is that this decomposition is more general and not specific to groups, so it can be done to any function on any sets, not only group homomorphisms

[glass sinew]

of course!

[crystal vale]

Is option c correct? Because how f(t) = 1/t^2 is fixed by G?

[velvet hull]

where is 1/t^2 coming from

[crystal vale]

So f(t) = 1/t^2 is an element of k(t), right?

[velvet hull]

okay the notation here is kind of dogshit

b) should be Prove that the element a = (t^2-t+1)^3/t^2(t-1)^2 is fixed by G

and c) should be prove that k(a) = Fix(G)

[crystal vale]

Oh I see

[crystal vale]

Let's p>3 then 2, 3 and 6 are distinct over F_p.

And say these polynomials have no solutions over F_p.

Then over the splitting field of (x^2-2)(x^2-3)(x^2-6) then F_p(√2), F_p(√3) and F_p(√6) are intermediate field of degree 2.

But we know there can be only one intermediate field of degree 2[ we are working with finite fields].

Hence, these fields are equal, and they satisfy the equation x^4-x, that's means they satisfy the equation x^2+x+1. But these all are distinct so this 2 degree polynomial has 3 different solution which is a contradiction.

Is it correct?

[quiet pelican]

Why do they satisfy x^4 - x?

That’s true if p = 2

[crystal vale]

Ah

Sorry

Okay let me do

[quiet pelican]

This doesn’t really need any Galois theory

[crystal vale]

Okay

Any hint?

[velvet hull]

I’m not sure I agree with your statement that there is only one intermediate field of degree 2

It’s true that there is one degree 2 extension of Fp up to isomorphism

But I’m not sure that necessarily means they are the same extensions

[crystal vale]

Ignore that statement

[crystal vale]

The thing is Galois group will be cyclic therefore there should be exactly one intermediate field of degree d

[velvet hull]

Ah right

Of course

[quiet pelican]

The unit group of a field is cyclic

What are the squares in a cyclic group?

[crystal vale]

Sorry i don't get your question

[velvet hull]

You can consider x -> x^2 as a group homomorphism of F^x

So then the claim amounts to showing that 2, 3 or 6 is always in the image of this homomorphism for all p

[quiet pelican]

More specifically, ||show that if both 2 and 3 aren’t in the image, then 6 is||

[crystal vale]

i am stuck

[glad osprey]

Quick question: the map K[x, y] -> K[x, y] that sends x to a constant a and leaves y is a K-algebra homomorphism, right? Or I guess more generally you can send x and y to whatever independently and still have a homomorphism, including swapping x and y for example?

[crystal vale]

say that group is generated by a then 2 = a^i and 3 = a^j

[velvet hull]

Yeah, to check k algebra homomorphisms it is equivalent to verifying that the base ring homomorphism commutes with the respective k-maps that give the algebra structure

[crystal vale]

so i have to check if 2 and 3 are not in the image then i + j is divisible by 2

yeah

if i is even then we will have root in F

similarly for j

so both has to be odd

hence the sum is even

so for 6 we can choose a^(i+j)/2

that will be a root

right?

[quiet pelican]

Yes

[chilly radish]

You can send x and y to any element you want and this will define a k-alg homomorphism. This is because k[x,y] is freely generated by x and y as a k-algebra

[crystal vale]

thank you @quiet pelican @velvet hull

what does it mean by biquadratic extension?

[swift tundra]

Iirc all finite subgroups of the group of units are cyclic, but not all subgroups. Q^* is definitely not cyclic. Though you can have infinite cyclic subgroups.

[crystal vale]

yes but in my question field is finite

so biquadratic extension means the Galois group is Z/2Z \times Z/2Z?

[chilly radish]

Extension generated by 2 square roots

[crystal vale]

i see

[chilly radish]

Yes as long as it's actually degree 4

Which is usually what people mean

[crystal vale]

so we need the extension has to be Galois extension

[chilly radish]

Well it'll still be a galois extension if it has deg 2

It just means one of the roots you added was already in the base field

[crystal vale]

but when extension has degree 4, it doesn't imply the extension is Galois

So when we change the base Q to F_p, Galois group does not change?

[chilly radish]

not every degree 4 extension is galois no, but every biquadratic extension of degree 4 is, I was just saying that if you write Q(sqrt(a),sqrt(b)) that's not guaranteed to be a biquadratic extension

[chilly radish]

That's not necessarily true, and also doesn't make sense generally (since the extension is defined using roots of elements of Q, you need those to have analogues in F_p).

It's possible to have smth that's not a square over Q but is over F_p. Many finite fields have square roots of -1

[chilly radish]

This exactly shows the galois group does change

[crystal vale]

then what is biquadratic extension?

[crystal vale]

do we need a and b such that they are squarefree

[chilly radish]

An extension generated by 2 square roots. Usually when people say "biquadratic" they're assuming that it's a degree 4 extension

[chilly radish]

Yea that works

[crystal vale]

so why Q(\sqrt a, \sqrt b ) doesn't implies it is biquadratic extension?

[chilly radish]

What if a=4?

What if a=2 and b=8?

[crystal vale]

oh yeah

[crystal vale]

i don't get it how do i use the biquadratic information when base field is F_p?

[chilly radish]

Every finite extension of finite fields is cyclic

[crystal vale]

yes

no no i know the Galois group can't be Klein 4 group in case of finite field

but how does the extension not changed in F_p?

[chilly radish]

Ok I'm actually not sure what they're getting at with the hint, but you already get all square roots over F_(p^2) since finite fields have unique field extensions of every degree up to isomorphism

[chilly radish]

It IS though? That's the point of the exercise, the extension becomes degenerate since the polynomial becomes reducible

[ripe harbor]

Is there a nice way to shift from normal subgroups to regular ones? We can prove that $$|NH| = \frac{|N||H|}{|N \cap H|}$$ with the third isomorphism theorem, but we need $N \unlhd G$.

[cloud walrus]

ILikeMathematics

[ripe harbor]

The same holds for regular subgroups though

I thought about maybe using some trick with the normalizer..?

[delicate orchid]

I have never heard of a regular subgroup outside of very technical results involving p-groups and as a synonym for normal subgroup

I’d be interested in knowing what property youre talking about

[ripe harbor]

Sorry haha, I meant just subgroups in general

Not necessarily normal

[delicate orchid]

Oh.

LMFAOOOO

Im confused as to what your question is asking still, are you asking if there is a way of finding new subgroups using normal subgroups?

[ripe harbor]

If we can go from
|NH| = |N||H|/|N n H|
with N normal in G, H subgroup in G quickly to
|UV| = |U||V|/|U n V|,
U, V subgroups in G

Like taking U, V arbitrary, then considering instead U and the normalizer of V, applying the first result, and somehow reverting back to V

[delicate orchid]

Yeah it’s still true but your UV might not actually be a subgroup

The elementary combinatorial approach still works

[ripe harbor]

Oh, there is such an approach? My approach to the first was using the second iso theorem

Oh

Basically by definitions

[delicate orchid]

You genuinely just count the elements brochaco. Consider the cosets of U in UV

[ripe harbor]

Yeah

Thats even more elegant than second iso theorem lol

[knotty badger]

it's cause size only needs a bijection of sets

not of group structures

[ripe harbor]

How would one write this down though? Is it possible to write it more precisely than

"|UV| = |{uv | u in U, v in V}| consists of all combinations of U and V, without double-counting the ones in the intersection; so we need to divide out |U n V|. Thus |UV| = |U||V|/|U n V|"

[delicate orchid]

You also consider the cosets of the intersection in U

This will be |U|/|U n V|, so assuming you can show the number of cosets of V in UV is |UV|/|V|, then construct a bijection between these two and multiply by |V|

Thinking about the map from the second iso might be a good candidate bijection

Alternatively, this is closer to your approach, consider the map f : U x V -> UV and show that for any element of UV there are precisely
|U n V| elements of U x V that map to it

This does require use of the group structure though which is a bit yuck

[ripe harbor]

Alright, thanks, I will try both

[ripe harbor]

My book already showed that $|G : H| = |G|/|H|$. Now, consider $$\varphi: UV/V \to U/(U \cap V), \quad uvV \mapsto u(U \cap V).$$ This is obviously surjective. Now $$x \in \ker(\varphi) \iff \text{for }x = uvV, \text{ u} \in (U \cap V) \iff x = uvV = V = 1$$ and so the kernel is trivial and we have injectivity

[cloud walrus]

ILikeMathematics

[ripe harbor]

Does that look right?

It follows that $|UV|/|V| = |U|/|U \cap V|$

[cloud walrus]

ILikeMathematics

[ripe harbor]

Oh wait UV is not necessarily a group so we cant directly apply [G : H] = |G|/|H|

Hm, I guess we just directly show that then by writing out the definition of UV/V

[chilly radish]

Ker doesn't make sense unless one of them is normal, since you don't have a group structure

[ripe harbor]

oh, right..

[ripe harbor]

Is the map I consider the right one though?

[chilly radish]

I would consider the map wew outlined above, but yours will work too. However, you need to show that it's well-defined first

[chilly radish]

You're gonna need to show that the proof of lagrange works even if UV is not a group, basically

[ripe harbor]

Maybe I'll try the second approach, this seems quite long

Let $$f: U \times V \to UV \quad (u, v) \mapsto uv.$$ This is obviously surjective. How do we show that for each $x \in UV$, exactly $|U \cap V|$ elements map to it?

[cloud walrus]

ILikeMathematics

[ripe harbor]

Let $a = xy \in UV$. Then $(x,y)$ maps to it. But also, let $i \in U \cap V$ be arbitrary. Then $i^{-1} \in U \cap V$ and we have $$a = xy = xi^{-1}iy = xii^{-1}y$$

[cloud walrus]

ILikeMathematics

[ripe harbor]

Ok well 2, possibilities but 2 of those will overlap, so |U n V| possibilities

That should be it. This seems more elegant than the first method

[unkempt comet]

To see irreducibility of integer polynomials a good criterium is mod p reduction

Is it interesting to embed Z-> O_K and doing mod ideals there and checking irreducibility?

[wraith cargo]

Uh I mean

Maybe?
It seems a lot harder to compute residue fields in O_K than it is in Z lol

[unkempt comet]

I assume that if p = fg with f,g in O_K[x] are irreducible then p should be irreducible if say f not in Z[x] (assuming monic or primitive or whatever)

Like a stupid example,
p(x) = x^2 - 2, this decomposes as p(x) = (x-sqrt(2))(x+sqrt(2)) as the polynomial x-sqrt(2) is not in Z[x] the polynomial is irreducible

by UFD on O_sqrt(2)

I meant something closer to this @wraith cargo

[ripe harbor]

Let $H \leq G$ and $[G : H] = 2$. I want to show that $H \unlhd G$. The conditions means that there are two cosets of $H$ in $G$, so $G = aH \cup bH$ with $aH \cap bH = \varnothing$. So, let $g \in G$. Then WLOG $g = ah_1$ and so $$ghg^{-1} = ah_1hh_1^{-1}a^{-1}$$

[unkempt comet]

didn't know unlhd did that

Yeah, it's standard notation the proof isn't too complicated

[ripe harbor]

We need to show this is in H

[cloud walrus]

ILikeMathematics

[ripe harbor]

Why is this in H..?

Maybe this is the wrong approach

[ripe harbor]

Got an idea on how to continue this approach?

Or would you do it differently

[glass sinew]

hint: consider exactly what the two left and right cosets must be

[ripe harbor]

$G = aH \dot \cup bH = Hc \dot \cup Hd$ for some $a, b, c, d \in G$.

[cloud walrus]

ILikeMathematics

[glass sinew]

bigger hint: 1H is always a coset of H, don't overthink

[ripe harbor]

So one of the two elements in G/H is 1H

The other one needs to be G \ H = aH for some a in G

[unkempt comet]

I'm sorry I was answering the delivery

It's literally just ||consider conjugation explicitly||

[glass sinew]

now consider two cosets aH, Hb where neither are H. what can you say about them?

[ripe harbor]

Uh, I did that..? We have $$gHg^{-1}$$ and the left coset is in $G$; thus either $1H$ or $G \setminus H$ as we established..

I feel like I'm somehow overthinking this

[cloud walrus]

ILikeMathematics

[glass sinew]

okay so the left coset is either H or G\H right

what about the right one

[ripe harbor]

Either H1 or G \ H

[glass sinew]

yeah boom

[ripe harbor]

Ok so if gH = 1H then this is 1Hg^{-1}. Now we have two possibilities

Hg^{-1} = H1 in which case we are done

Or Hg^{-1} = G \ H

[glass sinew]

if gH = G\H then Hg will be G\H proof over

[delicate orchid]

Mainly because Hg contains g, which isn’t in H

[ripe harbor]

Oh, gH = 1H => g in H

So the whole thing is H

[ripe harbor]

Wait why

g in Hg sure

g not in H because g in gH = G \ H

[delicate orchid]

Cosets are disjoint. H and gH are the two left cosets, H and Hg are the two right. By disjointness gH != H, and thus must be Hg and we are done

It’s essentially the pigeon hole principle with an extra step at the start

[ripe harbor]

Oh, if Hg^(-1) were H, then g^(-1) in H and so g in H, so gH would also be H. But it isnt. So Hg^(-1) = G \ H

So when gH is G \ H, then gH and and Hg^(-1) are both G \ H

[ripe harbor]

Ok so what if both are G \ H, isn't that a problem?

We need this to be H

[glass sinew]

(a) condition for normality is that for each g, gH = Hg

[delicate orchid]

A subgroup is normal if and only if Hg = gH for all g

Well looks like the nice guy has finished last ONXE AGAIN 💔💔💔

[glass sinew]

WELL

[ripe harbor]

Oh, right, my definition is gHg^(-1) = H but thats equivalent

[delicate orchid]

Right multiply by g twin ✌️

[ripe harbor]

Yeah

[glass sinew]

what are u studying from btw

[ripe harbor]

mainly just some lecture notes with problem sets and Artin's book

[delicate orchid]

Now let p be the smallest prime dividing |G|. Show that all subgroups of index p are normal

[ripe harbor]

could Sylow be helpful here?

[glass sinew]

as a warm up let p be a prime dividing |G|. show that G has a cyclic subgroup of order p

[delicate orchid]

Not really, the proof is a bit more complicated than the proof we have for p = 2

[wraith cargo]

This is a nontrivially hard thing to prove iirc

[ripe harbor]

Ah right Sylow could only give us a chain of normal subgroups at most but not really that any of those are normal in the actual one

[glass sinew]

shhhhhh

[delicate orchid]

My hint for my problem is to consider the action of G on the cosets of H

As a warm prove that all finite simple groups are 2-generated

[ripe harbor]

How do you come up with it/how do you know what action to consider or even considering an action in the first place? Is this just having seen this before? In the p = 2 case, we didn't need actions at all

[delicate orchid]

All normal subgroups arise as kernels of a group homomorphism, we attempt to construct a homomorphism with kernel H

All group actions give a map into a symmetric group, find one where it’s easy to see that the kernel is H

[glass sinew]

||just the projection G -> G/H right?||

[delicate orchid]

Yur

[ripe harbor]

Do you have some general tips on approaching these? Do you usually think of examples or do those not help much for the general case anyways?

[delicate orchid]

Honestly you just have to see a billion exercises and practice switching between concepts and getting a feel for general approaches

[glass sinew]

legally acquire from a reputable provider d&f and spam exercises until you dream about groups

worked for me

[delicate orchid]

It’s funny cause you eventually define quotient objects as equivalence classes of epimorphisms

[marble hinge]

haha, now I have exactly the same comment as I had about your other problem: this was proved in D&F before Sylow's theorem appears (I used this lemma in my proof of your previous problem about cyclic groups of pq order). It's done in the chapter about group actions, as suggested by @delicate orchid

[glass sinew]

literally just had a think about this that made 1st iso thm click

[delicate orchid]

1st iso theorem ❌
Admits an epi-iso-mono factorisation ✅

[marble hinge]

but to be honest, now I can't fully recall the proof, need to look at it again

[delicate orchid]

The proof of the classification of groups of order pq?

[marble hinge]

no, this current lemma: if p is a smallest prime dividing order of finite group G, then any subgroup of index p is normal in G

[delicate orchid]

Ah ok

[marble hinge]

(generalisation of the p=2 case that is much easier to prove directly)

[delicate orchid]

Consider that the second cohomology of a p-group with coefficients in a p’-module vanishes and the size of Aut(C_p) 🆘🆘🆘🆘🆘🆘

[delicate orchid]

It’s easier to prove because S_2 has no subgroups!

[ripe harbor]

Haha D&F has everything!

[delicate orchid]

“Follows from Schur-Zassenhaus”

[ripe harbor]

Alright, thanks

[delicate orchid]

Have you seen group actions before? I wasn’t expecting you to actually be able to solve it I was bringing it up just to tell you that this generalises!

[marble hinge]

Reading about group actions kinda changed how I think about groups now :)

[ripe harbor]

Yes I have

[delicate orchid]

Swagtastic

[marble hinge]

I also became excited about prospects of learning some geometric group theory, where groups acts on some spaces and we can study things like Cayley graphs, groups acting on graphs or trees and whatever

[noble nexus]

ggt is very cool

[ripe harbor]

In what way? Every proof I have seen that uses group actions uses |X^G| cong |X| mod p for G a p-group in the end

[noble nexus]

I want to learn more of it but its kinda intimidating ngl

[marble hinge]

I haven't properly started and I know little about geometry per se, but I think I can deal with graphs and trees first :)

[delicate orchid]

You WILL learn Bass Serre theory and yoy WILL like it

[marble hinge]

maybe just in general, that now I realise better that those group elements permute some sets

and induce some homomorphisms into S_A

[delicate orchid]

Does Cayley’s theorem use this?

[marble hinge]

so it helps to think about problems, maybe just indirectly

[ripe harbor]

That f: G -> S_G with g -> [f(g): h -> gh] is an injective group homomorphism? That's what I call Cayley's theorem

[marble hinge]

also I appreciated that concepts like normalisers and centralisers were finally unified under "stabiliser" umbrella :)

[delicate orchid]

Yeah. Essentially “the regular action exists” lol

[ripe harbor]

Ah, no, the proof of that was just .. trivial..?

[delicate orchid]

Yeah kind of lol

[ripe harbor]

Yeah it was just checking the definitions

[delicate orchid]

The only known proof that all groups of order p^aq^b are solvable uses group actions in a very non-trivial way LOL

[marble hinge]

Also things like (Not) Burnside lemma are quite useful to count things (and is based on group actions)

[ripe harbor]

The proof of Cauchy's theorem did some really neat trick with group actions

[delicate orchid]

once you get to group extensions and see semidirect products and how all groups are essentially just built out of simple groups “acting” on other groups I think you’ll see the true value

[ripe harbor]

considering the set of all elements of the group that whose product is 1

[delicate orchid]

Oh yeah I remember that

[ripe harbor]

and then of course |X^G| cong |X| mod p

[marble hinge]

nice, reminds me about similar trick in the proof of Wilson theorem in Elementary Number Theory...

[marble hinge]

is "group extensions" in D&F?

[delicate orchid]

I remember coming up with the proof of Cauchy for p = 3 and struggling to generalise it lol

[delicate orchid]

I have absolutely no idea

[ripe harbor]

Do you tend to think about multiple exercises at once? Or think about one until you're done and only then go to the next

[delicate orchid]

I switch if I get stuck for a while

But honestly I haven’t really done any exercises for like a year lol I’ve been too busy writing

[ripe harbor]

What's an example of a finite group $G$ and $H \leq G$ with $a, a', b \in G$ with $$aH = a'H \quad \text{and} \quad abH \neq a'bH?$$

[cloud walrus]

ILikeMathematics

[ripe harbor]

I was thinking about matrices, but there is probably a simpler solution

Something like Z/nZ doesnt work because it's abelian

[delicate orchid]

I actually don’t know off of the top of my head

Other than the free group on 2 points but that’s incredibly boring

[ripe harbor]

Oh, well I haven't had free groups yet anyways

[delicate orchid]

Imagine the most boring group possible

[ripe harbor]

That's like ({e}, *)

[delicate orchid]

That’s the free group on 0 points

[quiet pelican]

WLOG a’ = e

So we need exactly a \in H and b^{-1} a b \notin H

So like, G = S_3, a’ = e, a = (12), b = (123), H = <(12)> works

So any group with a non-normal subgroup works for this

[quiet pelican]

That’s interesting

[delicate orchid]

I meant as an example to this particular question. But the only pure group theoretic thing thats interesting about it specifically is that it contains all other countably generated free groups

At least Thats the only one I know of. My mind is open to being changed

And DONT give me that coset enumeration bullshit

[ripe harbor]

Oh, right, thats a nice simplification

G = S_3, a’ = e, a = (12), b = (123), H = <(12)> works
how did you think of this? Does this also work for many other things as counterexample/is it "famous" or did you just think of this for this specific exercise

[quiet pelican]

That’s the smallest example of a group with a non-normal subgroup, and a pair of elements exhibiting that fact

[marble hinge]

I think this H(ab) = (Ha)(Hb) is the feature that makes operation on cosets work (and it requires H to be normal). And in this example you are asked to find something where it doesn't work

so it's reasonable to look at things which are not normal

like <(12)> in S_3

[ripe harbor]

Ah, alright

[ripe harbor]

Let $G = \langle g \rangle$ be of order $n$ and $k \in \mathbb Z$. I want to show that $$\text{ord}(g^k) = \text{ord}(g^{\gcd(k, n)}).$$ We know that $\gcd(k, n) = ak + bn$ for some $a, b \in \mathbb Z$ and so $g^{\gcd(k, n)} = g^{ak}$ but that doesnt seem helpful.. We know that $$\text{ord}(g^k) = p \quad \text{with $p$ minimal so that $g^{kp} = 1$}$$

[marble hinge]

@ripe harbor I wonder, in which order are you studying things? When I first saw your messages, I thought that you are just starting group theory, because you were asking some question about subgroups or something relatively simple. Then, after a couple of days I've seen you proving something with Sylow -- I thought, wow, they progressed really quickly! But then I looked at your older messages and saw that you did some fields and rings previously. So now I am intrigued by the way you are studying!

[cloud walrus]

ILikeMathematics

[marble hinge]

Are you doing like multiple passes over abstract algebra?

Or studying things on demand once the need arises?

[ripe harbor]

Haha, if with older messages you mean like one or two years ago, yes, I self-studied some AA but I wasn't officially taking it. Now I am, I'm doing groups right now and I'm at the end of the chapter right now, so I'm doing a bunch of exercises about everything in it

[ripe harbor]

What about you? I saw you have a thread, are you doing a second pass or learning things now?

[marble hinge]

mostly learning this bit. But I am just studying math as a hobby (i.e. I already have a degree in applied math, but that was some time ago)

[ripe harbor]

Ah, nice

[ripe harbor]

We also know that ord(g^gcd(k, n)) = n/gcd(k, n) but that also doesnt seem helpful..

[candid patrol]

o(g^k) = o(g)/(gcd(o(g),k))

[ripe harbor]

I don't have that unfortunately

Only o(g^k) = n/k for o(g) = n and k | n

[candid patrol]

Like you don't wanna use it or you don't know it ?

[ripe harbor]

latter, we haven't had that

[ripe harbor]

Can we use this?

[something else we could use is g^m = 1 => n | m]

[and ord(gh) = ord(g) ord(h) for gcd(ord(g), ord(h)) = 1].

Thats pretty much all lol

[warm badge]

you should try proving it, then the required result follows immediately. what you know already is just a particular case of that identity, when k | n gcd(n,k) = k, so o(g^k) = n/k. you should try proving the stronger result.

[coral steeple]

Is there a word for a polynomial f in K[x] whose splitting field over K has degree (deg f)! ?

[rocky cloak]

I haven't heard of such a word, but normal could make sense as this would happen if K[x]/(f(x)) is normal (assuming f irreducible anyway)

[coral steeple]

Hm, ok, thanks

I'm asked to show that the splitting field of $2x^5-10x+5$ over Q has degree 5!. I know that this polynomial is separable, (and should be) irreducible, and I know that it is enough to show that adjoining roots one at a time gives a chain of field extensions with degree 5, then 4, then 3 and 2. But I'm not able to calculate the degree of these simple extensions. Here's my attempt. Label the roots $\alpha_1,\dots,\alpha_5$. To show the second extension has degree 4, we need to show the irreducibility of a quartic in $\mathbb Q(\alpha_1)\cong\mathbb{Q}[x,y]/(2x^5-10x+5)$. That seems infeasible.

[cloud walrus]

person2709505

[rocky cloak]

Oops, I missed the factorial.

Then I don't think I have a good name

[coral steeple]

I suppose that in light of my second question, my first was a bit misleading

[winter niche]

Are projections onto linear subspaces lipshitz?

[coral steeple]

Solved. I had the wrong approach entirely.

[small tendon]

Anyone know a good survey article on the monster group?

[azure cairn]

is this proof OK? this theorem isn't proven in my alg book

[spice whale]

yeah that's good

[spice whale]

it can be made even shorter

actually

[azure cairn]

makes sense i just wrote this out on a first pass

[spice whale]

you just note that the product has order rs (by set theory)

and then (x,y) has order rs so it must be cyclic

[azure cairn]

i see

[spice whale]

because it has an element of order |G|

[azure cairn]

indeed

[azure cairn]

does this thm have a name

[cursive spindle]

Internal direct product theorem?

[azure cairn]

looks right from googling

[cursive spindle]

lowkey would not call it a theorem

[azure cairn]

i dont really understand the point of this proposition

this is true, now what

[cursive spindle]

conditions to decompose G into a product I guess

play around with some examples

[south patrol]

I don't think it really has much of a name but is helpful to know for understanding direct products and recognising a group as a direct product. Later you will see variants about semi-direct products which are very useful and much more general

[coral steeple]

Say I have a group G defined by some generators and relations. Obviously, if H is isomorphic to G, then it has generators satisfying the same relations. Is there a converse statement? What if I impose extra conditions, like H and G both finite and of the same order?

[south patrol]

I am not sure what sort of converse you mean. It seems the converse is saying whether giving generators and relations determines the group, but this is true by what it means for things to be generators and relations for a group

[azure cairn]

what is a "direct product" (or if you have any reading on them)?

[coral steeple]

Ok, I have never seen a proper definition of "generators and relations"-- does it mean the group is the free group on n generators quotiented by the normal subgroup generated by the relations?

[south patrol]

Probably what you would call a product (of groups)

[azure cairn]

so like (a, b) * (a', b') = (ab, a'b') is a direct product?

[south patrol]

Yes. Or more precisely if you say a group G is generated by some elements (of G) subject to some relations, then the map from the free such thing to G is an isomorphism

[coral steeple]

Ah ok great

[south patrol]

This is the multiplication in the direct product yes

[azure cairn]

aa

lemme finish the product group section and see if i get anything more about this

[coral steeple]

How about this somewhat related statement: if G is given by n generators satisfying some relations, and H contains n elements satisfying those relations, then G is iso to H

This is false

But I think it may be true if G and H are finite of the same order. Or it may just be badly false

[somber goblet]

im pretty sure it's $(a, b) * (a', b') = (aa', bb')$

[cloud walrus]

lexi

[somber goblet]

but similar idea

[coral steeple]

It sounds fishier and fishier the more I think about it

[azure cairn]

for the law of composition on a product group?

[cursive spindle]

yeah it's the obvious thing to do

[somber goblet]

i could be wrong but im somewhat sure

[azure cairn]

artin defines it by this

maybe a textbook difference

[cursive spindle]

ah wait you wrote something else

in what you wrote it should be (a,b) * (a',b') = (a a', b b')

[somber goblet]

oh that makes sense

[azure cairn]

so wait is a direct product distinct from the law of composition on a product group?

ah wait yes i see

the direct product is how you would multiply cartesian-ly?

[cursive spindle]

yes

[somber goblet]

the "product group" is a little bit of an ambiguous term since there are multiple ways to form a product of groups, but usually it refers to the direct product

[cursive spindle]

also keep in mind that the operation may be different in each component

[azure cairn]

weird. i'll look into this later

[cursive spindle]

why is it weird?

[azure cairn]

i don't understand the point of a direct product nor why it's different from the group operation artin chose for product groups

[somber goblet]

it's not

[cursive spindle]

is this from artin?

this is a direct product

[azure cairn]

wait a minute

2wfgeuhodisjkpm i was mixing up letters LOL

[cursive spindle]

ahaha yeah I noticed

[somber goblet]

direct products are useful in the study of abelian groups

[azure cairn]

ok so yeah the product group usually uses the direct product for its group operaiton

[cursive spindle]

For abelian groups that are written additively you can refer to the direct product as the direct sum

[azure cairn]

is it cuz u can decompose them into

yeah

[cursive spindle]

but yeah the idea is still imitating cartesian

[azure cairn]

i learned before that cyclic groups of order rs can be decomposed into a product of cyclic groups of orders r and s

[somber goblet]

the direct product is the most natural way to get a product of groups

[cursive spindle]

yeah when you get to semidirect product you'll question why tf would someone define something like this

[somber goblet]

until you see dihedral groups

[azure cairn]

ok so the proposition above is basically just a way to show if G is isomorphic to H x K

and that's useful cuz yes

[somber goblet]

btw both G and G' are isomorphic to normal subgroups of G x G'

so it's very well-behaved

[azure cairn]

is this beyond what i've learned

cuz idr that result

[somber goblet]

wait have you done first iso theorem yet

if so this result is really easy

[azure cairn]

not in this book but im familiar with it (i learned it in linalg)

also i guess like technically yes ive done it i know [G:ker phi] = |im phi|

[somber goblet]

we have $G/\ker \phi \cong \operatorname{im} \phi$ right

[cloud walrus]

lexi

[somber goblet]

you've done isomorphism right

idk what order artin does things

wait idk if you even need the full first iso theorem for this

[azure cairn]

yes

[azure cairn]

at 2.11 rn

[somber goblet]

the map $\pi \colon G \times G' \to G$ where $\pi(x, x') = x$ and the map $\pi' \colon G \times G' \to G'$ where $\pi'(x, x') = x'$ are homomorphisms

[cloud walrus]

lexi

[azure cairn]

projection maps?

[somber goblet]

yes

$\ker \pi$ is the set of all elements $(0, x')$ where $x' \in G'$, and $\ker \pi'$ is the set $(x, 0)$ where $x \in G$

[cloud walrus]

lexi

[somber goblet]

homomorphism kernels are normal subgroups

[azure cairn]

yep

[somber goblet]

and it's pretty easy to set up an isomorphism $\ker \pi \cong G'$ and $\ker \pi' \cong G$

[cloud walrus]

lexi

[azure cairn]

indeed

yeah no i see yes

[somber goblet]

there's probably another way to show this but i think this way is the nicest

[azure cairn]

oh wait

did you mean a normal subgroup or every

[somber goblet]

with first iso theorem you can show that $(G \times G')/G \cong G'$ and $(G \times G)/G' \cong G$

you havent done quotient groups yet i just realized

[cloud walrus]

lexi

[azure cairn]

i understand the idea (i think?)

[somber goblet]

yeah

[azure cairn]

G/N for normal subgroup N is the set of cosets of N

[somber goblet]

yep

you're taking the group "mod N"

[somber goblet]

wdym

[azure cairn]

then the zeros are elements in the trivial coset N

[somber goblet]

yeah

[azure cairn]

so dividing G by N basically means N acts your zero elements

[somber goblet]

you might also be able to see how this relates to the kernel of a group homomorphism

[azure cairn]

are G and G' isomorphic to every normal subgroup of G x G' or just that specific normal subgroup?

[azure cairn]

yes

phi(ak) = phi(a)

[somber goblet]

just those specific normal subgroups

[azure cairn]

so G/ker phi distinguishes elements by eliminating the kernel component of them

[somber goblet]

yes

[azure cairn]

in linalg its the idea that u + v = u + w under V/kerT if v, w are kernel elements

[somber goblet]

this is the first isomorphism theorem

we have a sticker for it

[azure cairn]

so vectors that differ only by ker phi are equivalent

yea ik there's a sticker lol

it's cool this is true for structures less heavy than vector spaces

and also cool to see the commutative diagram

[somber goblet]

it's true for almost everything worth studying in algebra

[azure cairn]

for rings too?

[somber goblet]

groups, rings, fields, vector spaces, modules, algebras

[azure cairn]

(idk what a ring is )

[somber goblet]

you name it

[somber goblet]

abelian group with an associative multiplication operator that distributes over addition in the abelian group

[swift root]

generalization of numbers

i.e. a structure where you can add, subtract, multiply and the usual distributive rules apply

[azure cairn]

well i know the definition because im a popmath indulger but idk like the actual theory surroudning it

[somber goblet]

rings have an analog to normal subgroups as well, that lets the first iso theorem work

[swift root]

it's very rich

[azure cairn]

looking Forward to it (maybe..)

[somber goblet]

it sadly isn't as 'clean' as quotients in vector spaces where you can quotient any subspace

[azure cairn]

yea its normal subgroups only

since that makes left and right cosets equal

[somber goblet]

of course this makes sense in the context of vector spaces being fancy abelian groups

[azure cairn]

in subspaces all cosets are equal

[somber goblet]

and all subgroups of abelian groups being normal

[azure cairn]

and yes abelian-ness is sufficient for normality

[swift root]

important to note that ideals are not subrings, unlike normal subgroups

[azure cairn]

cuz gag^-1 = gg^-1 a = a

wow im so Good at group stuff!!

[swift root]

(unless you use the convention that rings may be nonunital, in which case you should reevaluate your life choices)

[somber goblet]

i left unital out of my definition up there for simplicity's sake

[swift root]

I wouldn't say adding a 1 makes it more complicated

[somber goblet]

but yes i usually take rings to be unital

[somber goblet]

i mean in relating ideals to normal subgroups

[swift root]

I suppose

but sometime you're gonna have to accept that all of these things are equivalence relations anyways

[somber goblet]

an ideal is just a normal subgroup with extra structure

anyway

the equivalence relation is exactly the coset relation on the additive group

algebras are awesome

im trying to understand galois theory

[cursive spindle]

peak

[somber goblet]

its cool that "polynomial evaluation" is a ring homomorphism

[next obsidian]

It’s kind of universal

[somber goblet]

im a little stuck on $F[x]/(p) \cong F(\alpha)$

[cloud walrus]

lexi

[somber goblet]

i understand it intuitively but idk how to prove the isomorphism

herstein very annoyingly leaves this as an exercise

[swift root]

well clearly F[x] has krull dimension 1 and (p) is prime and nontrivial, so F[x]/(p) has to be a field and then you're essentially done

[somber goblet]

i know F[x]/(p) is a field

p is assumed irreducible so (p) is a maximal ideal

[cursive spindle]

what are you stuck on exactly

[somber goblet]

this might sound dumb but

how do we know which root we get?

we only get the adjoin of one root right

when we quotient by the irreducible polynomial

and it's the one corresponding to the polynomial $x + I$ in the quotient field

[cloud walrus]

lexi

[cursive spindle]

okay think about this example: Take Q[x] and the polynomial x^2+1. What would happen when you send x->i? how about x->-i?

[somber goblet]

we get both roots in the extension field either way

but the degree of the splitting field is just the degree of the polynomial

what about if it's greater

[cursive spindle]

yeah either way the resultant field is the same up to isomorphism

[somber goblet]

but surely the extension doesn't contain all the roots under all circumstances

since we can have a degree 3 polynomial yielding a degree 3 extension

but that has a splitting field of degree 3! = 6

like $\mathbb Q[x]/(x^3 + x + 1)$

[cloud walrus]

lexi

[somber goblet]

this can't contain every root of that polynomial, because it has a splitting field of degree 6 but this field must have degree 3

[cursive spindle]

yes

[somber goblet]

so how do we get all of them

factoring the polynomial in the extension then quotienting whatever is irreducible left over?

[cursive spindle]

you construct the splitting field

[somber goblet]

ok

and we know this exists because we can always pull out roots using this method

[cursive spindle]

yep you keep adjoining roots till you get a field that has all the roots

[somber goblet]

ok that's reasonable

[cursive spindle]

each time you go up the ladder adjoining a new root you record the degree of the extension

[somber goblet]

that sets the boundary $\deg p \leq [E:F] \leq (\deg p)!$

[cloud walrus]

lexi

[somber goblet]

ok the simple element theorem

implies that if i have a polynomial with a certain set of roots, i can find a single element of order up to n! that 'generates' the splitting field

[somber goblet]

or whatever it was called

i don't really understand the proof for that

[cursive spindle]

yeah essentially the degree of a splitting field is a divisor of n!

[somber goblet]

oh

oh right yeah

because we're going down the definition of a factorial but skipping multiplying by some numbers if we get to a polynomial we can actually reduce in our current extensions

[fleet cairn]

An identity morphism A->A is obviously isomorphic
automorphism is an isomorphic arrow A->A

These look the exact same, the difference is one you start with A->A then conclude it's isomorphic, and the other is your premise being that the morphism A->A is isomorphic, these looks practiclaly same 😭 (im thinking that a morphism is an automorphism iff and only iff it is an identity morphism)

is there a difference?

[somber goblet]

an automorphism $\sigma \colon A \to A$ is an isomorphism of $A$ onto itself

[cloud walrus]

lexi

[somber goblet]

it does not need to be the identity

[fleet cairn]

Why is it not an identity, it looks the same

A->A and A->A

[tulip otter]

take A=C and the map to be complex conjugation

[somber goblet]

it maps the object A onto itself but it can move elements around

[tulip otter]

this is an automorphism of C but its not the identity on C

[fleet cairn]

sorry i dont understand this example

[tulip otter]

np, so do you know stuff around complex numbers?

just what is a complex number and what is complex conjugation

[fleet cairn]

i know that its root of i and it rotates on the complex plane thats kinda it

[glass sinew]

if f(x) = -x, f is an automorphism of Z

[somber goblet]

yeah this is the example i was going to use

if we're talking group automorphisms

[fleet cairn]

But -x is clearly not itself

[tulip otter]

ok so here is the misunderstanding that you have, an automorphism doesnt necessarily map an element to itself

[glass sinew]

we're saying that Z and -Z are isomorphic, but not equal

[tulip otter]

the only map that maps every element to itself is the identity map

[fleet cairn]

but an automorphism is still of form A->A?

[tulip otter]

an automorphism of A is a bijective homomorphism from A to itself

[glass sinew]

A -> A does not mean everything is mapped to itself, just that the map has domain A and image in A

[tulip otter]

sure

[fleet cairn]

im sorry im not familiar with homomorphism either?

[somber goblet]

maybe think of "set automorphisms," which are permutations

[glass sinew]

read up on the basics

[fleet cairn]

my book allufi doesnt cover it yet

[tulip otter]

well thats essential to know what an automorphism

[glass sinew]

he will soon

[tulip otter]

I have a hard time believing that, does he perhaps say morphism instead?

[glass sinew]

are you in the groups chapter yet? you should go through that one fully

[fleet cairn]

yeah

[cursive spindle]

?

homomorphism is half the answer

[fleet cairn]

he covers isomorphism on sets as bijection, then he teaches categories and morphism and now the next chapter is about ismorphism

[tulip otter]

ok so an automorphism of A is a morphism A->A which is bijective

[tulip otter]

alright then thats good, lets take an example of an automorphism of a set A which isnt the identity

[cursive spindle]

okay let me see what are you working with here?

[glass sinew]

man you're not even at the groups part finish that chapter first

[cursive spindle]

Groups?

[fleet cairn]

im simply following the order the book gave, idk im sorry

[tulip otter]

take A=Z and f:Z->Z defined by f(x)=x+1

[glass sinew]

it'll take you a hot second to get there then, yeah

[cursive spindle]

It's not a good idea to jump into categories like that

[tulip otter]

check that this map is a bijection from Z to itself

[fleet cairn]

do you suggest to skip automorphism to 2 chapters ahead?

[tulip otter]

so its an automorphism of Z as a set

[glass sinew]

aluffi runs categories-first

[tulip otter]

but its clearly not the identity map

[somber goblet]

eww

[glass sinew]

if this is your first scrape with abs alg i'd recommend d&f over aluffi probably

[cursive spindle]

now you know why I hate category theorists sometimes

[tulip otter]

start abstract algebra with infty categories

[fleet cairn]

Ohh yes I see but it's only on a particular structure on Category on Sets right?

if the morphism doens't behave like a function

then it wouldnt make so much sense

is my point

is that right?

[tulip otter]

an automorphism in general isnt necessarily the identity regardless of the category you are working in

[somber goblet]

let's say you have a set C representing deck of cards. if you shuffle the deck of cards, then that's a non-trivial bijection of C onto itself, which is an automorphism in Set

it's the same deck of cards, but you shuffled the elements around

[fleet cairn]

but if the categories are not objects of sets then these examples wouldn't work right?

[somber goblet]

automorphism sort of captures that concept

[glass sinew]

the morphisms are different in each category

[somber goblet]

complex conjugation is an automorphism of fields, and taking the negative of an integer is an automorphism of groups but not fields

[tulip otter]

I mean yea morphisms in a category depend on the objects of this category

[glass sinew]

what brings you to ask what an automorphism is? i don't recall that coming up so early in the book unless i'm misremembering

[fleet cairn]

i understand the difference now but im just asking for a bit more understanding because these examples on autmorphism looks like it relies on the concept of objects being sets

[fleet cairn]

its quite literally in the same chapter where isomoprhism is covered

wait let me show

[cursive spindle]

Please just drop categories for now and learn about groups instead. Use D&F

[tulip otter]

what you should do is move on and you will understand more of the category theory when you have more examples in the bag

[tulip otter]

no dont use D&F

[cursive spindle]

Whatever just use anything else

[tulip otter]

maybe lang is the way to go sotrue

[azure cairn]

is this aluffi ch0

[tulip otter]

:chad:

[glass sinew]

well there's your answer really

[fleet cairn]

yes

[fleet cairn]

i dont get it

😭

[cursive spindle]

Mamma mia

[somber goblet]

you need more examples

[fleet cairn]

groups is 2 chapters ahead

*3

[somber goblet]

learn about any algebraic structure

preferably groups

[tulip otter]

lol as i told you, dont stick to this rn

[fleet cairn]

ohh ok

[glass sinew]

do yourself a great favor and read the first few chapters on groups in d&f. much more straightforward

[cursive spindle]

How does aluffi introduce groups? Does he say "groups are groupoids with one element?"

[fleet cairn]

i definitely understood the autmorphism examples on sets/functions though

[tulip otter]

you will encounter automorphisms all over the place when you learn about any algebraic structure

[azure cairn]

an automorphism can also be defined on groups btw, this is just a step above

[glass sinew]

yeah.

[fleet cairn]

so can i just continue allufi with this much understanding

[azure cairn]

like categories are an abstraction of all the algebraic structures you will learn about (i think?)

[tulip otter]

thats good enough, I think thats what you can do for now with the examples you have

[cursive spindle]

It's kinda not funny anymore

[glass sinew]

categories abstract the relations in the class of all such algebraic objects

[azure cairn]

did you typeset this LOL

[glass sinew]

aluffi did

genuine starter to groups

[azure cairn]

that's an actual part of his book??

🐒

[somber goblet]

wtf

[tulip otter]

https://media.discordapp.net/attachments/917866853198594131/1471604215905259761/hmmmm.webp?ex=69ac8aba&is=69ab393a&hm=e2edca5fc18a493e47ca89d532c83cd44f0533323d430ef3c4ea2f4359a762ac&=&format=webp&width=72&height=72

[glass sinew]

yea he really explains it, super unfunny

[fleet cairn]

ok thank you for helping i didnt expect such a reaction on allufi 😭

[tulip otter]

I mean I didnt say anything about aluffi

[novel star]

i remember one of my friends was learning algebra from this

[cursive spindle]

No I learned something today. I won't recommend Aluffi anymore

[lusty marlin]

I hate this so much

[fleet cairn]

everyone else did lol

[novel star]

and he thought that was the funniest thing on the plannet

[glass sinew]

carl e linderholm moment

[tulip otter]

he also doesnt expect you to have many examples rn

[glass sinew]

what a fucking nerd

[azure cairn]

book of proof was pretty funny

[tulip otter]

so dont get worried lol

[novel star]

every single person he told that to said that as well

[tulip otter]

if you like the book continue with it, if you dont then ask for some other recommendations

[novel star]

aluffi is very contingent upon the exercises iirc

[somber goblet]

herstein doesn't touch groups until after introducing modular arithmetic and permutations and separately formulating group axioms on both

[glass sinew]

if you struggle there are easier ways to familiarize yourself with abs alg

[novel star]

theres 1000+ for a reason

[cursive spindle]

Inb4 someone defines the abelianization of a group as the first cohomology group