groups-rings-fields

[urban acorn]

yeah

so, for example, you have 1 in <r^2> in <f, r^2> in D8

a chain containing 4 subgroups nested inside each other, beginning with 1 and ending with D8

and all of the inclusion of each subgroup in the next are as normal subgroups

and the quotients are all C2

[red fox]

Yep

[urban acorn]

btw, so remember how I said the first non-solvable group is of order 60?

[red fox]

Mmhmm

[urban acorn]

there's this mathematician who has a bunch of numbers as his banner on his youtube channel

and they all have some significance, and people are supposed to figure those out on their own

60 is one of them, and I guarantee you it's because it's the order of the first non-solvable group

[red fox]

Nice

[urban acorn]

by the way, unfortunately, it is in general very difficult to understand the structure of a group given the structures of a normal subgroup and that of the quotient by it

but we do have a classification of all finite simple groups

[red fox]

My problem is, when you reduce D8 to C2s, where does the information in the semidirect product go?

[urban acorn]

so this very difficult problem is actually the barrier between our current mathematical knowledge and a complete classification of finite groups

[urban acorn]

you lose it

these representations are not unique

i.e. the elementary abelian group of order 8 has the same series

[red fox]

exactly

I'm trying to reduce D8 into C2s so that information is not lost

[urban acorn]

well, good luck

[red fox]

you were talking about "proving" this. What would I need to prove?

[urban acorn]

not so much proving, as making it formal

because right now, you have a vague idea, but you can't express it formally

there's not even a statement to try to prove yet

and I can warn you that what you're trying to do is very difficult, but I don't want to discourage you from trying, cause your mathematical understanding could still benefit from trying even if this approach is a dead end

[red fox]

I think I would want to prove the version of D8 in E16 has is isomorphic to the real D8

[urban acorn]

what do you mean by the "version of D8 in E16"?

you can't embed it as a subgroup

[red fox]

the C2 x C2 x C2 x C2 with 1 degree of linear dependence

which I am claiming represents all aspects of D8

[urban acorn]

well, it's not clear to me what that means

[red fox]

you would agree that D8 =/= C2 x C2 x C2, right?

[urban acorn]

yeah

[red fox]

I would be trying to prove something like
D8 ≅ (C₂×C₂×C₂) ∪ (C₂×C₂×C₂) ∈ C₂×C₂×C₂×C₂

the split of 2 (C₂×C₂×C₂) is because of the two kinds of cosets

[urban acorn]

that's not standard notation

∈ stands for inclusion as an element of a set, not a subset

and even if you did mean subset

you can't get D8 as a subgroup of C2 x C2 x C2 x C2

[red fox]

right, because you also have to add in that linearly dependent contraint

D8 =/= C₂×C₂×C₂×C₂

It is like how a circle can be draw in ℝ²

[urban acorn]

how do you use "linear dependence" to get D8 here?

[red fox]

so going back to this, you see how every element is both in a column and a row?

[urban acorn]

well, look, I'm probably going to sleep soon

so someone else might pick this up

I'm just gonna watch something on youtube first, probably

[red fox]

thanks for your help

[urban acorn]

nice to talk to you

[red fox]

If you are represent the column as a C₂×C₂ group, and the row as another C₂×C₂ group, every element is defined in the two contexts, and is the common factor that makes the two contexts linearly dependent

So an element would be represented by two linearly dependent klien four groups

[potent briar]

given the theorem
If S is any set of vectors in a vector space and if M is the subspace spanned by S, then M is the same set as the set of all linear combinations of the elements of S

prove the theorem
If U and W are any two subspaces and if M is the subspace spanned by U and W together, then M is the same set as the set of all vectors of the form u + w, with u member of U and w member of W

i have that so far

am i close? :x

[terse crystal]

I think you are proving span(the union of U and W)=U+W right? It’s directly derived from the definition

Any Linear combination of elements from the union of U and W is of the form u+w

So it should be union not sum in your picture

[potent briar]

i agree

but im not sure if that proves the theorem

[terse crystal]

?? Almost just definition

[potent briar]

i think it's wrong

[thorn delta]

Are you sure “span of U and V together” doesn’t mean span(U \cup V)?

[potent briar]

right

so the union of the span of two lines, for example

is that plane

not the two lines

[thorn delta]

Well why do u have span(U+V) written then?

[potent briar]

that's the theorem

that's the first theorem

[thorn delta]

The theorem you want to prove is span(U \cup V) = U+V

The theorem ur given is span(S) = {finite linear combos of elements of S}

[potent briar]

not the sum of the sets

the sum of all vectors

u + w

[thorn delta]

Huh? Yes that is what U+V means

[potent briar]

hmm the sum of two subspaces is defined after the theorem

in my book

[thorn delta]

Or wait what. No U+V is the set of all u+v with u in U and v in V

[potent briar]

yes

[terse crystal]

And you just go ahead check that span(U \cup W)=U+W

I really can’t see what’s the obstacle

[potent briar]

but i know that, since U and W are subspaces

their span = the subspace

no?

so from my picture im 1 step away?

[thorn delta]

U \cup V is not a subspace. U+ V is

[potent briar]

sure

[thorn delta]

Anyway, u just gotta directly apply the theorem ur given about span

In fact, depending on how u defined span, one of these inclusions might be obvious

Namely, if you define span(S) to be the smallest subspace containing S

[potent briar]

thats my definition of span yes

my problem isnt imagining the objects

its writing the proof

thats where im weakest

[thorn delta]

You’ll want to use the definition of span to show that span(U\cup V) is contained in U+V and use the theorem ur given for the other inclusion

If u need a hint lmk

[potent briar]

thank you

[red fox]

@potent briar I actually just had a homework problem related to this. This is my first proofs class, so I'm not sure how good the proof is.

[kind temple]

i would restate the claim as W1 U W2 is a vector subspace of V if and only if W1 subset W2 or W2 subseteq W1

[green locust]

Can someone please tell me if I’m going in the right direction for this? (Part a). My idea was to take a look at the intersection of the preimage of 0 for all the functions in M and pick one for our ‘c’ but I’m not sure how to show such an element would exist in the first place. I feel like I’m either missing something very straightforward or I’m thinking about it totally wrong. My other idea was to take a look at the quotient map R/M but I’m not sure where to go from there

[gritty sparrow]

Here is a hint, ||proceed via contradiction, and show that for every x in [0,1] there is a function f_x st f_x(x)/=0. Now by continuity show that we can even create an open set U_x st x is in U_x and f_x is never zero on U_x. Notice that U_x forms an open cover||

[green locust]

@gritty sparrow proceeding by contradiction makes sense, I’m a little stuck on the open cover part though

[gritty sparrow]

Which part are you stuck on, the existence of U_x or showing that it is a cover?

Oh I should have worded things a little carefully, the collection {U_x | x in [0,1]} is a cover, not just some particular U_x

[green locust]

Ok, so since it’s covering a compact set we can find a finite subcover, say U_{x_i} where i goes from 0 to n and we have each f_{x_i}(U_{x_i})/=0?

[gritty sparrow]

Yes

[green locust]

Do I make some new function with these f_xi’s

[gritty sparrow]

Yes

Just try to create a function that is never 0 for all x. Those sorts of functions are invertible in this ring

[green locust]

I’m thinking just adding them all together

[gritty sparrow]

A better idea is to take the square of each function and then add, that way we don’t get any accidental cancellations

[green locust]

Ohhhh ok, so that way we avoid cancellations, we know it definitely stays away from zero, and then we can say (x-c) times this new function would be zero at c and would be in M so we have a contradiction?

[gritty sparrow]

The contradiction is that this function is non zero everywhere, so it has an inverse. So when we multiply the two we get that the constant 1 function (the identity of the ring) is in the maximal ideal, which is impossible bc maximal ideals are always proper

[green locust]

That makes sense, thanks for your help I appreciate it

[gritty sparrow]

Np

[last cargo]

Guys, in relation to abstract algebra 1, what is the book which has the hardest questions?

[dusty river]

Looking at the shit narwhal asks here, I would guess d&f

[gritty sparrow]

I can’t guarantee that these hvse the hardest questions, but if you want advanced books for abstract alg1 you can try lang’s algebra or Jacobson’s basic algebra

[cursive temple]

Jacobson has some pretty awful exercises as well but i doubt they are on the level of d&f

[last cargo]

I mean, I'm taking aa1 on next semester

And my prof is lunatic, he pass master-level questions

So I intend to start studying it now.

Who is d&f?

[gritty sparrow]

Dummit and foote

[last cargo]

I see, thanks, I hope they better than elon lages

[vapid narwhal]

Trying to understand conjugation as a change in perspective from an intuitive, layman, visual point of view. I'm working off the below example I've found of rotation and horizontal reflection.

I understand why HA^-1 has the reflection along that point, but should then the next A in AHA^-1 be a clockwise rotation after that horizontal flip? Aren't we applying successive transformations which we then have to respect? Is this some sort of notational abuse?

Thanks in advance!

[gritty sparrow]

No matter what happens, A is always anti-clockwise rotation

The previous transformations have no effect on what A is going to be, so doing H before A really should change nothing

A will still be anti clockwise rotation

[hot lake]

I really don't understand what those pictures are trying to communicate

[gritty sparrow]

My guess is that he is acting some group on the pentagon, A is anti clockwise rotation and H is reflection across the line drawn

But I could be wrong

[hot lake]

but it says H is a horizontal reflection and the line drawn is not horizontal

[gritty sparrow]

Oh, I didn’t notice that

[vapid narwhal]

@gritty sparrow Working off an example I've found, which presents conjugation as change in perspective. But the application seems arbitrary. So that's the question: why is H modified by A^-1, but A is not modified by H...

[gritty sparrow]

If H is supposed to be horizontal reflection, then this diagram is wrong, H should not be modified at all

[vapid narwhal]

@gritty sparrow Thanks for answering. Then defined transformations do not affect each other regardless of order.

But would this still hold true for matrices? If A and H were matrices for example, and the transformations were defined numerically by those matrices?

I've seen this as applied basis change in linear algebra, again with the whole change in perspective approach.

[hot lake]

I'm not sure what you're asking but if A and H were matrices you would compute the matrix of AH by multiplying the matrices of A and H

[gritty sparrow]

I don’t think I fully understand the question

[vapid narwhal]

@gritty sparrow
So, in this particular example, the transformations are unaffected by one another.

But would this still hold true if that octagon was defined by a set of vectors in a 2D plane, and if those transformations were matrices defined numerically?

Would the end result be the same as the above?

I'm trying to grasp conjugation as a change in perspective, and the linear algebra equivalent seems to be the change in basis.

[gritty sparrow]

Yes they would be as above

[vapid narwhal]

@gritty sparrow Ah, so the example is correct in that case. Then my initial question still stands, but for matrices and vectors.

Why is the final A an anticlockwise when the reflection tranform should change it to clockwise?

[vapid narwhal]

Thank you for your input, but since it was established that example is incorrect, it's no longer useful to refer to it. See my last two comments.

[hot lake]

what are you learning out of ?

[novel parrot]

what does it mean that B is an extension of C by A?

having that exact sequence means that B/im(A) = C no?

C is just B but all Im(A) hmm?

thats what that means?

[hidden haven]

That's pretty much the definition, the existence of such a short exact sequence makes it an extension

I'm guessing it's called an extension of C because it's like taking C, and in place of 0, you put the whole module A and the each element of C becomes a coset of A

[novel parrot]

i see

[novel parrot]

if two maps composed are identity

why must they be isomorphisms?

f : A -> B , g: B -> A, fg = gf= 1 for example

g must be fs inverse

and f is g's inverse

mmm

[hidden haven]

I don't get your question, if gf=1 and if fg=1 then f and g are inverses of each other

But you can't say fg = gf = 1 because they are identities of different objects

1_A and 1_B

[novel parrot]

ahhh

i actually mean

if they are both identities on their respected objects

why are they isomorphisms

[hidden haven]

What definition of isomorphism are you using

[novel parrot]

are they not isomorphism?

[hidden haven]

They are

[novel parrot]

injection and surjection definition

[hidden haven]

That's the definition of an isomorphism

Hmm I see

Well having a 2 sided inverse implies bijectivity

And similarly bijectivity implies having a set map inverse, which you can then easily prove is a homomorphism

Usually bijective homomorphism is not a good definition though because it mostly only works for isomorphisms between algebraic structures, so it's better to think of invertibility as the definition of an isomorphism

But yeah in the case of rings/modules they are equivalent

[novel parrot]

how?

[hidden haven]

Forget about the module structure

[novel parrot]

ok

[hidden haven]

This is pure set theory

[novel parrot]

ohh

ok

[hidden haven]

Whenever you have an invertible map between sets it has to be bijective

[novel parrot]

yep

[celest mantle]

Hi : An exercise asks me to prove that if A is an abelian group of order M², and if every subgroups A_m = { P \in A : mP = 0 } of points of order dividing m has order m² with m a divisor of M, then A is a direct product of two cyclic groups of order m. I don't think I really understand the question, I think it's a typo and the author just meant cyclic groups of order M, could anyone confirm ?

[mild laurel]

Are you asking if A should actually be a cyclic group of order M? @celest mantle

[celest mantle]

no i'm asking if A should actually be a direct product of two cyclic groups of order M instead of m

[mild laurel]

Oh yeah that seems like a typo

[celest mantle]

yup that's what i thought, strange that I didn't find this on the book correction web page, thanks

[wooden ember]

ideas for this one?

im sick of skipping questions from this chapter lmao

[urban acorn]

easy, it's a finite question so use a computer to enumerate all possible groups of order 168

[wooden ember]

reeeee

this seems like it could have a slick solution

but idk

im trying make it act as automorphisms of an elementary abelian group of order 8 but that hasnt worked

then tried to give it a normal subgroups and maybe reach a contradiction but that hasnt worked either

[urban acorn]

idk any sylow theory right now so i can't help you

[wooden ember]

aight

tried constructing counterexamples but that hasnt really worked out either since i havent found motivations for these potential counterexamples

[narrow marten]

I'm really new to module theory. Let I be a proper left-ideal of the free R-module R, then is I not itself free?

[mild laurel]

No, in fact it's interesting to prove that if I is always free then R must be a PID, at least when R is commutative

So you can take a ring that's not a PID, like Z[x] for example, and take an ideal that's not principal like (2,x), then this ideal is not free

@narrow marten

[narrow marten]

@mild laurel thx for the exempel! Another question: if we again have ideal like previously constructed but also finite, then is the basis for the R-module R finite?

[dusty river]

Does R not have 1?

[narrow marten]

@dusty river yes it does, sorry for late responsen

[warm holly]

If we have sub spaces $V_1,…,V_n$ and they are in direct sun, I know that $V_i \cap \sum_{i \neq j}^{n} V_j={0}$ for $i=1,…,n$. Is it necessarily true that $V_1 \cap V_2+V_3={0}$ for examples ?

[cloud walrus]

𝔻аniil

[hidden haven]

Well then the R-module R has finite basis, {1}

wait is that not true if R is not a domain

nvm it is because 1 can't be a zero divisor

[gritty sparrow]

If your bracketing is like $V1\cap(V2+V3)$ then yes, If the bracketing is like $(V1 \cap V2)+V3$ then no.

[cloud walrus]

saketh

[warm holly]

Yep I meant the first one, thank you very much

[hidden haven]

But I'm not sure what the question is are you asking if having a non principal ideal implies having a basis?

[hidden haven]

That seems weird because {1} is always a basis, since it spans (r = r • 1) and is linearly independent (if r • 1 = 0 then r = 0)

@narrow marten

Not immediately, have never worked with non unital rings

Unital rings should still count as a counterexample though, because even if you define rings to not necessarily have a 1, you still call it a ring if it does have a 1. Unless the problem specifies that the ring doesn't have an identity, I'd say that a unital ring is a valid counterexample

[gritty sparrow]

Sorry, but what exactly is the question, I don’t understand what was written above

[primal pier]

what is the most efficient subgroup-enumerating algorithm? is it brute force?

[lethal cipher]

I could use some help trying to figure out how to start 59. I am having a hard time figuring out how to relate the order of a group with the order of its elements.

Oh wait, it's because of inverses. I see it now

[chilly ocean]

explain it to me

[lethal cipher]

Okay sure. First off, if G is a group, the identity element e is in said group

Then if the order of elements is 2k, for some natural number (not zero), then there are 2k-1 elements in G that are not e

Okay, well, every element has a unique inverse. But...you can only match all but one elements with a distinct inverse.

So there is some a so that a=a^{-1}

[lethal cipher]

Oh, to cap it all off, this means that aa^{-1}=a^2=e

[chilly ocean]

hm, can you elaborate on "you can only match all but one elements with a distinct inverse."? you're definitely on the right track, thinking about inverses, but i just wanna know how you got a = a^-1 here

at least a little more explicitly

maybe im just misreading

[lethal cipher]

Well, sure. Let's assume that for every b in the group (not the identity of course), that b≠b^{-1}

If this were true, then b,b^{-1} takes up even number of elements.

And if we do this for all elements, we should have an even number of elements....but we have 2k-1 elements that are not e itself, which is odd.

That is where the contradiction is.

Yea, pretty much slimvesus

[chilly ocean]

nice

[lethal cipher]

Yep. Although it seems trivial, we haven't 'proven' that yet. So this kinda also shows that as well.

Here's what I did. Thanks for the moral support TTerra 😁

[south patrol]

and this generalises to all prime divisors which is cool lol

i wonder how many other similar theorems work for all primes but are particularly cute for 2... the fact a subgroup of lowest prime index is normal comes to mind

[lethal cipher]

Really? How'd we go about it for 3?

[south patrol]

the general theorem is Cauchy's theorem, but the idea is similar

[latent anvil]

*of index the lowest prime dividing the order of the group

[south patrol]

yes, my bad sorry

[lethal cipher]

Cauchy's got a million theorems. Got to be a bit more specific :p

[latent anvil]

It's called cauchy's theorem

[lethal cipher]

Cauchy's theorem is also a theorem in complex analysis

[latent anvil]

Have you seen group actions dackid?

[south patrol]

Cauchy's theorem for groups should give you the right thing in any case

https://en.wikipedia.org/wiki/Cauchy's_theorem_(group_theory)

[latent anvil]

The proof I remember uses them

[south patrol]

same here.

[lethal cipher]

True. Thank you for catching that

[lethal cipher]

No, not yet.

[chilly ocean]

just wanted to check you actually got it

[south patrol]

Especially when it's the type of contradiction where you assume not p to prove p. constructive logic ftw

[lethal cipher]

I wasn't thinking about partitions here, but I do like that phrasing quite a bit. I'll keep it in mind for the future.

2k-1, but I'm following.

I'm with you.

[hidden haven]

If I understood it correctly: if R has a finite proper left ideal, then R is not free as an R-module

@narrow marten I hope this is correct

[hidden haven]

And this is what I think about the unital ring thing

[gritty sparrow]

Hmm, what a weird question. That seems like it would never be true

[hidden haven]

I could see it not being true if your ring doesn't have 1

Like 2Z is not free over 2Z

I think

[mild laurel]

That's so cursed

[hidden haven]

Ye lmao

But don't know enough rng theory to even try coming up with counterexamples for this

[gritty sparrow]

Is 2 not a basis? But I don’t doubt there is some weird example, rings without unity didn’t even cross my mind.

[hidden haven]

Not spanning lmao

[gritty sparrow]

Ah, I see

[wooden ember]

ended up producing $Z_3 \cross \left(E_8\rtimes Z_7\right)$ where the homomorphism definining the semidirect product sends a generator of $Z_7$ to a generator of any sylow 7-subgroup of $GL_3(\mathbb{F}_2)$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

this isnt simple and has more than one sylow 7 subgroup

and im pretty sure along with $Z_3 \rtimes (E_8\rtimes Z_7)$ they're the only non simple groups of order 168 with more than one sylow 7 subgroup

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

so that's additional info gained

hmm actually the later might have 3 forms im not sure

not bothered to verify frankly

[hidden haven]

So elegant

[chilly ocean]

[robust patrol]

I mean, I don't blame him, I can't either lol

[chilly ocean]

what is there to solve

#groups-rings-fields

[chilly ocean]

although maybe you're right slim

i've taken 3 differential geometry classes and i still don't know what there is to solve

so vacuously i cannot solve it

[maiden ocean]

@chilly ocean Solve it

[chilly ocean]

no

[maiden ocean]

I kinda regret making this joke because i am being reminded how gross this stuff was

i understand it slightly better now though

[wooden ember]

that looks atrocious

[simple mulch]

looks good?

[south patrol]

The inductive step doesn't quite fit, you need to show a_1w_1 + ... + a_{n+1} w_{n+1} is in W

in fact i believe your inductive step just shows that the case with n = 2 is correct

[simple mulch]

but then inductively doesn't it work?

[south patrol]

no, because all you've shown is that if you add any two vectors in W you get another vector in W

[simple mulch]

So its like I need to do induction for every case

[south patrol]

Your base case is fine (although I'd be a bit more clear with saying for any $a_1$) but your inductive step should start with smth along the lines of assume for for some $k < n$, $a_1,\dots,a_k \in \mathbb K$, $a_1 w_1 + \dots + a_{k} w_{k} \in W$

[simple mulch]

I see

[south patrol]

or perhaps k instead of n here since n has already been used, but yeah

[simple mulch]

this is weird, how do I know a_{n+1} is in F

and w_{n+1}

in W

o.O

[south patrol]

yeah, typo, use k

i'll change it

[simple mulch]

oh yeah I see

[cloud walrus]

potato

[prisma thunder]

Hey I had a quick question

I’m attempting to classify all groups of order 30, so I am trying to verify which groups are normal and which ones are not. I find N, which is of order 15, is isomorphic to Z_(15), and, by Sylow’s Theorem, P_(2) is my subgroup of order 2, so that, after some playing, I find that we have their semi-direct product Z_(15) \rtimes Z_(2). However, I’m trying to show by Sylow 2-subgroup P_2, which is isomorphic to Z_2, cannot be normal in G. I figured I could something like the class equation and end up with some undercounting issues but I’m not sure how to. Any ideas?

[tropic spade]

For a ring A and ax in A[x], then if n is an integer is there some rule or justification that tells me n(ax)=(na)x?

I feel like this should be true but I can't seem to find anything in particular in the algebra books I've poked around in that tells me so.

[small bison]

If you accept that ax + ax = (2a)x then it basically follows from induction

[tropic spade]

I see why ax+ax=2(ax) but not the other way.

[small bison]

I think it should just be the addition operation on A[x]

ax + ax = (a + a)x

[tropic spade]

I see.

Well wait maybe I don't.

So I guess the thing that bothers me is mixing up the unity from A with x^0 (or whatever we want to call the unity of A[x])?

So is this just a quirk of how people define x^0?

As in, we take x^0=1 where 1 is the unity from A?

[small bison]

usually x^0 doesn't need to be defined

i guess it depends on your definition of A[x]

like formally elements of A[x] are of the form a_0 + a_1x + ... + a_nx^n

[tropic spade]

I see what you mean now. I thought about in terms of sums of sequences and it clicked.

[fossil shuttle]

if A is an abelian group, can you think of any natural abelian group structure on the powerset of A?

[kind temple]

like cosets? or does it have to be defined on the whole power set?

[tropic spade]

There's the group (P(A),+) where B+C=(B-C)U(C-B) I think. But that doesn't seem like what you want.

[fossil shuttle]

yeah i guess i want the singleton map A -> PA to be a homomorphism

[gritty sparrow]

I don’t think so, bc then |A| would divide |P(A)|

Bc that map is injective so A will be a subgroup

@fossil shuttle

[fossil shuttle]

😠

bummer

[next obsidian]

Do you mean which groups are abelian / simple?

Normality isn’t a property inherent to a group, it’s a property of a subgroup

Anyway, any group of order pqr is solvable so none of them will be simple, this is… not too bad to prove but isn’t a simple element counting argument

[shell brook]

U guys know the statement thats in every first algebra assignment in the world

like for a in G a^2 = e implies G abelian

is there a cool way to solve this

like I know the symbol bashing way

but i mean. cool 😎

[kind temple]

is the symbol bash way just,
(ab)(ba) = a(bb)a = aea = aa = e = (ab)(ab)
so
ba = ab
?

[next obsidian]

Maybe a succinct way is to note that [a,b] = e for all a,b

But this is not really much better

[shell brook]

i think

or doing (ab)(ab)^-1

or contradiction

all symbol spam

[sturdy marsh]

that is not symbol spam

[robust patrol]

Each individual equation with variables

[scarlet estuary]

is there a not stupid proof of A_n being simple for n > 4

by "not stupid" i mean

• no unnecessary heavy machinery
• no dumb computational jamming (commutators, sylow, fixed point arguments...)

[robust patrol]

There as some cool proofs in books

[scarlet estuary]

what a helpful answer

[wooden ember]

im not sure there is, I was having a similar discussion with someone the other day

there are quite a few different proofs but I think most of them have some form of dumb computational jamming

what are the proofs you know of that you dont like?

[next obsidian]

The only ones I really know involve diving into how conjugacy classes split in A_n

[wooden ember]

there are a few based on the fact that A_n is generated by 3-cycles

i also know the one from dummit and foote which is kind of nice but i know nami wont like it either

I honestly dont know any slick proofs

wait would this be a proof actually

wait nvm

i think I figured out a different proof for when n is prime

no nvm it relies on existence

ill stop now lol

[wooden ember]

been working on a problem for 1h only to notice now i messed up right at the start

on something as stupid as permissible sylow numbers

[chilly ocean]

Can anyone help me with this, for an A-module M define the supp(M) to be the intersection of all f in A such that the localistion of M at f is zero,

You can show that if we have an exact sequence D -> E-> F then supp(E)\subset supp(D) union supp(F)

I think the following should also be true but i cannot show it

Let A->B->C->D and call the denote the map B to C by g

I want to make some kind of statement bout the kernel and cokernel of g having support contained in the support of A and D

the following would be enough

If the support of A and D is contained in some set K, then the support of the kernel and cokernel of g are also contained in K

[hidden haven]

The definition of support on wikipedia seems to be different it says it's the set of all prime ideals p such that localizing at p doesn't make the module 0

[chilly ocean]

yeah this is slightly different definition

[hidden haven]

What do you mean by intersection of all f, f are elements of the ring?

[chilly ocean]

sorry

my modules are modules over C[x1,..,xn]

not just an arbitrary A

[hidden haven]

Ok so supp(M) = {f ∈ C[x1,...,xn] | M_f = 0}?

Are you sure it's M_f = 0 and not ≠ 0? because support usually always has a non zero condition

[chilly ocean]

opps yes sorry you are right

[hidden haven]

Then if A → B → C → D is exact and B → C is called g, ker g should be a quotient of A and coker g should be contained in D. I think you should then be able to prove that if M ⊂ N are modules, then supp(M), supp(N/M) ⊂ supp(N)

So just establishing a relationship between the support of the module with the supports of its submodules and quotients (I might have written inclusions the wrong way, just check while you prove)

[chilly ocean]

why should the kernel kernel be a quoitent of A again?

oh

its ker(g)=im(A to B)

[hidden haven]

Yep

[chilly ocean]

and then this is a quoitent because

im(A to B) is A/ker(A to B)

[hidden haven]

Yep

[chilly ocean]

ty

damn it doesn't look like supp(M), supp(N/M) ⊂ supp(N) is true

certainly supp(N) ⊂ supp(M) union supp(N/M)

oh no nevermind

[hidden haven]

Yeah it should be true from what you said earlier about D → E → F

But the proof I had in mind for that sequence used this fact and the fact that localisation is exact

[chilly ocean]

using that localization is exact only gives that for M ⊂ N, supp(M) ⊂ supp(N)

im not actually sure that its true that supp(N/M) ⊂ supp(N) ?

but this is weird because N/M should be a submodule of N right?

[next obsidian]

No

If it is then you have a very special thing going on

Namely M is a direct summand of N

[hidden haven]

You can get a short exact sequence from that exact sequence as
0 → ker f → E → im f → 0
where f is the map E → F.
Now apply the localisation, localisation of E is 0 iff the other 2 are 0

And then use what I said about submodules and quotients

Because ker f is a quotient of D, im f is a submodule of F

[next obsidian]

If you’re dealing with the usual notion of support in terms of primes, given N < M you have

Supp(M) < Supp(N) U Supp(M/N), maybe you can adapt a proof of this to work with your different notion of support

[hidden haven]

Yeah same works here

[chilly ocean]

so the support of a submodule is indeed contained in the support of the module

but we cannot say that for a quoitent

[hidden haven]

If M ⊂ N, N_f = 0 → N_f/M_f = 0 → (N/M)_f = 0, so localisation of quotient is non zero implies localisation of N is non zero, so supp(N/M) ⊂ supp(N)

Did I make a mistake somewhere

[hidden haven]

And shouldn't this be an equality because localisation at p is an exact functor and so applying it to 0 → N → M → M/N → 0, you get a ses, and the middle term in a ses should be non zero iff at least one of the other 2 is?

[next obsidian]

No, it isn’t an equality in general

[hidden haven]

Wait so is localisation at p not exact?

I think the rest is fine

[next obsidian]

It is

Let me think of an example

Oh wait

Support Hurb

Yes

I was thinking about associated primes

[hidden haven]

Ah

[next obsidian]

You can get really weird stuff because of embedded primes there

[hidden haven]

Chmister commutative algebra 😌

[next obsidian]

But yeah for support I think it’s equality cuz of exactness

[chilly ocean]

sorry yes you are right

[steady axle]

I want to prove that if I have a countable group G such that for any finite subset of G, the subset is contained in an infinite cyclic subgroup then G is subgroup of Q. I want to check if the map I defined works. I map 0 to 0. and any element a_1 to 1. Then for any other element say a_2 I find unique positive integer n and integer m such that n a_2=m a_1 and n is minimal. I map a_2 to m/n.

[dusty river]

Yeah it should work

[prisma thunder]

All of them—abelian and non-abelian

[next obsidian]

Right so which class were you trying to look at first because you originally were asking which ones were normal

[prisma thunder]

I think they’re really tedious arguments imo—least favorite part of group theory imo

[next obsidian]

And without reference to a supergroup it doesn’t make sense to talk about normal subgroups

[prisma thunder]

Oh right, I was trying to determine which subgroups are normal in G, which is of order 30. By Sylow stuff, I find my single Sylow 3- and 5- subgroups are normal in G (including their inner direct product). What I was wondering if my Sylow 2-subgroup(s) are normal in G.

[next obsidian]

you can’t say in general

If G is abelian everything is normal

And… hmm

[prisma thunder]

That’s what I figured

:/ big sad

[next obsidian]

So just by number stuff

You can show Sylow 3 and 5 are normal?

[prisma thunder]

Yes

[next obsidian]

Thinky time

Hmmmm

Why can’t you have 10 Sylow 3s?

[prisma thunder]

Okay

Let’s go

Applying Sylow’s Theorems to both 3- and 5-, how many possible Sylow 3- and 5- subgroups can you get?

[next obsidian]

1,10 and 1,6

So you know one of them is definitely normal

[prisma thunder]

Exactly

[next obsidian]

And like you can do some work to always show 5 is normal

This much I do know

[prisma thunder]

Yes

[next obsidian]

Then how do you deal with 3?

[prisma thunder]

We apply the following: H be a subgroup of G and P be a Sylow p-subgroup of H. If P is normal in H and H is normal in G, then P is normal G.

Usually, normality isn’t transitive but this is one of those few instances we have that.

[next obsidian]

Yup I do believe this

It’s because P is characteristic

Woohoo

[prisma thunder]

Exactly

[next obsidian]

But you can’t shove the Sylow 3 into the Sylow 5

[prisma thunder]

You cannot

But you can shove it into something else

Lol

[next obsidian]

Also is this for a class because just from Sylow 5 normal you can classify them all

If you’re willing to use Schur Zassenhaus lol

[prisma thunder]

You know at least one of Sylow 3 and Sylow 5 are normal in G from our earlier arguments.

[next obsidian]

Yee

[prisma thunder]

Can you construct a group from Sylow 3 and Sylow 5?

[next obsidian]

I mean ideally you’d want to just bash the two together

But in general that isn’t a subgroup unless both are normal

Unless I’m being dumb and missing something

[prisma thunder]

You need normality in at least one of them to get it to be a subgroup—I think normality follows from both being normal but I’m not sure.

Otherwise, if neither are normal, I think it’s just a…set?

[next obsidian]

Wait lol right

You only need one to be contained in the normalizer of the other I think

This is like some Second Iso crap I think

[prisma thunder]

Lmao

Alright let’s move on

[next obsidian]

Okay right lol

Oh okay

I see right

Sylow 3 is normal in the order 15 subgroup

And you win

[prisma thunder]

Yup

You get both are normal in G

[next obsidian]

Okay I see I see

That is neat

[prisma thunder]

Yeah

I’m literally cataloging a series of theorems to simplify my semi-direct products as much as possible lol because brute-forcing it hasn’t been helpful

[next obsidian]

Right so

#groups-rings-fields message

That’s where this becomes relevant

[prisma thunder]

👀

[next obsidian]

Because if you assume this thm you necessarily classify them all using semi direct products

Schur Zassenhaus says if N is a normal subgroup of order coprime to its index (so |G/N|)

[prisma thunder]

Oh shit what the fuck

[next obsidian]

Then G = G/N semidirect N

And so it suffices to classify groups of order 6

Of which there are two

[prisma thunder]

Yes

[next obsidian]

Z/6Z

And then the weird semidirect

Now this doesn’t mean there’s just 2 options

Cuz like

[next obsidian]

This bit here

There’s multiple ways to make that

But it does give you the complete list

You just have to like… do cases I guess

When G/N is Z/6Z you can probably just count them

My gut says there’s two semidirect products of Z/6Z and Z/5Z

Because of coprimality

The direct prodict and then one more weird one

[prisma thunder]

I just talked to my professor

They said no we are not allowed to use it

[next obsidian]

Schur Zassenhaus needs group cohomology (okay it doesn’t need it because my TA did make a proof w/o it but it’s like always proved with it)

So that’s why I asked if it’s for a class haha

If it was it’s totally off the table

But wait like

No wait hold up this is dumb you don’t need Schur Zassenhaus

You’re already done

You have an order 15 subgroup right?

By just bashing your two normal subgroups

Like Sylow 3 and 5

Gives you a Z/15Z

Now you have G = Z/15Z semidirect Z/2Z

By just bashing your Z/15Z and your Sylow 2

One of them is normal so it’s an internal semidirect product

And this is way easier to deal with anyway because to classify these you only have to deal with the automorphism group of Z/15Z which becomes Aut(Z/3Z) x Aut(Z/5Z) = Z/2Z x Z/4Z

[next obsidian]

Also @prisma thunder regarding normal Sylow-2 I realized there’s a really simple counterexample, take the dihedral group of order 30. Then fr^k is order 2 for all k so there’s lots and lots of order 2 elements

[unique berry]

Can someone explain why you need to impose a riemannian metric on the upper half space to get a visualization of the group action of SL_2(R)\

[stone fulcrum]

SL2(R) is a group, not an action. What's the action we're seeing on SL2(R)?

I might not have the knowledge to answer this, haha

Is this an isometry?

[mild laurel]

I feel like you don't need a Riemannian metric, I mean you can just do it topologically if you care about a visualization

[mild laurel]

I mean, this action is super important for modular forms, but you could say that's a consequence of the Poincare metric too

I guess the way that it's motivated for modular forms is that {1,z} and {1,z'} generate the same lattice if and only if one is a Mobius transform of the other

[unique berry]

The action $z\to \dfrac{az+b}{cz+d}$

[cloud walrus]

KingArthur

[mild laurel]

does what we said clear up what you were asking or no @unique berry

[digital yoke]

hm I'm having a bit of trouble

how do I prove that p does not have an inverse in Fp^2 (or Z mod p^2Z) where p is prime

it's certainly true if you think of some examples, but I just can't think of a way to rigorously prove it for some reason

[hidden haven]

A zero divisor can never have an inverse

[next obsidian]

Just show it’s a zero divisor

Sniped

[hidden haven]

[digital yoke]

zero divisor?

[hidden haven]

a is a zero divisor if there is some b such that ab = 0

Non zero b

[digital yoke]

and if it's a zero divisor

it does not have an inverse?

[hidden haven]

Yeah try proving that

Assuming 0≠1 in the ring

[digital yoke]

yea that's what I'm going for

[next obsidian]

Actually

Can an element only have a left inverse in a ring?

I forget if for groups you require a left and right inverse then use associativity to show they’re the same or if one-side gets you the other. Maybe it’s something like if everything has a left inverse everything gets a right inverse?

[digital yoke]

oh wow

I did it

wtf this is

so non-obvious but

so simple

thank you

[hidden haven]

Np

:catKing:

[digital yoke]

wow

this is interesting

I never would have connected zero divisor with having no inverse

[hidden haven]

Yeah it's not obvious but multiplication with a unit u should be an injective function

But this is not true if you have a zero divisor

This is in fact true iff u is not a zero divisor

But there are things which are neither units not zero divisors so not a perfect test

[hidden haven]

I remember seeing this: "if an element has 2 left inverses, it has infinitely many"

Which is only somewhat related but interesting

And it feels like you could use this to cook up some counterexample

[next obsidian]

Hmm yeah

[hidden haven]

Because if everything has a left inverse, then all elements that are left inverses will be right inverses, so you want to show that not everything is a left inverse and that seems very much like a non surjectivity/repeated right inverses condition (even though they're not equivalent)

[digital yoke]

oh is there? can I have some examples?

[next obsidian]

2 in Z

[next obsidian]

I remember doing stuff like this, I forget if you had actual finiteness or like semi-finiteness conditions

[digital yoke]

a

right

lol

[next obsidian]

Tfw you helped someone with French math hw and it had u deal with “associative magmas”

[inner acorn]

The more I read Presentations of groups by Johnson, the more I like it

[inner acorn]

praise Bourbaki

[frank fiber]

let A be a ring with exactly one prime ideal, say I, then the nilradical is I

[next obsidian]

Yes

[frank fiber]

then every element of A which is not in I is unit?

[next obsidian]

Yes

[frank fiber]

why?

[next obsidian]

Because if (x) isn’t all of A then it’s contained in a maximal ideal

Aka I

But then x in I

This is just true for any local ring aka one with 1 maximal ideal

I guess to finish it up, so if x isn’t in I rhen (x) = A but that’s true iff x is a unit

[chilly ocean]

Noo why delete the question i wanted to think about it

[runic hemlock]

Let A be a f.g. algebra over a field k. Let K be its field of fractions. Let A' be the integral closure of A in K. Then A' is f.g. over A as a module.

I know this is true, but haven't been able to prove it myself or find a proof not behind a paywall. Could anybody help?

i copied it just in case

It's not really an exercise, more like a Theorem.

Too hard to be an exercise

I managed to find a proof online, if anyone's interested

[next obsidian]

If you want to meme really really hard this is because fields are Nagata

If you go onto Stacks Project and look up like “Japanese rings” you’ll find info on it

[runic hemlock]

lol thanks

[unreal portal]

Is the zero set different of a set of polynomials different from the set of points that are zero on each polynomial? I only ask because I've only ever seen this notated as V(S) to get a variety from a set of polynomials

[next obsidian]

What

Is the zero set different of a set of polynomials different from the set of points that are zero on each polynomial?

I dunno what you’re trying to ask here

[golden pasture]

id assume you mean

[\left{x|\forall p\in S\quad p(x)=0\right}]vs[\left{x|\exists p\in S\quad p(x)=0\right}]

[cloud walrus]

ari 亲

[golden pasture]

we rarely(if ever) use the second one

[chilly ocean]

yo would anyone here be interested in commutative algebra study group?

[potent briar]

can it happen that a non-trivial subspace has a unique complement?

my intuition is no but maybe with finite field idk?

[lethal cipher]

I am having a little trouble on part (ii) with the first direction.

Luckily, the other direction is extremely simple.

[chilly ocean]

have you tried contradiction?

[lethal cipher]

The issue I was having is I wasn't able to prove hk was not in H U K. There is no guarantee, but there is also no guarantee it's not.

I have an idea. Let me try something real quick.

[mild laurel]

I don't think so. If U is a complement of V, then U + v should also be a complement of V for any v in V

[potent briar]

yeah but with a field like Zmod2

would you only have one vector outside a subspace of dim 1 less than the space's dim?

[chilly ocean]

think about multiplicative inverse.

[mild laurel]

No you have 2

[potent briar]

U + 0 = U no?

and, v and -v span the same subspace

[lethal cipher]

Still not sure how this helps

[chilly ocean]

so uhh we want to get a contradiction

so take h \in H\K an k \in K\H. Take a look at hk

[lethal cipher]

Well, this just tells us closure is not guaranteed

[somber marsh]

what does the ; mean

[chilly ocean]

hmm?

we know hk is in H \cup K

[lethal cipher]

Ah, we are keeping that hypothesis. Yes.

[somber marsh]

what is set1 ; set2 mean

[lethal cipher]

Okay sure, hk is in the union. I am in agreement with that

[chilly ocean]

well, then hk has to be in H or K

[lethal cipher]

Agreed

[chilly ocean]

so let's say it's in H

hk \in H

this doesn't help

[lethal cipher]

Oh, disregard then

[chilly ocean]

but h^-1 is in H

[lethal cipher]

Ah, so h^{-1}(hk) is in H

[chilly ocean]

ye

and we chose k to be in K\H

[potent briar]

what about this @mild laurel, "Show that if M and N are three-dimensional subspaces of a five-dimensional space, then they are not disjoint"

[runic hemlock]

It's called "singular cohomology". Google it

[somber marsh]

ty

[lethal cipher]

Bam. Thank you very much Ledog

[runic hemlock]

This is wrong.

You assumed hk\in H.

Which may not always be true.

[lethal cipher]

I did WLOG, so the same argument holds if I say hk is in K

Instead it'd just say H is a subset of K

[runic hemlock]

Right, but it could happen that for some pairs the product is in H, and for some the product is in K.

[lethal cipher]

Sure, then "consider the case when..."

Is an easy fix

[runic hemlock]

What you proved is the following: For each h,k, either k\in H or h\in K. Do you agree?

[lethal cipher]

There. Fixed

[somber marsh]

i meant like what the ; means

not the H*

[runic hemlock]

While it is true that you can assume hk\in H, you cannot assume that hk\in H for ALL k. Only for a specific one.

So k\in H holds only for a specific k, not for all of them

Do you see the problem?

[runic hemlock]

The semicolon is used to differentiate between the topological space on the left and the coefficient group on the right

It has no special meaning

[lethal cipher]

Fine...then how do you propose I fix it

[runic hemlock]

If H is not contained in K, and K is not contained in H, then there are elements k\in K and h\in H, so that k is not in H and h is not in K. Now consider the product of those two specific elements

[scarlet estuary]

"oh #groups-rings-fields , please tell me what a 2-argument function is"

[untold cloud]

Hi, guys, in this solution, why it says Euclidean function here? I didn't see any relation here....

[still stream]

When you say Euclidian domain you have to define Euclidian function

[untold cloud]

Ok...But why do we need Euclidean domain?

[mild laurel]

It helps us classify the units

[untold cloud]

Sorry, could you be a little more specific.....I didn't see how it helps us classify the units. I mean if without saying the Euclidean domain, we can still say there exists y s.t. xy = 1 iff |x||y|=1.

[hidden haven]

How are you defining |x| in that case?

[untold cloud]

modulus?

[hidden haven]

You won't have that in an arbitrary ring

But even when you have a subring of C, the modulus need not give you a Euclidean domain

Eg Z[1+√-5] I think

And then you can still say that being a unit → modulus is 1

But modulus is 1 won't necessarily imply unit

[untold cloud]

so they might be i or something other than 1 when x and y multiply together?

[hidden haven]

No, if there is an inverse, then the modulus has to be invertible in N right? And the only invertible element in N is 1

Or ok not in N since you're not restricting to EDs

But here the modulus does go to N

So unit → modulus is 1

Converse isn't true in all subrings of C

And if you want to classify all units

You want an if and only if

Otherwise you're just eliminating some possibilities instead of giving a complete characterisation

[untold cloud]

sorry, what do you mean N here..

[hidden haven]

Natural numbers

[untold cloud]

why the modulus x has to be inverse if x is inverse?

[hidden haven]

This is assuming 2 things: |1| = 1, and |xy| = |x||y| for all x and y. Do you see it?

[untold cloud]

oh, I see...

So if it is not ED, then |x| is invertible does not imply x is invertible?

[hidden haven]

Yeah

[untold cloud]

Thank you!

[hidden haven]

[novel parrot]

So B is noetherian as an A-module

but allowed multiplication from B still lets it be noetherian?

[hidden haven]

All ideals of B are B-submodules, hence A-submodules, hence finitely generated over A, hence finitely generated over B by the same generators

By increasing the set of scalars you're allowed to multiply by, you are only increasing the span

[novel parrot]

got it

👍

[burnt pond]

Hi, let $A\subset B$ be an unramified extension of henselian DVRs (i.e. an uniformizer $\pi$ of A is also an uniformizer of B) and $k_A\subset k_B$ the corresponding
residue field extension, which we assume to be finite separable. Is $\operatorname{Hom}A(B,B)\cong \operatorname{Hom}{k_A}(k_B,k_B)$ via reduction mod $\pi$?
I know its true for local fields, i.e. when A and B are complete DVRs, but cant remember if completeness of A and B took any part in the proof. I think all we needed was henselian property? Can anyone confirm? Thanks a lot!

[cloud walrus]

mrvns

[novel parrot]

dont get that part

why does h have to be irreducible

[hidden haven]

h doesn't, but some factor of h is irreducible

Since k[x_1,...,x_n] is a UFD

And that factor cannot be the same as any g_j

[novel parrot]

right

but are we saying that h lies in F, the k-algebra

[hidden haven]

No, h is in k[x1,...,xr]

And in this ring it is coprime to all the g_j

So h^-1 in F, which is 1/h, cannot be obtained as a polynomial expression of the y_j's because the denominator of any such element will be a product of powers of the g_j

@novel parrot

[novel parrot]

are we trying to contradict that F is a field

[hidden haven]

Yeah

Or that the y_j cannot be the generators

[novel parrot]

as the k algebra

?

[hidden haven]

Yes

[novel parrot]

ok i think that makes sense

[novel parrot]

why is it written (n_i - 1) + 1?

[dusty river]

it is (sum (n_i - 1)) + 1

[novel parrot]

oh

would it not have worked with just summing n_i

[dusty river]

Yeah it would

this is just the smallest number for which this argument works

[novel parrot]

right

got it

[chilly ocean]

Wtf that is a cheeky little trick with the whole "if each of the n_i were even 1 greater than corresponding r_i then we break this constant I made up"

That's the type of shit they do in graph theory all the time that my tiny brain can't handle lmao

[hazy haven]

hello

I need help

[topaz leaf]

[barren sierra]

I'm struggling to show that this function is surjective

[dusty river]

instead try showing that there is a 2 sided inverse homomorphism

[barren sierra]

So show that some inverse is a homomorphism?

oh yea that's a thing

[dusty river]

well an inverse map, if it exists, is automatically a homomorphism

[barren sierra]

yea

[dusty river]

so you need to show that an inverse set map exists but in this case it will be obvious that it is a homomorphism too

[topaz leaf]

once you've done this, you can quickly see how to show the map is surjective in the "traditional" way too, if you wanted to

[barren sierra]

hm ok

[hazy haven]

I am stuck at this problem, "Let G be a finitely generated group, H a subgroup of finite index. Show that H is finitely generated", I mostly tried to look at examples of finitely generated groups to get a feeling of why this is true but it hasn't worked :(, I mean, I haven't (and never expected too) find a counterexample but I'm still unable to see the connection, of course the hypothesis that H has finite index is essential but I can't make a direct connection between the result and that, any tips or anything?

[topaz leaf]

i'm not sure, but have you had a look at previous results on subgroups of finite index? the problem you want to prove might follow from a previously proven result

[mild laurel]

Think about cosets of H and representatives of those cosets

[hazy haven]

well, apart from the meaning of H having a finite index, most previous results related to subgroups of finite index are for finite groups, there was a previous exercise that also applied to infinite groups which was Poincare Theorem, but I don't see how I could use that

[hazy haven]

I will think about that

[barren sierra]

oh we don't need to show that the inverse is also a homomorphism?

[dusty river]

So the definition of an isomorphism is homomorphism which has an inverse homomorphism

but for groups, any inverse function of a homomorphism is automatically a homomorphism

which isn't hard to prove

so yeah here just giving an inverse set map is enough

[south patrol]

and here if you've proven the map is a homo then the inverse function is automatically a homo by using g^-1 instead of g lol

[novel parrot]

dont get this

so sigma has a max

but i cant show that every max is prime

if was x,y was not in a and xy was in a, we wanna contradiction the fact that xy was not in a yeah?

[dusty river]

Yeah

"there exist x,y not in a but xy in a" is equivalent to "there exists x not in a such that (a:x) properly contains a"

They are just using this second statement

and the contradiction is that this allows you to write a as a sum of finitely generated ideals so it is finitely generated

[novel parrot]

ok

[upper pivot]

What Zoph said it’s probably what you wanna do. I wanted to mention a quick solution if you knew AT. Your group G is the fund group of some 2 dim CW complex with finitely many 1 cells, and a subgroup of index n corresponds to a n sheeted covering of this space, which obviously also has finitely many 1 cells and hence is finitely generated.

[hidden haven]

That's really neat

[barren sierra]

ngl I don't know what this question is asking

[scarlet estuary]

what are the possible groups G can be?

i.e. what groups can be the image of a homomorphism from S_3?

(up to iso)

[barren sierra]

||trivial group, alternating, and the whole subgroup?||

[chilly ocean]

up to 1st iso

[barren sierra]

wut

we haven't learned isomorphism theorems

I should probably google them I hear about them alot

[chilly ocean]

ok I thought youre supposed to use them

[scarlet estuary]

??

[chilly ocean]

some S_3/ker(phi) maybe I thought is doable

[scarlet estuary]

anyway yes spamakin thats correct

they have to be subgroups

hence youre stuck with A_1 and A_2, which are trivial, A_3, and S_3 itself

(you can verify very quickly that there do indeed exist homomorphisms from S_3 to these groups)

this is a corollary of the first isomorphism theorem

but you dont need it to prove this

[chilly ocean]

S2 is also a subgroup but need to be normal right

[scarlet estuary]

right

[barren sierra]

when you say A_1 and A_2 are trivial you mean that they're the trivial group since the identity is 0 transpositions?

[scarlet estuary]

yes

A_1 = A_2 = {e}

[barren sierra]

aight cool so I do understand this

neat

and the group used as the codomain of phi can't be anything with more elements than S_3 cause then phi may not be a homomorphism?

[scarlet estuary]

well to clarify

the group used as the codomain of phi can have more elements

but they wont be in the image of phi

since theres "not enough" elements of S_3 to map into them all

for example, we know theres a group homomorphism from Z_2 into itself, but we could also regard this as a homomorphism Z_2 into any group that has Z_2 as a subgroup, but the homomorphism is just defined to only map into the Z_2 part

so the codomain is the larger group but the image is just Z_2

[jagged field]

I am working on some problems and I have no idea how to do this one

can someone please help

[next obsidian]

This isn’t abstract algebra

[jagged field]

what channel do you think i should put it in?

[next obsidian]

#proofs-and-logic

[barren sierra]

wait then what is phi?

I'm confused

cause it says that phi maps from S_3 to a group G

but then it says "list possibilities of phi(S_3)"

[scarlet estuary]

consider the map $\varphi: S_3 \to S_4$ given by mapping each element to itself

[cloud walrus]

Namington

[scarlet estuary]

the codomain of this map is $S_4$

[cloud walrus]

Namington

[scarlet estuary]

but the image of $\varphi$ is only the $S_3$ subgroup

[cloud walrus]

Namington

[scarlet estuary]

@barren sierra

[barren sierra]

hm

got it

so the question is really asking about images

not codomains

[scarlet estuary]

yes

[barren sierra]

got it

[scarlet estuary]

there are infinitely many options for the codomain

[barren sierra]

right

so then the problem is trivial

[next marsh]

hey! im taking a group theory class and were doing field extensions and im a bit lost, if anyone is around who could explain it?

[mild laurel]

[next marsh]

fkjahsdf lol im just kinda lost, rn trying to find the degree of an extenion

[mild laurel]

what extension?

[next marsh]

im kinda confused on how to approach it

[terse crystal]

4

[mild laurel]

Do you understand the definition of the degree of an extension?

[next marsh]

lol i think thats where im getting stuck

i dont quite understand the definitions and so working from there is a mess

[mild laurel]

It might help to go back and review that and clear up your confusions about that first

It's hard to do a problem containing a definition that you don't really understand yeah

[next marsh]

any good resources for trying to understand this more?

[mild laurel]

Whatever your class uses is probably the best thing to do

[next marsh]

ah ok!

[mild laurel]

Dummit and Foote is a pretty standard algebra book that has Galois theory too

[next marsh]

ooo nice! i think my multi teacher recommended that for some fun reading (ive been using the art of problem solving textbook so far)

[mild laurel]

Oh what

I didn't know aops had Galois theory

[next marsh]

they have their group theory class! doesnt seem to get taught that often but its been really interesting

[mild laurel]

Oh interesting, they've grown a lot since I took their classes way back when

anyways, I can also talk a little bit about the degree of the extension too if you think that would be helpful

[next marsh]

very very possible! although group theory seems to be the last class they go to, which is kinda unfortunate for trying to do more abstract algebra things

yea! that would be really helpful

[latent anvil]

seconding d&f, I think their field stuff section is really good

[mild laurel]

yeah so, if F is contained in L and both of these are fields, do you see how you can consider L as an F-vector space?

[next marsh]

yea that makes sense

[mild laurel]

The degree of an extension is just the dimension of L as an F-vector space

[next marsh]

ahhhh ok wait that makes sense

[mild laurel]

so for your question, you can just calculate this directly

but another useful fact is that

if F is contained in L is contained in K

and we let [L:F] be the degree of the extension L over F and so on, then we have that [K:F] = [K:L][L:F]

So the degree of the extension is multiplicative in some sense

[next marsh]

ohh interesting

[mild laurel]

It'd be helpful to look at a proof of this fact from somewhere, but the proof is pretty intuitive

you just take a basis for L over F and a basis for K over L and multiply all the things together and you get a basis for K over F

[next marsh]

ooo nice!

[mild laurel]

but those ideas are all you really need to solve your problem

[languid moss]

I'm not even sure how to proceed with question 2. Can anyone help me guide through it?

[gritty sparrow]

Hint: ((ab)^n)a= a((ba)^n)

[languid moss]

Is that fit 1 or 2?

2 or 3

[gritty sparrow]

Question 3

Oh was that the wrong question?

[dusty river]

When you want to prove that |g| = n in any group, you have to show 2 things:

g^n = 1

and g^k is not 1 for any k<n

just try doing these, it should be straightforward

[terse crystal]

The intersection of <a> and <b> is {1}, therefore (ab)^r=1 implies that a^r=1=b^r

S_3 is a counterexample

[signal remnant]

Anyone feel like helping me with a little abs. alg. exercise I'm stuck on?

regarding external direct products

[hidden haven]

Post it, we can only tell if we can help once we see it

[signal remnant]

So I came a couple weeks back asking for a hint but I'm still stuck... I'm going to post the exercise and what I've written so far...

what i've written so far may be in the complete wrong direction but ty whomsoever helps me with this

I've done every exercise in the book up to this point... just been really stuck here with this...

[hidden haven]

My suggestion is work with small examples

[signal remnant]

the book is Gallian's Contemp. Abs. Alg., for reference

[hidden haven]

And you will be able to see how to find the subgroup

You just need to extend that pattern

No need to do it so abstractly

[signal remnant]

You are suggesting I try to find such a subgroup for some values of m,n,r,s?

[hidden haven]

Yes

[signal remnant]

Okay, so suppose I set m & n. Then it becomes find an element of the set {Z_r ⊕ Z_s | (r|m),(s|n)} that is guaranteed to be a subgroup of Z_m ⊕ Z_n?

[hidden haven]

Umm I can't parse that

An element can't be a subgroup

But I'm saying take some small values of m and n let's say 4 and 6

And some r and s which divide them, say 2 and 3

Now can you explicitly figure out a subgroup like this?

Think of what the direct sum of Z4 and Z6 looks like

And what it looks like for Z2 and Z3

(might be helpful to think of them as arranged in a 4x6 grid)

[signal remnant]

oh like a grid huh... so a subgroup in the grid model is a subgrid?

[hidden haven]

In this case yes

That is the intuition

Try and use this to get hold of a subgroup concretely, then prove that it is a subgroup

[signal remnant]

hmm... so then my brain goes:

I can go about trying to find a subgroup of Z4⊕Z6 that is isomorphic to Z2⊕Z3 by asking myself "What is some group which is isomorphic to Z2⊕Z3" and then showing that it is a subgroup of Z4⊕Z6?

I can answer to myself "Well, I can consider some other hypothetical group Z_a ⊕ Z_b such that Z_a is isomorphic (≅) to Z2 and Z_b ≅ Z3" (because I know from previous exercise #16 that this works) and so long as it is a subgroup of Z4⊕Z6 I am done? So I wonder "What's isomorphic to Z2?"

except now i'm feeling it going awry again

[scarlet estuary]

Z_2 is very easy to recognize up to isomorphism

its a group with 2 elements, an identity and another element

so that other element must be its own inverse

hence, if you have an element that is its own inverse

that (alongside the identity) gives a subgroup isomorphic to Z_2

[hidden haven]

I still feel you are working too abstractly. Here you can literally just list the elements of the group (at least in your head) and see what you can find

[signal remnant]

say... Z2 = {{e,a},*}

a set {e (for identity}, a (that self-invertable element)} with an operation *

am i then just interested in those operations?

like the first other one that comes to mind is the flipping group F.

that is, the group which flips a square

[hidden haven]

Can you identify any elements of Z4 ⊕ Z6 which are their own inverse?

To start with, what are the elements of this group?

[signal remnant]

ok one sec

elements of Z4 ⊕ Z6 are 2-tuples, so

(e,e), ofc... then elements (e,g) for g in Z_6 and then also elements (g,e) for g in Z_4

except, for instance, we have (e,g) = (e,g^3) for g in Z_4

[hidden haven]

You also have (g,h) for g in Z4 and h in Z6

[signal remnant]

and also we get (f,g) for f in Z4 and g in Z6

sure

[hidden haven]

And what's the group operation?

[signal remnant]

element-wise composition

[hidden haven]

Yep

[hidden haven]

So now try this

With this concrete description in mind

[signal remnant]

ok one sec

oh

what about

(a^2,g) for a in Z4 and g in Z6?

[hidden haven]

Is a some special element of Z4?

What if a=e

Or g=e

[signal remnant]

a is supposed to be e or aa

g is supposed to be whatever in Z6

[hidden haven]

hmm do you know how to represent Z4 using integers?

Like it is the group {0,1,2,3} under addition mod 4

[signal remnant]

you can't represent Z4 with only integers was what I was about to say

but yes

right so

sorry go on

[hidden haven]

Ye so let's use that notation

Similarly for Z6

And can you now say what you meant by (a,g) using these?

[signal remnant]

okay so we have Z4 = {0mod4, 1mod4, 2mod4, 3mod4}, sure

[hidden haven]

Yes

[signal remnant]

ok one sec

|0mod4| = 1 right?

[hidden haven]

Umm no

Unless you mean

[signal remnant]

i mean the order

[hidden haven]

Identity of the group

Ahh

[signal remnant]

|e| = 1 right?

[hidden haven]

Right yeah

[signal remnant]

ok so

|0mod4| = 1, |1mod4| = 4, |2mod4| = 2, |3mod4| = 4

[hidden haven]

Yep

So you've found an element of order 2 in Z4, which immediately tells you that Z4 has a subgroup isomorphic to Z2

Do you see this?

[signal remnant]

so by a^2 i mean (0mod4)(0mod4) = 0mod4, (1mod4)(1mod4) = 2mod4, (2mod4)(2mod4) = 0mod4, and (3mod4)(3mod4) =2mod4. that is, those elements that are guaranteed not to be of order 4.

oh uh

[hidden haven]

Hmm I see, but not all of those elements have the same order