groups-rings-fields

[next obsidian]

@chilly ocean

You’d have to be insane and it would probably take > a year I imagine

There is some fucking ridiculous exercises in there

[hidden haven]

didn't Narwhal do a huge portion of the book with all the exercises

[next obsidian]

Idk, I think Narwhal did

But I think Poros did even more

They did like every group theory exercise IIRC

This is coming from a guy that did every Matsumura exercise and wants to do every Hartshorne one

Aiming to do every D&F exercise is just absolutely insane

[hidden haven]

[wooden ember]

No lol I got to PID’s and that chapter was too boring for me and I haven’t continued since

Group theory was so much better

[next obsidian]

No like

I meant that you’ve done a large portion of the exercises

At least on the parts you covered

[wooden ember]

I think if you’re focused doing all of it would take a year

[next obsidian]

Doubtful

If you somehow could put like 12-14 hours a day for a year, sure

[wooden ember]

Did all except like 3 on some classification results of finite groups

[shell brook]

didnt poros do literally all of it besides last few chapters

[next obsidian]

But I’ve gone ridiculously hard on stuff for large periods of time

[shell brook]

like his solns were 500+ pages

[next obsidian]

I don’t think like any normal human could possibly keep it up

You’d have to have something in your brain that’s wired backwards or something

[shell brook]

wtf

[wooden ember]

Just the first 7 chapters and one section of the 8th

There are 15 chapters iirc

[wooden ember]

Kinda wanting to finish it before June and prove you wrong

[next obsidian]

I mean

[wooden ember]

But most definitely couldn’t lmao

[next obsidian]

If you want to sure

But like I don’t even see the point

[wooden ember]

Me neither lol

[next obsidian]

With Hartshorne and Matsumura like a lot of those results are useful

[wooden ember]

Those are alg geo though no?

[next obsidian]

With D&F like, the payoff quickly steepens off

Commutative algebra and AG

But like the point is there aren’t like 5

[wooden ember]

right

[next obsidian]

Prove a group of order 1628 cannot be simple

In each section

[wooden ember]

Those are fun okay

[next obsidian]

Almost every exercise is further results

No like I enjoy those

[wooden ember]

That’s more fun I agree

[next obsidian]

I love Sylow-bashing shit

But like, you aren’t gonna be an algebra GAWD

Cuz you did every D&F exercise

You’ll be super good at taking your algebra qual

[wooden ember]

Yeah they don’t teach you much I agree

[next obsidian]

But like after another 1 or 2 years a lot of that is kinda pointless

[wooden ember]

I don’t remember most of the results from the exercises lol

[next obsidian]

Whereas the other books I’ve done a lot of exercises from, the exercises form a large part of the core theory

I think even more than that, even if you did, a lot of them just aren’t things that will come up

If you enjoyed it, then more power to you

[wooden ember]

Yeah exactly

[next obsidian]

But like, doing every group theory exercise there isn’t really gonna help much in your career

[wooden ember]

There are a few fun ones but most are pretty pointless

[next obsidian]

So unless it’s fun and you have spare time it’s kinda dumb to do so

¯_(ツ)_/¯

[wooden ember]

I had fun through the group theory but the ring theory exercises are hella boring mostly

Which is why I stopped lol

Are there alg prerequisites to doing comm alg?

Like from harthstone or matsumara?

[next obsidian]

Yeah

I mean you should know rings and modules lol

But you don’t need too much module theory, that kinda gets developed as you go

[wooden ember]

In good detail or just the basics?

[next obsidian]

Eh

Basics is fine, like most textbooks start from a sorta elementary point

And build it up

Depending on your source you might need varying amounts of stuff tho

Like Matsumura assumes you know homological algebra

Tensor product, Ext, Tor, direct limits

There’s an appendix

But man, if you haven’t seen it before I have a hard time imagining you’ll have a fun time

[wooden ember]

Yeah no lol

[next obsidian]

Atiyah-Macdonald however like has an entire chapter on the tensor product

[wooden ember]

That’s a pretty large prerequisite id say

[next obsidian]

And I think has no Ext and Tor

Eisenbud I think builds up that stuff too

So like in terms of prereqs the subject itself doesn’t have much

You don’t need to know like all of a graduate algebra course

Just be somewhat comfortable with rings, fields, modules

It’s up to the book whether or not there’s more prereqs

[proud bear]

pain

[next obsidian]

LOL

Whoooooooopsssieeeeee

[chilly ocean]

those ones are next up for me

[next obsidian]

You can tell I hate A-M and have done like 1/2 of a chapter of it

Because I have no idea what’s in it

[wooden ember]

Oop

[chilly ocean]

there's nothing in it

[next obsidian]

It’s an 80 dollar problem book

[wooden ember]

So what should I look at before comm alg lmao

[next obsidian]

I mean if you don’t want to deal with tensor product stuff

You can try Undergraduate Commutatige Algwbra

I believe that book actually has very few prereqs

[chilly ocean]

tfw "undergrad" means "no tensor products"

[next obsidian]

I worked through a bit of it

[wooden ember]

I’m fine with dealing with tensor product stuff if the prerequisite book teaches it

[next obsidian]

I mean Atiyah-MacDonald doesn’t have too much even in-spite of what Stain posted haha

That’s… only in an exercise?

granted most of A-M’s content is in exercises but it won’t kill you to miss a few

[wooden ember]

Still I suppose it’s nice to know that stuff to get a broader view

[next obsidian]

Honestly if you are comfy with rings and fields and stuff

It wouldn’t hurt to try and jump into some basic homological algebra

[wooden ember]

Maybe I’ll skim through the rest of d&f while doing only a handful of exercises

[next obsidian]

Don’t do like the Weibel shot

With abelian cats

Or like chain homotopy blah blah

[wooden ember]

Or tbf I bought Lang I can work through it slowly

[next obsidian]

You can do a lot more adhoc treatment

I think D&F has like 1-2 chapters on it

And that section is actually okay

[wooden ember]

Yeah I heard they suck though lol

Oh alright

[next obsidian]

Oh really?

I referenced it a few times

[wooden ember]

I may be misremembering

[next obsidian]

Alternatively I learned a lot of it from Aluffi

I really like how he introduced Tor and Ext

[wooden ember]

On another note what are the prerequisite to representation theory

Cause that seems very interesting

[next obsidian]

Module theory

Maybe some field theory

Knowing what groups are

[wooden ember]

Right so I’ve got a good way to go then I only know very basic stuff on modules and fields

[next obsidian]

Yeah I mean it isn’t even like

That advanced stuff on modules

I wouldn’t say I “knew” module theory

Until after I learned rep theory

[wooden ember]

Module theory does seem pretty interesting in its own right though

[next obsidian]

You really only needed like some basic shit about them and definitions

I feel like module theory really isn’t something you study by itself

It’s developed in-tandem with either like the noncommutative stuff or with rep theory or something

Or alongside commutative algebra and algebraic number theory

I guess you get a hint of it when you do classification of fg ones over a PID

I think if you can get through that

You know enough to be able to go to rep theory

[wooden ember]

fg?

[next obsidian]

Finitely generated

[wooden ember]

Oh right

[next obsidian]

“Finite dimensional”

But for modules

I mean actually

[wooden ember]

Okay good to know

[next obsidian]

You probably covered the classification of finite abelian groups?

Maybe even finitely generated?

[wooden ember]

In d&f yeah

But only the proof for finite not for finitely generated iirc

[next obsidian]

Okay well

The classification of fg modules over a PID looks like that

There’s an infinite or “free” part

[wooden ember]

I think the proof for finitely generated is generalized from classification of fg modules over PID’s actually now that you mention it

[next obsidian]

Which for groups is just Z^n

Then the torsion bit which is finite

And looks like the sums of Z/nZ

Like there’s two ways to present it

Either prime powers

Or like increasing with divisibility

For a fg module over a PID R it looks exactly the same

Free part is R^n

Then direct sum the torsion part which is just R/aR for various a

Either prime powers

[wooden ember]

Huh

[next obsidian]

Or in increasing order by divisibility

[wooden ember]

That’s kinda cool

[next obsidian]

The proof goes through some module theory obviously

And introduces like canonical forms of matrices

[wooden ember]

I guessed that much lol

[next obsidian]

It’s the first sort of like

Linear algebra over a ring

You get exposed to

[wooden ember]

And so for abelian groups it’s the Z-module case then?

[next obsidian]

Yup

[wooden ember]

Nice

[near pecan]

Show that if $K$ is a ring then there is a bijection between congruences and two-sided ideals in $K$.

can anyone help me with this? i understand that congruences are equivalence relations that still "preserve" the ring structure on the equivalence classes, but idk how to start the proof

[cloud walrus]

sikfeng

[hot lake]

you could start by defining some function between the set of congruences and the set of two-sided ideals

[near pecan]

idk how exactly to do this...

i think i could have a function f, that could map like the identity relation to a zero ideal, and the congruence that has every element in the same congruence class to the ring itself(?), but i cant see how to get it to work

[hot lake]

or maybe a function from the set of two-sided ideals to the set of congruences

[near pecan]

hmm... i still dont get it, do u have more hints on how to construct this function

[hot lake]

if you have a two sided ideal, can you make a congruence out of it ?

(this should be a very easy thing to do)

[near pecan]

quotient of the ring?

[hot lake]

yes

so now you have a function f : { two sided ideals} -> { congruences } and you can start actually talking about whether it is injective or surjective or constructing an inverse

so first you try to show it's injective

then try to figure out, if you are given a congruence that you suspect is given by a two-sided ideal, how to get the ideal from the congruence

and finally try to show that this will work out for every congruence

[near pecan]

hmm ok thank you

[lethal dune]

Let G be a group of order n which has no subgroup of index 2. show that G is isomorphic to a subgroup of An. I kinda understand how to do it but not completely sure. Like if it is not a subgroup of An ( but is of Sn) then we can define the homeomorphism sending phi(G) to {-1, 1} giving us a subgroup of index 2.

[hidden haven]

Yeah that's it

You embed it into Sn and then kernel of sign homomorphism is a subgroup of index 1 or 2

[lethal dune]

ok

Show that $S_n$ is isomorphic to a subgroup of $A_{n+2}$. So for this do I just adjoin the cycle $(n+1 ; n+2)$ at the end of each odd permutations of $S_n$ and find the homomorphism that way?

[lethal dune]

Is that even a homomorphism

think yes, because (n+1 n+2) has order 2 and odd perm * odd perm is even

ig it will work

[hidden haven]

Yes

[lethal dune]

G be a group of order $2l$ where $l$ is odd. We have to show that $G$ has a subgroup of index 2. For this I know an approach that will work, looking for a better one. We will look at the permutation representation afforded by left multiplication. Now $G$ has an element of order $2$ call it $x$ then $\pi(x) \in S_G$ is a product of $l$ disjoint 2-cycles. So $x$ is an odd permutation. So $S_G$ contains an odd perm. Now we'll look at the homomorphism from $S_G$ to $\mbb{Z}_2$

[lethal dune]

thank you

[rustic crown]

that's the only proof i've seen really

but i didn't look further, so there might be something neater

i've seen a generalization of that statement tho,

Say G is a finite group and S is a sylow-2 subgroup,
if we assume that S is cyclic, then there is a normal subgroup of index |S|

[ancient rivet]

"Annihilating Radical Left Ideals and Blowing Up Points on a Plane" <-- I cant tell if this PDF is a math whitepaper or a terrorist handbook

[rustic crown]

sounds like some #algebraic-geometry

[ancient rivet]

yes... I cant think of anything else points could mean in the phrase "Blowing Up Points on a Plane", so it must be geometry.

[rustic crown]

yea there is a notion of blow-up and stuff relating to rees algebra

idk anything about it 🙈

[ancient rivet]

Its like a click bait title, but on a mathematical doctoral theses

[lethal dune]

[ancient rivet]

arn't all Sylow subgroups cyclic?

[lethal dune]

no?

[weak oriole]

Consider Z/2Z × Z/2Z

[hidden haven]

This is what my presentation was on

[rustic crown]

presentations are over 😌

raq tho

[lethal dune]

[hidden haven]

have fun

[chilly ocean]

Hi! I want to know if (P(e),∩) is a group, P(e) set of sets of e. I found P(e) as identity element ,[∀A,∈(⊂?)P(e) A∩P(e)=A] but does an inverse element exists or is this not a group?

[rustic crown]

the identity should be e and not P(e)

P(e) isn't a member of P(e)

[chilly ocean]

Alright

[ancient rivet]

additive inverse or multiplicative inverse? because it wont have a multiplicative inverse if its just a group

[viscid pewter]

??

there's no inverses

[rustic crown]

[∀A ∈ P(e), A ∩ e = A]
[∀A ⊂ e, A ∩ e = A]

so this is how you would write it

[viscid pewter]

intersection only ever decreases or at best keeps constant the cardinality of the elements

[rustic crown]

another way to see this is that in a group G, the equation x*x = x has only one solution, namely multiply by x inverse to get x = id
in our case, A ∩ A = A

so it's group only if P(e) is singleton

which happens when e is the empty set

[chilly ocean]

Ho yes e is not empty so it's not a group, thanks

[viscid pewter]

similarly union doesn't work either

what you want is symmetric difference, xor

[chilly ocean]

Yes i can see that with ∅ as identity element may be

[rustic crown]

yep!

[chilly ocean]

Alright thanks for your help!

[chilly ocean]

is it worth it to completely finish it? and not every problem, maybe like 90% of the problems

80?

because it'd take too much time to do everything? and it's not worth it to spend that much time?

so I should do like half the problems in D&F and move on to a graduate abstract algebra book?

it does seem inefficient since some problems might be really difficult

[woeful flint]

In need of some advice

im picking modules for my next semester and I want to study galois theory

I'm good at ring theory and field theory but I can't get my head around group theory that much, its just not something I like particularly

Should I take galois theory or not?

how group heavy is it

I guess my main question is if I'm not good at group thoery how will i fair in galois

That I quite like tbf

okok

I'll have a think about it

thx!

[waxen hedge]

Whatever you do it's probably better to know some group theory

[wispy scaffold]

Can someone recommend some of those books (#book-recommendations message )? Please

[lethal cipher]

It is probably a good call to get some experience with abstract algebra if you have not already

[chilly ocean]

btw

why are regular covers called galois covers

ig this falls under alg top sorta

but i dont see how a covering space being regular has anything to do with galois theory

[lavish summit]

Its because the spaces are part of a Galois correspondence

Which is more abstract than Galois theory the algebra field

[chilly ocean]

o ty

i didnt know it was a more general idea

i always thought it wad strictly fields and groups

[lavish summit]

Yeah its subtle and not clear unless you are shown that the correspondence is also it's own thing

[chilly ocean]

wikipedia page giving lots of insight

[lavish summit]

It even shows up in like data math stuff iirc

[chilly ocean]

woah

data math sounds like a fun buzzword

[lavish summit]

Yup it's a good one

[chilly ocean]

also do you know much about cohomology of groups?

[lavish summit]

Naw not really I just finished my first AT course

[chilly ocean]

same here

[lavish summit]

That might be a good question for Kanga Gang Made Man

[chilly ocean]

the big thing for me rn is finding motivation

there are so many topics

that i wonder why ppl care

stack exchange doesnt always help

[lavish summit]

In homology and cohomology or in general

[chilly ocean]

sort of in general

like yesterday learned about (co)homology of groups

and the takeaway is that group homology tells you how much the coinvariant functor fails to be exact

now i dont see why people care about this

[lavish summit]

Lol yeah I dont like that kind of motivation I work on applied math and algorithms mostly

[chilly ocean]

wow you are so cool

i want to learn some of that

[lavish summit]

Yeah it's cool stuff algorithmic analysis and complexity theory are cool and their motivations are clear

"Make program work" is a solid motivation

[chilly ocean]

yeah motivation always exists

and is clear cut to understand in those cases

atm my motivation for algebra is just the potential connections to geometry

which has inherent motivations

but im feeling this masculine urge to quit everything in math and just go deep into number theory

because number theory motivation is honest as heck

[lavish summit]

Yeah algebra and geometry sorta become this huge monstrous superfield which tbh scared me away

[chilly ocean]

yeah algebra is a monster

way too many connections

kind of scary at times

[lavish summit]

Yeah it drives me crazy

[chilly ocean]

like recently I was thinking about something

so like group homology and homology groups of lie groups

two different functors you can apply to this object

i wonder what im supposed to take away from their applications

in general objects like topological groups scare me, or objects that are a mixture of two very well structured things, but are seemingly unrelated

or ig to be specific objects that have a high potential to be abstract scare me

like R-Algebras

[lavish summit]

Yeah algebra gets really intense and becomes it's own pure math language thing

[chilly ocean]

yuh

and examples are always feeling sparse

but thats either because im young and dumb or because they really are sparse

[lavish summit]

You just get this crazy object in 3 different fields that you can do all the algebra to, I guess the takeaway with those topics is you need more algebra lol

[chilly ocean]

lol fr

[lavish summit]

The 100mile high mountain of advanced algebra

[chilly ocean]

then the idea of studying morphisms between objects

just spooks me

like category theory wasnt too bad for my first informal introduction, but it only gets crazier and crazier the more I learn

too many examples of categories show up

and i dont always have a nice interpretation of functors between categories and universal properties arent always super intuitive

[waxen hedge]

There is a categorical generalization of Galois theory and covers of nice topological spaces, which was invented by Grothendieck

[chilly ocean]

i hate that dude now

makes my life scarier

too abstract

[waxen hedge]

Also, maybe more "elementary" the analogy is concrete for Riemann surfaces

But I don't know much about it

[chilly ocean]

ok ill look into this

[waxen hedge]

There is a book by Szamuely

"Galois groups and Fundamental Groups"

Which covers this subject

[chilly ocean]

to be honest i have too many topics rn

im trying to go over 3 different books atm

commutative ring theory by matsamura, rotman introduction to homological algebra, dreaded atiyah mcdonald

only rotman’s isnt dense

i will put that into a book list tho

[languid moss]

Wow

[chilly ocean]

wuh

[languid moss]

Read Lang @chilly ocean

[chilly ocean]

ive done for a little

but it isnt my focus rn

commutative alg is my focus

lang isnt bad but its just standard grad alg stuff

[next obsidian]

With how many D&F has I feel like doing 50% is more than enough

I’d try to do problems until I feel like I could do the rest in a reasonable amount of time

Maybe go through a proof sketch in my head, see if I can argue through most of the problem and come up with something that seems@fruitful

[wind locust]

hey, so the order of an Abelian group (i.e. the number of its elements) does not necessarily coincide with the order of the corresponding $\mathbb{Z}-$module, right? As an example, the Klein 4-group $G=\mathbb{Z}_2\times\mathbb{Z}_2$ has order 4, whereas G viewed as a module has order two since $\text{ann}(G)={z\in \mathbb{Z} \mid zg=1_G \forall g \in G}=<2>$

[next obsidian]

No…

[cloud walrus]

iruneachteam

[next obsidian]

It’s exactly the same…

[wind locust]

huh

[next obsidian]

I mean

What is your definition of order of a module

Frankly I’ve never heard anyone use it

[wind locust]

the order of a module $M$ over a PID $R$ is the generator of the $\text{ann}(M)$

[next obsidian]

Wtf

Okay sure

Your example works

[cloud walrus]

iruneachteam

[wind locust]

alright, thanks a lot!

[next obsidian]

What the hell is this definition tho lmfao, i have never in my life heard of this

[plucky flicker]

Hi! What is the value of the transfer homomorphism on a central element?

[next obsidian]

Is this… used somewhere?

Rep theory…?

[wind locust]

haha apparently it makes sense, I encountered it in Advanced Linear Algebra by Roman

it is very useful for decompositions of modules over a PID

[next obsidian]

Oh sure, I’ve just never assigned this a name

[wind locust]

i've been pulling my hair out over something that has to do with this, if you know any other name for this notion i'd greatly appreciate it if you could tell me

[wind locust]

nvm figured it out

idk why modules are so ugly

[next obsidian]

😧

[simple mulch]

Hey

I have V,W finite dimensional vector spaces and T : V -> W linear. I want to show that if dim(V) < dim(W) then T is not onto.

Proof: Assume contradiction, T is onto. Then W = im(T). (can I assume the following: dim(W) = dim(Im(T))?) According with the dimension theorem, dim(V) = rank(T) + nullity(T) < nullity(T) = dim(W), thus implying nullity(T) < 0. Contradiction.

[viscid pewter]

your proof works

if W = Im(T), dim(W) = dim(Im(t))

this is really linear algebra, not abstract

[simple mulch]

thanks and sorry

[thorn delta]

no problem

[chilly radish]

can you have a commutative ring without identity but no nontrivial ideals?

if so, is there an easy example?

[viscid pewter]

0

[chilly ocean]

[thorn delta]

not sure if 0 counts since 0=1 in {0}

[chilly ocean]

rngs

[boreal field]

Hello, so I have been teaching myself abstract algebra for a little while now and I just learned about something called the commutator subgroup. My book says that the subgroup is normal in G and also that the quotient group formed from it is abelian but doesnt prove it so I tried to prove it myself and was wondering if my proof was mathematically sound. Thank you! ps sorry about how messy my paper is(my kitten really likes my math paper) 🙂

[lilac trench]

just had my abstract algebra final today, and I gotta say it wasn't very abstract, thank you all for the help!

[next obsidian]

Why does that H exist?

You can’t just magic in an abelian group H where you could define a map phi(a) = a

Well not necessarily

It would contain it

But yeah it’s essentially going in circles

[proud bear]

nah they defined phi(a)=a_1

[next obsidian]

Wat

[proud bear]

but they only showed that the kernel contains G' so still isn't complete

[next obsidian]

The fuck is a_1

That isn’t even a homomorphism then unless a_1 = e

[proud bear]

some element of H

[next obsidian]

It isn’t multiplicative

Yeah this is what you want to do

You I think need to sneak an e in

By multiplying by something and its inverse

It’s kind of tedious IIRC

Yeah

TRUE!

[wind locust]

actually hm

ill think about it a lil more

[boreal field]

Thanks everyone for the input. I've seen that other proof but I'm still learning and wasn't sure if it was ok to just define a different group that's abelian and define a homomorphism from elements of the first group to elements of the abelian group so that way the commutator subgroup would be the kernel of the homomorphism. My cat and I really appreciate it!

I believe the problem was that I can't just assume that H has a group structure, correct?

[viscid pewter]

you can't just assume that such a group exists

you have to define it

and then show that the homomorphism has such a kernel

[boreal field]

Ok, thank you so much! (:

[wind locust]

Hey, I'm trying to show that rational canonical forms are invariant under field extensions

So here's what I have in mind:

[next obsidian]

So I mean in my eyes the biggest issue was you said phi(a) = a

So it seemed like H would have the same objects as G, like they have the same set

But then it seems like H just is equal to G

But G isn’t abelian…

[wind locust]

damn actually i have no clue

so a good first step can be to show minimal polynomials are left invariant when we pass to a larger field

but idk how one would proceed from there

[boreal field]

My bad. I chose bad notation. I thought it would be more clear what I was trying to do if I chose my elements to be represented by symbols that looked similar so I choose a to map to some symbol called a1. I should have just chosen a more distinct symbol so people didn't think I was just trying to map to the same group

[next obsidian]

Oh I see

Well then your proof actually does show something very cool, but not what you want

There’s two issues

The first is, why does G ever have a map to an abelian group H?

What if somehow there just wasn’t one?

Two

This only shows the kernel contains

G’

But it might be bigger

What you did show though is any map G -> H when H is abelian factors through G/G’

Because the kernel has to contain G’

This is why G/G’ is referred to as the abelianization of G

It’s an abelian group which like, most closely approximates G as an abelian group

Because there’s a bijection of maps G -> H when H is abelian and the maps G/G’ -> H

I imagine what I’m saying doesn’t make too much sense right now, so don’t think too hard about it right now

But maybe later you’ll see this idea again and you’ll have already seen it before :)

[next obsidian]

Actually I can describe this bijection explicitly

Given a map f:G -> H when H is abelian, you showed that any element in G’ maps to 0

So we can define a map f’:G/G’ -> H by declaring
f’(g + G’) = f(g)

[boreal field]

That does sound really cool! I'm not in school anymore. Got my degree in chemistry and only a minor in math but I've taken math as a hobby and group theory always seems to make me really excited! Everything in it just seems so cool to me!

[next obsidian]

The issue is if this is well-defined, but g + G’ = h + G’ iff h = g + k for some k in G’ by definition

Then we see f’(h + G’) = f’(g + k + G’) = f(g) + f(k)

But f(k) = 0 since k is in G’, and f has to send elements of G’ to 0

So this assignment f:G -> H goes to f’:G/G’ -> H defines a map {maps G -> H} -> {maps G/G’ -> H}

To go the other way, we simply take a map g:G/G’ -> H and send it to the composition G -> G/G’ -> H

These two maps I described are inverse so the two sets of maps are in bijection

The benefit here is like, in terms of maps into abelian groups, G and G/G’ are the same because there’s this way to go between the two

The benefit is just that G/G’ is abelian which is nice

This exact thing actually shows up in representation theory eventually

[boreal field]

Cool! I've heard of representation theory and it seems really cool because it seems like it has a lot of applications! Can't wait to study it eventually!

[lethal dune]

for polynomial p(x)=x⁴-4x²+8x+2 in ℚ(√-2) do I have to prove it's irreducible the long way?

will eisenstein work or I'm missing something?

[lethal dune]

Am I right to call it false thinking about Sn group? given any 2 elements are conjugate means they have the same cycle type, or there is a semidirect product that makes them conjugate?

[lethal dune]

it is not necessary for a group of even order to have a subgroup of index 2. but then how do I separate the elements of G mod g?

[small bison]

g has even order so you can find an element that has order 2

that gives you a subgroup of order 2

[lethal dune]

hmm then?

[small bison]

then the answer is like 1 sentence away

[lethal dune]

I am given g specifically

If I could choose a g, it would be obvious like you said

[small bison]

hmm you're right

well just look at the cosets of <g> and see what happens

you have an even number of them

[lethal dune]

hmm like if there are even number of them I think it can be concluded

I'm getting confused by the ordering

like yaking the even powers and odd powers will work i think

[upbeat juniper]

where'd you get all these interesting problems from 👀

[lethal dune]

by the looks of it, you already know

[upbeat juniper]

this?

[lethal dune]

oh, you don't then

no these are from prev year entrance examination problems

[upbeat juniper]

oh lol

[lethal dune]

[upbeat juniper]

this one seems interesting but idk if I know enough to solve it

[rustic crown]

maybe think group actions?
<g> would act on G by left multiplication... if we show each orbit has even size then we're done

[lethal dune]

I was thinking of symmetric groups

[lethal dune]

yes that's what I ended up doing

but with 1 err

noh wait

[upbeat juniper]

that's equivalent to young_smasher's solution above which considers cosets right

[lethal dune]

hmm yes

[rustic crown]

number of cosets can be odd right?

[lethal dune]

yeah

[rustic crown]

this follows because the stablizer would be trivial as g * x = x implies g = e which has order 1 and not even.
so size of orbit = |g| which is even

[upbeat juniper]

what matters is the size of the coset (orbit), not the number of cosets though

[lethal dune]

right

[upbeat juniper]

same but I didn't make much progress

[rustic crown]

ah right

[upbeat juniper]

maybe you can consider smth like the double cover of Sn lol?

I barely know anything about that, just remembered it's a thing

[lethal dune]

I'm having trouble with orbit stabilizers again

[upbeat juniper]

|G| = |orbit of x| * |stabiliser of x|

[rustic crown]

<g> in our case

but yea

[lethal dune]

so {g, g^2, ... g^2n} acts on G by left multiplication. g^k*x = x means g^k=e. so |<g>| = 1 * |<g>| is even

now like for each x we do the separation by following {x, g^2x, g^4x, ...} and {gx, g^3x, ...}

[upbeat juniper]

it should be |<g>| = |orbit size| * 1

[lethal dune]

oh

yes

orbit size is even then

[rustic crown]

g^k*x = x
also this should be true for each k to be able to say x is stabilized by <g>

else k = 2n is sad

[lethal dune]

[rustic crown]

[lethal dune]

what about this one tho?

I strongly want to call it false

or like we do some shit like S_2n and adjoin cycles to make them compatible idk what I am saying rn

[sturdy marsh]

it's probably true

[lethal dune]

[sturdy marsh]

hint: what does the action of g and h on G look like

||they should have the same "cycle type"||

[lethal dune]

like (12) and (12)(34) in S4?

[sturdy marsh]

those arent of the same cycle type

[lethal dune]

they don't hv same type yet

no like we can make them of same cycle type

[sturdy marsh]

in a different symmetric group

let G act on itself by left translation

[lethal dune]

think I lost you there

[sturdy marsh]

what does the corresponding map G ---> S_G look like

i.e. what do the images of g and h look like

[lethal dune]

oh yes

product of n m-cycles

[sturdy marsh]

so they have the same cycle type in S_G

[lethal dune]

yes

[sturdy marsh]

and therefore are conjugate in S_G

[lethal dune]

so I was on the wrong track to begin with

[sturdy marsh]

yeah the only reason I initially said that it was probably true was because constructing a counterexample seemed too hard

and if it's a hw problem, it cant be too hard

[lethal dune]

so every element of same order in Sn are conjugates in Sn!

[sturdy marsh]

sure, but S_n doesnt sit in S_n! in the usual way

[lethal dune]

yes think that's where I messed up

[sturdy marsh]

so stating it that way can be slightly misleading

[lethal dune]

anyway cool problem

[sturdy marsh]

yeah it was pretty neat

thanks

[lethal dune]

thanks? i should be the one thanking you

[upper cape]

Has anyone got any tips on computing the maximal ideals of a quotient of a polynomial ring?

i.e. something of the form k[x_1,...,x_n]/(f_1,...,f_m)

I thought of using Nullstellensatz, to classify the maximal ideals of k[x_1,...,x_n], but since k is not necessarily algebraically closed, I'm not sure if that is a valid approach

[plucky flicker]

what's the connection between free modules and projective modules?

I mean I know there is some theorem about some embedding I guess, but I don't know it precisely

[cyan marten]

M is projective iff M + M' is free for some other M'. Over some nice rings, projective = free, but in general freeness is stronger (take M' = 0)

[plucky flicker]

So basically M can be embedded into a free module?

[cyan marten]

No, it has to be embedded as a direct summand, which is strictly stronger.

[maiden ocean]

Well maximal ideals of k[x1, ..., xn]/(f1, ..., fm) are maximal ideals of k[x1, ..., xn] containing f1, ..., fm ie (g1, ..., gk) with gi dividing fj and this is going to depend a lot on the field

like x^2 + 1 is maximal in R but not in C, so R[x]/(x^2 + 1) has no maximal ideals whereas C[x]/(x^2 + 1) has (x + i) and (x - i)

Basically theres not going to be a simple general solution to this because its very dependent on the base field

[glossy grotto]

In a free abelian group, does $x-y=a-b$ imply $x=a$ and $y=b$?

[cloud walrus]

blackiris

[glossy grotto]

With x,y,a,b in the generator of course.

[hidden haven]

If they are distinct generators then yes

assuming that the group is the free group generated by these

[glossy grotto]

Yeah, I can look at them as if they're basis elements, I see now.

[hidden haven]

yes

They are exactly basis elements

[simple mulch]

One question

G a group and S a subgroup of G

I want to show ${g \in G : gS = Sg}$ is a subgroup of G

[cloud walrus]

mns

[simple mulch]

I already proved closure under G operation

but now I want to prove $\forall x \in A, x^{-1} \in A$, where A is the set above

[cloud walrus]

mns

[simple mulch]

If this holds then $x^{-1}S = Sx^{-1}$. So I choose to start with $(x^{-1}S)(Sx^{-1})^{-1}$ and try to get e (identity)

[cloud walrus]

mns

[simple mulch]

So I did $(x^{-1}S)(Sx^{-1})^{-1} = (x^{-1}x)(SS^{-1}) = e(SS^{-1})$ but $SS^{-1}$ seems odd

[cloud walrus]

mns

[hidden haven]

You are multiplying sets here, you definitely won't get just identity

you will get a set

There is an easier approach

[simple mulch]

yeah

[hidden haven]

gS = Sg

multiply on both sides by g inverse

both left and right multiplication

[upper cape]

possible dumb question, are quotients of principal ideals principal?

[hidden haven]

Yes, the images of the generators under the quotient map are generators of the quotient

[simple mulch]

like $S = eS = g^{-1}(gS) = g^{-1}(Sg) = (g^{-1}g)S = eS = S$ ?

[cloud walrus]

mns

[hidden haven]

no I mean gS = Sg

now left multiply both sides by g inverse

then right multiply

and you get exactly the equation you want

[simple mulch]

g^{-1}(gS) = g^{-1}(Sg) ?

[hidden haven]

after the first step yes

and on the left side the g inverse g cancels

[simple mulch]

S = g^{-1}Sg

[hidden haven]

yes

[simple mulch]

what about?

I am sorry 😐

[hidden haven]

now multiply both sides on the right by g inverse

and the g g inverse on the right will cancel

[simple mulch]

oh Sg^{-1} = g^{-1}S

I see

[hidden haven]

I feel like you are not comfortable with this because we are multiplying sets

[simple mulch]

indeed

[hidden haven]

so just try to justify these equations set theoretically and it will make sense

like cancelling g and g inverse

[simple mulch]

Roger that, thanks man!

[median pawn]

https://math.stackexchange.com/questions/4324406/is-r-k-left-fracab-sqrtk2-a-b-in-mathbb-z-textboth-even-or-b

Feel free to try Q2, Q3 here lol

it has been unanswered for quite a bit

Personally, I think d = 1 mod 4 might have something to do with Q3, but not sure

[upper cape]

Hey could someone check my logic here? I have a ring A=k[x^2,x^3], and B=A/(x^4). Firstly, k[x] is an integral extension of A, so then (x)^c is a maximal ideal in A, then (x)^c/(x^4) is a maximal ideal in B

[rustic crown]

depends what P is

it can't be simultaneously an ideal of Z and Z[sqrt(-d)]

[dull root]

If I have a complex rep of a finite group, is it true that the corresponding character is multiplicative, that is is chi(ab) = chi(a)chi(b)?

[rustic crown]

idts

[cloud walrus]

det

[rustic crown]

which need not be true

chi(1) = dim V

chi(1)^2 = (dim V)^2 which is not chi(1 * 1)

[dull root]

hmm, the situation I have is the following. I have a finite grp G and an abelian subgroup A where the character vanishes outside of A. I want to take the all the elements of A which have nonzero character, and consider the subgroup generated by all those elements, say H. Certainly H is a subgroup of A and H is abelian. I want to show that H is normal in the G, but the issue is that the character of some element in H might vanish as well

Another approach I had was to take the subset of all elements in A with nonzero character. However since character is not multiplicative, I can not guarantee even that "group" is closed, and Im suspecting that it might not even be a subgroup

[rustic crown]

sowwy i'm too sleepy to think

[past temple]

how can i show that

if f \circ g = 0 => g = 0 for all g

then f is injective

like i kinda see how this might imply f has a left inverse

but im stuck on how to show it explicitly

[hidden haven]

Is this for maps between modules over a ring?

[hidden haven]

It doesn't imply left inverse

This is left cancellability (you can get from this that fg = fh implies g = h for all g and h)

Left invertibility is slightly stronger. For example the inclusion of 2Z into Z as abelian groups is left cancellable (follows from injectivity) but not left invertible (there is no map from Z to 2Z that sends 2 to 2, because 1 has to go to half of that but there is no such element in 2Z)

[past temple]

hmm okay

so im currently trying to figure out

let's suppose phi_1: M1 -> M2 is a module hom

apply the functor Hom(N, _) to phi_1

giving us a map

phi_1*: Hom(N, M1) -> Hom(N, M2)

given that phi_1* is injective, how do i show that phi_1 is injective?

@hidden haven

[hidden haven]

Yeah so this is the injectivity implies left cancellability bit

You can check this element by element pretty much

if phi_1 g = phi_1 h, then evaluate both sides at an element and see what happens

Oh wait you are going the other way

[past temple]

yeah

im going the other way

[hidden haven]

So if phi_1 is left cancellable, then suppose (going by contradiction) you have 2 elements x and y in its domain which map to the same thing

[past temple]

ok so phi_1(x) = phi_1(y)

[hidden haven]

well you don't actually need contradiction just assume non injectivity

Try defining maps into the domain so that the same thing gets mapped to x and y under the 2 maps

but then phi_1 puts them back together

to make the maps agree

that will make it not left cancellable

[past temple]

how do i explicitly define a map

that maps something to x

or maps something to y

like thats what i was stuck on earlier

like even withotu this contradiction stuff

[hidden haven]

Try the simplest module over a ring A

[past temple]

if we assume phi_1(x) = phi_1(y)

if we can create maps that like

map the same shit to x and why

then we get like

compositions on the left and right hand sides

[hidden haven]

x and why

[past temple]

ok so A itself?

[hidden haven]

yep

A is the free A-module on 1 generator

if you know the universal property of free objects

this should become obvious

or even without it its not so hard

This is a useful thing to keep in mind, because the universal property of the free object on 1 generator says exactly that giving an element of an A-module is exactly the same thing as giving a map from A to that A-module

so having elements x and y

you can immediately start thinking of them as their corresponding maps from A

If you haven't seen this, it isn't hard to prove, ||map the identity of A to whatever element you have||

[past temple]

so u consider the unique maps

that map 1 to x

and 1 to y

?

[hidden haven]

You might have also seen it written as Hom(A, X) is isomorphic to X

yep

exactly

[past temple]

and those maps are guaranteed to exist

by universal property

[hidden haven]

yes

also not hard to see directly

you are defining a maps to ax

[past temple]

right yeah

if we call the map

that maps 1 to x

x*

then x*(a) = ax

so then phi1(x) = phi2(y)

means phi1(ax) = phi2(ay)

means phi1 x*(a) = phi2 y*(a)

for all a in A

so then x* = y*

so x = y

[hidden haven]

yes

by putting a=1

[past temple]

right

tyty

ok so if we have

im phi_1 = ker phi_2

i want to prove that im phi_1* = ker phi_2*

is it true that if i have some f: N -> M2

such that phi_2 f = 0

then there exists some g: N -> M1 such that

f = phi_1 g?

i kind of feel like it directly follows from im phi_1 = ker phi_2

but im not 100% sure how to explicitly prove it

@hidden haven

[hidden haven]

phi_2 f right?

[past temple]

sry yeah

[hidden haven]

So f maps into the image of phi_1

and phi_1 is an isomorphism onto its image

so compose f with this "inverse" of phi_1

and you get the map you need

[past temple]

oh aight

[past temple]

are additive functors always right exact?

the definition of additive im using here is

that the functor is a group homomorphism

with respect to addition in a module

@hidden haven

[sturdy marsh]

no

Hom(A, -) for example

[past temple]

ok hmmm

so the problem im doing

gives additive functors F,G which are adjoint

and somehow theyre right exact and left exact respectively

and idk how to do this at all

[small bison]

It’s the dumbest thing

A left adjoint preserves limits (or colimits I can’t remember)

So assuming it preserves colimits it’ll preserve cokernels or something like that

And that’s right exactness

[sturdy marsh]

left adjoints preserve colimits

[rotund shoal]

Yo I can't seem to figure the problem out, even though it seems easy: say we have a group of order pqr with p > q > r all primes and G has no normal subgroup of order p. The question is how many elements of order p it has

So I figured out that the group needs to have order 2pq

but then idk where to go from here

[rotund shoal]

meant order p instead of "order H" lmao

Oh wait a second I might already be done lol

so I have that np = 2q, and so since p-groups intersect trivially or are the same group, and every nontrivial element in a group of prime order has order p, there must be 2q(p-1) elements of order p!

[past temple]

where doe the additivity come into play here

[small bison]

What’s the exact problem

[past temple]

ok

so F: A-Mod -> B-Mod

is left adjoint to G: B-Mod -> A-Mod

and F,G are additive

as in they're group homomorphisms with respect to addition

in the Hom module

and the bijection eta in the definition of adjiont functors

is also a gorup homomorphism

i want to show that

F is right exact

and G is left exact

[small bison]

Hmm if you’re not allowed to use continuity of adjoint functors then I guess you’ll want to show that F preserves cokernels directly

Cause it looks to me that these are extra assumptions since adjunction implies left or right exactness right away

[past temple]

yeah this is only the first part of the problem

wait so withotu

any of the additivity business

the fact that F is left adjoint to G

immediately means that F is right exact?

also what does it mean for a functor to preserve co kernels

[small bison]

Yeh

Hmm that’s a good question cause I’m just parroting what my algebra prof told me

But it should be like if C = M/im(f) then FC = FM/im(Ff)

[past temple]

hm ok

im still kinda lost tho

[small bison]

On which part?

[past temple]

like

how to even start tbh

like ok ig i start with an SES

say 0 -> M1 -> M2 -> M3 -> 0

and i want to show that F(M1) -> F(M2) -> F(M3) -> 0

is exact

but like im rlyyyy confused about what the additivity or adjointness

have anything to do with this

like im absolutley lost

at what the connection is supposed to be

[small bison]

Well Ff is in Hom(FM1, FM2) = Hom(M1, GFM2) so that’s a start

I’m just writing random stuff btw I don’t remember how the proof goes

[past temple]

wait does this follow from F, G being adjoint?

ive never learned this fact

[small bison]

Yeah

What’s your def of adjoint?

[past temple]

like

theres some bijection

[small bison]

The bijection should basically be the thing I wrote right?

It’s a bijection between two Hom things

[past temple]

yeah yeah

but the hom things are in bijection

but theres no equalities anywhere

ok this is a hint from stack

@small bison

is the adjunction isomorphism

the bijection between the homs?

[small bison]

I put an equals cause I’m lazy

But it should’ve been a bijection

Yeah should be

[lethal dune]

is there a formula of how many elements of order p a group can have? p being prime

[next obsidian]

Wdym?

This can vary wildly

What would you want this formula to depend on?

The order of G?

If so, it can’t exist

[lethal dune]

no like for example if 2| |G| then there are odd number of them

like now if 3 | |G| then is it possible that the number of elements of order 3 is ≡ 1 or 2 (mod 3)

[next obsidian]

Hmm

I feel like it should be 1 mod p

Like if x is order p

Then <x> is size p and contributes…

I dunno

[lethal dune]

maybe something can be said if the powers are taken to be pⁿ and not p? like using syllow's thms?

do syllow p-groups intersect non-trivially?

[hidden haven]

@past temple you have a short exact sequence and you apply G to it and you want to show that the result is left exact. Apply the functor hom(GX, -) (for an arbitrary object X) and you already proved that this won't affect exactness. Therefore it suffices to check left exactness of
Hom(GX, -) ∘ G
for any object X.

To do that, we have the natural isomorphism between functors
hom(GX, -) ∘ G = hom(GX, G-) ≅ hom(FGX, -)
And this last functor is known to be left exact.

[weak oriole]

If G is finite, then number of solutions to $x^n=1$ is a multiple of $n$, assuming $n$ divides |G|

[cloud walrus]

Noob666

[lethal dune]

any proof?

in S4, x³=1 has solutions as 1, and all the 3 cycles, i.e. there are 4+1=5 of them, and 3 doesnot divide 5

😶

[hidden haven]

Aren't there 8 3-cycles in S4

[lethal dune]

oh yes

it's true then

wow

[weak oriole]

The general thing is frobenius's result

The prime case can be done just using Cauchy

[past temple]

ahh ty

i eventually was able to come up with that strategy

[median pawn]

Not sure what to do here

[cloud walrus]

Hausdorff

[cyan marten]

Can you relate GL(n, A x B) to GL(n, A) and GL(n, B)? (I really don't know, this is not a hint)

[coarse storm]

Just going by intuition, I would think it is the direct product of the other two.

[lethal dune]

sanity check, a syllow p-subgroup is normal iff its unique

[rustic crown]

sounds right

[lethal dune]

[rustic crown]

[lethal dune]

tks

[upbeat juniper]

pillow p-subgroup

[lethal dune]

pillow subgroup

[lethal dune]

Is my line of reasoning all right for this one? given $|G| = 105=3\cdot 5\cdot 7$ and given it has a normal 3-subgroup. This means $n_3=1$ now for possible order of $Z(G)$ is 5, or we get an abelian group because otherwise we get a cyclic center. This means $n_5=1$.

[lethal dune]

I have to show that $G$ must be abelian, what do I do next? how do I show n7 = 1?

[cyan marten]

Cyclic center does not imply commutativity. But it's true that the order of Z(G) is either 1 or a prime, say p, and this would give n_p = 1.
Maybe a more fruitful approach would be to quotient by T, your given normal 3-group. Then G/T has order 35 and thus abelian (known result).

[lethal dune]

how do I go from G/T being abelian to G being abelian?

[viscid pewter]

if the quotient of the group by the center is cyclic, the group is abelian, is the one i remember

[lethal dune]

but T is not the center

[cyan marten]

Not directly. For example, if G/N is abelian, then G' is contained in N. So for example if you have G/N and G/K abelian for disjoint normal subgroups N, K, then actually G is abelian!

[lethal dune]

that's some high level explanation

ok can we do this? we do have subgroup of order 5 and 7, say H, K then HK has order 35 and hence index 3. So HK is normal. but HK is also cyclic. meaning abelian. Now G ~ HK x P_3?

[cyan marten]

Now, for n7, we have n7 dividing 15, so it's either 1 or 15. If 15, then we get 15 * 6 = 90 elements of order 7, and then maybe use the usual counting argument works.

[lethal dune]

yes with this we can show that HK is a subgroup

because atleast one of them is normal

[lethal dune]

this is a good argument tho, nice

[cyan marten]

Yeah, it's easy that at least n5 = 1 or n7 = 1, or else we would get too many elements

[cyan marten]

Ohh, I see it now

Well, we could say G/HK is abelian and G/T is abelian, and HK n T is trivial, so we are done.

[lethal dune]

[cyan marten]

It's not trivial that HK is cyclic, though. We will still need the "only one group of order pq" result

[lethal dune]

order of HK is 35?

so cyclic?

[cyan marten]

Yes

[lethal dune]

though not needed, all we need is the normality of HK

[cyan marten]

Can we prove this without the given assumption?

We still have HK normal

[lethal dune]

any subgroup of index 3 is normal anyway

[cyan marten]

Well, now G is a semidirect product between HK and T, and Aut HK has order 8, and thus no nontrivial homomorphism from T to Aut HK is exists.

[lethal dune]

all we needed was the normality of T

[cyan marten]

Here we actually prove the normality of T by proving that the semidirect product is trivial.

[lethal dune]

can you elaborate why Aut HK has order 8?

[cyan marten]

Aut HK is the unit group of Z/35Z

Which has order phi(35) = phi(7)phi(5) = 4 * 6 = 24

[lethal dune]

yes

why 8 then?

[cyan marten]

Oops, I was wrong

But on the bright side we have all the groups of order 105 now: There are two of them, Z/(105) and some semidirect product.

[cyan marten]

Wishful thinking. I thought of 21 for some reason haha

[lethal dune]

nice nice

[cyan marten]

After finding the automorphism of order 3, we will get an explicit description of this other group K!

[lethal dune]

will it give me all the groups of order 105?

[cyan marten]

Yes, because we started with an arbitrary group of order 105 and showed that it has a normal subgroup of order 35

I mean G = NT where T is a 3-Sylow subgroup

[lethal dune]

ok

[cyan marten]

K is generated by a, b, c, of orders 5, 7, 3, and a and b commute. Finally, c commutes with a and conjugation by c sends b to b^2.

[lethal dune]

semidirect

[cyan marten]

lol

Maybe try Keith Conrad's notes

[lethal dune]

where?

I read dummit foote and yeah it's hard to read

[lethal dune]

nice

[median pawn]

Yeah, seems like that to me too

[maiden ocean]

I think you can use the fact that G/C(P_3) is isomorphic to a subgroup of Aut(P_3), which has order 3 - 1 = 2, and since G is odd this has to just be 1, so that P3 is contained in Z(G)

@lethal dune

[median pawn]

but I haven't found an isomorphism yet

[maiden ocean]

Then since P3 is in Z(G) G/Z(G) is in G/P3

so it should follow that G/Z(G) is cyclic

bc |G/Z(G)| should divide |G/P3| = 35 and hence is either 5, 7, or 35

and any group of any of these orders is cyclic

[lethal dune]

yes thank you

so normality of P_3 is necessary for G to be abelian?

[maiden ocean]

Probably

since any group of order 105 should have a subgroup of index 3 but idt theyre all abelian

Probably some semidirect products

[lethal dune]

nice result and nice proof, tks

[cyan marten]

If so, we can just work out GL(2, Z/(3)) and GL(2, Z/(2)). I know the latter is S3.

[lethal dune]

A question from ring theory, I am given a ring of order 35. I have to find the number of rings upto isomorphism. I know the (R, +) is Z35 and we have two classes, one is zero ring and another is Z/35Z. Are there any other? If so is there a general way to count them?

[median pawn]

makes sense, let us think

[cyan marten]

Yes. There are two groups of order 105, one cyclic, and the other has the presentation above. I didn't work out n_3 in this case, though.

[cyan marten]

The GL(n, A x B) = GL(n, A) x GL(n, B) thing seems shady honestly

[lethal dune]

yes counting groups is kinda tedious

[kind temple]

satisfying tho

[median pawn]

maybe it works if A = Z/pZ and B = Z/qZ where p,q are both prime

[cyan marten]

Yeah maybe. But in general we have an embedding of the RHS in the LHS.

[coarse storm]

Not sure how helpful this approach is. There are way too many groups of order 48. (Edit: Nevermind. This is non-sense.)

Do the preceding questions provide any hints?

[cyan marten]

I remember reading somewhere that GL(n, Z/(m)) = GL(n, Z) / GL(n, mZ), where GL(n, mZ) is the ideal of n x n matrices whose entries are all divisible by 6. Using the Chinese reminder theorem, I think we can show that this follows.
I still don't know about GL(2, 3), though.

[maiden ocean]

I dont see what you mean exactly? like

what do you mean one is zero

The zero ring does not have 35 elements

[lethal dune]

no, i mean the product a*b=0 for all a,b

[maiden ocean]

Does your definition of rings require identity

[lethal dune]

yes, it does

[chilly ocean]

Id assume yes

[maiden ocean]

does it require associativity

Oh

[lethal dune]

also yes

[maiden ocean]

Well the zero ring structure has no identity

[chilly ocean]

ring homomorphism by definition maps 1 to 1

[lethal dune]

yeah, not that one then

[maiden ocean]

if a * b = 0 for all a, b, then a * e = a = 0 for all a

contradiction

in general our ring structure will be determined by our choice of identity here i think

since every element will have the form e, e + e, e + e + e, and so on

[lethal dune]

ok brain fart moment, I though we require associativity to be a ring

[maiden ocean]

Well ok lets not say its determined by the identity thats harder to show

But if e = 1 in Z/35Z then theres only one ring structure right

because xy = (1 + 1 + ... + 1)(1 + 1 + ... + 1) and then distribute

makes sense?

[lethal dune]

yes

[lethal dune]

also do we not require associativity for a ring?

by default?

[maiden ocean]

usually we do there are just a lot of competing defns of rings

[lethal dune]

ohh

[maiden ocean]

anyway so we just want to show that e has order 35 and hence = 1. order meaning as an element of the group. the order of e divides 35 so it is 1, 5, or 7. if e has order 5 then e + e + e + e + e = 0, so k + k + k + k + k = k(e + e + e + e + e) = 0, that is, every element has order 5. this obviously isnt true bc any group of order 35 has an element of order 7

for the same reason e cant have order 7

so e has order 35 and is 1

@lethal dune

[lethal dune]

yes reading

[rustic crown]

why not using some vector space stuff?

i think that makes the proof almost immediate

consider a ring R of order 35, look at the unique map Z --> R, say the kernel is nZ
so we have injective map Z/nZ --> R
which means n divides |R| = 35

n can't be 1, as that would force R = 0

n can't be 5 or 7 as then R is a Z/5Z or Z/7Z vector space
(i might need to verify we don't need commutativity)

so n = 35 done

[maiden ocean]

is this actually any shorter lol

[chilly ocean]

yes

[rustic crown]

well the thinking is... because Z --> R is a very common thing to do

[lethal dune]

ok very basic question, how does order or e=35 means that we have only one ring?

[maiden ocean]

i mean it is pretty much exactly the same thing

nZ being the kernel means exactly that e has order n

in this case

[rustic crown]

yea lol nvm

[maiden ocean]

e has order 35 implies e = 1 right

[lethal dune]

like we define a*b = (1+1+..+1)(1+1+...+1) = ...

[lethal dune]

yes

[maiden ocean]

so yeah ab = (1 + 1 + ... + 1)(1 + ... + 1)

and now we can just distribute

hence its determined

[lethal dune]

yeah nice

ok so I was trying your approach of using Aut(P3) for another problem

this time group is of order 1575 = 3^2 5^2 7

[maiden ocean]

[lethal dune]

similar argument shows |G/Z(G)| possible orders are 5, 7, 35, 5^2*7

for first 3, it's clear, what do I do for 5^2*7?

though I am supposed to use Syllow, aut approach seemed cool

[maiden ocean]

are you trying to show that G is abelian again

[lethal dune]

yes and I also realised that $Z_{5^2}\times Z_7 \simeq Z_{5^27}$

[lethal dune]

[maiden ocean]

yeah so the last case is going to be G/Z(G) = Z_175 or G/Z(G) = Z_5 x Z_5 x Z_7

[lethal dune]

yes another one too

was too fixated on using syllow that I forgot basic gcds

only have to deal with $Z_{35}\times Z_5$ now

[maiden ocean]

Uhhh obviously the former will just finish the problem, the latter... im not sure, maybe you can get a contradiction by looking at subgroups of Z_35 x Z_5

you probably need to appeal to sylow directly i think