groups-rings-fields

[next obsidian]

that fixes it for you

tfw tfw tfw tfw

[stoic rose]

The left Cayley graph with generators {s_1, ..., s_n} is isomorphic to the right Cayley graph with generators {s_1^(-1), ..., s_n^(-1)}

[coral shale]

So if we take our left graph, replace all the labels for the generators with their inverses
a -> a^-1

We will get the right graph essentially?

[stoic rose]

Or if you consider it as a nonoriented graph then no need to invert the generators

[coral shale]

After we change the direction of the arrows

We flip all the arrows and get the right graph

The vertices themselves may not necessarily be the same ofc (will be G^-1)

I think I can see it

[turbid pond]

so you'll probably want S to be symmetric too

[coral shale]

Left graph

1. reverse all arrows
2. replace all vertices with inverse
   Right graph

I will try it 🤔

[pastel cliff]

gm chillgebra

[hidden haven]

yooo

[lethal dune]

[frank fiber]

is it true that for any a,b algebraic numbers then Q(a,b)=Q(a+b)?

[hidden haven]

Not necessarily

But this does work in almost all cases

You need a condition that a - a' ≠ b - b' where a' and b' are any conjugates of a and b respectively

[cursive temple]

generally a similar result holds, called the primitive element theorem

[hidden haven]

See the proof of the primitive element theorem for details

Lol

[cursive temple]

sniped

[hidden haven]

[lethal dune]

[coral shale]

Finding $Gal(Q[x,y]/(x^5-2, 1+y+y^2+y^3+y^4) : Q)$.\\

The roots of $h^5-2$ are $xy^a, 0\le a\le 4$.\
The roots of $1+h+h^2+h^3+h^4$ are $y^b, 1\le b \le 4$.\\

Let $\sigma(a, b)(x, y) = (xy^a, y^b)$.
$$\sigma(a, m)\sigma(b, n)(x, y) = \sigma(a, m)(xy^b, y^n)$$
$$ = (xy^{a+bm}, y^{mn}) = \sigma(a+bm, mn)(x, y)$$\\
So
$$\sigma(a,m)f(b,n) = \sigma(a+bm, mn)$$

[cloud walrus]

Shuri2060

[coral shale]

That right 👀

$$\mbb F_5$$

[cloud walrus]

Shuri2060

[coral shale]

Then I can see we have the multiplicative group of F5 acting on the additive group

So it is the semidirect product of C4 acting on C5

@hidden haven did u do all this in your head D:

[hidden haven]

I have a lot of power in my hands

Power = theorems

[coral shale]

I see I see. Finally get it now tho lol

[hidden haven]

This looks kinda right ye

Nice

But how did you prove that these are automorphisms

[coral shale]

Which

the uh sigma?

[hidden haven]

σ (a,b)

[coral shale]

I did it in scratchwork

[hidden haven]

Nice

[coral shale]

Just enough to convince myself

[hidden haven]

There's a problem with that

So you are defining a map on this field by defining it on the generators

[coral shale]

im defining it on Q, x, y

[hidden haven]

Then extending it using the homomorphism properties

[coral shale]

and assuming that is enough

yes

[hidden haven]

Ye

This extension may not be well defined

Because there may be multiple ways to write the same element in terms of the generators

I think the most painless way to do this would be to just prove and use theorems about this stuff 👀

[coral shale]

All K-homomorphisms are automorphisms, is this not enough?

[hidden haven]

We don't even have a K-homomorphism yet

[hidden haven]

Because you are defining the function on the whole field by that last equation

[coral shale]

It definitely fixes Q. And I construct it so it follows the homomorphism rules

[hidden haven]

That depends on how that element is being written in terms of x and y

[hidden haven]

That construction is not well defined necessarily

[coral shale]

There is only one way to write each element mod the 2 polynomials, no?

[hidden haven]

You can add one of the polynomials to it

[coral shale]

🤔 ok ill think on this

but

[hidden haven]

This will be hard I think

[coral shale]

arent I defining this homomorphism on the field

Q[x,y]/(x^5-2, 1+y+y^2+y^3+y^4)

This one

and each element can be uniquely written as Qx^ay^b

[hidden haven]

Not necessarily

[coral shale]

unique up to modding?

[hidden haven]

You mean [x] and [y], where [] means equivalence class in the quotient

[coral shale]

yes

If I adjust by homomorphism defn using equiv classes

[hidden haven]

But then the issue is that you can instead just add [x⁵-2] to it

It's still the same element

But that changes its representation in terms of x and y

[coral shale]

🤔

I think I see and don't see hmmmmmm

So you're saying this is not a trivial issue that can be fixed with a little notation adjustment?

[hidden haven]

ie you are picking another element of [x]

[hidden haven]

No

You have to use

That should fix it

The problem is that you've defined a map on ℚ [x, y]

And you need a map on the quotient

So you use the first isomorphism theorem

[coral shale]

[q] -> [q]
[x] -> [x][y^a]
[y] -> [y^b]

[hidden haven]

But then this may not be well defined

[coral shale]

I := (x^5-2, 1+y+y^2+y^3+y^4)

[hidden haven]

Because [x] = [x⁵ + x - 2]

[coral shale]

[g] := g + I

Thinking and writing

[hidden haven]

[coral shale]

So it isnt 'obvious' apriori I can write everything as

sum Qx^ay^b

with 0 =< a =< 4, 0 =< b =< 3

[hidden haven]

Sums of that

But it is obvious

Uniqueness isn't

[coral shale]

Lets say I have f

My claim is I can find something in [f]

[coral shale]

that looks like this

[hidden haven]

Hmm ok

[coral shale]

And use that instead to represent

ig this is not easy to put into a proper proof

[hidden haven]

I'm not sure if that is true though

[coral shale]

eh?

[hidden haven]

How would you eliminate sums

[coral shale]

Can I use this twice?

That tells me the those powers of x and y combined will form a basis

[hidden haven]

Ye so you'll have a 20 dimensional space

[coral shale]

yh

this tells me everything can be written in exactly that form?

[hidden haven]

What I'm saying is that a general element doesn't just look like qx^iy^j

[cloud walrus]

Moldilocks1337 ✓

[coral shale]

Right

but they can be written with i and j within the ranges we want

[hidden haven]

Yes

Then once you look at this element in the quotient, this form need not be unique

[hidden haven]

Example

So this isn't directly a well defined map

[coral shale]

I am applying this theorem on the quotient

[hidden haven]

Because what it does to an element, depends on how that element is written, and there are multiple ways to write a single element

[coral shale]

Is that not enough ? It would tell me I can write that form uniquely

[hidden haven]

ohhh

Yes ok that will give you a vector space homomorphism because basis

And then you'll have to just check multiplication

[coral shale]

ok, I didnt write that explicitly at the start 😅

didn't realise i was leaning on this

thanks for pointing that out. Edit, probably missing that the roots cycle as well to conclude they act like C4, C5

[hidden haven]

[coral shale]

This feels like the 1st non-trivial galois group I've found 👀

My exercises are mostly like x^2-2, etc

[hidden haven]

[chilly radish]

Am I wrong or in Vn(x,G) are they not allowing for monomials here, or generally onoy allowing for multilinear polynomials where every factor has all n indeterminates (up to action of some g)? G is an automorphism and antiautomorphism subgroup and x^g:=g(x).

It's also not clear to me why the g_i are labelled (G is finite but not necessarily of order n) but ig it's just to emphasize that each entry has a distinct associated (anti)automorphism right?

[cloud walrus]

Shuri2060

[coral shale]

(or rather, how many)

[hidden haven]

Is that polynomial irreducible?

Usually you figure out whether you have all the roots or not by just doing subfield of ℝ inside ℂ memes lol

In some cases you can use trace and norm

[rapid bramble]

If you have to prove $0\cdot a=0$ for a in a ring, can you basically copy paste the proof for the statement in vector spaces?

[cloud walrus]

Mosh

[rapid bramble]

since vector spaces have the same structure (then some) as rings

[coral shale]

It is and ok hmmmmm. Looks like I will have to wait for my lecturer to post more content

[hidden haven]

Vector spaces don't have more structure than rings

You can't multiply any 2 vectors

But the same argument should probably work idk what argument you have in mind but you can write 0 = 0 + 0

[rapid bramble]

yeah that argument

then b = additive inverse of a

oh wait, no, 0+0a=0a+0a

0+0a+b=0a+0a+b then basically done

[hidden haven]

Ye

[pastel cliff]

my textbook talks about S_n before defining groups lol

this just keeps getting weirder

[hidden haven]

S_n is the prototype for groups

Rubik's cube enthusiasts in shambles 😌

So defining S_n first is somewhat reasonable

[pastel cliff]

just wanna make sure im getting something right: the sequence 3412 isn't itself an element of S_4, but if we consider it as a permutation of 1234 then it is

also would like to continue talking abt this if it's possible/worthwhile

[pastel cliff]

@delicate orchid👉 👈

[delicate orchid]

I'm gonna cry

[delicate orchid]

s4 is not a group of sequences it's a group of permutations

the permutation (13)(24) is the element of S_4 that corresponds to that permutation of {1, 2, 3 ,4}

{3, 4, 1 ,2} is the result of applying the group action of (13)(24) to {1, 2, 3, 4}

[pastel cliff]

that's what i meant! yay!

[delicate orchid]

how wholesome

[pastel cliff]

i will bother u about this later

i must acquire sustenance

https://youtu.be/HUT-hkmemMk

wha t

[delicate orchid]

classic

[pastel cliff]

i just had a stroke

[delicate orchid]

check out mlg group theory

[pastel cliff]

im

i

uhhh

hngh

[delicate orchid]

what DID he mean by this

[pastel cliff]

i know not

i am groping around in a fuckster truck of material that im probably overcomplicatiing

https://tenor.com/view/theworldmayneverknow-idk-goodquestion-gif-5415227

get this man a phd

[delicate orchid]

there's a nice presentation of S_n if that's what you want

[pastel cliff]

probably yes

[delicate orchid]

well nice in the fact it has two generators

the uhh relations are uhh

actually why am I even bothered, this presentation is beyond useless

you can generate it with "swaps" of elements i.e. permutations of the form (1a) for all a in {1, 2, ..., n}

although now I'm sure you want to see the presentation I found, so

$S_n = \langle t, c \colon t^2=c^n=(tc)^{n-1}=(tct^{-1}c^{-1})^3=(tc^kt^{-1}c^{-k})^2 = 1 : \forall k \in {2, ... \lfloor n/2 \rfloor} \rangle$

useless

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[lethal cipher]

I was wondering if I could get some help on part a). I am personally not even sure where to begin.

Here's what I do know. If we let F(x)=c_nx^n+.....+c_0, then the leading coefficient of f(x) will be c_np+1, which is not divisible by p. Lastly, p doesn't divide F(a), which means that p^2 does not divide pF(a), and that is somehow supposed to be correlated to the constant term. But I personally do not understand why this is, and I am not too sure how to show all non-leading coefficients are divisible by p.

[proud bear]

@lethal cipher try to apply Eisenstein's to f(x+a)=x^n+pF(x+a). if f(x+a) is irreducible over Q, then so is f(x)

[gilded gull]

How would I go about proving that every image of a cyclic group is cyclic?

Would I want to show that there's a generating element for the image and if so how would I get there?

[gritty sparrow]

The image of the generator

[gilded gull]

o

[delicate orchid]

homomorphisms are uniquely determined by where they map the generating set - which for cyclic groups is just one element

[gilded gull]

I mean it makes to sense to me I just don't know how I would start a proof

[delicate orchid]

I'd do it by contradiction, assume that the image isn't cyclic and see what breaks

[gilded gull]

o ok

I'll try that

[pastel cliff]

if elements of a group commute, must their inverses commute as well?

[thorn delta]

yes

[pastel cliff]

is that just bc closure??

[delicate orchid]

ab = ba
take inverses of both sides
(ab)^{-1} = (ba)^{-1}
=> b^{-1}a^{-1} = a^{-1}b^{-1}
no way jose

[pastel cliff]

ah so they dont

[delicate orchid]

what

[gilded gull]

no they do?

[delicate orchid]

I've literally written $b^{-1}a^{-1} = a^{-1}b^{-1}$

[pastel cliff]

oh i misread that

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[gilded gull]

lol

[delicate orchid]

rip

[gilded gull]

I did the proof differently

[delicate orchid]

how did you end up doing it?

[gilded gull]

Idk how you would do it by contradiction lmao

I'll just ss what I wrote

Honestly idk if it works

It feels too loose

[delicate orchid]

nah that's a nice proof and now that I think about it is way nicer than mine

[gilded gull]

Oh ahaha

cool

[delicate orchid]

f(a) generates the image of <a> yeah

[gilded gull]

I'm like two weeks behind in my class so

[pastel cliff]

also this is basically lin alg but in general, what kinds of linear transformations/matrices preserve the angle between vectors

[gilded gull]

I'm kinda rushing

[pastel cliff]

im tryna be lazy

[delicate orchid]

k*M where M is in O(n)

i.e., some orthogonal transformation plus some scaling

kind of cheating saying that "a matrix is orthogonal if it's in the group of orthogonal matricies"

so I'll say that a matrix is orthogonal if and only if it's transpose is it's inverse

[pastel cliff]

my line of reasoning for this so far is that since SO is a subgroup of SL, we know every matrix must have det=1, which means that orientation and area are preserved

so the only thing to throw on that would be to show that matrices in SO preserve angle of basis vec's too

[delicate orchid]

you can preserve angles without preserving orentation https://cdn.discordapp.com/attachments/359052581022203914/940656085893799936/unknown.png

and you can preserve angles without preserving area

trying to remember the proof I think it has a dot product in there

[pastel cliff]

https://tenor.com/view/jesus-ballin-mars-bars-gif-19910027

LETS GOOOOO

anyways

[pastel cliff]

i know this much i was hoping there might be a way to show that a matrix's transpose being its inverse preserved angle

[delicate orchid]

ahh yes... it's angle preserving if and only if $Mx \cdot My = x \cdot y$ yesss mmm yummy

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[thorn delta]

someone asked this exact question yesterday

classmates 👀

[delicate orchid]

$Mx \cdot My = (Mx)^TMy$ right? I haven't used a dotproduct in 142334 years

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[delicate orchid]

the dimensions work out I'm sticking with it
now $(Mx)^T$ is just $x^TM^T$ so we get $x^T(M^TM)y$ which must be $x \cdot y = x^Ty$ yesssss

[lethal cipher]

That helped a lot. It does most of the work for you. Do you have any tips for part b)?

[barren sierra]

don't Sylow theorems only give that $n_2(S_3 \times S_3) \in { 1, 3, 9}$?

[cloud walrus]

Spamakin🎷

[delicate orchid]

ok I'll write it out neatly now I've figured it out

[pastel cliff]

wait so that line of reasoning does work?? pog

[delicate orchid]

angles are preserved by a linear transformation if and only if $Mx \cdot My = x \cdot y$. Using the fact that $A \cdot B = A^TB$ we get $$Mx \cdot My = (Mx)^TMy = x^T(M^TM)y = x \cdot y = x^Ty$$$$\Rightarrow M^TM = I$$

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[delicate orchid]

I really hope this is right

[pastel cliff]

A dot B 👀

[delicate orchid]

sure why not

mf I've been using specific inner product spaces on FUNCTIONS for the past 2 months

I'll make EVERYTHING an inner product space

[pastel cliff]

ok oprah

OHHHH

i think i get it

[delicate orchid]

it's just using properties of the tranpose

and then left/right cancellability

[pastel cliff]

i forgot for a sec that dot product is related to angle between vectors kek

[delicate orchid]

yeah that's like, it's think

that's true for the cross product as well but I doubt anything useful comes from it

non-associative

[pastel cliff]

is AB = A^TB a different way of writing AA^T = i

[delicate orchid]

$$AB = A^TB$$
$$\Rightarrow ABB^{-1} = A^TBB^{-1} \Rightarrow A = A^T$$
$$\Rightarrow A(A^T)^{-1} = I$$

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[delicate orchid]

assuming A^T and B are actually invertible, of course

[pastel cliff]

but isnt the condition that A = (A^T)^-1

not that A = A^T

[azure seal]

If $n$ and $m$ are coprime integers, how can we show that $n$ and $1-\zeta_m$ are coprime in $\mathbb{Z}[\zeta_m]$?

[cloud walrus]

Pishleback

[delicate orchid]

yes?

which is... exactly what I got?

[upbeat swift]

If g has order 18, and G=<g>. What are all of the generators of G?

[next obsidian]

To be a generator is just to say it has order 18

Try to figure out a formula for the order of g^n in terms of the order of g, and of n

[upbeat swift]

it's all that has gcd (k,18)?

to be 1

[next obsidian]

Yup

[upbeat swift]

okay cool

[next obsidian]

Try to prove it too

[next obsidian]

You can just adapt that formula to this situation

[upbeat swift]

I just need a direct answer but thank you tho

more like, i needed to clarify if i was right or not

[open pilot]

Do you mean the possible subgroups of G? Since G is cyclic they all have to be cyclic and can therefore have orders equal to the divisors of |G|.

[rapid bramble]

For $a,b\in R$ (Ring), I want to show $(-a)\cdot(-b)=a\cdot b$, is this as simple as I think it is, or is there a reason I can't say $-1\cdot -1=1$?

[cloud walrus]

Mosh

[delicate orchid]

see if you can prove that $-1 \cdot -1 = 1$ for any ring

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[delicate orchid]

(hint, 0 = 0*0 )

[rapid bramble]

oh right

cause -1*-1 is a specific case of what i'm proving

[delicate orchid]

yeah

[rapid bramble]

so do that case 1st, then use it in the general case

[delicate orchid]

and then once you've done that you can adapt the proof very easily to show that (-a)*(-b) = a*b

or, if your ring is commutative, just factor out -1 times -1 lol

[rapid bramble]

yeah my proof rn has it being used, so just make 2 cases, a=b=1 and then a!=b neither 1

But I'm struggling to see where the hint of 0=0*0 comes in rn

[delicate orchid]

1+(-1) = 0

[rapid bramble]

0=(1+(-1))*0

[delicate orchid]

double the trouble

there's another substitution you can make

[rapid bramble]

oh

one sec

$(1+(-1))\cdot(1+(-1))=11+(1)(-1)+(-1)(1)+(-1)(-1)$

[cloud walrus]

Mosh

[rapid bramble]

Then pick the right associativities QED?

It also feels weird FOILing here but cest la vie lol

[delicate orchid]

you get a (-1)^2-1 = 0 => (-1)^2 = 1

[delicate orchid]

why lol it's just polynomial multiplication

[rapid bramble]

No?

• in rings aren't always commutative

wait no

[delicate orchid]

addition is always commutative

[rapid bramble]

already proved -a=(-1)*a

so it's fine

[delicate orchid]

🤨

[rapid bramble]

the cross terms give a problem no?

[delicate orchid]

you just cancel the $1\times1+1\times-1$

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[delicate orchid]

leaving you with $-1+(-1)^2 = 0$

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[delicate orchid]

add 1 to both sides

$(-1)^2 = 1$

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[rapid bramble]

Oh I was looking at the wrong term right right

ok yeah

[delicate orchid]

cool

now, is your ring commutitive

_ _

[rapid bramble]

Nope, any ring R

[delicate orchid]

all rings are commutitive aren't we lucky

wait is -1 in the centre of the multiplicative monoid for any ring uhhhh

[rapid bramble]

wut

[delicate orchid]

I'm trying to just factor out -1^2 from (-a)(-b)

[next obsidian]

It is

[delicate orchid]

yeah it is

write the element (-a) as -1*(1+1+1+...), distribute, commute with 1 (1 = -1^2 as we've proven and powers of elements always commute with each other), slam it all back into a(-1)

[rapid bramble]

$(-a)(-b)=[(-1)a][(-1)b]=[(-1)(-1)]ab=1ab=ab$

[cloud walrus]

Mosh

[delicate orchid]

so yeah now we can just do that

[next obsidian]

You can’t write -a as a sum of -1

Lol

[delicate orchid]

no I'm writing -a as -1 times a sum of 1

[next obsidian]

You can’t do that

[delicate orchid]

what

[next obsidian]

What if a = x^2

[delicate orchid]

god I hate ring theory so much

[next obsidian]

Inside of k[x]

[delicate orchid]

right I give up

[next obsidian]

But just do this

[delicate orchid]

just gonna say it commutes because it always does in any ring I can think of

[next obsidian]

-a•-b = -1•a•-1•b = -1•-1•a•b = ab

[delicate orchid]

"it commutes because it commutes"

great

[rapid bramble]

OH DUH, -1*1=-1, I just realized 🤦

[next obsidian]

-1 commutes with everything

[delicate orchid]

yes that's what I'm trying to prove, chmonkey

[next obsidian]

Oh lol

[delicate orchid]

well, was

I'll just take it as a fact because it's obvious lol

[next obsidian]

You want to use uniqueness of additive inverse

Show that -1•a + a = 0

And that a•-1 + a = 0

Then both of the things on the left are the inverse of a under addition

And those equalities follow from distributivitiy I think

[delicate orchid]

yeah that's what I was thinking

[next obsidian]

(-1 + 1)•a

[delicate orchid]

factor a out

[next obsidian]

Versus a•(-1 + 1)

[delicate orchid]

then a0 = 0a = 0

[next obsidian]

Yup

[delicate orchid]

so glad I've never actually had to deal with non-commutitive rings

and I never willlllll

[rapid bramble]

wait, this assumes commutativity

[next obsidian]

Of what?

You only need to commute -1 and a

[rapid bramble]

multiplication in the ring

[next obsidian]

And I proved above that -1 commutes with everything

You only switched where -1 and a are in the middle there

[rapid bramble]

Oh I missed that

[next obsidian]

Yee

[rapid bramble]

this right?

[next obsidian]

Yup

You multiply by a on the two sides

This shows -1•a and a•-1 are both the additive inverse to a

So they’re the same

[rapid bramble]

[next obsidian]

Swag

[chilly ocean]

Been working through the rings section of Artin's algebra (rings defined here as commutative with 1) with a friend, and I have some leftover problems I wasn't able to crack/want to verify. One important one is showing that a function $\varphi$ on $R\left[x,y\right]$ given by $x \mapsto x+f(y)$ with all else fixed is an automorphism. By the substitution principle it's sufficient to prove that this is a bijection. I believe it is enough to say that a well-defined left and right inverse exists, in this case $\varphi^{-1}(g(x,y)) = g(x-f(y), y)$, am I nuts?

[cloud walrus]

zd

[next obsidian]

You only need a set theoretic inverse yeah

That looks good to me

[chilly ocean]

Yeah mulling it over myself it makes sense, so now after that I am stumped on a problem that hints at using the result of the first problem: determining all automorphisms on $\mathbb{Z}[x]$. The coefficients of any polynomial are fixed, so it depends only on where we send $x$. My logic was that if we send $x$ to something of higher degree then the way we map it back can't be a homomorphism since the degree can't decrease, otherwise sending $x$ back gives you a constant and you lose the bijection. Hence the only options are $x \mapsto ax+b$, but if $a\neq 1$ then we would have something wacky like $x = \phi^{-1}(ax+b) = cax+bc+d = x$ for non-unit $ca$

[cloud walrus]

zd

[chilly ocean]

Therefore we have only $x \mapsto x+a$

[cloud walrus]

zd

[chilly ocean]

OH

[next obsidian]

[chilly ocean]

I guess the previous problem just finishes the whole thing then? Since it's just a special case of it that this is always an automorphism

[next obsidian]

Well this one is easy

[chilly ocean]

I'm wondering if there's some big brain way to crack this proof without the logic I used, using mostly the result of the previous problem

[next obsidian]

It has a really obvious inverse

x -> x - a

[chilly ocean]

yeah

[next obsidian]

No, what you did is good

[crisp badge]

You're missing a small bit

[next obsidian]

Noting that any automorpjism fixes Z is important

[crisp badge]

Why did you jump to a \neq 1?

[chilly ocean]

For that contradiction? Just non unit I guess works yeah true

-1 is a thing xD

[crisp badge]

Yeah

[next obsidian]

Oh I guess that’s right that’s right

[chilly ocean]

thanks for validating that there's probably not some super obvious way to use the previous problem besides what I already did, I was weirded out cause Artin writes "(see previous problem)" at the end of the problem description lol

[chilly ocean]

also thanks that means $x \mapsto -x+a$ works for sure too so I missed that case, slapping any unit there gives you a nice two sided inverse either way

[cloud walrus]

zd

[crisp badge]

Yeah

Now I'm gonna go back to crying about one of my lecturers

I'm not gonna be to harsh on them as they're a recent postdoc

And it's UK, so they won't have much experience teaching

But like, ree

Some bad decisions have been made

[barren sierra]

why is this called "Lattice"

like for example the diamond isomorphism theorem I get why it's called Diamond

[lavish nexus]

if you draw out the lattice of a group

[barren sierra]

what is the lattice of a group

[lavish nexus]

the lattice of a quotient group is basically that lattice chopping off the part under the quotiented group

[barren sierra]

ok I have googled what a Lattice is

and this makes sense

[lavish nexus]

the lattice of Q8/<-1> is where you chop off the 1

[barren sierra]

and <-1> is normal in all of those

right?

yea it is

[lavish nexus]

yes

[barren sierra]

neat

[gilded gull]

When we're trying to prove that any cycle is a product of transpositions

My textbook just uses the previous theorem that any permutation can be written as a product of disjoint cycles

But if we're using $$(a_{1}, a_{2})(a_{2},a_{3})etc.$$ then aren't we not using disjoint cycles

[cloud walrus]

Rishi

[gilded gull]

like $$(a_{1}, ..., a_{n}) = (a_{1}, a_{2})(a_{2},a_{3})...(a_{n-1}, a_{n})$$

[cloud walrus]

Rishi

[gilded gull]

this is probably a really stupid question but im blanking nr

[paper flint]

It is a product of transpositions, not necessarily disjoint transpositions

It can be expressed as a product of disjoint cycles though

Where cycles can have length greater than 2

[gilded gull]

Omg im an idiot

disjoint cycles

tysm

[coral shale]

Does using g' to denote g inverse clash with any common notation that uses ' in Algebra? (ik its not convention)

[chilly ocean]

Another problem in Artin I'm still stumped on, but I'll just mention the part I want to check so I can try the other part myself without spoilers uwu:
Trying to find the kernel of $\varphi: \mathbb{C}[x,y]\rightarrow \mathbb{C}[t]$ given by $x \mapsto t+1$, $y \mapsto t^3 -1$. Clearly $f(x,y) = (x-1)^3 - (y+1) = x^3 - 3x^2 + 3x -y -2$ is in the kernel, and viewing this as a polynomial in $y$ (isomorphism to $(\mathbb{C}[x])[y]$ formalizes this) we can see the remainder $r(x,y)$ by division of anything in the kernel by $f(x,y)$ gives us a polynomial in $x$ only. It's unclear to me to how proceed formally from here but intuitively if we send everything through $\varphi$, the remainder goes to $r(t+1)=0$. I guess now since $r(x,y) = r(x,0)$, just use that $x\mapsto x+1$ doesn't change the degree and we have $r(x,y) = 0$. Feel free to go error hunting in my sketch here, much appreciated 🙏

[cloud walrus]

zd

[chilly ocean]

To be clear you only get a polynomial in x since r must be less than degree 1 in C[x][y]

[chilly ocean]

Wait isn't that sometimes notation for taking the derived subgroup a certain number of times

[coral shale]

Haven't heard of this. Staring at your Q and no clue.

[chilly ocean]

Ah yeah so derived group/commutator subgroup of a group G is sometimes denoted [G,G] or just G'

[coral shale]

Ah ok didnt know

[chilly ocean]

the hyphen is useful cause you can keep going deeper and get a sequence of subgroups $G^{(n+1)} = [G^{(n)}, G^{(n)}]$

[cloud walrus]

zd

[chilly ocean]

so it's less clumsy notation than the brackets often

[coral shale]

ty

[chilly ocean]

agggh well I'm pretty convinced this works to show that this polynomial generates the kernel since we map to $\mathbb{C}[x][y]$, then we have an isomorphism (restricted to the constants in $\mathbb{C}[x][y]$) taking $r(x)(y)$ to $r(t+1)$, and since the latter polynomial is 0 so must be the first one

[cloud walrus]

zd

[chilly ocean]

it made sense intuitively already after just looking at the division argument, but I had a hard time coming up with a precise way to put a formal argument into words here with all this mapping business

Wait I think I have some more intuition I didn't realize fully earlier... say, more generally, we have $a$ and $b$ in the kernel of a ring homomorphism, and suppose there is sufficient structure to get a remainder by division of $a$ by $b$. So this isn't necessarily for all $a, b$ in the ring. Now we have $a=bq+r$, and yeet the homomorphism to get that $r$ is also in the kernel. If we can show $r=0$, we simultaneously get that the kernel is $(b)$ and we've lifted the curtain off to show that $b$ dividing $a$ and no nonzero $r$ existing is an if and only if kind of situation. This is in principle super obvious

[cloud walrus]

zd

[chilly ocean]

But keeping it in mind is useful when looking at divisions like these is nice because now I can go and say for this kind of setup "ah yes remainder always in kernel"

I'm leaving out a lot of exceptions but just the intuition is what I'm happy with here

even more obvious connection, without applying the homomorphism, a-bq = r, kernels are ideals boom

[prisma shuttle]

i think there's smth in napkin about this

can anyone give me list of the properties that can be inherited going from R to R[x]

[tranquil parcel]

It wildly depends on what R is, eg if R is a field, R[x] is a euclidean domain. But it usually isn’t when R is just a mere commutative ring.

[prisma shuttle]

just commutative rings

like i know noetherian and integral domainess is preserved

anything else

[hidden haven]

UFD too

[prisma shuttle]

oh nice

also can someone explain in R[x] why is ideal (2) different from (2,x)

[chilly ocean]

the latter contains elements of the form 2f(x) + xg(x)

[hidden haven]

In particular (2) don't have x

[prisma shuttle]

doesn't the first also though

oh wait shoot

u have to have a nonzero power of 2 right

[hidden haven]

Ye

[chilly ocean]

not if g(x) isn't divisible by 2 yeah exactly

[prisma shuttle]

oh i see ok thx so much

[chilly ocean]

speaking of ideals with two generators, I'm trying to show that any ideal $I$ containing that kernel from earlier, $K=(x^3-3x^2+3x-y-2)$, can be generated by two elements. I noted that if we take $i(x,y) \in I$, $i(x,y) = i(x,y) + k(x,y) - k(x,y) = i_2(x,y) + k(x,y)$ with $k(x,y) \in K$
maybe this is a red herring route but I figured making some kind of sum would help since I eventually want that there are $a_I(x,y), b_I(x,y)$ so that for all $i(x,y) \in I$, $i(x,y) = v(x,y)a_I(x,y,) + w(x,y)b_I(x,y)$

[cloud walrus]

zd

[chilly ocean]

v and w unrestricted (this is all in $\mathbb{C}[x,y]$ btw)

[cloud walrus]

zd

[chilly ocean]

pls no full spoilers but I think I need a lil push in a good direction

I remember looking at this last week and being like uhhhhh... yeah idk lol

well if you evaluate through the homomorphism, $i(t+1,t^3 -1) = i_2(t+1, t^3 - 1)$

[cloud walrus]

zd

[chilly ocean]

uhh nvm idk lol

[neat valley]

Suppose M is an abelian group, which also has a structure of a (finite dimensional) vector space over F. Is this vector space structure uniquely determined up to isomorphism?

[barren sierra]

ok so like

6ii, 7, and 8

I have no idea how to do these quickly

these are on a practice midterm for me and I understand in theory how to do them but doing them quickly I have no idea

[lavish nexus]

What is C(x) and Cl(x)

[barren sierra]

oh my bad

C(x) is the centralizer of x

and Cl(x) is the conjugacy class of x

and N( <x> ) is the normalizer of <x>

[chilly ocean]

you kinda a god for being able to do any of this quickly lmao

completely forgot how this works outside of the class equation

[lavish nexus]

Every cycle only commutes with its powers

or something that is completely disjoint

and all cycles of the same length are conjugates

That should give you 6)

[barren sierra]

the numbers I got

but like

the isomorphisms

so like the centralizer of (1 2)(3 4) has 8 elements

is that Z_8? D_8?

ok well

clearly not Z_8 cause nothing has order 8

but like you get the idea

or better

centralizer of (1 2)(3 4 5) has 6 elements

is that S_3? Z_2 x Z_3? D_6?

[next obsidian]

The middle one is just Z_6

[barren sierra]

sure

[lavish nexus]

well it’s two disjoint cycles

[barren sierra]

oh yea so no elements of order 6

[lavish nexus]

so Z6

[barren sierra]

oh wait yea

there are 💀

Hm ok

ok then what about 7?

some of it is easy-ish

like order 2, that's elements made of 1 or 2 disjoint 2-cyces

[lavish nexus]

orders can be 1 2 3 4 5 6

[barren sierra]

yea

I could have 5 choose 2 transpositions or 5 choose 2 * 3 choose 2 / 2 I think products of 2 disjoint 2-cycles?

wait no that sounds too big

[lavish nexus]

that’s about right

[barren sierra]

hm ok

[neat valley]

Anyone have any ideas on this?

Not a hw question btw

[next obsidian]

Hmm

[barren sierra]

I saw that

[next obsidian]

So first of all I assume the structure of vector space agrees with its abelian group structure

[barren sierra]

I want to intuitively say no

but I have no idea why

[next obsidian]

As in, the + is the same

[barren sierra]

that's what I presumed too

[next obsidian]

I also think no

So the structure of vector space

Is the same as a map from F -> End(M)

Where this is the group endomorpjisms of M

Literally all this is saying is, any element f in F determines a map on M

[hidden haven]

It should be true for ℚ I guess

[next obsidian]

Namely, x -> fx

Now, given a vector space structure on M

Aka a map F -> End(M)

We could compose with any automorpjism of F

To give a new vector space structure on M

I guess I didn’t have to phrase it this way but

[hidden haven]

So good

[next obsidian]

In terms of like elements

[hidden haven]

[next obsidian]

If you had the original scalar multiplication

[neat valley]

Yes, but lets say I have a vector space M_1 and M_2, each with the same underlying abelian group M, but different induced vector space structures F --> End(M_i). Are M_1 and M_2 necessarily isomorphic as vector spaces?

I agree there can be different ways to define a vector space structure, but I wonder if there can be non-isomorphic different ways

[next obsidian]

Oh non-isomorphic

I mean yeah omegalol

Here’s an extreme example

But this should probably convince you non-extreme examples exist

Just let V be a a vector space over F

Define a new structure by setting f•v = 0 for all f, all v

Oh this doesn’t work

MonkaS

[hidden haven]

Module moment

[neat valley]

lol

[next obsidian]

What about like

Yeah idk

Lol

[hidden haven]

It needs to be over an infinite field

There's my contribution

[next obsidian]

I feel like you can frame this in terms of the like F -> End(M) thing

Like maybe isomorphic structures on M correspond exactly to automorphisms on End(M) or something

And then you just need to find examples of those maps which don’t differ by any automorpjism

But I’m not sure

[hidden haven]

I don't think vector space maps correspond to anything nice on F → End V

[neat valley]

Damn. If it helps, the inspiration for my question is follows:

I'm being asked to show the Euler characteristic of a chain complex C of free Z-modules is the same as the alternating sums of the dimensions of the homology vector spaces of the chain complex C \otimes_Z F for any field F. We are specifically asked to do this via the universal coefficient theorem for homology.

My confusion is: the UCT gives the homology H_n(C\otimes F) as abelian groups, not F-vector spaces, so is there a way to naturally "recover" the F-vector space structure of the abelian group H_n(C\otimes F) given by the UCT?

[hidden haven]

Doesn't UCT also do it for F vector spaces

[next obsidian]

I think homology doesn’t care about what you’re a module over tho

[hidden haven]

I think it applies to any R-modules

[next obsidian]

Because it’s computed set theoretically

[hidden haven]

As long as you take Ext over that R

[next obsidian]

But if Exts come in, then yeah it matters

Also can you phrase this in terms of bases?

We know isomorphic iff same dimension

But yeah, idk

This is a question I don’t like

[neat valley]

I'm talking about UCT for homology, so no Ext stuff

Specifically, we are supposed to set R=Z, is what my professor told me to do

But then the splitting gives a direct sum of two Z-modules

So I don't really know how to talk about the dimension of the vector space, when the UCT doesn't even give a vector space!

[next obsidian]

Well

So

These have a natural structure as an F-vector space I assume

This is me spitballing here

If you can show those maps are F-linear

[neat valley]

Unless, given any abelian group M, every F-vector space structure on M has the same dimension

[next obsidian]

All you need is the fact that this sequence is exact

Then because it’s over a field, it’s split exact

And you can do your dimension calculations

I mean you can even use rank-nullity to get info about the dimensions as F-vector spaces

This is one way to maybe try and get at what you want, but I’m not sure about what the specific setup is and how it plays out so maybe it doesn’t work

[terse crystal]

I already answered that😂

Here

[barren sierra]

I thought I understood how you got that and I went back and realized I didn't

[neat valley]

I think this is the correct answer, yea

Namely, for any abelian group A and field F, we can give A \otimes_Z F an F-vector space structure by extension of scalars, and similarly Tor_1^Z(A,F)

The UCT is really frustrating imo, I can't find any resources that explicitly state anything about how the R-module structure is preserved if you compute using Z

[terse crystal]

Rank is defined that way at least on cohomology of groups by brown. He defines rank A to be dim(Q otimes A) on page 242.

[terse crystal]

this ? direct product of (S_λ_i Π(Z/iZ)^λ_i) in terms of i where λ_i doesn’t equal zero

For any b from S_n, b is from C(a) iff b maps any cycle of a to a cycle of a preserving the order. You are okay with this step so far?

Like a=(143)(25) ba=ab , b must map the 3-cycle of a to that 3-cycle so the b(1),b(4),b(3) can be 1,4,3 or 4,3,1 or 3,1,4. Similarly b(2),b(5)=2,5 or 5,2

[pastel cliff]

in the group (Z, -), every element has order 2 right?

[terse crystal]

When a has λ_i many i cycles, we only consider those i where λ_i is non zero. Then b maps cycles to cycles, so it first can be viewed as permutation of those λ_i many i-cycles, next if we consider the detail, there are i ways to map a i-cycle to another i-cycle, to (Z/iZ)^λ_i

Take direct product then it’s finished

[chilly ocean]

group? a-(b-c) = a-b+c

associativity moment unpogged

[pastel cliff]

https://tenor.com/view/flight-flights-flightreacts-scream-screaming-gif-23350989

im tired

the question is to prove that there exist infinite groups with no elements of infinite order

and i think i can just be like

consider a group such that the relation a^2 = id for all a in G

[barren sierra]

Consider infinite direct products of cyclic groups

[gritty sparrow]

I know that this isn't what you were after in the end, but i think I can put two vector space structures on the same abelian group as follows: consider C (complex numbers), By choosing a transcendence basis over Q (which will have uncountable cardinality) and mapping that basis into itself by an injective but not surjective map, we can get an inclusion of fields from C to C that is not surjective and we can even make the second copy of C be transcendental over the first. So that second copy of C has the structure of an infinite dimensional C vector space but the additive group is still that of C, so the additive group of C is such an example.

[chilly ocean]

this suffices as a proof that chmonkey fans are not necessarily isomorphic to a subchmonkey

[pastel cliff]

chmonkey group

[pastel cliff]

what if i choose not to tho

[chilly ocean]

too late you already did, if you mean these are relations on a free group with an infinite set of words xD

I think that makes sense though but I'm tired too

[pastel cliff]

,ti

[cloud walrus]

The current time for nitezba is 01:42 AM (EST) on Wed, 23/02/2022.

[pastel cliff]

[chilly ocean]

so you only get 1st powers of words, so you can string together stuff

[next obsidian]

Chmonkey

[chilly ocean]

that condition actually implies everything commutes I think

but you have infinitely many distinct elements at the very least so it works

[pastel cliff]

suck my dick and balls professro

[next obsidian]

Why does that group exist

Lol

[pastel cliff]

i really want it to

[next obsidian]

I mean you can take infinite product of Z/2Z

[chilly ocean]

I think "an infinite group" is too vague tho ;-;

yeah

[next obsidian]

And that does work for that

But like saying “take X such that…”

You need to know such an X exists

[gritty sparrow]

counter example to FLT: consider integers st x^3+y^3=z^3

[pastel cliff]

wdym infinite product

[next obsidian]

Like

It’s an infinite Cartesian product

[barren sierra]

Ye do what I said and you get your desired a^2 = id for all a in their group

[next obsidian]

And the operation is done component wise

$\prod_1^\infty Z/2Z$

[cloud walrus]

CHMOLUMBIA

[next obsidian]

This guy

[barren sierra]

\bigoplus

[pastel cliff]

havent seen this before

[next obsidian]

You can use either direct sum or direct product

It don’t matter

I mean just take countable tuples

And define (a1,a2,…)•(b1,b2,…) = (a1b1,a2b2,…)

[chilly ocean]

since everything commutes as well in the construction I was thinking of with infinitely many distinct words I think it's actually isomorphic to an infinite product of Z/2Z so we came up with the same thing xd

[pastel cliff]

i thought about Z\2Z but discarded it cuz it's not an inf. group

how is this infinite

[next obsidian]

Bro

[pastel cliff]

ik im being dense

[next obsidian]

Elements are infinite tuples

(a1,a2,…)

Where a_i are in Z/2Z

[chilly ocean]

for example an element would be (0,1,1,0,0,0,0,0,0,0,....)

[next obsidian]

Like you have the infinitely many elements which are

1 in the i-th place

And 0 everywhere else

Taken over all i in N

[pastel cliff]

so the group is all "combinations" of products of elements of Z\2Z (just 0 n 1 ofc)

[next obsidian]

Yeh

Take just countably many

[pastel cliff]

again, ik im being dense i just havent seen this before

[next obsidian]

To make writing stuff down easy

Uncountable tuples are

[pastel cliff]

okie

[next obsidian]

I mean have you seen Cartesian products before?

And like infinite Cartesian products?

You can think of these as countable sequences

And you just multiply (add) them like term-wise

[pastel cliff]

i mean yeah ive seen cartesian products before ofc, but i dont think inf. ones

[next obsidian]

Oh

Well they work… sorta the same lol

[pastel cliff]

id imagine

[next obsidian]

You probably have to start defining them in terms of functions but

Whatever

[pastel cliff]

just an inf large ordered "pair"

[next obsidian]

This will become

[pastel cliff]

i am tired.

[next obsidian]

Yeah

[pastel cliff]

wtf is GL(2,Q)

rational 2x2 matrices?

[proud bear]

yeah

with nonzero determinant too

[pastel cliff]

ye ik GL, the 2 in the parentheses just threw me off

merci

[proud bear]

de rien

[lethal dune]

what does it mean to generate ring over a ring?

[next obsidian]

As an algebra

Think of it as taking all polynomials in the generators with coefficients over the other ring

As in, if A is generated over R by say x1,…,xn

Then any element of A can be written as f(x1,…,xn) where that’s a polynomial in x1 through xn, with coefficients in R

If “polynomial in x1 through xn” doesn’t quite make sense, it means you take a formal polynomial f(X1,…,Xn) in R[X1,…,Xn], then plug in the (x1,…,xn) inside of it

[gritty sparrow]

this looks like it is for non-commutative rings, so you have to look at non commutative polynomial rings instead of the usual

[next obsidian]

Oh my god

[lethal dune]

[next obsidian]

I mean it’s whatever

[lethal dune]

non-commutative polynomial algebra

[next obsidian]

It works the same

Yeah it just means that like

[lethal dune]

sad

[next obsidian]

X1X2 ≠ X2X1

Oh but

Are a and b

Just formal things here?

Like did they get like,

Magiced into existence?

Or are a and b elements of some other ring

[lethal dune]

a, b aren't elements of R, I think

[next obsidian]

But like

Okay

Is this the entire statement?

Or is there more context?

[lethal dune]

then how do I construct the isomorphism

without using

[next obsidian]

Is there any more context

[lethal dune]

no

[next obsidian]

Okay

[lethal dune]

that's all I have

[next obsidian]

So actually what’s happening is

This is just the ring

R[x,y]/(x^2, xy + yx - 1)

a corresponds to the image of x

And b corresponds to the image of y

[lethal dune]

okay

[next obsidian]

What they’re writing there is like a presentation of an algebra

Which you’ve probably seen in the context of groups

[lethal dune]

this part makes sense to me, just the isomorphism is not trivial yet

[next obsidian]

Yeh

Idk the isomorphism

I bet x and y correspond to like

Nah idk

Lmao

Gl

Kekw

Okay well

You can at least use

[lethal dune]

like I was thinking $b = \m{ 0 & 0 \ 1 & 0}$ and $a = \m{0 & 1 \ 0 & 0}$

[lethal dune]

but idk this will help

[next obsidian]

I don’t think b should have the relation that b^2 = 0

[lethal dune]

we aren't allowed to use categories

[lethal dune]

oh true

then idk

[next obsidian]

You know that ab + ba = I

Yeah I’m not sure

[lethal dune]

yeah how do I choose b

doesn't look trivial

[delicate orchid]

Universal property of the free ring

[lethal dune]

I'm tempted to write it, trust me

[next obsidian]

I mean constructing the map isn’t hard

You just need to know what to send b to

[lethal dune]

that's like the hardest part

[next obsidian]

Also I think you need to include x somewhere

Inside one of what a and b goes to

Otherwise your map will never be surjective

I go to bed now

[lethal dune]

ok like are they even isomorphic?

[hidden haven]

wew?

[lethal dune]

free ring a thing?

[hidden haven]

Ye

Free commutative ring on a set S is Z[S]

Free ring is this but non commuting polynomials

[lethal dune]

oh

[hidden haven]

You can prove that this has the universal property

[lethal dune]

over Z

[hidden haven]

All rings are over Z

In the sense that a ring is exactly a Z algebra

More generally, the free R-algebra over R generated by S is R<S>

[lethal dune]

what is this?

[hidden haven]

Where by <> I mean non commuting polynomials

[hidden haven]

Variables don't commute

[lethal dune]

ok so we need one ring to define it

nvm

[hidden haven]

[lethal dune]

also

[lethal dune]

welp

[hidden haven]

R-algebra I suppose

[lethal dune]

how do I show the isomorphism

[hidden haven]

[lethal dune]

NO

I'm not allowed to use it

[hidden haven]

Says who

[lethal dune]

our course

[hidden haven]

dum course

[lethal dune]

ye ikr

[hidden haven]

How quotient when not first isomorphism theorem

[lethal dune]

how do I do it using

I still have to show it's iso to M2(R[x])

[charred flume]

Hi is a U(n) group the same as Z/nZ ?

this kinda confuses me

[charred flume]

hmmm

[hidden haven]

Find some surjection with that kernel idk

[chilly radish]

This book I'm reading claims that if $G$ is the grassman algebra generated by countably many basis elements $e_i$, and $G^{(0)}\oplus G^{(1)}$ is the natural grading into the span of even and odd length monomials, then $[G,G]\subseteq G^{(0)}$, but this seems wrong to me since it also claims $G^{(0)}G^{(1)}+G^{(1)}G{(0)}\subseteq G^{(1)}$

[cloud walrus]

ShiN

[lethal dune]

it isn't trivial

[chilly radish]

So if you take an odd homogeneous element and an even homogeneous element and take their commutator, the result would be an odd element for example

They use this to prove the commutator identity [[x,y],z]=0 is a polynomial identity for the grassman algebra (Since G^(0) is central), but this reasoning seems wrong

So what's going on here

Wait nvm

If u commute an odd and even homogeneous element u get identically 0

I guess it's enough to check this for monomials since the commutator is bilinear

[chilly radish]

Of course, because G^(0) is central...

[hidden haven]

bruh is this just exterior algebra

[chilly radish]

Yea

[hidden haven]

Who calls it grassmann algebra

[chilly radish]

PI algebra people

[hidden haven]

Private investigator?

[chilly radish]

Lmao

[hidden haven]

[chilly radish]

Polynomial identity

[hidden haven]

😵‍💫

[delicate orchid]

My polynomial identity is x^3-1

[hidden haven]

Shuri's is x⁵-2

[chilly radish]

This won't be a polynomial identity for any nonzero algebra

[hidden haven]

You can't reject someone's polynomial identity like that, please apologise

[chilly radish]

I'm not

I'm just saying it's trivial

[hidden haven]

oh ok

[delicate orchid]

I’m a trivial kinda guy

[chilly radish]

Commutator bracket of algebra is bilinear right. I don't feel like doing the computation lmao

Associative.algebra

[hidden haven]

AB - BA right?

Yes

[chilly radish]

Yea

Ok thanks lol

[hidden haven]

23 mins to verify a 23 sec check

[chilly radish]

I wasn't in a rush btw

I had to stop reading anyways

Otherwise id've checked on my own

[hidden haven]

Sure.

[regal tulip]

Hello fellow abstract-algebrists
Does anyone know if the group presented by

<a, b|aba(b^-1)>

Is a known group?

[coarse storm]

I am not sure how to evaluate the predicate as a boolean.

Or hmm... is it supposed to be <abab^{-1}: a,b in G>?

[hidden haven]

That's the presentation of a group by generators and relations

[regal tulip]

It's the group presented by the symbols a, b and by the relation
a b a b^(-1) = 1

I was calculating the foundamental group of the Klein bottle with SVK theorem and I got this

[delicate orchid]

It’s F(a, b)/<abab^-1> :troll:

[lethal dune]

technically the truth

[delicate orchid]

have I ever been wrong?

not counting messages you can't see without scrolling up or searching

[lethal dune]

✓

[regal tulip]

Well not in my professor notation apparently

[regal tulip]

Well if you don't answer my actual question we'll never know

[delicate orchid]

I literally just worked backwards from the definition of a presentation the statement itself is meaningless if you've actually got a presentation already

ab = ba^-1 gives me some kinda generalised dihedral group vibes ngl

[lethal dune]

b²=1 needs to be satisfied for that

[regal tulip]

Why do you say so? For example:

<a, b| ab(a^-1)(b^-1)>

Is just Z x Z

I was wondering if something similar might apply to the presentation I got

[delicate orchid]

hence, "generalised"

[delicate orchid]

a presentation <a_1, a_2, ... |R_1, R_2, ...> is naturally constructed via a quotient of the free group on a_1, a_2,... specifically quotienting by the relations

I'm confused as to what you're confused about

[regal tulip]

Yup, agreed

[regal tulip]

I was confused about why you said my statement was meaningless

[delicate orchid]

yeah so I was making a lighthearted quip by just saying that's what the group is called

no no no

sorry

MY statement is meaningless

[regal tulip]

Oooh, k
Sorry, I didn't understand😅

Well

[regal tulip]

At least now we know the answer

[delicate orchid]

I've seen many finite versions of this group with the same ab = ba^-1 relation

dihedral or otherwise - no clue what they're called though

[regal tulip]

I'll check them out then, thank you

[delicate orchid]

yeah, maybe you can generalise some stuff to the case where a and b aren't of finite order

[regal tulip]

The only thing I know about presentation groups is what is needed to understand SVK theorem (that's why our professor introduced them)

So basic stuff

Universal property

And not much more

[barren sierra]

Prayers up for me I got a midterm 🙏

[chilly radish]

Actually not technically true, should be quotienting by <<abab^-1>>, <abab^-1> might not be normal

<<>> is normal closure

Semantics

[proud bear]

Good luck 🙏

[wooden ember]

isn't that the fundamental group of the klein bottle?

i suppose that's well studied no?

[chilly radish]

Yes, they asked about it because they reached it after computing fund group of klein bottle

I think the question was more like 'does this have a name'

[wooden ember]

oh lol okay

also ShiN be yellow

[spice whale]

how long did the isomorphism theorems take to prove

[delicate orchid]

how long? with relation to when?

like how long after the concept of a homomorphism was established?

[chilly ocean]

Still lost on this, any hints appreciated 👍💩

[tender mist]

Use the fact that $\frac{I}{K} \vartriangleleft \frac{\mathbb{C}[x,y]}{K} \sim \mathbb{C}[t]$ is an ideal of a PID

[cloud walrus]

Mat

[chilly ocean]

Thanks a ton for the hint I'll give it a try!!

[coral shale]

isnt that slightly diff from what you wrote though

You take the smallest normal subgroup containing the relations

not just the smallest subgroup generated by them

?

[regal tulip]

Basically

So my guess is that it's not anything known
A part from being the fundamental group of the Klein bottle
Damn it

[coral shale]

it probably does

[delicate orchid]

yeah shin corrected me on that
it's just a little notation trolling we do a little trolling it's called we do a little trolling

[coral shale]

ah oki

[chilly radish]

ok so just a notational question, in the context of graded rings/algebras, does R_mR_n mean simply the set of all products of elements, the linear span of all products or the algebra generated by all products

[coarse storm]

Typically, finite sums of products.

I... think you usually have commutativity with this?

At least, I had only came across it in such contexts. In which case, we have finite sums of r_m r_n terms.

[chilly radish]

You don't have to have commutativity

[tribal moss]

You definitely don't want the algebra generated by all products, that would make the main example of the polynomial ring graded by monomial degrees not work.

[chilly radish]

Yea I figured

I think in the case of algebras we take the linear span of products

I mean tbh for the graded structure it doesn't really matter because if all the products are in certainly their linear span is in

[pastel cliff]

is this a valid proof of the statement "A group $G$ that satisfies the relation $\alpha^2 = 1$ for all $\alpha \in G$ must be abelian."

[cloud walrus]

μ₂

[pastel cliff]

i could be more explicit but im trying to work on my verbosity, i tend to write proofs that are too long lol

[coral shale]

It can be very short symbolically

I would rather write symbols but justify each line on the right

saying exactly which law is used

you are also trying to prove ab = ba

Not your final line

[tribal moss]

Not necessarily: consider R[X,Y] graded by total degree. Then R_1 contains just linear combinations of X and Y, but R_1·R_1 = R_2 contains things such as X^2+Y^2 which is not itself a product of R_1 elements.

[coral shale]

small technicality

[next obsidian]

Doing it with inverses is completely fine

[tribal moss]

(R is here the real numbers).

[pastel cliff]

this is what i was worried about

[next obsidian]

They justified it because enumerating over inverses is the same as enumerating over all elements

[coral shale]

I mean since the proof is so simple, might as well do it properly

[next obsidian]

Just take a = a^-1

And b = b^-1

This is a proper proof

It’s equivalent to enumerate over the inverses

[pastel cliff]

that's what i figured

[chilly radish]

I meant the condition RmRn\subseteq Rm+n, but not necessarily strict equality

[next obsidian]

Like if you want to be super anal

[tribal moss]

Ah, right.

[next obsidian]

Just take a = x^-1 and b = y^-1 for arbitrary x,y

Woah now you proved the statement

[coral shale]

Let $a,b\in G$
$$ab = a^{-1}b^{-1}$$
(a and b are self-inverse)
$$= (ba)^{-1}$$
(property of inverses)
$$= ba$$
(ba is self inverse)