groups-rings-fields

[next obsidian]

The entire point of the project

Is to come up with a closed form function?

[toxic zephyr]

for a major portion of it ye

[next obsidian]

Yeah I mean why does it have to be closed form lol. Like are you trying to use the function to do something

Or is that just the goal

[toxic zephyr]

pretty much this. it would also make other portions of the project easier.

[next obsidian]

Okay well good luck then, I really don’t see how you can do anything other than just try things ad nauseum

[toxic zephyr]

thanks. i was able to do it for the dihedral group, and it wasn't that bad since sr^a is a self inverse. but once the powers of s get crazy like this, it's a huge pain.

[next obsidian]

Yeah I just… I don’t think you can be clever

The fact it doesn’t respect the group structure means you basically can’t use algebra at all

[formal ermine]

write a computer program to find it for you

[toxic zephyr]

wdym by not respecting group structure?

[next obsidian]

It isn’t multiplicative

Like this question isn’t a group theory question

Because you can’t use any group theory

[toxic zephyr]

word maps typically aren't multiplicative, unless I'm misunderstanding what you're saying

[next obsidian]

I don’t know what a word map is

I’m just saying f(xy) isn’t f(x)f(y) so your function won’t be a group homomorphism

So it leaves the things which group theory can be applied to

[toxic zephyr]

basically a string of variables and constants/group elements.
like f(x)=x^k1 g1 x^k2 g2...
it isn't typically a group homomorphism, yeah

[toxic zephyr]

thanks anyway. i'll keep trying

[swift totem]

Let $G$ be a finite group with an automorphism $\sigma$ such that the only fixed point of $\sigma$ is $1_G$. I claim that every element of $G$ can be written in the form $x^{-1}\sigma(x)$ for some $x\in G$.
$\~\$
Consider the map $\psi\colon G\to G$ defined by $x\mapsto x^{-1}\sigma(x)$. If $\psi(x)=\psi(y)$, then $x^{-1}\sigma(x)=y^{-1}\sigma(y)$, then $1_G=xy^{-1}\sigma(y)\sigma(x)^{-1}=xy^{-1}\sigma(yx^{-1})=(yx^{-1})^{-1}\sigma(yx^{-1})$, or that $\sigma(yx^{-1})=yx^{-1}$. But since $\sigma$ only fixes $1_G$, we must have $yx^{-1}=1_G$ or that $x=y$ thus showing $\psi$ is an injective map on $G$. But since $G$ is finite, $\psi$ must be surjective as well.
$\~\~$
Hence, every element $g\in G$ can be written as $g=\psi(x)=x^{-1}\sigma(x)$ for some $x\in G$

[cloud walrus]

Anon581

[swift totem]

Can someone proofread this? And perhaps even suggest some simpler argument to show the same if there is?

I was actually wondering if there's an argument to directly show that the map \psi is surjective without having to show \psi is injective and then exploit the fact that a map X->X is injective iff it is surjective for a finite set X

[sweet echo]

What is this $\wr$ symbol denoting? This is just the table of finite Coxeter groups on wikipedia. Semi-direct product?

[cloud walrus]

ΣAC

[hot lake]

it's a wreath product

[formal ermine]

TIL

[sweet echo]

interesting thankyou!

[coral spindle]

If you’re somewhat familiar with reflection groups, $C_2 \wr S_n$ is isomorphic to $G(2,1,n)$

[cloud walrus]

Boytjie

[prisma ibex]

GO_n
mom pick me up I'm scared

Lie theory should study groups like SO_n not PGSp_2g

[chilly ocean]

what are some major combinatorial/geometric group theory applications in computer science and where can I find info about it?

[tender wharf]

google [arbitrary search engine] is your friend

[formal ermine]

how do we show that $\bZ[i\sqrt{3}]$ is not a ufd?

[formal ermine]

ah

we use that $p$ prime $\iff p$ irreducible

[formal ermine]

so we gotta find a prime element that is not irreducible?

or the other way around

[dull nacelle]

There are quite a few use-cases in cryptography. E.g. elliptic curve cryptography is beginning to play an important role for communication due to its benefit of having a shorter key size as well as the fact that the patents on ECC are coming to an end. The ideas behind using elliptic curves for cryptography were in part derived from the elliptic curve group

[formal ermine]

2 is not prime

cuz we have 2 | (1 + isqrt(3)(1 - isqrt(3)) but not 2 | 1 +- isqrt(3)

ok so uhh

if we look at it under the norm

4 = (a^2 + 3b^2)*(c^2 + 3d^2)

then b = 0or d = 0

so a = +- 1 or c = +- 1

so one of those is a unit

so 2 is irreducible

ok epic

[chilly ocean]

Google is my biggest enemy

[tender wharf]

what about duckduckgo tho

[chilly ocean]

where can I read more assuming I only know undergraduate level stuff?

[coral spindle]

I'm afraid the semigroup theorist in my office didn't know what N_5 and M_3 were, or what modularity/distributivity meant. I'm happy to try and ask some of the professors, but would you mind sending a reference or some text where this occurs? It just makes it easier to approach the higher-ups if I have that

[chilly ocean]

It's on lattice page on wikipedia

And should be on distributive lattices page too

[coral spindle]

OK thanks, I'll give it a look and send an email or two

[chilly ocean]

Someone from universal algebra should know

[coral spindle]

I'm not familiar with it myself, but someone I know is looking into an application of ggt to neural networks. The idea is that certain functions they want to train for should be symmetry-invariant, so if they average the weights over the action of the group of symmetries of their input space, they can train their net for free.

[formal ermine]

having a bit of a notational issue

I saw that shuri

so S_n acts transitively on the roots { alpha_1, ..., alpha_n } of an irreducible polynomial f because for any alpha_i, alpha_j we can take sigma = (ij) where sigma . alpha_i = sigma(i) = j = alpha_j by definition

the last part looks wrong

how do I fix this

ignore the content

might be wrong idk

this is from a very different thing that would require way more context

this is just a question about the notation :p

[coral shale]

if u tex it

[rotund aurora]

you dont

need to

press enter

every two words

[tender wharf]

[formal ermine]

I like posting incoherent thoughts as individual messages >.<

[coral shale]

tex this notation

[formal ermine]

so $S_n$ acts transitively on the roots $\Set{ \alpha_1, \ldots, \alpha_n }$ of an irreducible polynomial $f$ because for any $\alpha_i, \alpha_j$ we can take $\sigma = (ij)$ where $\sigma . \alpha_i = \sigma(i) = j = \alpha_j$ by definition

[coral shale]

whats sigma . alpha i

[formal ermine]

the last equation looks very wrong to me

idk how to fix the notation

[formal ermine]

sigma operated on alpha i

[coral shale]

arent u using sigma to mean 2 different things.

[formal ermine]

like the group action

[coral shale]

u have sigma on roots

and then u decided to have sigma on indices

choose one.

[formal ermine]

S_n acts on the roots by sigma . alpha_k = sigma(alpha_k) or sigma(k)?

yeah idk

[coral spindle]

Usually we would define $\sigma . \alpha_k = \alpha_{\sigma(k)}$

[cloud walrus]

Boytjie

[coral shale]

,,\sigma = (\alpha_i,\alpha_j)

or u can do this

[formal ermine]

yes

that's what I was trying to say lol

ok thanks!

[formal ermine]

$\sigma . \alpha_i = \alpha_{\sigma(i)} = \alpha_j$

epic

[formal ermine]

thanks boytjie and shuri

[coral spindle]

no worries

[white yoke]

How can I prove that these representations are not isomorphic?

I tried proving that they have different character but it turns out they have the same character.

[coral spindle]

I would assume the existence of an isomorphism and derive a contradiction

I haven't worked this through fully, but I think that works. (Edit: I checked it and it works.)

I think you will find that compatibility with the action of x and the action of y simultaneously is not possible

[formal ermine]

Let $f \in \mathbb{Q}[x]$ with $\deg(f) = n \geq 3$ and $\operatorname{Gal}(f/\mathbb{Q}) \cong S_n$. Let $\alpha$ be a root of $f$. Show that there is only one $\mathbb{Q}-$Automorphism of $\mathbb{Q}(a)$.

---

$f$ is irreducible, as $S_n$ acts transitively on its roots. Thus, $f/\mathbb{Q}$ is a Galois extension.

Let $\lbrace \alpha_1, \ldots, \alpha_n \rbrace$ be the roots of $f$. Let $\alpha := \alpha_k$ for some $1 \leq k \leq n$. Asking how many $\mathbb{Q}-$Automorphisms of $\mathbb{Q}(\alpha)$ there are is the same as asking how big the Galois group of $\mathbb{Q}(\alpha)/\mathbb{Q}$ is. We know that $\mathbb{Q}(\alpha) = \mathbb{L}^{\operatorname{Fix}{S_n}(k)}$ where $\mathbb{L}$ is the splitting field of $f$ over $\mathbb{Q}$. Using the fundamental theorem of Galois theory, we get that $\operatorname{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q}) = \operatorname{Gal}(\mathbb{L}^{\operatorname{Fix}{S_n}(k)}/\mathbb{Q}) = [S_n : \operatorname{Fix}{S_n}(k)] = \frac{|S_n|}{|\operatorname{Fix}{S_n}(k)|}$. For this to be $1$, we would need $|S_n| = |\operatorname{Fix}_{S_n}(k)|$. Where is my mistake?

[coral shale]

(I hope you don't choose this font for your papers)

[formal ermine]

of course not because I can actually use comic sans instead of comic neue in latex documents

[toxic zephyr]

what is a transversal?

[agile burrow]

A (left) transversal is a choice of representatives for (left) cosets of a given subgroup

[simple mulch]

Suppose $x$ has order $p$ and $y$ has order $q$ with $\gcd(p,q) = 1$. We have $|\langle x \rangle \langle y \rangle| = \frac{|\langle x\rangle||\langle y\rangle|}{| \langle x \rangle \cap \langle y\rangle|} = \frac{|\langle x \rangle| |\langle y\rangle|} = pq$

[cloud walrus]

mns

[agile burrow]

Sorry, I forgot to reply. See my above response

Hall's marriage theorem is cool. One consequence of it is that given a finite index subgroup H of G, there is a subset of G which is simultaneously a left and right transversal

[coral spindle]

Now I want to think of a counterexample in the infinite-index case

[agile burrow]

i'm not actually sure if it's true or false in the infinite-index case, so please feel free to ping me if you think of one

[toxic zephyr]

that's really interesting! thank you 😊💖

[formal ermine]

.

[woeful flint]

Given a field tower $\mathbb{Q} \subset L \subset K$ does knowing the ring of integers $\mathcal{O}_L$ tell you anything about the structure of $\mathcal{O}_K$. For example, if $[K:L] = 2$ then $K = L(\sqrt{\alpha})$ for some $\alpha \in L$ what can we say about $\mathcal{O}_K$ by knowing $\mathcal{O}_L$.

[cloud walrus]

Kraft Macaroni

[next obsidian]

Well O_K is also the integral closure of O_L inside of K

This might make it easier to compute O_K

But this might be better for #advanced-number-theory since I bet this is really more of a number theory thing

[woeful flint]

Ah fair I'll ask there

cheers man

[formal ermine]

ok so I found another proof for it that works but I still couldn't figure out where this breaks

[chilly radish]

Q(a) is not necessarily galois since it's not necessarily a normal extension, so its automorphism group isn't gonna follow the fundamental theorem in terms of the number of elements

The formula you wrote is still true for the degree of Q(a) over Q (in fact you can see it's n because you adjoin a root of an irreducible polynomial of degree n) by the fundamental theorem, but the automorphism group might be smaller

All you can say in the general case is that the size of Aut(Q(a)/Q) is bounded above by n

[formal ermine]

I was trying to look at Q(a) as an intermediate field extension of L/Q

[chilly radish]

Yea

Again, the formula for the degree is true

But that's the size of the aut group.iff Q(a) is normal

[formal ermine]

Ah

got it

[chilly radish]

👍

[formal ermine]

thanks shin

[chilly radish]

Np

I hope this also makes clear why they take n>=3

In the case n=2 it's obvious that this CAN happen (That Q(a) is normal)

[formal ermine]

yeah quadratic extensions are always normal

[chilly radish]

Yee

[formal ermine]

we also in theory would need the extensions to be separable, but that's given because Q is char 0, right?

[chilly radish]

Indeed

[formal ermine]

why does f(x) = k(x^n - m) for some k,m imply that its galois group over Q is at max of order n^2

[south patrol]

well you can kinda write down explicitly a field extension in which f splits

have a go at that

ofc k is irrelevant if it's non-zero

[formal ermine]

ah and then we just get the degree of that and use |gal| <= |extension|

ok thanks

[simple mulch]

This is in the proof of the classification of groups of order 12, why don't we consider the case $a = b = -1$ and $ab = 1$?

[cloud walrus]

mns

[simple mulch]

oh, nvm

two of them are qual to -1

[dull nacelle]

Introduction to Mathematical Cryptography by Hoffstein should be good

There's an elliptic curve chapter in there

[chilly ocean]

thanks

[south patrol]

Btw where should I look for seeing the link between Ext^n and equivalence classes of extensions? I've seen n=1 but not sure how it generalises (i want to look at Yoneda products)

[rustic crown]

i think i heard Ext^n is isomorphism classes of extensions of length n

so exact sequences like
0 --> M --> E_1 --> ... --> E_n --> N --> 0

and the operation Ext^n(N, P) x Ext^m(M, N) -->Ext^n+m(M, P) is then splicing up two such sequences

never verified that tho

this also tells you why Ext^0 = Hom

[south patrol]

Oh I mean I know all that stuff I just meant like how do I prove the relation to the typical Ext in terms of derived functors

(thank tho)

I guess maybe the best way to do it is to use derived cats

[rustic crown]

i have a feeling i probably got the order of M, N wrong again

[bleak abyss]

My channel now

Not going for understandability here jsut blasting through

If you're curious about what my notation means look up "Spherical Functions on a Group of p-adic Type" by Macdonald

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Horrible notation actually, call it p

[cloud walrus]

Sloth King Daminark

[bleak abyss]

And that should be fine by the above

Moving on

[cloud walrus]

Sloth King Daminark

[molten viper]

Hey folks, I’m looking at a proof in a text. I’ve got 2 rings A and B. I know A isnt the zero ring, and the only principal ideals of A are (0) and (1). My text claims that if I have a homomorphism phi: A —> B, then Ker(phi) ≠ (1). Why is this? Isn’t phi(x) = 0 a perfectly good homomorphism?

[chilly ocean]

your book might require ring homomorphisms to preserve 1

[bleak abyss]

God I hate macdonald so much lol

[molten viper]

I missed this, that’s correct. Thanks!

I’m still in group/vector space mindset

[molten viper]

My qualm with it currently is notation being horrendous

[bleak abyss]

Not Atiyah-Macdonald if that's what you're thinking of

[molten viper]

Oh

That’s what I thought of haha

I am in fact working through Atiyah Macdonald so it’s on the brain

Question the second: this refers to a “natural homomorphism” A -> B with kernel (x). Where x is a non-unit in A. What is that natural homomorphism?

Better way of asking maybe, can I guarantee the existence of a homomorphism with kernel (a) for any a?

[chilly ocean]

quotient map

[molten viper]

So specifically the homomorphism A -> A/(a)

Which I know exists as an article of faith (I believe this will be proven in a future algebra lecture)

[chilly ocean]

this is just a general thing about ideals

it's completely analogous to what you know for groups and normal subgroups

ideals are kernels of homomorphisms are the things you quotient by to get new rings

[molten viper]

Yeah I think I’ve hit the extent of my current understanding

I recall that we have an isomorphism from the quotient group given by a homomorphism to the image of that homomorphism, so this is really the exact same thing but with rings?

[chilly ocean]

yes, rings have a first isomorphism theorem too

if f: A -> B is a ring homomorphism then ker f is an ideal in A and A/ker f is isomorphic to im f via a + ker f -> f(a)

[molten viper]

Yeah that’s what I was referring to

I’m gonna try proving that

[chilly ocean]

it's pretty much the exact same thing as the group version of the theorem, but you have an extra operation to check

[molten viper]

I don’t quite recall the proof for the groups

[chilly ocean]

every condition should be straightforward to check

you just... check it

no tricks

[molten viper]

So now, to check that something is a subring, I need to check closure for both operations, the presence of 0 and a multiplicative identity, right?

Not necessarily 1, could be something different

[chilly ocean]

why not check a&m and see what they say?

i don't see where you need to check anything is a subring here. maybe if you really want to check the image is, you could

[chilly ocean]

hmmm

[molten viper]

I’m checking that the image is

[chilly ocean]

that's not really important

but you can if you want, i guess

[molten viper]

In this case it’s basically automatic from the definition of homomorphism

The wording of subring in AM is quite terse

[chilly ocean]

if you can say that confidently, then you definitely don't need to write out a careful proof that the image is a subring

[molten viper]

Well I say it confidently because i just checked it haha

Now, the elements of Ker(phi) are cosets x + Ker(phi) I believe. So am I on the right track if I try to map f(x+Ker(phi)) = phi(x)? Checking ofc that this is well defined?

[chilly ocean]

you can check if you're on the right track by continuing the proof

[molten viper]

Alrighty

[molten viper]

Alrighty, I believe I’ve completed that proof. Thanks for walking me through this/ letting me talk @ you @chilly ocean

[formal ermine]

might be a dumb question but how do we show that a galois extension is finite

[chilly radish]

depends on your definition of galois extension

like the usual first definition you see requires the extension to be finite, normal and separable

one can generalise this but that generally comes later

[formal ermine]

we defined a galois extension as just normal and separable

[chilly radish]

and the fundamental theorem in that case needs to be amended so that it still holds for infinite extensions

[formal ermine]

ah

yeah we did the fundamental theorem for finite normals so I was a lil bit confused about the online ones omitting finite

but it's just a different definition of galois extension then

oke thanks

[chilly radish]

the fundamental theorem of galois theory isn't exactly true for the infinite case as stated in the finite case. You introduce a topology on the galois group and then the correspondence is between intermediate extensions and closed subgroups

under this topology

[formal ermine]

haven't done topo yet

it's next on my list

[chilly radish]

if the sources you're looking at don't mention this, then they're implicitly assuming the extensions are finite

[formal ermine]

yeah

this might be a bit off-topic but if I use the formulas for the discriminant that use the coefficients of the polynomial instead of its roots (derived using something like vieta), would I have to prove that in my homework?

[chilly radish]

that's a question for your professor/TA

if you haven't seen it in class presumably you'd have to prove it

but we can't answer for someone else's grading policy

[formal ermine]

H is the galois group of f over Q. why do we know that its order is divisible by p exactly?

[chilly radish]

f is irred of degree p hence Q(a) is of degree p. By the tower law, the degree of the splitting field over Q is divisible by p, hence so is the order of H

[formal ermine]

ah

making a polynomial monic doesn't change the degree

right

thanks

sigma and tau are permutations, sigma_n and tau_n are their cycle decompositions, what does the thing on the right mean?

[elfin furnace]

size of the cyclic subgroups I think

[formal ermine]

wdym

[elfin furnace]

the langle rangle is the cyclic subgroup generated by whatever is in between them

[formal ermine]

ohh lmao

I literally didn't see those angles

thank you!

[coral shale]

cyclic in this case, not if theres more than 1

[elfin furnace]

yeah

[astral galleon]

Dumb question is the klein 4 group not a toplogical group ?

[runic hemlock]

no, because it isn't equipped with a topology

you could make it into a topological group by giving it the discrete topology, for instance

[astral galleon]

That makes sense tho one question still somewhat bothers me it seems anything with a discete symmetry cannot form a topolgovial group

[runic hemlock]

any group can be given a topology which makes it into a topological group

for example, the discrete one

[astral galleon]

Kk makes sense 👍

[formal ermine]

what's a p-cycle?

[chilly radish]

Context?

If it's in the context of permutation groups, probably a cycle of length p...

[next obsidian]

p-cycle is something like

(a1a2…ap)

A cycle of length p

[formal ermine]

ah

what does a^k = (01...) mean

specifically the ...

[next obsidian]

01234…p-1

Do you see the pattern?

[formal ermine]

yes

what do they mean with "we can reindex the other elements"

[next obsidian]

If it was like

03829 or something

You can just relabel the stuff

The numbers are arbitrary

But once you relabel them once you can’t relabel them again

Think of it as just composing with some auto morphism of S_n

[formal ermine]

ah ok

[chilly radish]

As opposed to a manual morphism

[formal ermine]

let $K \subseteq \bR$ be a field and $f \in K[x]$ with $\deg(f) = p$ prime. how do I show that if $f$ has exactly $p - 2$ real roots, then $\on{Gal}(f/K) \cong S_p$? I've been trying to solve this myself but didn't get anywhere, I then looked on the internetz and found a couple of solutions (see above) but all of them were talking about Q instead of R and were using symmetric group properties and stuff like that (is it even possible without those)?

[astral galleon]

Also would be helpful to think of toplogical groups as topological spaces that admit a group structure ?

[runic hemlock]

idk what would be helpful. I have 0 understanding in pedagogy

[astral galleon]

I ask because it seems its more intuitive when trying to build non topological groups or toplogical groups

[lime badge]

Question about applying the classification of finitely generated modules over PIDs to studying linear transformations on finite dim vector spaces over a field k. The image is from Chapter 6 of Aluffi's algera book and concerns using the decomposition of our vector space in terms of cyclic modules k[t]/(p_i(t)^r_ij), where p_i are the elementary divisors

My question is why the minimal polynomial has max_j(r_ij) in the exponent. Wouldn't we expect it to have min_j(r_ij) instead?

[ruby sundial]

top spaces with symmetries is the motivation

specifying the group structure tells you the symmetry you care about

[coral shale]

whats the motivation of top again then?

[lime badge]

To answer my own question, I just realized the minimal polynomial corresponds to the highest degree invariant factor

[coral shale]

why put one on a group

[ruby sundial]

yeah thats it

[ruby sundial]

having a notion of convergence or closeness of points

thats why suggestive terminology like neighborhoods and limits are used

[formal ermine]

.

[astral galleon]

Makes sense now I get it :D

[ruby sundial]

ig important structure is that multiplication and inversion are cts maps

[primal tusk]

I proved part 1

But I’m confused why part 1 doesn’t imply part 2 since a and b are arbitrary

[next obsidian]

I guess it wants you to show that there’s r,s such that rm + sn = 1

This is probably more suited for #elementary-number-theory though

[primal tusk]

Oh okay

Yea sorry this is in a modern algebra class so I just thought I should ask here

[untold basin]

Hello, if I know all the subgroups of a group G and the same thing for a group H, do I also know all the subgroups of G x H ?

[coral shale]

prove if A sub G, B sub H then A x B sub G x H

now the converse hmm is this true

interesting, idk.

[untold basin]

Maybe could you help me for my main question

I say this because I'm doing an exercice :
G(K/Q) is isomorphic to Z/2Z x Z/6Z (I've done this)

I'm asked to find how many subfields there are of K that [F : Q] = 2

where K = Q(zeta_7 * sqrt(2))

[coral shale]

this isnt galois is it

[untold basin]

it is, over Q

Because K = Q(zeta_7, sqrt(2)) and the roots of the minimal polynomial of zeta_7 over Q and the roots of the minimal polynomial of sqrt(2) over Q are included in K

[coral shale]

so it is.

so remind me, youre looking for all subgroups of index 2?

[untold basin]

exactly

[coral shale]

well index 2 subgroups are easier. i think

[untold basin]

they are normal

[coral shale]

C2 x C6
Theres the obvious C6. Then C6 has C3 subgroup. I can't think of anything else

[untold basin]

what is C2, C6 ?

[coral shale]

cyclic 2 cyclic 6

[untold basin]

Yes it is all the subgroups I have too

But for example, if I take a subgroup H of order 4 of Z/2Z x Z/6Z

(if H is order 1,2,3,12 I found that he has to be one of the subgroups I know)

What can I do with 4 ?

[coral shale]

im lost.

it has to be the C2 x C2 copy

[untold basin]

In my explanations or you don't know what to do ?

[coral shale]

yeah i feel like the converse of this has to be true

All subgroups of G x H are of the form A x B for A, B subgroups

[untold basin]

mmh I can't use a property we haven't proved in class

[coral shale]

well prove it.

[untold basin]

Okay for now lmao but I need to prove it in class if I use it

[coral shale]

Ive only stated it as a guess, i would prove it.

shouldnt be too hard

[untold basin]

I'm tired I will see tomorrow

00:30 here

thanks though

[coral shale]

Oh.

google says no @untold basin

stack exchange

[untold basin]

oh sht

xddddd

[coral shale]

yikes

maybe works for cyclic

[untold basin]

I wonder how can I do my ex then

for cyclic I don't need that

[coral shale]

or abelian to be more specific

i mean a direct product of abelian, i think it might hold here

[untold basin]

there exist a unique sub-group of order d where d divides the order of the big group

[coral shale]

yh sry i dont have much kf a clue

[untold basin]

No problem

I asked some friends, I will see tomorrow

[coral shale]

tldr, subgroups of a direct product is a non trivial problem

https://math.stackexchange.com/questions/361620/how-to-find-all-subgroups-of-a-direct-product

And 'goursat's lemma' is relevant

but yh idk

I suppose for small enough groups u can get away with explicitly laying out the entire structure

[vagrant zinc]

how do I create the multiplication table for any group, I have for U(12)

Some clue

[delicate orchid]

have you tried multiplying stuff

[vagrant zinc]

I found something about U(n) is given for Z(n) modular for multiplication and takes the relative primes.

If n is a relative prime then I select all elements e.g. U(11), 1....10 and make the table

[delicate orchid]

that sounds correct if I'm understanding your notation correctly

$U(n) = (\bZ/n\bZ)^{\times}$ and $Z(n) = \bZ/n\bZ$ I'm presuming

[cloud walrus]

Wew Lads Tbh

[vagrant zinc]

look U(12)={1,5,7,11}

[meager jolt]

Does anyone know how to prove that separable extensions are unique up to isomorphism? Is it similar to the uniqueness of algebraic closure?

[next obsidian]

Wdym?

Do you mean a separable closure?

Because any extension is separable in char 0, and there’s definitely non-isomorphic extensions of Q

[meager jolt]

oops typo

I do mean separable closure haha

[next obsidian]

Okay in that case

I think you can proceed as following

Fact: if F is the separable closure then any algebraic, separable extension of K has an embedding into F

Proof: by being algebraic the extension, let’s call it E, has an embedding into K-bar

This will be a separable extension so it lands inside of F

Then doing this for two different separable extensions give you joint embedding

You now I guess want to say something about uniqueness to get that these define an iso…

🤔

I guess it works like maybe like umm

Hmmm idk how to explain this hahahaha

But like fix two algebraic closures K’1 and K’2

And then you have two separable closures F1 and F2

You can embed F2 < F1 as I said before

Damn I should learn this proof

Time to look it up

[meager jolt]

Yeah I couldn't find it anywhere 😢 but I probably wasn't looking hard enough

[next obsidian]

I mean okay like

I don’t even think it’s unique iso?

Like that’s not true for algebraic closures

Cuz like Galois groups tell you

Exactly how non-unique it is

[glossy crag]

Is this a valid proof:
Consider the reciprocal polynomial $\Psi(x)=\Phi(\tfrac1x)x^n$, then it has as its roots all of the primitive n-th roots, since $\Psi(x)=\prod(1-x\zeta)$, so $\Phi\mid\Psi$. For $n>1$, $\Psi$ has leading coefficient 1 (since $(-1)^{\phi(n)}\prod\zeta=1$), so $\Phi=\Psi$ and symmetry of coefficients follows (and the $\pm$ is unnecessary).

[cloud walrus]

Ocean Man

[chilly ocean]

Why does this hold

[tender wharf]

what context is this

[chilly ocean]

The dot is action..

[tender wharf]

can you post more

those 2 lines literally tell me nothing

[tender wharf]

I feel like someone was talking about this a while back

lemme see if I can pull it up

im too lazy to get my pen and paper

[tender wharf]

is this (a, x)(b, y)(c, z)

[tender wharf]

ah

that's what I wanted you to post

im hella confused sorry

[chilly ocean]

can someone elaborate what this exercise wants?

[rustic crown]

what are generalized free products?

they wanted you to give a universal property, something like maps from this generalized free objects correspond to maps on the generators or something

[untold basin]

Hello again, can someone check my question here ?
I need to find all the subgroups of Z/2Z x Z/6Z but I struggle if a subgroup H is of order 4 or 6

[rustic crown]

https://math.stackexchange.com/a/488222
here is a characterization of subgroups of a direct product

but your case has small enough orders, so better to just bash it up

[rustic crown]

also for your original question, it suffices to only look for index 2 subgroups

[untold basin]

That's what I have

i.e subgroups of order 6

but I don't have them all

[rustic crown]

the group in question is Z/2 x Z/6

[untold basin]

I have all the subgroups generated by 1 element

[rustic crown]

and if you want a subgroup of order 4, it must contain an element of order 2 and an element of order 3

(since the whole group is abelian, this is actually implies the these two elements will generate a subgroup of order 6)

[untold basin]

order 6 you mean?

[rustic crown]

ah right

.<

if you think of the group as Z/2 x Z/2 x Z/3 this shouldn't be too long to bruteforce

wait so an abelian group of order 6 will have a unique order 2 element, right?

[untold basin]

if the group is cyclic yes

but this isn't the case here

[rustic crown]

abelian implies cyclic for order 6

there are only 2, Z/6 and S3

and there are only two order 3 elements in Z/2 x Z/6 where one is inverse of other

[untold basin]

Wait I'm lost with all these informations

:c

[untold basin]

How do you get this ?

[rustic crown]

oh have you not seen like cauchy/sylow theorems?

[untold basin]

No

[rustic crown]

(i assumed that since your original question was galois theory)

[untold basin]

Last year, my prof saw these but not for my year

[rustic crown]

hmm okie, lemme rephrase then

okie how about this, notice that Z/2 x Z/6 has a unique subgroup of order 3.

[untold basin]

My profs said that we won't see exercices implying Sylow theorem but can I find them without the theorem?

[untold basin]

Again, I have a unique subgroup generated by 1 element of order 3 but maybe I didn't collect all the subgroups maybe one is missing

[rustic crown]

(so two things, our groups are abelian and have super small orders, so using these theorems is an overkill anyway. and second you could very well verify the conclusion of these theorems for our case directly)

[rustic crown]

okie how many order 3 elements do you think are there here?

[untold basin]

I'm wrong here

I have (0,2) and (0,4)

[rustic crown]

(they generate the same subgroup)

[untold basin]

right

I have one (only if the subgroups are of the form <g>)

but maybe there is another subgroup of order 6 of the form <g,h>

[rustic crown]

it is order 3, any group of order 3 is cyclic

[untold basin]

right xd

okay so there exist a unique

[rustic crown]

right, so do you believe that there is a unique subgroup of order 3?

[untold basin]

yep

[rustic crown]

now given a subgroup of order 6, it will contain a subgroup of order 3.

[untold basin]

By Lagrange I know that a subgroup can be of order 3 but do I have the existence of it

the only one is the unique one

[rustic crown]

okie let's bash it completely, forget everything i have done above.

i'm not gonna use any theorems

[untold basin]

lol

[rustic crown]

except this one

[untold basin]

okay

[rustic crown]

abelian group of order 6 is cyclic

let's quickly prove this maybe?

[untold basin]

I would like

ooooooooooooooh

S3 and Z/6Z

[rustic crown]

so let's pick a non trivial element g

by lagrange, g has order 2, 3, or 6

if 6 we're done

[untold basin]

ok

if 2 or 3 then prime so cyclic

but I saw that : group of order 6 -> iso to S3 or Z/6Z but it is abelian so impossible to be iso to S3

So it is iso to Z/6Z so cyclic

[rustic crown]

wait so you know this, then great, i won't have to bash this then :3

so you know subgroups of Z/2 x Z/6 are automatically abelian, and so a subgroup of order 6 would be cyclic and hence it reduces to look for elements of order 6.

[untold basin]

And for order 4 ?

it is iso to Z/4Z or Z/2Z x Z/2Z

[rustic crown]

nope, only the latter

there are no order 4 elements in your group

it is iso to Z/2 x Z/2 x Z/3

[untold basin]

does it imply that there exist no subgroup of order 4 ?

[rustic crown]

there is a unique subgroup of order 4, namely
{(0,0), (0, 3), (1, 0), (1, 3)}

this is iso to Z/2 x Z/2

[untold basin]

I found the same by accident

How did you find it so quickly ?

[rustic crown]

cause i wrote the group like this lol

that Z/6 and Z/2 x Z/3 are isomorphic

in this isomorphism 3 in Z/6 corresponds to (1,0) in Z/2 x Z/3

so i just wrote Z/2 x Z/2 replacing the 1 on the right with a 3

[untold basin]

iso because 3 is of order 2 and (1,0) too?

[rustic crown]

after a while you'll be very comfortable with groups, so don't worry so much about this >.<. you wouldn't have to think while using theorems like cauchy or thinking why groups of order 6 are Z/6 or S3

[rustic crown]

that's one reason

the above iso, Z/6Z --> Z/2 x Z/3 is given by sending n to (n mod 2, n mod 3)

[untold basin]

yep I had forgotten Chinese theorem xd

Ok I get it

[untold basin]

I'm still searching for this

[rustic crown]

wanna bash it again?

[untold basin]

Wait a bit pls

[rustic crown]

okie :3

[untold basin]

I'm searching how you get easily the elements of order 4 for Z/2Z x Z/2Z x Z/3Z

[rustic crown]

ping me up if you want a hint

[untold basin]

I can't fin

I think that the last component has to be 0

because 1 -> 2 -> 0 -> 1
2 -> 1 -> 0 -> 2

but if I put (1,0) or other combination I have order 2

[rustic crown]

yep that's right

[untold basin]

so there is no order 4 ? 😂

[rustic crown]

another way to say this is by noticing that 6 kills every element of your group

so if 4 kills some element, then 6-4=2 does as well!

[rustic crown]

yee

[rustic crown]

[untold basin]

Bruh I knew this already

why did I search again

[rustic crown]

[untold basin]

I'm still confused on how you get the order 4 group

[rustic crown]

if you bash completely, this would be the argument,
If H is a subgroup of order 4, then 4 kill every element in H. If you think of the whole group as Z/2 x Z/2 x Z/3, then the stuff killed by 4 is exactly Z/2 x Z/2 x 0

which leaves just enough space for H

so H has to be this Z/2 x Z/2 x 0

[untold basin]

I just noticed a thing

H subgroup of order 4 -> iso to Z/2Z x Z/2Z or Z/4Z

If it is iso to Z/4Z then it is cyclic

But I have no element of order 4

so it is iso to Z/2Z x Z/2Z

And this group has only elements of order 2 except (0,0)

So i can build a subgroup of order 4 in Z/2Z x Z/6Z with the elements of order 2

So there exist a unique subgroup of order 4

because i have 3 elements of order 2 in Z/2Z x Z/6Z

[tender wharf]

wrong channel

[untold basin]

wrong server

[rustic crown]

yep, that works

[untold basin]

I'm searching for order 6 now

I have 6 elements of order 6

[rustic crown]

<@&268886789983436800>

[chilly radish]

Holy.shit again

Same image too

[rustic crown]

[untold basin]

xd

[rustic crown]

notice that there are only two generators in cyclic group of order 6, namely the generator and its inverse

so you get 3 order 6 subgroups from those 6 elements

[untold basin]

Why do you treat the case of cyclic group of order 6 ?

[rustic crown]

since you already saw abelian group of order 6 is cyclic

[untold basin]

Ok 1 min, you gave me an idea

[untold basin]

Ok so

Let H be a subgroup of order 6

if H is abelian then H is cyclic then H is of the form <g> which I know
if H is not abelian then H is iso to S3
furthermore, S3 has three elements of order 2 and two elements of order 3
so I can build H because in Z/2Z x Z/6Z I have exactly three elements of order 2 and two elements of order 3

I think I solved my problem with that

@rustic crown thanks a lot :p

(0,2) and (0,4)*

LOOOOOOOOOOOOL

[rustic crown]

but yea, you can't build S3 from those elements

[hot lake]

Are you trying to find a copy of S3 inside an abelian group

[untold basin]

Which abelian group are you talking about ?

[rustic crown]

Z/2Z x Z/6Z

if you have two groups G and G' such that they have number of elements of each order, then that doesn't imply that G and G' are isomorphic

(it does if you further assume that both G and G' are abelian)

[untold basin]

I know

[untold basin]

I said this because H is of order 6 and not abelian

So it has to be iso to S3

Wait I'm confused

[rustic crown]

furthermore, S3 has three elements of order 2 and two elements of order 3
so I can build H because in Z/2Z x Z/6Z I have exactly three elements of order 2 and two elements of order 3
i was confused with this statement

[untold basin]

where am I wrong

[rustic crown]

are you trying to build S3 from those elements of order 2 and order 3 in Z/2Z x Z/6Z

[untold basin]

I'm building H from S3

yes

[rustic crown]

if H is a subgroup of an abelian group, then it is abelian

[untold basin]

that's what I thought of

my second case doesn't exist then

so it is easier

[rustic crown]

oh okie

[untold basin]

I missed trivial things

thanks again

you're a cool person

:p

[rustic crown]

arigatou

oh is here :3

[next obsidian]

[rustic crown]

[next obsidian]

I am always lurking in the trees

[rustic crown]

cute

[sly rain]

Im looking at Z/nZ for the first few n to see which are a field or not.
Regarding Z/1Z, this is a Ring with only one element? And this element is the neutral element of addition AND multiplication and hence its not a field because for it to be a field we require 0=/=1?

And it only has one element, because every integer = 0 mod (1) and belongs to the same equivalence class?

This is probably very basic, if any thing I said can be worded better please do so.

[cosmic hull]

How do beachy&blair compare to d&f? I’m learning abstract algebra in a 3000 introductory course then taking advanced linear in an honors 4000 course; the two courses use these two books, respectively. Do you guys like the two books?

[next obsidian]

This is right

[next obsidian]

Try #book-recommendations

Also I’ve never once heard of the former

D&F is fine imo

[glossy crag]

Anyone?

[rustic crown]

yep it's valid, just a typo Psi(x) = Phi(1/x) * x^(phi(n))

[glossy crag]

How could this possibly hold when $\Phi_q(x)=\sum_{i=0}^{q-1}x^i$, you'd necessarily have $qe_i=m$, no? Could it be a typo and $\Phi_m(x^q)$ was meant?

[cloud walrus]

Ocean Man

[compact needle]

No, you have that $qe_i + i = mj$ for some $0 \leq j < q$. The problem is stated as intended.

[cloud walrus]

Turgul

[glossy crag]

You're right lol, I multiplied instead of adding.

[formal ermine]

I just found a way easier proof assuming that L/L^H is galois: by the fundamental theorem we know that Gal(L/L^H) = H => |Gal(L/L^H)| = |H| because L/L^H is galois we get that [L : L^H] = |H|

[untold cloud]

Hi, guys, is it true that if f,g\in k[x]\subset L[x], f and g are coprime in L[x], then f and g are coprime in k[x]?

Is it possibble that f and g are not unit in k[x], but they are unit in L[x], for example L[x] = (k(x))[x]

[formal ermine]

I would think so, but don't take my word for it

[formal ermine]

yeah, take k = Z and L = Q

and any constants for f,g

[untold cloud]

oh, sorry, i did not make this clear, k, L are fields, and k is a subfield of L

[formal ermine]

hmmm

what if we take something like L = Frac(k[x]) = k(x)

[untold cloud]

Great! then we have a counter example, Thank you!

[molten viper]

Are "idempotent" elements of rings something that are spoken about in literature/textbooks? They're defined in a homework I'm doing and it feels like something my prof didn't just make up for a problem

idempotent being an element where e^2 = e

[lusty marlin]

Yes, for example, a ring in which all elements are idempotent is known as a Boolean ring, and these are intimately linked with Boolean algebras.

[long geyser]

a projection in linear algebra is a lot of the time defined as an idempotent transformation

[lusty marlin]

Where is the ring here? I am unable to make the connection.

[long geyser]

endomorphism ring?

[elder wave]

endomorphisms of a vector space

[lusty marlin]

Ah right, thanks

[agile burrow]

idempotents are important when talking about decompositions of rings. Unfortunately I'm not sure of any good expositions of this in the literature, but I also haven't looked very far. There's a brief description of this on the wikipedia page for idempotents and I'm sure Dummit and Foote talks about it a bit in the section on Wedderburn's theorem

[pastel swift]

warwick timo?

[elder wave]

?

[pastel swift]

did you go warwick uni?

[elder wave]

No

[pastel swift]

ok you not timo i know

[sly rain]

In class we introduced the following Algorithm to calculate the basis of image and kernel
Say A is a m x n matrix
Then you put the n x n identity Matrix below A
And basically bring A into column echelon Form and apply the same column operations to the identity matrix.
Then the first r (rank) vectors of Matrix A are a basis for the image and n-r vectors of the bottom matrix are a basis for the kernel

Now why does this work out so nicely? What mental model should i have, because its Not clear to me why it works.

[formal ermine]

#linear-algebra ?

[sly rain]

Lol sorry, yeah wrong channel

[glossy crag]

To any who are interested, https://www.calstatela.edu/sites/default/files/users/u1061/cyclotomic.pdf has a neat elementary proof why for $n=2^ap^bq^c$ the coefficients of $\Phi_n(x)$ lie in ${0,\pm1}$ (was just reading it myself).

[cloud walrus]

Ocean Man

[untold basin]

Hello, I'm searching all the subgroups of order 4 of Z/2Z x Z/2Z x Z/2Z

I determined that if H < Z/2Z x Z/2Z x Z/2Z is of order 4 then it is H is isomorphic to Z/2Z x Z/2Z because there isn't any order 4 elements in Z/2Z x Z/2Z x Z/2Z

So H = {neutral , order 2, order 2, order 2}

But I'm struggling to find a formal argument on how to say there is 3 groups of order 4 because intuitively it's okay, but I can't really write a proper argument

[formal ermine]

nvm my stupidity then

sorry

[coral shale]

The way I'd approach is write it in terms of generators maybe? hm

{(1, 0, 0), (0, 1, 0), (0, 0, 1)}

Then your subgroup has to be Z2 x Z2 structure
Not formal yet, but you can 'see' in this particular case u have to pick 2 out of 3 generators because of the limitations involved

[cloud walrus]

Ocean Man

[smoky ivy]

Is C[x]/R[x] isomorphic to some nice ring?

[chilly ocean]

R[x] isn't an ideal in C[x] so this quotient isn't gonna be a ring with the "obvious" operations

[smoky ivy]

sorry, i meant

module*

i’m considering this as an R[x]-module

[rustic crown]

C = R⊕iR
C[x] = R[x] ⊕ i R[x]

[smoky ivy]

so isomorphic to iR[x]?

i see

okay, so my initial problem was to compute the length of C[x]/R[x], but iR[x] has infinite length no?

[brazen furnace]

Hi can someone help me with my Basic Linear Algebra I Question?

[next obsidian]

That’s one dimensional as an R[x] module

It’s isomorphic to R[x]

[next obsidian]

#linear-algebra

Wait am I dumb

[smoky ivy]

yeah

is it not?

[next obsidian]

Yeah it is but

Uh

That has sub modules

Lmfao

[smoky ivy]

but the length would still be infinite, no?

[rustic crown]

R would be a simple R[x] module right

[smoky ivy]

has infinitely many submodules/ideals

yeah

R[x]/(x)

[next obsidian]

I mean it’s not about the amount

It’s about chains

[smoky ivy]

yeah

[next obsidian]

But yeah it’s infinite length

[smoky ivy]

you find an infinite chain

by powers of x

[smoky ivy]

does that help?

[rustic crown]

Probably not lol, I was wondering if it has like infinite length decomposition. But I think it's easier to show it has arbitrary length thing

[smoky ivy]

okay, yeah, i wasn’t rlly sure if it rlly has infinite length, i guess it does then

since i assumed it would be finite for the exercise i’m doing

thanks

[rustic crown]

What's the chain you have in mind?

[smoky ivy]

powers of x

no?

[rustic crown]

Is that even an R[x] submodule?

[smoky ivy]

oh

hmm

okay, wait, if C[x]/R[x] is isomorphic to R[x]

[rustic crown]

That's what I was trying to say here

[smoky ivy]

the length of R[x] is just defined by the maximal length of chains of submodules in R[x]

but submodules in R[x] are just ideals, no?

[rustic crown]

Wait maybe I'm dumb and I thought by powers of x you mean R+...+Rx^n

[smoky ivy]

oh nono haha

i meant (x), (x^2), etc.

[rustic crown]

Yeah makes a lot more sense now xD

[smoky ivy]

yeah hahah

[rustic crown]

Idk why I thought that :p

[smoky ivy]

it’s infinite, should be good then

[sonic coral]

is the order of a group defined by the number of elements in the group?

because we were talking about the order of specific elements, where the order was how many times the operation must be done for the element to return to itself

or are the equivalent somehow?

[next obsidian]

R[x] isn't artinian so it wouldn't be finite length

that's maybe the highest level way to see it as finite length = Noeth + Artinian, and any Artinian ring is dim 0

[next obsidian]

yes

you can define the subgroup generated by an element g as say, <g> = {g^n| n is an integer}

then it follows that the order of <g> is the same as the order of g

because if g has order n then <g> = {e,g,...,g^n-1}

[sonic coral]

so a subgroup of g has the same order as g

[next obsidian]

no

[sonic coral]

we looked at Z mod 4 under addition and not all of the elements had the same order

[next obsidian]

g is in a group G

then the subgroup of G generated by g, which I denoted <g>, has the same order as g

[sonic coral]

and the order of <g> is how many times the operation needs to be carried out repeatedly to return to itself

[next obsidian]

well it's the size of {g^n| n is an integer}

but one can see easily that this is just {e,g,...,g^n-1}

[coral shale]

could be infinite

[next obsidian]

so it has n elements if the order of g is n

[tribal niche]

g has to be a generator

oh yeah nvm u specified that

[sonic coral]

we havnt used that term yet

[sonic coral]

i can believe this though it makes some sense

[next obsidian]

don't believe it, prove it

it's just what you get taking any integer mod n and then using the fact that g^{a + bn} = g^a(g^n)^b = g^ae^b = g^a

[tribal niche]

the order of a group G is the number of elements it has. the order of an element g is the number of times you have to multiply the element with itself to get the identity. if g is a generator of G, then the order of the group and the order of g correspond. note that you can find subgroups of G generated by g (i.e. a subgroup of G, say H, that has the same order as g), even if g does not generate G.

[sonic coral]

okay that makes alot more sense up to the generator and subgroup stuff

i have those listed on things to know for my first test but him and i havnt gotten to that yet

[lavish nexus]

so I'm not a number theorist
someone told me that he can guess what an algebraic number x is once he knows what it is mapped to in F_p^k for enough k's
does anyone know how this works

something he calls lifting an algebraic number from finite fields

[glossy crag]

It's likelier you'll find an answer in #advanced-number-theory.

[sonic coral]

how’s this so far? i’m kinda stuck from here on how to manipulate the information im given to proceed

i’m thinking maybe a * a^2b is b since a^3 is e so e*b = b

[celest furnace]

this is correct

[quiet pivot]

a² is not e

[sonic coral]

oh right a^3 is

[quiet pivot]

yes, also a thing that may help you to verify that it is correct is that there should be only exactly one identity element in each row and column

[celest furnace]

and you know things like a^{-1} = a^2

[sonic coral]

yes we talked about that

[quiet pivot]

o sorry I didn't read above

[sonic coral]

and then if the group table is symmetric about the main diagonal it is albelian

[sonic coral]

where does this come from

[celest furnace]

a^3 = e, multiply both sides by a^{-1}

[sonic coral]

yeah that’s my issue i’m not good at manipulating the information i’m given

but i see that now

[celest furnace]

you get better at it with time

but just try to do stuff like that

[sonic coral]

gonna try to fill this out it’s like a sudoku lol

this one seems to be albelian though

i also have one on dihedral 8 but i don’t think that one is

these four i circled i didn’t really know how to show i just took my best guess

just by looking at the diagonals

[celest furnace]

well $a^2ba^2 = a(aba)a = aba = b$

[cloud walrus]

Geopchad

[celest furnace]

(rearrange the second equation given to aba = b)

also the group shouldn't be abelian

as $a^2 \cdot a^2b = a^4b = a^3ab = ab$

[tender wharf]

the rearrangement here is aa(a^-1 b)a

[cloud walrus]

Geopchad

[tender wharf]

to make it clear

dihedral groups are not abelian (unless you look at D_1 )

[celest furnace]

or D_2

[sonic coral]

since a inverse is a^2 then a has order 3 right

which is equivalent to what i was given with a^3 = e

[tender wharf]

yes a has order 3 (you can already see this by the group presentation)

[sonic coral]

yup okay

[tender wharf]

probably D_3 in this case

[celest furnace]

it says it is D_3

[tender wharf]

I am blind

[celest furnace]

lol

[sonic coral]

because dihedral is 2n

so it is also notated D3?

[tender wharf]

depends

some people use D_6

[celest furnace]

its not always consistent

[tender wharf]

notation is always contentious

you follow the notation that your course wants you to follow

[celest furnace]

my favorite is tan^{-2}(x)

[sonic coral]

fair enough, need to see how gallian doing it

[tender wharf]

gallian notates it D_3

[sonic coral]

tbh my prof let me pick the book just bc he was going to teach me from memory i think

[celest furnace]

what a goat

[sonic coral]

he didn’t use any notes today other than notation

i was impressed

he’s a group theorist so i think he knows his shit

[tender wharf]

your prof is tutoring your personally?

that's crazy

[celest furnace]

what

[sonic coral]

i’m the only one in the class

[tender wharf]

oh

[celest furnace]

😮

[tender wharf]

there are a few typos in Gallian btw

see the messages I sent here like a couple months ago

[sonic coral]

we just changed the day the class meets and go sit in a room somewhere

it’s kinda cool tbh i love it

[tender wharf]

I really should build a collection of typos and email gallian himself

that's kind of cool

[sonic coral]

we got through all of chapter 2 today in an hour

which would never happen with other students i don’t think

for the orders of e a a^2 b ab and a^2b i got 1, 3,3,2,6,6 respectively

if there’s an easy way you guys know to check without going through the table

or maybe you have D6 memorized somehow

[celest furnace]

you shouldn't get anything of order 6 (or for that matter, anything >= 4)

[sonic coral]

i messed up i think i was checking to make sure it wasn’t albelian but my table says it is

[celest furnace]

well at least a^2 and a^2b should be two different things when multipled in different orders

that's what we showed up above

[sonic coral]

i follow this

[tender wharf]

the rotations are in a cyclic group so the orders are easy

[sonic coral]

thinking about it geometrically?

[tender wharf]

order of a = order of a^2 which is 3 yes

ab shouldn't be order 6 I believe

otherwise this implies D_3 is cyclic

[sonic coral]

i asked ab that today and he said we will get to that soon but not yet

[tender wharf]

cyclic groups?

[sonic coral]

when i think of cyclic i think of something coming back to itself

yes

[tender wharf]

anyway nothing should be 6 yeah

<a> = {a^k | k in \bZ} this is a cyclic group

[sonic coral]

so a^2 ab =a^3ab =eab=ab

[tender wharf]

wait

where did you magically get another a from

a^2 ab = a^3 b = b

[sonic coral]

thin air i guess

[tender wharf]

that's probably how you accidentally got an order 6

[sonic coral]

yeah i need to think about this some

[tender wharf]

the best place to get things from

[sonic coral]

well a^2 a^2 = a^3 a = ea = a

so then i can just use the fact that each can only appear once in a row and column and i’m done

since i only had 2 in each of the unfinished rows and columns left

that still gives me order 6 for ab though what

[tender wharf]

er

I doubt that

ab should have order 3 at most

[sonic coral]

where is the mistake in the table then

[sonic coral]

i don’t see how this is true also

maybe that’s my issue

[tender wharf]

well this is a bit cheaty but

D_3 is a triangle

[sonic coral]

yes

[tender wharf]

a is the rotation by 120 deg and

[sonic coral]

yup

since a^3 = e

[tender wharf]

b is the reflection

[sonic coral]

i agree since b^2 = e

[tender wharf]

so just

[next obsidian]

You messed up the b column

For example

You put ab in the position (a,b)

[tender wharf]

oh it should be ba

[next obsidian]

But that’s not consistent with the rest of the table

[tender wharf]

im guessing

if the table is defined as taking the column and it goes on the left

[sonic coral]

so ab = ba depending on if we use the row or column?

[tender wharf]

no

[next obsidian]

I mean

[tender wharf]

they can't be equal?

ok how does your prof define this multiplication table

[next obsidian]

If you have x in the row

And y in the column

You’re putting yx there

[sonic coral]

then wouldn’t all of my entries be backwards

as in written out of order

[tender wharf]

most definitely

[next obsidian]

Like these are not consistent

The first circle says

The element in the column goes first

And then the thing in the row

[sonic coral]

the one furthest to the left.?

[next obsidian]

The other circle says the opposite

[sonic coral]

okay i didn’t know that mattered

[tender wharf]

it does

[sonic coral]

i need to ask him about that

[tender wharf]

because D_3 is not abelian

ba != ab

[sonic coral]

how does that help me with my orders then

and must i flip everything else?

[tender wharf]

you need to flip those that you did wrongly

[sonic coral]

but how do we know which are wrong? any that have both an a and a b?

like if i define column first, then would a^2 a^2b = ab

and the other way be ba?

[next obsidian]

No

Because

the first part of your table says if you have x in the column and y in the row

You place xy there

You’ve randomly decided to put yx in that spot instead like halfway through

Like imagine I gave you a 10x10 table

With the numbers 0 through 9 at the top and to the side

And say to put x - y in the spot with x marking the column and y in the row

If you randomly decide to change this and put down y - x then it’s suddenly wrong

[sonic coral]

good analogy

[next obsidian]

Like in the spot with 7 in the column and 3 as the row you put -4

[sonic coral]

they should be opposite signs

but i put them with the same sign

[next obsidian]

Well no

Not even

You’ve just put something backwards

[sonic coral]

i feel like i did the whole table that way though

[next obsidian]

Like for groups there is no concept of like - like this

Well you clearly didn’t

[sonic coral]

so everything should be wrong

[next obsidian]

I drew counterexample a

No

If you’re consistent then you’re fine

It doesn’t matter, the two conventions differ by a transpose of the table

You just can’t swap

[sonic coral]

well when i was filling it out i wasn’t thinking ab the column or rows first

so idk what happened

[next obsidian]

Well idk

[sonic coral]

how does this effect my order 6 issue though

[next obsidian]

Idk man go fix it then figure it out

Lol

[sonic coral]

fair enough

i need to think about this some i should sleep on it

[tender wharf]

are you lacking sleep

[sonic coral]

depends how much sleep you think is enough lol

[tender wharf]

actually

it depends on how much sleep you think is enough

[sonic coral]

i’m running on 4 hours right now

feel fine, nothing abnormal for me

[celest furnace]

that's a gift tbh

20 (hopefully) productive hours is insane

[tender wharf]

you can feel fine but be not really fine

afiak current research shows that 6h is roughly the minimum you need

but it depends on the individual

[sonic coral]

all of them weren’t productive, but id say about 15 of them were

slept at 5, class from 9-2 and work from 3-6, came home and working on homework

[celest furnace]

why slept at 5 lol

[sonic coral]

that’s where the nonproductive part comes in

[celest furnace]

i see

[charred crescent]

can someone let me know if the mapping from G = ({1,-1},x) to the finite subgroup of SL2(R), (namely the 2x2 identity matrix and its negative) constitutes a faithful representation of G?

the subgroup of SL2(R) is clearly a subgroup of GL2(R), and the mapping is an isomorphism, so if i understand correctly it should be a faithful representation

i have a proof outlined, which i can send if needed (although i suspect that this could be considered a trivial statement)

[lethal dune]

Faithful means G-> GL(V) is injective. Is your map injective?

[charred crescent]

what do you mean by GL(V)?

im a noob

[lethal dune]

GL_2(R) in your case

[charred crescent]

why the difference in notation, if i may ask ?

is GL(V) a more general statement

[lethal dune]

It’s just notation, one requires basis other doesn’t

[charred crescent]

gotcha

and yes my mapping is bijective

[lethal dune]

Bijective?

How?

[charred crescent]

yeah injective and surjective

[lethal dune]

Surjective how?

[charred crescent]

cuz there are 2 elements in each group

my mapping is phi(x) = x multiplied by the 2x2 identity matrix

[lethal dune]

SL_2(R) is not { I,-I}

[charred crescent]

it's a subgroup of it tho

or nah

[lethal dune]

Yes but you should say bijective with this sub group not SL R

[charred crescent]

im not sure i understand

[lethal dune]

Also all you need is injectivity

[charred crescent]

may i ask why ?

[lethal dune]

that's the definition of faithful representation

[charred crescent]

yeah but why is that the case

[lethal dune]

that's just how the definition is

I don't have a proper justification

[charred crescent]

fair enough

[lethal dune]

But it says that if two element act identically, they must be the same

that is the kernel of the homomorphism is 0

so you won't loose any info about the group

maybe that's why it's called faithful idk

[charred crescent]

okay i think that makes sense

thank you for your time

ooh i think i know why the definition only requires injectivity; if the mapping is injective, then the image of the mapping under GLn(R) is by definition surjective, hence the mapping is an isomorphism

[somber sleet]

hey guys, is a ring always abelian?

[rustic crown]

if R is a ring, then (R, +, 0) is an abelian group

but (units, *, 1) may very well be non-abelian

[somber sleet]

yeah sorry, I meant under multiplication

[rustic crown]

under multiplication, (R, *, 1) isn't a group, just a monoid

and that usually isn't required to be abelian

matrices are nice examples of non-commutative rings

[somber sleet]

yeah well you're right

I asked because I was demonstrating that the unit ring is an abelian group under multiplication

but I think that in this case we assumed that R is abelian, because otherwise I don't know how to do it

[rustic crown]

yea, if you don't assume R is commutative, then in general it's not going to work

if R = M_n(k) then the units are invertible matrices GL_n(k) and it's not hard to see that for n>=2 this is non-abelian

(sorry i'm a little sleepy >.<)

[somber sleet]

no worries, I'm already grateful for your answers

[somber sleet]

then it makes sense, thank you very much.!

[chilly ocean]

You often define rings to be abelian with a unit

But some authors require none of these things

There's lots of chaos in ring theory