groups-rings-fields

[chilly ocean]

oh

[delicate orchid]

also I'm saying character instead of rep again, sorry

[chilly ocean]

ok

[delicate orchid]

force of habit

[chilly ocean]

all good

so what does endomorphism-algebra mean?

[delicate orchid]

so if you have a module U over the group algebra, you can look at algebra maps from U to itself, this forms a new algebra End(U)

it's important to know this dimension for decomposing the group algebra

[chilly ocean]

right

[delicate orchid]

I can't recall any quick tricks for calculating it, unfortunately

[chilly ocean]

would you be able to help with this proof of Maschke's? I've tried some things but I'm not sure if they are allowed

My proof for 1 and 2 just both feel super iffy

[delicate orchid]

well what's your proof for them

[chilly ocean]

My idea for 1 was to show that we can use this isomorphism

but then I ran into issues of trying to show surjectivity

so I wans't sure if it was the correct approach

[delicate orchid]

this looks promising

[chilly ocean]

I could show injectivity

but then I tried to show surjectivity by saying like
phi(v) = k x w
But then I wasn't sure how to actually always write v so that it works

k = v - phi(v)
and w = phi(v) so v = w?

[delicate orchid]

v = k+w perhaps, which we can write because we know the map is injective
so V is isomorphic to a subspace of ker(pi) (+) W

[chilly ocean]

I don't know if that assumption is correct

oh

[delicate orchid]

lemme think about how I'd do it

[chilly ocean]

so

then we would have like

(k + w - pi(k + w)) x pi(k + w)

is pi(k + w) = pi(k) + pi(w) allowed?

[delicate orchid]

let v in V, then pi(v) = w, so write v = (v-w)+w, this shows that ker(pi)+W = V, now to show the sum is direct we just need to show that ker(pi) intersects w trivially, so assume w is both non-zero and in the kernel, then this is an immediate contradiction as pi(w) = w \neq 0, but pi(w) = 0 by assumption it's in the kernel

[delicate orchid]

why wouldn't it be?

[chilly ocean]

linear map is the same thing as a homomorphism right?

so that's fine

[delicate orchid]

it's a homomorphism of modules yes

[chilly ocean]

ah ok

cool ok

so my idea for part 1 is fine right?

[delicate orchid]

yeah it seems to work

[chilly ocean]

part 2 I was a bit lost in

uh

lemme see what I did

oh

I didn't use the G-invariant fact

or

G-equivariant fact

[delicate orchid]

yes you need to use that fact to show that it plays nicely with scalar multiplication

[chilly ocean]

I just said let's look at pi(a) + rpi(b) = pi(a) + pi(rb) = pi(a + rb) so it's in the image

I won't be able to say rpi(b) = pi(rb) irght?

where r in k[G]

[delicate orchid]

not immediately

but you can conclude it from G-equivariance

[chilly ocean]

ah

so like

an element in k[G] can be written as like

k + kg + kg^2 + ... +

right?

(k + kg + kg^2 + ... +) pi(b) = kpi(b) + kgpi(b) + kg^2pi(b) + ... +

we want this to be equal to pi(kb + kgb + kg^2b + ... + )

since it is G-equivariant that means that the first line is

kpi(b) + kpi(bg) + kpi(bg^2) + ... +

k(pi(b) + pi(bg) + pi(bg^2) + ... +) = k(pi(b + bg + bg^2 + ... + ))

am I allowed to just take the k in now?

[delicate orchid]

this assumes g is cyclic

[chilly ocean]

oh

but g is finite but not necessarily cyclic?

[delicate orchid]

$z \in k[G] \Rightarrow z = \sum_{g \in G} k_gg$ for some $k_g \in k$.

[cloud walrus]

WewGhostTbh

[chilly ocean]

right

[delicate orchid]

I think your logic is correct though

[chilly ocean]

ok

I'm curious for question 3, do you think I can assume that pi is g-equivariant?

because I get like

and if I can just assume that pi is g-equivariant then the proof is done right?

[delicate orchid]

but \pi isn't g-equivariant

[chilly ocean]

oh

it isn't?

[delicate orchid]

the key thing here is that you're summing over the entire group

so when you multiply p(v) on the outside by some g' in G you're just rearranging the terms of the sum via the bijection g |-> g'g

hmm

[chilly ocean]

oh

[delicate orchid]

yeah it does just say projection, nothing about G-equivariance

[delicate orchid]

see if you can use this idea to prove it, I've gotta go to bed lol

3am here

[chilly ocean]

oh ahah

thanks

[chilly radish]

Uhh isn't that the assumption??

[chilly ocean]

?

[chilly radish]

Part 2 says pi is G equivariant

[chilly ocean]

yea, but I wasn't sure if we could assume that to be true for part 3

[delicate orchid]

Read the first sentence of 3

[chilly radish]

Oh ok

I missed the part where you moved to q3

My b

Go to sleep wew lads

[delicate orchid]

Maybe

[chilly ocean]

wait why does the first setnence mean that pi isn't g-equivarient?

[chilly radish]

It means you can't assume it

[chilly ocean]

oh ok

[chilly radish]

Because you weren't given this fact

[chilly ocean]

I see

[chilly radish]

And in fact you don't need it

Like wew said, this is a very common trick, multiplying by an element just permutes all the group elements

[chilly ocean]

right

so the summation is the same

wait

ok so like from here

we have this idea of a bijectino from g |-> g'g as wew pointed out

why does this mean they are equal

[chilly radish]

So you can put h into the sum

And then you get hg

So now set g'=hg and sum over g' instead

[chilly ocean]

ah I see

ok I think I kinda understand parts 1-3

but I don't get how it ties to part 4

[chilly radish]

(Note the h appears inside pi cuz you can write g=h^-1 hg, so g^-1=g'^-1 h, verify this)

[chilly radish]

If V is not simple, there is a nontrivial k[G] submodule.
You can always project onto W. You learned how to turn this projection into a G-equivariant one, such that V decomposes into smaller k[G]-submodules. Continue by induction.

[chilly ocean]

oh I See

man this stuff is so hard

[chilly radish]

It gets easier

[chilly ocean]

I don't understand how to do it

[chilly radish]

It's always hard the first thing

Time

But a lot of these techniques are standards

You'll get the hang of it

[winter yew]

My task is two find all invertible elements in F_4[x]/(ax^2+bx+c) where the poly has two different roots . I thought I could get my life easier if I show the isomorphism above and find all invertable inside F_4 x F_4 but then I realised that I don't know how to work with F_4 x F_4 in this context, like what operation, since it is not a field.

[eager willow]

That'd be the direct product of rings.

Maybe start by seeing which elements are zero divisors

Just to get a hang on the operation.

Oh rereading maybe I should ask you what do you mean with \cong R^2? Is that you observing that you know the dimension is 2 or do you know how the isomorphism to the direct product of rings works already?

[winter yew]

the isomorphism is f(x) ->(f(a), f(b)), where a, b are the roots

[eager willow]

Iso as rings or vector spaces?

[winter yew]

as rings

[eager willow]

Ok gotcha. So if it's two fields product together, what's the unit?

[winter yew]

unit x unit? in our both cases it is 1x1

[eager willow]

Yeah that seems right. It's like ptwise multiplication of the functions which are only well defined on the set of your two points

So given you know the unit, what's the inverse of (x,y) supposed to look like?

[winter yew]

There is the problem. how multiplication is defined for FxF? is it ab x cd or something else?

[eager willow]

How would you know the multiplicative unit if you don't know the multiplication?

[winter yew]

this is the second problem 😀

I understand what is on the left of the iso, but don’t what is on the right

direct product of two rings, yes, sum must be the same as usual, but what about multiplication?

[eager willow]

Ok let's start over. If it's an isomorphism it's surjective. What's an element of the polynomial ring that is taken to (1,0)?

[winter yew]

the second one is root, the first one I need to calculate, since we must have zero divisors, so the first could be not 1

Can we assume it is (1, a root)

[eager willow]

Is f->(f(a), f(b)) surjective?

[winter yew]

with zero divisors I doubt that

[eager willow]

So it's not an isomorphism?

[winter yew]

yes, then yes

[eager willow]

Idk you had me convinced it was an isomorphism before.

I mean there's clearly an isomorphism of vector spaces because you know the dimension.

So how did you know that one was an isomorphism?

I think the space of functions from a set of two points to F_4 has the structure of a ring with ptwise multiplication. Its also a 2 dimensional vector space over F_4 with ptwise addition and can therefore be identified with a ring structure on F_4\times F_4. The element (1,0) representing the function taking a to 1 and b to 0, and the element (0,1) taking a to 0 and b to 1

[winter yew]

We can take mapping R[x] -> RxR as i mentioned above. Its kernel is <(x-a)(x-b)> and it is surjective, then by the first iso theorem R[x]/ker -> RxR must be an iso

[eager willow]

Now is that the first isomorphism of groups, vectors spaces, or rings????

[winter yew]

not fields for sure, the are not fields

we did’t study vector spaces iso so far

[eager willow]

You claim it is surjective. Which polynomial is taken to (1,0)?????

[rapid junco]

For the second claim why do we require kerpi \subset kerpsi?

I never used it, but this means my proof is probably wrong?

[next obsidian]

You haven’t formulated b) correctly

When did you ever show the existence of delta only from the fact that ker pi < ker psi?

Your entire proof only ever deals with a delta which was already assumed to exist

@rapid junco

I strongly urge you to prove these as a) and then as b) both for clarity to the grader and for yourself

Or err… I have tripped myself up a bit too

[rapid junco]

take delta to map h in H to $\psi(\pi^{-1}(h))$

which can be done since pi is surjective

[next obsidian]

You have done that, but also just random thing you said g < ker pi, you want g in ker pi

[cloud walrus]

Brayden

[next obsidian]

Your map is not guaranteed to be a homomorphism

But also psi(pi^-1(h)) is not an element

That’s a set

pi^-1(h) is an entire set of preimages

[rapid junco]

oh so take a representative of it and show it doesnt depend then?

[next obsidian]

Yes

[rapid junco]

ah I see

[next obsidian]

And this is where your hypothesis comes into play

And this lets you also show that delta is a homomorphism

[rapid junco]

can you clarify this?

[next obsidian]

(You need to use that pi is a homo morphism)

[next obsidian]

You used the symbol for subset, g is not a subset of ker pi it’s an element

So you want to use \in

[rapid junco]

oh no thats element

[next obsidian]

Lmfao

I see it now

[rapid junco]

bad quality pic

[next obsidian]

Had to zoom in xD

[rapid junco]

Right okay thanks

so my proof of hom is incorrect?

[next obsidian]

Anyway, do you know the first iso????

[rapid junco]

Yeah

[next obsidian]

Yeah you absolutely need the hypothesis about the kernel

[rapid junco]

why where does it fail

[next obsidian]

Oh sorry do you mean for a)?

[rapid junco]

suppose there is a function that is well defined

for b

then is my proof correct?

[next obsidian]

Oh, yeah

This is just surjectivity of pi

[rapid junco]

oh okay sounds good

[next obsidian]

Yeah?

[rapid junco]

yes

[next obsidian]

Sorry for my critique being a little incoherent lol

But I hope I got across the fact that you need the hypothesis for existence

👍

[rapid junco]

I am slightly confused as to how one used kernels?

@next obsidian

[next obsidian]

Well okay

So if you took two preimages of the same guy

Call these preimages g and g’

[rapid junco]

Yup.

[next obsidian]

You want to compare psi(g) and psi(g’) to make sure these are the same

[rapid junco]

Oh but psi is a hom

[next obsidian]

Right

[rapid junco]

so psi(g) = psi(g') iff psi(g)psi(g')^{-1} = 1

iff gg' in ker psi

[next obsidian]

gg’^-1

But right

So if ker pi < ker psi all you would need to know is that gg’^-1 in ker pi yeah?

[rapid junco]

ah right

because g is in G

wait

oh I see

okay

iff pi(gg'{-1}) = 1

[next obsidian]

g’^-1

Yup

[rapid junco]

iff hh^-1 = 1

ahh

Okay

Thanks

[next obsidian]

So like

More generally

Or like, okay what everyone says

Is that g and g’ “differ by something in ker pi”

This is saying their difference (gg’^-1) is in ker pi

The point is that if f is a homomorphism, then f(g) = f(g’) iff g and g’ differ by something in ker f

If the word “differ” here doesn’t make too much sense, if G was abelian we’d be writing g - g’ so that’s why it’s their “difference”

Anyway, so the point is just, if g and g’ were preimages of h, their difference lies in ker pi < ker psi, so that means psi(g) = psi(g’)

This is literally what you just proved

But this wording of “differs by something in…” is a really good way to express these things

In words

Ya dig?

[rapid junco]

right I see

sort of

[next obsidian]

I’d try to prove this real quick

This is a fact u will use a lot

[rapid junco]

this good now?

thanks for the help

makes sense now that I think abt it

[tribal drum]

I want to show G = HK where H = {g^m | g in G}, K = {g^n | g in G} and mn is order of G. I’ve shown HK is a subgroup of G, how do I show G is a subgroup of HK?

Or is there an easier way

[coral shale]

Is this really generally true

[swift raptor]

Are m n coprime?

[coral shale]

im thinking surely not if mn = 8 = 4 x 2

[delicate orchid]

or mn = 2*2

if they're coprime then I believe it

[tribal drum]

M n are Coprime

G abelian

[glossy crag]

How come adjoining a single root of an irreducible polynomial in some cases automatically adjoins all (e.g. cyclotomic polynomials or x^p-x-a in characteristic p) and hence yields a normal extension, while in other cases it doesn't (e.g. Q(cube root of 2))? I'm trying to remember whether there were definitions/theorems in field theory that make sense of this phenomenon, but I don't think there are any. Does it have to do with the polynomial having a cyclic Galois group? The question may be rephrased as "when are roots of an irreducible K-polynomial representable as K-polynomials of a single root".

[coral shale]

"why"...

I feel like this can be understood in terms of quotient groups, using the galois correspondence

i dont feel like cyclic is rlly related

[rocky cloak]

The Galois group being cyclic implies this, but the converse isn't true.

For example the Galois group of the cyclotomic polynomial x^4 + 1 is C2xC2.

I'm not sure that there is another nice characterization of when this happens.

[coral shale]

Is the primitive element thm relevant here? the single root part

[rocky cloak]

Sort of I guess. If you take a Galois extension, then the primitive element theorem tells you it's the splitting field of some polynomial with this property. But I don't see that it says anything about the polynomial you started with

So I guess a take away would be that there are plenty of such polynomials

(and that there really is no restriction on the Galois group)

[tawdry plover]

What are all the things that we can say about a group with odd squarefree order

Do they always have a nontrivial centre

[swift raptor]

What is the center of S3?

[abstract rock]

i was gonna give the example of an abelian group too

[tawdry plover]

I see so if I add an extra condition that p_i does not divide p_j -1 for i not equal to j

[abstract rock]

that leaves you with square free odd ordered groups, no?

you know any fun facts bout those

[tawdry plover]

I should have mentioned it like that

[tawdry plover]

No

[abstract rock]

there's a rather difficult to prove theorem for that class of group, hopefully it wont be required for whatever you're looking to show here

[tawdry plover]

I would still like to know

[abstract rock]

look up feit-thompson theorem

its also known as the odd order theorem

took about 51 years to prove after burnside posed it

and requires a lot of knowledge bout characters n representation

[tawdry plover]

Oh ah it's about solvability

Well I am not concerned with solvability for now

So I'll pass

[tawdry plover]

Anything more for this

[tawdry plover]

that leaves you with square free odd ordered groups, no?

This is not equivalent to the above condition entirely i guess

[abstract rock]

yeah i suppose not all pair of odd primes fulfill that criteria

[tawdry plover]

I agree so I have a stronger restriction

[abstract rock]

perhaps come up with a more concrete set in then and see what kinda groups live there

[tawdry plover]

Okay I'll try that for now 👍 thank you

[rotund aurora]

in proving this, I just proved that the numbers are all congruent to 1 mod p by using n_p(G)=1 mod p. Is it posssible to do the problem without proving this stronger claim? In particular, without using n_p(G)=1 mod p

Using only the second Sylow (the Sylows are conjugate) I proved that n_p(G)K=n_p(G)L mod p where K denotes the number of subgroups of order p^a in a fixed Sylow S and L denotes the subgroups of order p^a in G

[rocky cloak]

If a group has order pq (p > q) and q doesn't divide p-1, then the only such group is the cyclic group of order pq

[rotund aurora]

In particular, n_p(G)^2=n_p(G) mod p

[rocky cloak]

In fact all such groups must be cyclic:

Let G be the smallest non-cyclic group of this type. Since every group of square free order is solvable G has Cp = G/N as a quotient group. Since N is smaller N is cyclic, so G is the semidirect product of two cyclic groups Cp and N. Since p doesnt divide phi(|N|) the product is direct, so G = NxCp which is cyclic, since |N| and p are relatively prime.

[tardy hedge]

Dude my prof writes stuff down that is LITERALLY WRONG

He just wrote “subgroups of order 2 are normal”

Da fok?

{e, (1,2)} order 2 subgrp of S3

Its not normal

[rocky cloak]

Subgroups of index 2 are normal though

[tardy hedge]

Yea thats what he meant

[rocky cloak]

We are all human

[coral shale]

um.

[tardy hedge]

Yeah true, but tbh im just a bit salty cause he always does this sort of thing which makes it very hard to learn from him

[rocky cloak]

All hail shuri, out great robot overlord

[coral shale]

take it as a challenge to pay attention and correct sht when its off

[tardy hedge]

Yeah thats a good point! Its good in that sense but only if i pre-studied the material

[open sluice]

that’s something you should be doing anyway tbh

[tardy hedge]

I do when i can

[delicate orchid]

this is true as long as you can't count higher than 5

[chilly ocean]

anyone have any hints for this?

[crystal turtle]

Induction.

[chilly ocean]

ah

that's what I was thinking

induction on m?

no

on i right

[native sequoia]

Hello please I have more difficulty to get something about group and rings , how can I give me some advice for have intuitive notion about it

[coral spindle]

Do lots of exercises 👍

[native sequoia]

recommend book for it

[chilly ocean]

am i insane or is this what i should do

[crystal turtle]

Fix a k and m and induct on i.

[chilly ocean]

gotcha

thanks

okay so

[native sequoia]

// please

[chilly ocean]

dummit and foote or serge lang, or herstein

[coral spindle]

#book-recommendations

[crystal turtle]

Alright we can't all type that fast

[chilly ocean]

trivially $\sigma^1(a_k) = a_{k+1}$ because that's the definition of the permutation

[cloud walrus]

z

[crystal turtle]

sniped me

[chilly ocean]

now suppose $i = r$ and proposition is true for some $r \in \mbb{N}$

[cloud walrus]

z

[chilly ocean]

in other words $\sigma^r(a_k) = a_{k+r}$. Then comes the induction step. we can re-write $\sigma^{r+1}$ as \sigma(\sigma^r)$ right?

[cloud walrus]

z
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[chilly ocean]

since its composition of functions

[chilly ocean]

ok I finished the proof but do I have to mention that this is reduced mod m

or prove anything regarding its reduction modulo m

[barren sierra]

What is sigma^m?

[chilly ocean]

1

[barren sierra]

There’s your modulo argument

[chilly ocean]

identity permutation

wdym

[barren sierra]

sigma^i = sigma^i+m for example

[chilly ocean]

oh

[chilly ocean]

$\sigma(a_k) = a_{k+1}$ which is trivially true from the definition of the permutation. Now comes the inductive hypothesis. Let $i = r$. Assume $P_r$ is true for some $r \in \mbb{N}$. That is, $\sigma^r(a_k) = a_{k+r}$. Consider $i = r + 1, \sigma^{r+1}(a_k) = a_{k+(r+1)}$. But we can use the proposition to achieve this result too:
$$\sigma(\sigma^r(a_k)) = \sigma(a_{r+k})$$
in other words $\sigma^{r+1}(a_k) = a_{r+(k+1)}$

is this ok

[cloud walrus]

z

[chilly ocean]

does that make sense or

[crystal turtle]

You're missing the part about reduction mod m. What if r+k+1 > m [or for that matter, even k + 1]? What does a_{r+k+1} mean?

[chilly ocean]

yeah thats what i thought. if r + k + 1 > m then we take r + k + 1 mod m

[barren sierra]

But why does that work is the key point

(use the definition of mod plus what you said about identity)

[chilly ocean]

If we take the morphism $\mathbb C [x,y] \to \mathbb C [\cos, \sin]$ that sends $x \mapsto \cos$ and $y \mapsto \sin$, then how can I show that the kernel is $(x^2+y^2-1)$. One inclusion is ok, and for the other, we can divide some $f$ in the kernel by $x^2+y^2-1$ and take the image of this to get $\sin \cdot p(\cos) + q(\cos) = 0 $, but i don't see how to show that $p = 0 = s$.

[cloud walrus]

zxxz21

[abstract rock]

Cmon you didnt need to give it away!

[rocky cloak]

Sorry

[abstract rock]

its alright they probably would have realized it by the time they came back

[tardy hedge]

I got 98% on the group theory midterm lol

It wasnt hard though

[delicate orchid]

I hate to be that guy but what was the 2%

[tardy hedge]

I calculated phi(100) wrong lol (euler totient function) and my answer for the biggest order of a permutation on 12 elements was wrong

I just wasnt careful on finding the partition that makes the biggest product

For the order

[delicate orchid]

simple mistakes lol. That phi(100) calculation sounds annoying

unless you do the ol phi(n) = \prod_{d|n} phi(d) thing

[tardy hedge]

Yeah. You know this does make me think though, how do grad schools interpret grades from undergrad? Like, at the uni i got my bachelors from, their abstract algebra classes are TOUGH and the averages are very low

Thats why i didnt do them cuz i didnt want to tank my goa

Gpa

And i wanted to solely focus on it without other bs classes

Which is why im doing it now

But this uni’s class is clearly much easier

[white oxide]

i'm having a little trouble seeing the equality underlined in blue for if $i \neq j$; for then we would have $\iota_j^*(a_j \otimes b) = (0, 0, \dots, (a_j \otimes b), \dots, 0)$ but ${\pi_i\iota_j(a_j) \otimes b}{i \in I} = {0 \otimes b}{i \in I} = {0 \otimes 0}_{i \in I}$ no?

[cloud walrus]

okeyBOOkay

[white oxide]

i mean i understand that equality holds if we take $i = j$ and it makes sense that it's generated by all elements of the form $\iota_j^*(a_j \otimes b)$

[cloud walrus]

okeyBOOkay

[white oxide]

nvm

i got it

[rapid junco]

it is always true that gH = g^{-1}H correct?

oh wait this reduces down to showing that H = 1 H

[rocky cloak]

No, that's rarely the case.

[rapid junco]

oh I see

my actual question was if g_1H = g_2H then g_1^{-1}H = g_2^{-1}H

here is the work I have done so far. and its in the first paragraph but missed a lot of class so wanted to confirm if I was indeed correct

[delicate orchid]

gH -> Hg^-1 is well defined bijection because it's left multiplication by g^-1 and then right multiplition by g^-1

the actual proof looks fine

[rocky cloak]

Your condition that g1H = g2H iff g1g2^-1 is in H, is reversed. It should be g2^-1g1.

Also not sure why you want to prove that since you don't use it for the proof.

[rapid junco]

Right

well [G : H] is the number of left cosets

we havent established that the number of right cosets equals the left number

[delicate orchid]

they both partition G into sets of size H

it's pretty obvious they're the same number

[rocky cloak]

Well you're proving that in your paragraph.

The inverse maps gH to Hg^-1 so that's a well defined bijection. It never comes up wheter g1^-1H equals g2^-1H

Maybe you just meant to write Hg1^-1 = Hg2^-1...

[white oxide]

i'm a little bit confused as to how this is defined LOL. like are they showing that $[(\alpha f)(a)]$ is the homomorphism in $\text{Hom}_R(A, \text{Hom}_S(B, C))$? if so, then why does it have input $b$?

[cloud walrus]

okeyBOOkay

[rapid junco]

OK

[rapid junco]

yeah

[rocky cloak]

alpha f is the homomorphism in Hom(A, Hom(B, C)). Then you evaluate that at a and you get something in Hom(B, C)

[white oxide]

ohhh, and then $[(\alpha f)(a)]$ is in $\text{Hom}_S(B, C)$

[cloud walrus]

okeyBOOkay

[white oxide]

so they're just defining it's action on an element of $b$

[cloud walrus]

okeyBOOkay

[white oxide]

i guess that makes sense yeah

[crystal turtle]

AAH! Your name scared me 😭

[delicate orchid]

JESUS

[rocky cloak]

A name for the season

[white oxide]

ikr it's pretty fucking terrifying

[delicate orchid]

jumpscare warning

AA

[white oxide]

they never see it coming

[crystal turtle]

anyways uhh yeah if you know currying for set functions (like given a function f : A x B --> C, fixing an a gives a function f(a, -) : B --> C), this is basically the same thing

[white oxide]

the term currying is very new to me

[crystal turtle]

🍛

https://en.wikipedia.org/wiki/Currying

[white oxide]

computer science

i'm not a computer scientist sadly, so don't think i'll read

[delicate orchid]

ok then buddy. wanna play silly games?

https://ncatlab.org/nlab/show/currying

you win silly prizes

[coral spindle]

Currying is just the cartesian product-hom adjunction

[delicate orchid]

I mean

I mean

[coral spindle]

don't be mean

be nice :)

[delicate orchid]

;3

[crystal turtle]

:3c

[coral spindle]

:3c

But like fr

$\operatorname

fuck

[delicate orchid]

\text

[coral spindle]

$\newcommand\Hom{\operatorname{Hom}}\Hom(- \times S, -) \cong \Hom(-, -^S)$ is usually how it's written

why

[delicate orchid]

oh my god

Hom$(- \times S, -) \cong $Hom$(-, -^S)$.

[cloud walrus]

Ramify it (extensions) down

[coral spindle]

I will die on this hill

[cloud walrus]

WewGhostTbh

[delicate orchid]

perfect

[white oxide]

ah much better

[white oxide]

bro why do you do \operatorname

[crystal turtle]

PUONTEHWIGS?

[coral spindle]

Why do you

[white oxide]

true

[coral spindle]

Idk it's the ostensibly right way and I am a massive fucking pedant

[delicate orchid]

guys it's HALLOWEEN??!!? lets not fight during the holidays please

okey is right though

[coral spindle]

Oh no I was hoping if I did it slow it would come out nicely but no

discord bad

[crystal turtle]

oh

:uponthewitnessing:

[coral spindle]

:uponthewitnessing:

[delicate orchid]

🇼 ℹ️ ✝️ ♑ 3️⃣ 🇸💲 🇮 🇳🗜️ _ _

[white oxide]

where am I going wrong in checking that $\beta g$ is indeed a homomorphism? $(\beta g)(a_1 + a_2 \otimes b_1 + b_2) = [g(a_1 + a_2)](b_1 + b_2) = [g(a_1) + g(a_2)] = [g(a_1)](b_1 + b_2) + [g(a_2)](b_1 + b_2)$ but $(\beta g)(a_1 \otimes b_1) + (\beta g)(a_2 \otimes b_2) = g(a_1) + g(a_2)$

[cloud walrus]

okeyBOOkay

[white oxide]

i can't really do any fancy things in the tensor product since $R$ and $S$ are not necessarily unital

[cloud walrus]

okeyBOOkay

[peak root]

$(a_1+a_2)\otimes (b_1+b_2)$ should be $a_1\otimes b_1 +a_1\otimes b_2+a_2\otimes b_1+a_2\otimes b_2$, not $a_1\otimes b_1+a_2\otimes b_2$.

[cloud walrus]

Dong_Valentino

[delicate orchid]

it's like expanding out (a+b)(x+y)

[white oxide]

oh huh

i've been adding them component wise

bruh

well ig that makes sense

since we can kind of think of tensoring as multiplication right

cuz distributivity

associativity of scalars

very naive question: why is it that not all modules M over a ring R are flat? the map in the second entry is the identity, which is injective, and similarly $\varphi$ in the first entry is given to be injective. why then is $\varphi \otimes_R id_M$ not automatically injective?

[cloud walrus]

okeyBOOkay

[white oxide]

is there a counterexample?

[elder wave]

there is

try to come up with one

hint: flat modules are torsion free

[white oxide]

ooh ok i'll try it out, thx

[rotund aurora]

How many solvable groups are there?

Asking about the quotient (# solvable groups of order <= n)/(# groups of order <= n) as n goes to infinity

[crystal turtle]

p-groups are solvable. Most finite groups are 2-groups (https://math.stackexchange.com/questions/241369/more-than-99-of-groups-of-order-less-than-2000-are-of-order-1024)

[rotund aurora]

I had the impression that it was going to be positive

thanks

waah this is Chmonkey or Ryx

🕵️ moment

[coral spindle]

I should hope it's positive, considering both parts of the quotients are positive...

[rotund aurora]

I meant in the limit

could be zero

[coral spindle]

Lol yeah

[crystal turtle]

Oh it is listed there that it's still open whether "almost all finite groups are 2-groups"

[tribal moss]

I wonder if it makes a difference whether we count isomorphism classes of groups or, say, binary operations on {1,2,..,n} that happen to satisfy the group axioms.

[crystal turtle]

The former seems morally correct, but I think the latter would make it depend almost entirely on the largest order considered (since more elements = more "groups" by just relabelling)

Like there's one group of order 7 say, but now there'd be 7! relabellings of it that you'd count

[tribal moss]

But divided by how many automorphisms it has, I think.

[crystal turtle]

I think that works then? Dividing by n! I mean, not Aut(G)

[rotund aurora]

mmmh are there nice bounds on the size of Aut(G)

nvm

Aut(S_n)=S_n I think

oh wait but thats not too bad, because then they have the same size

I think the automorphism group of (Z/2Z)^n has size n! ? because its essentially permuting coordinates?

[tribal moss]

Except for n=6.

[rotund aurora]

and n=2

[tribal moss]

Okay right.

[rotund aurora]

because of this, the upper bounds will be shity

[crystal turtle]

I would anticipate that's about as bad as it gets though

log_2(|G|)!

[rotund aurora]

you have to take the log of |G|, tho

[tribal moss]

It's worse that that: the automorphism group is GL(n,2).

[rotund aurora]

right

[crystal turtle]

oops

[tribal moss]

And the order of GL(n,2) is between 2^(n^2/2) and 2^(n^2), which makes it asymptotically something like |G|^(c log |G|).

[rotund aurora]

I found that its greater than 2^(n(n+1)/2)

[tribal moss]

Sounds plausible; 2^(n^2/2) was very crude lower bound.

Much cruder than I intended it to, in fact. It should be 2^(n(n-1))

[rotund aurora]

yes I realized

I took out just a 2, when I had to take 2^(n-1)

you have f(n)=(2^n-1)2^(n-1) · f(n-1), right

[delicate orchid]

$GL(n, 2) = \prod_{k=0}^{n-1} (2^n-2^k)$

lower bound is tricky

[cloud walrus]

WewGhostTbh

[tribal moss]

Right, that's what I was trying to bound with some simple closed expressions.

[rotund aurora]

yes, trying to estimate that in terms of like 2^(c n^2+...) where c is some constant

[delicate orchid]

this sounds fun

[rotund aurora]

Yes, I got that too

I used (2^n-1)2>=2^n many times, so there is considerable room for improvement

[wraith cargo]

In certain cases you can choose End(G) to be arbitrary so if it's the same case as for Aut then bounding it could be a problem

[delicate orchid]

I've got this as $2^{\frac{n(n-1)}{2}}\sum_{k=0}^n \begin{pmatrix} n \ k \end{pmatrix} 2^{nk}(-1)^{n-k}$

[cloud walrus]

WewGhostTbh

[rotund aurora]

I think I can see why that will be the worst for abelian groups, but idk for nonabelian ones

[white oxide]

so I'm trying to show that if $R$ is an integral domain, and $I$ is a proper nontrivial ideal, then $R/I$ is not flat. is this because if we let $\iota: I \to R$ be the injection, then $\iota \otimes_R R/I: I \otimes_R R/I \to R \otimes_R R/I$ is not injective since for $a, b \in I$, $ax \otimes r + I \mapsto ax \otimes r + I = x \otimes a(r + I) = x \otimes I$ and similarly $bx \otimes r + I \mapsto x \otimes I$?

[cloud walrus]

okeyBOOkay

[white oxide]

since $ar$, $br \in I$?

[cloud walrus]

okeyBOOkay

[next obsidian]

This isn’t gonna pan out for you

[white oxide]

:(

[next obsidian]

You really aren’t going to be able to argue about injectivity like this using simple tensors like this because it’s nearly impossible to say that two simple tensors are not equal

[white oxide]

what would be the best approach then?

or could you give me a hint?

[next obsidian]

The map you will tensor with which will become noninjective is a map R -> R

[white oxide]

hm ok, thanks

[white oxide]

does this really boil down to there being so many manipulations you can do with simple tensors that it's very easy to show that they're equal or smt

[next obsidian]

Yes

The relations you quotient by are insane

It’s incredibly difficult to show any particular simple tensor isn’t 0

[white oxide]

wow okay, i'll keep that in mind moving on lol

[next obsidian]

Legit

The only way I know how to show x (x) y is nonzero when you aren’t in a very particular tensor product (like with R or R/I when you can rewrite it)

[white oxide]

so what would be the best way to demonstrate non-injectivity then lol if you can't take two "distinct" elements or rather it's very difficult to take two elements that you think are distinct and show that they map to the same thing

[next obsidian]

Is to take a bilinear map M x N -> L such that (x,y) maps to something nonzero

Because then the universal property gives a map M (x) N -> L sending x (x) y to something nonzero

[white oxide]

oh yea that makes sense

[next obsidian]

Because tensoring with R/I against R just gives you R/I

[next obsidian]

That’s the answer to this

[white oxide]

yeah ok

[next obsidian]

So you aren’t looking at a map on some tensor product with fucked up tensors

It’s just a map R/I -> R/I

Which you can decipher what it does if you know what the isomorphism M (x) R -> M is

On simple tensors

[white oxide]

oh is that just because for any R-module A A(x) R is iso to A

[next obsidian]

Yes

[white oxide]

checks out

[next obsidian]

Or A (x) R/I = A/IA

You can use either

[white oxide]

hm ok

cool thanks, i'll attempt

hmm so basically if we can find a map $f: R \times R/I \to R \otimes R/I$ that has nontrivial kernel and use the fact that [\begin{tikzcd}
{R \times R/I} & {R \otimes R/I} \
& {R \otimes R/I}
\arrow["f", from=1-1, to=2-2]
\arrow[from=1-1, to=1-2]
\arrow[dashed, from=1-2, to=2-2]
\end{tikzcd}] commutes we'll automatically get an induced homomorphism with nontrivial kernel from $R \otimes R/I$ to itself

[cloud walrus]

okeyBOOkay

[white oxide]

well i guess we have to produce an explicit injective map first

uhhhhh

[delicate orchid]

Lemme try and remember how Tor works again

[white oxide]

bro's tryna browse the dark web

[delicate orchid]

Tor is overkill here + annoying to think about

You must know about exact sequences right

[white oxide]

yeah lol

i know that M is flat if and only if it preserves exactness right

with a 0 on the left

[delicate orchid]

Ok my proof strategy is a little weird, it essentially finds a condition which must follow from R/I being flat that obviously not every ideal satisfies so idk if it’s the best approach

Or rather, the best approach for an introduction

[white oxide]

perfect

[delicate orchid]

Assume R/I is flat and then apply - (x) R/I to the sequence 0 -> I -> R

[delicate orchid]

Then apply this

[white oxide]

okay i'll try that, here - (x) R/I is the tensor functor right or whatever it's called

[delicate orchid]

Yeah just tensor the sequence

Lmk when you’ve simplified stuff down

[white oxide]

oh sorry i was planning to work on this tomorrow, i'm really bad at math at night lol

but i appreciate it i'll def come back to it

[rapid junco]

Slightly lost how does one show this?

We can assume there exist 2 integers since 0 is easy.

But then if g^k is in there then g^-k is in there

[rapid junco]

Oh wait

If G is finite and G has order p or 1 then H is cyclic

So wlog G has order m where m is not prime

Then if S has elements that all divide G then we are done

So WLOG S has an element that doesnt divide G and thus Z/m is generated by that element

Wait a sec.

H is abelian.

[alpine island]

a cyclic group is by definition a group that can be generated by a single element, and H is generated by g by definition. I don't see what the issue is here.

[rapid junco]

a has to be in Z

and g is not necessarily in the subgroup

[alpine island]

Oh I see the issue then

take gcd of all a in S

that should generate it

no there's an easier way

look at $H' = {g^z, z\in\mathbb{Z}}$

[cloud walrus]

nHail

[alpine island]

that's a (cyclic) subgroup

H is a subset of H', and is known to be a subgroup of G

so H is a subgroup of H', and every subgroup of a cyclic group is cyclic

[rapid junco]

Ah

[tawdry plover]

There are a lot of small results whose proof I need to figure out but thanks 👍 this is very strong

[tulip glacier]

Isomorphic structures intuitively represent two structures which are the same (the only thing differing being the "labeling" of their elements). However, I cannot connect this intuition to the formal definition of an isomorphism (which states that an isomorphism is a bijective homomorphism). Further, I have no intuition for what a homomorphism should represent.

Any help would be appreciated.

[cobalt heath]

To make matters worse, not all bijective homomorphisms are isomorphism

It is a theorem for certain categories, like category of groups/rings/modules.

[peak root]

Well, we usually talk about isomorphism in some given settings, or isomorphism as a given structure. For example, we say A and B are isomorphic as a set when we can find a bijective map between them. We say G and H are isomorphic as groups when we can find group homomorphisms composing as identity. Homomrophism refers to structure preserving map, like group homomorphism needs to respect group operation, it is not necessarily an isomorphism (as groups).

A good way to think about isomorphism between A and B is that you can always find a structure preserving map from A to B and B to A, they compose as identity maps on A and on B.

It may require bijiectivity or may not, depends on the structure you want to preserve.

[tulip glacier]

Hmm, I see. In the context of groups, for example, how does $f(x *_A y) = f(x) *_B f(y)$ mean that $f$ "respects the structure" of the groups $A$ and $B$ ? I think this is exactly what I struggle with.

[cloud walrus]

Math Is Life

[cobalt heath]

It sends *_A to *_B.

Like maybe you can write this way

(Nvm trying to find it)

[peak root]

You could try deducing from this law the following:
A group homomorphism f: G->H sends the identity element in G to identity elements in H.

[cobalt heath]

f maps like $x *_A y \mapsto f(x) *_B f(y)$

[peak root]

And f sends an inverse a^{-1} to f(a)^{-1}

[cloud walrus]

Absta

[cobalt heath]

You can see how f maps *_A to *_B, as x, y are mapped to f(x), f(y).

[tulip glacier]

Hmmm, I think I understand.

I'll think about it more.

[feral timber]

Might be a very basic question, but how do I say that the free group generated by two elements a+b-c and c is an index 2 subgroup of Z+Z (direct sum)?

[delicate orchid]

I'm assuming you want to actually prove this rather than state this and that you mean "free abelian group" instead of "free group"?

[feral timber]

Yeah, I mean free abelian group. I don't really want to prove it rigorously, but get an idea as to how we claim that it is an index 2 subgroup?

[delicate orchid]

I don't really believe this result, or at least I can't see it either

It's much more natural to consider G = <a+b-c, c> = <a+b, c> as a subgroup of Z^3 = <a,b,c> which spans a plane as a lattice (which looks like a series of points on a plane containing the c-axis and sitting between the a,b-axes) we have that the horizontal points are spaced sqrt(2) apart by Pythagoras.

this isn't sufficient distance

if it was 2a+b I'd believe it

[feral timber]

Essentially I have to show that <a+b-c,a+b+c> is an index two subgroup of <a+b,c>

[delicate orchid]

but... that's a completely different question?

[feral timber]

Can't I say that <a+b,c> is isomorphic to Z+Z?

[delicate orchid]

<a+b-c, a+b+c> = <-2c, a+b+c> = <-2c, a+b-c> = <2c, a+b-c>
<a+b, c> = <c, a+b-c>

[wraith cargo]

Very classy

[delicate orchid]

so the quotient is <c, a+b-c>/<2c, a+b-c> = <c>/<2c> = Z/2Z

hence index 2

[feral timber]

Yeah, makes sense. But how do I know what manipulations I have to do? Would that be just trial and error?

[delicate orchid]

since we wanted to show it was index 2 I knew I needed to get a coefficient of 2 in front of something

and we had a c on it's own in <a+b, c> so I tried to put it infront of the c

[feral timber]

That is something I didn't know. I just wanted the quotient <a+b-c,a+b+c>/<a+b,c>. I don't have the information that it is an index two subgroup. How do I proceed without this information?

[delicate orchid]

You’d want to get the presentations in as similar of a form as possible

There might be some algorithms that’s eluding me like coset enumeration but that seems very overkill here

[wraith cargo]

Wew are you gonna come up with an algorithm to always know what group a certain presentation corresponds to 🤨

[delicate orchid]

That’s undecidable in pretty much all interesting cases

[delicate orchid]

Todd-Coexter enumeration is what I’m referring to here if you want to Google it

[feral timber]

So like in general, it's mostly trail and error?

[delicate orchid]

Nah there’s got to be a clever way of doing it

Because what I was essentially doing in my head there was row reduction

Have two matrices with columns corresponding to generators and rows corresponding to the coefficients of a,b,c in each generator

Then do elementary row operations to try and get the matrix into some nice form… hmm

I’ll have a think, I’ve gtg

[feral timber]

Ah, alright. Thanks for helping out

[wraith cargo]

what a cop out

[delicate orchid]

Been thinking about this. It’s correct but you destroy a lot of information, for example <a+b-c, a+b+c> is also isomorphic to Z^2, but it’s clearly not an index 2 subgroup of itself - the generators are very important!

[delicate orchid]

it it makes you feel any better it's pretty easy to get reasonably good at doing it quickly

for common classes of groups anyway

[wraith cargo]

I did enough calculating of homology back in my day to get good at it 🚬
(8 months ago)

[delicate orchid]

alright then, what is <a, b, c | a^3 = b^8 = c^2 = [a, b] = [a, c] = 1, cbc = b^3> isomorphic to

[wraith cargo]

itself

what have you been up to recently Wew
I've been gone from here for a bit

and why is my very active roll gone

[delicate orchid]

C_3 x SD_16 but yeah

[delicate orchid]

PHd shite

[tawdry plover]

If H is a subgroup and aHa^-1 is a subset of H then does this imply aHa^-1 = H

And I mean there exists an a such that this happens

[delicate orchid]

are your groups finite

[tawdry plover]

Does it matter

Okay yes they are

[delicate orchid]

no you're right, it doesn't matter

or maybe it does

cause if they're finite then |aHa^-1| = |H| and aHa^-1 subset H implies aHa^-1 = H

but I can't think of an example in an infinite group where this fails

[tawdry plover]

What if H is a nontrivial centre

[delicate orchid]

I don't see why that matters at all

you have a bijection from a finite set to a finite set

this isn't even group theortic really

[tawdry plover]

Nvm

I see

[chilly radish]

In the infinite case just conjugate by a^-1 for the converse inclusion

[untold basalt]

I'm working on a piece of code related to baic group theory and I have a little problem with the python interpreter

>>> import numpy as np
>>> np.linalg.det([[2,1],[1,2]])
2.9999999999999996
>>>

can someone explain to me why this happend and how I can avoid it? I tried sending this in the computer science discord but my message gets deleted by some bot and I really don't have time for that kind of crap.

[delicate orchid]

floating point rounding errors

[untold basalt]

but there's nothing to round

[delicate orchid]

until I see the entire source code of np.linalg.det I wouldn't assume that

anyway just tack a round(...) on the start and you're be right as rain, bobs your uncle, cheap as chips, if you will

[untold basalt]

ok thanks

[elder wave]

there's always rounding

[daring nova]

it might even be that numpy automatically converts to float64 from the get go when creating a numpy array

I think by default it takes dtype=np.float64

so it's not even det's fault

[daring nova]

also that belonged in #computing-software

[daring nova]

did you need to float that ?

[coral spindle]

Dunno I just don't know if python will automatically make it a double. I don't really care though, this illustrates the problem just fine.

[daring nova]

python's floats are double

[coral spindle]

That's nice

[daring nova]

there's no float32

[daring nova]

you had more than 15 zeroes anyways

(idk how many bits are used for the mantissa)

but with more than 25 0s, the size of the mantissa doesn't matter

64 bits can't do that

[coral spindle]

OK

[elder wave]

[daring nova]

is that passive aggressive ?

[elder wave]

this reminds me of the horrors of my numerics courses

[scarlet field]

about this one my idea is take an arbitrary permutation in S_n to show \phi(\sigma)=\psi(\sigma), and we can write the permutation as a product of transpositions if we can show \phi((ij))=\psi((ij)) for any i,j then we can prove this. But how to get this from the given condition because not all permutations all transposition have to do with (1i), are there any properties of S_n i am missing? or it has something to do with automorphism?

[next obsidian]

(1i)(1j)(1i)= (ij)

[scarlet field]

ah thanks

[flint cave]

Anyone else feel like basic field + Galois theory is much easier than the rest of algebra to think about? Like I find the theorems about fields much more natural than theorems about modules and stuff

Like for example intuition for Galois subgroups correspondence with subfields makes much more sense than structure of modules over PID to me

[delicate orchid]

do you know about smith normal form

[rustic crown]

hi wew

[delicate orchid]

because to me structure of modules over PIDs is way more intuitive than the Galois correspondence, which is actually the whole reason why we care about Galois theory - the lattice isomorphism is not obvious at first!

lots of areas of maths are just exploiting a single unexpected isomorphism - character theory comes to mind

[delicate orchid]

hello discord user det

[rustic crown]

sounds like a case of "lack of experience makes things seem harder than they actually are"

[summer path]

det

[rustic crown]

tubu

[elder wave]

exact opposite for me

[flint cave]

Oh yeah lol you might actually be right, I am just now learning Galois theory whereas I was learning the module over PID stuff almost a year back.... but still I somehow feel like I have the most intuition for field theory stuff, I am able to prove most of the theorems in Lang myself whereas in modules I would be drudging through proofs from Hungerford for a long time

[crystal turtle]

det

[rustic crown]

[hot lake]

it did something like exp(log2 + log3 - log2) cuz it computes log of the determinant so that you still have something if it underflows or overflows

[night onyx]

if you type 2.0 * 2.0 - 1.0 * 1.0 in the interpreter what do you get?

[abstract rock]

you're going to want to cast your matrix as numpy integer matrix

looking at the source it'll cast whatever matrix you're inputting as _commontype of the matrix, which i'd hazard a guess as being your matrix of floats

[rocky cloak]

If you want symbolic computation, use sympy. Otherwise there is always things to round

[untold basalt]

thanks, I did it with sympy in the end

[white oxide]

i'm kinda confused as to how we have m \geq n; I wrote f = (b_0, b_1, \dots, b_m, 0, ..., 0) = (a_0, a_1, \dots, a_m, a_{m + 1}, ..., a_n, 0, 0, ...) and am trying to derive a contradiction (supposing m < n) - is it because it would contradict a_n being the largest nonzero coefficient? but then isn't that assuming we write f = (a_0, a_1, \dots, )?

[daring nova]

Yes, you need to assume an != 0
Then any bk, k > n must be 0, and m >= n because otherwise you miss some unmissable coefficients

[white oxide]

got it, thanks

[last spoke]

My number theory take home just dropped. I'm fucked

[feral timber]

Then why did my book consider it to be Z+Z only? The context is that this is supposed to be the group generated by boundaries of two simplicial 2-simplexes.

[delicate orchid]

that certainly seems like a particular set of generators!

[feral timber]

Specifically, it is supposed to be in RP^2

[white oxide]

is this the usual construction for the ring of polynomials in n indeterminates?

[coral spindle]

Yes

A lot of textbooks will just call them formal symbols but behind the scenes this is the standard construction.

[white oxide]

cool stuff

[chilly ocean]

what are some applications of third isomorphism theorem of groups?

(first and second too if there are good ones)

[coral spindle]

Which one do you call the third one – there really isn't a universally agreed-upon numbering

But it's a bit weird to talk about 'applications' because it's just something you use whenever you have to. I use the first isomorphism theorem all the time without even thinking about it, and whenever I use any of the others it's similar.

[misty cosmos]

how would we go about describing bases for fields?

wait nvm

[coral spindle]

1 :)

[misty cosmos]

haha i forgot that the scalars are from the field itself 😂

[white oxide]

is induction the best way to go about this? it's obviously true for k = 1, and i'm assuming it's true for k and trying to show it's true for k + 1. but this is annoying since xi is not a homomorphism or anything like that

assume $x_i^k(k\epsilon_i) = 1_R$, then $x_i^{k + 1}\bigl((k + 1)\epsilon_i\bigl) = x_ix_i^k\bigl((k + 1)\epsilon_i\bigl)$ and i'm stuck

[cloud walrus]

okeyokay

[white oxide]

oh

nvm i guess it's not composition

[rapid junco]

is there a group that contains Per(X)?

that is not Per(X)

[crystal turtle]

Per(X) = permutations of X?

For a set X

[rapid junco]

yes

[crystal turtle]

Sure, pick your favorite group G and consider Per(X) x G. Or, pick your favorite superset Y of X, and consider Per(Y). These (and many other groups) will have an isomorphix copy of Per(X). Namely, Per(X) x {e} for the first and the permutations of Y which fix every element not in X for the second.

[rapid junco]

right but I mean as sets Per(X) and Per(X) x G are not subsets

I suppose Per(X) \subseteq Per(Y)

[crystal turtle]

It's just a relabelling of them, so for all intensive purposes they're the same

[rapid junco]

because the restriction of a bijection is a bijection

[crystal turtle]

I could simply relabel (f,e) as f in the first one, for any f in Per(X)

That would make it a set inclusion

[rapid junco]

I meant more like something like Inn(X) <= Aut(X) <= Per(X)

something that changes the functions themselves

not the sets

poorly posed question

[topaz solar]

Per(Y) for X < Y adds info since those functions differ?

Since ya know, being defined on Y instead

[rapid junco]

right okay

thanks

[sinful nebula]

How can I find a group G in which every proper subgroup is cyclic even though G is not cyclic.

anyone have any hints, would be much appreciated?

[crystal turtle]

Try small groups.

Like, start at n = 4 (1, 2, and 3 obviously won't work) and trial and error.

[topaz solar]

How many proper subgroups do you have for a group of order p^2

[sinful nebula]

I'm imagining in my head like the rotational symmetries of a cube but we haven't done anything like that in class yet

[crystal turtle]

I believe that has non-cyclic subgroups.

Too large.

[rustic crown]

4! > 4

[crystal turtle]

(I mean, you can make it work with arbitrarily large finite groups. But there are some simple, small examples)

(hint hint ||if you know Lagrange's theorem and the fact that all groups of prime order are cyclic, think about how that you can use that to limit the subgroups to being cyclic||)

[rustic crown]

(you essentially told the whole solution )

[crystal turtle]

(hidden )

[sinful nebula]

Hmm I don't know lagrange's thm yet

I think maybe working of the rotational symmetries of a cube intution could I have the set of ordered pairs from {1,-1}?

under the binary operation that just multiplies the corresponding elements?

[crystal turtle]

Yes that works. You might also hear this called the "Klein 4-group" V_4

[sturdy mirage]

Can anyone help me understand this? what exactly does it mean that i is an inclusion when N' is just some arbitrary A-module.

(rings are commutative, with identity)

[torn warren]

Here, the author said "unfortunately", why does he say this? Is this not the typical way?

[sturdy mirage]

looks like in general it doesnt look like inversions. as per the next sentence.

[torn warren]

in that sense, he should say "fortunately"?

[sturdy mirage]

because he tries to teach math, it is unfortunate that the example makes something seem true generally, when it isn't

thats unfortunate for him as an educator, and for students who might be misled

[torn warren]

then why not pick a different example? to avoid this coincidence.

[sturdy mirage]

ah, that i cannot tell you!

[tribal drum]

if a group G has order n, does that mean any g in G has g^n = 1_G ?

[torn warren]

true

[reef trench]

Yep

that's because the order of g (the element) divides n

And every element of a finite group has an order

(ie the cyclic subgroup)

[tribal drum]

so for this question i am showing any element of H has order dividing n and K dividing m. is saying let h in H have order k, h^k = g^mk = 1_H = 1_G = g^mn => g^k = g^n = 1_G => k divides n a valid proof? ( i have already shown H and K being subgroup of G)

because i thought g^mn doesn't necessarily = 1_G

[torn warren]

g^mk = 1_H = 1_G = g^mn => g^k = g^n this step is invalid

for example, in Z4, 1^2=2^2=1, but you can't say 1=2

[tribal drum]

oh i dont mean like they are equal i mean since they both equal the identity then k must be a multiple of n

[alpine island]

do you mean "for all intents and purposes"?

[crystal turtle]

You understood what I meant

[delicate orchid]

it's irrelevantgardless.... these sayings are a diamond dozen in this doggy dog world....

[white oxide]

is there any nonpainful way to prove (iii)

[white oxide]

is that you in ur pfp

for some reason i wouldn't be surprised if it was

[delicate orchid]

no this is "the nostalgia critic"

[white oxide]

i see

good man for that

[chilly radish]

Wew lost no nostalgia November.... Like he does every year

[white oxide]

oh shit i didn't even know it was november

well ig it doesn't count then

[delicate orchid]

this is actually true, I failed it the past 2 years

[white oxide]

it's aight champ

you'll get 'em next time!

[broken quartz]

does anyone know if a field is closed under multiplication

[white oxide]

it is indeed closed under multiplication

[broken quartz]

so this proof essentially answers this question, no?

since q = p^f, the field_q has prime characterisitc

i got it from my textbook, just wanted to check if it does or not.

[fair lantern]

Does anyone have an example of a finite-rank module that contains a submodule of non-finite rank? And a basic question: Let M be a module and N be a submodule of M, Is it true that rank(M/N)=rank(M)-rank(N)?

[chilly radish]

Take Z[x1,x2,...] I.e. ring of polys in inf many variables, and consider the module (x1x2,x1x3,x1x4,...) Inside (x1)

I think that's the standard example

The key point is lack of noetherianity

Also the rank of a general module over a general ring is not a well defined notion

What you said is true under suitable assumptions on the ring and the modules

[white oxide]

if $f = \sum_{i = 0}^m a_ix_1^{k_{i1}} \dots x_n^{k_{in}}$, $g = \sum_{i = 1}^t b_ix_1^{j_{i1}} \dots x_n^{j_{in}}$, then how can you give an expression for what $f + g$ is? because this will be a hassle and involve combining like terms and stuff no?

[cloud walrus]

okeyokay

[white oxide]

cuz i'm trying to verify this and.... it's a pain

[solemn hollow]

a and b can be 0

rewrite your polynomials so you can add them

easily

[white oxide]

Wdym

[broken quartz]

thats actually so weird

literally same question on my pset

[fringe heath]

hi! how would i go about approaching this / these types of questions?
my first instinct would to be list all of them and check and applying some small generalizations to save checking them one by one
but clearly, the question does not want you to do this
thus, is there a better way other than trying different combinations of linear polynomials multiplied together?

[delicate bloom]

I'd try enumerating the complement

[fringe heath]

what do you mena?

mean

[delicate bloom]

count the reducible polynomials

[fringe heath]

oh right yeah
thats what i was thinking of doing but it seemed long
which is what i was implying by trying different combinations of linear polynomials multiplied together

i mean it could be feasible if theres a way to count faster without trying everything

[delicate bloom]

not sure we're on the same page exactly

let's just think about monic polynomials for a sec, how many ways can I pick the roots a,b here: (x-a)(x-b)

I suggest solving this by thinking about (Z/p)[x] in general

then plug in p=5 in the end

[fringe heath]

for distinct a,b in z/5 its 5C2
if not distinct its 5C2 + 5 surely, unless im not understanding what you mean?

[delicate bloom]

yeah

so we're almost there

# reducible polys + # irred polys = # all polys

[fringe heath]

yeah

oh i think i see where this is going

monic polynomials can be defined by their roots
the number of monic polynomials in z/5 of degree 2 is 5^3 = 125
number of reducible monic polynomials in z/5 is nothing but 5C2 + 5 = 15 thus there are 110 irreducible monic polynomials in z/5?

[delicate bloom]

the number of monic polynomials in z/5 of degree 2 is 5^3 = 125
not quite

[fringe heath]

oh wait

just 25

dont have to enumerate x^2 coeff

so 25-15 = 10

[delicate bloom]

depends on which you're wanting to count, monic or not

[fringe heath]

yeah yeah i brain farted
25 monic, 125 total

[delicate bloom]

you missed some

[fringe heath]

hmmm

[delicate bloom]

or I missed some lol

[fringe heath]

probably me

i was thinking like

5 ways to choose constant, 5 ways to choose coeff of x

[delicate bloom]

nah you'r eright I'm just tired

[fringe heath]

ok ok

[delicate bloom]

there aren't 125 degree 2 polynomials though

[fringe heath]

100

125 - 25

[delicate bloom]

I was thinking slightly differently, 4*5^2

since 0 can't be a choice, there are 4 options

[fringe heath]

yeah hahah thats the same thing

[delicate bloom]

but yeah both work, just wanted to give a diff perspective

[fringe heath]

yep yep

im not sure how to generalize after that because i was going to say like

thinking about (ax-b)(cx-d) next then either a,d = 1 with 25 choices for pairs of b,d
or otherwise, we want a|b and c|d which doesnt particularly work the way you want it to since
3 doesnt divide 2 for example, but 3|12 (since 12 \equiv 2 \mod 5)

with my first thought at least

[delicate bloom]

no need, just the roots is enough in a field

then worry about fixing the leading term afterwards

there will be p-1 ways to choose what to multiply (x-a)(x-b) by to get a different leading coefficient

[fringe heath]

yeah, that makes sense
and we dont need to think about cases where we have (ax-b)(cx-d) since at some point with c(x-a)(x-b) it will be equivalent? (the a,b,c,ds arent the same between the two expressions)

[delicate bloom]

yeah

[fringe heath]

right

so really, since each constant must be a unit in Z/p (which is probably why you were insisting on looking at prime p in the first place), then we can just multiply everything by p-1

from the monic case

wait

[delicate bloom]

yup

your phrasing seems slightly off but maybe you mean

[fringe heath]

surely though if k(x-n)(x-m) is equivalent to (ax-b)(cx-d) and k is a unit, the number of reducible polynomials must be fixed?

for the reason that we dont care about multiplying about units when considering irreducible elements ?

or am i mixing somethign up

[delicate bloom]

I'm sorta unsure what you mean, trying to count by (ax-b)(cx-d) isn't clear to me cause it mixes up the coefficient on the x^2 term up with the roots now

so it's a pain to try to count that way as far as I can tell

[fringe heath]

hmmmm

alright alright

[delicate bloom]

there are p-1 quadratic polynomials that have the same roots, they just differ by a different leading coefficient

I think you got that earlier but just reiterating it to be safe

[fringe heath]

yeah yeah

[delicate bloom]

or like are you wanting to try to count it a different way

[fringe heath]

no no i understand what youre asying its fine hahaha

monic case:
25 polynomials
15 reducible
general case:
100 polynomials
15*(p-1) = 60 reducible

huge, thanks for the help

[delicate bloom]

yeah, you're welcome

although I'd suggest working out the entire thing in terms of p, not just that last little piece lol

[fringe heath]

yeah yeah hahaha

i dont know why i put that

thank you thank you

[delicate bloom]

it will make the relation between the monic and non monic case much clearer that it is what you'd think it'd be

[fringe heath]

nah you explained it pretty well in my opinion dont worry

[delicate bloom]

haha I wasn't, I'm just saying seeing plain numbers is lame and to do it more generally 😛

[tardy hedge]

Intuitively, what makes a subgroup normal? I think its something to do with the normal subgroups group structure sort of "builds" the bigger group structure , where as regular old subgroups may not

[rocky cloak]

A subgroup is normal iff it is the kernel of a group homomorphism, i.e. iff the quotient G/H is also a group in a natural way.

[tardy hedge]

ahh yea i saw that theorem in the book but havent properly read through it and thought about it yet

[rocky cloak]

So in this sense if H < G is a normal subgroup you can "build" G out of H and G/H

[tardy hedge]

Ok thanks, that part is kind of what I was thinking

Still seems like for me it would be hard to look at some subgroup and intuitevely tell without checking that its normal

its quite an interesting and fun idea to me though

[rocky cloak]

It can indeed be difficult to determine if a subgroup is normal, depending on what you know about it

[tardy hedge]

i see

[cloud walrus]

Couldn't find an attached image in the last 10 messages.

Nov 2, 2023, 2:07 PM

[tawdry plover]

How to do number 20

I was trying to construct a homomorphism to a abelian group but the determinant one isn't just working

And I couldn't come up with more ideas

Okay I just noticed that N is a elementary matrix

So uh maybe we could make some combinatorial arguments

[daring nova]

((a, b), (0, d)) ((a', b'), (0, d')) = ((aa', ab'+bd'), (0, dd'))
So A N_b' = ((a, ab' + b), (0, d)) = A + ab' E12
So it looks like the classes can be chosen to have representants ((a, 0), (0, d)), i.e. the diagonal matrices ?

from which you should clearly see that it's an abelian group

[tawdry plover]

What do u mean by N_b'

[daring nova]

N_b = ((1, b), (0, 1))

[daring nova]

well you could also have seen that by interpreting it using column operations, but either way, you find this result, right ?

[tawdry plover]

I am bad at seeing things 😭

I need some time to process your ans

[daring nova]

I just computed what the class of A looks like

and if you choose b' = -b/a (hence ad != 0 is useful) you find a diagonal matrix in its class

[tawdry plover]

Ah

Okay very cool

Thank you

[daring nova]

then this follows from multiplication between diagonal matrices being commutative

[tawdry plover]

I can accept that

[daring nova]

you should convince yourself of it

[tawdry plover]

It's easy to check

[daring nova]

yes

[tawdry plover]

Also it's just scaling operations in terms of elementary matrices

[daring nova]

and that this means the quotient group is commutative

[tawdry plover]

Or linear transformation

Which is convincing enough on why they should commute

[tawdry plover]

I was trying to construct an explicit homomorphism so that I could kill two birds with one stone by using first isomorphism theorem

But this was easier

[daring nova]

what

[tawdry plover]

If I could make N a kernel

[daring nova]

that would make it normal

[tawdry plover]

Yeah and if it's image group was abelian

Then I would get b

[daring nova]

since G/N is isomorphic to the image

[tawdry plover]

Yeah

Couldn't come up with one 😞

[daring nova]

well once you found the classes, mapping the class to the diagonal as a subset of R*² gives a commutative group with kernel N

seems quite natural

[tawdry plover]

Right

[daring nova]

and this formula guarantees it's a morphism

[rocky cloak]

Intuitively G is controlled by a, b and d, and N is controlled by b. So you would expect G/N to look like something involving a and d. So the first thing you might check could be diagonal matrices with ad nonzero. And indeed, in that case, there is a homomorphism with kernel N

[white oxide]

just to verify: all R-modules are Z modules right?

[agile burrow]

Yes, because every R-module is an abelian group and abelian groups are Z-modules

[white oxide]

ah right

[agile burrow]

Equivalently, there's a unique map Z -> R which lets you pull back the R-module structure to a Z-module structure

[white oxide]

so i'm a little bit confused as to why precisely $A \otimes_R B \neq A \otimes_{\mathbb{Z}} B$. because for instance, if we consider the left hand side and let $A$ be a right $R$-module and $B$ a left $R$-module, don't we get equality?

[cloud walrus]

okeyokay

[white oxide]

oh hm neat

[crystal turtle]

They're entirely different constructions. Don't let the notation trick you into expecting them to be the same.

[white oxide]

oh so for example one element of tensor product may equal zero on the lhs

but be nonzero on the rhs

because that's the only distinct relation you're quotienting by

like lhs you quotient by $(ar, b) - (a, rb)$ and rhs it's $(an, b) - (a, nb)$ for $n \in \mathbb{Z}$ and $r \in R$