groups-rings-fields

[keen sparrow]

yea it is

[scarlet estuary]

sorry i dont understand what the confusion is then

[frank lake]

it's gonna take a while for me to adjust to this stuff, when you're used to doing HS algebra

😂

[scarlet estuary]

a subgroup needs to be closed under multiplication and inverses

[frank lake]

mhm

why only multiplication though?

[keen sparrow]

What does xa^-1 = a^-1x

[frank lake]

why not other operations

[scarlet estuary]

"multiplication" means the group operation

[frank lake]

oh

[scarlet estuary]

common abuse of terminology, sorry

these are the 2 parts

[shrewd junco]

you have coordinate spaces like R^3 right? like where you plot 3D points. Is it an algebraic structure?

[scarlet estuary]

pink: closed under the group operation

[frank lake]

mhm

[scarlet estuary]

blue: closed under inverses

[scarlet estuary]

vector space

[shrewd junco]

R is a field but idk about R^n

[keen sparrow]

@shrewd junco yes for one it's a vector space

[shrewd junco]

but it doesn't have addition and scalar multiplication

[scarlet estuary]

if you just mean the set R^3 with no operations

[shrewd junco]

yes

[keen sparrow]

That's weird

[scarlet estuary]

then it cant be an algebraic structure as it has no structure

its a set

[shrewd junco]

ah right just a set

thanks

[scarlet estuary]

in algebra, we dont care about the sets, just structures on them

[shrewd junco]

right like what algebraic operations are defined on them

[scarlet estuary]

the sets we use are just convenient ways to label elements in a structure

if you have no operations you have no structure

[shrewd junco]

ah okay

[frank lake]

thanks for the help @scarlet estuary @keen sparrow

[scarlet estuary]

(this is why isomorphism means two structures are "the same")

[shrewd junco]

so a vector field is a mapping from a set to a vector field right?

[scarlet estuary]

to a vector space

[keen sparrow]

It's also important to consider maps from and to our algebraic structure

[scarlet estuary]

formally we say to a tangent bundle

[shrewd junco]

right space

[scarlet estuary]

which in practice means that the base fields need to "agree"

like we wouldnt call a map from Z/5Z into R^3 a vector field usually

we'd say "a vector-valued function"

but "vector field" is a bit more restricted

[shrewd junco]

I see

[scarlet estuary]

that said, such examples dont really come up naturally anyway

so thats more semantics than anything

(i say "agree" but that doesnt necessarily mean "the same", just like... compatible)

(like people would be fine if you mix and match R and C)

(they play well together)

[shrewd junco]

isomorphic?

oh nvm

[scarlet estuary]

"vector field" is a term that mostly comes up in applications or physics so like

just imagine "does this assignment of vectors to each point in a space seem to carry some intuition"

if so, calling it a vector field is probably fine

but if youre taking a subset of R^2 and your output vectors are from the set of continuous functions on [0, 1]

calling that map a vector field is a stretch

you cant really imagine drawing tangent arrows or whatever, after all

again though those dont really come up naturally anyway

so the point is kinda moot

[shrewd junco]

right yeah

[shrewd junco]

what exactly is it? I've seen that a lot

"tangent bundle"

it's from differential geometry right?

[scarlet estuary]

yes

the definition isnt too bad though

as long as you blackbox what a manifold is

(a tangent space at a point is a vector space holding all the vectors tangent to that point when embedded in euclidean space)

[shrewd junco]

what's that square union symbol?

[scarlet estuary]

disjoint union

basically we need to "distinguish" tangent vectors from different tangent spaces (ie associaited to different points)

we do this by making them actually ordered pairs

where the first entry in the pair is the point in question

[shrewd junco]

im too tired get this rn im just gonna go eat ice cream and come back to this

[lavish nexus]

say $G = G_1 \times G_2$ and $H \leq G$. Is it true that $ H = H \cap G_1 \times H \cap G_2$

[cloud walrus]

Iteribus

[lethal dune]

H \meet G1 not defined

[trim grove]

is there any non trivial group homomorphism form S3 to (Q,+)?

[next obsidian]

No

Anything in the image of S3 has finite order

Or well “the image”

[white nymph]

how to do this problem? it's been a few years since ive taken a course in linear algebra... in a.) if phi invertible then Im(I) = Im(phi^0) = V = Im(phi), but if phi not invertible i don't know how to proceed rigorously (i have some example operators in my head that satisfy both this and part b, like the derivative operator...)

also, if this should instead be in the linear algebra tab, let me know

[rustic crown]

notice that as $n$ varies, $\operatorname{im} \phi^n$ gives you a decreasing sequence of subspaces of $V$.

[cloud walrus]

det

[rustic crown]

What does finite dimensionality of V tell you?

[white nymph]

its not necessarily strictly decreasing though, right?

like it could stablizie before zero

[rustic crown]

yep! the thing is it has to stabilize!!!

[white nymph]

but in either case it stablilizes, yes?

then in part b we show it for the two cases?

case 1 phi^{m-1} = 0, case 2 stabilize before zero?

[rustic crown]

there's no need to take any cases.

[white nymph]

ok

[rustic crown]

when it stabilizes, we get an m such that im phi^{m-1} = im phi^{m} = im phi^{m+1} = ...

that's what part (a) asks for

for part (b) try taking a vector in the intersection, and argue if it has to be 0 or not.

(lemme know if you want me to say a little more)

[white nymph]

thanks - im trying to put it together myself right now

[trim grove]

i am confused in this question , please give me a hint

[sturdy marsh]

Try looking at b = -a to get rid of all but one option

and prove the correct option

[white nymph]

yeah @rustic crown , theres something im missing here. letting v in intersection we get that v = phi^{m}(x) for some x in V... but v being in ker(phi^m) gives us that 0 = phi^{m}(v) = phi^{2m}(x) and im not seeing something (or going about this entirely wrong)

[rustic crown]

oh, you're almost there. notice that if images stabilize, then so do the kernels because their dimensions are just dim V - dim images.
ker phi^m = ker phi^{m+1} ....

[white nymph]

ahhh. finite dim vector space so pleasant

[trim grove]

thankyou that was easy

[white nymph]

thanks much det

[rustic crown]

[white nymph]

i knew i was missing something semi obvious 😄

[lavish nexus]

I'm using $H = (H \cap G_1) \times (H \cap G_2)$ and since both $H \cap G_1$ and $H \cap G_2$ have to be normal they could be either 1 or $S_4$ or $A_4$ or $V_4$ so considering the order of the product they cannot be the last two?

[cloud walrus]

Iteribus

[hidden haven]

I don't think that first equation is true

H could be the diagonal

all elements of the form (g,g)

[lavish nexus]

hmm

still they have to be normal

the intersections normal to H

and because they are contained in H $(H \cap G_1)(H \cap G_2) \leq H$

[cloud walrus]

Iteribus

[lavish nexus]

but the intersection of the intersections is {1}

so the order argument still works?

[hidden haven]

They could both be V_4?

or maybe like V_4 and 1

[lavish nexus]

right but there should be some order 3 elements I think

[hidden haven]

They could be somewhere other than the axes

[lavish nexus]

only V_4 and 1 is possible

by lagrange

[rustic minnow]

Is this correct for the two binary operations defined?

Here this holds for all x,y in R for the binary operations

[pastel cliff]

is this just... a fact? or is there some way to arrive at it? or rather is it just by the nature of e?

[lavish nexus]

so if I take $(H \cap G_1) = V_4$ and the other to be {1}, all order 4 elements are of the form (g, g'), g, g' order 4, since if g is not order 4 it has to be order 2, but then (g,g')^2 = (1, g'^2) so $H \cap G_2$ is nonempty
then suppose g = (abcd) = (ad)(ac)(ab). multiply (g,g') by (ab)(cd) $\in H \cap G_1$ we get ((ac), g') of order 4 which is again a contradiction

[worthy haven]

$e^{ix} = \cos(x) + i\sin(x)$ is a fact that you can prove in several ways! Do you know any calculus?

[cloud walrus]

Nick

[worthy haven]

and the fact I stated is a special case of that with x = 1

[pastel cliff]

can u gimme a hint without giving it away

[worthy haven]

well do you know any calculus

that's question 1

if you do, taylor expand both sides and see what happens

[pastel cliff]

yeah i can do that ~~after going back through my analysis notes ~~

actually i might remember it off the top of my head but i dont have pen and paper on me

will try later, merci for tip

[cloud walrus]

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

[worthy haven]

there's also a nice geometric interpretation that you can see here @pastel cliff . The idea is that the derivative of e^(ix) is ie^(ix), which is perpendicular to e^(ix), and so you can geometrically argue that e^(it) parametrizes a circle, from which the result follows, since points on the complex unit circle are in the form cos(t) + isin(t

https://www.youtube.com/watch?v=v0YEaeIClKY

[rustic minnow]

could someone just check that I showed associativity for the two binary operations. I believe I got it right, just unsure since I havent solved many of these problems.

[hidden haven]

ye it looks fine

[rustic minnow]

ok thanks

[worthy haven]

Quickly skimming this looks correct to me

[pastel cliff]

god i need to start using LaTeX again

[pastel cliff]

based 3b1b

[rustic minnow]

I do all my homework in latex, so its somewhat more inspiring to read through before exams

[pastel cliff]

so do i but not while im self studying

which i should

also is there usually a visual aspect to complex numbers? ive never formally worked with them but the section in my book that covers complexes has more pictures than expected

[lavish nexus]

yes

[pastel cliff]

idek if this is the right channel but im going thru my algebra book so

[pastel cliff]

dare i ask... why

probably a loaded question but anything you can elaborate would be cool

[worthy haven]

complex numbers are often represented with a 2D plane, where one of the axis is the real numbers and one is the imaginary numbres

so you'll often think of 1 + 2i as being the point that's one unit right and 2 units up in a 2D plane

[pastel cliff]

that much i do know

so perhaps im just not used to it

[lavish nexus]

complex plane

and if you add infinities you get a sphere

the top is infinity

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

[wooden ember]

would someone have some intuition on why a short exact sequence $$1\to K\to G \to H\to 1$$ that left splits must right split?

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

like i see that there being a left split is a strong condition on K, and that varying K also has an impact on H but i still dont quite see the full conclusion

[hidden haven]

A left splitting is like a projection from G to K, in the sense that if there is a left splitting then G becomes isomorphic to K x H with the first map looking like the inclusion of K as K x 1, and the second map looking like the projection onto H. Then the right splitting is given by inclusion of H as 1 x H

[wooden ember]

hmm i see

but in the case where H is a subgroup of G

does a left split force normality on H then

(i guess K would have to be a subgroup too for that)

[hidden haven]

I haven't worked much with sequences of non abelian groups so I would have to think about this 🥴

[wooden ember]

that's aight

[hidden haven]

K can be assumed to be a subgroup, the first map is injective

[wooden ember]

im just trying to reconcile my understanding of direct products and semidirect products with short exact sequences

seems like an important perspective

[wooden ember]

true

[hidden haven]

And H is a subgroup as 1 x H, and that is normal in K x H

[wooden ember]

yeah

[hidden haven]

Wait does splitting even imply that G is isomorphic to K x H for nonabelian stuff

I can only see K being normal

lol

[wooden ember]

it just feels weird because it doesnt patch up with any standard group theory theorem. Like how have i never heard: if $K$ is a normal subgroup of $G$ such that $G$ retracts onto $K$ then $G\cong K\rtimes G/K$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

i also just dont see how H acts on K

[hidden haven]

So if the maps are f: K to G and g: G to H, and the left splitting is h: G to K so that hf = 1

Then we are identifying G with something like

[wooden ember]

nvm this this would be for a direct product, semidirect would be a right split

[hidden haven]

x maps to (fhx, x(fhx)^-1)

I think

This is what we do for modules

Does this work here

if not then I have no clue

[wooden ember]

hmm

trying to wrap my head around this

[hidden haven]

We are saying G is product of fhG and 1(fh)^-1 G

[wooden ember]

ah i suppose that makes some sense

[hidden haven]

fh is idempotent, so just exploiting that

[wooden ember]

like if we have the retract $\rho : G \to K$ and we want to see "how" an element of $x$ of $G$ maps to $K$ we'd look at $x\rho^{-1}(\rho(x))$ right

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

wait i have to go eat get back to this later

[white nymph]

since the above conversation is taking a break... could someone help me get started on this problem? my linear algebra is a bit rusty

[hidden haven]

Try taking a basis of the kernel, extending it to a basis of V. Then choose basis of W in the only way that makes this work

[lavish nexus]

It's clear that the intersection $H_1 \cap H_2H_3....H_n$ where they are Sylow subgroups is trivial but how do I show the product is the whole group?

[cloud walrus]

Iteribus

[hidden haven]

Look at the order

[lavish nexus]

oh duh

thanks

also I'm still not sure how to work this one

[hidden haven]

uh are these part of some test

given the rapid fire

[lavish nexus]

nah I'm prepping for quals

[hidden haven]

oh alright

[lavish nexus]

these are ancient quals

[next obsidian]

The assumption here says that (I + J)/J is the unit ideal

This should imply that (I + J^2)/J^2 is the unit ideal

Well… that’s what this problem is asking you to prove

[lavish nexus]

what is the unit ideal

do you mean the whole ring

[next obsidian]

I guess what I should clarify is, you just need to show 1 exists in the ideal

And yes

In fact by the correspondence theorem

You don’t even need to mod out

Just take I + J = (1)

And show I + J^2 = (1)

This is “easy” in the sense that it’s kind of a standard result

In fact, you can kinda cheese it lol

Using ideal arithmetic

(I + J)^2 = I^2 + IJ + J^2 < I + J^2

But the left hand side is also (1)^2 = (1)

[white nymph]

what is "the only way that makes this work"? im rusty on this... do i just apply L to every vector I made in the basis for V? If so, is this a spanning set in W? Then i guess i could reduce it to a basis, but how would I know that i would get the desired matrix?

[hidden haven]

Yes apply L to every vector in this basis. The basis elements in the kernel go to 0. For the other ones, prove that the images form a linearly independent system. Extend that system to a basis, and you get a matrix in exactly that form

[white nymph]

ahhh, thanks much!

[wooden ember]

so back to this, how would one describe the homomorphism from H onto automorphisms of K when a semi direct product arises from a right split

[lavish nexus]

this is quite some cheese
thank you

[next obsidian]

I think doing it with ideals is actually the only way to do it that isn’t omega painful

[next obsidian]

You track how things multiply

You know that it works something like (it depends on which goes where) but roughly like

(g1,n1)•(g2,n2) = (g1phi_n1(g2), n1n2)

So you need to determine what phi_n1 is

So take g1 = e and n2 = e

Then your product gives you
(phi_n1(g2),n1)

Range g2 over all g in G

Then this tells you what phi_n1 is because it gives you ever output

[wooden ember]

you mean g in K?

[next obsidian]

Sure

I don’t know what your setup is

Ppl do different ones, which group is on left or right, etc

This is for G semi direct N

[wooden ember]

im not sure how i get phi_n1(g2) though

idk how the semi direct product looks in this case

[next obsidian]

Oh I see

Well this is given by conjugation sorry

[wooden ember]

it's for a right splitting short exact sequence 1->K->G->H->1

[next obsidian]

Like the way it works is

[wooden ember]

ohhhhh

[next obsidian]

This is an internal semi direct product

[wooden ember]

like you take the preimage of H through the right split

[next obsidian]

Yeah

Or no

[wooden ember]

and the inclusion of K

and conjugate?

[next obsidian]

You take its image

Under the splitting map

Or umm

[wooden ember]

yeah that's what i mean

[next obsidian]

Yeah

Then you do the internal semi direct product

[wooden ember]

ah i see

[next obsidian]

I forget what conjugates what

you can look this up I bet

But yeah internal semidirect products are conjugation

So it’s probably like umm

[wooden ember]

well it should be $\phi(h)=\operatorname{Ad}_{\rho(h)}^{\iota(K)}$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

where rho is the splitting map and iota is the inclusion

[next obsidian]

g1n1•g2n2 = g1n1n2(n2’g2n2)

But this doesn’t seem right

Idk

Lol

I forgor

Maybe

[wooden ember]

since K is isomorphic to i(K) and H is isomorphic to rho(H) i should just be able to take conjugation representation in G no?

[next obsidian]

Yeah but I forgot like what conjugates what

[wooden ember]

wdym

[next obsidian]

Like

Where do you put the inverse

Blah blah

was it ^-1 on the left

Or the right

Etc etc

I forgor

[wooden ember]

oh yeah it's a right action i think

so ^-1 on the right

hmm ill think over this again tomorrow

[simple mulch]

Hey I have a question

V a vector space

[cloud walrus]

mns

mns

[simple mulch]

I feel this map is not yet correct

[viscid pewter]

oh, second iso theorem again lol

[simple mulch]

Hum, I have this

oh I could map A+B -> A as a+b -> a

[next obsidian]

You want a map the other way

A -> A + B

You can’t define a map A + B -> A in any meaningful way

[simple mulch]

Could I map A -> A + B as a -> a + b

[next obsidian]

That’s not a valid element

B isn’t an element of V, it’s a set

What is b

[simple mulch]

what about

a -> a + e

[next obsidian]

What’s e

[simple mulch]

since B is a subspace it has e the identity

[next obsidian]

Okay let’s call that 0 lol

[simple mulch]

yeah

0

lol

[next obsidian]

Yeah so it’s just the inclusion

This is the right map

[simple mulch]

A -> A + B as a -> a + 0

[next obsidian]

Now take quotients

Just say a -> a lol

This map is injective

Then you compose with A + B -> (A + B)/B

So on elements this is a map sending a to a + B

When is a + B zero?

[simple mulch]

when a + B = 0 + 0

[next obsidian]

Also A -> (A + B)/B is clearly surjective since we’ve modded out by B

[next obsidian]

No…

This is an element of the quotient

When is an element of a quotient zero?

[simple mulch]

good question, i don't know

[next obsidian]

You should review the definition of a quotient before continuing

It’s not going to be productive to try to prove an isomorphism theorem relating quotients if you’re unsure what the quotient really is

[simple mulch]

I mean a quotient has equivalence classes

[next obsidian]

You want to determine when a + B = 0 + B

What’s the definition for equality mod B?

[simple mulch]

I mean, I see a = 0 ?

[next obsidian]

.

[simple mulch]

I don't know

= mod b ?

this?

[next obsidian]

This is why I say you need to review what quotients are

You’ve constructed equivalence classes, so what’s the equivalence relation

This is what determines when things are equal

[simple mulch]

oh so a + B = a' + B when a - a' in B

[next obsidian]

So apply that here

[simple mulch]

when a - 0 in B

[next obsidian]

Which says?

[simple mulch]

a in B

[next obsidian]

So going back to here

Your map is A -> (A + B)/B

Sending a to a + B

So when is a + B zero?

you answered this above, I just want to go back to that point

[simple mulch]

I don't understand what you mean with when is a + B = 0

[next obsidian]

a + B is an element of (A + B)/B

I’m asking when it’s 0

You have to figure out the kernel of this map

[simple mulch]

so B + a in (A+B)/B

when b = 0

[next obsidian]

This doesn’t make sense

Do you mean a = 0?

[simple mulch]

I have a question

[next obsidian]

Okay

[simple mulch]

We know that a in A and we've checked that a in B, right?

[next obsidian]

This is what’s supposed to happen, yes

But I’m not sure that you are able to justify it yet

[next obsidian]

That would follow after you can answer this bit

[simple mulch]

(A+B)/B has quotients of the form B + (a+b) where a in A and b in B

[next obsidian]

Yes

Well no

“Cosets”

[simple mulch]

yeah

cosets

[next obsidian]

Is the term for those but yeah

[simple mulch]

so since a in A and a in B, what follows is that B + (a+b) = B + a is possible for a = 0 or b = 0 (according with above) ?

[next obsidian]

I mean yes

That would result in a + b + B being zero

But that’s not the only way

First of all, mod B you can just drop b

a + b + B = a + B since their difference lies in B

Then a + B is zero if and only if a is in B

[simple mulch]

I see

[next obsidian]

This point shows the map is surjective

[next obsidian]

This shows the kernel is A\cap B

The map here being A -> (A + B)/B

a maps to a + B

[simple mulch]

I don't think I've proved this before

[next obsidian]

I literally just justified it

I took an arbitrary element of the quotient and showed it was equal to something in the image of the map

[simple mulch]

I will recapitulate

Define f : A -> A + B as a -> a

Our goal is to show (A+B)/B is isomorphic with A/(A cap B)

one way is by defining a function between the two sets and show it is well defined

[next obsidian]

I think this is the only way lol (nvm I guess you can show they have the same rank but it eventually reduces to a certain map. Ignore me)

[simple mulch]

So we define g : A/(A cap B) -> (A+B)/B

as what?

[next obsidian]

We use the first isomorphism theorem

So it suffices to find a map A -> (A + B)/B that’s surjective with kernel A\cap B

[simple mulch]

oh

[next obsidian]

I mean you don’t have to do it that way but

You’re basically in the process reproducing the first isomorphism theorem

On the quotients themselves it looks like

a + (A\cap B) -> a + B

[simple mulch]

I've already proved the first theorem but I don't see this one

yeah

I was thinking about that

[next obsidian]

I mean this is what the first isomorphism theorem gives you

[simple mulch]

more like a + (A \cap B) -> f(a) + B

[next obsidian]

Like that’s the exact map you get from A -> (A + B)/B

You don’t need to use f

a is a literal element of A + B

[simple mulch]

well yeah

[next obsidian]

But if you’ve proved the first isomorphism theorem you should just do this

Manually showing the map there is well-defined is pointless

Since the first isomorphism theorem gives you that it’s well-defined

[simple mulch]

I just don't see how you relate this with the first theorem

[next obsidian]

Because then the first isomorphism theorem tells you that A/(A\cap B) is isomorphic to (A + B)/B

Write down what the first isomorphism theorem says and this is immediate

The image is (A + B)/B because the map is surjective

And the kernel is A\cap B

[next obsidian]

Do you know what surjective means?

[viscid pewter]

first iso says: with a homomorphism h, G/kerh is isomorphic to imh

G = A. if there exists h such that kerh = A cap B and imh = (A+B)/B, then:

G/kerh is iso to imh

ie. A/(A cap B) is iso to (A+B)/B

[simple mulch]

onto

[next obsidian]

Yeah

So if you have a surjective map A -> (A+B)/B

That tells you A/kernel is isomorphic to (A + B)/B

I specified that the kernel is A\cap B

So plug that into the theorem

[simple mulch]

oh

lol

I see it now

I had my arrows reversed

[next obsidian]

I have to go now

Everything you need to prove this is in here

[next obsidian]

I produced a map A -> (A + B)/B

And have proved it’s surjective and has kernel A\cap B

try to understand what’s written there, best of luck

[simple mulch]

f : A -> (A+B)/B as a -> B + a

[next obsidian]

Yes

Proving surjectivity and computing the kernel only requires understanding the definition of equality in (A + B)/B

Anyway I gtg

[simple mulch]

If I prove the kernel of the function above is A cap B then it follows

Thank you very much!

Cya!

Basically now I just have to prove that Ker(f) = A \cap B

[lavish nexus]

First two are done. Then either there is 1 group of order 5 or 31 of them

so there is 1 group of order 155 or 31 of them and they are all conjugates?

and let H be a group of order 155, so |G/H|= 13 and it is impossible for all of them to be conjugates?

[willow mason]

just use sylow

[lavish nexus]

sylow gives either 1 or 31

[willow mason]

the 2nd and 3rd theorems

and then case bash

the other

or use group actions

[next obsidian]

You can’t use Sylow for 155

It isn’t a prime power lol

You have to first prove there’s a subgroup of order 155 although this is very simple

[lavish nexus]

it's a product of some group of order 5 and the normal subgroup of 31

[next obsidian]

Yeah exactly

[lavish nexus]

do I map G_13 into Aut(G_155)

[next obsidian]

I’m not really too sure how to show it’s normal

It’s either index 13 or 1, so the normalizer is either itself or the entire grip

So you assume the normalizer is itself

[lavish nexus]

it seems there's only trivial hom mapping this way

so G_13 and G_155 should commute?

[next obsidian]

I suppose so

You’re letting G_13 act via conjugation right?

[lavish nexus]

yeah

[next obsidian]

But why do you think there’s only the trivial homomorphism?

G_155 is a semidirect product of Z/5Z and Z/31Z

[lavish nexus]

hmm how do I calculate |Aut(G_pq)|

I remember I'm seen something like this in the books

[next obsidian]

I mean okay

There’s a thing about groups of order pq

Was it like for p < q

p not dividing q-1

?

If that’s the case it’s just the product in which case the automorphisms is the product cuz coprime stuff

But I think here it’s either the product or the single non-trivial semidirect product

So in one case the order is 120 so there’s no non-trivial map by divisibility

But idk how the semidirect product works

Man my group theory skills have atrophied hard

:(

Wait

Wait wait wait

Oh, no it’s 5 and 31

Not 31 and 13 sad

[lavish nexus]

I mean there is a semidirect product still

5|30

[next obsidian]

Yeah

I don’t know how to compute the automorpjism group of a semidirect product tho

@lavish nexus I asked a friend

And their solution was so simple

I am kinda mad I didn’t see it earlier lol

You do it entirely differently

Look at G/G_31

This has order 5•13

Which is Z/65Z

This has a normal subgroup of order 5

So just use the correspondence theorem, this gives you a normal subgroup of order 5•31 = 155

[lavish nexus]

🤦‍♂️ 🤦‍♂️ 🤦‍♂️

ty

[next obsidian]

Same

[lavish nexus]

I wish my qual is like this
so much easier than the previous ones

[next obsidian]

My school’s quals are so funny

The old ones from like 2010 are so much easier

I did half of one with a friend after a single quarter of group theory

As a freshman undergrad

The newer ones meanwhile are like 5x harder lol

[lavish nexus]

like the first one here doesn't really need any abstract algebra

[next obsidian]

I didn’t even look at it lol

Umm

So the constant is odd

And the sum of coefficients is odd

Do you just use the rational root theorem or vieta’s formulas or something?

[lavish nexus]

I don't think I need anything
sum of a_n to a_1 is even
so there are even number of odd coefficients
then if x is even a_nx^n+...+a_1x is even
if x is odd it is still even...

[next obsidian]

Why are all the coefficients odd?

You can only assume a_0 is odd right?

[lavish nexus]

there are an even number of odd coefficients, 0 is also even so it includes where a_n to a_1 are all even

[next obsidian]

Ohhhhhh

I see I see

You’re saying the odd coefficients have an even number

Or ugh

[lavish nexus]

yeah

[next obsidian]

I see

I thought you were saying all coefficients are odd

And thus there’s an even number

But I see what you mean now

Yeah, that seems good lol

[sonic coral]

what are the typical topics covered in abstract algebra

i’m looking to self study it to prepare for an entrance exam

a lot of the practice test i have looked over talk about rings and isomorphism and order

if anyone responds, tag me

[next obsidian]

Groups, rings, fields. Sometimes more linear algebra, modules over rings, maybe Galois theory, representation theory @sonic coral

It depends on how long the course is, how in depth, what level, etc etc

[lethal dune]

which notations do you guys prefer? \times or \oplus for direct product?

[next obsidian]

\times

Use \oplus for direct sums

They’re different in infinite cases

[trim grove]

just curious what is main difference between direct sum and direct product.

[next obsidian]

Product let’s all of the terms be non-zero

For a direct sum only finitely many can be non-zero

So if you think of an element of a product like Prod_i in N G_i as a tuple

(g1,.g2,…,)

For the direct product all the g_I could be non-identity

For the direct sum only finitely many

[lethal dune]

[dark oxide]

Are these ideals maximal in Q[x,y] and C[x,y]?

[lethal dune]

what are the suggested books/vids for module theory?

[covert perch]

is the product group of conjugacy classes, the product of conjugacy classes of individual groups?

[proud bear]

should be. if x\in K_1 and K_1 is a conjugacy class of G, y\in K_2 and K_2 is a conjugacy class of H, then for any (g,h)\in G x H, (g,h)(x,y)(g^-1,h^-1)=(gxg^-1,hyh^-1) so (x,y) is in K_1 x K_2

[hidden haven]

In each case, use the fact that R/(p, q) ≅ (R/(p))/((q)/(p))

[hidden haven]

I like Jacobson but it also depends on what kind of module theory you wanna do

Atiyah MacDonald for commutative stuff lol

Lang is decent but has hard exercises

[dark oxide]

I came that far, but couldn't understand how that helps me

[hidden haven]

In all 4 cases, R/(p) becomes a polynomial ring in 1 variable, then you can apply irreducibility theorems for q

This should be easy to see in the first 2

If you are familiar enough with quotients

[dark oxide]

guess not

:/

[hidden haven]

The try proving that R[x,y]/(x) ≅ R[y]

[lethal dune]

oke

is DF an option or should I look somewhere else?

[hidden haven]

I haven't read many books

[lethal dune]

I'll look into those, tks

[next obsidian]

D&F works

[dark oxide]

I think I sort of get it maybe 😅

The first one is not maximal as y^5+45+6 is not irriducible in Q[y]. But then the next one...

You would need to check irreducibility of xy-1 in Q[y]...

[hidden haven]

y^5 + 4y +6 should be irreducible

[hidden haven]

nope, x gets killed

[delicate bloom]

eisenstein p=2

[hidden haven]

you have to compute (q)/(p) before you can quotient by it

(xy-1)/(x) in this case

[keen sparrow]

Aluffi is the New Testament of the Algebra bible

Dummit and Foote is Old Testament

I like Aluffi because category theory

[next obsidian]

Lmfao

Is Lang the book of genesis?

Or is that Old Testament lol

[dark oxide]

x just disappears?

[keen sparrow]

Book of genesis would be boubarki

Made the mistake of reading that

[hidden haven]

yeah so under the isomorphism R[x,y]/(x) = R[y], the polynomial xy-1 maps to -1

and the notation (xy-1)/(x) means that we take the image of the ideal (xy-1) under the quotient map

[keen sparrow]

Boubarki can be summed up as too general too early too fast

[dark oxide]

so definitely not maximal then?

[hidden haven]

yep, the quotient is just the 0 ring

(by yep I mean yes not maximal)

[dark oxide]

haha, oke thnx

i was confused :p

And for the third we can fill in y=x^2 and find x^3+10x-5. We then see that that is irriducible

and thus maximum

thnx man

[next obsidian]

I like Bourbaki

Blasna reference

[iron vessel]

D&F is the Bible of Algebra

[lethal dune]

can't seem to find the book, can you write the full name

[hidden haven]

Basic Algebra

There are 2 volumes

I think modules are in volume 2

volume 2 ch1 is cat theory so you might as well read that while you are at it

All a plot to catpill all the UGs

[lethal dune]

what the hell is cat theory

how to pet a cat? lol

[wooden ember]

If only

[hidden haven]

universal cats

https://tenor.com/view/dafuq-cat-flower-head-the-prophecy-is-true-gif-16939346

[iron vessel]

CATernions

[lethal dune]

oh lol categories

[cyan raft]

moldi can you explain what this means

[hidden haven]

Do you know what a functor is

lol it won't be possible without that

[cyan raft]

i'll try to learn that soom

[hidden haven]

😌

[willow mason]

lang's book is a gem

never read anything better

🌹

[wooden ember]

If i have a polynomial in $\mathbb{Z}[X]$ that when evaluated on the algebra of reals has a root at $\alpha$ can i construct another polynomial in $\mathbb{Z}[X]$ with $\alpha^k$ as a root?

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

ie if $\alpha$ is an algebraic number can i say that $\alpha^k$ is an algebraic number for all integer k

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

i see you can do it for k=-1

but im failing to construct a general polynomial for the other cases

and i also see you can do it if we drop k being an integer and take k=1/m for natural m

this probably doesnt relly belong in this channel but i didnt really know where else to post

ig it's #prealg-and-algebra

[next obsidian]

Algebraic numbers are closed under multiplication

The only way I know how to do this is to pick up a book on commutative algebra and look up “integral extensions”

And then note that algebraic numbers are the integral closure of Z

[wooden ember]

to be clear im asking about constructing the polynomial

[next obsidian]

Oh

[wooden ember]

i probably shouldnt have asked can you but how can you

sorry

[next obsidian]

It… might get spat out from the proof that it’s closed under multiplication

Maybe…

I dunno

[wooden ember]

ill give it a look later thanks

[final pasture]

Yeah

You can get it pretty explicitly if you do the proof with elementary symmetric polynomials

But I forgot the details so I'll leave it to you to work them out. Lmk if you need a ref for that proof but you should find it online pretty easily @wooden ember

[final pasture]

(also pretty sure it belongs here)

[hidden haven]

Must not have read much else then

[lethal dune]

[worthy haven]

In one of my courses I saw a proof that the maximal ideals of $\mathbb{C}[x_1, ..., x_n]$ are in the form $M = (x_1 - a_1, ..., x_n - a_n)$, and the proof first showed that if $M$ is a maximal ideal then $\mathbb{C}[x_1, ..., x_n]/M \cong \mathbb{C}$, and from there it concluded that $M = (x_1 - a_1, ..., x_n - a_n)$. I know the first part but I don't remember how the second part follows, nor do I have it in my notes. Can anyone fill it in?

[cloud walrus]

Nick

[worthy haven]

the second part being $\mathbb{C}[x_1, ..., x_n]/M \cong \mathbb{C} \implies M = (x_1 - a_1, ..., x_n - a_n)$

[cloud walrus]

Nick

[rustic crown]

x_i = a_i mod M

moldi nice nickname lol

[hidden haven]

[worthy haven]

...

[rustic crown]

if x_i + M on the left corresponds to a_i on the right, then we must have x_i = a_i mod M, this shows that (x1 - a1, ..., xn - an) is contained in M.

[worthy haven]

wow I'm fucking stupid

thank you det

[worthy haven]

I looked at this for like 10 seconds and was like "what" then it hit me haha

[boreal lion]

Can I not just say straight away here that any composition of bijective functions is bijective?

σ is a permutation too, so obviously they're all bijective, right?

[devout crow]

I would've thought so?

[rustic crown]

they're just proving that composition is bijective

[pastel cliff]

why does my algebra textbook go through integer operations (modulo, euclidean algo type stuff), randomly introduce rings and fields briefly, then go into analysis building up the rationals and reals, then polynomials, and then two more chapters before finally defining a group

[boreal lion]

Right, gotcha

[pastel cliff]

but artin starts with groups

[boreal lion]

What the

[rustic crown]

basic stuff about Z is used all the time

for instance if you wanna prove that order of an element will divide n if a^n = e

[devout crow]

constructing the reals in an intro algebra book

[sharp sonnet]

euclidean algo and modulo stuff very important, will be generalized in ring theory

the construction of the reals, less so

although the construction of the rationals is important

[pastel cliff]

my book goes 1. integers 2. integers to the complexes 3. polynomials 4. homomorphisms and quotient rings 5. field extensions 6. groups

doesnt end there obv but i was under the impression that groups were the most "basic" or at least fundamental

[pastel cliff]

i'll pay some more attention to this then but i remember most of the construction stuff from analysis

[devout crow]

teaching groups after rings is definitely less common but I found a note which gives some kind of reason

(from the infinitely large napkin)

[sharp sonnet]

it sounds kinda cool

[pastel cliff]

idk if i can appreciate that yet

[sharp sonnet]

mainly because ring theory is actually good and finite group theory is bad

[pastel cliff]

ik my class next semester is gonna do groups on day 1

so hopefully what im doing now counts for something

[devout crow]

I think (from experience) the more common way is to introduce vector spaces, then groups, then rings

[iron vessel]

I’m not sure I agree with rings having “more vivid first examples”, for example I think dihedral groups help a ton when learning about groups and it is easier to define symmetry when talking about shapes

[pastel cliff]

well lin alg is a prereq at my institution so i think we're expected to already have an idea of vector spaces in general

but that aligns with other things ive heard yeah

[sharp sonnet]

yeah i wonder what kind of examples the author is thinking of

in my experience one of the main issues in intro algebra is having good examples of rings that "appear naturally"

its hard to define them without some algebraic number theory or geometry background

[wooden ember]

honestly the only good examples of rings i have are rings of functions or polynomials

but both can be extended to algebras

so it's kinda sad

[sharp sonnet]

(tbf i dont think there are many interesting finite groups either, but they are easier introduced/motivated)

[rustic crown]

what about quotients of those?

[wooden ember]

imo there are plenty of interesting examples of finite groups

[rustic crown]

like Z[i] and stuff

[wooden ember]

my point is seeing it as a ring instead of an algebra isnt immediately advantageous

but tbf to understand an algebra you gotta understand a ring so...

[devout crow]

it's a cool perspective though, I remember enjoying the first time I realised C \iso R[x]/(x^2 + 1)

[iron vessel]

My Rings and Fields professor gave an example called “the drunk knight “ and its basically the pattern that the elements of Z[i]/(y^2+y+1) make

Not sure if that is the actual name tho

[devout crow]

also I don't even know the definition of an algebra so I assume most students starting AA won't either

[wooden ember]

yeah fair enough

it's just a ring acting on a ring

so yeah you do need to know rings for it

it's just i have trouble seeing when rings on their own are the advantageous perspective rather than some more powerful structure like an algebra

[rustic crown]

also algebras are kinda general, people like to think about say Lie algebras which aren't like rings, as in the product isn't associative

[iron vessel]

And homorphisms as with modules an R-algebra is a ring A together with a homomorphism that maps the identity of A to the id of R

[wooden ember]

fair enough

[rustic crown]

but you're right, rings are just Z-algebras so weird to think about them separately on their own

[wooden ember]

idk i just personally have found rings less interesting than finite groups for now

though quotients of polynomial rings are pretty chad

[rustic crown]

the factorizations stuff are pretty nice about rings

[wooden ember]

yeah

it's just groups (not necessarily finite) i find a lot more fun

idk

[iron vessel]

I like rings because of ideals

[wooden ember]

i think im just lacking enough experience with rings to judge

(and groups for that matter)

[iron vessel]

What is your favorite group?

[south patrol]

group hugs

[pastel cliff]

i have a problem that asks me to write gcd(a,b) = d in the form d = ma + nb where m and n are integers

for 56, 77, gcd is obviously 7, but brute forcing the desired form is proving tedious

however 14 = 3(56) + (-2)77

can i use that in any way since 14 is a multiple of 7

[viscid pewter]

nrly

why are you brute-forcing it

use the extended euclidean algorithm

[pastel cliff]

[viscid pewter]

it's like 3 steps with EEA

[pastel cliff]

yup i knew that

[terse crystal]

You just need to find one pair (u,v) such that ua+vb=d,then {(u-bs,v+as):s is from Z} gives you all pairs having this property

[viscid pewter]

what

look

i'll go through it

[terse crystal]

Let a=a’d, b=b’d then you need you find u, v such that ua’+vb’=1 when a’ and b’ are coprime

[viscid pewter]

77 - 1(56) = 21 [call this A]
56 - 2(21) = 14 [call this B]
21 - 1(14) = 7 [call this C]
14 - 2(7) = 0

therefore:
7 = 1(21) - 1(14) [by C]
= 1(21) - 1(56 - 2(21)) [by B]
= 3(21) - 1(56) [simplifying]
= 3(77 - 1(56)) - 1(56) [by A]
= 3(77) - 4(56) [simplifying, done]

[terse crystal]

Then you can take u to be a’ to the power of φ(b’)-1 where φ is the Euler function

[pastel cliff]

well that was easy

merci kai

[viscid pewter]

you should look up extended euclidean algorithm

[pastel cliff]

currently doing so

im just used to writing it out as a = qb + r

becomes obvious when you write it as a - qb = r

[viscid pewter]

mmm

[terse crystal]

Give me two numbers that are coprime

[viscid pewter]

wait i fucked it up

wait

shit

[pastel cliff]

42 and 35 i think are coprime

nope nvm that's dumb

wow we're doing great here

[viscid pewter]

ok i fixed it

[terse crystal]

They are not comprime

[viscid pewter]

1 and 2

[pastel cliff]

i realized immediately yes

e and tau

[terse crystal]

Anyway 6 and 5 are ,

(a,b)=(6,5) u=a^(φ(b)-1)=6^3=216,v=(1-ua )/b=-259

See just one step

[viscid pewter]

you're overcomplicating lol

[terse crystal]

One step is complicated? Lol

[viscid pewter]

there are commas in so i wouldn't call it one step, and also yes it is complicated, you don't need euler's totient or w/e

[pastel cliff]

they probably mean using euler's function is unnecessary

i appreciate both methods but im practicing euclidean algo rn anyways

[white nymph]

in part b, for the case where no eigenvalue in R, how to take the fact there is an eigenvalue (in C) and translate that to a dim 2 subspace of R^n?

[terse crystal]

T(u+iv)=(a+ib)(u+iv) then

(Tu Tv)=

(u v)

(a b

-b a)

[white nymph]

wut?

[terse crystal]

？

[white nymph]

i have no idea what you typed after the first line

[terse crystal]

You have a basis of R^n (e_1,…,e_n) right? Then T(e_1,…,e_n)=(e_1,…,e_n)A where A is a real matrix right?

Consider A as a complex matrix therefore it has a complex eigenvalue λ=a+bi and a corresponding eigenvector x=u+iv

(e_1 … e_n)u=r (e_1,… e_n)v=s then Rr+Rs is the subspace you are looking for

The corresponding equation of matrices is AX=XB

Where X=(u v)

B=

(a b

-b a)

[unique berry]

What is the latex command for that funny looking variable?

[chilly ocean]

\vartheta

[unique berry]

huh interesting

[spice whale]

is it possible for an algebraic structure to not be associative, but be commutative

[chilly ocean]

a lie algebra over a field of characteristic 2 would probably do it

[spice whale]

oh interesting

[delicate bloom]

$x \star y = a(x+y)$ is commutative but not associative for $a \ne 0$ or $1$

[cloud walrus]

Merosity

[regal arrow]

I read that a lattice over C is an Abelian group L such that L \otimes \mathbb{R} is the set of complex numbers

is this the tensor product, this \oplus?

*additive discrete group, not Abelian group lol

[cloud walrus]

Alison40

[spice whale]

nvm not a loop

[foggy merlin]

I'm currently taking a commutative algebra course and I'm not sure if every commutative algebra course is like that, but it is basically Ring- and Ring module theory for the most part. My question was is that so far I am not really aware where the less abstract history of it comes from, or what a more simple problem that is more accessible to the public is for which this theory had been developed. With field theory for example one is able to prove that trisecting an angle is impossible and then with galois theory the 5th degree polynomial stuff.
And I know both of these fields also use results from ring theory, but those results have already been presented in my first algebra course. This commutative algebra course mainly presents other results, for which I have no idea what they will be used for/ why they had been developed in the first place

I know it's a bit of a redundant essay, but I'm essentially asking for applications of ring and module theory that aren't just results from field or galois theory.

[vast quiver]

algebraic geometry! check out Gathmann's notes on commutative algebra, chapter 0. He gives a tiny introduction to algebraic geometry and the translation between it and commutative algebra

https://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013.pdf

[foggy merlin]

Bruh phone says all the links to gathmanns notes aren't safe for god knows whatever reason

[vast quiver]

oof

here I just uploaded the whole thing on discord lol

[foggy merlin]

Wow thanks. I read through the first couple of pages and this seems very cool, makes me appreciate the topic a lot more

[shrewd ravine]

can someone help me understand part a?

C6 = {e, g, g^2, ... ,g^5}, and H = {e, g^3}, so as H has index 3 in G there should be 3 left and right cosets right?

how do I know this quotient group is a normal subgroup of C6?

and is it abelian, and if so, why?

[lavish nexus]

I guess we can consider g to be just a size n Jordan block

not sure how to proceed

[lavish nexus]

cyclic groups are all abelian

any element in any group commutes with its own power

so every subgroup is normal

[lavish nexus]

got only Z5*Z55, Z5 *Z5 * Z55, Z11 * Z55

[chilly ocean]

sus

[long obsidian]

Does anybody know the origin of the word "torsion" in algebra? I saw someone say that when Emmy Noether first defined torsion she was studying the "geometry" of things like the Mobius band. My best interpretation of what they meant is that she was studying the fundamental groups of things and twisting motions corresponded to elements of finite order. But I'm not entirely sure I am interpreting it right.

Any help making sense of the word torsion in algebra and torsion in mechanics I'd appreciated

[chilly ocean]

i thought physics was origin

[long obsidian]

But why did the word pop up again in algebra

[chilly ocean]

its descriptive probably?

like you can visualize finite cyclic groups as wheels

and applying enough torsion to that wheel gets you at starting point?

[long obsidian]

Hmm that's fair. I can see that

[lavish nexus]

my professors said it was from some topology stuff

I don't really know topology so I don't remember the exact thing

[wooden ember]

The name comes from poincare I think

I think there were some twisted spaces or smth that had torsion elements in some related group

Lemme check I don’t remember

So if I understood it correctly, m-1 torsion in a fundamental group is only present in a nonorientable m-manifold?

Someone more knowledgeable than me should be able to clarify

[chilly ocean]

Hello there, Is there anyone available to tutor Galois theory with me for the next 5-6 weeks? I’m really committed to learning with someone, even if they are wanting just to refresh themselves on group and field theory. Please message me if you are interested 🙂

[long obsidian]

That somewhat agrees with what I heard because when I heard about it being related to geometry they mentioned the Mobius strip, which is nonorientable

[wooden ember]

Yeah

[spice whale]

are invertible matrices a field under multiplication
or do fields require commutativity

[terse crystal]

You just said under multiplication

So it’s only a group

Ring structure isn’t even given

[spice whale]

oh wait

but

under addition and multiplication

[terse crystal]

The sum of two invertible matrices isn’t necessarily invertible

(1 0

0 1)

(0 1

1 0)

=

(1 1

1 1)

Not invertible

[long obsidian]

Can someone help me see why if $G/Z(G)$ is cyclic then G is abelian?

[cloud walrus]

fajitas

[hidden haven]

Everything is in the center

oh

I read it as the converse nvm lol

[terse crystal]

Any element is of the form ax^i where a is in the center and G/Z(G) is the union of x^iZ(G)

Clearly ax^i*bx^j=abx^(i+j)

[long obsidian]

Thank you

[spice whale]

ah this makes sense
but in general do fields require commutativity

[hidden haven]

Yes

Otherwise is division ring

[terse crystal]

Finite division ring is always a field though

[spice whale]

oh

[dawn kiln]

so im trying to work out the algorithm for converting a matrix to smith-normal form

the wikipedia page is not particularly helpful

does anyone have any pointers for this type of diagonalisation?

[hidden haven]

Do row operations to get an entry which is the gcd of all entries in the upper left corner

[terse crystal]

Jacobson basic algebra volume 1 has it

Chapter 3

[dawn kiln]

haven't been able to find a pdf online of it, could you maybe send a ss?

[terse crystal]

http://library.lol/main/939A680C1971BB1FFA6FC690EB1294F9

[dawn kiln]

thank you!

[lavish nexus]

this is like the general way to do it in a PID
in a field it is easier with standard row and column ops
alternatively just find gcd's of n-rowed minors

gcd of 1-rowed minors is d1
gcd of 2-rowed minors/gcd of 1-rowed minors = d2
proceed

[dawn kiln]

im working with integers exclusively, so it needs to be in a PID rather than a field

[dawn kiln]

what do you mean by 1-rowed minors?

[lavish nexus]

determinant of 1*1 minor
basically every entry in A-λI

usually turns out to be 1

this is also in Jacobson 3.7

although I have some strong reservations about that book...

[terse crystal]

I think it’s great, containing almost everything I needed to know in the beginning, except exercises are sometimes too easy but I don’t do much exercise anyway …

[dawn kiln]

lol tbf im not planning on learning this stuff in detail, this is actually for a compilers project

[terse crystal]

Don’t worry it’s just several pages without the need of reading the previous chapters

[robust magnet]

Hey, what is the necessary condition in order to have an isomorphism from group (Z/aZ,+)x(Z/bZ,+) to group (Z/cZ,+)x(Z/dZ, +)?

[long obsidian]

If H and K are subgroups of a group G them there is a set HK that may or may not be a subgroup of G with elements of HK of the form hk.

Is there a name for this "product set"?

[viscid pewter]

product of subgroups

[lavish nexus]

they should have the same invariant factor decomposition?

[terse crystal]

Iff matrices diag{a,b} and diag{c,d} are equivalent I think

Meaning gcd(a,b)=gcd(c,d) and ab=cd perhaps

[robust magnet]

Thank you very much!

[viscid pewter]

actually

could just look at it through fundamental theorem of abelian groups

[terse crystal]

Yeah finitely generated modules over pid like they discussed, they were talking about smith normal form

[viscid pewter]

yeah idk that stuff lol

[lavish nexus]

is Q/Z artinian?

every finitely generated subgroup is cyclic

and is of the form <1/r>

if r has prime factorization p1^n1p2^n2....pm^nm

[viscid pewter]

silly question, but why are you considering finitely generated subgroups

[lavish nexus]

because the only nonfinitely generated subgroup seems to be Q/Z itself

[viscid pewter]

consider the entire group

then consider the subgroup that's all elements without a factor of 2 in denom

then the sub of that without 2's or 3's

?

[lavish nexus]

I see thank you

[viscid pewter]

oh sick i wasn't sure if i was missing something or what

[lavish nexus]

there are infinitely many primes
not artinian

[viscid pewter]

mmm

[terse crystal]

？

Q/Z isn’t even a ring

[viscid pewter]

who cares about rings

imagine having two operations

wait

[terse crystal]

Oh my bad… artinian module I see

[viscid pewter]

oh yeah no 1

[dawn kiln]

sry im not quite following the explanation in the book

so take the example of:

$$\begin{pmatrix}3&5&11\ -5&7&9\end{pmatrix} \equiv \begin{pmatrix}1&0\0&1\end{pmatrix}\begin{pmatrix}3&5&11\ -5&7&9\end{pmatrix}\begin{pmatrix}1&0&0\0&1&0\0&0&1\end{pmatrix}$$

[cloud walrus]

maximwebb

[dawn kiln]

so ive worked out by hand that we can rewrite this as

$$\begin{pmatrix}1&0\ -17&1\end{pmatrix}\begin{pmatrix}1&0&0\ :0&2&0\end{pmatrix}\begin{pmatrix}3&5&11\ 23&46&98\ 4&8&17\end{pmatrix}$$

[cloud walrus]

maximwebb

[dawn kiln]

by manually applying row operations and updating P and Q accordingly

but i dont quite understand how we jump directly to this, with the GCD method

the chapter's explanation was a little confusing

[long obsidian]

Why is the definition of a group homomorphism the way it is?

The best explaination is that if $f:G\to H$ is a group hom. then it's kinda cool that the image of the product is divisible by each of the images of the factors of the product i.e. f(a) divides f(ab) and f(b) divides f(ab). This seems kinda nice but it's not clear otherwise why functions such that $f(ab)=f(a)f(b)$ perverse "algebraic structure"

[cloud walrus]

fajitas

[viscid pewter]

it's nothing to do with factors or whatever

it's just that a homomorphism is something that preserves structure

in some sense

it preserves the ways in which elements interact

even if those elements are all the same element after the mapping

[thorn delta]

A homomorphism imprints the structure of the domain into the codomain. You can see this is as the image of a group hom is itself a group

[viscid pewter]

when a whole bunch of maths is just about certain structures, a map from one structure to another that preserves how the stuff in the first thing works is Quite Important

[long obsidian]

Is it something like "the homomorphism commutes with the group operations"?

As in I can operate in the domain then apply the homomorphism, or I can apply the homomorphism first then apply the codomains operation? Is that how it preserves "algebraic structure".

I just hate such vauge notions lol

[viscid pewter]

u wot

the definition of a homomorphism is a map f: G -> H such that:

for all a, b in G, f(a)f(b) = f(ab)

clearly (a)(b) = ab

it's like

the mapped versions of the elements interact in exactly the same way as the originals

the homomorphism preserves all those relationships

it preserves that structure

[long obsidian]

That's scary

[dawn kiln]

Lol I think you have the right idea here, but commutative is a bit of a tricky word here, so I'd just stick to thinking of it in terms of the actual definition

But you can see empirically why homomorphisms are useful

[viscid pewter]

it's like, say you have a bunch of people moving around

and then you have their shadows

when people stop moving, so do the shadows

[dawn kiln]

They preserve lots of helpful properties of groups (and other algebraic structures once you come round to them), like identities, inverses, generators

[viscid pewter]

shadows move in an analogous way to the objects

[dawn kiln]

Lol I see what you're trying to say, but I think that could make things more confusing

[viscid pewter]

ok

i mean i think more experience will help here, yeah

they're very natural

[dawn kiln]

Try proving that identities and inverses are conserved under homomorphisms, that's a good start

[viscid pewter]

^

[chilly ocean]

Homomorphisms are kind of like immigrant communities
Like imagine a bunch of Italians come over from G (Italy) via a ship (a homomorphism phi) to NYC (a different group H) in the early 20th century
They can do all sorts of things with the New York natives, but if they are by themselves they can always adopt the conventions of their original culture that they're used to

So for example if H is not a commutative group, but G is commutative, then a*phi(g) might not commute, but phi(g1)phi(g2) =phi(g1 g2)
We can pull them back under the hood they came into H from without any repercussions

And since g1 and g2 commute now we can send g2 g1 back through phi to get a final result of phi(g2)phi(g1)

Like how if a 1910s Italian asked some random New Yorker how to do a certain dance they might not be able to
But if you know that New Yorker also came over from the same part of Italy then those two will be goin "AYYY"

[wise igloo]

YO

I love this analogy

[dawn kiln]

This might be the weirdest fucking thing I've read on this server lmao

[tribal moss]

I just hope the homomorphism in question is injective.

[viscid pewter]

this is a crap analogy ngl

i preferred the shadows thing

[chilly ocean]

if it's not injective then you have to start merging people into one

[chilly ocean]

ok 😊

[viscid pewter]

it's about the map

not the groups

[tribal moss]

So a couple who were married in Italy are still spouses when they arrive in NYC and their kid is still theirs. Phew.

[subtle ivy]

i just want to say that i really love all the creative interpretations here. it's so lovely seeing what crazy ideas different people come up with about such abstract stuff!

[dreamy fiber]

For the field of complex numbers, if I have a linear transformation T from V to W, V, W being finite dimensional inner product spaces.

The adjoint of a linear transformation T, T^*, has the relationship that:

M(T^*) is a conjugate transpose of M(T)

But what if I switch to a different field? Or even to a general ring? Is there any specific generalizations of this?

[lethal dune]

they are conjugate transpose to each other wrt an orthonormal basis

[chilly ocean]

How do you do part (iii)?

[hidden haven]

Mega Euler

Degree of extension = degree of the minimal polynomial of generator

[hidden haven]

They said field other than ℂ lul

[lethal dune]

no idea how u define inner products on other field, ex finite fields

[rustic crown]

what about ordered rings?

[lethal dune]