groups-rings-fields

[elfin wraith]

To make this more clear, if this were not the case the group generated by x would not be a group, and you kinda want Z to be a group. Inverses are in the definition

[thorn jay]

(Fourier analysis jumpscare in Serre's book abt group representations)

[delicate orchid]

I'm talking about the finite group case....

you seem to be reading my messages out of order

it's not a group

not under addition anyway

[elfin wraith]

Famously closed under multiplication though

[delicate orchid]

I was covering my ass for some C_21 structure bullshit

[thorn jay]

Under exponentiation too!

May form just a quasigroup though..

[delicate orchid]

partial quasigroups 💔

[thorn jay]

There will be no partial blasphemy outside of partial monoids in this channel

[elfin wraith]

Yes

[delicate orchid]

yeah

[elfin wraith]

This may or may not work for infinite groups however

[delicate orchid]

correct

[delicate orchid]

is appending "partial" idempotent btw? are partial partial groups just partial groups

I think they are

[thorn jay]

Partial is just when the operation is not defined on the whole codomain

so yes idempotent

[elfin wraith]

Can I have a semi demi locally simply connected space

[thorn jay]

You just get an averagely smaller codomain

[delicate orchid]

the memihemidemisemi T_0 space 💔

[thorn jay]

Memiminminiminedemi space

[elfin wraith]

I’m adopting American topological notation, it’s now a half locally simply connected space

[delicate orchid]

you should see a doctor about that I think

[elfin wraith]

Unironically musical notation is the one thing Americans do right

[delicate orchid]

they know how to make a beef brisket sandwich

[thorn jay]

What the FUCK do you mean "half locally"

[delicate orchid]

union of R_<0 and a ball centred at 0

this is bullshit btw

[thorn jay]

Center Deez nuts at 0

[elfin wraith]

My bad I mean sixteenth locally, got my American conversions wrong in my head

[thorn jay]

Dying

[elfin wraith]

Well what is the group of rigid motions of a tetrahedron?

A good starting place is to just forget groups and think about how you can move around a tetrahedron

[cedar vault]

Its most probably asking for the symmetries of a regular one

Yes, the number will not be fixed if we allow arbitrary tetrahedroms

[delicate orchid]

it's giving C_2

although it's hard to tell because it's flat

is the purple one in the same (orthogonal) plane as the green one

then yeah it's 1

[elfin wraith]

Integers mod 2

The cyclic group on 2 elements, you’ll get to it

Unless does D&F use funky notation for it? I feel like they do

[delicate orchid]

they always do

[cedar vault]

C2,Z2,D2,U(Z)

[delicate orchid]

U(Z) is beyond a crash out

firstly group of units is * secondly U(Z/3Z) is right there

[cedar vault]

You mean, Z* to denotè units of Z?

[delicate orchid]

well in tex I'd do Z^\times but that's a pain to write in plaintext

[elfin wraith]

Z^x is right there sweaty

[delicate orchid]

me and you are going to have words

yup

[elfin wraith]

It’s isomorphic to that at least yeah

[delicate orchid]

there is another sneaky 180 degree rotation you can do

along this line

granted, this is the product of two of the permutations you said but it's important to visualise

[tribal moss]

But that is among the ones they've already counted, so it's not really additional.

[delicate orchid]

they're still at the stage where "generated" is a new concept. It's better to just find them all

[tribal moss]

Wasn't the task just to count the symmetries?

[delicate orchid]

what do you mean by this exactly. I can always draw a line between two points? And the definition of rigid motion is exactly what you think it is: a linear (more generally affine but don't worry about it) transformation that preserves distances between all points

[delicate orchid]

ok fair point I was thinking about actually finding the group

[thorn jay]

Sweaty 💦💦

[vast verge]

I don't really see how they got the solution here

[barren sierra]

So all maps $\phi$ from $\mathbb{Z}{10}$ to $U(11)$ are determined by $\phi(1)$ since $\phi(n) = \phi(1)^n$ using additive notation for $\mathbb{Z}{10}$ and multiplicative for $U(11)$. So we want to find an isomorphism $\phi$.

Then the order of $1$ in $\mathbb{Z}_{10}$ is $10$ right? So then the image of $1$ under our isomorphism $\phi$ must be an element of what order? Can you find such an element? Then just argue directly that this map is indeed an isomorphism (it's a homomorphism, it's injective, it's surjective)

[cloud walrus]

Spamakin🎷

[vast verge]

2 has order 10 in U(11), so this means Φ(1) = 2, and thus Φ(n) = 2^n

[barren sierra]

ok so you've given me a map, and it seems to be a homomorphism

can you prove it's surjective and injective?

and thus an isomorphism?

[vast verge]

I have to do everything in modulo 11

So it'll be trickier

[barren sierra]

It's not that tricky.

One thing that may help is noticing that the two sets are both finite

so to show that your map is a bijection, it suffices to show just one of injectivity or surjectivity

(this is a general fact of maps between finite sets, good to prove on your own)

so pick either injectivity or surjectivity to prove, whichever one seems easier to you

[vast verge]

Is it true that each isomorphism Φ is uniquely determined by its choice of Φ(1)?

[barren sierra]

yes, any homomorphism from a cyclic group is uniquely determined by it's choice of $\phi(1)$. So what I specifically mean is that if $G= \langle a \rangle$ is a cyclic group and $\phi, \psi$ are two group homomorphisms $G \to \overline{G}$ for some other group $\overline{G}$ such that $\phi(a) = \psi(a)$, then $\phi(a^k) = \psi(a^k)$ for all $k$ (written multiplicatively here in this example, you can also write it additively it doesn't matter it's just notation)

[cloud walrus]

Spamakin🎷

[vast verge]

This is what I've done @barren sierra

[barren sierra]

You don't know a priori that phi is an isomorphism

Nor does every isomorphism from Z10 to U(11) map 1 to 2

So what you've written at the start is wrong, you've assumed what you want to show

You want to show that phi mapping 1 to 2 is an isomorphism

[candid patrol]

2^10 is of course equal to 1 by Langrange, and you can only shows the injectivity by cardinality

[barren sierra]

So you can't say that "since phi is an isomorphism, the order of phi(1) in U(11) is equal to the order of 1 in Z10"

also you forgot k = 10 when in 2nd picture

[vast stump]

in general a homomorphism from G to H is determined by the image of any generating set of G (assuming the resulting map actually is a homomorphism)

1 generates any cyclic group so to define a map out of Zn it suffices to specify the image of 1 (but again the resulting map might not be a homomorphism)

[vast verge]

Because G = <g> if and only if phi(G) = <phi(g)> correct?

[fervent meteor]

I spent an hour yesterday losing my shit trying to find a contradiction in the field axioms that would derive from trying to let a field have 0=1

I was over joyed to realize that the definition I was working with required that a field have two elements

Which means that if 0=1 then the field cannot have two elements

How do you prove that 0\neq1 in a field if you don't place a lower bound on the cardinality

It's easy to say "a field is a non-trivial commutative division ring" but what if you just assume a set with operations and you list the algebraic laws

[thorn jay]

You don't really

It's an axiom

Personally, I like the definition of a field as a commutative ring satisfying
x ≠ 0 <=> ∃y : xy = 1

This excludes the zero ring

[fervent meteor]

Fair enough

[knotty badger]

well $\sigma$ is a function from $\Omega$ to $\Omega$

[cloud walrus]

Pseudonium

[knotty badger]

so it can accept inputs that lie in $\Omega$

[cloud walrus]

Pseudonium

[knotty badger]

unsure what you mean there, but no, you can apply $\sigma$ to individual elements of $\Omega$

[cloud walrus]

Pseudonium

[knotty badger]

you don't need to supply the whole set at once

yes, this is true

but a function always just takes inputs to outputs

and you're allowed to plug in a specific element of the set as input

which will give you a specific element of the set as output

it just so happens that the "global" structure of the function is to re-arrange the set

for example, you could take $\Omega = {1, 2, 3}$, and have your permutation be swapping $1$ and $2$

[cloud walrus]

Pseudonium

[knotty badger]

you can describe this by the function $\sigma : \Omega \to \Omega$ defined by $\sigma(1) = 2, \sigma(2) = 1, \sigma(3) = 3$

[cloud walrus]

Pseudonium

[knotty badger]

kinda? remember that elements of sets aren't intriniscally ordered

you could equally well say $\Omega = {3, 1, 2}$

[cloud walrus]

Pseudonium

[knotty badger]

and $\sigma$ would be defined the same way

[cloud walrus]

Pseudonium

[knotty badger]

if you wanted to swap 1 and 2, that is

essentially my point is that, generically, there's no "first element" of a set

mhm!

applying it twice would be considering the function $(\sigma \circ \sigma) : \Omega \to \Omega$

[cloud walrus]

Pseudonium

[knotty badger]

in this case, that'd be the identity $\text{id}_\Omega$

[cloud walrus]

Pseudonium

[knotty badger]

i think you mean apply sigma?

$\Omega$ is the name of the set, $\sigma$ is the name of the permutation

[cloud walrus]

Pseudonium

[knotty badger]

also do note that not all permutations go back to the identity after being applied twice

for example, consider $\tau : \Omega \to \Omega$ defined by $\tau(1) = 2, \tau(2) = 3, \tau(3) = 1$

[cloud walrus]

Pseudonium

[knotty badger]

in this case $(\tau \circ \tau)$ is still not the identity, but $\tau \circ \tau \circ \tau$ is

[cloud walrus]

Pseudonium

[knotty badger]

ah, this is not how function equality is defined in maths

two functions $f , g : A \to B$ are equal if and only if $\forall x \in A, f(x) = g(x)$

[cloud walrus]

Pseudonium

[knotty badger]

it is not enough to simply have $f(A) = g(A)$

[cloud walrus]

Pseudonium

[knotty badger]

well, you don't typically input entire sets to functions

they're equal if, for each input of the set, they give the same output

this is called "function extensionality"

[knotty badger]

it is defined, implicitly you have $f(4) = 4$ using this cycle notation

[cloud walrus]

Pseudonium

[knotty badger]

also !nogpt

[thorn jay]

You don't know what you don't know..

[sour rain]

oops

[alpine island]

if you identity sets up to set equivalence (bijections), math becomes really boring really quickly. Every group would be the same as every other group, since a group is just a set {G, *} of the underlying set and a function on it, and that's the same as any other two-element set.

[knotty badger]

Lollll I didn’t think about it this way but that’s genius

[alpine island]

It's fun to think about the basic set definitions of things sometimes. You get some weird results. Like that a seperable hausdorff topological space has cardinality at most 2^2^N.

[keen badge]

I have this cube and there is this "transpernt axis" that goes between the middle of the edges BC, A'D'.

If we rotate the cube 180 degrees w.r.t that axis, what will be the image of each vertex?

(I'm asking in this section because this is a group action, and I am trying to find its orbits)

[velvet hull]

A' to D', A to C', D to B', B to C

source: my rubiks cube

[keen badge]

I also tried to do it with my rubiks cube lol.
But I couldn't figure out how to rotate the cube

Because it is not as easy as if it were between the middle of the faces BB'C'C and AA'D'D

[velvet hull]

yeah the rotation motion is really janky

[keen badge]

How did you take the picture? With your third hand?

[keen badge]

Yes, that's the point

Ofc, I need to rotate w.r.t this edges

[velvet hull]

yes with my third hand

[keen badge]

but how the motion looks like

[velvet hull]

no I just propped up my phone and set a timer

[keen badge]

is there a way you could(somehow) explain to me what is the motion of the action?

[velvet hull]

the top and left faces swap

equally the bottom and right faces swap

and the front and back faces swap

[keen badge]

ohh

that's it?

Damn

[velvet hull]

and then there's the orientation

but thats how the faces move

[keen badge]

ok

thank you 🙂

[cedar vault]

what is the definition of an m-cycle?

[cloud walrus]

mqinsweden

[tough raven]

Another way to see this: the midpoints of AC', A'D', BC, B'D all lie on the axis by symmetry considerations.

Pick any x ∈ {1, ..., m} and look at x, s^i(x), s^i(s^i(x)), ... and think about when it repeats.

Oh, I didn't see this.

Why x-1+i?

and why m+1

Ah, so you mean s^i(x) = (((x-1)+i)MOD m)+1

Easier to just say s^i(x) ≡ x+i (mod m) and 1 ≤ s^i(x) ≤ m in my opinion

but yes, that's correct.

[kind temple]

prove the more general fact that |a^n| = |a| / gcd(|a|, n) for any element a of finite order in a group G

[cedar vault]

you can show first that if i is not coprime, order of sigma^{i} will be atmost m/gcd(m,i), but an m cycle must have order m. if i is coprime, we want to prove that we form an m cycle, so that means the repeated image of 1 must cycle through all elements 1,2,3,...,m. i.e 1 != 1 + ki (mod m) for k in 1,2,...,m-1

if x is an element of a group and has order m that means x^m = e(identity) and it is the smallest such positive number (equivalently the order of the cyclic subgroup generated by x)

[velvet hull]

you can

you just compose the functions

(1 2 3)(1 2 3) = (1 3 2)

[cedar vault]

For some integer k, (sigma^(i))^(m/gcd(m,i))=sigma^(m*k) =(sigma^m)^k=e, so order of sigma is atmsot m/gcd(m,i) < m so sigma^i is not an m cycle when i is not coprime

[vast verge]

Where does the assumption in this question come from? It doesnt make sense to me

[cloud walrus]

mqinsweden

[valid fox]

in the case there is an element of order 33, there is also an element of order 3. So we consider the case where there is not an element of order 33

[vast verge]

Ohh ok

[valid fox]

I agree the solution is written weirdly

[cedar vault]

Yes, now you should prove that if gcd(m,i) =1, then sigma^i is an m cycle

[vast verge]

I thought they were saying that there was no element of order 33 😅

[cedar vault]

There could be, for example: Z/33Z

You can show by counting the elements that there must be one of order 3

[cedar vault]

I would say not much, but a little. the hint is given here

Sorry, i dont understand. The original question says sigma is just the m-cycle (1 2 3 .. m)

[vast verge]

Can someone explain where the boxed inequality comes from?

The questions in this chapter (Cosets and Lagrange's Theorem) have really ramped up in difficulty, most of the questions I can't do without looking at the answers, and even then the solutions don't help me understand much. Maybe I should go back in the book to do the more easier questions...

[thorn jay]

Because H1 • H2 must be contained in G, so their order is ≤ |G| = 2m

[glad osprey]

I know the feeling, but I don't think it's a good idea to go back unless there are specific things from previous chapters you need to understand. When you're struggling as you are right now is when you're learning the most, so if you just keep grinding I think it'll get easier

[vast verge]

This is my first time doing self-learning so I don't really know what to do if I feel like I'm not progressing

It's no fun being unable to engage with any interesting questions if you don't know where to start each time and just end up reading the full solutions

[glad osprey]

I guess you could try some easier questions in the same chapter, like computations instead of proofs for example. But keep in mind that proofs are hard for everyone, particularly when it's a new subject, so you're not used to the way of thinking. Don't think of it as a failure to not be able to prove something

[vast verge]

Ok thank you

I'll try spending more time understanding the questions in this chapter

[glad osprey]

Btw, Fraleigh has a good mix of questions of various difficulty, it might be easier than jumping straight into proofs

[vast verge]

I'm doing Gallian, but I wasn't a big fan of how the questions are done

As someone who's used to university problem sheets, it's a bit jarring

But then again I haven't done many algebra courses

[glad osprey]

Gallian is good too 👍 when self-studying it can be hard to pick which problems to solve tho, I think some of them are much harder than others

[vast verge]

What's the advantages and disadvantages of Fraleigh compared to Gallian in your opinion?

[glad osprey]

I haven't read Gallian, so I can't really say, but if you like Gallian then just stick with it, I don't think there's a huge difference between the two. If there's some explanation in Gallian you don't understand you could take a look at Fraleigh tho - it's always good to have multiple sources to draw from

[karmic moat]

do atiyah-macdonald for rings

and modules

[keen badge]

Let $F$ be some subfield of $\bC$ and let $\alpha\in\bC$. Is it true that

$$
\deg(\text{irr}(\alpha / F) | \deg(\text{irr}(\alpha / \bQ))
$$

the fact that $\text{irr}(\alpha / F)$ divides $\text{irr}(\alpha / \bQ)$ is clear to me, but they should be both irreducible, so it means that they are the same polynomial up to scalar, which I don't understand how it's true.

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[tidal schooner]

By irr$(\alpha/F)$, do you mean the minimal polynomial of $\alpha$ over $F$? If so, then irr$(\alpha/F)$ is irreducible over $F$, and irr$(\alpha/\bQ)$ is irreducible over $\bQ$. But that does not mean that irr$(\alpha/\bQ)$ is irreducible over $F$. For example, take $\alpha=\sqrt{2}$ and $F=\bR$. Then irr$(\alpha/\bQ)=x^2-2$ is irreducible over $\bQ$, but it's not irreducible over $\bR$ since it factors as $(x+\alpha)(x-\alpha)$. (Edit: typo)

[cloud walrus]

harmacist

[tidal schooner]

So "they should be both irreducible" isn't really the full statement — they're both irreducible over their respective ground fields, but they won't generally be irreducible over the same ground field

[keen badge]

For some reason I thought that thinking of irr$(\alpha / F)$ over $\bQ$ is appropriate.
Go figure.

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[keen badge]

So if I say that $\text{irr}(\alpha / F)$ divides $\text{irr}(\alpha / \bQ)$ it means there exists a polynomial over F such that $\text{irr}(\alpha / F)\cdot p(x)=\text{irr}(\alpha / \bQ)$, right?
and how can I use this to prove the degrees divide each other

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[keen badge]

if that's even true tbh

[violet spade]

What about:

Take $\alpha=\sqrt[7]{2}$, then $deg(irr(\alpha/\bQ))=7$, and if $F=\bQ(\sqrt[3.5]{2})$, then $deg(irr(\alpha/F))=2$ which is clearly a counterexample

[cloud walrus]

joel

[violet spade]

@keen badge

[sour rain]

this is criminal from dummit and foote

why would R be the subring and S be the ring 😭

it should so obviously be the other way around

[uneven jackal]

to be fair

it's because we often write R -> S for a ring map (alphabetical order)

and this is the special case where the map is injective

[sour rain]

i can understand it, but i wont accept it

[south patrol]

This

S being a subring of R is more perverse to me

[uneven jackal]

Lie algebra question for those knowledgeable: the wikipedia article says that we need the PBW theorem to see the correspondance between representations of g (as Lie algebras) and representations of U(g) (because we can extend the scalars simply, or restrict them to g). For me, it's rather because of the universal property of U(g) : a map g -> gl(V) is a map g -> Lie(End(V)) which is the same as U(g) -> End(V) (general nonsense about "alpha-applications" in Bourbaki or whatever). Am I right ? Does that secretely need PBW to work ? Am I missing a step ?

[uneven jackal]

sure

p(0) is always 0

if you have an additive group morphism 0 goes to 0

why would p(1) be equal to p(1)p(0) ?

no

it's an additive morphism yes

and even then p(1*0) is p(0) not p(1)

because there's no reason for it to be correct

I agree that p(0) = 0 so that p(1)p(0)=0

I would suggest reviewing the definition of a morphism

you should be able to find it on wikipedia or somewhere

but the point here is that "additive morphism" means it is compatible with addition

I believe you're mixing up addition and multiplication somehow

yes ?...

by morphism I mean homomorphism yes

sure, but what are the operations here ?

answer is addition

so I would suggest not writing addition as multiplication when you're dealing with numbers

yes that's better

if you try to put x=1 and y=0 now, do you get anything interesting ?

yes you get p(1) = p(1), not much help

[cedar vault]

I would simply note that ||there are subgroups of Q such as Z where Q/Z has elements of all orders but there is no subgroup of S of Z such that Z/S has elements of all orders||

[south patrol]

Another is ||Z is divisible whilst Q is not. Or as a special case, multiplication by 2 is surjective on Q but not on Z||

[thorn jay]

Also, ||Q is not generated by one element, as <a/b> will never contain 1/d where d is coprime to b||

Also ||doesnt Q have uncountably many subgroups? At least one unique one for any { 1/p_i | i ∈ I } where p_i is prime||

[south patrol]

||Aut_Z(Z) = {+-1} but Aut_Z(Q) = Q^x||

[glad osprey]

Also Z is isomorphic, but Q isn't

[thorn jay]

That last thing seems fairly nontrivial

Mm nvm I think I see it

[south patrol]

Nah it is straightforward cause ||an auto of Q is determined by where you send 1, e.g. as f(1) = q then f(m) = mq and b f(a/b) = f(a) = aq ||

[thorn jay]

Yeah exactly

Man it's late I should go to bed

Just one more reason why Q isn't isomorphic to Z I swear

[south patrol]

||Hom(Q, Z) = 0 but Hom(Z,Z) isnt||

[tardy hedge]

what does divisible mean here? Divisible as Z-module? (i just heard about what it means for R-mod M to be divisible)

[cedar vault]

G abelian group is divisible if for all n , nG=G

[south patrol]

Yeah sure, that is equivalent, though personally I heard of divisible abelian groups before any modules lol

[tardy hedge]

tbh im sad i didnt have a good group theory class or never taken ring theory etc

[south patrol]

It's divisible if this is true for all n

[tardy hedge]

i got into math later on

[tardy hedge]

so Z is divisible because nZ = Z for all n?

group isomorphism ?

[cedar vault]

No, but Q is divisible

[tardy hedge]

so this is typo

[south patrol]

Yeah sorry

Lol

[tardy hedge]

oki

[thorn jay]

Would be insane because it would mean that Z was a field

[tardy hedge]

lol

[tough raven]

ℚ is not isomorphic to ℤ because if it were, then ℚ and ℤ would be isomorphic, which is clearly absurd.

[tribal moss]

Repugnant to the nature of a number line.

[tough raven]

Indeed the PBW theorem has nothing to do with the correspondence of representations, which follows entirely from the way U(g) is defined (exactly as you said).

[last spoke]

notation question: for some indeterminate (\alpha), is (\mathbb{Q}(\alpha) = {a + b\alpha \colon a,b \in \mathbb{Q}})

[cloud walrus]

StarvinPig

[velvet hull]

When you use parenthesis it denotes the field generated by Q and a.

When you use square brackets it’s the ring generated by Q and a, which is the set of all polynomials in a with rational coefficients

[twilit wraith]

i have to find the group of units of the ring Z[i]

i suspect its powers of i but idk how i prove this

[topaz solar]

Well, why do you suspect this

[twilit wraith]

i know that if some $a + bi$ is a unit in Z[i] then $\frac{a}{a^2 + b^2}, \frac{b}{a^2 + b^2} \in \mathbb{Z}$

[cloud walrus]

hiidostuff

[twilit wraith]

i suspect it because i need the denominator to divide both a and b

but i dont think that can be the case if both a and b are nonzero

[topaz solar]

So, how did you derive your division thing

[twilit wraith]

i looked at 1/(a + bi) and multiplied by the complex conjugate

on the top and bottom

[kind temple]

Q(𝛂) is the smallest field containing Q and 𝛂 where 𝛂 is indeterminant - essentially meaning transcendental - over Q. such a field would contain Q[𝛂] and 1/f(𝛂) for all non-zero polynomials f. Q(𝛂) being a field would then contain all rational functions over Q evaluated at 𝛂.

for this reason, Q(𝛂) can’t be the set you described, since for instance, if 𝛂^2 = a + b𝛂 for rational a and b, we would have 𝛂^2 - b𝛂 - a = 0, implying that 1 = 0 (since 𝛂 is indeterminant over Q).

Q(𝛂) is exactly the field of fractions over Q[𝛂], that is, the field of rational functions over Q in 𝛂, denoted Frac(Q[𝛂]).

[topaz solar]

Mhm I see

Now, there’s a very dumb way to see the answer

You can go through a lot of effort about this divisibility and such

Or

What if you shoved it in C

[twilit wraith]

wdym

[topaz solar]

Z[i] -> C

Draw it vro

[twilit wraith]

oh its just like a lattice

[topaz solar]

Yeah

[twilit wraith]

i know what it looks like in my head

[topaz solar]

Ok, now

Take x in your lattice

[twilit wraith]

wait i think i know now

[topaz solar]

And you know 1/x is in C

(x nonzero)

[twilit wraith]

each element of Z[i] has a modulus at least 1

[topaz solar]

Yeah lmao

[twilit wraith]

but for some element of Z[i] with modulus x, its inverse would have to have modulus 1/x

but that cant be the case unless x = 1

duh

[topaz solar]

And that leaves 1, -i, -1, i

[twilit wraith]

so the group is generated by i

i dont even think i need to prove that this is a group tbh

weve seen this group before in class

[topaz solar]

Now, you can do this more generally, as an aside, if you can define norm-like functions on your ring

But you’ll probably see something on that or related later

[twilit wraith]

maybe not, theres only one week of class left

[topaz solar]

A nice algebraic kind of norm is a very special thing though, and is stronger than controlling ideals

The whole Euclidean algorithm type of thing

[twilit wraith]

yeah

is that what a euclidean domain admits?

[topaz solar]

But if you’re char 0 and integral? You fit into C

[twilit wraith]

interesting

[velvet hull]

Couldn’t there be cardinality issues?

[topaz solar]

It has some special nice thing to let you induct on things

[topaz solar]

True, but you morally fit into C anyway by how alg closed is just the cardinality and characteristic when uncountable

The point was more these countable rings but you are correct

[velvet hull]

Well if a ring isn’t even uncountable no one cares anyway so it is true

[last spoke]

I have a feeling that only working with (\mathbb{Q}(\sqrt{k})) may have fucked my intuition a tad

[cloud walrus]

StarvinPig

[kind temple]

what is k here?

[last spoke]

Some non-square integer, usually 2 or 3

[kind temple]

in the case that 𝛂 is algebraic over Q, Q(𝛂) is isomorphic to Q[𝛂] as fields.

this is because of some minimal polynomial jazz + bezout’s lemma

[twilit wraith]

i have to prove that for a ring where all elements satisfy x^n = x, we have that xy = 0 implies yx = 0

i was thinking about expanding (x + y)^n but it seems like its too much

im wondering if it boils down to a counting argument

there are 2^n terms in that expansion but if we ascribe 0 to x and 1 to y then we have that any binary expression with "01" anywhere is 0

so we count all such terms

then theres something about that which forces all the expressions with "10" to be 0 as well

nvm it was just a specific manipulation

im a bit worried that my proof here is not valid

i know the inverse stuff i used works with groups but isnt there weird stuff with rings that makes invertibility a bit harder

[kind temple]

if every non-zero element of a ring has an inverse, it’s a field.

[twilit wraith]

yes so then it has to be commutative right

[kind temple]

you say in your proof: since a and b are non-zero, they have inverses

[twilit wraith]

what im worried about tho is that a and b could have different left and right inverses

[kind temple]

but that’s not true

take for example, R = Z

and a and b any two elements not equal to 1 or -1

[twilit wraith]

oh i misinterpreted what "ring with unity" means

this is a bit harder then

[kind temple]

you are essentially proving this statement

[twilit wraith]

well if ab = 1 then aba = a

oh then a(ba - 1) = 0

so ba - 1 must be 0

[kind temple]

yup!

[knotty badger]

it is true in any category that if f has both a left and right inverse, then they coincide~

[knotty badger]

though that doesn’t quite help here but does help for this

[twilit wraith]

rings are definitely more frustrating than groups but i am somewhat familiar with what they do

and its cool

[knotty badger]

I think I actually prefer rings to groups :P

[twilit wraith]

theres another problem on the homework that im not gonna submit but i kinda wanna ask about how i should approach it

i dont see a way to do so that isnt tedious

i have to prove that Z[\sqrt{d}] is an integral domain

for some integer d

you get a system of two equations and 5 variables if you just multiply stuff out

im wondering if theres just some trick im not seeing bc thats what algebra feels like in these first introductory parts

[kind temple]

you’re saying treat the ring as a category?

[knotty badger]

so the categorical thing doesn’t actually help here

all I mean is that if f has a left inverse g and right inverse h

[twilit wraith]

why can i do that

[knotty badger]

Then $g = g \circ (f \circ h) = (g \circ f) \circ h = h$

[cloud walrus]

Pseudonium

[kind temple]

okay, so you were more commenting on their doubt here

[knotty badger]

Mhm

[kind temple]

whenever R is an integral domain, R[x] is also an integral domain.

[twilit wraith]

how can i prove this

[kind temple]

contradiction is the way that i see doing it

[twilit wraith]

i just have no clue since i just started ring theory today

[kind temple]

what is the product of two polynomials

like, their coefficients?

[twilit wraith]

depends on if theyre polynomials of a commutative ring right

[kind temple]

just assume the ring is commutative

[twilit wraith]

ok then its integers

[kind temple]

Z is commutative

[twilit wraith]

idk if you want further detail than that

its choose function stuff which is integers

[kind temple]

maybe let me phrase it like this: $\newline$

suppose $p(x) = \sum_{k = 0}^n p_kx^k$ and $q(x) = \sum_{k = 0}^m q_kx^k$ are non-zero elements of $R[x]$ (with $R$ an integral domain) of degree $n$ and $m$, respectively. this means $p_n$ and $q_m$ are non-zero. $\newline$

what is the leading coefficient of $p(x)q(x)$?

[cloud walrus]

c squared

[twilit wraith]

p_k q_k

[kind temple]

p_n q_m, but yea

can this be zero?

[twilit wraith]

no

[kind temple]

so can p(x)q(x) be the zero polynomial?

[twilit wraith]

no

i see

it just takes one

[kind temple]

right

so we actually just showed via contrapositive that R[x] is an integral domain if R is

in your specific case, you can assume that d is square free, since otherwise the square part gets absorbed into Z

the most extreme case of this being when d is a square, and Z[sqrt{d}] is just Z again

but it doesn’t matter really because of the fact we just proved

[manic cairn]

Me when Z[(1+\sqrt(-19))/2]

Definitely one of the rings

Whag was that ones special thing again

UFD but not PID?

Or was it PID but not euclidean
Man i forget my rings

[night tartan]

it's PID but not ED

[coarse prairie]

Hello hi i have a question about polynomial roots
While doing some excercises about field/galois theory i saw excercises where my professor went like:
Let f be x^4-4x^2+2 in Q[x]; its known that its irreducible and thus E=Q[x]/<f> is a field
Let α be [x] in E

Show that the roots of f in E are α, -α, α^3-3α and -α^3+3α

Now its easy to show that those are the roots of f but my question is how does one find those expressions for the roots of the polynomial?
First of all the poynomial has to have the property that its splitting field is obtained by just adding one of its roots to Q and that doesnt happen all the time even with separable irreducible polynomials (take x^3-2 in Q[x] and Q(cbrt(2)))
Then, even if we know that this polynomial has this nice property, and thus if we know that all of its roots can be written down as polynomial expression of just one of its roots, how does one go about finding them? I couldnt figure out how to make that α^3-3α from the question i put in the beginning of this message pop up...

Is there a characterization for when a poynomials splitting field is obtainable by just adding one of its roots? Is there then a algorithmic way to find polynomial expressions of the roots of a polynomial in terms of just one of its roots? If there isnt, are there some common techniques to go about doing that?
Thanks in advance! :P

[glad osprey]

I don't think there's a simple characterization for when a splitting field is obtained by adjoining a single root, at least judging from this post: https://math.stackexchange.com/questions/454130/is-kx-langle-fx-rangle-always-the-splitting-field-of-an-irreducible-pol

[keen badge]

Denote $p_n=|\mathcal{P}n|$, and define a set $\mathcal{L}_{n+1}={L(z_1,z_2): z_1\ne z_2\in\mathcal{P}_n}$ where $L(z_1,z_2)$ is the line that goes through $z_1,z_2\in\bC$(does not matter for my question).

The book gives an upper bound $|\mathcal{L}_{n+1}|\le \frac12 p_n (p_n+1)$.

Why is this the bound, and not $|\mathcal{L}{n+1}|\le\frac12 p_n (p_n-1)$, as an element in $\mathcal{L}{n+1}$ is by choosing two distinct points in $\mathcal{P}_n$, so it should be $\binom{p_n}2=\frac12 p_n (p_n-1)$.

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[keen badge]

(This is from Ian Stewert's book Galois Theory -- Chapter 7: Ruler&Compass constructions)

[kind temple]

what is b?

[keen badge]

I think he meant to write "by"

We can, it will just be 0.

I guess it depends if you think of 0 as a scalar or as a vector

You CAN'T think of 0 as a vector, only as a scalar.

[kind temple]

the multiplication for a field is defined on all of F. the multiplicative group of a field is F - 0

[elfin wraith]

You definitely can. The real numbers are the canonical example of a field

[south patrol]

Idk what you mean by this

And also yeah it is s part of the definition of a field that you can multiply any two elements by one another

[modest basin]

But is mq talking about a group or a field, cause groups are indeed closed by their operation like a composed with b must be in G if a and b are in G.

[kind temple]

i made that distinction already

[elfin wraith]

I recommend just looking at the field axioms again, I’m not quite sure what the confusion is

[south patrol]

Well one thing is you can frfine a field F as a set multiplication and addition operations such that x restricts to turn F \ 0 into an abelian group and F under addition is an abelian group (+ distributivity and 1 isn't 0). But it is important that the statement about F \ 0 is about a restriction of an operation which is defined on all of F

At least that's how I think the confusion may have come about lol

[kind temple]

if you are doing some specific analysis with the group F - 0 and kind of forgetting that it sits in a field, then it wouldn't make sense to multiply by 0, since that's simply not an element of F - 0.
but if you are doing something within the field F, its perfectly fine to multiply by zero

[glad osprey]

Maybe it's less confusing to think about rings in general? You can multiply any two elements of a ring R, but the multiplicative group may be much smaller than R \ {0}. The multiplicative group is just the set of invertible elements

[crystal vale]

So there is good news for me, i have to share with you guys, i got selected for the master program in ISI( Indian Statistical institute) and HRI (Harish Chandra Research institute).

So in this my journey you guys helped me a lot specially @rocky cloak , @south patrol, @faint forge , @kind temple @tough raven
Thanks a lot ❤️, apologies if this message should not be here, I will delete if you want

[kind temple]

good luck!

and congratz!

[crystal vale]

Thanks man ❤️

[south patrol]

Congratulations!

[crystal vale]

Thank you ❤️

[tardy hedge]

Congrats man, you're a hard worker.

[queen quarry]

Suppose we have a finite index subgroup H of the free product G * Z, where G is a finite group. The Kurosh Subgroup Theorem tells us what H looks like. Can we say something more detailed, for example that the free component of the free product is Z or that the indexing set of subgroups of G is finite?

(i mean more detailed relative to this, which is the wikipedia version of the theorem)

[thorn jay]

The humble zero vector:

[keen badge]

Thats ehat Im saying. You cant treat 0 as a vector.

[thorn jay]

??

Zero "vector"

[elfin wraith]

Yes you can?

A vector space is a module over a field, a field is a module over a it’s self, 0 is an element of every field

[queen quarry]

but a vector is an element of a vector space, not of a module ☹️

[elfin wraith]

“A vector space is a module over a field”

[queen quarry]

a vector space is precisely a module over a field is what you need to say

[thorn jay]

I think that was implied

[queen quarry]

well, "a field is a module over a field" has the same sentence structure, but here the precisely would be wrong

[thorn jay]

??

Tf

It's also just a wrong sentence

Because that's not what a field is

[elfin wraith]

Are you happier if I clarify a field is a module over itself

[thorn jay]

A field can be seen as a module over a field (namely, itself)

[queen quarry]

Well, isn't a field just a free module with dimension one over a field?

[elfin wraith]

[thorn jay]

No, because you lose the data of the constant 1 and the multiplication

[keen badge]

Denote $p_n=|\mathcal{P}n|$, and define a set $\mathcal{L}{n+1}={L(z_1,z_2): z_1\ne z_2\in\mathcal{P}_n}$ where $L(z_1,z_2)$ is the line that goes through $z_1,z_2\in\bC$(does not matter for my question).

The book gives an upper bound $|\mathcal{L}_{n+1}|\le \frac12 p_n (p_n+1)$.

Why is this the bound, and not $|\mathcal{L}{n+1}|\le\frac12 p_n (p_n-1)$, as an element in $\mathcal{L}{n+1}$ is given by choosing two distinct points in $\mathcal{P}_n$, so it should be $\binom{p_n}2=\frac12 p_n (p_n-1)$.

I am asking in this section because this is from Ian Stewerts book: Galois Theory -- Chapter 7: Ruler&Compass constructions

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[dull marsh]

The proof is correct, but I would like you to notice that the assumption g != e_G in the first part was never used, so you didn't really need to do it by contradiction there

[tardy hedge]

If M is an R-module are there necesary and sufficient conditions for the ring map R->End(M) so that M is a vector space? that map being injective is necessary of course

idk was just thinking about this when i was sleeping last night

[dull marsh]

Isn't an R-module a vector space iff R is a field?

Oh, I guess you want to determine that from only the map somehow

[delicate orchid]

Assuming these are k-element subsets of a set of size n, then S_n is acting via xU = {xu : u in U} for a k-element subset U

the way I'd do it is by showing the homomorphism S_n -> Aut(X_k) is injective aka has trivial kernel

where X_k is what you think it is

I would've done like, suppose x in S_n not the identity that fixes everything, then x(u) != u for at least one u, then for all k < n there will exist a subset that contains u but not x(u) and so x acts on that set non-trivially. I think that's basically the same as what they've done

[kind temple]

it’s not sufficient though. if R is an integral domain and M = R, then the ring map is injective since ax = 0 means that a = 0 for any non-zero x.

also, the map R —> End(R) is injective if R = Z/4Z, yet R is not an integral domain.

[tardy hedge]

Yeah

[kind temple]

i think its true that M is torsion free iff 𝜇(a) is injective for each non-zero a, where 𝜇 : R —> End(M).

so this is also necessary if M is a vector space over R

i thought of this it’s desirable to not have your ring elements kill any of the vectors if they aren’t zero

[tardy hedge]

yeah makes sense

[queen quarry]

But don't you get those out of the scalar multiplication?

[tough raven]

Lean enjoyer detected

[tough raven]

"Vector space over R" is only defined when R is a divison ring, and when R is a division ring "vector space over R" and "module over R" mean the same thing.

[kind temple]

but are there any conditions on the maps R -> End(M) for R-modules M that would enforce this?

[tough raven]

Frankly, I'm a bit skeptical that that's what the question meant, but if it did, the condition is that R is a field. 🤷‍♂️

I find it unlikely that a random ring homomorphism from R to some random ring (even if it's given to be the ring of endomorphisms of an abelian group) can tell you whether R is a field.

Perhaps one can say "if M has a maximal submodule the quotient by which R maps injectively into"?

[tardy hedge]

that is indeed what my question meant

[tardy hedge]

oh ok good point. I guess I was thinking the ring wasnt so "random"

[tough raven]

I guess there are possibilities.

[tough raven]

E.g. this

Or it's imaginable that all subrings of End M of a certain cardinality are division rings so it would suffice that R have that cardnality and the module be faithful.

[tardy hedge]

Thats cool

im not good enough at math to come up with things like that

I dunno i guess for me its been hard to see if im actually improving my understanding of things or not

[lapis trail]

Let me ask a really stupid question but the dihedral group of symmetries of a regular polygon with 2 vertices, aka the Klein 4 group... basically it's a line with the identity one symmetry, rotating the line 180 degrees around the midpoint like a propeller is another symmetry, flipping the vertices about a perpendicular bisector is Third symmetry, but what's the fourth symmetry?

[velvet hull]

doing both

[prisma ibex]

tfw groups have an operation

[vast verge]

I'm not sure where to start with this to be honest

[quiet pelican]

Whats exercise 87 of chp 4?

[violet spade]

Use that |{x\in G : x^d=1 for some d | n}| = n

where n = |G|

What does phi(d) calculate?

[vast verge]

Just to prove n=sum(phi(d))

[vast verge]

Number of coprime numbers less than n

[elfin wraith]

No no, just a module supremacist

Vector space? Ideal? Special cases mate

[tough raven]

Based

[violet spade]

Well... try to lay a connection with the order of an element inside of the group with being coprime

this is an equivalent definition

So try to prove: $\varphi(d)=|{x \in G: order(x)=d}|$

[cloud walrus]

joel

[violet spade]

Where we have G a specific kind of group...

(take $G = \mathbb{Z}/n\mathbb{Z}$)

[cloud walrus]

joel

UGOBEL

[coral spindle]

No they are completely different things

[crystal vale]

Oh you changed your username

[coral spindle]

It’s for a joke

[woeful sage]

Ducktjike

[crystal vale]

Actually this message is also for you @coral spindle

[coral spindle]

Oh congratulations!

Well done man

[crystal vale]

Thanks man ❤️

[woeful sage]

Centralizer of a subset A of G is the set of all elements in a group G (it actually forms a subgroup) that commute with every element of A

the centralizer of the entire group is the center (i.e the set of all elements that commute with every element of G)

normalizer of a subset A is a bit different (it is more "coarse" I guess), but it is of the same spirit instead of gag^-1 for all a \in A, we have gAg^-1 as a whole

[keen badge]

I want to prove that if ${1,\alpha,\beta,\gamma}$ is a basis for a field over $\bQ$, then so is ${1,\alpha,\beta,\alpha\beta}$.

I notice that $\alpha\beta=q_1+q_2\alpha+q_3\beta+q_4\gamma$ for some $q_1,\cdots,q_4\in\bQ$(not all 0).

Then I know what to do if $q_4\ne 0$ but what can I do if $q_4=0$? I'm trying to somehow prove that this leads to a contradiction that ${1,\alpha,\beta}$ is linearly dependent, but I can't do it.

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[woeful sage]

centralizer of G in G

just open up the algebra bible (D&F) they explain this stuff clearly section 2.2 iirc

[coral spindle]

So you were not talking about centralisers/normalisers in general but rather the centraliser/normaliser of a group in itself.

You should have specified this in your question, since otherwise you were just asking definitions!

The normaliser of a group in itself is always the whole group

But groups can be non-Abelian

So typically this is very different from the centre

Anyway I am sure Neam will elaborate

[woeful sage]

N_G(G) is just the entire group G yea, since gGg^-1 = G always because a group is closed under group operation

I don't understand what you mean by this

[knotty badger]

Isn’t the centre the intersection of all centralisers?

[woeful sage]

I think that's equivalent? but I've never heard of this characterization

[knotty badger]

Ah my course had a different def

[velvet hull]

I mean, an element commutes with everything iff in it's in everything's centralizer

[woeful sage]

pretty cool

right yeah

makes sense

[knotty badger]

Centralisers were only defined for individual elements of a group

Rather than for subsets

[woeful sage]

Herstein moment

[coral spindle]

You can define it for subsets

And it’s just the intersection of all the centralisers

This is pretty important for e.g. double centraliser theorems

[knotty badger]

Mhm, makes sense

[woeful sage]

me when studying analysis: I miss algebra
me when studying algebra: I miss analysis

[knotty badger]

me when studying analysis: I love analysis

me when studying algebra:

[keen badge]

Just pinging the question

[coral spindle]

A centre just takes one argument, yeah. Just a group. The others need two arguments

[woeful sage]

The center tells you how close a group is to being Abelian

Niels Henrik Abel, he is based, that's why everything is named after him

[keen badge]

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

[kind temple]

he was a savant

[woeful sage]

Yes

[velvet hull]

I don't see a nice way of proving it, I think your best shot is to concretely describe the algebraic structure of the field

[woeful sage]

What is the definition of a center?

[keen badge]

For the case where q4=/=0 or for the general case?

[glad osprey]

Hmm, is this actually true? Let w be a third root of unity, then consider {1, w, w^2, sqrt(2)} as a basis of Q(w, sqrt(2))

[velvet hull]

well it's not a basis for the field

[glad osprey]

What do you mean?

[woeful sage]

Here is where it says

G is an arbitrary group

[velvet hull]

Q(w) is degree 3, Q(sqrt(2)) is degree 2

so the composite extension is degree 6

[glad osprey]

Ah, lol, good point 👍

[velvet hull]

I mean, that that F is either Q adjoin two square roots, or it is Q adjoin the root of some irreducible deg 4 polynomial

I don't see a way of "abstractly" proving your question without revealing that

[long obsidian]

He was a great Norwegian mathematician

[woeful sage]

"a e A"

[glad osprey]

Okay, I may be missing something again, but what about letting w be a primitive 8th root of unity, ie w = exp(2pi*i/8). Then Q(w) has degree 4 over Q, with basis {1, w, w^2, w^3}. But {1, w, w^3, w*w^3} is not a basis

[woeful sage]

just write a \in A

But yeah that is true

[velvet hull]

wait I think you're onto something

@keen badge

[glad osprey]

If it's hard to prove, it may be because it's false

[woeful sage]

these functions a --> gag^-1 are called inner automorphisms btw, they are quite cool and very important in group theory

[keen badge]

damn

[woeful sage]

But don't fret over the normalizer too much for now, you will understand it more when you study the next chapter (normal subgroups)

[keen badge]

So I guess this theorem works, as long as q_4==0 which is equivalent to αβ being independent of {1,α,β} in which case this is just trivial

[woeful sage]

an automorphism is just an isomorphism from a group to itself

Auto - self

[glad osprey]

I guess you could try to prove it in the case that a and b have degree 2

[vast verge]

Got any hints how I can approach this?

[sonic coral]

Do you know that A_5 is simple?

[vast stump]

if you don't know that a_5 is simple then you can use cauchy's theorem to start a proof

[vast verge]

I haven't covered simple groups yet

I'm mainly just using lagrange's theorem and cosets

[keen badge]

Do you think there is a GENERAL way to find a fourth element for the basis, given the elements {1,α,β}?

[woeful sage]

use Cauchy's theorem to prove that A_5 is simple?

[velvet hull]

well if you know alpha and beta then it has to be some polynomial in 1, alpha and beta

[woeful sage]

polynomail, we saw that

[vast stump]

well the task is to show that a_5 has no subgroup of order 30

[velvet hull]

no you didnt

[woeful sage]

Oh for that particular fact, okay

[keen badge]

Is there a generaly polynomial that would work?

[woeful sage]

yea if you knew A_5 is simple, then since a subgroup of order 30 would be of index 2...which would be normal and...

[velvet hull]

no, never

[keen badge]

Is it proven?

[south patrol]

I guess there is the meme thing that uh

[velvet hull]

by explicit counter example, if p(x) worked then you let alpha be a root of p(x)

something like that

[south patrol]

well actually what i said was just going to be reproving that A5 is normal, lol

[keen badge]

I need to learn about radicals... Maybe there is a way to prove it w/ it

[south patrol]

Like you can work out the sizes of conjugacy classes

[keen badge]

or that will do

Can I dm you? @velvet hull

[velvet hull]

I think you have before

[woeful sage]

so I was asking whether you meant "show A_5 is simple using Cauchy's theorem" or "Show that A_5 has no subgroups of order 30 using Cauchy's theorem"

[velvet hull]

but yeah, do you see a more straightforward way then reproving the simplicity of A5?

cause I dont lol

but thats a (somewhat) lengthy proof

[south patrol]

Honestly i haven't thought about it yet lol hm

[woeful sage]

I always leave the thinking for tomorrow's me

[south patrol]

Oh yeah I know a proof

Here we go

||A5 is generated by 3 cycles. If N is an index 2 subgroup then it must contain all 3 cycles (since the order in A5/N must divide both 3 and |A5/N| = 2||

[velvet hull]

yeah that's just reproving simplicity lmao

[south patrol]

Why?

[velvet hull]

hmm, actually it does skip the latter half of that proof

it's better

[south patrol]

The only bit I skipped details in was the first bit but I don't think that's hard (just check a couple of cases)

[velvet hull]

yep

[vast verge]

I haven't covered Cauchy's theorem or simple groups

[velvet hull]

yeah, without all of that its going to be a somewhat roundabout proof, see potato's message

[vast verge]

I'll check the solutions later seeing as everyone seems rather stumped on it

[wicked patio]

Can you find the sizes of the conjugacy classes?

A subgroup of order 30 would need to be normal, meaning that it has to be some subset of those conjugacy classes

If you can prove that there's no way to make the sizes add up to 30, you're done

[velvet hull]

computing the conjugacy classes of A5 using only langrage sounds like tough work

[jade mason]

Is there a deeper reason for why monomorphisms in Ring are injective but epimorphisms aren't necessarily surjective?

[wicked patio]

Is orbit-stabilizer stuff allowed?

[jade mason]

I've seen the proof of the former via the universal property of Z[x], so it sounds like there's something Yoneda going on

[knotty badger]

There is a categorical reason I guess

[jade mason]

oo

[velvet hull]

their words, they only know of lagranges theorem and cosets, you tell me if you think thats allowed lol

[jade mason]

I'd like to know more

[knotty badger]

Monomorphisms can be expressed as a kind of limit

And right adjoints preserve limits

The forgetful functor from Ring to Set has a left adjoint

[wicked patio]

Well they probably know things other than those

[knotty badger]

Hence it is a right adjoint, so the forgetful functor preserves limits

[velvet hull]

no they just started out learning group theory the answer is no

[knotty badger]

In particular this implies it preserves monomorphisms

[wicked patio]

It would be hard to make a comprehensive list of all your knowledge that might be relevant

[knotty badger]

So it takes monomorphisms in Ring to monomorphisms in Set

Which just says every mono in Ring is injective

[wicked patio]

For example, do you think they've heard of an isomorphism before?

[mighty spade]

what are some equivalent conditions for an arbitrary (commutative) ring to be a euclidean domain?

[knotty badger]

It doesn’t have a right adjoint though, so it won’t necessarily preserve epimorphisms

And indeed in Ring, there are non-surjective epimorphisms

It is a nontrivial theorem that in Grp, every epimorphism is surjective

[knotty badger]

This approach does it a bit more directly by adapting the proof of monomorphism => injective for Set

[jade mason]

guess I'll have to finally learn about adjoints then

[knotty badger]

You should!

[velvet hull]

is there a categorical proof of that? or is it algebraic

[jade mason]

I saw the definition once and it didn't feel enlightening

[knotty badger]

I don’t recall exactly how this is proven

[velvet hull]

I guess you would necessarily have to rely on the group axioms somehow

[knotty badger]

https://ncatlab.org/nlab/show/epimorphisms+of+groups+are+surjective

[velvet hull]

since the existence of non-surjective epis in Ring kind of rely on the ring axioms to do the rest of the "leg work", is how I understand them

[knotty badger]

I think it’s algebraic

I believe this is called Schreier’s theorem

Actually there are a few results of that name so maybe not

[velvet hull]

mfw mathematicians have more than 1 notable result

[jade mason]

we should just force mathematicians to retire after their first notable result to not clutter the namespace

[karmic moat]

we could make it a paid service

your first two theorems are free and the rest you have to pay for per citation

[velvet hull]

or, much better: force them to change their names

[jade mason]

in the spirit of #groups-rings-fields , the new name should be a permutation of the old one

[long geyser]

could you prove that using category theory? does category theory "see" the distinction between surjection and epimorphism? I feel like you'd have to descend to the actual sets involved to make the argument

but I'm just guessing

[velvet hull]

I think no, because such a proof most definitely needs to invoke the group axioms at some point

because the ring axioms have enough "degrees of freedom" per say to allow for non surj epis

[knotty badger]

Yeah the proof is definitely algebraic and not categorical

[lean sail]

quick question. reading section 9 of Fraleigh's a first course in abstract algebra. at the beginning of the section, he says ``We have the group $U$ of complex numbers of magnitude $1$ under multiplication, which is isomorphic to each of the groups $\mathbb{R}_c$ under addition modulo $c$ where $c \in \mathbb{R}^+.$''

I vaguely remember $\mathbb{R}_c$. I looked through my notes and skimmed back through the previous sections, I can't find any mention of it. Internet searches turn up "oh you mean $\mathbb{Z}_c$"... no i didn't mean that! i think Fraleigh is using some nonstandard notation here... but what is this group? what is addition modulo "a real number"?

[cloud walrus]

proofman

[velvet hull]

you can just take c to be equal to 2pi.

then the claim amounts to saying that multiplication on the unit circle is isomorphic to addition mod 2pi, namely through the isomorphism e^{2pi i theta} ---> theta

and by scaling you can see that you can in fact choose c to be any nonzero real number you like

[lean sail]

i think it's coming back to me... i am pretty sure i remember seeing this exact example now somewhere in the textbook...

[cloud walrus]

Slomenist

[lean sail]

quick follow up, fraleigh says that $\mathbb{R}_c$ is an "infinite group". it's not immediately obvious to me why that would be the case. if we're adding modulo $2 \pi$, wouldn't it loop back around? or is it because you can choose any element in $\mathbb{R}$ to do the addition with?

[cloud walrus]

proofman

[velvet hull]

there are infinitely many real numbers between 0 and 2pi

[lean sail]

did you by any chance use Fraleigh?

[velvet hull]

no

[lean sail]

@velvet hull ah ok. is $\mathbb{R}_c$ standard?

[velvet hull]

no, a more fomal way to write it is R/cZ

[lean sail]

ok, that is one thing I saw. that it was a ring. i know nothing about rings yet. we haven't gotten there.

[velvet hull]

you don't have to think about it as a ring

just think about it as a group

[vast verge]

Ok the solution was really long

I could never have come up with this on my own...

[coral spindle]

R/cZ is not a ring actually

Unless you are using the word ring informally, to mean (topologically) a circle

[lean sail]

any ideas on how we can show that $\mathbb{Z}_2 \times \mathbb{Z}_2 \cong K_4$? i think i have an idea for how i can create a mapping, but how would we show the homomorphism property holds generically?

[cloud walrus]

proofman

[candid patrol]

I mean K4 = Z2 x Z2

[lean sail]

i guess i don't really understand this. i've seen a statement that, there are 2 groups up to isomorphism of order 4. i am guessing that one is cyclic, and the other one is not. so i guess all of the groups that are cyclic and order 4 are isomorphic to one another, and all the groups that are not-cyclic and order 4 are isomorphic to one another (like the Klein-4 group and other non-cyclic groups of order 4)?

[tribal moss]

Yes.

Showing that the Klein 4-group is isomorphic to Z2 × Z2 would be by defining an actual isomorphism. (The details will depend on how you have defined the 4-group -- but, as it turns out, every bijection between those two groups that maps the identity in one to the identity in the other will turn out to be an isomorphism ...)

It's also pretty common to think of "the Klein 4-group" as simply an alternative name for Z2 × Z2 -- but in that case there wouldn't be anything for you to prove, so I assume that's not the definition you're working with.

[lean sail]

thank you for jumping in and elaborating on this!

i am new to all of this. i've kind of started working out the details on what could be an isomorphism between the two groups (even though technically they are the same):

Consider the group $K_4$. We will define a map $\phi$ such that
\begin{align*}
(0,0) &\mapsto e \
(0,1) &\mapsto a \
(1,0) &\mapsto b \
(1,1) &\mapsto c.
\end{align*}
Let $a,b \in \mathbb{Z}_4 \times \mathbb{Z}_4$, consider
\begin{equation*}
\phi(ab)
\end{equation*}

getting stuck at the homomorphism property aspect

[cloud walrus]

proofman

[tribal moss]

Basically at this level I'd just go through all 16 possible combinations of a and b one by one.

(which you'll probably want to call x and y here, to avoid confusion with the a and b that name elements of your K4).

[lean sail]

i was thinking the same. i am taking a class, but was never asked to do this exercise. i was just noodling out how this could be done on my own.

[sonic coral]

and also take them out of the right group

[tribal moss]

7 of the 16 combinations are just composing with the identity from one side or the other, so those are quickly handled.

For the remaining 9, you can avoid some tedium by noting that in both groups, the non-identity elements behave as

• an element composed with itself gives the identity.
• two different non-identity elements give the third.
  and you can satisfy yourself by looking at the two group tables that this recipe works at both ends of your map.

[lean sail]

thank you for explaining all of this, and going into great detail! also, is it safe to say that at this point where i am at, there is no "generic" way to do this?

[tribal moss]

The only "generic" way is brute force. You'll pretty much never need to do it that way, though -- there are almost always shortcuts, but they're specific to what you already know about the groups and their definitions.

[glad osprey]

Also both groups are abelian, so you only need to check half the combinations

[twilit wraith]

i have to show that if R is a ring with unity and there exists a ring homomorphism from R to S then S has a unity

the thing i did first is show that for any x in R, we have that phi(x) = phi(x)phi(1) = phi(1)phi(x)

but this doesnt prove anything unless phi is surjective right

[quiet pelican]

I think that’s false
Should you have the word surjective in it?
Notably, if you take the 2x2 matrices with entries only in the top row, this is a ring, and it’s easy to see there’s no unit as (a, b)(c, d) = (ac, ad)

[tribal moss]

It proves something, not just the thing you need to prove ...

[quiet pelican]

Then inject R into this in the obvious way, with the top left component

[twilit wraith]

no im pretty sure its right

at least on wikipedia it states that ring homomorphisms send unities to unities

im working with the definition that doesnt include this though

[quiet pelican]

That’s an assumption we make on ring homs
Not something that follows from a non-unital defn

[tribal moss]

Note that (1 0 / 0 0) actually works as a unit in that ring.

[quiet pelican]

It’s only one sided

[tribal moss]

Oh foo.

[twilit wraith]

wait so my homework question is asking me to prove something false?

[tribal moss]

Perhaps rings are commutative in Hiidostuff's context?

[quiet pelican]

This doesn’t feel like something commutativity should fix

[tribal moss]

Just grasping at straws.

[twilit wraith]

"Let R be a ring with unity, and let φ be a ring homomorphism from R onto S. Prove
that S has a unity." verbatum

fuck it says onto

damn it

[tribal moss]

🤣

[twilit wraith]

if i had a dollar for every time i didnt read the word onto id be rich

well now this is seriously easy

all elements of S can be written is phi(x) for some x in R

then phi(1) fits as the unity

[quiet pelican]

Yeah just intuitively here, if we don’t have surjectivity, then we can’t really say stuff about the ring outside the image

Oh wait, for a commutative example, we have Z/2Z injects into Z/2Z x 2Z

[twilit wraith]

thats what i was thinking

but at this point in time i question how much i really know about algebra

so i wasnt confident

[tribal moss]

Delightfully simple. I was trying all kinds of weird stuff with polynomials, and running into a wall.

[open owl]

I have a problem here, anyone can assist?

[twilit wraith]

i cant tell what the problem is

it looks more like linear algebra than it does abstract algebra

[karmic moat]

Go to #linear-algebra

[static mauve]

Thinking about Free Abelian groups, {1} is a basis for 2Z, yes? (Though maybe its a bit weird that {1} generates elements that dont belong in 2Z)

Ah nevermind... 1 is not a element in 2Z so this doesnt work.

[kind temple]

2 generates 2Z

abstractly though, it doesn't matter what you call the generator. 2Z is, up to isomorphism, the free abelian group generated by a single element

[hidden wind]

yeah {2} is a basis for 2Z

idk why you’d be talking about 2Z and not just Z though

note that {1} is a basis for Z, but {1} is not a basis for Z/2, even though 1 generates Z/2

because Z does not give unique coordinates for elements of Z/2, the relevant map obviously being many-to-one

[cobalt heath]

Is Q(zeta_3) = Q(zeta_6) ?

[mighty kiln]

Yes

[cobalt heath]

Is Q(zeta_n) = Q(zeta_2n) in general, for odd n?

If so.. I forgot how to show this

[mighty kiln]

Yes

Consider generators of (Z/2n, +)

[cobalt heath]

Additive group? I can only think of 1 and -1

Ah wait, nvm. I guess all integers coprime to 2n are generators

[mighty kiln]

*generating sets

[cobalt heath]

Damn

n + 2 ((n + 1) / 2) = 1 in mod 2n

So I guess
zeta_2 zeta_n^((n+1) / 2) = zeta_(2n)

Now I feel dumb..

[cobalt heath]

How did you know this right away?

[rustic crown]

picture ig

imagien a cricle and 3 dots on it

[cobalt heath]

Am I just having skill issues

[mighty kiln]

Just considered this

[rustic crown]

then imagine 6 dots

[cobalt heath]

I see, simply that I got skill issue

[cobalt heath]

Yeah, I only thought of how zeta_3 + 1 = zeta_6

[rustic crown]

(zeta_n)^((n+1)/2) is the next dot after -1. so -1 * (zeta_n)^((n+1)/2) is a dot between 1 and zeta_n.

[mighty kiln]

So any for any collection of integers with gcd 1 this holds as well

[cobalt heath]

Yeah, now I see that works.

Guess additive (!) Z/kZ is still hard for me

[rustic crown]

we're looking at the generators, so it's just the multiplicative group

mu_{2n} = mu_n * mu_2

where mu_k is the subgroup of k-th roots of unity of say C*

.< the multiplicative group mu_n is isomorphic to the additive group Z/n >.<

[rustic crown]

here tehse are normal subgroups since everything is abelian and intersection is {1}.

in particular zeta_2n lives in Q(mu_n, mu_2) = Q(zeta_n, -1) = Q(zeta_n)

[cobalt heath]

Do you also happen to know:
Is
Q(zeta_m) \otimes Q(zeta_n) = Q(zeta_mn)

[cobalt heath]

If this is true, set m = 2 and I get the zeta_2n result right away. Dam

[rustic crown]

true when m and n are coprime

[cobalt heath]

How do you show this?

[rustic crown]

because you have an easy map in the forward direction, and by the stuff we were saying, it is surjective, and both sides have same degree/dim_Q

[tribal edge]

notice that zeta_n = (zeta_2n)²
so zeta_n is in Q(zeta_2n)
hence Q(zeta_n) ⊂ Q(zeta_2n)
the other inclusion is satisfied as well since Q(zeta_2n) and Q(zeta_n) have the same degree phi(2n) = phi(2)phi(n) = phi(n) over Q where phi is the euler's totient function

[rustic crown]

ig here we're using that [Q(zeta_n) : Q] = phi(n) = #(Z/nZ)*. which relies of irreducibility of the cyclotomic poly. dunno if we can avoid it

[cobalt heath]

Yeah this is fine for me

[cobalt heath]

You mean the map from the tensor product to Q(zeta_mn)?

[rustic crown]

yee

[cobalt heath]

Yea the map is there by tensor product property

[knotty badger]

universal property? 👀

[cobalt heath]

Surjective is easy, dimension is also easy

[cobalt heath]

Pushout 😉

[knotty badger]

Cool

[inner owl]

Ok so I had this problem where I have to prove that x^4+1 factors in F_p[x] for all prime p but I am not making much progress can someone give some hints

[rustic crown]

i think you mean F_{p^2}[x]

[inner owl]

Nope

F_p[x]

[rustic crown]

ah maybe then not completely factor ig?

[inner owl]

I mean I have to show that the polynomial is reducible in all F_p[x] where p is prime

[rustic crown]

idk if there are general ways to see it, but
i would just write
(x^2)^2 + 1 = (x^2+1)^2 - 2x^2 = (x^2-1)^2 + 2x^2

p = 2 is obvious. for odd p, oen of -1, 2, or -2 must be a square.

so it's a difference of squares and thus factors

[velvet hull]

it doesn't always split completely, as x^2 + 1 = 0 has solutions mod p iff p = 2 or p = 1 mod 4

as det says, they're probably just looking for a factorization

[inner owl]

Yes

Sorry

[inner owl]

Is this some known result?

[rustic crown]

yeps, but we're not using anything very deep here. the multiplicative group (F_p)* is cyclic of order p - 1, so therefore if we quotient by the subgroup of squares, we get (F_p)*/((F_p)*)^2 = {+-1} is a group of order 2. in pariticular this isomorphism shows that multiplying two things which are not squares mod p gives you a square!

[quiet pelican]

That sort of “add and subtract, then dots” can be a relatively common way to factorise x^4 + a, as well

[rustic crown]

this character (F_p)* --> {+-1} is called the legendre symbol, maybe you've heard of it :3

[rustic crown]

one can give a slightly more down to earth argument if you don't like me using F_p* is cyclic

[cobalt heath]

Thanks! Nice, now I can understand the hypercube space acted by Galois group

[rustic crown]

ooh what's that?

[quiet pelican]

How does that argument go?
I can’t see anything that doesn’t go through “F_p^* is cyclic”

[cobalt heath]

Basically, you can think of
Q(zeta_m) \otimes Q(zeta_n) as rectangular Q-vectorspace spanned by zeta_m^k zeta_n^l.

Gal(Q(zeta_m) \otimes Q(zeta_n)) is (Z/nZ)^× * (Z/mZ)^× , where each component act on each "side" of the rectangle.

If m, n are odd prime power, e.g. (Z/mZ)^times is cyclic, so it gives rotation on the side of rectangle!

So Galois action is like, rotating the hypercube along certain axis!

Welp, I messed up slightly. To give rise to the hypercube structure, I should talk about R \otimes Q(zeta_(m n)) or Z_t \otimes Z[zeta_(m n)]

[rustic crown]

ah so idea is that everything in F_p* is a root of x^(p-1) - 1 = (x^((p-1)/2) - 1)(x^((p-1)/2) + 1), and all squares are also roots of x^((p-1)/2) - 1. but for i = 1, ..., (p-1)/2 the values i^2 are distinct so these must be all the squares. and all the roots of x^((p-1)/2) + 1 must be non-squares. this shows the euler's criterion: the map F_p* --> {+-1} given by a mapping to a^((p-1)/2) is a character which maps to 1 on squares and -1 on non-squares.

[cobalt heath]

P sure R \otimes Q(zeta_(m n)) is isomorphic to C^(phi(m)) \otimes C^(phi(n)), which is the actual 2-dim'l hypercube where Galois group acts by rotation.

[quiet pelican]

Ah right
bleh, elementary NT

[rustic crown]

maybe you mean C at the front instead or R? me not sure in what sense this is seen as a rectangle.

initially i was thinking it as a discrete rectangle of size phi(m) x phi(n). but the last part says the sides of the rectangle are C-vector spaces?

[cobalt heath]

Ah, yeah that should be C

It is discrete rectangle of size phi(m) × phi(n) where you can put complex number in each "slot"

In other words, C^(phi(m)) \otimes C^(phi(n)) is like a complex matrix of size phi(m) × phi(n)

From the Galois group, (Z/mZ)^× portion acts by vertical sliding (yeah it is called "rotation" but it looks more like sliding),
and (Z/nZ)^× portion acts by horizontal sliding.

[rustic crown]

is Q(zeta_n) generated by zeta_n^k for k, n coprime as a Q-vs?

if not this, then the Galois group doesn't act on the discrete rectangle right

okie this false for n = 4

Q(i) is not Q{i, -i}

so really need to think of the side as something like C ⊗ Q(zeta_n)

[inner owl]

So here if we assume 1 -2 and 2 are simultaneously not squares then 1×2 or 1×-2 are squares thus a contradiction

[fading acorn]

it’s way more than typical undergrad algebra course

Undergrad algebra courses are toward to doing some basic galois theory

and dnf does way more than that like it contains some basic com,homalg and rep theory

[velvet hull]

part 1 (group theory), 2 (ring theory) and 4(field theory) is the syllabus of a typical undergraduate algebra sequence

part 3 (module theory) is a slice of the module theory one would encounter in a graduate algebra course

part 5 is a crash course on commutative algebra and homological algebra, it's a bad idea to learn about those topics here lol it looks way too rushed, there are whole textbooks dedicated to each of the two topics

and then part 6 is representation theory, which is kind of its own thing

[vast stump]

they are covered in 1 year, i.e., in a 2 semester or 3 quarter sequence of courses

[velvet hull]

depends on how the university organises their classes

but the content is the same

[fading acorn]

depends on uni but for example some unis are like

3-1 grp theory
3-2 ring theory/modules theory
4-1 fields theory

other example is just they do all of them in just one year course

[velvet hull]

If you haven't done any university math before there isn't really anything good to compare this to

but the standard learning time is 1 school year (9 months)

[rustic crown]

yeps

-1*

1 is always a square, 1 = 1^2 :p

[velvet hull]

every pure math major, yes

it's a mandatory requirement for a pure math major

that's pretty normal, if anything half a semester sounds a bit low

of course you should probably be doing other things at the same time as well

you'll have to show the syllabus for that class, it does not sound standard

it's not typical, but it certainly looks like an interesting class

[grim bridge]

sounds fun. you will probably know whether it would be fun for you by the time you take ring theory

[velvet hull]

no harm in taking it, if you can

[grim bridge]

well, you would be taking it at least the semester after you take group theory, ideally the semester after you take ring theory

oh, i see. if i were you, i would take group theory and ring theory, then if you like it you can take this class in a later semester

what is this one you mentioned?

so, you are taking an abstract algebra course right now? or is abstract algebra the group/ring theory course

are you well through your current course or just starting? like you know groups and rings and stuff or no

these courses look like they would probably not use D&F, they are graduate level

you will probably cover D&F section 1 in your current course, if i had to guess

[fading acorn]

noethern and artianian rings are usually taught in grad courses like commutative algebra

[grim bridge]

maybe ring theory but it seems rushed if so

wow, that's a lot. if you are enjoying it so far, i would think about adding those courses. maybe look ahead in the book a bit?

specifically can look at chapter 6, see if it's interesting. but may be hard since you are still early in your current course

from my experience i feel these will be somewhat present throughout all abstract algebra. they will probably get easier as you go through your current course but i'm in the same boat that they are not my favorite. if it's any consolation, i would still be happy to take the two courses you mentioned even with that considered

yeah, it seems like that is the best option. since it's decently likely you will like them, but not 100% certain

yeah it does seem very fast. i don't know exactly how long your semester is but the other two seem fast as well, though probably less fast than your current

keep in mind i only know a little of what those courses cover

ah, they probably will be a lot slower paced then

[barren sierra]

How do I find such elements x and y?

I don't really understand this hint (I know how to solve the problem in a different way not using the hint so I actually want to understand this hint)

[south patrol]

nilponent 🥀

[barren sierra]

lmao didn't even notice

[south patrol]

I guess I has to be principal so you can consider a generator as your x

[barren sierra]

principal?

how?

[karmic moat]

maybe by minimality not sure tho

[south patrol]

Well like if aI is nonzero then there's x in a and y in I with xy nonzero

So (y) is an element of the set of ideals, and is contained in I, so it is I

[barren sierra]

ah yea

ok cool I'll try to go from there

[south patrol]

I think yeah the point is just that you want to go from talking about ideals to talking about elements

[barren sierra]

I don't really see what $xy = x$ gets me then. So $y$ is in $\mathcal{N}^n$ for some $n$. I'm not sure what contradiction I'm supposed to find here. Maybe somehow an ideal which is smaller than $I$?

[cloud walrus]

Spamakin🎷

[barren sierra]

Like I can't say anything as far as I can tell from x(1 - y) = 0 because we don't know anything about zero divisors i think

[tribal edge]

no
notice that x = xy^j for all j ≥ 0 by induction

[barren sierra]

ohhhh

ty

I got as far as (xy)^n = x^n y^n = 0 😵‍💫

[tribal edge]

about the zero divisors, if a ring is artinian and an integral domain it would be a field, Z/nZ for non prime n is artinian but not a domain

[barren sierra]

Right

[south patrol]

Also like 1 - y is invertible cause y is nilpotent but ye

[tardy hedge]

Lol

[south patrol]

Lol

[karmic moat]

i can never remember the millions of nakayama's lemmas

[south patrol]

There is only one

[south patrol]

Wdym by this tho lol

I mean okay I know of a few forms

[karmic moat]

like if you held me at gunpoint and told me to state one form of nakayama's lemma

i'd just say pull the trigger

i've used them so rarely that i have to look it up everytime it's mentioned

[karmic moat]

the 1-y thing made me think of nakayama's lemma tho

ohhhh shit wait yeah that makes sense to me now

[south patrol]

I think the form I always have in mind is M fg and IM = M => there's x in I with (1+x)M = 0

Cause strong and immediately gives you M = 0 if I is contained in jacobson radical, or if M is over a local ring

[karmic moat]

oh yeah that's nice

actually i can't even remember the last time i use nakayama

[south patrol]

Me neither lol

[karmic moat]

oh actually i think the last time i saw it was in a class

something to do about grothendieck groups

[serene acorn]

i have a homework question from group theory that i have turned in alr but i couldn't really find a nice answer to:
Let $H$ be a subgroup of $G$ so that $[G:H] = n$ is finite. Show that there exists some $N \unlhd G$ such that $[G: N] \leq n!$ and $N \leq H$.

[cloud walrus]

haseeb

[rustic crown]

have you learned about group actions already?

[south patrol]

Cayley

[serene acorn]

oh. ._.

[thorn jay]

Kernel of the group action of G on G/H

[rustic crown]

well it's been answered now

[south patrol]

Yes considering group actions and knowing that you need n! to appear lol

[serene acorn]

i did figure it had to do with perm representations but i wasn't sure how

[thorn jay]

The n! was a dead giveaway lmao

[thorn jay]

I'm sorry 😔

[rustic crown]

it okie

[serene acorn]

i have, but the cayley's part was not obvious to me

[south patrol]

Is there another way hm

And also is this optimal

[serene acorn]

for some reason, applying it to G/H exactly, i kept trying to apply it to G or H

[rustic crown]

hmm getting the n! might be hard

[serene acorn]

wait but we don't have that $H \unlhd G$ so that $G/H$ is a group?

[cloud walrus]

haseeb

[rustic crown]

but getting the N can be said without that language

[thorn jay]

Set of cosets

(left)

[rustic crown]

yea because you're just intersecting all the conjugates of H.

[thorn jay]

You're letting G act on the set of left cosets on H, which there are [G : H] = n of

[south patrol]

Well I guess you can also just form the intersection and then bound the index somehow

[rustic crown]

i didn't see any cleaner way than just this lol

~~i'm not doing an induction ~~

[serene acorn]

im not doing one either

[serene acorn]

that does make sense, but what if G/H isn't a group?

[rustic crown]

yee so you look at the set G/H := {aH | a in G} as the set of left cosets of H. and let G act on it by g * (aH) = (ga)H.

and in general if G acts on a set X, you can interpret the action as having an automorphism g : X --> X for every g in G which behaves well with the group operation of G.
i.e. you have a group homomorphism G --> Perm(X).

[rustic crown]

and the kernel of this homomorphism is what was suggested here

[serene acorn]

i think im still missing something. if |G| = nm, say (where m = |H|), then G is isomorphic to Perm(nm), not Perm(n)

where am i wrong

[rustic crown]

G is iso to a subgroup of Perm(|G|) not equal to it

also G in the question could be infinite for all you know :3

the finiteness of G/H is what is needed anyway

[serene acorn]

and we're picking the kernel of the action anyway

[serene acorn]

whoops you're right! that makes a lot more sense

i think i will go renew my understanding of actions and Cayley's Thm

thanks friends :)

[rustic crown]

good luck

[thorn jay]

I love doing the "kernel of permutation representation" trick it works everytime you want to generate a normal subgroup