groups-rings-fields

[chilly ocean]

no

[elfin patrol]

How? Isn’t trivial homormorohism just mapping each element to itself

[chilly ocean]

i mean the homomorphism G -> H taking everythingin G to the identity element of H

[viscid pewter]

trivial homomorphism maps everything to identity

identity homomorphism maps everything to itself

i believe

[elfin patrol]

Oh I see. That makes sense then thanks

[tawny sphinx]

any ideas how to prove that φ : R-> Frac(R) is surjective?

[chilly ocean]

that phi being?

[hidden haven]

and what is frac(R)

[sharp sonnet]

you take an element of Frac(R) and find a preimage under phi

[chilly ocean]

Why is ideal generated by content of a polynomial the ideal (1)=R

uhh dont u need R to be a field or sth lol

[tawny sphinx]

the field of fraction

yh R is a field

sorry forgot to mention

[chilly ocean]

like c(2x) = 2 this is not whole ring

[chilly ocean]

R is any commutative ring

I think

in Z[x]

[hidden haven]

lol that was very important

yeah so given a/b in frac(R), find an inverse image

[chilly ocean]

frac(R)phi(b) should be right b cosets of R which should contain everything in fraction field besides elements looking like a/b

right?

frac(R)phi(b)

[chilly ocean]

this is wrong lol

i forgot phi was there

[hidden haven]

it won't be cosets, it would be the ideal generated by phi(b)

[chilly ocean]

idk what phi is too

[hidden haven]

in frac(R)

I am assuming that phi is the natural map R → frac(R)

[chilly ocean]

oh

sending a->a/1

[hidden haven]

yes

[tawny sphinx]

yeah it is

[chilly ocean]

good question im not sure

like I said its not true

oh f needs to be primitive

in general rings

oh lol

[hidden haven]

lol

[thorn delta]

by definition this is true when ur poly is primitive

[chilly ocean]

only true in UFDs

[thorn delta]

what is?

[chilly ocean]

what is example of primitive and gcd not equal to 1

[thorn delta]

there is none

[chilly ocean]

I have a theorem here saying if R is a UFD and f is primitive iff content is 1

[hidden haven]

what is definition of primitive and content

because the definition I see is primitive iff content 1 lol

[chilly ocean]

A polynomial is primitive if for all principal prime ideals p, f not in pR[x]

[thorn delta]

busting out hungerford. are you talking about this?

[chilly ocean]

Yea

no

[thorn delta]

yea ok your definition of primitive is weird

[chilly ocean]

Said it isnt usually presented in this way

idk a good example

uh

Z[sqrt negative 3]

what is an example of a primitive polynomial in it

x

fuck

So say we have $(\sqrt{-3})Z[\sqrt{-3}]$

[cloud walrus]

mMahael

[chilly ocean]

where () is ideal generated by

actually fuck this

ill come back to this on tuesday

Well I guess I want to find a polynomial f such that for any principal prime ideal p in Z[-3], f not in pZ[x]

Im starting to forget why I should care about this

oh

uh

pZ[x] are in general the set of polynomials with coefficients in p

So I just need a polynomial with coefficients not in every prime principal ideal of Z(sqrt-3)

[chilly ocean]

how about 2x

[coarse stag]

What does it mean for conjugates to intersect in the identity subgroup

[next obsidian]

When two things intersect at the identity it just means H\cap G = {e}

[warped herald]

(i already asked this in the help chats but they could not answer)...this is a question about fermats last theorem, can someone please explain to me how you link elliptic curves to modular forms through the use of L-functions (modularity theorem) because i dont really get it...

[chrome hinge]

Hi. Can anyone give me a nice reference to study a little bit of the Heisenberg group?

[hidden haven]

You might wanna delete and post in #advanced-number-theory, this will probably just get buried here

[warped herald]

tyy

[chilly radish]

do nonunital group rings not contain the group elements themselves (i.e. the elements generating the module)?

also if rg=rh, and r is not zero, this means that h=g right? We didn't even formally learn about group rings over rings lol, only over fields and I know modules are funky w.r.t like independence and stuff

but this is true right

or is this only true if R is unital

[lethal cipher]

Hey y'all, so I am having a hard time

1. processing the fundamental theorem and
2. Determining how to actually find what the direct product is based on the finite group G

Moreover, what does "prime-power order mean"?

[coarse stag]

You should be able to decompose it if you know the order of each element

[chilly ocean]

oh

gcd isnt defined for all integral domains

you need a ufd or a gcd domain

so thats why content is defined weirdly

[next obsidian]

Yup

Definitions of primitive vary based on who’s saying it and they aren’t always the same

[prisma thunder]

For what $p$ does $f(x) = x^5 + x + 1 \in \mathbb{F}_p[x]$ have distinct roots over its splitting field?

[cloud walrus]

eM

[prisma thunder]

Let $L$ denote the splitting field of $f$. Before we proceed, observe that $0$ is not a root of $f$ since $f(0) = 0^5+0+1 = 1 \neq 0$ for any characteristic $p$. We will consider two cases: $p = 2$ and $p \neq 2$.

Suppose $p\neq 2$; observe that $f$ has a multiple root $\alpha$ if and only if $\alpha$ is a root of $f'$, i.e., if $\alpha$ is a multiple root, it satisfies $\alpha^5+\alpha+1 = 0$ and $5\alpha^4+1 = 0$. From this, we determine that $4\alpha = -5 = (p-5)$, which, by Fermat's little theorem, since $4$ is coprime to $p$, implies $\alpha = 4^{p-2}(p-5)$. Observe that if $p=5$, $\alpha = 0$, which cannot happen. Since $4^{p-2}(p-5)$ is non-zero in $\mathbb{F}_p$ for $p>2$ and $p\neq 5$, we have $f$ has multiple roots in $p> 2$, $p \neq 5$.

However, if I check $p = 11$ and see its splitting field, I find that $f$ does not have multiple roots, so I'm wondering what I'm doing wrong.

[cloud walrus]

eM

[next obsidian]

A finite field is perfect so any extension is separable. This means if f is irreducible it will have distinct roots

So you only have to analyze what happens when p is such that this is reducible

[dire bramble]

am I being stupid or does this problem have a typo

wouldn't this mean that every galois extension is trivial lol

[next obsidian]

Trivial in what sense?

[dire bramble]

like, if K is already the splitting field for every element of L then isn't L/K just K/K?

[next obsidian]

But K ain’t

Cuz alpha_i isn’t in K

The product of this magically has coefficients in K

Take like R and then consider a third root of unity

The Galois conjugates are the other third roots of unity

[dire bramble]

wait give me a second to parse this

I've been ass at this part of galois theory

[next obsidian]

Or take C over R this is easier

The degree is 2

Generated by complex conjugation right?

Let alpha = i then the alpha_i are i and -i

Then the min poly for i is (x - i)(x + i) = x^2 + 1

[prisma thunder]

Perfect fields are top tier

[next obsidian]

I am learned field theory for comm alg

[next obsidian]

Do you want me to walk you through how to prove this?

Whoops should’ve replied to the original problem

[dire bramble]

Just give me a moment

I am reviewinf notes

[white nymph]

is this an appropriate place to ask a question about some (basic) field theory? I'm having a hard time with an exercise and trying to understand my instructor's hints/comments.

[next obsidian]

Yes

[white nymph]

my instructor wrote me back and I have better grasp on the problem. Thank you though.

[prisma thunder]

It reduces to $(x^2+x+1)(x^3-x^2+1)$ in any field.

[cloud walrus]

eM

[prisma thunder]

BUUUTTtTTTtT realizing that helped me solve another problem completely 🙏

[next obsidian]

Oh fucking rip lmao

Something something the different factors can’t share roots, so only have to see if the two factors are reducible?

I think if both are irreducible then they can’t share roots

Because like if alpha was a shared root both of these would be its min poly… but they’re different degrees lol

[prisma thunder]

I was told to look at this polynomial and compare it to its formal derivative since if it had any multiple roots, its formal derivative would be zero there as well.

[next obsidian]

Yeah like

alpha is a multiple root iff alpha is a root of f and f’

this is why char 0 fields are perfect

Why did I forget this fact kekw

[prisma thunder]

From earlier, I found if f had a multiple root alpha (nonzero since zero is not a root ) that’s not characteristic two, I found that, comparing f and f’, we have alpha satisfies 4*alpha = -5.

I wasn’t too sure what to do after that

[neat valley]

What of interest can I say about two homomorphisms with the same kernel, domain, and image?

Of course, they aren't necessarily the same homomorphism, consider two automorphisms on $\bZ/p\bZ$ (for $p$ prime with $p>2$), the first of which maps $1$ to $1$ and the second of which maps $1$ to $2$.

[cloud walrus]

Isaiah

[dire bramble]

Ok, so is the proof just that m_alpha splits into a product of linear factors in L, the orbit of Gal(L/K) of an algebraic element is the set of roots of m_alpha, so m_alpha is the product of (x - s(alpha)) over s in Gal(L/K)?

Is it really that simple?

The coefficients being in K just follows from K[x] embedding in L[x] I guess

[next obsidian]

This is the idea basically

The reason the coefficients are in K is cuz when you apply any g in G to the polynomial, all you’re doing is swapping around where the alpha_i are

So the polynomial is fixed by every g in G

Then as K is the fixed field of G, all the coefficients are in G

So you have to now show that the product of the (x-alpha_i) is actually of minimal degree, but this is true cuz the minimal poly of alpha necessarily has g(alpha) as a root for all g

[dire bramble]

Okay perfect

Now i have another question

Is this not just false

Sorry, take F_p(x)(y)

[next obsidian]

What is a simple extension again?

Like generated by one thing?

Anyway I think this is supposed to say algebraic

Cuz this is the primitive element theorem

Or I guess a finite extension

Perhaps “extension of finite fields” means a finite field over another finite field which is equivalent to just any finite extension of a finite field

[hidden haven]

An algebraic extension is simple iff it has finitely many intermediate extensions, and if you have a finite extension of finite fields then this is obviously true

[next obsidian]

This image is good

It can only be shown in small durations

[hidden haven]

[next obsidian]

Maybe it should live in category theory actually

[hidden haven]

Oh even easier lol

[next obsidian]

Just use the primitive element theorem

[hidden haven]

Finite subgroups of mult group of a field are cyclic

So extension is generated by a generator of multiplicative group

[next obsidian]

How do you get from fact 1 to the conclusion?

[rustic crown]

what's fact 1?

[neat valley]

Can anyone give me any hints?

Let $(x,y)$ and $(x',y')$ be elements of $\bZ_p \times \bZ_p$ that are not the identity. Show that if $(x',y')$ is not an element of $\langle(x,y)\rangle$, then there exist some integers $a,b$ such that $(ax + by \equiv 0\ \mathrm{mod}\ p)$ and $(ax' + by' \not\equiv0\ \mathrm{mod}\ p)$

[cloud walrus]

Isaiah

[neat valley]

p prime

[rustic crown]

p-adics or just Fp?

if it's F_p then, notice (F_p)^2 is a vector space over F_p and the condition then says that (x, y) and (x', y') are linearly independent. we can define a linear map which sends (x,y) to 0 and (x',y') to 1. but linearly forces it to look like (s, t) maps to as+bt mod p.

[neat valley]

If by F_p do you mean cyclic group of order p, then yes

[rustic crown]

yep, F_p is used to like stress out it's a field, rather than just a group.

[neat valley]

oh cool, TIL, that makes me think of the free group with p generators tho :P

[rustic crown]

oh lol

[neat valley]

This linear map you're defining, what is its codomain?

[rustic crown]

F_p^2 --> F_p

so you can find a, b such that ax+by = m (mod p) and ax'+by' = n (mod p) for any choices of m and n

[neat valley]

I'm unclear on how this map extends to the rest of F_p^2 outside of elements generated by (x,y) and (x',y'). Would you mind revealing a bit more? FWIW, I'm trying to prove that the subgroups of Z^2 of index p correspond exactly to subgroups of F_p^2 of order p, and this is one way of formulating the last thing I need for my proof.

[rustic crown]

oh i see, this is what i was using...
if V is a vector space over a field k, and B is a basis of V, then you can extend any set function B --> W to a linear map V --> W.

if you haven't seen much linear algebra, i can probably remove this language

[neat valley]

Right, sorry, that makes perfect sense. Would you mind hinting as to how I can do this using just group theory?

I would need to prove multiple things, such that any element of F_p^2 can be written as a(x,y)+b(x',y') for some a,b, and I feel there should be a more direct way then converting the problem to linear algebra, no?

[rustic crown]

right, lemme think if i can directly do it with group theory

okie this should work

if H is a subgroup of ZxZ of index p, then since everything is abelian i can talk about (ZxZ)/H which will be a group of order p. in particular if r is an element of ZxZ then p * (r+H) = H meaning p * r is in H.
therefore subgroups of index p will contain the subgroup (pZ) x (pZ)
now look at the quotient map
(Z x Z) --> (Z x Z)/(pZ x pZ) =~ (Z/pZ) x (Z/pZ)
by correspondence theorem subgroups of right correspond to subgroups of left containing pZ x pZ. just restrict this correspondence to subgroups of ZxZ with index p.

since size of (Z/pZ)x(Z/pZ) is p^2, index p and order p are same thing.

[neat valley]

I would need to show that subgroups of index p are exactly those containing pZ x pZ as a subgroup, right?

Wait that's not true

Wait is it?

[rustic crown]

nah, you only need index p subgroups contain pZ x pZ.
{subgroups of ZxZ containing pZxpZ} and {subgroups of (Z/pZ)^2}
now restrict the left to subgroups of ZxZ with index p.

the correspondence sends H with index p to H/(pZ x pZ)

but H has index p in Z x Z, so H/(pZ x pZ) has index p in (Z x Z)/(pZ x pZ)

[neat valley]

Right, thanks

This was very helpful, I really appreciate it

[rustic crown]

so start with the correspondence
{subgroups of ZxZ containing pZxpZ} and {subgroups of (Z/pZ)^2}
and restrict to get the correspondence between index p subgroups

[neat valley]

Wow, that's an amazing proof haha, thanks lol

My original one involved enumerating all surjective homomorphisms from Z^2 to Z/pZ, and then associating each homomorphism f to the pair (f(1,0), f(0,1)) in Z/pZ x Z/pZ, and then showing that two homomorphisms had the same kernel iff they belonged to the same proper subgroup of Z/pZ x Z/pZ

And I got stuck on what I posted lol

[hidden haven]

did you just number someone else's facts?

[next obsidian]

No

[hidden haven]

[next obsidian]

Fact 1 was the

Finite subgroup of field’s multiplicative group is cyclic

Lmfao

[hidden haven]

lol

[next obsidian]

I was asking how you get from there to your conclusion

[hidden haven]

lol

[upper cape]

Anyone know if the quotient of a principal ideal is still principal?

i.e. let A be a ring, I an ideal of A and A/I the quotient ring. If J is a principal ideal in A, is J/I a principal ideal in A/I?

[hidden haven]

J/I is generated by the image of the generator of J

[upper cape]

thanks!

[winter thorn]

Hey yall, I need to show that if G is a finite group, then the number of x in G s.t. x^3 = e is odd, and y in G s.t. y^2 =/ e is even.

[weak oriole]

If x has order 3, then so does x^2

[chilly ocean]

A cute group theory problem

Show that every finite abelian group is isomorphic to a subgroup of a U-group

U-groups are U(n)={k>0 : gcd(k,n)=1 } under multiplication mod n for n>0

[hidden haven]

just reverse the arrows and apply yoneda

[chilly ocean]

[woeful flint]

If ${b_1,\cdots,b_n}$ is a basis for a ($n$-dimensional) Lie algebra $\mathcal{L}$ what can we say about the basis of a subalgebra $\mathcal{L}^\prime \subset \mathcal{L}$

[cloud walrus]

Kraft Macaroni

[woeful flint]

I have a feeling that the existence of certain dimensional subalgebras in some way has to restrict our basis because the lie bracket applied to the basis of the subalgebra must give elements contained within the subalgebra's basis

[winter thorn]

Can closure be thought of as a symmetry of sorts (not exactly fitting the mathematical definition)?

[vocal wolf]

what kind of closure?

[winter thorn]

algebraic closure

[hidden haven]

what kind of symmetries are you looking for?

like why do you think that they would be linked

you could always study the aut group of the extension cl(k)/k for a field k

[fading wagon]

hmm, cool, by classification we get that it's a product of cyclic groups, then for each cyclic group of order k, find another prime of the form mk+1, then multiply the primes together to form n

though is there a less whack solution?

[rustic crown]

that is pretty neat tho... I saw that argument when my book proved that every finite abelian group appears as a galois group of some extension over Q.

[elfin patrol]

If I have homomorphism phi: G to H, phi_m(x)=[mx], G=(Zn,+) how can I prove phi_m(G) is equal to <[m]>? It feels like it’s by definition and not sure how to show rigorously

[chilly ocean]

Ye

dirichlet theorem was the key to this problem

[chilly ocean]

and uh

I guess you didn't include it but

we are using full dirichlet theorem

not just existence of one prime

since $G\approx Z_{p_1^{n_1}}\oplus Z_{p_2^{n_2}}\oplus\cdots\oplus Z_{p_k^{n_k}}$

[cloud walrus]

CityHunter

[chilly ocean]

p_i need not be distinct

we consider primes of form $r_i=s_i p_i^{n_i}+1$ such that all $r_i$ are distinct

[cloud walrus]

CityHunter

[chilly ocean]

and then consider U group of product of r_i s

which is direct sum of U group of r_i

and that is $\approx Z_{s_ip_i^{n_i}}$

[cloud walrus]

CityHunter

[rustic crown]

i was wondering if we could avoid using the special case of dirichlet's theorem by using something like for a > 1order of a modulo a^n - 1 is n. This solves it for cyclic groups, but didn't find a nice way to patch up things.

[long obsidian]

If the kernel of a linear map is nontrivial can that say something about linear dependence?

[chilly ocean]

T: V -> W is one-to-one iff it takes linearly independent sets in V to linearly independent sets in W

[long obsidian]

Ahh I was actually looking for this result. Does anybody know of a way to think about why this might be true?

[prisma thunder]

@next obsidian Hey, I found a neat way of doing those problems I wrote yesterday.

Since we know f is separable if and only if gcd(f,f’) = 1, I look at my polynomials like they are in Q and find polynomials g and h for which gf+hf’ = 1 and look at the content of g and h. Any prime factorizations will immediately tell me which F_p f is separable or not

[chilly ocean]

consider $T\colon V \to \bR^n$ given by $$T(v) := (L_1(v), \dots, L_n(v)).$$ this is linear and $\ker T = \cap \ker(L_i) \subset \ker L$, so by linear algebra, there is a functional $f\colon \bR^n \to \bR$ such that $L = f \circ T$. such a functional is of the form $f = t_1e^1 + \cdots + t_ne^n$ for some scalars $t_i\in\bR$, which gives the result

[cloud walrus]

TTerra

[long obsidian]

When you say "by linear algebra" what precisely are you using?

[chilly ocean]

something non trivial i didn't feel like proving lol

[long obsidian]

Fair enough lmao

[chilly ocean]

but since you asked ill elaborate

[long obsidian]

😋

[chilly ocean]

by the first isomorphism theorem, we get a linear isomorphism $\tilde{T}\colon V/\ker T \to T(V)$ such that $\tilde{T}\circ\pi = T$. since $\ker T \subset \ker L$, we have an induced map $\tilde{L}\colon V/\ker T \to \bR$ such that $\tilde{L}\circ\pi = L$. so $\tilde{L} \circ \tilde{T}^{-1}\colon T(V) \to \bR$ is a functional. we can extend it to a linear functional $f\colon\bR^n\to\bR$, and it shouldn't be hard to see that $f\circ T = L$

[cloud walrus]

TTerra

[chilly ocean]

something like that

eh i probably could have included it in the original answer

didnt wanna get sidetracked though

[oak grove]

i want to ask just a dumb definition question. I'm having a hard time figuring out exactly what ring with unity means

my book implies that it means that you have a multiplicative identity for all elements

but then other sources say that it means that there exists an inverse for some elements?

it should be the first right?

[chilly ocean]

indeed

the second sounds like "unit group"

[tropic spade]

Let $p(x)\in F[x]$ be irreducible over $F$, and let $c$ be a complex root of $p(x)$. Let $h:F\to\Bbb{C}$ be a monomorphism. If $deg p(x)=n$, prove there are exactly $n$ monomorphisms of the form $F(c)\to\Bbb{C}$ which are extensions of $h$.

[cloud walrus]

DootDooter

[tropic spade]

In this wording, are they meaning to assume F is a subfield of C?

I'm not exactly sure what to make of "let c be a complex root of p(x)" when it seems like F could be some really weird field.

Nvm, I'm dumb they hid the part where they said that F is a subfield of C somewhere else lmao.

[coarse storm]

I think the monomorphism does imply an embedding of fields.

[tropic spade]

So you wouldn't even need to state F is a subfield of C?

[coarse storm]

I think so, yea. Assuming it is a monomorphism of fields.

[tropic spade]

Yea it is a field monomorphism

How does that implication work? The defn they gave for a monomorphism was just that it's an injective homomorphism.

[coarse storm]

It implies that F is isomorphic to a subfield of C.

[tropic spade]

Yeah I see what you mean. So in the absence of explicitly stating F is a subfield of C you'd take c being a complex root of p(x) as something like c is isomorphic to a complex root of h(p(x))?

[past temple]

are module homomorphisms from Z[x]

uniquely determined by their action on 1 and x?

[coarse storm]

Hmm... I am not sure how to parse that. h does not really act on x.

Been too long since I have done Galois theory...

[tropic spade]

Sorry, h(p(x)) was just notation from my book for the polynomial you get by applying h to every coefficient of p(x).

[coarse storm]

I want to say you are right, but part of me is... unsatisfied by that answer.

[tropic spade]

No worries, we might just be taking the blurb I mentioned too far out of context lol.

[coarse storm]

I have a slight feeling that I had been tricked by this exact same thing in the past...

[next obsidian]

Which ones?

Z[x] homomorphisms are defined by where 1 goes

Z homomorphisms are determined by x^k for all k >= 0

As Z[x] is the free module with basis {1,x,x^2,…}

[winter mesa]

how do i figure the solution out for this question?

[scarlet estuary]

hint: $(fg)^{-1} = g^{-1}f^{-1}$

[cloud walrus]

Namington

[hollow imp]

for (a), use the fact that a n-cycle $(a_1\ a_2 \ \cdots \ a_n) = (a_1\ a_n)\cdots(a_1\ a_2)$

[cloud walrus]

Alphyte

[scarlet estuary]

this is a fact of groups as well as functions (when the domain makes sense)

[hollow imp]

oh wait

part a was already answered

[winter mesa]

i got part a wrong

it allows multiple attempts. i wasnt able to figure out what i was doing wrong

[hollow imp]

you decomposed (3472615) as (34)...(35), where it should've been (35)...(34)

same with (734)

[winter mesa]

ooooh okay i got it right when i submitted it this time

i still dont get how to figure out part b tho

[hollow imp]

use namington's hint

the inverse of a 2-cycle is itself

[winter mesa]

im still not getting it 😭

I also needed help with these

im not even sure what even permutations are and what it means

[lethal cipher]

Ah, even permutations are permutations that can be decomposed into an even number of transpositions, also known as 2-cycles.

Hope that helps @winter mesa

[candid maple]

hi not sure if this is the right channel since i'm super new to group theory, but i have some uncertainties about some fundamental concepts. firstly, is the symmetric group S_1 considered a part of the symmetric group S_2?

[tropic spade]

Nope.

[kind temple]

there are copies of S_n inside of S_{n+1}

[tropic spade]

So for ex take any elt of S_1 and take any elt of S_2 then compare their domains.

Their domains are not the same.

[candid maple]

yeah thats what i was thinking when i was asking this question

i see

yeah i was unclear about that

thanks so much

what about the cyclic group on a sequence (x1, ..., xd)

vs the symmetric group on the same sequence

is the cyclic group a part of the symmetric group?

since they seem to be on the same domain

[kind temple]

wait, what do you mean "a part of"

[candid maple]

hmm

i dont really know myself

[kind temple]

theres different ways to interpret this then

[candid maple]

i see

[kind temple]

like, there is a subgroup of S_2 that is isomorphic to S_1, so in a sense, S_1 is a part of S_2. on the other hand, you have DootDooter's argument

[candid maple]

hm i see

lets say like we have a sequence (x1,...,xd) that is invariant to the symmetric group

then does that automatically imply it is invariant to the cyclic group?

or is this an ill defined question

probably is

in multiple ways lol

[lethal cipher]

No. For starters the symmetric groups are highly non-abelian

S_2 is the only commutative symmetry group

[viscid pewter]

oh yeah? what about S_1?

checkmate

[kind temple]

lol

[lethal cipher]

Damn! Ya got me

[viscid pewter]

what do you mean invariant to the symmetric group

[kind temple]

only non-trivial commutative symmetry group. that fixes it

[lethal cipher]

The main way to relate two groups is through an isomorphism. However, since S_n is non-abelian, you can't find an isomorphism with an abelian group

Abelian-ness is an invariant of isomorphisms

[candid maple]

well im not too sure myself cause im not even taking group theory, but our deep learning homework has suddenly transformed into advanced mathematics which the professor didnt teach lol...

so ive been trying to self teach myself group theory

[viscid pewter]

sure

i just don't understand what you mean by invariant in this context

can you explain what you mean without using that word specifically

[kind temple]

hmmm. there are abelian subgroups in S_n. i think a better comparison would be like an injective homomorphism, since i dont think in the context of the problem, we only want to worry about comparing via isomorphisms

[candid maple]

yeah im not sure how but this is the line involving invariance in the hw if it helps: "The study and design of generic invariant function approximations starts with a good understanding of the set of invariant polynomials (since any continuous function can be well approximated by a polynomial of a sufficiently large degree)"

[lethal cipher]

Subgroups, sure (powers of an element are obvious contenders)

[candid maple]

i know i shouldnt be pasting hw problems here

[viscid pewter]

invariant under what, a group action?

hmmm

[candid maple]

the question then asks to give an example of a polynomial such that it is not invariant to either of the groups

the groups being

[lethal cipher]

Also, I said isomorphism c-squared, not homomorphism.
But I cam see why you'd bring it up if I said homomorphism.

[candid maple]

the cyclic and symmetric group

[tropic spade]

Did your professor define invariance that you know of?

[viscid pewter]

oh, that sort of polynomial

the crap kind

it's gonna be like x1 + x2 + ... + xn, right? that sort of weird polynomial? or

[candid maple]

this is the definition he uses

[candid maple]

this kind of polynomial

[viscid pewter]

yeah it's one of those weirdos

[candid maple]

i feel bad for just pasting everything from the hw lol

[viscid pewter]

and then the group action is mapping them to each other very simply

ok

[candid maple]

been trying to figure out the definitions and stuff myself for a while 😂

[viscid pewter]

ok so

there are cyclic permutations of order n in the symmetric group of order n!

[viscid pewter]

so i think the answer is yes

but it's vacuously true

[candid maple]

yeah that makes sense to me

[viscid pewter]

i'm not sure there's anything that's really invariant under the entire symmetric group other than dumb things like 0

i mean actually hmm

no i'm a fool, in this context we can just do like every single coefficient is the same and then it should work?

[candid maple]

that would be an example that would be invariant to both groups correct?

[viscid pewter]

yes

[candid maple]

the question asks for a polynomial that is not invariant to either group, so in that case i could just choose any polynomial with coefficients like 1,2,3,4,... then?

in that case its a really weird question...

[viscid pewter]

yes

i think so

[candid maple]

ok

thanks Kaisheng

and everyone else

this server is super helpful wow lol

i hope to one day be as good as you guys at math lol

[lethal cipher]

Me too

[winter mesa]

if H is a subgroup of S7 and H has 24 elements in it, does that mean S7\H would have 210 elements in it?

[viscid pewter]

no

S7 has 5040 elements

so S7\H has 5016 elements

@winter mesa

[lethal cipher]

Yea, S7 is wayyy larger than you are giving it credit :p

Here is a question for you Hazzy: If H is a subgroup, is S7\H a subgroup as well?

[winter mesa]

i would think yes but im not very smart lol

[lethal cipher]

Well, do we agree that the subgroup H has the identity in it?

[winter mesa]

yeah

[lethal cipher]

Okay. Does S7\H have the identity?

[winter mesa]

no its doesnt then

[lethal cipher]

Correct. And one of the requirements of a subgroup is to contain the identity. So S7\H can't be a subgroup

[viscid pewter]

oh

well why does that matter

[lethal cipher]

Oh, it was just a question to test their understanding.

That's all

[viscid pewter]

oh i missed the part where you actually asked that lol

gods i'm tired

[lethal cipher]

I feel that

[viscid pewter]

anyway i was expecting them to give the obvious followup clarification that oh they meant / not \ but they didn't

i'll just write what i would have written

S7/H is not a group

as H is not normal in S7, as only A7 is normal in S7

@winter mesa

[lethal cipher]

I think quotient groups come a bit after ya first learn symmetric groups

[viscid pewter]

yes, but 5040/24 = 210

can't be a coincidence

[lethal cipher]

Hmm, you may be right

[winter mesa]

nah i was being dumb. i was saying since S7= 7! and S4=4! i could just take out the 4 times 3 times 2 times 1 part out of the 7!

to get 7 times 6 times 5

[lethal cipher]

Oh. Not quite how that works

It is literally 7!-4! here

[viscid pewter]

7x6x5x4x3x2x1 - 4x3x2x1 = (7x6x5-1)x4x3x2x1

[long obsidian]

Do people really refer to non unital rings as rng?...

If so that's disgusting ...

[winter mesa]

is there a quick little trick to figure out how many subgroups are in the cyclic group Z35?

[vocal wolf]

theres a unique one for each divisor of 35

[winter mesa]

ah okay ty

so 4 subgroups since there is 1, 5, 7, and 35 right

[vocal wolf]

yep

works for every finite cyclic group btw

[median pawn]

hey, is this the right place for homological algebra?

[regal carbon]

I kind of stuck at algebraic manipulations. Are there any resources where i can learn and master it? And is it too important for abstract algebra. Please anyone help me!

[untold cloud]

Hi, guys, is there any clue how to prove "<--" direction? i was trying to differentiate until g' in f'=(gk)'=g'k+gk'=gr=f' becomes a constant

[rustic crown]

you probably want some sort of irreducibility of g

f = x^3 and g = x^2 won't work

you also probably need char R to be zero

otherwise R = F_p[t] with f = g = x^p - t

[untold cloud]

oh, yeah, i see, if g is reducible, but it only says R is integral domain rather than UFD, so how can i say reducible g can be factorized into irreducible elements

[rustic crown]

that may not be true for general integral domains, existence of factorization is equivalent to ACCP. which in simple terms means that you can't keep on taking non-trivial factor of something and never stop.

[rustic crown]

but like this example shows it's not enough. You need char = 0. g here is irreducible that can be seen in multiple ways, one is simply by using eisenstein with the prime ideal (t). but here f' = 0 and so f itself manages to divide both f and f', but clearly f^2 doesn't divide f

[untold cloud]

Thank you!

[coarse storm]

Where did the proposition come from?

[winter thorn]

I need spoilers. How important are GL(n, Z/Q/R) groups?

[untold cloud]

it comes from my commutative algebra lecture note, just did not know how to prove this more general statement

[robust pollen]

Why?

[hidden haven]

what do we use when we want to also remove 0?

[coarse storm]

rng* !!

[robust pollen]

What about a ring without 1 and 0, but with a star structure? A -rng?

[woeful flint]

is it true that any non-nilpotent element has a non-zero eigenvalue

[robust pollen]

Dang discord. *-rng*

[woeful flint]

or rather non zero eigenvector

[hidden haven]

non nilpotent element of?

[robust pollen]

Zero is never an eigenvector

[winter thorn]

Are there any groups whose order is equal to the cardinality of their underlying set?

[coarse storm]

What definition of group order are you using?

[winter thorn]

The number of elements in a group.

wait, I think I'm being stupid here.

[chilly ocean]

I guess you perhaps mean like, group generated by some set X?

[winter thorn]

Yes, that exactly.

[chilly ocean]

{e}

[winter thorn]

That's the only one?

[chilly ocean]

hmm probably not, like perhaps look at rationals?

[winter thorn]

Good idea.

[chilly ocean]

yeah, rationals work. Just have to prove you cant finitely generate them.

[winter thorn]

What if it's not?

oh

[coarse storm]

I guess it means, like, some minimality in the generating set.

[robust pollen]

btw, no. $\begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & -1 \ 0 & 1 & 0 \end{pmatrix}$ is not nilpotent, but the only eigenvalue over $\mathbb{R}$ is 0

[cloud walrus]

expectTheUnexpected

[woeful flint]

Ah thanks

Also I had another questio

what exactly is the definition of the normalizer of a lie subalgebra

our professor gave us a question on his sheet that asks us to find the normalizer of a subalgebra but he doesnt define it in the notes

[chilly ocean]

set of x such that [x, y] is in subalgebra for all y in subalgebra

[quaint tree]

Sorry if this is a dumb question, but suppose that $I=\langle xy-1\rangle\subseteq\mathbb{R}[x,y]$ is an ideal. Then is $I\cap\mathbb{R}[x]=\emptyset$? But $I\cap\mathbb{R}[x]$ should be an ideal, and I don't think $\emptyset$ is an ideal, so I think I'm doing something very stupid right now.

[cloud walrus]

e^(𝜋√163)∈ℤ

[quaint tree]

I see now I should have gotten $I\cap\mathbb{R}[x]={0}$.

[cloud walrus]

e^(𝜋√163)∈ℤ

[rustic crown]

intersection is trivial because R[x,y] is a UFD and (xy-1) is irreducible. if it divides something in R[x], then either that is 0, or has non-trivial y-degree which isn't possible.

[quaint tree]

Yeah I was just being a dummy and forgot that 0 is in all ideals, and got the empty set by mistake.

[rustic crown]

happens lol

[sly crescent]

Is there a generalization of Borel-de Siebenthal theory to arbitrary semisimple algebraic groups?

[chilly ocean]

Let R be a ring and I and ideal of R

certainly for every prime ideal P in R, the image of P under the quoitent is a prime ideal in R/I

its it true that all prime ideals of R/I arise in this way

[runic hemlock]

Yes, and it's not hard

[chilly ocean]

i thought weird things might happen like

(0) is a prime ideal

but I need not be prime

[runic hemlock]

0 isn't a prime ideal in every ring...

only in integral domains

and if R/I is an integral domain, I is prime

[chilly ocean]

ahh

[runic hemlock]

To prove it, if p is a prime in R/I, consider the set of all elements of R that represent an element of p in the quotient. Try to prove this set is a prime ideal.

[rustic crown]

image of a prime need not be a prime, you need to have the prime P contain the ideal I.
else the image is (P+I)/I and P+I may not be prime in general

[chilly ocean]

ah thank you that is a good thing to lookoutfor

[rustic crown]

also i think it might be just better to use the correspondence theorem to prove this, than working with elements

there is a inclusion preserving bijection between
{ideals of R containing I} and {ideals of R/I}
the correspondence is J --> J/I
in particular all ideals of R/I look like J/I for some ideal J containing I. This correspondence can be restricted to primes or maximal ideals via the third isomorphism theorem.
(R/I)/(J/I) = (R/J) and using that P is prime iff R/P is integral domain and M is maximal iff R/M is a field. (i'm assuming commuative rings)

[runic hemlock]

isn't the simplest proof of this theorem looking at elements?

[rustic crown]

@hidden haven's fault. he made me cat brained.

[hidden haven]

🤓

There's an adjunction here, might as well use yoneda

[rustic crown]

what's the adjunction

[hidden haven]

The one you get from univ prop of quotients

Call category of pairs (R,I) Ring-2 or something

R → (R,0) and (R, I) → R/I is adjunction

morphisms of Ring-2 are ones that map ideal of domain into ideal of codomain

[chilly ocean]

i do find that

the "simplest" proof tends to be the least enlightening

[languid moss]

How can I improve after getting a zero on my graduate abstract algebra midterm?

[chilly ocean]

lol are you me?

[languid moss]

you got zero too?

[chilly ocean]

same happened to me, but did pretty well on the final

I realized you have to study or sth

[languid moss]

I usually do better on the finals

Yea, at the moment I'm watching a youtuber who helped me pass linear algebra and so far is being great 🙂

[chilly ocean]

like, algebra midterm was the only 0 ive gotten ever, but somehow its the subject I ended up studying the most

[languid moss]

He also done Algbera videos so that is why I'm watching him

wait, you're me now. I'm starting to like the subject

wth

I never taken algebra as an undergrad

[chilly ocean]

I really do recommend Dummit & Foote book, if I liked it you will too I guess. From youtubers: Borcherds

[languid moss]

I'm using his book but is just not my tastes.

Borcherds videos are long. I don't have time to watch his videos

[past temple]

how does one take graduate algebra without having taken undergrad algebra

[chilly ocean]

yeah also what do you cover in 'grad algebra'?

[languid moss]

No idea mate. the masters program I am doesn't offer a lot of courses for masters students. It was either Algbera or stats. I couldn't take stats because prof said I never taken an undergrad version of stats.

fast pace content. in 7-8 weeks you bascially cover group theory

we just started ring theory on Wednesday

[past temple]

that just sounds like an undergraduate algebra class

doing group theory for two months and then doing ring theory

[chilly ocean]

It obviously depends on the uni

[languid moss]

Yea it depends on the uni.

[chilly ocean]

but if this is your first time doing this stuff

[languid moss]

Yeah first time

[chilly ocean]

you will need to chew through exercises

[past temple]

did u major in math before your masters?

for undergrad

[languid moss]

Yup, but my undergrad is consider weak in terms of math. Alright, back to watching videos. Later

[chilly ocean]

now

i do think you should

focus on doing exercises from the book

Yeah, even if you dont like D&F, at least check out the exercises, those are good

watching youtube videos on this stuff really won't get you to a graduate standard

if you know that your undegrad was maybe a bit weaker than others

you will unfortunately have to work much harder

[past temple]

kind of only tangentially related

[chilly ocean]

which means doing exercises

lots of them

[past temple]

but there also seems to be a lot people on this server doing a masters in pure math

like

what is the point of doing that

if u want to go into academia and continue doing pure math

then you need a phd

if u want to go into industry

then it'd be better just to do a masters in whatever field is most closely related

to that industry

maybe it just differs by country?

[chilly ocean]

some people just want to learn man

[chilly ocean]

this

[past temple]

its a lot of money and debt

[chilly ocean]

lol

[past temple]

to just learn math out of pure interest though

[chilly ocean]

eu free uni in most countries

a masters in europe is about the same as the first 2 to even 3 years of grad program in america

[past temple]

so do most europeans get their masters in preparation for their phd?

[tribal pasture]

This is also a roadway to PhD as well. Some countries require you to do a master before you can do PhD. Also sometimes people aren't able to land a grad school right after undergrad so they have to settle for doing Masters and then applying for PhD again

[chilly ocean]

its pretty much mandatory

in europe, essentially you cannot do a phd without a masters

[past temple]

i kinda had some questions about this too

if you apply to phd programs as a masters student

will graduate admissions committees

put u to a higher standard?

[tribal pasture]

The competition is tougher

Yeah

For a masters student, having some sort of research work already under your belt is a must whereas that isn't the case with undergrad standards.

[past temple]

in pure math?

[tribal pasture]

Yeah

[past temple]

okay also

what about if u apply to grad schools

after you graduate undergrad

but u dont do a masters program or anything

will grad admissions committees still put u to that s tandard?

the higher standard for masters students

[tribal pasture]

In that case, the standards are lighter because you aren't expected to have publications or research work. It is strongly recommended to have those, however.

[past temple]

my issue right now is that

[thorn delta]

are yall talking about american schools or european/elsewhere here?

[past temple]

america

[tribal pasture]

I am mainly talking about America

[past temple]

^

im worried because

im a third year undergrad rn

in america

and i feel that i d ont have sufficient background

to apply to grad schools next year

which is the cycle im supposed to apply

[tribal pasture]

Why do you feel like that?

[past temple]

the competition for math grad school seems very tough

[thorn delta]

same. I really need to get into doing research next semester

[past temple]

i feel that i'll be more prepared

after finishing my fourth year and graduating

and then applying

but the catch is

im worried that

they'll put me to a higher standard

if i apply after i graduate

compared to if i apply next year

which would then make this whole plan pointless

[chilly ocean]

Ok so i looked a little bit into functors and rep. theory, but i didnt quite understand that much honestly. But I think its not really necessary actually to know the background stuff, i can just think about Sym²(V) like this: So if V is a K-Vector space with basis x,y, then we can describe Sym²(V) as the K vector space with basis vectors x², 2xy, y², where the "²" is just good notation basically, so we can for example describe the veronese map in a coordinate-free way by $v_d : \mathbb{P}V \rightarrow \mathbb{P}(Sym^d)(V), [v] \mapsto [v^d]$, so for example if we consider the image of the projective line $P^1$ under the quadratic veronese map, we have $[v] \mapsto [v^2]$, so if $v = ax + by$, then $v^2 = a^2x^2 + 2abxy + b^2y^2$, so $[a,b] \mapsto [a^2, ab, b^2]$ which is precisely what the veronese map does

[tribal pasture]

but if you take a gap year, the standards are a bit higher because then they are expecting you to show some work you might have done in that year.

[cloud walrus]

Gewisser Fler

[past temple]

but why would i already have work during that year?

like

im not applying after i take the gap year

im applying during my gap year

technically toward the beginning of it

as most grad apps are due in december

[tribal pasture]

So you mean instead of applying before you graduate (so you can have a position right after you graduate) you are saying you will apply at the point of graduation (so you will have a position sometime after graduation)?

[past temple]

no okay

so i am a third year right now

right

the cycle im supposed to apply

is by december of next year

[tribal pasture]

Btw you might wanna take this to another channel

[past temple]

we can also DM?

if thats okay with you

[tribal pasture]

Sure

[languid moss]

@past temple If I can get in to grad school with the sufficient background I had like very weak. I'm pretty sure you can get in too. don't give up

[oak grove]

looking at number 3 now

my thought is yes, but im unsure how exactly to justify

[hidden haven]

An ideal is maximal iff quotienting by it gives a field

[oak grove]

im not sure weve seen fields yet

id have to check

[hidden haven]

oh

Interesting

[oak grove]

i dont think shes asking for a proof

just a justification

[hidden haven]

Ok

[oak grove]

i was going to say that any group that <x,2> is a subgroup of is just the entire ring right?

[hidden haven]

Yeah

[oak grove]

it could only be a subring of <x,1>

[hidden haven]

So this is an index 2 subgroup of the ring

And index 2 subgroups are maximal

[oak grove]

oo

[hidden haven]

🎉

[oak grove]

e z p z

thanks 🙇‍♂️

[chilly ocean]

yo ive never seen this I think

[hidden haven]

It's because the index can't be any smaller

Without being 1

[chilly ocean]

ye

honestly I dont like this argument

I dont like the word 'index'

[bleak abyss]

There is a very low tech way of saying this

Namely, give me an ideal containing I and containing any polynomial g such that g(0) is odd

Well g-1 is in I, so your ideal contains that too

But if you contain g and g-1, you contain 1

So you're the whole ring

This is if you want to really avoid saying the word quotient

[chilly ocean]

ye

I dont want to avoid the word 'quotient', I want to avoid the word 'index'.

[bleak abyss]

I guess by "you" I mean @oak grove

Point is this is the lowest tech way to say it

[chilly ocean]

R a ring, I and ideal, the prime ideals in R/I are exactly the prime ideals in R that contain I

Consider now Z[x,y]/(xy-1)

i want to know the prime ideals of this

so i want to know the ideals in Z[x,y] that contain (xy-1)

the zero set of xy-1 is

so any idea of the form (x-a,y-1/a) should contain (xy-1)

but i cant seem to find any other prime ideals

[next obsidian]

Z[x,y]/(xy-1) = Z[x]_x

So you’re looking for prime ideals of Z[x] which don’t contain x

These look like either 0, (p) or (p,f(x)) where x is not divisible by x and f is irreducible mod p

So there should be more

In particular 0 here corresponds to (xy-1)

[chilly ocean]

im a little bit confused

p is not divisible by x?

[next obsidian]

Oh whoops

I meant f(x)

Is not divisible by x

Or really just that x isn’t in (p,f(x))

Maybe this is a little stronger than f not being divisible by x

Oh

I think the ideals you’re missing are just (p,xy-1)

[chilly ocean]

this seems weird?

xy-1 is just zero?

[next obsidian]

I mean that if you pullback 0 in Z[x]_x all the way into Z[x,y]/(xy-1) it corresponds to the prime (xy - 1)

When associating Spec Z[x,y]/(xy-1) with a closed subset of Spec Z[x,y]

[chilly ocean]

ah okay yes

so in Z[x]_x

the prime ideals end up just being (p) where p in not just x

[lunar lintel]

Trying to prove G=Z_n1 x Z_n2 x ... x Z_nk is cyclic iff n1,...,nk are pairwise coprime with n_i>1

so far I have proved LCM(n1,...,nk)=lg=0 but am unsure

[next obsidian]

No this is not true. I listed them out, it’s (0),(p), and (p,f(x)) where x not in (p,f(x)) and f is irreducible mod p

It’s 2-dimensional

[next obsidian]

Eh?

Which direction are you doing

[lunar lintel]

teacher assigned this problem related where l=LCM(n1,n2,...,nk) and i proved lg=0 for g in G. i assume u can use it somehow

not sure which direction its for

[next obsidian]

What is lg

Oh

[lunar lintel]

l=LCM(n1,n2,...,nk) g in G

[next obsidian]

You’re using additive notation

[lunar lintel]

yes sorry should have mentioned

[next obsidian]

So again

Which direction are you trying to do?

=> or <=

Well actually both directions basically should follow from this lemma

[lunar lintel]

well i cant do either lol but i can try <=. if n_1,n_2, n_k paiwise coprime then LCM = n1n2nk

[next obsidian]

If you have (g1,…,gn) in G1 x … x Gn

Then the order of (g1,…,gn) = lcm(|g1|,…,|gn|)

From this the <= directions is very easy

Since the order of the group you’re looking at is n1…nk

And the lemma let’s you grab an element with that order

The other direction follows from the lemma but it isn’t as immediate

You should do something like if |(g1,…,gn)| = n1…nk

Then because |gi| divides ni that all ni have to be pairwise coprime

And this is possible just requires playing with some stuff

[lunar lintel]

alright ill try to write something up formally and see if i can figure out other diretion

thanks

[next obsidian]

Np

[elfin patrol]

If I have homomorphism phi: G to G, phi_m(x)=[mx], G=(Zn,+) how can I prove phi_m(G) is equal to <[m]>? It feels like it’s by definition and not sure how to show rigorously

[elfin patrol]

Any ideas?

[lunar lintel]

if i got homomorphism phi, G to H, and G abelian is ker(Phi) always abelian

[thorn delta]

yes, ker(phi) is subgroup of G, and any subgroup of an abelian group is abelian

[thorn delta]

show that phi_m(G) = <[m]>. There are two inclusions you have to show

[lunar lintel]

didnt remember that ker(phi) subgroup, thanks

[thorn delta]

np

[elfin patrol]

Does inclusion mean subsets? If so then phi(G) subset of <m> seems simple but by definition

Group generated by m is just m, 2m,…, km for k integer

[thorn delta]

yea, like they're both subsets of each other. So let $x \in \phi_m(G)$. Then $x = \phi_m(y) = [my]$ for some $y \in G$. So why is $[my]$ inside $\langle [m] \rangle = {k[m] : k \in \bN}$?

[cloud walrus]

kxrider

[elfin patrol]

Seems like u can just have y=k since y in G and G just Z_n which all positive integers

[thorn delta]

i mean yea that's the idea

[my] = y[m]

[elfin patrol]

And that should work both ways right? Just reverse signs it looks like

[thorn delta]

i mean formally speaking, to show that if x \in <[m]>, then x is in phi_m(G), you have to show that there exists y in G such that phi_m(y) = x.

[elfin patrol]

Appreciate it, I think I have something coherent. There’s another part where I have to prove \varphi=\phi_m for some m 0 \leq m \leq n-1 for arbitrary homomorphism \varphi G to G. Any tips on this one?

[tame grove]

https://cdn.discordapp.com/attachments/897284761763082310/906751795055951872/unknown.png

Im not sure where to start on this one. I think I might need a theorem I am not aware of, potentially a corollary of Lagrange theorem?

Or maybe I can use the definition of order to say that for all $n,m \in \bN, o(a)=n$ and $o(b)=m$ where e is an identity $e=a^n=b^m$ Hence $o(ab)=nm=e=mn=o(ba)$

[cloud walrus]

CaesiumIsFake

[tame grove]

does that make sense?

since e^2 = e because it is the identity

[thorn delta]

this isn't true. Instead, think (ab)^n = ababab...ab and see what happens depending on where you place parentheses

[tame grove]

hmm, so can I take (ab)^lcm(n,m) is equal to the identity then write out (ab)^lcm(n,m) and (ba)^lcm(n,m) to show they are the same?

[thorn delta]

where is lcm(n,m) coming from?

[tame grove]

I was thinking about the order of a product of disjoint cycles and then applied it to the element (ab) and (ba)

actually that might be assuming commutativity.

[thorn delta]

G is an arbitrary group. I don't think it really makes sense to be thinking about cycles here. Anyway, o(ab) = lcm(o(a), o(b)) is only true when your group is abelian

[tame grove]

Yea that makes sense, im not sure where to start then. Is there a trick using basic stuff or is there a theorem I am missing?

[thorn delta]

Quick counterexample: take D_8: If r is reflection and s is rotation by 90 degrees, then o(s) = 4, o(r) = 2, while o(rs) = 2 != lcm(o(r),o(s))

[thorn delta]

the trick is (ab)^n = ababab...abab = || a(baba...baba)b ||

[tame grove]

Alright, thanks. I think I just need to stare at it for a while until it clicks how these statement can be put into a way that clearly shows o(ab)=o(ba)

Oh wait, do I then show that (ab)^n divides (ba)^m and (ba)^m divides (ab)^n, thus by lagrange, the order n=m implies ab has the same order as ba

well I guess I wouldnt need lagrange actually

[thorn delta]

If n = o(ab) and m = o(ba), you need to show that m divides n and n divides m. Is that what you meant?

[tame grove]

yep, sorry for the mess im still trying to figure it out so it is ugly as of right now.

[thorn delta]

np, anyway yea that's the idea

[tame grove]

Alrighty, thanks again

[waxen hedge]

You need to have pgcd(m, n) = 1

[dull root]

Is it necessarily true that if $G$ is a group, $N$ an abelian normal subgroup of $G$, and $H \leq N$, that $H$ is also normal in $G$? I think this is the case, but I'm having trouble showing $H = g H g^{-1}$. Or maybe someone has an easy counterexample

[cloud walrus]

Mr.Hahn-Banach

[plucky flicker]

Hi! In the ring category every epimorphism has a right inverse. Is this true?

[hidden haven]

no

Z → Q is epi

[plucky flicker]

Sorry I have to reformulate my question

So I have a ring R with identity, and I consider the left regular R-module. If I take a surjective endomorphism of this module, then there exists a g R-module endomorphism such that fg = 1.

Is that true?

[rustic crown]

looks like it

the map f : R --> R is just right multiplication by a = f(1). By surjectivity, there exists b such that f(b) = ba = 1. Consider the map g which is right multiplication by b.
f(g(r)) = f(rb) = r(ba) = r

[rustic crown]

what about A4 and N = klein 4 group?

H = {(1), (12)(34)}

this isn't normal

[dull root]

Yea -- I realized what I wrote was not correct. So the question I have is that: let $G$ act on some set $X$, and $N$ is an abelian minimial normal subgroup of $G$. I want to show that if $N$ acts transtively on $X$ for any $x \in X$, $G = N \cdot Stab_G(x)$.

[rustic crown]

don't think that's true either

G = Z/nZ acts on n-things by cyclically permuting them. stablizer of any x is trivial.

[cloud walrus]

Mr.Hahn-Banach

[dull root]

Sorry, I forgot an additonal condition

So by transitiveness and $N$ abelian, I see that all stablizers of $N$ are the same set

I also certainly know $N \cdot Stab_G(x) \leq G$, so all I need for equality is to show that $\lvert N \cdot Stab_G(x) \rvert = \frac{\lvert N \rvert \lvert Stab_G(x) \rvert}{\lvert N \cap Stab_G(x) \rvert} = \lvert G \rvert

[cloud walrus]

Mr.Hahn-Banach
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rustic crown]

wait is G finite?

[dull root]

The issue is that I am seeing anyway to relate the order of those two -- the orbit stablizer gives $\lvert X \rvert = \frac{\lvert N \rvert}{\lvert Stab_{N}(x) \rvert}$ which doesn't seem to be giving me any helpful information

[cloud walrus]

Mr.Hahn-Banach

[dull root]

yes $G$ finite

[viscid pewter]

the most unnecessary latex

[rustic crown]

it's helpful actually

stab_N(x) = stab_G(x) n N

which is exactly what you need

[viscid pewter]

stop wasting latex, there are children in africa struggling to do subscripts

[dull root]

wow, i can't believe i didn't see such a trivial statement.... I think that's a good hint for me

But it seems then this does not require that $N$ is abelian

[rustic crown]

don't see why you had minimal/abelian/normal

[dull root]

It was a small part of a larger question, so it was unnecessary for this part

[rustic crown]

oh okie

[trim grove]

i have given a quotient ring $$\frac{\mathbb{Z}_{3}[x]}{\langle x^2+x+1 \rangle}$$

i have to find the number of units of this

as i can see this is same as $$\frac{\mathbb{Z}_{3}[x]}{\langle (x+2)^2\rangle}$$

but here i can't apply the CRT, any hint , will be appriciated.

[cloud walrus]

Algebra

det

[rustic crown]

Your work will then easily generalize to any R[x]/(x^n)

[wooden ember]

isnt this just a field and so necessarily a PID?

[rustic crown]

yea if that prime is non-zero 😛

[trim grove]

okk i will try

[wooden ember]

yeah lol but then it's trivial in that case

[trim grove]

so i have to chek is this prime ideal

[wooden ember]

no no i wasnt mentioning your exercise haha

[trim grove]

ok

[wooden ember]

your ideal isnt prime and Z_3[x] isnt a PID anyways im pretty sure

[rustic crown]

Here's a hint if you get stuck.
||Factorize (u^m - r^m), and use r^m = 0 for some large enough m||

[trim grove]

it is PID, field [x] is always a PID, i think.

[rustic crown]

usually Z_p is used for p-adic integers...

else yea, k[x] is a pid for k a field

[trim grove]

but how can i use this here?

[rustic crown]

we wanna show u-r has an inverse, that factorization pretty much tells you the inverse

[chilly ocean]

let R(G) be the representation ring of a group G

if H is a subgroup of G

then there should be some map from R(G) to R(H)?

is it just restricting the representation?

[rustic crown]

what's a representation ring?

i've heard of group rings

[chilly ocean]

same as the character ring

[rustic crown]

okie seems like i dunno this stuff

[cloud walrus]

Mr.Hahn-Banach

[rigid cave]

is it okay if I ask a basic question real quick or should I wait?

okay math general moment

[hidden haven]

Just make a thread smh

[rigid cave]

already aksed lmao

[dull root]

I was thinking about this some more based on the hint that was given eariler, but I am having trouble finishing. Let $G$ be a (finite) group that is acting on a set $X$. Suppose $N$ is a normal subgroup of $G$ that acts transitively on $X$. Then I want to show for any $x\in X, G = N \cdot Stab_G(x)$. By Orbit Stablizer, we have
$\lvert X \rvert = \frac{\lvert N \rvert}{\lvert Stab_N(x)}$ Moreover, noting $Stab_N(x) = N \cap Stab_G(x)$, we can say
$\lvert N \cdot Stab_G(x) \rvert = \frac{\lvert N \rvert \lvert Stab_G(x) ) \rvert }{\lvert Stab_N(x) \rvert} = \lvert X \rvert \lvert Stab_G(x) \rvert$. I would need to show that is equal to $\lvert G \rvert$, but I don't see why it holds. It seems that there is some basic counting here that I am not seeing.

Also, there are some additional conditions namely that $N$ is a minimal normal subgroup and that $N$ is abelian, but it doesn't seem like those two components are necessary for showing this first part.

[cloud walrus]

Mr.Hahn-Banach

[rustic crown]

if N acts transitively so does G

[dull root]

In general, if a subgroup acts transitively, the whole group acts transitively

yea

[rustic crown]

so orbit stablizer on whole group says?

[dull root]

|X| = |G| / |Stab_G(x) |

[rustic crown]

[dull root]

so we are good

yea

Wait, does this still hold if G is infinite?

[rustic crown]

[trim grove]

How can i get the number of conjugacy class of group of order $5^4$ given that its center have order $5^2$ elemnets

Solution i tried : here center have 25 elemnts so definitely 25 conjugacy classes of order 1, but rest i can't understand how to find. Any hint

[cloud walrus]

Algebra

[rustic crown]

right, so pick an element g outside the center. Now it's centralizer contains g as well as all of Z(G), so what can its size be?

[plucky flicker]

To prove that a given map is a group representation I just have to check if the map is a group homomorphism?

[dull root]

So if $G = A \cdot B$ where $A$ is normal in $G$ and $B$ is just some subgroup. If $C > B$. ie $C$ contains $B$ as a proper subgroup, why is it true that $A \cap C < A$. This feels trivial, but I can't show it...

[trim grove]

P+1?

[cloud walrus]

Mr.Hahn-Banach

[rustic crown]

well centralizer of any element is a subgroup...

[trim grove]

Hmm then it must divide p^n ,hmmmm.

[rustic crown]

hello carla

[rustic crown]

yee p^4

but it properly contains Z(G) with size p^2

[trim grove]

Then p^3

Definitely

What is next step,please

[dull root]

Wait, nvm it was indeed trivial lol

[rustic crown]

so for each g in Z(G) you know the conjugacy class is just {g}

for each g outside Z(G) it contains p things.

p^2 classes are singletons, this leaves us with p^4-p^2 elements.

[trim grove]

I am confused on this

[trim grove]

I got this one

[rustic crown]

i'm using orbit stablizer. size of conjugacy class is p^4/size of centralizer

[trim grove]

I think i got it

[rustic crown]

?

[trim grove]

I think i got it , but

Calculation was wrong

[rustic crown]

p^2 singletons, and (p^4-p^2)/p with size p

[trim grove]

Ohhhh

[trim grove]

how you got this because acc to class eqauation $$o(G)= Z(G)+\sum_{i=1} \frac{O(G)}{C(a)}$$

[cloud walrus]

Algebra

[rustic crown]

yea so each o(G)/|C(a)| = p, so that sum is p * (number of conjugacy classes with size p)

which should be o(G) - |Z(G)|

or just say directly like i did, p^4-p^2 elements arrange themselves into stacks of p, so number of stacks = p^3-p

[trim grove]

okkk , now i got this, thankyou so much for you time

[rustic crown]

[chilly ocean]

[rustic crown]

Hello

[plucky flicker]

Supposing we don't know character theory, how can we check that a group representation is irreducible? By definition it is irreducible iff the only invariant subspaces are the two trivial subspace

[rustic crown]

you'll just have to show that you can reach any non-zero vector with any other non-zero vector by some element of kG

but that's practically the definition

[trim grove]

I have a question
:
let F be a finite field and $F^{}$ be the group of all non-zero elements over multiplication,If $F^{}$ has a subgroup of order $17$ then what is possible minimum order of $F$\

Solution i tried : order of field will be $p^n$ and we know that order of $F^{*}$ will be $p^n-1$ and will be cyclic, and subgroup of order $17$, that means \

17 will divide $p^n-1$ so $p^n = 1 mod(17)$, but after this i have no idea how to procede ,please give me a hint

[cloud walrus]

Algebra

[rustic crown]

look for the smallest prime power which is 1 modulo 17

[trim grove]

does , eulers theorem can give any idea.?

[rustic crown]

not a lot

but you can look at 2^8

8 is the order of 2 modulo 17

so you know one prime power that works

which means the answer is definitely less than that

and 2^8 is not that big

so just check all numbers which are 1 modulo 17

[rustic crown]

this is easy to see because 2^4 = 16 = -1 (17)

its just 256 lol

[trim grove]

answer is 103

[rustic crown]

yea...?

just look at 1, 18, 35, ... , and see which one is a prime power

you don't need to go beyond 256

that like just 16ish numbers to check

[trim grove]

please sugget me any alertanative

if you can

[rustic crown]

the answer looks pretty weird lol

[trim grove]

yes 103 can be power of some prime?

[rustic crown]

103^1

[trim grove]

my bad

[plucky flicker]

I have a group G acting on a finite set omega. We consider the elements of omega as a basis for an |omega| dimensional vector space, and consider an |omega| dimensional representation of G. When is this action irreducible? My intuition says that this is true if omega has only one element, i.e. this is a one dimensional representation but no clue how to prove this

[rustic crown]

say k was the field look at the subspace k(w1 + ... + w_n) where omega = {w1, ..., w_n}

[plucky flicker]

so if I want irreducibility then this subspace should be invariant

[rustic crown]

nah, other way... this subspace is actually invariant...

so what does irreducibility say?

G acts on omega so it pretty much permutes it's elements, but then it has to fix the element w1 + ... +w_n

[dull root]

So I am convinced the following statement is true even if $G$ is an infinite group. Let $G$ act on $X$, and $N$ be a normal subgroup of $G$ which acts transitively on $X$. Then I want to say $G = N Stab_x{G}$. The way to do it in the finite case, where we could exploit the orbit stablizer, no longer works here since in the finite case, we simply needed that $\lvert G \rvert = \lvert N Stab_x{G}\rvert$

[cloud walrus]

Mr.Hahn-Banach

[plucky flicker]

it says that this subspace is either 0 or the whole space

[rustic crown]

yep, but it's clearly not 0, so this must be the whole space. so dimension is 1

[plucky flicker]

ohh jeez

obviously

thanks 😄

[untold cloud]

Hi, guys, what does this definition mean? i mean, i know this is just a way to define something, but for example if i know a_0, how do i find a_1? a_1 = a_0+p·c, for constant c?

[sharp sonnet]

yeah and you can choose that constant in 0, 1, ..., p-1 if you want

[untold cloud]

Thanks!

[toxic mortar]

help

[lethal cipher]

I seemed to have messed up in my algebra here (The 2abcd terms for N(zw) should cancel out for these to be equal). Can someone help me figure out where I messed up?

[lethal cipher]

Fine, new question.
If you have a Ring with the property that (xy)^2=x^2y^2, does this imply R is communicative.

It does for groups, but usually we work with inverses to get that result.

[lethal cipher]

So I am at this part, how can this let me conclude xy=yx w/o inverses?

[proud bear]

just expand out (x+y)^2 and use that that equals x+y

[lethal cipher]

So does it make sense to say (x+y)^2=x^2+2xy+y^2? Or does that 2 not make sense here?

[proud bear]

(x+y)^2=(x+y)(x+y)=x^2+xy+yx+y^2

oh wait

[lethal cipher]

Oh true

[proud bear]

actually this might not work

nvm

[lethal cipher]

Yeah, it does not seem that way sadly

Another attempt I was going for was to say x(yx-xy)y=0. But we don't know if R is an integral domain, so we can't conclude xy=yx (provided x,y aren't 0).

On the other hand, if R is an integral domain, the only thing R can be is {0,1}

[chilly ocean]

x + y = (x + y)^2 = x^2 + xy + yx + y^2 = x + xy + yx + y, so xy = -yx. this implies that xy = yx because z = -z for any z in the ring

stain's proof works

[lethal cipher]

Oh, that's smart

Why does z=-z?

[chilly ocean]

xy = -yx for all elements in the ring, so in particular if you put in x = y = z you get z^2 = -z^2 or z = -z

[lethal cipher]

Ah, I see. Thank you Terra and Stain

[chilly ocean]

thank stain

[lethal cipher]

Also, do you mind looking at this? I am pretty sure my algebra is correct, so I think the claim is wrong

[lethal cipher]

My professor's formula was wrong. N(...) was supposed to equal a^2+Db^2

[coarse forge]

if you've done a bit of work on embeddings, then the norm is defined by the product of the embeddings of your quadratic number field. The embeddings can be uniquely defined by where sqrt(-D) is sent, so the two embeddings in this case are sigma_1 : sqrt(-D) -> sqrt(-D) and sigma_2 : sqrt(-D) -> -sqrt(-D).

[winter thorn]

just curious, but isn't additive notation for groups just taking the logarithm of the multiplicative notation, just with some caveats like shoes-socks-boots-shit?

[deep nova]

can someone explain the difference between homomorphism and isomorphism? I dont quite understand it because they both seem to have the exact same definition

[viscid pewter]

isomorphism is bijective

why am i getting pumpkinned

[deep nova]

my textbook really doesnt do a good job at explaining it

made me real confused

because the definition was the copy paste of isomorphism

[chilly ocean]

post a picture liar!

[tight flare]

[deep nova]

well i mean the only thing missing was the bijection but when i first looked at it, i was confused why its even a thing since it looked like iso @chilly ocean

[tight flare]

did you define isomorphism before homomorphism?

[deep nova]

yeah

[stoic grotto]

Nice

[viscid pewter]

bruh bijection is what makes and breaks it

[deep nova]

as i said, i didnt look at it properly

do you knnow how it feels like taking 5 courses and sleeping 4-5 hours a day does to you

[tight flare]

[viscid pewter]

seriously what do the pumpkins mean

are they good, are they bad

[deep nova]

[viscid pewter]

ok if communicating unclearly is the point whatever then

[tight flare]

[hidden haven]

[plucky flicker]

offtopic:

lmao

[hidden haven]

I just have a problem with isomorphisms being defined as bijective homomorphisms but I understand that that is a necessary thing when defining them for someone encountering them for the first time

[tight flare]

[lethal dune]

[stoic rose]

This made me laugh way too much