groups-rings-fields

[next obsidian]

So…

[dusty river]

ah I see

[next obsidian]

You can’t rule out the identity plus 32 order 3 elements

At least not with this proof

[mild laurel]

No, because there's no non-trivial morphism from Z5 to Aut(Z7) = Z6 and no morphism from Z7 to Aut(Z5)=Z4, which is what you'd need for a semidirect (and not direct) product

[dusty river]

Ye, ty

[next obsidian]

Shut the up?

[lethal cipher]

I'm not sure I've seen this.

[next obsidian]

I mean

I didn’t expect you had

I just don’t know there’s a good proof without more machinery

Either Sylow or the classification

[mild laurel]

schur zassenhaus go brrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr

[sturdy marsh]

yeah that works too

but you need to assume abelian

[lethal cipher]

Ah okay. Well, this case by case proof was definitely helpful.

[next obsidian]

So

At least it didn’t ask if the same is true for 33

[sturdy marsh]

if you dont assume abelian, it again reduces to showing that there are no nontrivial semidirect products

[next obsidian]

It asks if your proof works, and it doesn’t

[lethal cipher]

No it does not

[next obsidian]

:(

[lethal cipher]

Because 2 does divide 32

[next obsidian]

Tfw

[lethal cipher]

I think that's why they chose that number, to make the readers cry inside

[sturdy marsh]

Here's a fun problem

assume you have an inverse system of sets

such that the cardinality of the images of all the structure maps are uniformly bounded

[next obsidian]

Shut upppppppp

[sturdy marsh]

prove that the inverse limit is a finite set

[next obsidian]

Isn’t the inverse limit in set just like

[dusty river]

what does uniformly bounded mean here?

[next obsidian]

An explicit construction as a quotient of a product

This seems like it’s just an annoying thing to show there’s finitely many equiv classes

[sturdy marsh]

There is some N such that the image of each map f_i has cardinality less than N

[sturdy marsh]

there's a slick way of doing it

[next obsidian]

Use Yoneda

[sturdy marsh]

[next obsidian]

Hey man

I like working in Set okay

[sturdy marsh]

here's another fun problem

let p,q>0

p odd

show that S^p x S^q is parallelizable

[dusty river]

structure map refers to any map within the inverse system right? Or are they the maps from the limit to something in the diagram?

[sturdy marsh]

the maps in the inverse system

but this also implies that the maps from the limit have image bounded by N

as they factor through the structure maps

Here's another one

Let f: C ---> C be a nowhere 0 continuous function

show that f = e^g for some function g

[next obsidian]

Hurb

Just use log

[sturdy marsh]

the last one is a one-liner

C ---(exp) ---> C* is a covering map

just lift

[mild laurel]

you can also just look at f/f' and say its analytic so the derivative of some g so that f is e^g

[sturdy marsh]

f is continuous

no differentiability assumption

[mild laurel]

o sadge

[sturdy marsh]

@hidden haven if you want a hint, replace the inverse system with a nicer one

[dusty river]

oh i was just away to eat lol

but I'll think about this

[sturdy marsh]

here's another fun one

let G be a finite group

assume G admits an irreducible faithful representation over C

prove that the center of G is cyclic

[next obsidian]

Isn’t this something like

Finite subgroups of a field are cyclic

[sturdy marsh]

yeah

[next obsidian]

I swear I did this before

Swag

The center embeds into the matrix group in the diagonal yeah?

Or something like that

[sturdy marsh]

you need to prove that

[next obsidian]

Yes but that’s why it’s restricted to the center

And not just G

[sturdy marsh]

there's a way to do it without using matrices

[next obsidian]

I believe it

I forget if this was the same problem I did that

I don’t remember the rep theory part of my class very well

[sturdy marsh]

Z(G) embeds into the automorphisms of the representation

the rep is irreducible, so the automorphisms form a division ring

over C

finite dimensional

so it is C

[next obsidian]

Hurb

[sturdy marsh]

as C is algebraically closed

[next obsidian]

Really using that theorem

I proved this for clsss and used it 0 times

[sturdy marsh]

I used that a lot

[next obsidian]

Fucking when

I have never been dealing with division rings

But okay you haven’t only dealt with commutative stuff for

194939294939 years

[sturdy marsh]

I remember there was a problem like

assume V is a finite dimensional vector space over an algebraically closed field

let I \subset Aut(V) be a subalgebra

such that V is a simple I-module

prove that I = Aut(V)

or something like that

I think I had to use Schur's lemma (automorphisms of a simple module is a division ring) multiple times to do this

[dusty river]

how do you do the inverse system problem?

[sturdy marsh]

the inverse system is the set {(a_i): f_{i+1}(a_{i+1}) = a_i}

right?

[dusty river]

[sturdy marsh]

?

[dusty river]

wait i+1?

is it indexed by N?

[sturdy marsh]

yes

it's an inverse system

[dusty river]

oh I thought that was just a dual of directed system

[sturdy marsh]

ok fair enough any directed poset

choose better notation

the solution doesnt change

[dusty river]

let me think

[sturdy marsh]

it's the set {a_{\alpha}: f_{\beta}(a_{\beta}) = a_{\alpha}}

[dusty river]

ok so I think you are replacing each set with the image of everything before it right?

and then it must eventually stabilize

[sturdy marsh]

that's the right idea

replace each A_{\alpha} in the directed system by B_{\alpha} = intersection of the images of all the maps to A_{\alpha}

[dusty river]

oh intersection

[sturdy marsh]

yeah and now all the structure maps are surjections

[dusty river]

right

[sturdy marsh]

as the image is bounded, they must eventually be isos

and that's the limit

[dusty river]

wait

ohh

wait I am not sure why that is the limit

[sturdy marsh]

try proving it

[dusty river]

because the limit also has to map to the sets where those things are not identified right?

hmm alright

The arrows in an inverse system go n+1 to n right? just want to confirm

[sturdy marsh]

yeah

[dusty river]

I think I might still be confused with the definitions though
1 ← 2 ← 2 ← 2 ← ... should be an inverse system where n means the n-element set and the maps between 2s are all identity. Then this stabilizes at 1, so the claim is that 1 is the inverse limit?

[sturdy marsh]

2 is the limit

[dusty river]

oh eventually as in going to the right 🤦‍♂️

[sturdy marsh]

ye

[dusty river]

ok yeah then I see it lol

neat

[sturdy marsh]

:petthecat:

my roommate took a bunch of quals this week so I have a large collection of problems lol

[dusty river]

I was wondering where you were getting so many of them

what are quals anyway I hear a lot about them

[sturdy marsh]

theyre a bunch of exams that you need to pass

otherwise they kick you out of grad school

[dusty river]

wtf that sounds scary

[sturdy marsh]

yeah

you do get multiple tries tho

you dont have to pass them on your first try

[dusty river]

phew that's nicer

[sturdy marsh]

you need to pass them in two tries

I think

it depends on the university ofc

[dusty river]

damn

[terse crystal]

If your derivation means k-derivation then any k-derivation from R’ to R’ has the form d mapping a polynomial P to Σ (partial derivative of P with respect to x_i ) multiplied by d(x_i) and d is an extension of R iff you choose n symmetric polynomials as d(x_1),…,d(x_n) respectively

By structure of finitely generated abelian group, the only abelian group of order pq is the direct sum of Z/pZ and Z/qZ which is isomorphic to Z/pqZ

[eager willow]

yes I know that's just the cotangent bundle

and that's also not true about choosing n symmetric polynomials for what I was asking, I gave a counterexample

the equivalent question in terms of restrictions is how to find k derivations k[x,y,z]->k[x,y,z] which when restricted map symmetric polynomials to symmetric polynomials, but may map nonsymmetric polynomials (such as each indeterminant) to anything.

I've actually done some work since and have begun to answer my own question using localizations

for a glimpse of higher order derivatives, the laplacian is a good example of an operator with the extension/restriction property I have in mind, while not being a derivation

[terse crystal]

My bad. d(x_i)=P_i doesn’t necessarily need to be symmetric. It just should satisfy that any permutation g of {1,…,n} P_i(x_g(1),…,x_g(n))=P_g(i)(x_1,…,x_n)

I thought it implies being symmetric but it doesn’t 😂

[eager willow]

yeah for example in 3 variables dx =x, dy = y, dz = z works, and each elementary symmetric polynomial is an eigenvector with eigenvalue the (total) degree

this is also true of any number of variables with the derivation mapping each indeterminant to itself

[terse crystal]

I see. It satisfies the above equation indeed

P_2(x_2,x_3,x_1)=x_3=P_3(x_1,x_2,x_3)

[eager willow]

it's a nice condition for determining which derivations restrict, but not illuminating on the question of extensions. Starting with a derivation on the ring of symmetric polynomials, the cotangent bundle is defined similarly under isomorphism by partial differentiation of elementary symmetric polynomials, so if you know where each elementary symmetric polynomial is mapped, can you determine whether or not it has an extension?

I've figured this out for 2 variables, almost done on 3. It comes down to linear algebra over localizations of k[x,y,z]

it's also highly constructive and I'm wondering if there's some magic to generalize the way I'm creating a basis

[terse crystal]

I see I see so we need to find d’(x_i) such that d(e_i)=Σ(partial derivative of e_i with respect to x_j) multiplied by d’(x_j)

If d can extend to a d’

[eager willow]

I've written some notation for partial differentiation of a symmetric polynomial with respect to an elementary symmetric polynomial that might be helpful here

[terse crystal]

So it’s a linear equation with the matrix A where A_ij=e_(i-1)(x_1,…,x_(j-1),x_(j+1),…,x_n)

[eager willow]

I'm a little lost with the formatting of underscore

[terse crystal]

Oh so d has extension if and only if (d(e_1),…,d(e_n)) this vector with all elements being symmetric polynomials, belongs to the {linear combinations of columns of A ie (1,x_2+…+x_n,…,x_2…x_n) ,…, (1, x_1+…+x_(n-1),…,x_1…x_(n-1)) with coefficients being polynomials}

[eager willow]

Well I'm definitely working with that same vector (...d(e_i)...)

I haven't been looking at it with a matrix though I'm still a little lost

[terse crystal]

One way of such linear combinations gives one extension. By defining d’(x_j) to be the coefficients of that linear combination

[paper flint]

Are you working through Gallian?

[lethal dune]

Anyone follows Dummit Foote?

[chilly ocean]

ye

you need their addresses?

[lethal dune]

[somber marsh]

how do exact sequences exist

as in like

0 -> G_1 -> G_2

if thats the case then,

G_1 is a subset of the kernal of G_1 -> G_2, implying G_2 is 0 and is 0, so it must be 0 -> G_1 -> 0?

[next obsidian]

Wut

Why is G_1 a subset of the kernel?

[somber marsh]

well if there is a map

0 -> G_1

lets cal that tlike

[next obsidian]

The image is 0

[somber marsh]

?

[next obsidian]

There’s only one map from 0 -> G_1

And the image is gonna be the trivial sub-thing

Idk if these are groups or modules or whatever

[somber marsh]

groups

thats why G

but im confused how come image is 0?

[next obsidian]

Uh…

[somber marsh]

isnt image meant to be every possible output

[next obsidian]

What

No that’s the codomain…

The image is the elements which are mapped to by elements of the domain

The only element which is mapped to by anything from 0 -> G_1 is 0

Because the first is a singleton snd 0 has to map to 0

[somber marsh]

oh.

i read wrong apparently ive thought image was all possible outputs for like

10 months

[next obsidian]

Bro…

[somber marsh]

💀

ah i see now

i thought image would be G_1 itself

but since 0 is the domain then the image must be {0} since (assuming 0 is {0}) 0 gets mapped to 0 in G_1

the image is G_1

that makes a lot more sense.

[lunar spruce]

In the example I assume there are only two elements in U right?

How can I make sure there exist such coprime orders d1' and d2'. The German version even says you can use u1 and u2' = u2 ^(d2/gcd(d1,d2)), but u1 and u2' order is not neccessarily coprime to me.

[chilly ocean]

yeah I dont think they are coprime because order of u1=d1, order of u2= gcd(d1,d2), coprime only if d1 and d2 coprime

Oh ok so rather like, first find powers of u1 and u2 such that their orders are coprime and to those apply the german argument

[next obsidian]

Suppose that u1 and u2’s gcd is n

Then n divides the order of u1, so let k be |u1|/n

Now u1^n has order k

Now k and the order of u2 share no factors

Because you basically ripped the gcd out of the order of u1

If you don’t believe that k and the order of u2 are coprime, write the order of u1 and u2 as a product of primes using the fundamental theorem of arithmetic

The gcd is gonna be the product of the primes they share, then when you raise u1 to that order you end up with something whose order is that divided by the thing you raised it to

[lunar spruce]

do you mean |u1| order of u1?

[next obsidian]

Yes

[lunar spruce]

I think this is similar to my first idea

I made the bad example: let u1^4=1 and u2^8=1

we just construct u1'=u1^4, so order of u1' is 1.

and 1 is coprime to any number.

so it doesnt matter what u2 is, its order is coprime to u1.

ANd then I apply the gcd trick thing to construct my alpha.

[chilly ocean]

I don’t understand why I can assume G to be a Elian

[terse crystal]

G has an abelian tower G=A_0>=A1>=…>=A_n={e}

And next the author replaced G with A_i/A_(i+1)

Clearly A_i/A_(i+1) is abelian and finite

[chilly ocean]

I don’t understand how we can assume G to be abelian

Sorry it was autocorrect

[hot lake]

they never assume that G is abelian, only that it has an abelian tower

[chilly ocean]

It said "it suffice to prove that G is abelian ...."

[hot lake]

???

[terse crystal]

I already answered. you are considering A_i/A_(i+1)

[hot lake]

that if G is finite, abelian then it admits a cyclic tower

yeah they pick a nonabelian G with an abelian tower

then say that if the result is true for abelian groups then they can apply it to all the quotients

which then makes the result true for a nonabelian group with an abelian tower

(I don't remember, what's the difference between a finite groupe with an abelian tower and a solvable group ?)

[chilly ocean]

Ohh now I understand what you mean

It took me a second

[chilly ocean]

The solvable group has an abelian tower that ends with {e}

[prime cloak]

Burnside theorem has to be known guys ?

[mild laurel]

Has to be known for what

[obsidian sleet]

hello friends

i would like to see if my thinking is in the right direction

let me post my problem

[cloud walrus]

MetalNinja27

[obsidian sleet]

So

i was thinking that if H is a proper subgroup of G then there is an element in G that is not in H

call this element m, maybe

so then we can try generating a new subgroup $\langle H, m \rangle$ (this might not be the right notation ill check later)

[cloud walrus]

MetalNinja27

[obsidian sleet]

so we have two cases

this new subgroup is either going to generate G or it isn't

then in the case that this new subgroup generates G

H is the maximal subgroup of G containing H?

and then in the second case if this new subgroup does not generate G, i.e. is another proper subgroup of G, we can repeat the same thing by finding another element of G and generating another subgroup until we generate G, but we just take the one subgroup before G is generated

poorly worded but i hope this makes sense

oh please ping me if you have feedback thank you

[hidden haven]

This is not necessarily true if you just test a single element m

If you check all, then this would be true

[hidden haven]

The idea is correct, but the problem is that you may not generate G even after arbitrarily many steps

[obsidian sleet]

oh

but G is finite?

[hidden haven]

So for example, in the additive group of rationals <1> is contained in <½> in <¼> ...

[obsidian sleet]

i am confused

[hidden haven]

oh

Right yeah then it works lol mb

[obsidian sleet]

yeah i can see in the case for infinite groups this might not terminate

for sure

okay

[hidden haven]

Yeah so to be more formal, you can say that you extend the group at each step until you can't extend it any more by any choice of m

And then it must be maximal

[obsidian sleet]

ah okay

[hidden haven]

I mean not that different from what you said lol

[obsidian sleet]

right just more like

clear

like not to just test a single element m

[hidden haven]

Yep

[obsidian sleet]

but more like to take as many elements of G that are not in H and keep forming larger subgroups until i can no longer take anymore (no longer guaranteed by G being finite)

or something

where no longer means that if i took any more elements this subgroup is G itself

ill keep thinking of ways to improve the wording

but thank you for the advice

[hidden haven]

Yeah pretty much

[next obsidian]

Phrase it in terms of pigeonhole or sometbinf

[cyan marten]

This problem was posted days ago here: Prove that <x, y | x^2 y^2> is not isomorphic to the free product of C2 and C2

Or rather "prove or disprove"

[cyan marten]

Is the first the fundamental group of a space of relatively simple description?

[barren sierra]

I am dumb

part 2 I can't think of counterexamples for c and d

[gritty sparrow]

well, part c) doesn't really make sense because inverses are not defined in those cases. For part d) think of how the trivial group injects into any monoid, and how any monoid has a surjection to it.

[barren sierra]

well I'm thinking of preimage, not inverse

[gritty sparrow]

but preimage is not a map from B to A

[barren sierra]

?

it isn't?

oh it's from B to P(A)

right?

[gritty sparrow]

yep

[barren sierra]

ok yea

cause it may not be a well defined function right

[gritty sparrow]

or more generally even from P(B) to P(A) and we can restrict this map to B by the inclusion of B into P(B)

[barren sierra]

yea

[chilly ocean]

monoids

[obsidian sleet]

hello

i have returned with another question about my ideas

[next obsidian]

They’re all wrong okay next

[obsidian sleet]

trueee

alright for c

so one direction i suppose H is <x^p> for p dividing n

can i just choose p to be the smallest prime dividing n

and so the order of this subgroup is going to just be the largest divisor of n and so its maximal

[next obsidian]

the problem is worded so it enumerates over any prime

So no

[obsidian sleet]

or am i not allowed to do tht

okey

but it wont work for any prime dividing n tho will it

[next obsidian]

It will

When n is composite

You kind of get these different pillars

I mean they aren’t all disjoint

Put the tops are

But*

[obsidian sleet]

oh

[next obsidian]

Like take Z/6Z

[obsidian sleet]

A maximal subgroup

[next obsidian]

And look at 3 and 2

[obsidian sleet]

not The maximal subgroup

[next obsidian]

Yes

[obsidian sleet]

okay excellent i understand now

[next obsidian]

If there’s a unique maximal subgroup then you’re cyclic of prime order (for finite groups, I’ve thought about the infinite case and don’t think I came up with a conclusion)

[terse crystal]

What is C_2? Z/2Z? Anyway I think <x,y| x^2y^2=e> is neither abelian nor finite

[sly crescent]

Is Q(√5) sometimes referred to as the golden field?

[obsidian sleet]

and in the other direction if i suppose H is maximal, i can write H as <x^k> for some k and to show that its prime i can suppose via contradiction that its not prime and show that there is a subgroup of G containing H that isnt G or H

[cyan marten]

Yes, it is.
And yes, it is neither abelian nor finite. But neither is the free product <x, y | x^2, y^2>.
I computed the abelianizations of both and they turned out to be isomorphic to C2 x C2.

[terse crystal]

So it is x^2 , y^2 not x^2y^2?

[cyan marten]

These are two groups. Call the first G and the second K. Prove that G and K are distinct (G is the free product).

[cyan marten]

Yeah, you can also just consider the order of the quotient group: A normal subgroup M of G is maximal iff G/M is simple, so we need the order to be prime. Now the order of G/H is k, if H = <x^k>.

[terse crystal]

So G=<x,y|x^2=e=y^2> K=<x,y| x^2y^2=e> G and K are not isomorphic but their abelianization are isomorphic to Z/2Z*Z/2Z?

[cyan marten]

Yes (if you mean Z/2Z x Z/2Z)

Now G is the fundamental group of the wedge sum of two projective planes

It'd be interesting (but not necessarily useful) to find a simple space whose pi1 is K

[terse crystal]

The abelianization of G is the direct sum of Z/2Z and Z/2Z but is the abelianization of K too? I calculated It’s the direct sum of Z and Z/2Z

[cyan marten]

Well, K^ab has 2(x - y) = 0

Oh, my bad

You're right

That proves it then, G =/= K

[uncut girder]

Why is this bilinear?

[next obsidian]

Wtf is this map

I don’t get it

Okay more like

I can’t read it

[uncut girder]

its the supposed bilinear form associated to the supposed quadratic form

i'm trying to prove the above is a quadratic form on R^2

[latent anvil]

is z some fixed complex numebr

i assume so

write $z = x + iy$ and let $f(c,d) = |cz + d|^2$, then $f(c,d) = |(xc + d) + iyc|^2 = (xc + d)^2 + y^2 c^2 = (x^2 + y^2) c^2 + 2xcd + d^2$

[cloud walrus]

goblin shamrock

[latent anvil]

let $a = x^2 + y^2$ and $b = 2x$, so $f(c,d) = ac^2 + bcd + d^2$

[cloud walrus]

goblin shamrock

[latent anvil]

the associated bilinear form is defined by $2 B(v,w) = f(v+w) - f(v) - f(w)$. if $v = (c_1,d_1)$ and $w = (c_2, d_2)$ then \begin{align*}f(v+w) &= a (c_1 + c_2)^2 + b(c_1 + c_2)(d_1 + d_2) + (d_1 + d_2)^2 \ &= a (c_1^2 + 2c_1 c_2 + c_2^2) + b(c_1 d_1 + c_1 d_2 + c_2 d_1 + c_2 d_2) + (d_1^2 + 2d_1 d_2 + d_2^2) \&= (a c_1^2 + b c_1 d_1 + d_1^2) + 2 a c_1 c_2 + b(c_1 d_2 + c_2 d_1) + 2d_1 d_2 + (a c_2^2 + b c_2 d_2 + d_2^2) \&= f(v) + 2 a c_1 c_2 + b(c_1 d_2 + c_2 d_1) + 2d_1 d_2 + f(w)$. hence $2 B(v,w) = 2 a c_1 c_2 + b(c_1 d_2 + c_2 d_1) + 2d_1 d_2$. is it clear that this map is bilinear?

[cloud walrus]

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

[latent anvil]

rip

[proud bear]

the associated bilinear form is defined by $2 B(v,w) = f(v+w) - f(v) - f(w)$. if $v = (c_1,d_1)$ and $w = (c_2, d_2)$ then 
\begin{align*}
f(v+w) &= a (c_1 + c_2)^2 + b(c_1 + c_2)(d_1 + d_2) + (d_1 + d_2)^2 \\
 &= a (c_1^2 + 2c_1 c_2 + c_2^2) + b(c_1 d_1 + c_1 d_2 + c_2 d_1 + c_2 d_2) + (d_1^2 + 2d_1 d_2 + d_2^2) \\
&= (a c_1^2 + b c_1 d_1 + d_1^2) + 2 a c_1 c_2 + b(c_1 d_2 + c_2 d_1)  + 2d_1 d_2 + (a c_2^2 + b c_2 d_2 + d_2^2) \\
&= f(v) + 2 a c_1 c_2 + b(c_1 d_2 + c_2 d_1)  + 2d_1 d_2 + f(w).
\end{align*} hence $2 B(v,w) = 2 a c_1 c_2 + b(c_1 d_2 + c_2 d_1)  + 2d_1 d_2$. is it clear that this map is bilinear?

[proud bear]

[latent anvil]

ty

[wooden ember]

not sure about this one

[kindred mist]

Having a lot of trouble proving that for a finite group G, there is a left transversal of H< G which is also a right transversal (without combinatorical theorems like Hall's Marriage theorem)

made an MSE:

https://math.stackexchange.com/q/4254720/677593

[terse crystal]

G is the disjoint union of subsets HxH for some x from G. Each such HxH is disjoint union of subsets h_1xH for some h_1 from H also is disjoint union of subsets Hxh_2 for some h_2 from H . Now all those h_1xh_2 are what you need

[terse crystal]

There are infinitely many commutators I think. The length of words can go as large as it can

[wooden ember]

That doesn’t really help for a proof though: plus I’m look for finiteness in the generating set not in the group itself

[terse crystal]

It’s generated by commutators

x^ny^nx^-ny^-n has length 4n so there are infinitely many commutators

[wooden ember]

That doesn’t show it’s not generated by a smaller set though

[terse crystal]

So maybe we need to show that any commutator isn’t the multiplication of other commutators

[wooden ember]

That’s not true though

[terse crystal]

Yeah…

Not true

[terse crystal]

Wait I think I got it

Consider the canonical homomorphism from F=F(x,y) to F/[F,F] =Direct sum of Z and Z mapping x and y to (1,0) and (0,1) respectively

There exists a 1-1 correspondence between element of [F,F] and closed path in lattices of Z*Z with starting point and ending point being (0,0)

[wooden ember]

I saw an answer like this on stack

Didn’t really understand

[terse crystal]

Suppose that [F,F] is generated by finitely many elements then

[wooden ember]

I don’t even know why F/F’ is Z direct Z

[terse crystal]

any element of F/F’ has the form x^iy^jF’

[wooden ember]

why x^iy^j

Shouldn’t it be x^iy^j…

[terse crystal]

xF’yF’=yF’xF’

[wooden ember]

🤔

[terse crystal]

Then any closed path is the composition of those finitely many paths therefore bounded which is a contradiction since you can take a closed path large enough

[wooden ember]

Oh right

I’m stupid

Okay can you explain the 1-1 correspondence part

Nvm I get it

Except why are endpoint at 0?

[terse crystal]

For example xyx^-1y^-1 means you start from (0,0), go right, up, left, down, then as a result you returned to (0,0)

x and y represent going right and up respectively

[wooden ember]

Oh wait éléments of [F,F] are the ones in correspondence I misread

Okay I see that

Now there’s just the final part left?

Oh no I see

It’s a very neat argument

It’s often that to argue about free groups we appeal to geometry no?

[terse crystal]

Probably, using algebraic topology

[wooden ember]

Yeah there was another answer using covering spaces

[terse crystal]

But I haven’t figured out why that map mapping elements from F’ to {paths whose ending points are still (0,0)} is surjective but it doesn’t effect the proof since you can find a path being large and being an image at the same time

[wooden ember]

It doesn’t have to be surjective though right?

Just 1 to 1

[terse crystal]

Yeah but I think it’s surjective

[wooden ember]

Any path that is closed corresponds to an element in the kernel of the canonical homomorphism I think?

Hence the set of closed paths is a subset of the image of F’

(And so equal)

[terse crystal]

Oh yes.so it’s bijective

[wooden ember]

Yeah

That’s why really cool honestly

[terse crystal]

And [a,b] behaves kind of like a parallelogram

[wooden ember]

If I come across a riddle involving paths in ZxZ I’ll be sure to think about F/F’

Yeah

[terse crystal]

So any closed path can be divided into small parallelograms which corresponds to decompose an element of F’ into multiplication of commutators

[wooden ember]

Ah yeah

That’s dope

[terse crystal]

Lol

[wooden ember]

I feel like I’d really enjoy alg top

[terse crystal]

Me too. A lot of interesting stuffs waiting.

[fierce perch]

Let $L$ be a finite separable extension of $k$. We know that $|Hom_k(L, \bar k)| = [L : k]$. But why it also that $|Hom_k(L, k_{sep})| = |Hom_k(L, \bar k)|$?

[cloud walrus]

Lejoon

[fierce perch]

This should be stupid easy

[kindred mist]

This doesn't assume that H is normal right?

[fierce perch]

Is the image of the morphisms L \rightarrow \bar k inside k_{sep}?

[terse crystal]

Yes

[terse crystal]

chapter 10 of Rotman’s algebraic topology contains lots of stuffs about covering space. I just thought if you are eager to know covering space it’s okay to start from that chapter. Previous chapters don’t matter anyway

[terse crystal]

For any r from L, homomorphism f from L to k bar. The minimal polynomial of r is also the minimal polynomial of f(r) which is separable

[lethal cipher]

Gonna be honest, I have no idea where to go with this.
All I know is that if we look at integers (under addition), m and n, then the intersection of <m> and <n> is <mn>.

But that is only a specific example.

[wooden ember]

Use lagranges theorem

[lethal cipher]

Haven't talked about it yet

[terse crystal]

{e}. Since the intersection is also a subgroup whose order divides both orders of <a> and <b>

[lethal cipher]

So no, we cannot

[wooden ember]

Oh what

That’s going to be difficult then 🤔

[terse crystal]

And it’s not mn but the LCM of m and n I think

[wooden ember]

This is using Lagrange

[terse crystal]

😂

[wooden ember]

Well you know it has to be cyclic

Since it’s a subgroup of both

And the order of a cyclic group is the order of its generator

[lethal cipher]

Okay, can we step back a bit. Yes, both of them are indeed cyclic, and so the intersection ( a smaller set) is also cyclic.

[wooden ember]

But the order of a subgroup of a cyclic group must divide that of the cyclic group

Complete from there

Lagrange is simple in the case of cyclic groups ig

[lethal cipher]

Oh wait, I also proved last week that the intersection of two subgroups is a subgroup

That's useful

[wooden ember]

Yeah

So as you said it’s cyclic

And now write it as generated by a^i=b^j for some i,j

The orders of these elements must divide those of a and b

But those are coprime

[lethal cipher]

Okay, let me get back to you. I'm going to see what I can do on my own first (not really paying attention atm). I have ideas.

[wooden ember]

Okay cool

[lethal cipher]

Wow, that was easier than I made it out to be :p

[kindred mist]

never mind, it makes sense to me now, thanks

[lethal cipher]

So I got to this part of the proof and I am having a hard time showing that the function is surjective.

I know I'm basically there, I'm just having a hard time putting it into words

[lavish nexus]

if gcd(n, k) = 1 there exist a and b such that an+bk = 1

so if you mod both sides by n

[lethal cipher]

Then bk is congruent to 1 modn

[lavish nexus]

then you know g^b is mapped to g

but G is cyclic

[lethal cipher]

How does that tell us g^b is mapped to g?

[lavish nexus]

g^(bk) = g^(1+an)

G has order n so for any g g^n = 1

[lethal cipher]

Oh, so that's basically g^(bk)=g

[lavish nexus]

yes

then you can find explicitly which element maps to g^r for every r in {0, 1, ... n-1}

[lethal cipher]

Is this necessary? I'm pretty sure I'm almost done. This seems a bit excessive

[lavish nexus]

Oh it's assumed that G=<g> right

[lethal cipher]

Yep

It is cyclic, so I just chose an element g to be the generator

[lavish nexus]

because g^b is mapped to g, for every g^r g^(rb) is mapped to g^r
□

anytime gcd shows up think about linear combinations

[lethal cipher]

I believe this does the trick

For each y, I found an element in the domain so that y=f(x)

[chilly ocean]

looks good

[lethal cipher]

How is this direction?

[lavish nexus]

because it's onto let G = <g> then there is h = g^m s.t. h^k = g

then h^k = g^(mk) = g
i.e. mk = 1 mod n

conclude (k,n) = 1

[lethal cipher]

Please stop answering with full proofs. I would really appreciate you not do my homework for me.

[lavish nexus]

he's using the order of g^k is n/(k,n)
I prefer linear combinations that's all

[lethal cipher]

Well, more or less.
I just don't like being in a position where it feels like I'm just copying what you said.
That's why I tend to refuse that kind of help because it feels dishonest.

On the other hand, the mod arithmetic you have done definitely comes in handy. Although, I have a bit of trouble identifying when I could use it.
Any pointers on recognizing these things?

[wooden ember]

Mod arithmetic always comes in handy for cyclic groups

Since integers mod n represent all the cyclic groups up to isomorphism

[lavish nexus]

I'm not sure whether cyclic groups are denoted Zn because they are isomorphic to Z/nZ or because cycle in German is Zyklus but yeah

one nice point
(m, n) = 1 also means can find a, b s. t. am+bn = k for any k in Z.

[lethal cipher]

Ah true. Speaking of, Bezout's Lemma feels a bit like the Cauchy Schwarz Inequality: the use of it is unclear, and when you do use it, it seems like a magical fluke.

Of course, it's not, but I'm having a hard time 1) recognizing when to use Bezout's Lemma, and 2) how to actually make it useful

[lavish nexus]

idk which textbook you're using
if it's some established thick volume like dummit and foote it's got tons of exercises
and people crowd sourced a solution manual for it basically

it is a bit harder to find now but look hard enough and you'll find it
then you can just blast through the exercises and check your solutions

[lethal cipher]

I'm using Joseph Gallian's, but I do have Dummit and Foote

[chilly ocean]

What chapter of Gallian are you currently at?

[lethal cipher]

We finished cyclic groups and we are starting on permutation groups

[chilly ocean]

oh I think @oak grove has actually just started that chapter too

[lethal cipher]

Sweet! The good news is I learned a bit about permutation groups when I was trying to read Dummit and Foote.
So I have an edge w/ that, that's for sure

But that was very early in my math education when I was still trying to figure out how to read math books and write proofs. :p
So we'll see how much I remember.

[oak grove]

Oh, yea

well we don't have a book

so idk if chapter is fair

we have an exam on wednesday im kind of dreading but whatever happens 🤷

honestly basic probability is harder than anything in AA

[south patrol]

anything?

[oak grove]

hmm maybe not anything

[lethal cipher]

You and I are having very different AA experiences Jan :p

[oak grove]

I think my class might just be easy

[lethal cipher]

That's fair. My prof always tries to push us a bit by giving us some tough questions.

[oak grove]

i think my teacher is afraid to

although we had a question on this last HW that sorta looped me

[lethal cipher]

Out of curiosity, do you remember what it was?

[oak grove]

easy but it took me a bit

ledog had to help

[lethal cipher]

What is A_n?

[oak grove]

A_n is the subgroup containing all even permutations from some symmetric group

I guess thats a horrible definition

if you take some S_n then A_n is the subgroup with all even permutations of that S_n

[lethal cipher]

Oh boy. I get the feeling this is gonna come up in our homework too.

Either that or something harder

[oak grove]

probably harder

[lethal cipher]

So how'd you do it ?

[oak grove]

[lethal cipher]

Good answer xp

[oak grove]

im still workin on it

[lethal cipher]

Ohhhh, disregard then :p

[oak grove]

but i think the pathway i had at first sorta works

well its interesting that it seems like uhh

well it needs to be generalized

like im pretty sure if you exclude the trivial symmetric group

which makes this trivially true

s_n has an equal number of even and odd permutations

so uhh

idk lemme spend a minute on it here

itd be nice to get the correct answer to a problem today considering i think i got like 30% on my probability homework

[lethal cipher]

Sure. Btw, can you give an example of what the set A_n looks like?

Say, for S_4

[oak grove]

yea, its sorta large

[lethal cipher]

I am having a hard time conceptualizing it

[oak grove]

the order is n!

[lethal cipher]

Well, that was a mistake 😆

[oak grove]

s_3 is easier

[lethal cipher]

Sure, S_3

[oak grove]

do you know cycle notation

[lethal cipher]

(1,2,3) something like this? So 1->2->3->1

[oak grove]

its (), (12), (13), (23), (13)(12), (12)(13)

yea

then A_3 is (13)(12), (12)(13), and ()

[lethal cipher]

Wait, why are (1,3) and (1,2) separated by a paranthesis in 5th element?

[oak grove]

theyre 3 cycles

(123) and (132)

writing them as 2-cycles helps get the order out more quickly

well imo

idk if theres a faster way to get it

maybe you just know by inspection

but it should be clear that those are the two 3 cycles, both going one orientation/direction

[lethal cipher]

So how does (13)(12) translate to (123)?

[oak grove]

1 goes to 2, 2 goes to 1 which goes to 3, and 3 goes to 1

[lethal cipher]

But doesn't 2 go to 3, not 1?

[oak grove]

it goes to 1, then to 3

1 is the only place theyre not disjoint

so (12) sends 2 to 1

then 1 gets passed to (13)

[lethal cipher]

Oh, I see. It's just function composition

[oak grove]

which i guess is how it usually happens? or idk

its a tactic we saw

(1234) = (14)(13)(12)

there must be other ones

well i guess you could pick any ordering as long as you do it right

(21)(24)(23) etc

[lethal cipher]

Alright, I'm with you.
So (123) is an element of S_3

So the elements of the permutation groups are these cycles?

[oak grove]

yea

well i guess theyre uhh

idk this is where im probably not rigorous enough

S_3 i am almost certainly wrong but would be any permutation of 3 elements

so (123) is encoding 123, 231, and 312

everything of that direction of shifting

[lethal cipher]

I know these permutations are functions. So really the elements of the group are the functions

[oak grove]

yea

but (231) and (312) are the same permutation so

(123) captures it

[lethal cipher]

And is (123) the identity element here?

[oak grove]

no, () is

[lethal cipher]

Oh, it's just (), got it

[oak grove]

123 would be the identity ordering i guess?

thats probably the wrong terminology

[chilly ocean]

Dackid so every element of Sn can be written as product of 2 cycles. An is the subgroup containing element that can be Expressed as even product of those

[lavish nexus]

odd order means no element of order 2

[oak grove]

👀

i dont think thats how theyre using odd here

but i could be wrong

[lethal cipher]

Ah true, because the order of each element must divide the order of the group.

[oak grove]

so wait

[lavish nexus]

also every group of even order contains an element of order 2
although this is not relevant here

[lethal cipher]

The first claim is not confined for cyclic groups, but the 2nd one is.

[oak grove]

take S_3 then, H could be all the odd cycles in S_n

[lethal cipher]

Ah true!

[oak grove]

then it wouldnt be a subgroup of A_n

right

and itd be odd order

H could be (12) (13) and (23)

oh but it wouldnt be a group

[lethal cipher]

Oh, A_n doesn't mean it has even order 🤦‍♂️

[oak grove]

not always, but i think theres only one case where it doesnt

if S_n only has ()

otherwise S_n is even order, then half of its elements are even, which is an even number of elements\

[lethal cipher]

Wait, so hold up. If we look at (12)(13) for example. This takes 3 elements to map back to the identity.

So this has odd order and is in A_n

[oak grove]

idk im more confused than when we started talking about this

[topaz leaf]

the alternating group isn't about the odd or even-ness of the order, it's about the parity of the permutation in a different sense

[lavish nexus]

every element in Sn can be expressed as a product of transpositions

[lethal cipher]

Well, I think I have it.

[lavish nexus]

count the number of transpositions to determine whether it is odd or even

it is not related to the order of the element

[oak grove]

okay

[topaz leaf]

i think the confusion comes from this:
a cycle is even if it has even length (and/or is of even order - same thing)
a permutation is even if it is composed of an even number of even cycles
e.g. (12)(13) and (123) are both even permutations

[lethal cipher]

So going back, how would we write (12345) as a product of two even cycles?

[lavish nexus]

(15)(14)(13)(12)
(145)(123)

[lethal cipher]

Okay perfect. I know how to do this now

[oak grove]

im confused about this question again

[lethal cipher]

For starters, given (123...n), it can be written as (1n)(1(n-1))...(13)(12)

[lavish nexus]

no the other way around (1n) goes first

[lethal cipher]

Ah, my bad

Since there are n-1 pairings, this has even order if n-1 is even

Which implies n is odd if n-1 is even.

And actually, that's the gist of it. Just have to generalize it for cycles of the form $(a_{\sigma(1)}...a_{\sigma(n)})$.

[cloud walrus]

dackid

[oak grove]

sigma?

[lethal cipher]

You can just say a_1 to a_n. I don't actually think sigma (the permutation function) matters in the subscript here.

But yeah, I believe this solves the problem

[oak grove]

im confused about what they said about permutations and cycles

so we have that the order of every element is odd

since every element is a permutation, and it's odd, it must have possible to write them all as an odd number of 2 cycles?

[lethal cipher]

So let's generalize it. If we take any element in H, say (a_1,...,a_n), it can be written as (a_1 a_n)(a_1 a_(n-1))...(a_1 a_2).

But this is an n-1 number of pairings, where n is odd

And we know n-1 is even, so the number of pairings is even.

Which means that element is in A_n

[oak grove]

[lethal cipher]

Also yeah, I played hypocrite a bit. I know.

[oak grove]

but doesnt a permutation being odd mean that you can write it as an odd number of even cycles

[lavish nexus]

This needs to be added
every permutation can be expressed as a product of disjoint cycles
Now we have all elements are of odd order
So take any element s and write it as a product of disjoint cycles
the order of s is the lcm of the orders of the disjoint cycles individually which means it is a product of disjoint cycles of odd order
and that means the length of these cycles are odd

[lethal cipher]

As long as they've learned to express it as a product, is it really necessary? Again, seems excessive

[oak grove]

im just trying to track the hint about oddness and what it means about the elements in H

based on the distinction you made about the order of cycles and permutations

sorry, that zain made

[lavish nexus]

note that (12)(345) has even order

[oak grove]

oh, sure

then that definition is wrong that i have in my head

no wait why does it have even order

[lethal cipher]

True, so I did not cover every cycle. :(

[oak grove]

google here i come

[lethal cipher]

Every 2 cycles (12) gets back to the identity, and every 3 cycles (345) gets back to the identity

The first time they are both at the identity is the lcm of those cycles which is 6.

[oak grove]

our definition doesnt make a lot of sense from class

"if a is even every decomposition of a into 2-cycles has an even # of elements"

[lavish nexus]

if a is an even permutation*

[lethal cipher]

Well, H is of odd order-> no element has even order.
Remember the order of an element is how many times you have to operate it with itself to get e.

[lavish nexus]

all we know is odd order means even permutation from the argument above
but even order doesn't mean anything

[oak grove]

sure, H is odd -> every h in H is odd

thats a statement about the order of permutations

so every permutation in H has an odd order

[lethal cipher]

So if x^n=e and n is the smallest positive integer which is true, then ord(x)=n

Order of an element is precisely this.
So if we say an element has order 3, it takes three times to get to the identity.

[oak grove]

right

[lethal cipher]

In this case, the operations are functions. So for (123), 1->2->3->1. It takes 3 operations for 1 to get back to 1. So the order of (123) is 3

[oak grove]

in my notes my teacher said (123) is in A_3

because we can write it as (13)(12)

[lethal cipher]

It is indeed

But A_3 does NOT mean the order of the element is even

A_3 does not say anything about the order

[oak grove]

so "even permutation" doesnt mean "permutation of even order"

[south patrol]

It means this

[lethal cipher]

Correct. Based on what y'all said, an element is in A_n if there is an even number of pairings that make that element.
Again (123)=(12)(13) is even since there are two pairings

[oak grove]

geez what a mess

there are so many other words in the english language they could have chosen

[lethal cipher]

On the other hand, (12)(345) can be written as (12)(35)(34) which is an odd number of pairings.

[oak grove]

okay

[lethal cipher]

So it is not in A_n

[oak grove]

i dont really understand this definition

of elements here means # of k cycles right

where k is even

[lethal cipher]

It's literally this idea, but generalized.

[south patrol]

No, the number of 2 cycles in the decomposition

[oak grove]

thats the same thing isnt it

[lethal cipher]

No, the elements are the function composition.
(12)(345)(67) is an element in itself.

[oak grove]

which would be even but not even order

well it is even order

but it would be even i mean because it has 4 2-cycles

[lethal cipher]

Yep, this one has even order.

Good catch

[oak grove]

no, its even

👀

oh

[lethal cipher]

I meant to say even, went on autopilot xp

[oak grove]

you could have fooled me lol

im considerably lost atm

i need to write some stuff down

[lethal cipher]

I have a question. Mind if we switch over to it while you think about this more?

[oak grove]

yes please do

[lethal cipher]

I need to prove that for n>=3, U(2^n) is not cyclic

Honestly, I need a starting point. U(n) is a weird group and is tough to think about for me.

Btw, if the notation is not consistent, U(n) is all the positive integers less than n that are relatively prime.
U(n) is under multiplication modular n

[south patrol]

Oh wow

I assumed you meant unitary matrices

lol

[lethal cipher]

Lol, definitely not xp

[south patrol]

I assume you mean that are relatively prime to n?

the typical notation for that would be (Z_n)*

but cool

[lethal cipher]

Yes that is right

2^{n-1}

Which is just all odd elements, and also the number of elements in U(2^n)

I'm definitely with you so far.

I'm just not sure where this takes me

Lol, nice xp

Well, I guess the first step is to assume it is cyclic

[lavish nexus]

OK so there's some conclusions you can draw about p-groups but there must be some more elementary way to do this...

[lethal cipher]

I can share my notes if you'd like to see the toolbox I have available to me

Ohhhh, well this is interesting.
All elements of U(8)={1,3,5,7} have order 2 (except identity)

Woah woah woah, definitely not in my toolbox

[lavish nexus]

alright so

if it is cyclic there is only 1 element of order 2

[lethal cipher]

Why?

[lavish nexus]

use the order formula

[lethal cipher]

Ah true

[lavish nexus]

now if an element has order 2 let it be k then k^2 = 2^n* something + 1

also (a+1)^2 = a^2 + 2a + 1

if this is enough hint

this is also where the n>=3 comes in

[lethal cipher]

I'm not entirely sure where this is leading, but it seems like progress, so that's nice.

But no, I am not sure where this is going

Maybe an extra hint would be nice

[lavish nexus]

so the way I phrased it is not very good

I want to solve x^2 = 1 mod 2^n

there's one obvious solution that isn't 1

-1

so find one more

[lethal cipher]

Oh shit! You right!

[south patrol]

I think your hints are pretty useful iteribus. I'd use this as a bit of a heuristic and use it to find candidates order 2 :)

Although it seems we've gone separate paths still

[lethal cipher]

Well, actually hold on. We found 'an' element of order 2, but we didn't necessarily show it is unique

[south patrol]

we want to show it isn't unique actually

[lethal cipher]

Yea, that's what I mean. My b

[lavish nexus]

we want to show we have more than 1 so it is not cyclic

[south patrol]

spoiler (don't view dackid xd; for iteribus) ||not sure what you're thinking iteribus, but (2^(n-1) ± 1)^2 = 1 mod 2^n :) iis that what you did?||

[lethal cipher]

So we just need to show that there exists another element that is not 2^{n}-1 where that holds.

[south patrol]

yes

[lethal cipher]

1 has order 1, so that doesn't cut it.

[lavish nexus]

manipulate it a bit

to make it a perfect square

[lethal cipher]

Make 1 a perfect square or 2^{n}-1?

[lavish nexus]

so the square is k* 2^n + 1
make this an explicit perfect square by adding or subtracting something that still makes it = 1 mod 2^n

[lavish nexus]

yes

[lethal cipher]

Ummm, here is something

[lavish nexus]

congratulations you solved it

should be + though since you put +

[south patrol]

I wonder what the nice way is to show the actual isomorphism

[lethal cipher]

True true. I guess my only issue is that this is bigger than 2^n

[south patrol]

not necessarily

oh

[lavish nexus]

k = 1

[south patrol]

well that doesn't matter anyway ye if the square is bigger

[lavish nexus]

then it's smaller

k = 2 you get exactly -1

[south patrol]

Seems like 3 always has order 2^(n-2) and then I suppose <3,2^(n-1) -1> must generate the group right? idk

some nice link to nt here apparently from wiki aha

[lavish nexus]

it's Z2*Z2^(n-2)

[south patrol]

yeah

[lethal cipher]

So why does n>=3 play a role here?

[lavish nexus]

if n = 2 what is 2^(n-1)-1

[south patrol]

(Z^4)* is order 3, hence cyclic

oops

[lethal cipher]

Oh, duh :p

[south patrol]

the case n = 1 will be left as an exercise

[lethal cipher]

I don't know, that seems too hard xp

[oak grove]

i think i figured out my problem

[south patrol]

sweet.

The way my notes did the parity of a permutation actually used determinants which has advantages and disadvantages i guess lol

[lethal cipher]

Thx for the help! This was also some good practice with modular arithmetic. So thanks for that.

[lavish nexus]

this is a very interesting problem
unfortunately I don't remember much from number theory or it would've been easier

[lethal cipher]

Hey, you've taken number theory. I have not. So this is a new way of thinking for me.

[lavish nexus]

number theory is obsessed with solving congruence

they have a whole bag of tricks

[lethal cipher]

Number theory feels like magic sometimes.

[golden pasture]

look Fp is perfect

||sorry for being cringe ><||

[fallow loom]

Does a subgroup share the same binary operation as the group that it is a subgroup of?

Because the general linear group is a subgroup of the affine group

And the operation on the affine group is composition, but for the general linear group its matrix multiplication

(but this could be because this general linear group is invertible linear operators and not the definition that uses nxn matrices)

[next obsidian]

This is the definition of subgroup

You’re a subgroup of a group if you’re a subset where the operation restricted to this subset actually produces a group

It doesn’t make sense to talk about say for example R\{0} under multiplication as a subgroup of R under addition since the operation is different even though set theoretically the former is a subset of the latter

Here R is the real numbers

[warm saffron]

right and R\ {0} doesn't have an identity element anyway.

[next obsidian]

Yes it does

Under multiplication it’s just 1

[warm saffron]

not the additive identity

oh whoops yeaa I thought you said under addition

my b

[fallow loom]

Ah ok thanks

[next obsidian]

I mean that’s a proof it isn’t a subgroup

[fallow loom]

I have another q

[next obsidian]

But it kinda is tangential to the point I was tryna illustrate

[fallow loom]

If you have an identity which is the composition of 2 things (in a group that has composition as its binary operation), then is the identity shared between all of its subgroups, and would each of the 2 things being composed be in the group

[next obsidian]

Wat

[fallow loom]

so if $a\circ b$ was the identity for a group under composition, then would $a\circ b$ also be the identity for the subgroup

[cloud walrus]

Marikyuun

[warm saffron]

yes

[next obsidian]

Idk why you’re using \circ but yes

[fallow loom]

And would $a ,b\in \text{the group}$?

[cloud walrus]

Marikyuun

[warm saffron]

yes

[next obsidian]

I mean…

They have to be right?

[warm saffron]

it only make sense for a and b to be in the group

[next obsidian]

Like a and b are elements of your group

[warm saffron]

otherwise the operation doesn't apply

[next obsidian]

So they are in your group yes

[fallow loom]

Ok right thank you

I probably sounded so dumb

But thank you

[warm saffron]

you asked good questions.

you're welcome!

[next obsidian]

I have the impression that you have a group you’re looking at

Like a specific one

[fallow loom]

Yep

[next obsidian]

And just looked up what a group is

I’d recommend just

Picking up an intro book on algebra

And just read maybe the first bit about groups and subgroups

[fallow loom]

Ive just been going through the coursebook

[next obsidian]

I think it will clarify a lot of these questions

[lethal cipher]

So I tried to make this if and only if proof into one big argument. Can someone check to see if this is alright. I don't think I see any problems here, but I want to make sure.

[next obsidian]

What’s a ginite cyclic group

[lethal cipher]

Woah, good find :p

[next obsidian]

So this is insanely pedantic but I’ll only say it because you chose to do symbolic logic

You need another for all y in G predicate or w/e before there exists t in Z

I mean this is totally just pedantry but if you’re gonna do one like this I think it makes it a little clearer

[lethal cipher]

Oh sure, that's not a big deal at all

[next obsidian]

It looks right to me

[lethal cipher]

It also puts gcd(n,k)=1 in one line, so I am more than happy with that change xp

[next obsidian]

The only thing I’m unsure how you’re justifying is the equality in the second to last bit

With the order of <a^k>

But I think it’s probably just via

Order of a^k in terms of order of k and order of g

[lethal cipher]

Oh sure. I can clarify. It also equals n because it is the same set as <a>

[next obsidian]

Oh yeah

Consider the function given by… be a surjevtive map

Anyway the only qualms I have are non mathematical in nature

[lethal cipher]

Ah yeah, I just need to write let instead of given

[next obsidian]

The logic goes through

I use iff

\iff

I think it’s the right way to handle stuff like this

Arguably you can like

Separate each line

Use a like \align*

Then put \iff at the start of each line

[lethal cipher]

How's that?

better?

Hmmm, I agree. Not sure how to fix that while keeping the iffs to the left of each line though.

[next obsidian]

Yeah

You will do it like

The part to the left & \iff blah blah\\
& \iff next line \\
& \iff

I would type an example but I’m on my@phone lol

[lethal cipher]

Ah that's write. This function is unique.

Just 2 am slipups, that is all

[next obsidian]

He’s just saying you need the word “be”

There’s nothing about uniqueness

[lethal cipher]

Just ignore it

that looks much better

I sure did. Thank you for catching that

[novel parrot]

is xy is in q

x is not in q

y^n is then, so y nilpotent in A/q

what about x

how would x be nilpotent

[gritty sparrow]

if y is not in q, x is nilpotent. If y is in q, x is not necessarily a zero divisor in A/q (as y is zero)

[novel parrot]

alright, got it ty

[novel parrot]

when it says all their powers are primary

is that for every i

i can only do it for p_n

[hidden haven]

For all i

hint ||in your previous screenshot||

[novel parrot]

ok

[novel parrot]

for this: x in SP(0), x/1 = 0/1 so yx = 0 y is not in P, but yx = 0 (inP) so x is in P

that works?

[gritty sparrow]

yes

[novel parrot]

nice

i found another solution online that was alot longer

after doing this, Sp(0) would actually be P?

nvm

[quaint tree]

(reposted from #help-8 because I don't know why I asked this there initially, I think I'm just dumb):
I'm trying to figure out how to use Macaulay2 to calculate S-polynomials, and for some reason I can't find the correct command. For instance, I might do something like:

i2 : f1 = x^2 * y - 1
i3 : f2 = x * y^2 - x```
What I'm trying to do is find S(f1,f2) as defined on page 85 of *Ideals, Varieties and Algorithms* by Cox, Little and O'Shea 4th edition using a Macaulay2 command. I've tried just doing 
```i4 : S(f1,f2)```
But that didn't work, and I can't think of what else it could be.

[oak grove]

so i deleted earlier but i still cant quite figure this out

[next obsidian]

Well having an order 7 subgroup means the group hs order a multiple of 7 right?

So the generator has order 7k

And we know k > 1

Then what’s the order of g^7 when g is the generator?

[oak grove]

i believe you but what is the justification for the first part

sorry thats a dumb question, i got it

[south patrol]

There's a nice generalisation of this actually, lol, which i can share in a second

[oak grove]

the order of g^7 when g is the generator

just n/7?

where n = |G|

or is that dumb

oh you mean a different generator

[south patrol]

k in chmonkey's notation

[oak grove]

k isnt |G| i dont think

[south patrol]

It isn't

[oak grove]

oh youre saying n/7 = k

[south patrol]

Yeah

So what is <g^7>?

we have three possibilities for what it can be

[oak grove]

im sorry i dont follow the relevance of g^7 here

[south patrol]

Well it's sort of to force us into finding the order of the group. We don't know much about G's structure, but we do know it has order 7k, where k is the order of g^7

So we need to find the order of g^7, equivalently the size of <g^7>

But we can work out what that is from the given information

[oak grove]

this is where I'm sort of confused

im sorry these are all silly questions

[south patrol]

Nah don't worry, feel free to ask

[oak grove]

n=7k with k > 1 seems to me say like

[south patrol]

Sure

[oak grove]

we'd expect 4 subgroups

1, 7, some other prime, and then G itself

im being sloppy but i hope thats clear

[south patrol]

Well why another prime?

[oak grove]

🤔

[south patrol]

In fact, the lack of there being another subgroup of prime order tells you something important.

[oak grove]

i guess not another prime necessarily

but some other non-1 order subgroup

[south patrol]

yeah, exactly

well

Why that?

[small bison]

prob Z/49Z right?

[south patrol]

don't spoil it

[oak grove]

we havent gotten to that

[south patrol]

I know the answer, I'm guiding jan...

[small bison]

well the damage is done

[south patrol]

lol

[oak grove]

i dont know what it means

[south patrol]

Well er

[oak grove]

maybe weve gotten to it but whatever that is we dont use that notation

[south patrol]

Anyway, consider <g^7> again

we know it's a subgroup of G and it's either trivial, the whole of G or order 7

Use process of elimination

[oak grove]

since we know its a generator of some subgroup

since 7 is a divisor of the order of G

okay

[south patrol]

Well every element always generates a subgroup

by definition of < >

[oak grove]

oh this is special about cyclics right

[south patrol]

No

[oak grove]