groups-rings-fields

[trim grove]

yes i think if i replace x with x+1?

[iron vessel]

If you think that is simpler than doing the rational root theorem then you should try it

[trim grove]

by replacing x with x+1 i get x^4+5x^3+10x^2+10x+5

[iron vessel]

That does look simpler.

[trim grove]

but i am doing this whole thing in Z2, does this answer for Q? , i am confused here

[iron vessel]

In Z2? Then after you substitute x+1 you get x^4+x^3+1

In Z2 it is far simpler to (use the rational root theorem)<—NO NEED. Because if there is no linear root then you have that f(x)= (x^2+ ax+1)(x^2 +bx + 1) with a and b in Z/2Z

And you only have a few options to check

[trim grove]

okk , let me try this one more time

[next obsidian]

Why use the rational root test?

there’s only two numbers to plug in lol

[iron vessel]

Lol true

I was mentioning it mainly for part 2 and 4

[next obsidian]

Fair enough

[trim grove]

but the degree here is grater then 3

[white nymph]

just to make sure.... if Z mod pZ isomorphic to K contained in F (field), then F has characteristic p, yes? since 1+1 + ... + 1 (p times) = 0 in K, for a in F we have pa = (1+1+ ... +1) a = 0a = 0. i get tripped up on somewhat trivial things and worry about taking things for granted which are not necessarily true

[next obsidian]

Yes

There’s a sort of converse to this. Every field contains a prime subfield, which is one of F_p or Q. It contains F_p iff it is char p and contains Q iff it is char 0

[white nymph]

yeah im familiar with that. i was unsure if my partial converse stated above was true 🙂

thanks

[trim grove]

Thank you , i got it

[rustic minnow]

Is $\equiv_6$ supposed to denote a modulo $6$ relation in general?
I am given $x \equiv_6 y$ iff $x-y = 6k$ where k is in the integers.

[cloud walrus]

Fredrikpiano

[pastel cliff]

i havent seen that notation but it's probably textbook dependent

[rustic minnow]

given the relation specified it makes sense

Also, I showed that this is an equivalence relation, but when finding all the equivalence classes Im unsure if there are $6$ or $11$? I.e do I need to give the sets $[1] = {\dots, -5, 1, 7, 13, \dots }$ or $[1] = {1,7,13, \dots }$ and $[-1] = { -1,-7, \dots }$ separately ?

[cloud walrus]

Fredrikpiano

[next obsidian]

[-1] = [5]

[rustic minnow]

Since k is in Z it makes sense that there should be 6 equivalence classes no?

[next obsidian]

yes

[rustic minnow]

ok, thanks

could also use the notation $[1] = {6k+1 | k \in \field{Z} },...$ I guess. But in the book Im using they just list the elements

[cloud walrus]

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

[pastel cliff]

in proving that C is field, how do you prove commutativity of addition and simple stuff like thst

is literally writing out just... enough

[oblique river]

what is your definition of C

[pastel cliff]

the complexes

[oblique river]

yes... but how did you define them

[pastel cliff]

ordered pairs of real numbers (a,b) = a + bi, i having the obv definition

[oblique river]

use the fact that addition in R is commutative

[pastel cliff]

ahhh that's what i was weary of

[oblique river]

weary? that makes it sound like a bad thing lol

[pastel cliff]

nono i mean like if i just expand (a,b) + (c,d) it's a + bi + c +di, then i thought moving things around would be using the commutativity im trying to prove

but a and c are just reals so

[oblique river]

use the commutativity on the ordered pairs side

(a + bi) + (c + di) = (a,b) + (c,d) [by definition of C]
= (a+c, b+d) [by definition of addition of ordered pairs]
= (c+a, d+b) [since addition in R is commutative]
= (c,d) + (a,b)
= (c+di) + (a+bi)

[pastel cliff]

or that too yeah

[oblique river]

i mean i think that is the right way to do it

like you said you have to be careful

that you dont use teh commutatitivity youre trying to prove

[pastel cliff]

what i did was that (a,b) + (c,d) =
a + bi + c + di (definition)
= a + c + bi + di (commutativity of reals)
= (a + c) + (b + d)i (associativity of reals - this is the one im unsure about bc of the 'factoring')
= (c + a) + (d + b)i
...
= (c,d) + (a,d)

[sharp sonnet]

bi is not real, so why bi + c = c + bi
similarly you used the distributive law for complex numbers, which you probably don't have

[oblique river]

wasn't this exactly your point?

that you have to be careful to not use a commutativity law that you havent proven yet

[pastel cliff]

it was indeed, that's what i first had, hence i asked thanks for clarifications yall

[tawdry apex]

Hello there, I am stuck on part c) of this atm I'm struggling to prove the ring is closed under multiplication. More specifically I think the roots of f(x) cycle round for higher powers but I am not sure how to prove that without directly calculating the roots.

[oblique river]

a^4 + a^3 + a^2 + a + 1 = 0

and therefore a^4 = -(a^3 + a^2 + a + 1)

[tawdry apex]

oh my god

[oblique river]

this isn't special to this polynomial

by which i mean like

it could be any irreducible degree 4 poly

and this would still be true

[tawdry apex]

for god's sake

thank you so much lmao

[oblique river]

np :)

[tawdry apex]

I can't believe I missed that

anyway

swiftly moving on, hope to pretend this exchange never happened goodbye

[oblique river]

haha it happens to everyone

[wise igloo]

yo
so for the reals, multiplication is thought of as repeated addition, clearly a solid relationship between the two operations. however, in abstract algebra, they're treated as completely separate. someone explain pls

[oblique river]

im not sure i even agree with your first sentence

how do you interpret e*pi?

how do you add pi to itself e times?

[wise igloo]

oh that's right

only for integers then

so for reals, is there a connection between addition and multiplication?

because there definitely is in the discrete world but now im thinking maybe not in the continuous world

[oblique river]

yes: if n is an integer and r is a real

then n*r = r + r + ... + r

n times

and that's true in any ring

[wise igloo]

oh okay

alright ty

[oblique river]

👍

[sterile garden]

I've never really given this much thought until now. Maybe I did when I was a lot younger and I've since forgotten about the difficulties of interpreting such a statement.

[oblique river]

it was meant as a rhetorical question

you can't "add pi to itself e times" because that doesn't really have any actual meaning

the point is that "multiplication is repeated addition" only really makes sense with natural numbers

[sterile garden]

No, I get that it makes no sense. It was just something I haven't really thought of.

[oblique river]

ah okay yeah

I know what you mean like

we are taught that "multiplication is just repeated addition" from such ayoung age

[sterile garden]

It fails to extend to things like the complex numbers and matrices too.

[oblique river]

but really it doesn't make sense beyond basic arithmetic with naturals

yeah

[sterile garden]

I wonder if many kids get confused by this when they're introduced to rationals and irrationals.

[oblique river]

i'm sure a nonzero number of them because ive seen questions about it in this server

[wise igloo]

really it starts with square roots

that's when I first remember dealing with non integer stuff

of that sort

like "wtf is a half power, multiplying it 1/2 of a time?"

[oblique river]

yeah i think that's the exact same issue

"exponents are repeated multiplication" only works for natural number exponents

[wise igloo]

exactly

[serene radish]

I should prove that G acts primitively

I don’t understand why do we need this pair statement at all

I’ve tried assuming that it doesn’t act primitively, supposed that there are two disjoint blocks B and B^g but nothing leads me to a contradiction

[oblique river]

are you sure you have that hypothesis right

because right now it's meaningless

do you maybe mean "for each pair (x,y) and (x',y') of elements in X^2 where x \neq y and x' \neq y' there exists a g such that x' = x^g and y' = y^g"

?

[serene radish]

Yeah sorry, that’s the right one

(The one that you’ve sent)

[oblique river]

non-primitivity basically means that G moves X around in "blocks"

but the hypothesis is kind of telling you that G moves each element around completely independently

and those are at odds with each other

If X = X1 U X2 where X1 and X2 are disjoint blocks

(assuming the action is imprimitive)

and if x, y are elements in X1

if x^g is in X2 you know that y^g must also be in X2

[serene radish]

Yup

[oblique river]

but our assumption on the group action basically tells us that we should be able to move x and y around independently from each other

i.e. we should be able to find some g in G such that x^g is in X2 but y^g stays in X1

and that would violate imprimitivity

see if you can make that precise

[serene radish]

I mean proving this would be enough

[oblique river]

that's the definition of imprimitivity

[serene radish]

We’ve defined imprimitivity as blocks being non trivial

[oblique river]

yes

[serene radish]

Oh okay

[oblique river]

X1 and X2 are your nontrivial blocks

[serene radish]

Got it

[oblique river]

assume the action is imprimitive, therefore X = X1 U X2 where X1 and X2 are nontrivial

and that's where they come from

[serene radish]

Thanks!

[oblique river]

np

[trim grove]

I want to make a non abelian finite group of order 55 , any hint?

[rustic crown]

do you know about semidirect products?

[trim grove]

okk i will try , now i have a hint.

[upbeat juniper]

random interesting question I thought of: what is the transcendence degree of the hyperreals over R?

||so far I have trdeg >= aleph_1, so if CH is true we are done since trdeg <= |R|. not sure if we can do any better though||

[trim grove]

ok i check the definition of semi direct product , i found out that
55 =5 x 11 , so it will have normal subgroup of order 11, and also there will subgroups of order 5 ( by sylow), and both of them will be cyclic , also they will have only identity common because of order of subgroups , and also product of these two subgroups will form original subgroup , because product will have 55 , so we can say original group is semi direct product of Z_ 5 and Z_11?

[rustic crown]

yep, so what you have proven is that if you have a group of order 55 then it has to be a semi-direct product of Z5 and Z11.

[trim grove]

shouldn't be a non abelian?

[rustic crown]

direct products are special cases of semidirect products >.<

there is a way to do semi-direct product externally as well!

you take groups H and N and a map H --> Aut(N) and use this data to make a nice group!

so if you can guarantee a non-trivial map H --> Aut(N), then you have a non-abelian semi-drect product

[trim grove]

This looks tough one , but i will try , btw Thankyou

[rustic crown]

the idea behind semidirect products is actually super simple.

[trim grove]

and what is that idea?

[rustic crown]

lets look at the internal situation

say G is a group, N is a normal subgroup, H is some other subgroup such that N and H intersect trivially and NH = G

what we know is that any element of G can be written as g = n*h for some n in N and h in H

and this decomposition is unique.

so the question is how do you multiply together n1*h1 and n2*h2

one answer is just put then together

n1 * h1 * n2 * h2

but we want to write it as an element in N times an element in H

that's where we use normality!

n1 * (h1 * n2 * h1^{-1}) * (h1 * h2)

so if you wanted to perform this computation, all you need to know is how h1 conjugates n2, right?

[trim grove]

that's why normality is important while driving semi direct product?

[rustic crown]

so if someone gives you the map H --> Aut(N) which sends h to (conjugation by h) then we have all the data

that's how the construction practically goes

you look at pairs (n, h) and say we have the map phi : H --> Aut(N)

then (n1, h1) * (n2, h2) = (n1 * phi(h1)(n2), h1 * h2)

this is the definition of the multiplication map

all the usual verification are carried out that this indeed forms a group and stuff

there is another way to look at this, and it's like matrix multiplication

so in our case, say we wanted to make a group of order 55

notice that Aut(Z11) =~ Z10 and so it has a subgroup of order 5

[trim grove]

yes

[rustic crown]

it's {1, 3, 4, 5, 9}

in (Z11)*

(all powers of 3)

so we can look at matrices of the form

$G = \left{\begin{bmatrix}a & b\ 0 & 1\end{bmatrix} : a \in {1, 3, 4, 5, 9}\text{ and } b \in \mathbb{Z}_{11}\right}$

latex is so hard

[cloud walrus]

det

[rustic crown]

(it's sort of like the group of affine transformations)

a corresponds to some scaling and b corresponds to some translation!

[lethal dune]

btw is there a std notation for this group?

I don't know any or what it's actually called

[iron vessel]

Which group?

[lethal dune]

$G= \left{ \m{ a & b \ 0 & 1} : a \equiv 1 (\mod p^2), b \in Z_{p^2}\right}$

[rustic crown]

some version of this
https://en.wikipedia.org/wiki/Affine_group

[cloud walrus]

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

[lethal dune]

yes it's a subgroup of thr Aff(Zp²)

but any std notation for this?

[rustic crown]

where does a live in this?

[lethal dune]

Zp²

but condition that it's ≡ 1 mod p

[rustic crown]

so a = 1?

[lethal dune]

no like 1+mp form

[rustic crown]

oh 1 mod p?

[lethal dune]

mod p, my bad

[rustic crown]

i haven't really seen a name for this one

but there is like another group of order p^3 which has a name

[lethal dune]

then there isn't ig

yeah one is Heis(Zp)

and the other one is this one

[rustic crown]

yea

[lethal dune]

so no notation?

strange

[rustic crown]

ig no one cared because it can be written as a semi-direct product

[lethal dune]

they haven't bothered to name the 2nd non-iso group of order p³

[lethal dune]

true

[trim grove]

i tried this and get that only identity , and 2cycle 2cycle 1cycle , satisfies this, but is there any mathematical way, because i solved this with just guess?

[rustic crown]

if we put j = sigma(i) then we want i <= sigma^2(i) for every i. this probably implies sigma^2 = id

and if sigma^2 = id then sigma^-1 = sigma and for this sigma^-1(j) = sigma(j) which is satisfied

so you might be missing the cycle type(12)(3)(4)(5)

[trim grove]

my bad

[rustic crown]

issokie

[wooden ember]

The notation i know is $AGL_n(K)$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

which is pretty ugly but not much uglier than PGLn i guess

[lethal dune]

[trim grove]

i tried to solve this, first option is wrong , but i have doubt in 2,3,4, for second option should i have to show that C[0,1]/<x> iso to some field?

[simple mulch]

hey

V,W vector spaces. T : V -> W is an homomorphism. Then T is an isomorphism if T is injective (why we don't need to check surjective?)

[paper flint]

That doesn't sound right

[lethal dune]

either it's assumed to be surjective or it's talking about T(V) and not whole W

[paper flint]

I guess a counterexample could be id: Q->R?

Or huh maybe not

Same fields

[lethal dune]

IG they assumed it to be surjective anyway, because you can always replace W by T(V)

[simple mulch]

ok, so T must be bijective to be an isomorphism..

[lethal dune]

ye

[paper flint]

Yeah, it must be

[simple mulch]

another thing

We have V_n the vector space of polynomials of degree less than N and F^n the vector space of tuples with n entries (both vector spaces are over F)

We want to show V_n is isomorphic to F^n

The injective part is easy as have two vectors, v,w in V_n with v = ax^(n-1)+bx^(n-2)+...+c and w = ... such that T(v) = T(w). then they both have the same coefficients and we would see that v-w = 0 so v=w

The surjective part is easy as well but I have some problem wording it

I could let (a,b,...,c) be an arbitrary n tuple

Then since F^n is over F, there must be a vector v with coefficients a , b , ... c

How would you do the surjective part?

[lethal dune]

$(a_0, a_1, \cdots, a_{n-1}) \mapsto a_0+a_1x^2+\cdots + a_{n-1}x^{n-1}$

[simple mulch]

so you did T^(-1)

[lethal dune]

you can call it S if you want

because you can't call it T^-1 yet

[simple mulch]

but what next?

[trim grove]

isn't he proved the surjective part?

[simple mulch]

where did he?

just by defining S?

[trim grove]

for every tuple of type (a_0,a_1.......,a_{n-1})there is a polynomial in V_n?

[simple mulch]

yes

[trim grove]

one and onto homomorphism === isomorhpism

[weak oriole]

Herstein has this notational issue

Its clarified at the end of that paragraph

[latent night]

Let R be a commutative ring viewed as an R-module over itself, and let M be an R-module. Show that Hom(R, M) is isomorphic to M. Do anyone here have any idea how to do this?

If I understand correctly Hom(R, M) is a module over R, and it inherits addition from M, but I don't know how to identify the elements together.

hmmm

[weak oriole]

Can you do it if R is a field

[simple mulch]

I see it but

[latent night]

yeah, then M would be a vector space. so the set of linear transformations from F to a vector space over F would just be F^n, right? The set of linear transformations from F to a vector space over F is isomorphic to the vector space itself.

n dimension of M

[weak oriole]

Right so
Can you think of an explicit isomorphism

without directly using dimension I mean

[latent night]

pick any element m of M, and then phi(r) = rm?

[weak oriole]

Yup

[latent night]

aaaaah. i understand now.

i thought about basis vectors in the vector space, but this is how it would be for a general module

okay thanks

[simple mulch]

uh I still don't understand how should I write the proof for the exercise above

I've already showed phi is an homomorphism

Moreover, for any v,w in W such that phi(v) = phi(w) we must show v = w

oh

another way to prove is to show ker(phi) = (0)

I am looking for all the possible ways to do this exercise tho

[keen sparrow]

What is phi?

[simple mulch]

oh sorry, I mean $\phi : W \rightarrow V$

[cloud walrus]

mns

[keen sparrow]

Think about the case when F is algebraically closed

[simple mulch]

what's that?

[keen sparrow]

When all polynomial factor into linear factor

[simple mulch]

wdym?

[hidden haven]

How does that simplify things?

[simple mulch]

I am interested in knowing what you would do after defining S

[hidden haven]

Prove that TS and ST are both identities on their respective domains

So T is an invertible homomorphism ie isomorphism

[simple mulch]

I am sorry, I didn't understand what you meant with TS and ST being both identities on their respective domain

[hidden haven]

T ∘ S = identity homomorphism

[simple mulch]

oh

interesting

[rustic crown]

why not just show W has dimension n by exhibiting a basis of size n? general stuff takes care of the rest.

[simple mulch]

general stuff

[simple mulch]

[hidden haven]

Det is talking about "V ≅ W iff dim V = dim W"

[simple mulch]

I mean, I am asking because I don't see how someone would do the proof by showing phi is bijective

[hidden haven]

What definition of isomorphism do you have?

[simple mulch]

I see the injective part tho

[trim grove]

this condition follows when field is same? or for different fields?

[simple mulch]

uh phi is a bijective homomorphism

[hidden haven]

Same field

[keen sparrow]

Can you show phi is surjective ?

[hidden haven]

Invertible maps are bijective

[simple mulch]

I can't

[hidden haven]

So TS and ST being identity → T invertible → T bijective

[simple mulch]

interesting

do you approach every exercise like this in that way?

[keen sparrow]

I mean x^i to e_{i+1} and extend linearly

[simple mulch]

would this suffice?

[hidden haven]

Very often invertibility is easier to show/phrase than bijectivity arguments

[simple mulch]

The goal is badly written tho

[hidden haven]

And also isomorphisms are actually defined as invertible homomorphisms, in the case of vector spaces bijectivity → invertibility but this is not always the case so the invertible definition is better to get used to

[hidden haven]

You shouldn't say TS = id = ST

Because they are equal to different identity maps

[simple mulch]

yeah

id : V_n -> V_n
and id : F^n -> F^n

[hidden haven]

Yep

[simple mulch]

the only definition of invertibility I know is that the function must be injective and surjective

that's why I am used to show both things

[hidden haven]

Invertibility means having an inverse

So T is invertible if there is some S such that ST and TS are identities

[simple mulch]

but in this exercise I was having an hard time showing T is surjective because after taking (a,b,...,c) in F^n I didn't know what to do

[hidden haven]

(a,b,...,c)

[simple mulch]

xd

(is the exercise above correct?)

[hidden haven]

The preimage of an element of F^n is S applied to that element

Because TS is identity

[keen sparrow]

Here since they’re both finite dimensional, the dimension is everything you need

[hidden haven]

Yep

[hidden haven]

Dimension is everything you need even in infinite dimensions if you are trying to show an isomorphism

[simple mulch]

yeah but I didn't defined S so I was kinda stuck

[hidden haven]

Right

You would have to come up with the definition of S pretty much

[simple mulch]

So TS and ST equal id of domains it means T is invertible and therefore T is bijective

[hidden haven]

When trying to show that there is a pre image

Yep

[simple mulch]

interesting

can this fail?

[hidden haven]

Nope, not for vector spaces

But here you can say T is invertible as a set map to make your life slightly easier lol

Because otherwise you would have to show that S is linear

Because invertible homomorphism means homomorphism that has an inverse homomorphism

But it is enough here that T has an inverse set map, because that alone guarantees bijectivity

[simple mulch]

hum

[hidden haven]

Basically I'm saying that here it doesn't matter whether S is linear or not

[simple mulch]

why?

[hidden haven]

Linearity of S is enforced by being the inverse of T

[simple mulch]

I see

[hidden haven]

Inverse of a linear map is linear automatically

[simple mulch]

where can this fail then?

[hidden haven]

And that is exactly why for vector space maps, invertibility = bijectivity

[hidden haven]

Inverse of a continuous map is not necessarily continuous, so this can fail in metric/topological spaces where we talk about continuous functions (and isomorphisms are continuous functions with continuous inverses)

Or inverse of an order preserving map between partially ordered sets need not be order preserving

So for partial orders too, invertibility is stronger than bijectivity

In those case checking bijectivity is not enough to check if something is an isomorphism

[simple mulch]

makes sense

nice, such motivation ahah

thank you!

[hidden haven]

[white nymph]

in this problem, im not sure whether the statement is true or false... i.e. i dont understand something going on here/dont know how to begin... and any help would be appreciated.

in this problem i see that im trying to get that all the roots ought to be roots of unity (else the poly would have infinite roots), but im not sure how to begin showing part a.)

[terse crystal]

Let L be the splitting field of f over F, L’=F(sqrt(m)) you have that

dim(L’/L)dim(L/F)=dim(L’/F)=dim(L’/K)dim(K/F) therefore dim(L’/K)=3dim(L’/L) and dim(L’/L) is either 1 or 2

[white nymph]

im having a hard time understanding your notation @terse crystal
in the dim notation, sometimes you have the smaller field on top and sometimes the smaller on bottom and im confused.

[terse crystal]

?

[white nymph]

dim(L’/L)dim(L/F)=dim(L’/F) i dont follow

[terse crystal]

F is a sub field of L and L is a sub field of L’ then the dimension of L’ over F equals the multiplication of the dimension of L over F and the dimension of L’ over L

[white nymph]

the splitting field is a subfield of L'?

[terse crystal]

Yes by construction

[weak oriole]

Perhaps u meant L'=L(sqrt(m))

[white nymph]

that makes more sense

i will try to re-parse using that edit

[terse crystal]

Oh yeah sorry my bad

Yeah L’=L(sqrt(m))

[terse crystal]

Thanks

[white nymph]

why dim(L’/K)=3dim(L’/L) @terse crystal ?

i see now the two towers of field extenstions you made

[terse crystal]

2 dim(L’/k)=6 dim(L’/L)

[white nymph]

so saying dim(L'/K) = 3 or 6 is equivalent to saying that f(x) is either irred in K or factors as two irred cubics?

[terse crystal]

Let g(x)=f(x^2) since g(a)=0=f(a) therefore f divides g, there exists h such that f(x^2)=f(x)h(x)

[terse crystal]

Yes

[white nymph]

thanks much!

last question i have for now is: if i have field F (not characteristic 2) and K an extension of F of degree 2. how to show that K = F(\sqrt{a}) for some a in F? i know how to show that if K = F(\alpha), the min poly for alpha over F splits in K, but then how to show that alpha = sqrt{a}

[hidden haven]

Try using the quadratic formula

[terse crystal]

you choose any x from K\F

[hidden haven]

alpha won't necessarily be sqrt a

[terse crystal]

F<=F(x)<=K therefore [F(x):F]|[K:F]=2 therefore K=F(x)

[hidden haven]

Yeah but x may not be the square root of something in F

[white nymph]

to emphasize, i just want to show that K = F(\sqrt{a}) for some a in F. the other stuff i typed may not be relevant

[hidden haven]

Yes, try looking at the quadratic formula. Basically take a generator and replace it with some quantity that is the square root of something

But still gives the same extension

[white nymph]

so youre saying since K = F(alpha), we have alpha^2+a*alpha+b = 0 (for some a and b in F). use quadratic formula here to get something that shows alpha is a sqrt? if so, this is what i tried first and failed.... i can try again

[hidden haven]

No, alpha won't be a sqrt

[terse crystal]

?

[hidden haven]

You replace it with some other generator that is a sqrt

[terse crystal]

a^2+sa+t=0 then you already said that characteristic is not 2 therefore (a+s/2)^2+t-s^2/4=0

Then let b=a+s/2 then F(a)=F(b)

Where b^2 is from F

[hidden haven]

Yeah exactly, you're replacing alpha with sqrt of the discriminant of its minimal polynomial

[white nymph]

okay so lemme see here... how on earth to know to rewrite a^2+sa+t as (a+s/2)^2+t-s^2/4? also, how do we know that b^2 = (a+s/2)^2 = a^2+as+s^2/4 is in F?

[hidden haven]

Completing the square

Expand out the (...)²

[white nymph]

yeah but is this just a common technique to use in such problems?

these are old exam questions showcasing common techniques/stuff to know... and id need to be able to do these problems quickly/recognize how to approach

[hidden haven]

It is a technique you use to deal with quadratics a lot

That's how you derive the quadratic formula too

[white nymph]

kk

[hidden haven]

Also the second part is true because that polynomial is 0 when you substitute a, sand so that squared things is t-s²/4 which is in F

[terse crystal]

Yeah s and t are elements of F

[white nymph]

oh, since a^2+sa+t = 0, a^2 + sa = -t so (a+s/2)^2 = a^2+sa+s^2/4 = -t+s^2/4, and since s,t in F, yes?

[hidden haven]

Ye

[white nymph]

thx

both of you

[keen skiff]

Sorry if this has been asked before. I'm learning about finite subgroups $\Gamma$ of rigid plane transformations. There's a proof that says any such subgroup has at least one fixed point. The proof creates a barycenter by taking any point "s" and applying and averaging all the group elements g: $p = \sum_{g'} g'(s)$. It then uses linearity of any rigid motion m to show that $m(p) = 1/n * \sum_{g'} m\ o\ g'$. Letting m be another element of the subgroup, we get that $m\ o\ g'$ is some third element of $\Gamma$, and so the sum just gets reordered - and since vector addition is commutative we get $p$ back. However, how are we guaranteed that $g\ o\ g'$ for $(g,g') \in \Gamma^2$ leads to a unique element in $\Gamma$? In order to "reorder" our sum we'd need a guarantee that we are simply relabeling, no?

[cloud walrus]

yablak

[keen skiff]

Hmm; i suppose if $g1 * g2 = g1 * g3$, then that must imply that $g2=g3$ otherwise we can't have inverses and nice things like this 🙂

[cloud walrus]

yablak

[keen skiff]

sorry for the silly question!

[devout crow]

yeah, multiplication by an element of the group is a bijection

btw I found this statement on https://en.wikipedia.org/wiki/Tensor_product

isn't it false for the vector space with two elements?

i.e. iso to F_2

because then any sum that equals the nonzero element must contain the nonzero element

[oblique river]

yes

[devout crow]

nice

[cyan raft]

dumb question: if there is an element with order = order of the group that element is in, does that mean that the group must be cyclic?

[viscid pewter]

yes

[cyan raft]

oh okay

thanks kangaroo

[delicate bloom]

now prove it 😎

[cyan raft]

uhhhhhhh

that element would be a generator of the group, since its order is the order of the group?

but how can it be the only one

wait a sec i'm dumb

so since the group has 1 generator, it's syclice?

*cyclic

am i right?

[viscid pewter]

basically

[delicate bloom]

the group could have more than one generator

err, well I think maybe I'm misreading what you meant by that

[viscid pewter]

i think they mean that it can be generated by a single element

it has at least one generator?

[delicate bloom]

yeah, a cyclic group is generated by a single element

since the order of that element is equal to the order of the group, all the elements are distinct

so that must be the full group

[cyan raft]

uhhh merosity wdym?

[delicate bloom]

I just meant g and g^-1 generate the same cyclic group but are not necessarily the same, that's why I said this

I misinterpretted you as saying that it was uniquely generated

[trim grove]

some one would like to give me any hint?

[chilly ocean]

show that C[0, 1]/{f in C[0, 1] : f(0) = 0} is a field

[keen sparrow]

Nice exercise

[trim grove]

,trying to find that isomorphism

[chilly ocean]

hint: ||consider the map f -> f(0)||

[trim grove]

yes it is isomorphism ( pointwise multiplication and addition) , and kernel is {f : C[0,1] : f(0)=0}?

[chilly ocean]

you're right about the kernel, but that map is not an isomorphism

[trim grove]

sorry a , surjective map to R?

[chilly ocean]

yes

[trim grove]

my bad

[chilly ocean]

so what do you get when you quotient C[0, 1] by its kernel?

[trim grove]

R

[chilly ocean]

and what does that say about the ideal {f: f(0) = 0}

[trim grove]

maximal

[chilly ocean]

bingo

[keen sparrow]

Was his namo

[trim grove]

can you please guide me for other two parts?

[keen sparrow]

If R/I is integral domain, then what can be said about I

[trim grove]

I is prime ideal

okk got it , let take x and x-1 , there product x(x-1) is in given ideal but individually they are not in that ideal , which contradict the definition of prime ideal?

[keen sparrow]

Right

If you learn some algebraic geometry, then #4 is not true in general (over a scheme) that sometimes functions are not determined by their values at points

[trim grove]

is this a direct result form algebraic geometry ? can i try this without algebric geometry bcs i don't have any knowledge of AG?

[keen sparrow]

You don’t need AG for this

[trim grove]

okkk, then i will try

[red fox]

Are the real numbers mod 1, R/Z, a field?

[hidden haven]

Non integer rationals are zero divisors so no

[oblique river]

R/Z isn't even a ring

multiplication isn't associative

(21/2) * 1/2 = 01/2 = 0

but 2*(1/2 * 1/2) = 2*1/4 = 1/2

[delicate bloom]

is there a more general perspective on this, like if we drop the requirement of a ring homomorphism to send multiplicative identity to multiplicative identity, then having 0 and 1 having the same image ends up losing associativity or something?

[red fox]

Sounds like if I want to represent the circle group using linear algebra, I'd need it to be over the field of complex numbers then?

[next obsidian]

No

Or hmm

You could take R^2\{0} and let the operation be multiplication (like for the complexes). So the units of C, then mod out by the span of a single vector. But this is as an abelian group, not as a vector space

Actually I think the circle can’t be a vector space really since idk what 0 would be

[sharp sonnet]

is it not just SO(2)

[next obsidian]

It just doesn’t seem right

Idk, maybe I’m being a

[delicate bloom]

what feels wrong about using 2x2 rotation matrices

[next obsidian]

Idk, something was telling me it isn’t a vector space

But they are I think

Bad intuit

[hidden haven]

Isn't the circl group too 1 dimensional to be a vector space over ℂ

[next obsidian]

Seems like it yeah

[wooden ember]

what you can at least say when dropping that requirement is that if the multiplicative identity is not preserved it becomes a zero divisor but idk about associativity

[wooden ember]

i mean the circle group is definitely not iso to $\mathbb{C}^n$ for any $n$ so i dont see how it could be a finite dimensional vector space (im not sure how you would argue against infinite dimensional)

[cloud walrus]

𝓛ittle ℕarwhal ✓

[hidden haven]

If it were a complex vector space it would be a >=2 dimensional real vector space but there is no way to define a real scalar action on it because for a non zero vector v in it (the zero vector has to be 1), rv = v^r has to be true for all rational r (this is true for r = 1) so take v to be -1 and r to be 2 and then r has to act like 0

[wooden ember]

Why is rv =v^r

Nvm

Right that makes sense

[hidden haven]

it just looks weird because the multiplication of the circle group is the addition of the vector space

[wooden ember]

Yeah

[rustic crown]

another way to rephrase the argument is by saying R/Z as an abelian group has torsion elements, say 1/2. can't be vector space at all.

[wooden ember]

Why do torsion elements prevent it from being a vector space

Well as a vector space over C that makes sense

But over some finite field

[rustic crown]

vectors spaces are free and torsion free right?

[wooden ember]

I don’t fully know what that means 😔

[hidden haven]

vector spaces are flat so can't have torsion

my brain after doing the grad alg assignments

[rustic crown]

lol

[wooden ember]

But what about F_p over itself or am I saying nonsense

[hidden haven]

Torsion is when non zero element kills it

[rustic crown]

and fields have inverse so 1 will kill it, which is bad

[hidden haven]

p is 0 in F_p

[wooden ember]

Alright this stems from me misunderstanding the word torsion lemme look it up

[hidden haven]

yeah torsion as an abelian group is any non zero element of Z killing it tho

[wooden ember]

Right okay

[rustic crown]

yea right, we probably need to say both 1/2 and 1/3 are torsion elements, so field can't have char 2 or 3 as well

[hidden haven]

det's original phrasing works for char 0

[wooden ember]

I got confused with this

[hidden haven]

down with det

[rustic crown]

bob told me that axler thing has a happy ending

[hidden haven]

who the heck is bob

[hidden haven]

oh that bob

yeah he doesn't say det bad

he say linear alg bad if taught with det

so det should not be allowed in intro linear alg classes

[wooden ember]

Honestly though

[hidden haven]

good thing there was no det in my linear alg classes

[rustic crown]

what's bad about it tho?

[wooden ember]

I agree with the viewpoint that it’s weird to use it

[hidden haven]

read paper

[wooden ember]

Like

[hidden haven]

I have the pdf open on my laptop since then

[wooden ember]

My teacher developed a bunch of intro multi linear algebra and even mentioned what I think was character theory to define the determinant

[hidden haven]

studying more important stuff since then

like clerk's old rants

[wooden ember]

And then the only real application was as an invariant polynomial and a criteria for inversibility

[hidden haven]

axler does all the char polynomial stuff without det too

[rustic crown]

woah

[wooden ember]

We didn’t even have time to look at the characteristic polynomial lol

I still need to read through that part of the notes

[rustic crown]

so like does he use smith normal form?

[wooden ember]

There was a very nice seminar though a few weeks back about classifying all invariant polynomials

[hidden haven]

lul no idk

haven't read

[rustic crown]

wait you can probably do stuff if you assume alg closure and stuff with triangularizability

[wooden ember]

Axler based

[hidden haven]

he does define generalized eigenvalues right at the beginning

[rustic crown]

i remember that's how we proved cayley hamilton in sem1

[hidden haven]

I properly learned how minimal polynomial of matrix works and how to properly use jordan form while preparing for endsem of alg4 with saketh lol

[rustic crown]

alg4 had endsem is what surprises me more

[hidden haven]

endsem was 3 hour conversation with upendra

[wooden ember]

What do you do in alg4

[rustic crown]

ah yes, makes a lot more sense now xD

[hidden haven]

galois theory and modules over PID

[wooden ember]

I see

[hidden haven]

Jordan form came up when doing modules over PIDs

[rustic crown]

we proved jordan form by induction in linear algebra

every lecture was basically induction

[wooden ember]

[rustic crown]

(then we met toc 😭)

[wooden ember]

Toc?

[hidden haven]

theory of computation

[wooden ember]

Like complexity classes and stuff?

[rustic crown]

automata, grammar stuff as well

[wooden ember]

Yeah okay

I don’t think we have that sadly

Just some really introductory stuff on Turing machines and complexity classes in our ICC course from first sem

[hidden haven]

in proof theory endsem we had to prove that some system NK(→, \bot) has weak normalisation

it was so fuckal

more than 10 induction cases

[wooden ember]

Ouch

Proof theory sounds whack

[hidden haven]

Even the hint for that problem was a page long

[rustic crown]

[wooden ember]

Lmao

[hidden haven]

the problem was just "prove weak normalisation for this"

[rustic crown]

moldi will you TA model theory?

[hidden haven]

No it was 2 pages

1 line problem, then 1 page of definitions, then 1 page of steps

[hidden haven]

if they have a TA yes

I will have to study lol

[wooden ember]

[hidden haven]

They are doing a lot more stuff than last year

last year was scam

[rustic crown]

kummini tho

[wooden ember]

Are you two in the same school

[hidden haven]

yes

[wooden ember]

Ah i see

[hidden haven]

I taught det cat theory

[rustic crown]

what's with stog >.<

[wooden ember]

It would be healthy to stop posting elephant shit

[rustic crown]

i'll have to redo those string diagrams and unit-counit adjunction stuff, got pretty complicated towards the end

[wooden ember]

I’d argue none except some experience in proof writing

But linear algebra can’t hurt

[rustic crown]

i'd say knowing some stuff like det and a little elementary number theory would help, as they give some nice examples

[wooden ember]

Fair enough

[hidden haven]

Sadge

I used string diagrams in a gal assignment last week lul

Needed the fact that right adjoints preserve limits so included the proof

Try proving that maybe

It was tricky using string diagrams

[willow mason]

whats gal?

[hidden haven]

Graduate algebra course at det and my uni

[willow mason]

oh

what sorta stuff does a graduate algebra course cover?

[hidden haven]

This sem was galois theory, category theory, modules

tensor and flatness and projective/injective modules stuff

[willow mason]

nicee

[iron vessel]

I have Galois Theory in about a month. Simultaneously with topology, complex analysis and dynamical systems

Not gonna be a very healthy semester

[hidden haven]

oh I had the same except diff eqns instead of dynamical systems

[lethal dune]

I have diff geometry, diff manifold, rings modules and statistical inference this sem

pray

[kind temple]

10

[latent night]

i have galois theory, topology and complex analysis as well starting on monday

might drop the complex analysis, I think it will be too much

[serene radish]

Hi, so I’m trying to compute the cyclic index of a dihedral group, my first guess is to see how permutations look like, but I’m having a trouble finding their cyclic decomposition, can anyone help me seeing how they look like or if there is any alternate approach to finding the cyclic index

I’m computing cyclic index for the general Dihedral group, not for some particular case

I know how rotation looks like obviously, but I have no I idea how to find how r^2 looks like for example

[rustic crown]

what's the cyclic index?

[serene radish]

Or cycle index

[rustic crown]

oh i haven't seen this definition yet.

for r^k, you can break it into two parts. Say gcd(n, k) = d. Then r^d will be product of d disjoint n/d-cycles. And now we need to take that to the power of k/d. since gcd(n/d, k/d) = 1, they'll remain n/d cycles.

so r^k is product of (n, k) many cycles which of which has order n/(n, k)

(i'm assuming r is the n-cycle (123...n))

there are phi(n/d) many values of k for which (k, n) = d, like we said, each one has d disjoint n/d-cycles

[cloud walrus]

det

[rustic crown]

need to handle the reflections now...

[serene radish]

Yeah

[rustic crown]

need to take cases tho, for odd n they look like bunch of 2-cycles and a fixed point. but for even n, you either have no-fixed points or two fixed points.

depending on whether reflection is through a pair of sides or through a pair of opposite points

[serene radish]

Thanks, that was really helpfull

[pastel cliff]

reading about complexes - it said that any complex num z = r(cos(theta) + i sin(theta))

does this apply to something like z = e^i as well

im trying to think of complexes not of the form a + bi is all

though ig e^i could be manipulated into that form...?

[pastel cliff]

do i need to be super familiar with polynomials to study homomorphisms and quotient rings

there's a chapter on polynomials in my book that i really wanna skip, it goes through the euclidean algo for polynomials, rational root theorem, fundamental thm of algebra (which i have some familiarity with already), and some other stuff

[hidden haven]

No, polynomials are one of the reasons for studying ring theory

So are usually discussed first

But it is good to have some examples that you can work with

[pastel cliff]

sounds like im skipping the chapter

[worthy haven]

e^i = cos(1) + isin(1)

this holds for any complex number, and in fact, you'll most commonly see cos(theta) + isin(theta) written as e^(i theta)

[worthy haven]

knowing the euclidean algorithm exists is important, knowing the rational root theorem exists is important, knowing the fundamental theorem of algebra exists is important. Being able to compute the euclidean algorithm for polynomials is not important

[simple mulch]

Hey

[cloud walrus]

fajitas

[simple mulch]

I also want to question something about defining operations on Hom(U,V)

U,V vector spaces

So, we want to define an operation on Hom(U,V) namely for T,S in Hom(U,V), we have (T+S)(x) = T(x)+S(x). First of all I want to show this operation is well defined.

To do so, should I take T,S,R,Q in Hom(U,V) such that T+S = R+Q and demonstrate that (T+S)(x) = (R+Q)(x) right?

I sense something is wrong here, namely in the T+S = R+Q step

[proud bear]

i think you just have to show that T+S is actually an element of Hom(U,V). so just show that it's a linear map

[simple mulch]

but

uh

so (T+S)(x+y) = (T+S)(x)+(T+S)(y) = T(x)+S(x)+T(y)+S(y) = T(x+y) + S(x+y)

?

[proud bear]

well technically you have to say (T+S)(x+y)=T(x+y)+S(x+y)=T(x)+T(y)+S(x)+S(y)=(T+S)(x)+(T+S)(y) but yeah

[simple mulch]

oh ok

😐

for the scalar

T in Hom(U,V)

a in F

aT(x) = T(ax) ?

I mean

[proud bear]

(T+S)(ax)=T(ax)+S(ax)=a(T(x)+S(x))=a(T+S)(x)

[simple mulch]

Ok I have some questions

The initial goal was to define addition and scalar multiplication on Hom(U,V) as to make it a vector space

Define addition as (T+S)(x) = T(x)+S(x)

Now we proceed to show this operation is well defined; To do so we need to check linearity of T+S. I.e, (T+S)(x+y) = (T+S)(x)+(T+S)(y). Also (T+S)(ax) = a(T+S)(x). Hence linearity is verified and T+S in Hom(U,V).

Now shouldn't we define scalar multiplication? I.e, for any a in F and T in Hom(U,V) we must have aT in Hom(U,V)

This is the exercise tho

[oblique river]

everything youve said is correct

the only thing missing is that i dont think you've said what aT is

i.e.

how do you define (aT)(x)

for x in U

[simple mulch]

Ok, so define scalar multiplication on Hom(U,V) as for any a in F and T in Hom(U,V): aT(x) = T(ax) (correct?)

[oblique river]

you should use parentheses when you write aT(x)

so that it's clear that you're talking about (aT)(x)

and not a * T(x)

[simple mulch]

Correction: (aT)(x) = T(ax)

[oblique river]

you can define (aT)(x) = T(ax), yes

you can also define it as (aT)(x) = a * T(x)

which are the same because T(ax) = a * T(x) since T is a linear map

[simple mulch]

indeed

Now we should check this operation is well defined

[oblique river]

yes -- is the map x --> T(ax) linear?

i.e., show that (aT)(x+y) = (aT)(x) + (aT)(y)

and (aT)(bx) = b * (aT)(x)

[simple mulch]

(aT)(x+y) = (aT)(x)+(aT)(y) = T(ax)+T(ay).
Also, (aT)(bx) = T(abx) = bT(ax)

[oblique river]

that first line is bad

you need to prove that (aT)(x+y) = (aT)(x) + (aT)(y)

so you can't start witht hat as your first equality

[simple mulch]

Oh I see (aT)(x+y) = T(a(x+y)) = T(ax+ay) = T(ax)+T(ay)

[oblique river]

and then = (aT)(x) + (aT)(y)

yes

[simple mulch]

Interesting, so the exercise is finished?

[oblique river]

i mean, this is what it asked you to do

i guess technically to prove it's a v.s. you need to check all the axioms

like associativity and distributive property

but like

that's annoying

imo this is enough

[simple mulch]

+1

Thank you very much!

[oblique river]

np

[neat valley]

Quick question: Let $H$ and $K$ be uncountably infinite subgroups of an uncountably infinite abelian group $G$. Furthermore, assume $H$ and $K$ intersect nontrivially.

Why is it true that $(H+K)/H\cong K/(H\cap K)$?

Note that the $+$ is not a direct sum. Furthermore, the assumptions regarding the intersection of $H$ and $K$ and the cardinality of $H$, $K$, and $G$ may be completely irrelevant

[cloud walrus]

Isaiah

[neat valley]

My professor took this as a "simple fact" in lecture, I'm struggling to see it

[oblique river]

isnt this just the nth isomorphism theorem for either n = 2 or n = 3?

[neat valley]

huh you're completely right

Lmao, thanks

[oblique river]

np

and i just looked it up, i think it's the second one

which doesn't make any assumptions on cardinality

[neat valley]

yep, second seems right

just normality

[unique berry]

How would y'all recommend preparing for a course in commutative algebra/algebraic number theory?

[sturdy marsh]

make sure you remember everything from your algebra class ig

[uneven folio]

@boreal lion another example is the mod n map

from Z to Z/nZ

then n gets mapped to 0

[boreal lion]

Ooooh that's right, I see

So like Merosity said, it does hold if φ is injective?

[uneven folio]

yes, it does

[boreal lion]

Perfect, thank you

[south patrol]

Are there any nice examples of vector spaces which aren't isomorphic to their double dual? The simplest I know is set of finite sequences in R

[sturdy marsh]

anything infinite dimensional

[south patrol]

Oh wow yeah i just found that now whilst researching, cheers :)

[unique berry]

I'm gonna go over the giant list of potential problems they have for the algebra prelim exam

It covers basically every bit of algebra from undergrad

[coarse forge]

I recently finished a course on algebraic number theory so I can weigh in a bit on what knowledge we needed. Our course began with a discussion on field extensions because the primary algebraic structure you'll be learning are number fields which are finite extensions of Q. You'll also need to familiarise yourself with ideals as the topic of norms (and norms of ideals) remain a fairly big chunk of the course. Ramification theory also covers a bit of the isomorphism theorems

[upper cape]

is the splitting field of f(x) the same as the splitting field of xf(x)?

specifically, why do we say that F_(q) is the splitting field of X^q-X? Isn't this the same as the splitting field of X^(q-1)-1?

[unique berry]

Idk how deep my class will go but we're starting with field extensions and some Galois Theory then moving to the algebraic number theory stuff

idk how quickly we'll go through the field extensions and galois theory stuff though

[delicate bloom]

splitting field just means smallest field where it factors completely into linear factors, the extra x is a linear factor that doesn't affect this so it doesn't matter

[upper cape]

Is it just written like that out of convenience then?

I mean with the extra x factor

[delicate bloom]

yeah

x^q - x has derivative -1 in a field of characteristic p, which is convenient for showing it has no repeated roots

although you could just as well use x^(q-1)-1 to show this too

just seems cleaner that way

[upper cape]

Thanks! That makes sense

[delicate bloom]

plus it kind of looks nice when talking about the frobenius automorphism, x^q=x fixes elements

x^(q-1)=1 being true only for the multiplicative group

not too serious really, yup, you're welcome

[upper cape]

Ahh I did use the frobenius a few months ago, so that adds up 🙂

[carmine fossil]

If p and q are irreducibles in R which is a Euclidean domain,is it true that |R/<pq>|=|R/<p>| |R/<q>| in general?

[chilly ocean]

I think so yeah! R is a commutative ring with 1 so <pq> = <p><q>, and then just use Chinese remainder theorem to get R/(<p><q>) iso R/<p> x R/<q> and boom

It works cause we're in a PID

and the irreducible elements give you maximal ideals, so they're coprime ideals

Otherwise you can't use CRT necessarily

[trim grove]

some one please help me with this, i have to find the generator of U(25), but operation is multiply so multipliying a number and then check is very dificult,any hint?

[chilly ocean]

i think its everything coprime to 25

the goal is that you want g^n=e only if n = 25*k for any natural number k and g an element of the units

and you know multiplication operation is a*b - 25*k for a,b in G and k a natural number

[delicate bloom]

not everything, for instance 24 doesn't, 24^2=1

[final oasis]

remove the perfect squares and elements with 5th roots

[delicate bloom]

my hint is, you need it to generate (Z/5Z)* at a minimum

otherwise it has no chance of generating (Z/25Z)*

that automatically rules out 3 options

[final oasis]

why can you say that

[delicate bloom]

because if x^n=a mod 25 then it must reduce to x^n=a mod 5

[final oasis]

4^2

[delicate bloom]

yup since that's 1 mod 5, you have no chance of representing 2 or 3

it helps if you write numbers in base 5 to see this

if you say 4^n = a+5*b mod 25 with a and b both digits picked from {0,1,2,3,4} (with a !=0) then 4^n = a mod 5

[trim grove]

why we restricting a and b here?

[delicate bloom]

just because it's the base 5 representation

we don't have to

[trim grove]

looks like this way can work for any perfect square type ?like 25,49,..

[delicate bloom]

yeah, or any product really

although you have to use a mixed radix basis to represent it

but like if we were looking for generators of (Z/30Z)* then it has to be a generator for (Z/nZ)* for each of n=2,3,5

I guess that's not the same cause that's CRT

well maybe I should just say that, when doing stuff in Z/nZ you split across powers of primes so that you have the problem in Z/p^kZ then you try to solve it in Z/pZ and then you lift back up with Hensel's lemma to Z/p^kZ and then combine the results to Z/nZ with CRT

that's a decent strategy to break down problems

[trim grove]

Thankyou @delicate bloom , for such wonderful explanation.

[delicate bloom]

heh you're welcome

[carmine fossil]

Anyone familiar with RSA?

I was just thinking about Eisenstein integers and RSA using those instead of regular integers just seems better

[chilly ocean]

uh

you know order of k doesnt divide order of G/H doesnt divide

[unreal lodge]

sorry if it's obvious, I'm just trying to see some worked examples (this is my first abstract proofs class, could you please elaborate)?

[chilly ocean]

yeah idk lol

im thinking of why

its not immediately obvious for me

[weak oriole]

What can you say about the group KH

[chilly ocean]

what can be said?

something about the order?

[unreal lodge]

KH <= G?

[chilly ocean]

everything im writing is orders

G = [G:H]*H right

Multiplying by order of K on both sides we have

|G||k| = [G:H]*|H||k|

lol

can do some algebra since these are all numbers

this is silly

[weak oriole]

Are you sure these aren't reversed?
order of H and index of K?

[delicate bloom]

are you trying to show |G| = [G:K]|K| = [G:H]|H| and so |K| divides |H| because it can't divide [G:H]?

[chilly ocean]

oh

that would make more sense

but idk how it would be a subgroup

[unreal lodge]

yea I think so (that's what it shows on the zoom recording) but I wasnt in person so maybe they did something on the board but this is what the thing says :/

[chilly ocean]

lemme think if i can find counter example

ok

Z24

taking H to be multiples of 8 and K to be multiples of 6

thats a counter example

[weak oriole]

Ya I think the situation is reversed

[chilly ocean]

my bad lol

Choose K to be 3

wait did i choose correctly

bruh

[weak oriole]

In which case consider xK for x lying in H

[chilly ocean]

H has size 3

[hidden haven]

No

[chilly ocean]

K has size 4

nah im good

[hidden haven]

H has index 8

[chilly ocean]

oh fuck

[hidden haven]

Not coprime to 4

[chilly ocean]

index yeah

lol

choose H as its cofactor then

uh

H is 3

yeah

H multiples of 3 and K multiples of 6

K has size 4

G/H has size 3

ok

idk why i didnt look for an example first

[hidden haven]

Then K is a subgroup of H lol

[chilly ocean]

is it?

[hidden haven]

Example has to be non abelian

[chilly ocean]

wait fuck

[hidden haven]

Otherwise Lagrange

[chilly ocean]

wait

its true for abelian groups?

how do you prove?

[hidden haven]

I think so

[chilly ocean]

i dont thinks so

[hidden haven]

Apply Lagrange on G/H

[chilly ocean]

yeah it has subgroups with order dividing G/H

so k isnt none of em

oh wai

brih

order 1

is that it?

oh wow lol

[hidden haven]

Ye if was not in H, it would be partitioned into cosets of K ∩ H

All equal cardinality and their number dividing index H

So H has to be not normal in the counter example

[chilly ocean]

If K not in H then G/H can be partitioned into cosets of K intersect H each of equal order?

what if K intersects one coset more

[hidden haven]

Cosets always have the same size

[chilly ocean]

yea

[hidden haven]

That's how you prove Lagrange

K ∩ H is a subgroup of K

[chilly ocean]

wait hold up

why

i cant prove that

[hidden haven]

The cosets of this as a pair on its own or as a pair inside G are the same

[chilly ocean]

i remember this vaguely though

[hidden haven]

Intersection of subgroups is always a subgroup

And if you have 2 subgroups, one contained in the other, then that one is a subgroup of the other

[chilly ocean]

yea

ok

[subtle ivy]

this is an exercise in dummit and foote i think

i just did this like a week ago >.<

yeah i think they were switched

[subtle ivy]

why did you delete your question -_-

[unreal lodge]

reposting lol

after I read more of the notes

i missed like a week of class cause covid and ngl these online recordings of their in-person sessions are not great so im just trying to piece together whatever they covered

[hidden haven]

Actually I think this is fine

That statement should be true because we have something like Lagrange for G/H

Even if H isn't normal

We can look at the left action of G on G/H (the set of left cosets of H)

And each orbit should have the same size here

And if K is a subgroup of G, then K/(K ∩ H) should have a natural non zero map to G/H

And since all orbits have the same size, any G-subset of G/H should have size dividing index H

But this assumes that you've seen group actions

Have you?

There should be a simpler phrasing though

(assuming this argument works because a couple of these claims are just from intuition lol)

[unreal lodge]

no group actions yet

[subtle ivy]

hm the dummit and foote problem is true because HN has to be a subgroup, since N is normal, and then we have that |HN|=(|H||N|/|K|) and then the index of HN is ([G:N]|K|)/|H|

i think nyway

[wooden ember]

What’s the question?

[hidden haven]

H and K are subgroups of G, K is normal, |K| and index H are coprime, prove that K is contained in H

[subtle ivy]

HK is a subgroup, and the index of HK is (|G| |N|)/(|H| |K|)?

where is N is the intersection

use euclids lemma to show K divides the intersection

K=N, K is a subroup of H

[trim grove]

just need a little bit hint , S_n is isomorphic to subgroup of A_{n+2} , under what isomorphism?

[rustic crown]

so there is the usual way to think of S_n as a subgroup of S_{n+2}, only problem being that stuff like (12) are still sent to odd permutations. so how do we fix that given we have two extra letters , n+1, n+2 to play with?

[trim grove]

Things seems complicated to me

[rustic crown]

we need a way to convert an odd permutation in S_n to an even permutation in S_{n+2}, what's one thing you can try?

[trim grove]

okk so odd permutation means it will have odd number transposition, so we can add extra (n+1,n+2) to make that even ?

[rustic crown]

yep!

now just quickly verify that this prescription is indeed a group homomorphism.

[trim grove]

okk, so here even permutation is mapped to ?itself

[rustic crown]

so the map is S_n --> A_{n+2}
given by sending a permutation p to p if it's even else to p * t if it's odd. t is the transposition (n+1, n+2).

[trim grove]

got it , but first i have to check is this a isomorphism? ( yes it is, what still have to show)

[rustic crown]

no not iso, it's definitely not an iso. left side has cardinality n! and right side is (n+2)!/2

you probably meant iso to the image, which is same as saying it's an injective homomorphism.

[trim grove]

Got it , and we have to use the fact that subgroup ( which is Sn here) under homomorphism is mapped to subgroup (in phi(A_{n+})?

[rustic crown]

yep

[trim grove]

Thankyou @rustic crown

[cyan raft]

is the group generated by ${a_1, a_2, a_3, ..., a_n}$ isomorphic to $\mathbb{Z}^n$

[rustic crown]

what's F_n ?

[cyan raft]

hi beans

[rustic crown]

oh you mean the free group?

[cyan raft]

uh... yeah

probably

[rustic crown]

isn't that going to be infinite, and Z/nZ is a finite group

[cyan raft]

oops i meant Z^n not Z_n

[cloud walrus]

bob (bean toucher)

[cyan raft]

sorry det

[rustic crown]

oh issokie

so if you're talking about free abelian groups, then that's right.

[cyan raft]

...right, forgot about that condition-

anyways

thanks det!

[rustic crown]

but for just free groups, not so much. free groups have no relations between the generators, so the generators don't commute with each other. which means F_n isn't commutative for n >= 2

[dull marsh]

took me a while to remember how to type like this

[keen sparrow]

Free ( in the context of group) means free from relation

[spice whale]

do matrices over finite fields form permutation groups?

[keen sparrow]

You mean are they isomorphic to a permutation group?

[lavish sigil]

Define formal power series $\exp(x),\log(1+x)\in\mathbb{R}[![x]!]$ just like the Maclaurin series of the corresponding real functions: $\exp(x):=\sum_{j=0}^\infty\frac{x^j}{j!}$ and $\log(1+x):=\sum_{j=1}^\infty\frac{(-1)^{j+1}x^j}{j}$. Since the constant term of $\log(1+x)$ is zero, the composition $\exp(\log(1+x))$ is a well-defined power series. I am pretty sure that this composition equals $1+x$ as a formal power series (as would be expected). How can this be shown?

[cloud walrus]

gustavn64

[lavish sigil]

I mean, we know that the composition $\exp(\log(1+x))$, when viewed as a composition of real functions, is identically equal to $1+x$. But I'm not sure exactly how to show that composition of functions corresponds to formal composition of power series...

[cloud walrus]

gustavn64

[spice whale]

yeah

sorry

[lavish sigil]

@spice whale Every group is isomorphic to a permutation group, by Cayley's theorem.

[spice whale]

oh

[frank lake]

I don't understand this proof 😦 can somebody break down the steps or something

[keen sparrow]

Which steps do you not understand?

[frank lake]

mainly the algebra

"Informally all we need do to obtain the second equation from the first one is simultaneously o bring the a's across the equal's sign"

[keen sparrow]

This is mainly showin that the center of a group is a subgroup

You have the equation ax = xa

[frank lake]

mhm

[scarlet estuary]

it shows how to do it formally

here

the "informally" part is just explaining what the goal is

[frank lake]

how did it turn ax = xa to xa^(-1) and a^(-1)x

that's really my only problem

[keen sparrow]

It's just multiplying by the inverse of a

[frank lake]

that would give you x = x? wouldn't it? $ax = xa$ multiply inverse of a on both sides $a^{-1}\cdot ax = a^{-1} \cdot xa \implies (a^{-1}\cdot a)x = (a^{-1}\cdot a)x$

[cloud walrus]

ChubbyMuffins

[frank lake]

is that assuming commutativity?

[scarlet estuary]

yes

or rather

its valid but getting x = x isnt the goal

[frank lake]

I see

[scarlet estuary]

like you can take any equation a = b

divide both sides by a

and get 1 = 1

[keen sparrow]

It's a^{-1}(ax)a^{-1} = a^{-1}(xa)a^{-1}

[scarlet estuary]

this is utterly useless

[frank lake]

I see

[scarlet estuary]

but you can

its irrelevant to the proof

in the proof, they took ax = xa

[frank lake]

mhm

[scarlet estuary]

multiplied by a^-1 on the left and the right

and got this

do you understand the steps there?

[frank lake]

why is there 2 inverse's of a's multiplied on both sides

[scarlet estuary]

because thats necessary for the proof

[frank lake]

OHH

WAIT

[frank lake]

is it because of this theorem?

[scarlet estuary]

no?

[keen sparrow]

Do you understand what it mean for H to be a subgroup?

[frank lake]

for H to be a subgroup it means that H is closed under some operation G

[scarlet estuary]

oh yeah thats the theorem its using

[frank lake]

given G contains the element H

[scarlet estuary]

thats why the goal is to show that ax = xa implies xa^-1 = a^-1x

[frank lake]

I see

I think that makes a little more sense

[scarlet estuary]

actually i dont think its using that theorem

isnt it just checking the definition