groups-rings-fields

[left monolith]

The right hand side, because of how sum and product is defined

[solemn rain]

yea cool

[left monolith]

So that would complete my proof

[solemn rain]

so to make this proof look 100% better

write this using sum notation

sum(a_ib_i)

thats x^

but u get the idea

got it?

[left monolith]

Got it

[solemn rain]

cool

[left monolith]

Thanks very much :)

[solemn rain]

np

[left monolith]

Go ahead with ur question

[solemn rain]

thanks

Prove that the ideal pZ/p^mZ is a nilpotent ideal in the ring Z/p^mZ

so id guess pZ/p^mZ = {pa(p^mZ) | for integer a}?

i dont know how that even looks like tbh

[final dove]

Can someone sanity check my exact sequence? https://math.stackexchange.com/questions/3704264/exact-sequence-of-ideals-intersection-product-direct-sum-sum?noredirect=1#comment7612698_3704264

I'm really not sure I'm doing it right...

[random ravine]

Is anyone able to do 3b/c

This is how far I’ve got I just don’t know how to finish it

[golden pasture]

@solemn rain you can just think of Z/p^mZ as taking mod p^m
What are the elements of (pZ/p^mZ)^n for some n?

[solemn rain]

(ap^(m),bp^(n)) ??

@golden pasture not sure at all

where a and b are integers

wait wait thats for n =2

so (a,b,c,....,n)

[golden pasture]

@final dove image of g is just (0,0). (-i,i) is in the kernel of h but not image of g

yea

smt like that

[solemn rain]

where a = px for some x in mod p^m?

and so on ..

??

[golden pasture]

Like for example 2Z/8Z, this consists of elements (0,2,4,6)

[solemn rain]

okay so yea i think what is aid is right then

where a = px for some x in mod p^m?
[11:42 PM]
and so on ..

[golden pasture]

wait as in?

[solemn rain]

elements of (pZ/p^mZ)^n

[golden pasture]

ah yea i meant that as ideal multiplication

[solemn rain]

ohhh

[golden pasture]

cuz we're trying to show it's nilpotent

[solemn rain]

is it always ideal multiplication

in th context of nilpotency

lmao yea

okay

okay soo

so any element in (pZ/p^mZ) would be a number that is in mod p^m and a multiple of p right?

[golden pasture]

yup

[solemn rain]

okay so that to the n is

(pZ/p^mZ)^2 for example = {ab | a in pZ/p^mZ and b in same}

right?

just making sure

[golden pasture]

@random ravine maybe move to a analysis channel? it's more of solving a ode

yea

[solemn rain]

i think

(pZ/p^mZ)^m should be 0?? not sure at all

cuz like

its the set of all products

they all have p a common factor

so p(abcd....) where all these are mod p^m

[golden pasture]

yup

[solemn rain]

so id guess i want abcd... = p^m

how XD

[golden pasture]

if a_i are in pZ/p^mZ, then a_i=pb_i for some b_i in Z/p^mZ
Product of a_i from i=1 to k is p^k <product of b_i>

[solemn rain]

okay

yea so its to the m

right?

[golden pasture]

yup

[solemn rain]

cool

tysm

[old nimbus]

are you in a group?

[mild laurel]

You can take G = Z_2 x Z_4

and the two isomorphic subgroups generated by (1,0) and (0,2)

but the first quotient is Z_4 whereas the second quotient is Z_2 x Z_2

[woven delta]

Lol

What do you mean quotient by the action of H?

What is the action?

[wind steeple]

R/Z, [0,1]/Z

[woven delta]

How does Z act on [0, 1]?

[wind steeple]

you can define a stupid group action imo

mh

no it doesn't work x)

[woven delta]

It shouldn't be

Hmm can you take 2 copies of R and define a funny Z action which sends (1, r) to (2, r+1) and (2, r) to (1, r+1)

[wind steeple]

you can find for diffeomorphisms, like that

[woven delta]

Would that give you S^1?

[wind steeple]

and take the symetry action for both

it will make the same space, but these aren't diffeomorphic

yeah your example seems to work fine @woven delta

[woven delta]

So the homeomorphism to R/Z is just [(i, r)] \mapsto [r]

You have to check this is a homeo but it's not too bad

[split knot]

Can anyone help with this question? I've been stuck on it for a while

I'm not entirely sure where to start; my groups&rings lecture notes are quite bad

[elder valley]

part a?

[split knot]

we'd start with a ye

[elder valley]

well (i) is just a computation, nothing to do with groups really

[split knot]

ah sorry I meant ii and beyond

I was given a definition of an isomorphism in my notes but I don't really understand bi/in/surjections well enough to actually understand isomorphism

[dawn kiln]

aight, so injections are functions where if f(a) = f(b), then a = b

[elder valley]

what answer did you get for (i)?

[dawn kiln]

ie. when no two items map to the same output

[split knot]

what answer did you get for (i)?
pi*i

[dawn kiln]

and surjections are functions where for any element k in the output set, there exists x such that f(x) = k

[split knot]

(simplifies to ln(-1))

[elder valley]

that can't be right. theta maps into R+

[split knot]

oh wait

nah ignore me

[elder valley]

i haven't done the computation but i'd wager it's ln(1) = 0

[dawn kiln]

no

it actually helps with the second part, since you don't really need to show theta is a bijection in the usual way

it evaluates to y

[elder valley]

sorry, i mean y

yeah which gives a hint for showing its a bijection

[dawn kiln]

@split knot so what does it mean if $\theta\left(\frac{e^y-1}{e^y+1}\right) = y$?

[cloud walrus]

maximwebb:

[split knot]

uh

[golden pasture]

it is a lot of like brute forcing here but it is easier if you consider the map by tanh, so x=tanh(u), y=tanh(v)

[split knot]

I may be out of my depth here

[dawn kiln]

alright, i'll give you a couple of pointers

so the first part basically shows that $\theta$ has an inverse

[cloud walrus]

maximwebb:

[dawn kiln]

ie. if we define a new function $\phi(x) = \frac{e^x -1}{e^x+1}$, then $\theta(\phi(x)) = x$

[cloud walrus]

maximwebb:

[dawn kiln]

(which is the definition of an inverse)

now bijections have two definitions - one is that they are functions that are both surjective and injective (which i defined previously)

the other is that they have an inverse

[split knot]

presumably they need to satisfy both

[dawn kiln]

nah, they are equivalent definitions

so if a function is surjective & injective, then it is invertible

and vice versa

so we can write $\theta^{-1}(x) = \phi = \frac{e^x-1}{e^x+1}$, and conclude that $\theta$ is both surjective and injective

[cloud walrus]

maximwebb:

[split knot]

So with injective and surjective, distinct x's correspond to distinct y's, every x corresponds to a y, so that means all x's have distinct y pairs which would make it bijective/invertible
And with the new function, you prove that it's the inverse of the original which means it's bijective, which then means it's also in+surjective

[dawn kiln]

yeh that's right

it's worth writing out the full proof, as an exercise

but that aside, look back at the question

do you remember the other requirement for a function to be an isomorphism

[split knot]

theta : G -> H being bijective means theta is an isomorphism from my notes, our H here being this I think

(Sorry, idk how to type in latex)

[dawn kiln]

an isomorphism is sometimes called a "structure-preserving bijection"

which means informally "things behave the same inside the map as they do outside the map"

[restive sundial]

$\theta: G \rightarrow H$

[cloud walrus]

Eddie:

[dawn kiln]

in symbols, that means for $a, b \in G, ; \theta(a * b) = \theta(a)\theta(b)$

[cloud walrus]

maximwebb:

[split knot]

I think I see where you're going with that

[dawn kiln]

for example, if you define a map $f: (\mathbb{R}, +) \rightarrow (\mathbb{R}, \cdot)$ by $f(x) = e^{x}$, then you can see that it preserves structure

[cloud walrus]

maximwebb:

[dawn kiln]

since $f(a + b) = e^{a+b} = e^a e^b = f(a)f(b)$

[cloud walrus]

maximwebb:

[dawn kiln]

functions that have that property are called homomorphisms. If a function is a homomorphism, and is bijective, then it is an isomorphism

[split knot]

so I need to prove that theta(x * y)=theta(x) + theta(y)?

(uhhh not * as in multiply, I mean the operation at the top of the Q)

[dawn kiln]

yeah exactly

@split knot so could you show that alright?

[split knot]

ah sorry, yeah I got that they're both equal to:

oh that's not rotated properly

but ye I think that's correct

[dawn kiln]

,rotate

[split knot]

thanks haha

[dawn kiln]

yeah that looks fine i think

[split knot]

so in conclusion, because they're equal theta is a homomorphism and because it's also a bijection, theta is therefore an isomorphism

[dawn kiln]

it's normally a bit neater to have one chain starting from f(ab) and ending with f(a)f(b) (or vice versa), but this is a bit of an annoying operation to work with so that's fine

[split knot]

yeah I was thinking about doing LHS = RHS downthe page, I just didn't have much space left

[dawn kiln]

yup, but make it clear that (i) shows it's an invertible function, and therefore it must be a bijection

[split knot]

that's how I used to do trigonometric proofs

ah yeah above that I discussed how it's invertible

[dawn kiln]

perfect

[split knot]

(((please ignore my handwriting lol)))

[golden pasture]

kinda ugly, slightly nicer solution:

Let $x=\tanh u$, $y=\tanh v$. Then $x*y=\tanh(u+v)$.

Recall $\tanh x=\frac{e^{2x}-1}{e^{2x}+1}$ and $\tanh^{-1}x=\frac12\log\frac{1+x}{1-x}$

[cloud walrus]

ariana:

[split knot]

been a while since I've seen tanh surprisingly

[golden pasture]

same

[dawn kiln]

such a neglected function :(

[split knot]

ngl I haven't touched hyperbolic trig in about 2 years i.e. not while I've been at uni

kinda sad :/

[dawn kiln]

btw, which textbook are you using?

[split knot]

I'm not

[dawn kiln]

oh right, so lecture notes?

[split knot]

ah sorry, ye

the module is called Algebra and Number Theory so it's split into groups, rings and number theory

[dawn kiln]

http://abstract.pugetsound.edu/download/aata-20170805.pdf

[split knot]

the question comes from a specimen paper for an upcoming exam, I have 3 more on the other topics

[dawn kiln]

this is what i used, it's generally pretty good either as a reference or for learning from

[split knot]

343 pages of loveliness huh

[golden pasture]

i used jacobson algebra

[split knot]

ahh I love it when PDFs have links

[dawn kiln]

tbf that's 343 pages to go all the way up to galois theory

[split knot]

ah that's a module i can take in third year

[dawn kiln]

v cool area of maths

[split knot]

I'll cross that bridge when I come to it heh

[golden pasture]

i need to figure how to add links to pdf manually

it's so uwu

[split knot]

my lecture notes usually don't ariana 😦

[golden pasture]

idk how some libgen scanned books have

[split knot]

good ol libgen

none of my modules have textbooks so I never need it heh

aight so

I think I'm slightly more confident with part b because I actually know what cyclic means

[dawn kiln]

spoilers lmao

[golden pasture]

oh oops

[split knot]

I saw e's

I am perturbed lol

[dawn kiln]

lol its fine, for a sec i thought you posted the actual answer lol

yeah e is sometimes used as the identity element

[split knot]

ah ok

I feel like this is probably useful but ngl my confidence is gone rip

turns out I'm not very good at this module : )

[dawn kiln]

algebra is hard, dw about it

that theorem certainly helps

have you come across langrange's theorem?

[split knot]

That'd be this guy

[dawn kiln]

yup

it's the first really important theorem you prove in group theory

[split knot]

ngl, I don't really understand how the fact that something divides something else is useful outside of proving something's prime

so the # of elements in A divides that of B... so what?

[dawn kiln]

you can flip the way you think about it

we use divisibility results to ascertain facts about groups

so for example, say if I give you a group G of order 37, and I ask you to find all the subgroups of it

you might start by saying that the identity has to be included, so that leaves 36 elements to either include or exclude - so 2^36 different potential subgroups to check

or about 7x10^10 subsets

haha no i was making a slightly different point, but yeah that result is cool as well

but basically, lagrange tells you that the subgroup order has to divide 37 - so the only subgroups have size 1 or 37 (since 37 is prime)

which is a substantial improvement over checking 70 billion sets

[split knot]

smiley face 🙂

[dawn kiln]

lmao

groups are great, embrace it 😂

[split knot]

well, I took the module because proving that a thing is what it is is cool

however

it's hard and hurts my brain

true lol

I haven't really focused on this module in my revision so it's definitely my weakest, plus this specimen paper is due in a day and a half

gotta crack on lol

[dawn kiln]

rip

[split knot]

least I've still got a couple weeks till the actual exam

[dawn kiln]

definitely spend some time on the basics once you've done this hw

[split knot]

for sure

[dawn kiln]

ie. just bash out as many exercises from that textbook i linked as you have time for

[split knot]

I will mate, cheers for that link it'd be useful to have textbooks to go alongside my lecturer's notes

as I said the notes for this module aren't great, it's a pretty big module that got interrupted by 2 strikes and the coronavirus

[dawn kiln]

yeah ig you

i do cs, and had a similar thing for our operating systems course lol

lecture notes with rhetorical questions on most slides, and blank spaces where i guess he would have elaborated if he'd shown up for the lectures lol

[split knot]

exactly ;_;

this occurs 4 times in the groups notes alone

[dawn kiln]

yeah that can be annoying if you just want to get through it

[split knot]

that's just because it's not examinable but like, other lecturers include stuff like that for interest

sometimes there's whole sections on non examinable but cool stuff

right anyway lol I wanna see how far we can get through this Q before I need to go to bed

ok so

To get the cartesian product of A and B in part b, I'm presumably gonna need abs(A)* abs(B)=abs(A* B), then because A=B, abs(A) divides abs(B) aaaaaaaaaand I've said a lot of things but idk how to actually use said things

[dawn kiln]

the cartesian product is just the set of ordered pairs from the two sets

if X={a, b} and Y={c, d}, then X x Y = {(a, c), (a, d), (b, c), (b, d)}

|X x Y| = |X||Y|, which you should be able to see somewhat intuitively

so there are 9 elements in A x B

[split knot]

ok ye I see that

so AxB={(e,e), (e,x), (e,x^2), (x, e), (x, x), (x, x^2), (x^2, e), (x^2, x), (x^2, x^2)}

I think

maybe

[dawn kiln]

yep

so, what does lagrange's theorem tell you about the size of AxB's subgroups

[split knot]

well

the sizes of A and B divide each other

[dawn kiln]

don't get bogged down with A and B

|AxB| is 9

if H is a subgroup of AxB, what are the possible sizes H can be?

[split knot]

well |H| divides 9 so |H| is 1, 3 or 9

[dawn kiln]

nice

[split knot]

is it correct then that the subgroup list is just [every combo of 1 element], [every combo of 3 elements], [the 9 element combo]

[dawn kiln]

wdym by "every combo of 1 element"

[split knot]

ah sorry

I mean like {(e,e)}, {(e,x)}... then {(e,e), (e,x), (e,x^2)}, {(e,x), (e,x^2), (x, e)}... then {(e,e), (e,x), (e,x^2), (x, e), (x, x), (x, x^2), (x^2, e), (x^2, x), (x^2, x^2)}

[dawn kiln]

an important thing with lagrange's theorem is that the converse isn't truee

[split knot]

err if that makes any more sense lol

[dawn kiln]

"If H is a subgroup of G, then |H| divides |G|"

the converse doesn't hold though: "If |H| divides |G|, then H is a subgroup of G"

in your list, {(e,x)} is not a subgroup, because it doesn't contain the identity (e, e)

lagrange only tells you if a set isn't a subgroup

[split knot]

oh wait so

ok

{(e,e)} and {(e,e), (e,x), (e,x^2), (x, e), (x, x), (x, x^2), (x^2, e), (x^2, x), (x^2, x^2)} are the only valid order 1 and 9 subgroups because only they have (e,e) in them

...but by that logic there's like a l o t of order 3 subgroups, I think

[dawn kiln]

there are a lot of candidates for order 3 subgroups

84 of them

well, technically only 28, as we know that they have to contain (e,e)

[split knot]

that's not with the (e,e) restru

ye

you posted before I finished lol

that;s a lot to write out

please tell me there's a shorthand/something I've overlooked ;_;

[dawn kiln]

so remember the other criteria for groups

they need to be associative, invertible and closed

the third one is the most helpful here for ruling out potential options

[split knot]

that being with a and b being in AxB, a ∘ b is also in AxB

[dawn kiln]

just write ab

[elder valley]

all of the order 3 subgroups are cyclic since they are prime order. so you can just look at each element

[dawn kiln]

so <(e, e)> = {(e, e)}, <(e, x)> = {(e, e), (e, x), (e, x^2)}, etc

[split knot]

is that notation the same as the symbols around a and b in the question?

[elder valley]

throwing out the identity, you have 8 elements that generate subgroups. some of them will generate the same subgroup

[dawn kiln]

<x> is defined as all of the elements that can be generated by repeatedly multiplying by x

so <x> = {x^0, x^1, x^2, ...}

(where x^n is just x∘x∘...∘x)

[split knot]

ah so here I'd just go <(e, e)> = {(e, e)}, <(e, x)> = {(e, e), (e, x), (e, x^2)}, <(e,x^2)> = {(e, e), (e, x^2)}, and then <(x, e)> and <(x^2, e)> are sets 2 and 3 but flipped

(saves me writing them out if you get me)

[dawn kiln]

careful though

because if we look at <(e, x^2)>

we have (e, x^2)^0, (e, x^2)^1, (e, x^2)^2

which is (e, e), (e, x^2), (e, x^4) = (e, e), (e, x^2), (e, x)

[split knot]

ahh sorry I misinterpreted your notation, right ye so the multiplication applies to the whole of (a,b) not just the part that's x

mb

[dawn kiln]

ye

(a, b)(c, d) = (ac, bd) in this case

[split knot]

yeah

[dawn kiln]

or rather, that's how multiplication is defined in the cartesian product of groups

[split knot]

so because the group is cyclic, (e, x^2)^2=(e, x^4) would cycle through (e, e) to (e, x) right? That's how you got to (e, x)?

am I right in thinking it's like going from (e, x^2) back to (e, e) then (e, x) that is

[dawn kiln]

yeh

if you know how mod arithemetic works, this is the same idea

4 = 1 mod 3

[split knot]

yeah because it's like going 3 steps through 4 (1,2,3) then ending up at the first step again (1)

that's how I think about it anyway

so to correct myself (hopefully lol):

<(e, e)> = {(e, e)}, <(e, x)> = {(e, e), (e, x), (e, x^2)}, <(e,x^2)> = {(e, e), (e, x^2), (e, x)}, then <(x, e)> and <(x^2, e)> are sets 2 and 3 but flipped

[dawn kiln]

sets also have no ordering on them, so sets 2 and 3 are the same

[split knot]

oh true

which would mean the latter two are also the same

(<(x, e)> and <(x^2, e)>)

[dawn kiln]

yep

[split knot]

so actually I only have 3 order 3 subgroups

[dawn kiln]

where'd the 3rd come from?

[split knot]

oh wait I included (e,e)

2*

[elder valley]

there are more than that

[split knot]

1 order 1 (identity), 2 order 3 ({(e, e), (e, x), (e, x^2)}, {(e, e), (x, e), (x^2, e)}) and 1 order 9, no?

[elder valley]

you only checked 4 out of the 9 elements to see what subgroup they generate

5*

[split knot]

ahhhhh I forgot the ones without e in them

smh

[dawn kiln]

well have you done them now lol

[split knot]

for some reason I chose to do this in notepad

uh ignore the 1's

[elder valley]

perfect

[split knot]

sexy

that was a whole lot of tedious writing for a 5 mark question though

is there no way of speeding that up? Or like some notation that summarises it?

[dawn kiln]

yeh, i mean with practice you'll get quicker

[split knot]

(well, the Q asked for a "list" so maybe not)

true

plus I was writing on keyboard not pen and paper

[dawn kiln]

also, once you're more used to the idea of cyclic groups and how they're generated, you'll spot straight away that things like <(e, x)> and <(e, x^2)> are the same group

[elder valley]

yes, in a group of prime order, any non-identity element will generate the whole group. so <x> = <x^2> in your case, which implies <(x,e)> = <(x^2,e)>

[split knot]

yeah, now that we've gone through that question it seems pretty trivial to see how to cycle through the set now

like well yeah of course x^4 leads round to x

[dawn kiln]

so for this next question, do you know what the order of an element in a group is

[split knot]

that's the same as the period right? I've heard that

[dawn kiln]

yep

so what's the order of (e, x) in the previous question?

[split knot]

(e,x)^1=(e,x), (e,x)^2=(e,x^2), (e,x)^3=(e,x^3)=(e,e) so 3

[dawn kiln]

so an immediate result from lagrange's theorem is that the order of any element |g| divides the order of the group |G|

[elder valley]

the order of an element is also equal to the order of the subgroup the element generates

[split knot]

the order of (e,x) being 3 means that ye it divides the order of the group, 9

and (e,x) generated a subgroup of order 3, so that works too

[dawn kiln]

yep

[split knot]

(side note: sick album choice maxim haha)

[dawn kiln]

lmao thanks

[split knot]

ok so now we test with (e,x)^2 I suppose

[dawn kiln]

v chill listening late

[split knot]

for sure

[dawn kiln]

i mean we're doing question c

[split knot]

bit more chill than TPAB lol

ah well I mean for like the understanding of what c is asking

[dawn kiln]

ah right

[split knot]

oh actually it probably doesn't want examples from b when it says "give examples"

nvm back to generality lol

[dawn kiln]

so i would start by assigning some letters

say |g| = n, |G| = m

[split knot]

alright

[elder valley]

it doesnt say G is finite

[split knot]

just that it's some group

[dawn kiln]

ah shit true

[split knot]

Lagrange's theorem needs it to be finite, that seems like an issue

[dawn kiln]

it's still worth doing the finite case though

[split knot]

can we then just extend the conclusion to infinite groups?

or would it require another method

[dawn kiln]

just thinking it through

i mean i think it's actually reasonably straightforward

so the order of g is the smallest n such that g^n = 1

suppose n is finite

and consider (g^2)^n = (g^n)^2 = 1^2 = 1

so the order of g^2 is at most n

so no lagrange required actually

[elder valley]

yeah the order of G doesn't really matter because you're only working in the subgroup <g>

[dawn kiln]

should probably go to sleep, it's half 3 where i am lol

[split knot]

same for me^^

[dawn kiln]

but yeah, hope some of that was helpful

[split knot]

ah why does g^n need to be 1? why not e?

[dawn kiln]

two main bits of notation that are relaxed with groups

a o b is just written ab

and the identity is just written as 1, unless there's a good reason to otherwise (say when talking about addition)

[split knot]

notation virgin vs group theory chad

gotcha

[dawn kiln]

anyway, night dude

[split knot]

night mate, thanks so much for your help! 😄

[dawn kiln]

nw, hope it helped clarify things a bit

[split knot]

definitely did

o and thanks to @elder valley and @chilly ocean heh

[elder valley]

no problem

[split knot]

I'll definitely be here tomorrow with more from this paper so keep an eye out

[split knot]

and consider (g^2)^n = (g^n)^2 = 1^2 = 1
hm so by this logic the order of g and of g^2 is at most n, but I'm asked to give an example where the order of g > the order of g^2

[elder valley]

right

[split knot]

i.e. if they both have the same order, how can one be greater than the other?

[elder valley]

if |g|=n, then that equation shows |g^2| <= n

because order is the smallest positive integer that gives 1

[split knot]

I'm not sure I understand why it has to be "at most 1"

is it just that we don't know what we're squaring so somehow it might be that the order of g^2 might be <n?

[elder valley]

because all you've showed is that raising g^2 to the power n gives 1. there could be a smaller number that also gives 1

it will be more clear when you see an example. so try looking at the groups you know of and see if you can find one

i think that was part of the problem anyway

[split knot]

turns out it's quite hard to find a > example from part b ;_;

[elder valley]

true, that one doesn't work

[split knot]

eh

it's 4:30, think I'll sleep on it and hope I find an example tomorrow

[elder valley]

it shouldn't be too hard to find one. get some rest

[final dove]

@golden pasture is it possible to fix the exact sequence in (https://math.stackexchange.com/questions/3704264/exact-sequence-of-ideals-intersection-product-direct-sum-sum) wrt g?

[leaden finch]

anyone know calc 3

[final dove]

Is there a good reference for the representation theory of the symmetric group "from the ground up"?

I want the book to ideally assume I don't know what a representation is, and end with me knowing how to compute the irreducibles of the representation of Sn

[rich patrol]

@leaden finch wrong channel #calculus

[golden pasture]

@final dove i doubt so tbh

[final dove]

@golden pasture is it because IJ is "multiplication-like", which does not work well module-wise?

like, what is your intuition on this

[golden pasture]

not rlly like mapping I intersect J to I oplus J is kind of unnatural

going from a single element to 2 eleemts

[final dove]

hmm

but we do use f: R -> R oplus R; f(r) = (r, r) often enough right?

so that didn't seem too weird to me?

[golden pasture]

hm true

actually
maybe can try (a,b)->a-b

but then the image of f is not in the kernel of g

since g has trivial kernel

[final dove]

a - b is the same as a + b, no?

[golden pasture]

the kernel is not

[chilly ocean]

Let Z[X] be the ring of polynomials with integer coeficients.
How is the ideal (2, X) of Z[X] defined?
I don't understand this notation.

[upper pivot]

elements of the form 2a+Xb with a,b \in Z[X]

[chilly ocean]

Ty!

[steep hull]

Try proving it’s not principal to see if you understand

[chilly ocean]

I can't prove it

That's what I've done @steep hull

[solemn rain]

@chilly ocean

what do you need help with proving

ping mr

me

[upper pivot]

@chilly ocean what you want to do is use the other inclusion

[cloud walrus]

JohnDoeSmith:

[chilly ocean]

JohnDoeSmith:
@upper pivot
I see, and proving this implication as false would be it, right?

[upper pivot]

yeah show both cannot happen

[chilly ocean]

So, using my notation:
d(x) = c0 + c1·x + ... + cn·x^n
Therefore:
2 \in d(x) => c1 = 0
x \in d(x) => c1 ≠ 0
Therefore we reach a contradiction.

Is this it?

[upper pivot]

the first one is wrong

x \in (d(x)) doesnt mean c1 != 0

particularly x\in (1)

(but thats the only exception, otherwise its fine)

[chilly ocean]

I don't get you.
My reasoning is as follows:
x \in d(x) = c0 + c1·x + ... + cn·x^n => c0 = 0, c1 ≠ 0, c2 = ... = cn = 0
I'm treating the inclusion as an equality so surely thre's my mistake, but I don't understand

[upper pivot]

umm (d(x)) is the ideal generated by d(x)

so its not equality

[chilly ocean]

So I'm saying x is a multiple of c0 + c1·x + ... + cn·x^n

Therefore c1 != 0

Because if c1 = 0, a multiple of c0 + c1·x + ... + cn·x^n will never be x

[upper pivot]

well as i said there is one exception, i.e 1

[chilly ocean]

Got it!
2 = d(x)·p(x) => d0·p0 = 2, d0p1 + d1p0 = 0
x = d(x)·q(x) => d0·q0 = 0, d0·q1 + d1·q0 = 1
Which is a contradiction

I was misunderstanding the (d(x)) concept

Thanks @upper pivot

[upper pivot]

uhh whats your contradiction, btw you could just look at divisors of 2 and divisors of x

[left monolith]

||The contradiction is that d(x) has to be ±1 and therefore ±1 = 2a(X) + Xb(X) with a(x),b(x) \in Z[X] => ±1 = 2a0 which is a contradiction because a0 has to be in Z||

Is that reasoning correct?

[upper pivot]

yeah thats the idea i was going for

[left monolith]

Thanks!!

[upper pivot]

btw you should spoiler that/delete cause i want him to work it out too

[vital quail]

@chilly ocean you can see this easier as follows: 2 in (d(x)) means 2 = P(x)d(x) for some P, and then 0 = deg P(x) + deg d(x) => deg d(x) = 0. and then since x in (d(x)) it must be +-1

[left monolith]

Nice alternative reasoning, thanks

btw you should spoiler that/delete cause i want him to work it out too
@upper pivot
Omw sorry

[split knot]

For this question, I've got an answer which is correct (99% sure), but my mate's method was different to mine, he wrote out a massive grid and somehow made conclusions from it whereas I used mod logic

I don't think I understand how to instantly draw the conclusion that 3 and 15 are irreducible from that mess of a table - are both methods equally valid? If so I'm going with mods 👀

[stone fulcrum]

Christ please don't write out a grid

It's pretty obvious which are which lol

It's a unit if it's coprime to 18

[split knot]

Ye I understand how you'd quickly find the units and non-units

I think the grid is for the purpose of finding irreducibles, though

[stone fulcrum]

You know, I said it was obvious, but now I'm having trouble sorting them out lol

I'm sure there's a few ways to make this quick. Let me think

[split knot]

sure

the units are easy, {1,5,7,11,13,17} and non-units {2,3,4,6,8,9,10,12,14,15,16}, I like the modulo method for separating (ir)redecibles I think

[stone fulcrum]

Quick way to get a few is the mapping Z → Z/18Z. Anything irreducible stays irreducible

By "a few" I mean "literally just 3 lol"

Actually also 2

[split knot]

are you saying you found 2 to be irreducible?

[stone fulcrum]

Oh wait fk no I did my logic backwards

[split knot]

should be 3 and 15 pretty sure

[stone fulcrum]

Things will stay reducible

Which, duh lol

[woven delta]

What do you mean "things will stay reducible"?

I don't think that's quite true

[stone fulcrum]

Hah you're right, factors become units

[elder valley]

@split knot do you know about maximal ideals and quotient rings?

what theorems do you have available to use?

[sterile garden]

How do I give the addition and multiplication table of a field having 8 elements?

[cloud walrus]

IamDerek:

[sterile garden]

I'm not sure how I'd even come up with that hint to begin with :/

[bleak abyss]

IamDerek: the field of 8 elements is the splitting field of that polynomial over F_2

[sterile garden]

@bleak abyss So can I just say that 'a' is a root of the above polynomial in F_2 and use elements 1,a,a^2,...,a^6, as my non-zero elements?

[bleak abyss]

Well, that's using one representation of the elements

But then the addition table becomes trickier

[sterile garden]

Yeah I'm seeing that

[bleak abyss]

Try factoring this first and see what happens

[sterile garden]

I get x(x^7-1)

Or do you mean to factor into all linear factors?

[bleak abyss]

Well, you won't do it into all linear factors

But factor over F_2

Into irreducibles

Alternatively

I think you can cite a theorem which says something like

x^{p^n} - x is the product of all monic irreducible polynomials whose degree divides n

Over F_p

So in this case

x^8 - x = x(x-1)(x^3 + x^2 + 1)(x^3 + x + 1)

If you didn't wanna cite this theorem you'd pull out a factor of x and x-1, and then you'd say okay, this polynomial is separable so there are no more linear factors. Well, the guy left over has degree 6 so that splits either into 3 quadratic factors or 2 cubic factors. But there aren't 3 irreducible quadratics over F_2

So it has to break down this way

The point here is that you can write F_8 = F_2[x]/(x^3 + x + 1)

Or F_8 = F_2[x]/(x^3 + x^2 + 1)

And now you have something concrete you can work with, just write out that

Makes sense?

[sterile garden]

I don't know how to factor polys over finite field :/

😅

[bleak abyss]

I mean so, I gave the sort of sneaky ways to do it using higher up facts

In one case you're really just using that x^8 - x is separable over F_2, which is in principle a computation

And then just being clever

In another you're using a bit of a more high powered fact that x^{p^n} - x is the product of monic irreducibles over F_p of degree dividing n

But if you don't want either you can just do it by force

Like okay linear polynomials over F_2 are just x and x-1

[sterile garden]

I haven't covered separable polynomials yet. Does that just mean it's reducible?

[bleak abyss]

Separable means that in an algebraic closure, it doesn't have multiple roots

Equivalently, coprime to its derivative

But yeah so there's a direct way

Let's start from 0

The linear polynomials over F_2 are just x and x-1

Both are obv irreducible, not much to say there

Okay let's look at quadratics

Well, if a quadratic polynomial factors, it has a root. But the possibilities for the root here are just 0 and 1

So really you can just say hmm, 0 is a root of a polynomial over F_2 iff the constant term is 0

1 is a root of a polynomial over F_2 iff the number of terms is even (since it's just 1 + ... + 1 mod 2)

So okay we need a quadratic with non-zero constant term and an odd number of terms. x^2 + x + 1

That's our only choice

Same reasoning for cubics

[sterile garden]

Wait

Okay, so you're saying that x^2 + x + 1 is irreducible over F_2?

(I think that's what you're saying anyway)

[bleak abyss]

Yes

The proof is

[sterile garden]

try 0 try 1 lol

[bleak abyss]

Assume it factored, it'd have a root

0 isn't a root, 1 isn't a root

gg

And that's the only irreducible quadratic over F_2

Now let's see cubics

x^3 + ax^2 + bx + c where (a,b,c)\in F_2

Well again, the cubic is irreducible iff neither 0 nor 1 are roots. Well, for 0 not to be a root, c=1

For 1 not to be a root, the number of terms must be odd. So either a=0 and b=1 or vice versa

So the irreducible cubics over F_2 are x^3 + x^2 + 1 and x^3 + x + 1

[sterile garden]

Ah

So then I get x(x-1)(x^3+x+1)(x^3+x^2+1)

and I've factored that polynomial into irreducible factors in F_2

[bleak abyss]

Yeah to factor x^8 - x, in principle the reasoning would be like

Okay without even thinking you can turn it into x(x-1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)

[sterile garden]

YEah

[bleak abyss]

And now the idea is to say hmm, well if I factor a degree 6

I either will factor it into 1+5, 2+4, or 3+3

This guy doesn't have a root so 1+5 is out

[sterile garden]

It's either a 1/5, 2/4, or 3/3

[bleak abyss]

Well, in the 2+4 case, does x^2 + x + 1 divide it? You can try it and see that the answer is no

So it's gotta be 3+3

[sterile garden]

So then it must be a 3/3 if it's reducible.

[bleak abyss]

Yup

So you can just try stuff until you're done

That's kinda the long way but it works

[sterile garden]

Okay, so back to constructing this field of 8 elements

I can take F_2[x] and quotient out by one of my degree 3 irreducibles

[bleak abyss]

Yup

And then do arithmetic

[sterile garden]

Ahhhhh

In the previous problem they ask for doing something similar, but with x^4-x over F_2. I now see I did it not as nicely 🙂

Thanks for your help

[kindred mist]

Please correct me if I'm in the wrong chat for this, but is there a canonical way of generalizing the notion of a norm on a vector space, to any algebraic structure? Namely one that can be embedded in a structure that has an identity in each operation (I suppose, one that is a monoid in each operation in particular) (idk if it would be say: triangle inequality and positive definite in a fixed operation for instance), I mean to include even when the structure only has 1 operation. What about for 2 operations but not necessarily a vector space (I assume the canonical definition can just be "grabbed and repurposed" in that case, no?)

I just figure if I try to employ something like this ^^^ I should first decide if the canonical definition fits my purposes, then stray from that if not.

[split knot]

Sorry to cut you off, hoping this'll be a quick one - can anyone help out with this?

I have conclusions for A and B but I'm not 100% with them

[acoustic pine]

Am I doing something wrong here?

Its in the field of quaternions, where F has characteristic 3

Idk how to go from what I have to "correct" values in F

[stone fulcrum]

Consider Z2×Z2, which has a normal subgroup Z2. It's clear that Z2×Z2/Z2 = Z2. Here we take the quotient of a non-cyclic and get a cyclic

[sterile garden]

Wouldn't 1/2 in R/Z under addition be an element of order 2?

Or any n for that matter

[stone fulcrum]

R/Z is best thought of as the circle. Going half way around is indeed an element of order 2

And yeah by that logic you can get any n order

[kindred mist]

shouldn't it be Z2 x Z2/ 0 x Z2 ?

[sterile garden]

I think that's the same thing.

[kindred mist]

ya there isomorphic, I was just asking for clarification

since we need a subset to mod out

[stone fulcrum]

Both notations are cheating in the sense that "we take whatever the subgroup is, we'll just call it Z2"

[kindred mist]

ya, makes sense, I like that notation, I just wanted to be sure I wasen't missing something

[split knot]

this is what I did for B and then A, seems I might've overcomplicated 👀

didn't do C or D yet

at least, my A is certainly over complicated

[stone fulcrum]

No need to explicitly compute Z2×Z2/Z2, you know it's a group of order 2, and thus must be Z2

[split knot]

yyyep that makes intuitive sense

plus it's a lot less work

[stone fulcrum]

Oh, and Z2 is a normal subgroup cuz abelian

[split knot]

ye

well, is it that Z2 is cyclic => Z2 is abelian => Z2 is normal?

[stone fulcrum]

Z2×Z2 is abelian → all of its subgroups are normal → Z2×Z2/Z2 is defined and is Z2

[split knot]

ok, gotcha

How would you show statement C? I'm not sure how to extend this to any n

R/Z is best thought of as the circle. Going half way around is indeed an element of order 2

(tbh I'm not sure I can picture it as a circle)

[stone fulcrum]

Basically R/Z is the fractional part of any number. Go past 0.99999 and you get back to 0.0001, where the circle loops

Anyway the element of order 4 is 1/4, because 1/4 + 1/4 + 1/4 + 1/4 = 0

[elder valley]

Just extend your method for order 2 to order n. It's the same idea

[stone fulcrum]

And you can keep going with that pattern

I shouldn't say "the", there's more than one.

[split knot]

so 1=0
1/2+1/2=0
1/3+1/3+1/3=0
.
.
.

[stone fulcrum]

Yaya

Easy way to generate some element of order n

[split knot]

I feel a sum notation coming on lol

I guess that's how you'd give a single statement to show it

[elder valley]

I shouldn't say "the", there's more than one.
@stone fulcrum is there?

[stone fulcrum]

Ya, 3/4 is also an element of order 4

Think generators of Zn

[split knot]

sum from 1 of n/n+1 I think?

I feel a sum notation coming on lol

[elder valley]

Oh, yeah you're right

[stone fulcrum]

Rotating the opposite direction pretty much haha

[split knot]

oh actually maybe I don't need to sum because I just need to show that sequence of adding fractions goes on forever, I feel like induction might be too much for the marks tho

[elder valley]

@split knot just do it for general n. What's an element of order n?

[split knot]

n(1/n)...?

well so n/n

which makes sense because using the "1=0" logic, it loops back round so anything summing to 1 is 0

[elder valley]

You're confusing the element itself with the procedure for checking it's order

[split knot]

oh true

so the element 1/n has order n

[elder valley]

That's it

[split knot]

oh that does it, yeah 😄

ok nice

alright so for the last statement

not sure I recognise Ln(R)

[stone fulcrum]

GLn(R)
Is the general linear group of size n real matricies

So n×n matricies with real entries

[split knot]

So presumably S in SLn(R) means it's a subgroup

[golden pasture]

S=special

Determinant 1

[stone fulcrum]

SL is the special linear group. Same thing, except only matrices with determinant 1

[split knot]

ah

[stone fulcrum]

Maybe you've proven in class that SL is normal in GL?

[split knot]

is this what we're looking for?

[stone fulcrum]

Yes exactly

[split knot]

This was a while ago so I'm not sure what that <-but-not-< means

oh nvm tiredbrain

normal

this is the part of group theory where I started to not understand much

I understand them equating det(ABA^-1) = det(B) = 1 but I don't see why it's useful to know ABA^-1 is in SLn(R)

[elder valley]

It's showing SL is normal using the definition

[stone fulcrum]

Definition of normal. If you take an element of SL, and conjugate it with any other matrix, you're still in SL.

[split knot]

ohh from this

but matrix-ised

[stone fulcrum]

SL is closed under conjugation and thus normal

[split knot]

so this subgroup SLn(R) is normal => GLn(R) is abelian?

[stone fulcrum]

Matrix multiplication is not commutative! So GLn(R) is not abelian

SL is, in this sense, special

[split knot]

sorry, I might be tiredbraining

but where is commutativity relevant in what we;ve done so far

[stone fulcrum]

Abelian and commutative are really the same word

For groups, anyway

[elder valley]

Isomorphic you might say 😂

[split knot]

I'm still a lil lost
like, where exactly has we shown that GLn(R) isn't abelian? My only thought it that we kinda rearranged the determinants however we wanted because we knew det(B)=1 and that det(A) and det(A^-1) would cancel each other out so it feels a bit cheaty to move them like that

[stone fulcrum]

det(A) is a real number, not a matrix

And of course, real numbers are commutative

[split knot]

ok, true

[stone fulcrum]

Matricies themselves don't commute. In other words, GLn(R) is not abelian

[split knot]

oh hang on sorry

right so

big picture - because GLn(R) is a group of matrices, it doesn't have commutativity by definition

so even if you take out all the det 1 matrices, you've still got a group of matrices, so it still isn't commutative

[elder valley]

You haven't really shown anything yet. The question is asking about the quotient group, not GL

[split knot]

so even if you take out all the det 1 matrices, you've still got a group of matrices, so it still isn't commutative
this is what we're after no?

if I'm understanding what a quotient group actually is, there

[elder valley]

Quotienting isn't really taking out the subgroup, it's collapsing it to a single element

[stone fulcrum]

Yes SL is also not an abelian group

[elder valley]

You can quotient a nonabelian group and end up with an abelian one

[split knot]

;_;

[stone fulcrum]

Take a group X, and one of its normal subgroups Y. Then the cosets generated by Y themselves form a new group we call X/Y

Where the group operation is coset multiplication

[solid drum]

Specifically take a group $G$ and mod out by the commutator subgroup $G'={xyx^{-1}y^{-1}\mid x,y\in G}$

[cloud walrus]

doctorkelp:

[solid drum]

The commutator always exists, but it isn't always interesting 😛

[split knot]

my brain hurts

group theory's definitions make my life harder than it needs to be ;_;

[golden pasture]

you'll get comfortable with the definitions after playing with examples

[solid drum]

I would sincerely recommend thinking about symmetric groups

they'll help you understand groups at large in my experience

[elder valley]

You have 2 ways to approach the question: if you think it's true try to prove it, if not find a counterexample of matrices not commuting in the quotient

[solid drum]

Braid groups are a pretty place to go after that 🙂

[split knot]

ehhh I think I'll go back to calculus 😅

[solid drum]

There are groups in calculus, too 🙂

[split knot]

ohno

pleaseno

[solid drum]

Solutions to homogenous linear differential equations form a group (underlying a vector space) 😛

[split knot]

next you'll tell me that ring theory is useful in ODEs

[golden pasture]

you'll eventually feel comfortable with definitions in abstract algebra after working through examples and theorems

it takes time

[split knot]

yeah I'd imagine i will

[solid drum]

Yeah it actually is depending on how you'd like to approach the theory.

[split knot]

took a while to get used to "clopen" being a thing

[golden pasture]

yea point set topology is similar too

[split knot]

right sorry I inadvertently derailed a bit - so using that neither GLn(R) nor SLn(R) are abelian, can I say that the resultant quotient group also isn't abelian? Or does this mean that there's examples where non/non -> abelian?

You can quotient a nonabelian group and end up with an abelian one

[stone fulcrum]

There are examples

[split knot]

this question has driven me insane lol

I don't think I can actually get an answer to this

or well like

you said it's non-abelian but I still don't fully understand why

[solid drum]

What's the simplest group that isn't abelian you can think of?

[stone fulcrum]

So abelian and commutative mean the same thing

An abelian group does xy = yx for any x and y

[split knot]

I guess just (G, ∗) with a,b in G where a∗b =/= b∗a

[solid drum]

Yes, but What is G 🙂

[scarlet estuary]

thats the definition of a nonabelian group

not an example

[split knot]

uh

[golden pasture]

what types of groups do you know of

[split knot]

my brain has imploded I think

[golden pasture]

maybe what is a example of a finite group

[split knot]

I guess a cyclic group would be finite

so like C2

[golden pasture]

ok yup cyclic groups are a example of finite groups

[solid drum]

and don't go crazy! You might want to consider counting up from a group of order 1, 2, 3, and ask about commutativity (abelian)

I'll give you a hint, you'll hit an example of order where things get interesting before 10 🙂

[golden pasture]

<i saw it mentioned here>

[elder valley]

Maybe, depends if he's seen such a group yet

[split knot]

aren't all cyclic groups abelian tho?

[golden pasture]

yup

[solid drum]

Soooo do you know of a group that isn't cyclic, and is finite?

[split knot]

oh I guess Z2 x Z2

[solid drum]

Nice. So you're familiar with products.

[split knot]

guess so lol

[golden pasture]

theres a whole family of groups that are finite and mostly nonabelian

[split knot]

so we've established that there are non-cyclic, finite groups

[solid drum]

Yes, and in particular we're hinting at that there are non-commutative (non-abelian) finite groups

[stone fulcrum]

I can't imagine trying to guess these if you don't know haha. The smallest non-abelian group is S3, it's of order 6

[golden pasture]

uh

sh

spoiler

lol

[solid drum]

Haha

[split knot]

symmetric?

[stone fulcrum]

Then all Sn where n≥3 are non-abelian

[golden pasture]

yes

[elder valley]

Like he said, he can't really guess it if he doesn't know. And it's not even the question he's trying to answer

[solid drum]

Maybe more familiar is that $S_3$ is the same as $D_6$ or the group of symmetries of an equilateral triangle.

[cloud walrus]

doctorkelp:

[split knot]

I recognise S and D

[stone fulcrum]

I definitely didn't learn that fact by guessing it

[split knot]

think we looked at them last year though

[solid drum]

So try this experiment out

Let $G=S_3,$ that is, the set of all bijections from ${1,2,3}\longrightarrow{1,2,3}$, and compute $G'={xyx^{-1}y^{-1} \mid x,y\in G}.$ Show that $G'$ is a normal subgroup of $G.$ Find its quotient.

[cloud walrus]

doctorkelp:

[solid drum]

If this isn't helpful anyone here can tell me to gtfo 😂

I just haven't talked about math in a red hot minute and I'm really excited.

[elder valley]

I think he should focus on his assignment right now lol

[stone fulcrum]

No no I get it haha. Group theory is bae

[split knot]

kelp

that looks like it'd probably be useful to do

however 😛

I have 3h left to do this and a couple more Q's, I need to sweat

[stone fulcrum]

You have MORE?!

[solid drum]

Didn't know the context.

[split knot]

well

[solid drum]

Sorry!

[split knot]

the one in #advanced-number-theory

np kelp, I'll look into what you posted tomorrow when the deadline's gone

but ye no more group theory questions after this I think

so uh

GLn(R)/SLn(R), huh

that's a group

[solid drum]

Yep!

[split knot]

lol

[stone fulcrum]

Let's just keep it basic. GLn(R)/SLn(R) is what happens when you take every element of SLn(R) and say "why don't we just make that the identity?"

[split knot]

I reeeeally want to just say that they're both groups of matrices which are non-abelian, so taking a quotient doesn't really change that every element is still a matrix

uh

[stone fulcrum]

Actually haha

Funny you should say that

[solid drum]

Retro, try it for GL2 and SL2 first

Get a good reason there.

[stone fulcrum]

The problem is that Retro doesn't know how to take a quotient, and this isn't a good first example

[split knot]

^

[solid drum]

Ohhhhh

[split knot]

I'm not entirely sure what a quotient actually does

except for when Auvera said it collapses stuff into one element

[solid drum]

Yes

[split knot]

hang on you said it kaynex lol it's just setting everything to the identity ye?

[solid drum]

You want to understand the kernel of a homomorphism

[stone fulcrum]

Take a group X, and one of its normal subgroups Y. Then the cosets generated by Y themselves form a new group we call X/Y

[cloud walrus]

doctorkelp:

[solid drum]

That's intuitively true.

[stone fulcrum]

There's two ways to think about this process lol, which is the first isomorphism theorem

[sturdy marsh]

Speaking of abelianizations, I'm trying to abelianize $G(Q_p^{tame}/Q_p)$. I'm looking at the tower $Q_p \subset Q_p^{un} \subset Q_p^{tame}$, which gives the s.e.s $1 \rightarrow G(Q_p^{tame}/Q_p^{un}) \rightarrow G(Q_p^{tame}/Q_p) \rightarrow G(Q_p^{un}/Q_p) \rightarrow 1$. I'm trying to show that it is right split and then write it as a semidirect product. My guess is a lift of the Frobenius elt. maps isomorphically onto $G(Q_p^{un}/Q_p)$. Is this true?

[cloud walrus]

Brofibration:

[split knot]

;_;

[stone fulcrum]

So the determinant happens to take matrices of SL to the identity of which group?

[split knot]

well det = 1 in SL

so

uh

the group that has numbers in it 👀

[stone fulcrum]

That's fair! So the determinant maps GL into multiplication on the reals. It also maps SL into the identity, so this is what we want out of GL/SL

[split knot]

oh the reals

obviously yes

smh

this is what happens when I do maths at 6:30am

[stone fulcrum]

Basically if SL is squished to identity, the only remaining info about a matrix is its determinant

[split knot]

sure

well

the only remaining info about matrices with det = 1 is their det

oh I guess you meant just considering SL

[stone fulcrum]

Basically GLn(R)/SLn(R) becomes R

[woven delta]

lol

no it doesn't

[stone fulcrum]

Multiplication on R I mean. Without the 0

[split knot]

wait but what about the non-det=1 matrices in GLn?

[stone fulcrum]

No? Do I have that wrong?

[woven delta]

no you are correct that GL_n(R)/SL_n(R) is isomorphic to R^x

[mild laurel]

(Interesting related question, is there a field F and a normal subgroup N of GL_n(F) that's not SL_n(F) such that GL_n(F)/N is isomorphic to F^x)

[woven delta]

I think R should be F?

[elder valley]

@split knot any two matrices with the same determinant are equal in the quotient even though they may not be equal as matrices

[split knot]

Yeah I understand that i.e. "squishing" SL to the identity

[solid drum]

Think about elementary row and column operations having determinant 1.

[split knot]

but I mean that GL has more matrices than just those with det=1

[stone fulcrum]

That's a good way to say that, yeah.
[1 0]
[0 2]
and
[2 0]
[0 1]

Are the same element in GL2(R)/SL2(R)

[split knot]

now I thought this question was about showing that the quotient group isn't commutative

and I don't exactly see how we're going to get that from this

[elder valley]

I thought it was true or false

[stone fulcrum]

No the question asks if it is true that it is non-abelian

Which, the answer is that it is false. That is an abelian group

[solid drum]

My bad.

I think you'll have to show that $SL_2(\Bbb{R})$ is the commutator of $GL_2(\Bbb{R})$ 😓

Actually, no.

Just show that it's isomorphic to $\Bbb{R}\setminus{0}$ for any $n,$ by showing that the kernel of $\det$ is $SL_n(\Bbb{R}).$

[cloud walrus]

doctorkelp:

[elder valley]

You can really just use the definition of abelian

[solid drum]

It doesn't seem easy to me to show commutativity using the quotient group directly.

[elder valley]

It's pretty straight forward

[split knot]

can I not literally just say that GLn(R)/SLn(R) still contains matrices of det =/= 1 so commutativity still isn't achieved

pretty sure this should be a ~1 mark question

[mild laurel]

(except this is wrong because it is commutative lmao)

[bleak abyss]

Hmm, let's F is an algebraically closed field of characteristic 2

Oh wait no

Nvm nvm nvm

[split knot]

I think this is 1 mark I'm ok with dropping

if I don't understand after almost 2h of work on it it's a lost cause methinks

[bleak abyss]

Okay so uh lemme drop some random ideas Zoph

So let's say F is a field

And we have a surjective but not injective homomorphism F^{\times} -> F^{\times}

That'd do it

[elder valley]

The discussion here side tracked you too much. If you still want to work on it, try using the definition of abelian to prove the quotient is abelian

[split knot]

oh I forgot to work out examples for 1c

this one, we worked on it yesterday

[elder valley]

You should be able to find an example using small groups

[split knot]

er

how small 👀

[mild laurel]

two elements

[split knot]

oh fractions might work

I think I'm tiredbraining again tho

[split knot]

right so apparently I can find a solution in Z/pZ

@solid drum could you help me finish this one if you're around?

[golden pasture]

isnt the easiest way to define a quotient group for $N$ normal in $G$ is to define $G/N={gN|g\in G}$ with $aNbN=(ab)N$ due to $N$ being normal

[cloud walrus]

ariana:

[golden pasture]

with $aN={an|n\in N}$ and $AB={ab|a\in A,b\in B}$

[cloud walrus]

ariana:

[upper pivot]

isnt that the standard way to define quotient groups?

[golden pasture]

yea

[upper pivot]

@inland otter ok i have researched and i found a normal subgroup

basically the subgroup of functions that permute only finitely many points

[inland otter]

Ah yes that seems to work

[upper pivot]

nice, may i ask what this is for?

[inland otter]

Nothing, I was reading Algebra by Artin and got really curious

[upper pivot]

oh i see

also nice artin is my favourite introduction book to the topic

[inland otter]

It does read really nicely yes

I was wondering what happens if you look at differentiable functions on $[0,+\infty[$ * and invertible

I was wondering if Schwartz functions maybe form a normal subgroup there

[upper pivot]

oh hmm

[cloud walrus]

dadaurs:

[upper pivot]

i mean i would imagine not

because conjugation by something thats not C^infinity might make the functions not C^infinity

(but im just guessing so)

(i do encourage for you to research this further btw, analysis is definitely not my strong suit so good possibility my intuition could be wrong)

[inland otter]

No what you say does seem correct, I'm playing around with chain rules to look at what conditions I could put on the group to make it well behaved

[chilly ocean]

when you apply modular arithmetic m(x) on polynomial f(x) in a polynomial field F[x], is it just like finding the remainder of a polynomial?

like finding r(x) when f(x) = m(x)q(x)+r(x)

where deg(r)<deg(m)

[solid drum]

You mean quotienting by the ideal generated by that polynomial yeah?

[chilly ocean]

I mean like in this problem

So for instance would part i) be x^3+2 = (x^2+2)*x+(x+2), so f(x) congruent to x+2 in mod m(x)?

[solid drum]

Yep!

[chilly ocean]

ok, thanks

[shy bluff]

Why is the centre of GL_n this?

Is this because the only matrices that are commutative are diagonal matrices where each entry is the same?

[mild laurel]

I mean, that's what this statement says yes

[shy bluff]

hrm

What is R/Z?

[scarlet estuary]

where R and Z are the rings?

[olive mirage]

must be groups, since Z is not an ideal of R

[scarlet estuary]

additive groups?

$\bR / \bZ$ is the group formed by equivalence classes of reals under the equivalence relation $a \sim b \iff a - b \in \bZ$

[cloud walrus]

Namington:

[scarlet estuary]

that is to say, it's the group where you take the real numbers, and say all the reals that are "an integer apart" are actually the same

one could think of it as the additive group of real numbers where we only care about "the part after the decimal"

(as well as the sign)

alternatively, you could think of it as what you naturally get if you say "take the reals, but now all integers = 0"

it's worth noting that there's a very cool connection between this quotient group and the unit circle

(in fact its isomorphic to the complex unit circle!)

(a proof is given here https://proofwiki.org/wiki/Quotient_Group_of_Reals_by_Integers_is_Circle_Group)

[shy bluff]

Wait what do you mean by "Take the reals but now all integers = 0"?

As in all integers are in the kernel?

[scarlet estuary]

the kernel... of what?

we're not dealing with a function here

[shy bluff]

Err

Wait sorry

[scarlet estuary]

we're dealing with a group

[shy bluff]

Sorry

Yea I think I see what you mean

Like if we had a map to go from R to R/Z

The quotient map

The kernel of the quotient map from R to R/Z is all the integers?

[scarlet estuary]

uh sure

(usually the term "quotient map" is reserved for a concept from topology but yeah)

[shy bluff]

Idk

We had quotient map in this course

[scarlet estuary]

oh interesting

[scarlet estuary]

i mean formally yeah this is a quotient map

i just dont usually see that terminology in an intro algebra class

it works though

[shy bluff]

Oh ok

Another question: If N is a normal subgroup of G, does that tell you antyhing about the number of subgroups of G that contain N?

[scarlet estuary]

not based off that information alone, no

there's stuff like the sylow theorems

[shy bluff]

What information do you need then?

[scarlet estuary]

cardinality of N and G would let you make arguments from sylow

[shy bluff]

Do indices have relationship to normal subgroups?

[scarlet estuary]

if N is a normal subgroup of G, then the index of N is the order of G / N

[left monolith]

if N is a normal subgroup of G, then the index of N is the order of G / N
@scarlet estuary
Lagrange Theorem

[scarlet estuary]

this is broader than lagrange's actually

lagrange's only applies to finite groups

but this statement is true of all groups

[left monolith]

Mmm.. you're right

[scarlet estuary]

but youre right that there's a very direct connection in the finite case

[sterile garden]

F a field, and p(x) in F[x] is irreducible if and only if p(x+c) is irreducible where c is some constant.

How do I show this?

I have that f(x) -> f(x+c) is an isomorphism

Is it enough to say that F[x]/(p(x)) is isomorphic to F[x]/(p(x+c))?

Is that even true?

[shy bluff]

What are they asking for in part b here? Are they aksing gme to show that if |G| = 4, then G is isomorphic to either either (Z/4Z) or (Z/2Z)^2? How do I show this?

[solemn rain]

can you prove that if |G| = p^2 then G is abelian?

and then use fundamental theorem of abelian groups?

@shy bluff

im p sure thats overkill tho 😦

[shy bluff]

We havn't learnt that yet

I think

We only learnt up to first isomorphism theorem?

[solemn rain]

do you know the fundamental theorem?

[shy bluff]

Parts c and d are related to what you just said though I think lol

What is f undamental theorem?

maybe I learnt it

[solemn rain]

a finitely generated abelian grouip is a direct product of just Zs

[shy bluff]

Nope

[solemn rain]

okay

[shy bluff]

This is what we've covered

[solemn rain]

can you show that if |G| =4 then G is abelian?

without the general case

( thinking about orders of elements )

[shy bluff]

How would you do that?

Oh wait question

If a group has finite order then that means that it can be generated right?