groups-rings-fields

[rapid junco]

How come?

[coral spindle]

Typically it will not act on Z(G), unless of course Z(G) = G

[coral spindle]

If I multiply something in the centre with something outside of the centre...

[delicate orchid]

z in Z(G) does not imply gz in Z(G)

[rapid junco]

ah

Im preparing for an exam, what are the bread and butter for actions to know

[coral spindle]

Conjugation

[rapid junco]

G acting on itself by conjugation, left mult,

[coral spindle]

Action on cosets

That's about it

[rapid junco]

By left mult?

[coral spindle]

Yes.

Verify for yourself that this works.

[rapid junco]

Theres also a lemma that I saw once but what does the Permutation group tell us about the existence of a normal subgroup?

it was something about dividing n!

Non trivial normal subgroup

[next obsidian]

Let G be a simple group with a subgroup H of index n

acting on the set of cosets via left mult gives you a map G -> S_n (this is the action)

Because G is simple this is either trivial or injective, it’s not trivial if n > 1 so it is injective

Thus G is a subgroup of S_n and has order dividing n!

This works for G acting on any set really, but this is the most common way to get the action

You can use this to show groups can’t be simple by finding a low index subgroup, so you have a group of order 20 injecting into S_3 or some shit

[rapid junco]

Because G is simple the kernel of phi is either G or 1.

[next obsidian]

Yes

[rapid junco]

Ah

Kernel being one is injective

[next obsidian]

Yes

[rapid junco]

If n > 1 why injective?

[next obsidian]

It can’t be G because if so the action is trivial as in gH = H always which isn’t true

Just take a g outside H, you can do this as long as H is index > 1

[rapid junco]

Ah

So low index subgroup implies generally not simple

[next obsidian]

Yeah

It’s relative

[rapid junco]

Right so this is like an extension of index 2

[next obsidian]

No

This is different

[rapid junco]

In a sense

Oh okay

[next obsidian]

That involves conjugation

There is a generalization of that

This is basically how you show a subgroup of index the smallest prime dividing |G| is normal

Or well

You do that via group actions as well

Is the point

The point is the normalizer is either H, or is G

The normalizer N, contains H

If it’s larger, then it has smaller index, but this forces it to be 1, because the index always divides the order of G (orbit stabilizer)

So if it’s larger than H it’s G, which means H is normal

So you have to rule out N = H

In that case, by orbit stabilizer the orbit of H has p elements, so acting on it gives a map G -> S_p

This gives an injection G/ker phi < S_p

Now G/ker phi can’t be trivial

This would say phi is trivial, as in gHg^-1 = H for all g

Okay, but then H is normal, which we assumed isn’t true

So G/ker phi has order dividing |G|, and dividing p!, and is not 1

Okay so, what primes can divide |G/ker phi|? Because |G/ker phi| divides |G| the primes have to be >= p

Because |G/ker phi| divides p! They have to be <= p, so the only prime that can divide |G/ker phi| is p

And then I forgot the last step

Maybe you can figure it out

Wait maybe you’re supposed to look at left mult

I forgot lmfao

Okay yeah you’re supposed to look at left mult

The point is that you get the kernel of the map G -> S_p has to be equal to H so H is normal

[rapid junco]

What is H

Oh

[rustic crown]

yep

if H has index p, then the G-set G/H has p-elements, so gives you a hom G --> S_p. the kernel of this consists of elements g such that gaH= aH for each a. in other words, g is in aHa'. which means the kernel is intersection of all conjugates of H.
the image of the map has order dividing both |G| and |S_p| = p!, so divides their gcd which is p. this means the index [G:ker] divides p, and since the kernel is completely inside H, [G:ker] = p[H:ker] so [H:ker] divides 1.

so ker = H is normal

[rustic crown]

ig this also works, since ker phi is still contained in H. here ker phi is intersection of all conjugates of the normalizer of H, but since you assume H = N at the start, the same thing is true although a lil more messier

oh det is very active

🔵

[warm wyvern]

I first read that as "oh det is very cute"

[rustic crown]

[celest furnace]

Can anyone give me a tip to showing that k -> k^a is bijective, i was able to show it when phi_1 and phi_2 are injective but not otherwise

I tried for a while by compraing the sizes of kernels but it just didnt get anywhere

[rustic crown]

det too tired to read that ><

[warm wyvern]

chad algebra arc

[tardy hedge]

Any field contains a subfield isomorphic to Q or Zp

Idk why thats cool but it is

[undone ledge]

what's a readable exposition on the quantale of ideals

~ introductory level

[tardy hedge]

I feel like im still shaky on the concept (similar one for normal subgroups) on how kernels are ideals and what the implications are

Does it mean the information on the types of ring homomorphisms that can come out of R is found in the ideals of R?

[coral spindle]

Every ideal gives a quotient; every quotient gives an ideal. That is what it means that kernels are ideals and ideals are kernels.

[undone ledge]

might be wrong but i think wlog a can be chosen to be coprime to the order of K (?)

for the finite case

[rustic crown]

okie a little less tired now, the point is that choice of a is in your hands. if you choose a bad a, this won't work for finite K.
a is only unique modulo |phi1(K)| = |phi2(K)|, and is invertible modulo that. When |K| is finite, choose a so that it's also invertible mod |K|.
this can be done, since a surjection Z/nZ --> Z/mZ induces a surjection (Z/nZ)* --> (Z/mZ)* on unit groups. this is easy to see when n is a prime power, for the general case you get this by chinese remainder theorem.

[tardy hedge]

Bro. I was just about to write something out then i realized im just stating the homomorphism theorem

So i guess it means what i was writing is correct lol

I was gonna say something like for any ring homomorphism out of R into some ring J, if the map is covering all of J then J can be written as some factor ring of R.

And .. that is just the theorem lol

[rustic crown]

seeing the word for the first time 🙈

[undone ledge]

for context i was looking at bpit -> pit

[undone ledge]

if G -> H is a surjection, is it not true that G* -> H* is a surjection?

[rustic crown]

you means R --> S?

[undone ledge]

oh right

[rustic crown]

yea, in general not true.

Z --> Z/5Z is a minimal example

since {1, -1} --> {1, 2, 3, 4} isn't surjective

[undone ledge]

aa

[celest furnace]

So at least 1 of {a + l|phi1(K)| for l in Z} is invertible mod |K|...

[rustic crown]

yep

[celest furnace]

So the general claim here is that if d divides n and a is invertible mod d, then there is some k so that a + kd is invertible mod n?

[rustic crown]

yee

[median rock]

we have a group G with |G|=p^n , p prime show that G has a normal subgroup with order p^k ,0<=k<=n. is it a good idea to show that with induction?

[rustic crown]

this is easy for prime powers, because you can choose any lift. but CRT messes up for general n.

[celest furnace]

Yes it is. Hint: for a p-group ||Z(G) is nontrivial*||

[rustic crown]

||every group is non-empty, non-trivial you mean||

[celest furnace]

Lol

[rustic crown]

[celest furnace]

Anyways this is def going on my note sheet for tomorrow

4 hour final from 8:30am to 12:30

[median rock]

is it necessary to use that Z(G) is non trivial? i mean i worked on the proof and it seems fine to me and i did not use that

[rustic crown]

det's prof said that he didn't find any significant difference between performance whether the exam was 1 hour or 3 hours. so might as well make it 1hr to make everybody's work less

[celest furnace]

To be honest i do kind of like not being rushed on time but i also dont like that its so long

I am conflicted

[rustic crown]

to induct, you just need a nice normal subgroup. doesn't matter how you get it.

Z(G) prolly the simplest?

[celest furnace]

If you use Z(G) then the proof also works almost immediately when you do it for nilpotent groups

[rustic crown]

nice = non-trivial, proper

[celest furnace]

Any chance I can have a hint on how to do this

Looking at CRT but it looks like the directions are messed up

[median rock]

i said that for m=0 e is normal then i said |G|=p^m with m<=n-1 has normal sungroup with order p^k k<=n-1 and then i show that for |G|=p^n G has a normal sungroup with order p^k+1

is it wrong?

[rustic crown]

if n = prod (p_i^a_i) and d = prod (p_i^bi)
then each Z/p_i^a_i --> Z_p_i^b_i induces a surjection on unit groups. take product and use the unit groups behave nicely with products.

the map on the products will be same as Z/n --> Z/d, cause there are no other ring maps.

1 --> 1 forces everything

(R x S)* = R* x S* is an easy check

?

[rustic crown]

and how did you do it?

[median rock]

i am do blind i read C(G)

i used Z(G)

i said G is a p-group so Z(G) is not trivial ....

thanks

[rustic crown]

:3

[celest furnace]

Was able to verify all the claims! Thanks for the help!!

[rustic crown]

[tender mist]

I don't understand, if there is a prime p such that p^k divides n but p doesn't divide d at all, we don't have the corresponding surjection to show that a is invertible mod p^k.. right?

[rustic crown]

p^0 divides d

[celest furnace]

I learned that i had messed up... What do you mean by you can choose any lift? If ab = 1 + p^mk (m < n) how would you lift a to be so that ab = 1 + p^nk? Or might b change too?

[tender mist]

Mm, although Z_p^0 is just {0}, and here 1=0, I think this doesn't allow us to derive that a is invertible mod p^k
I mean, we have the surjection Z_9 -> Z_1={0} but even though 3 is sent into 0("=1"), it's not invertible mod 9

[celest furnace]

I found something online about Hensels lemma but i think this is too complicated

[rustic crown]

ah my bad, i can't choose it arbitrarily here when the target is teh zero ring, but still on the units group the induced map is (some G) --> 1 which is always surjective and that's what we needed

[rustic crown]

Say you have Z/p^n --> Z/p^m, then if m = 0, then it's automatically surjective on unit groups as (Z/1) = (Z/1)* = {1} = {0}.

so assume m > 0, now units in Z/p^m are stuff that's non-zero mod p.

many ways to see this, one way is to notice is that it's a local ring with maximal ideal (p)

so anything outside the maximal ideal is invertible

so to see (Z/p^n)* --> (Z/p^m)* is surjective, pick anything on the right, and pick any lift using the map Z/p^n --> Z/p^m

[celest furnace]

WOW!

[rustic crown]

since the lift is automatically going to be non-zero mod p, it's unit

[celest furnace]

Thats awesome

[rustic crown]

sorry for just giving a proof sketch

<

okie i go again, will be back in 25min

oh it only took 15min

[white oxide]

why exactly is pi nontrivial? isn't F contained in IF?

[rustic crown]

if F = R, is free of rank 1, then IF = I, and certainly this is proerly contained in F=R

(if you know about tensor products, you're tensoring by R/I, so it becomes a R/I module and since tensor product commutes with direct sums, it's free)

(if not, do it by hand)

[white oxide]

hmmm ok

i'll try that, thanks

[tender mist]

Ok @rustic crown , sorry for being so slow, I get it know, this fact about non common prime factors between n and d initially surprised me because that indeed gives somewhat of an arbitrary way to actually compute the unit in Z_n which is sent to a chosen unit in Z_d, via chinese remainder theorem, after chosing arbitrary units in the Z_p^k factors in the product for primes p not dividing d

That's really nice, thank you!

[rustic crown]

[white oxide]

oh wait i'm dumb this was based off of the assumption that the ideal had a unit element which entirely defeats the purpose of ideals 💀

so now it makes sense lmfao

im' tired

[rustic crown]

yea but i should have been careful in saying "aribitrary" when p didn't divide d. If one had a unit [a] in Z/d for some a in Z, then just blindly looking [a] in Z/n doesn't work because though it gives the same value mod p for each p dividing d, for the p's that don't divide d, sadly this a could be 0 mod p. and that's exactly what caused the problem

thanks for the catch :3

[tender mist]

Ahah yes that's what threw me off originally

[rustic crown]

ig another way to prove this would be to choose k such that a+dk is not divisible by any p dividing n, this is clear for p that's already dividing d, and for the ones that don't, we choose k = (d^-1)a mod p. some CRT magic gives a nice such k. and then [a+kd] in Z/n works as a lift for [a] in Z/d

but i usually like the structure to carry most of the proof and only check smol details by hand

[icy bear]

idk if this is the place to ask this but
I'm trying to prove that a permutation is even iff the permutation has an even amount of inversions
I already proved that the sign of a permutation makes sense (aka any decomposition in transpositions is either even or odd)
but I can't get any further than that

anyone has any hint?

[rustic crown]

aka any decomposition in transpositions is either even or odd
but not both, right?

[icy bear]

yes

uniqueness of the sign

it's easy using determinants

[rustic crown]

well depends

[icy bear]

I mean, given that you know you can decompose any permutation in transpositions

[tender mist]

I agree on that view. Well, now I have to go, have a nice day/night

[icy bear]

(which I proved by realizing you can do that with a sorting algorithm)

[rustic crown]

proving det is multiplicative might or might not require that sgn is multiplicative

[rustic crown]

(makes sense)

[icy bear]

weeell you can prove that it's multiplicative for each transposition

which is kinda true by definition tbh

then since the determinant of the final matrix is the same anyways the sign has to be the same

[rustic crown]

are you multiplying the corresponding permutation matrix and calculating the det?

[icy bear]

ye

[rustic crown]

one just needs to be careful to not make any circular arguments, as det is apriori a more complicated object than sgn.

to show that det exists, you might be implicitly using that sgn is multiplicative

[icy bear]

ym like, by that 3 characteristics definition?

[rustic crown]

cause one standard (formula) definition of det is
sum of all permutations sigma of the product of a_{i, sigma(i)}}

[icy bear]

det is the only function that is linear in rows and so on and so on

[icy bear]

never heard of that one

[rustic crown]

well, one still needs existence

[icy bear]

yeah but if you prove (which is a famous result) that exchanging two rows negates the determinant it does the thingy

and I was already reading an abstract algebra book which proved that and other things

the chapter right after that was talking about permutation

and it said without proving that the sign is even iff the permutation is even

and didn't prove a permutation can't be even and odd at the same time

so I did it

now I was reading a diff geo book and was trying to do that thing of proving each theorem before reading the chapter and one of them was that the sign of the permutation is (-1)^#inversions

and this is such an interesting thing I don't want to spoil the proof

just want some hint

[delicate orchid]

so you have that every permutation can be written as a product of 2-cycles and that sign is multiplicative? or do you want to show that sign is multiplicative

[icy bear]

my current idea is trying to use the inversions as like, the first permutations you do, and then build the rest of the permutation

[icy bear]

I showed this already

[delicate orchid]

ok then we're basically done

[icy bear]

I'm talking specifically about inversions

how

[delicate orchid]

decompose your permutation into 2-cycles, 2-cycles are odd

(-1)^k is 1 if and only if k is even

[icy bear]

sure but like

how do you prove the number of inversions doesn't change

like

(1, 2, 3, 4)

it has one inversion

[delicate orchid]

what is an inversion

[rustic crown]

ig pairs i<j such that s(i) > s(j)

[icy bear]

one number that has gone to a lower index

in that case, (4, 1)

[delicate orchid]

I don't see how this is relevant

[icy bear]

4 has gone to a lower index

that's one inversion

which would mean the permutation is odd

[delicate orchid]

(1234) = (12)(23)(34)

(123) = (12)(23) also has an "inversion"

every non-identity finite permutation will have an inversion I still don't see the relevance

[icy bear]

yes obviously

but

[delicate orchid]

they're equal permutations, their actions of whatever finite set are equal

[icy bear]

how would I prove that the permutation is even iff it has an even number of inversions

[delicate orchid]

but that's just false

(123) once again

[icy bear]

oh yeah

I'm confused now tf

[delicate orchid]

pick any (2n-1)-cycle if you're not happy with just (123) lol

[icy bear]

yeah it was being hard to prove something false

this is a hard kind of theorem to prove

the false ones

crazy difficult

maybe the theorem was stated poorly

and it meant that if you decompose the thingy in inversions (every transposition is an inversion), so on and so on

[delicate orchid]

not sure why your source is mentioning these inversion things at all, just think in terms of transpositions

[rustic crown]

nuuu

[rustic crown]

it's correct

[delicate orchid]

how is that correct

[rustic crown]

(123) has two inversions

1<3 and 2<3

[delicate orchid]

1 -> 2, 2-> 3, 3 - > 1

only one of these isn't order preserving

[rustic crown]

1<3 goes to 2>1
2<3 goes to 3>1

[icy bear]

what

[delicate orchid]

oh we're considering an action on X^2 now?

what

this is really stupid

[rustic crown]

yea pairs i<j such that s(i)>s(j)

[delicate orchid]

why are we proving this like this?

I agree with you now det but this seems completely moronic.

[rustic crown]

why tho

[icy bear]

I just did not understnad

undestand

[delicate orchid]

2-cycles are odd. Decompose whatever you have into 2-cycles, sign is multiplicative. Therefore the sign of whatever you have is (-1)^(#transpositions in the decomposition) = 1 if and only if the number of tranpositions is even. No need to consider some weird action on pairs

[rustic crown]

idk, my prefered proof of multiplicativity of sign uses the vandermonde polynomial

and that gives sgn s = (-1)^inv(s) on the nose

[delicate orchid]

I quite literally asked at the beginning if we had sign was multiplicative a priori and gabi said yes

do we or don't we have that

[icy bear]

so now is that description just wrong?

like, that algorithm for finding the inversions

[rustic crown]

true, what wew said proves one thing that gabi wanted

[delicate orchid]

if we don't there's a nice proof using just basic word cancellation

[rustic crown]

but she also wants to prove the statement about inversions

oh wait nvm

i'm sleepy

she wanted inversions from the start

<

[delicate orchid]

lo

[icy bear]

ohhh I think I understood

[rustic crown]

anyway, show that multiplying a permutation by a transposition flips the parity of number of inversions. that's one by hand way to do it, and so i hate it

[delicate orchid]

yeah that'll work

[icy bear]

the (123) "matrix" is
[123
231]
the pairs (a, b) with a > b with a to the left of b is 2 1 and 3 1

[icy bear]

hum

okok

I'll try

[rustic crown]

my prefered way is this:
||consider the ring Z[x1, .., xn], the group S_n acts on this by permuting the variables. define Δ = prod_{i<j} (x_i - x_j)
then one sees that for any s in S_n, s(Δ) = +-Δ. And we define
s(Δ) = sgn(s) * Δ.
this immediately gives both sgn is multiplicative and that sgn s = (-1)^inv(s)||

[icy bear]

put it on spoilers

can't read

the proof

ty

[delicate orchid]

there's a nice proof in Humphres' reflection groups that does this for all coexter groups at once

[delicate orchid]

this one

sorry not coexter groups - real refleciton groups, slightly less general

[icy bear]

oh I know about coxeter stuff

I watched two long videos classifying coexter systems (if that's what you talking about)

[delicate orchid]

ok then S_n+1 is just A_n lel

won't help you for the inversions though noooooooooooooooooooooo how unwholesome

[icy bear]

which motivated representation theory

[rustic crown]

det no know coxeter stuff :c

[icy bear]

it's cool

I may have had an idea

seeing the permutation (1,2,3,4)

the inversions are (2, 1) (3, 1) (4, 1)
if you do the transposition (4, 1) on 2 3 4 1, it does 2 3 1 4
if you do the other transpositions but like not considering the positions but the numbers, you make everything go back

so maybe

I can translate inversions into transposition deecomposition

I'll try that until I can't anymore and maybe try the hint

[rustic crown]

inversions are (1<4),(2<4) and (3<4) right

[icy bear]

ye

wrote them int the reverse order because

.

apparently

using this, in the right order, will always get me back to the start

[icy bear]

this

I may have cooked!

if you do that one sorting algorithm I forgot the name, you're basically reverting the inversions

I mean, adjacent inversions

and reverting adjacent inversions will only change one inversion at a time

can't affect two

so like, ig the algorithm will transpose as much numbers as there are inversions

[rustic crown]

and swapping two things far apart is odd number of adjacent swaps

[icy bear]

wt

[icy bear]

that is, decompose the permutation in as much factors as there are inversions

[rustic crown]

(you could ignore me, i'm too sleepy)

[icy bear]

is that logic right?

I think I kinda understand what you said

[icy bear]

happens

[icy bear]

yeah I mean, the algorithm only subtracts one inversion at a time and it can't add any

so yeah

I forgot the name of that sort aaaa

[delicate orchid]

escape the algorithm!!!!! Slave to demonic simulacrium!!! RAAAAAAAAARRRRRRRRGHHHHHHHHHHHHHH

[icy bear]

hum

[icy bear]

I'm happy pretty sure that's true

[rustic crown]

sounds like bubble sort

[icy bear]

YEAH

THAT ONE

[delicate orchid]

lol

[icy bear]

you go scanning left to right and swap adjacent inversions

[rustic crown]

if i sleep now i have little under 4:30 hours to sleep

[icy bear]

I see nothing wrong in my logic and since I'm a perfect being I must be right

[icy bear]

doubt you sleep right now!!

you'd never do that!!!

[rustic crown]

[icy bear]

never seen someone sleeping my entire life so you must be amazing to do something like that!!!!

I mean, if you did!!!!

[rustic crown]

okie i should at least close discord

https://tenor.com/view/pokemon-evee-anime-sleeping-sleepy-gif-5740828

[icy bear]

gn

[white oxide]

silly question, but if we're able to construct a line segment of length r\sqrt{pi}, how do we know that we're able to construct the other three line segments of length r\sqrt{pi}?

[icy bear]

wdym "other 3"?

like, from the square?

[white oxide]

yeah

i was picturing something like this

assuming the line segment Pr\sqrt{pi| is constructible

[icy bear]

yeah like

what you can do is

extend the first line segment

then construct a perpendicular from each of the extremities using the circles thing

then use a compass to transport the length to both of the sides

that's it

[icy bear]

pretty much

well it'd be better if you did the perpendicular already in P and Q lmao

[white oxide]

i see, okay

yeah i think that makes sense

thank you

[icy bear]

👍

[white oxide]

anyone know what it means to transfer to R

also, how is this a valid proof strategy? 4 and 5 just say that additive inverses of complex numbers constructible from M are in M, and that the sum of two numbers constructible from M are constructible from M

but i can't see how any of those imply that if z is constructible from M and i is constructible from M then zi is in M

anyways i shouldn't take this shit too seriously i guess the point is just that they form a subfield of C lmfao

[white oxide]

what

how does w^5 = 1 and w not equal to 1 shows that it's a root of the equation

[prisma ibex]

what is (x^5-1)/(x-1)

[white oxide]

mathway says x^6 - x^5 -x + 1

wait hold up

😹

LMFAO

ahh got it

[cobalt heath]

This is a basic thing that I somehow got stumped on. Say, R = k[x, y] / <xy>. Clearly R is not a domain. How can I compute which modules are projective over R?

[cobalt heath]

(Nvm, I do not need answer about this now - it seems like we should deal with nicer cases than this mess, after all)

[brave agate]

confused on euclidean algorithm and how to work with fractions, i got gcd=3 when it's actually 1

for example in the second line when you do the division it's this, so i rewrote 7x+6 as 2x+1

[chilly ocean]

can someone give me two presentations of the same group, so that one has a finite number of generators, and the other one has infinite number of generators

or is it not possible?

[dull marsh]

The integers are generated by 1 and also by every integer

[lusty marlin]

Any finitely generated infinite group works

[chilly ocean]

no wait

lemme clarify

an infinite generating set such that we can't throw away any generators

it shouldn't be possible right?

[sharp sonnet]

why not try to find a proof? 😄

[sinful kite]

gcd 1 and 3 are equivalent as 3 is a unit in Z_5. Conventionally, might report gcd as monic so 1 instead of 3.

[cobalt heath]

I got curious again, how does one classify modules over any commutative ring? Especially k-algebras which is not a domain.

[rocky cloak]

What do you mean by that? Just classifying all modules over all possible commutative rings? That's definitely impossible

[cobalt heath]

Like I want an example for how to do it.

My original question was about A = k[x, y] / < xy >, but any similar ring will do.

[elder wave]

The question doesn’t make sense

[cobalt heath]

Huh

[elder wave]

You gotta specify what you’re looking for

If you’re looking for something like the classification of fin gen modules over a PID see jagrs message

[rocky cloak]

So I believe k[x, y]/(xy) is a tame algebra. So then there is a 'reasonable' classification of the finite dimensional modules.

[cobalt heath]

I see, sorry.

I should have been more specific then;
Let A = k[x, y] / (xy).
What does projective A-modules look like?

[rocky cloak]

I would think projective modules are free in this case

[chilly ocean]

I see it now, thanks

[sharp sonnet]

https://www.sciencedirect.com/science/article/pii/S0021869396902951 this seems to give an algorithm for f.g. A-modules for your A

[cobalt heath]

Thank you, will look into it!

[sharp sonnet]

in the references is a more general paper on "dedekind-like" rings

that doesnt provide an algorithm though

if you look at papers that reference this, you will also find an even more general paper that uses different techniques

https://arxiv.org/pdf/1308.6410.pdf

[median rock]

if |G|=kp^m with gcd(k,m)=1 and p^k does not devide (m-1)! show that G is not simple. my thought is that gcd(k,m)=1 and |G|=kp^m so every sylow p-subgroup has order p^k and then i said that if i show that i have one sylow p subgroup then it is normal to G and it has order of p^k so G is not normal is it ok as a thought?

[chilly ocean]

Let M=[1,inf). Find f:M->M s.t. (M,•) is a monoid where x•y=f(x)f(y)-f(x)-f(y)+2 for every x,y in M.

This is what i got

Where e is the identity of M

It is ok?

[icy bear]

found out why inversions are cool that facilitated some proofs

like, that the wedge product is anti-commutative

that made that trivial

[delicate orchid]

is this not just by definition

[icy bear]

no

not in that context at least

in geometric algebra it is for example but in diff geo it isn't

[delicate orchid]

you quotient the tensor algebra by v (x) w + w (x) v

and geometric algebra is borderline crankery

[rocky cloak]

Did you mean for G to have order m*p^k?

[icy bear]

why

[delicate orchid]

anyway, glad to know there are applications

[rocky cloak]

And did you perhaps mean that gcd(m, p) = 1

[icy bear]

that's the wedge product

where f is k-linear and g is l-linear

[rocky cloak]

What is A?

[icy bear]

it's summing up all of f but permuting the entries

and multiplying by the sign of each permutation

[rocky cloak]

I see, so then it's anticommutative by definition I guess

[icy bear]

almost ig

this is the actual anti-commutativity

[median rock]

yeap you are right

[icy bear]

it takes some small work to actually go ahead and prove it

[median rock]

exactly🥲

[rocky cloak]

So thing you can do is consider the action of G on G/P for P a sylow subgroup. This doesn't even use that P is sylow, just that |G| doesn't divide m!

[median rock]

so my thought is wrong?

[rocky cloak]

I'm having some issue parsing your thought, but if there is only one sylow subgroup then it's normal. So yeah if there is exactly one sylow subgroup the group is not simple.

But there can be more than one, in which case that doesn't help

[median rock]

but we know that if it is a P SYLOW subgroup then r is 1modp so r is 1 or r is m

and somehow i want to show that m modp is not 1

and i think with this p^k does not devide (m-1)! i can somehow show it

[rocky cloak]

So you can have more than one sylow subgroup. Like for example D6 (dihedral group of order 12) has three 2-sylow subgroups. But 4 doesn't divide (3-1)!

[median rock]

so mu thought is wrong

damn\

what should i do

[rocky cloak]

So one thing to keep on mind is that G acts (transitivitely) on the sylow subgroups. And what is a way to think about a group acting on n elements?

[median rock]

should i use the correspondence theorem theorem or something like that?

just a quick thought

[delicate orchid]

new isomorphism theorem just dropped

[median rock]

correspondence theorem

[delicate orchid]

and no, you should use the cayley embedding given by the group action on n-elements

[untold basalt]

irreducibility question

[teal vessel]

FINALLY after several days of work:
The union H across (inclusion) chain C of set S of all proper subsets of finitely generated group G=<A> is a subgroup of G.

1. by the union axiom, all elements in H are contained by an element somewhere in C.
2. by the definition of S, all elements of S are groups, therefore every element in H also has an inverse (from the same element somewhere in C)
3. because chains are totally ordered, and this ordering is by inclusion, for any elements x,y in H, they belong to elements X,Y of C, respectively, and by that inclusion ordering, one must be contained entirely within the other, therefore either X or Y contains both x,y and their inverses and (by being a member of S) xy^-1.
4. XUY (the set known to contain that xy^-1) is a subset of H, and x and y were originally arbitrary elements of H, therefore H must be a subgroup of G (H is subgroup iff H is non-empty and for all x,y in H, xy^-1 is also in H).

it's also a proper subgroup, and therefore by Zorn's lemma, S has a maximal element, and thus G has a maximal subgroup, but those parts were pretty easy to prove afterwards. This was the monster.

[round hull]

why

[coral spindle]

Because cranks talk about it a lot

[delicate orchid]

it's literally just the study of a single clifford algebra over R. What's the point.

it's an excuse to invent a shit ton of words that describe something which can really be described far more generally using just a little bit of module theory

[coral shale]

💀

i hope theres room for topological algebra

[delicate orchid]

#point-set-topology lol

[round hull]

topology and algebra don't commute

just like geometry and algebra

[languid trellis]

can someone please give a small hint with this exercise? I'm not even sure of a plan that would get me anywhere

[coral shale]

attempt to construct i presume

[delicate orchid]

the product of those two subgroups is another subgroup of G

then I'll just say the term "inner direct product" if you're still stuck

[languid trellis]

Okay I'll give this a go

I tried to show that H x K (the normal subgroups) and G were iso but that might be too strong

[rocky cloak]

Even more concretely I guess would be to let x be an element of order 3 and y of order 5 and try to find the order of xy

[delicate orchid]

D_30 satisfies this property so that statement isn't true

[languid trellis]

That's why I said 'I tried' lol

[delicate orchid]

but coming up with counter examples is FUN

RAAAARRRGHHH

[languid trellis]

Oh every group of prime order is cyclic

By some theorem in artin, HK is a subgroup of G if H is normal wrt to G

consider xy in HK (x and y are the elements that generate the cyclic sgroups)

This clearly has order 15 because lcm(3, 5) = 15

And xy is in G

I was sat thinking to myself 'this would be easy if the normal subgroups were cyclic'

And I reminded myself lol

Thanks guys

[rocky cloak]

You have to be a little carefully with your 'clearly' there.

For example S3 has an element of order 2 and one of order 3, but none of order 6.

What you need to finish your argument is that x and y commute

(or alternatively that the only group of order 15 is cyclic)

[languid trellis]

I'm struggling to see why I need commutativity sorry, I though i'd just take the cyclic subgroup generated by xy

[rocky cloak]

Right, but why does this have order 15?

[languid trellis]

i suppose because I can "identify" in some sense C_3 x C_5 with <xy>

that's sort of my intuition

and C_3 x C_5 has order 15

[rocky cloak]

Yeah, so if you have this characterization of internal direct sum and how it corresponds to external direct sum, then sure.

But this is saying the same thing as x and y commuting. Since the elements in C3xC5 commute with each other

[languid trellis]

sorry I don't know what you mean when you say "interal direct sum" and "external direct sum"

[rocky cloak]

I meant direct product, but what I'm saying is

How do you know that HK is C3xC5?

[languid trellis]

I think I can see an iso

xy -> (x,y)

i'd need to show that

[rocky cloak]

So to expand on what I was saying before. There is a notion of internal direct product and a notion of external.

If two subgroups H and K are such that they have trivial intersection and both H and K are normal in HK (and HK is a group), then HK is called their internal direct product.

You also have an external direct product HxK which is the set of pairs with pointwise operation.

Now it is true that the internal and external direct products are isomorphic by
hk |-> (h, k)

So if that is something you've already shown, then you're done.

But otherwise it's somewhat subtle. And it comes down to showing that H and K commute.

[languid trellis]

I'll go down the route of showing that HK and C_3 x C_5 are iso I think

because I'm still a little unsure how the commutativity of H and K relate to this problem

I'm pretty sure they do because of normality

artin gives an argument as such in the book

[rocky cloak]

Yes, being normal is a key property, and the intersection being trivial is also important

[languid trellis]

the intersection is trivial because these are cyclic subgroups of different order

[rocky cloak]

Indeed

[languid trellis]

and this is what artin writes wrt to commutativity

I'll flesh this out and latex it up. thanks for the help jagr

[supple vortex]

What is the order of $\langle g \rangle\cap\langle h \rangle$?

[cloud walrus]

splotchvan

[supple vortex]

g,h in a group G

|g|=15, |h|=16

[rapid junco]

gcd of these is 1

both cyclic subgroups

[coral shale]

bruh

no sols

if ur gonna help they should do the work themselves

[supple vortex]

Why don’t you help shuri

You’re quick to say no solutions but you don’t help

[coral shale]

i dont have to if i dont feel like it

[supple vortex]

Thanks Brayden

[coral shale]

i dont generally like assisting low effort Qs

[glossy crag]

What do you care?

[coral shale]

because, i dont like seeing low effort being rewarded

Show what you have tried if ur gonna ask

[supple vortex]

What do you mean low effort

[rapid junco]

Just try to compute |<g><h>|

[supple vortex]

If someones stuck theyre stuck

[coral shale]

cant tell how ur stuck if u dont say so

[rapid junco]

Try to show |<g><h>| has an element of order 15*16.

[supple vortex]

Thanks

[glossy crag]

That's hardly your problem, isn't it? It's between the person asking and the person answering (if any so wishes), I'm not sure how you fit in in that transaction.

[rapid junco]

And this follows from what divides your orders

[supple vortex]

He has an anime pfp

[coral shale]

Huh. If it was somewhere else sure

But its in the rules/guidelines not to give sols

[supple vortex]

Ignore the anime pfp

[glossy crag]

I guess I didn't know it's a thing, then.

[coral shale]

its like - why encourage low qual posting even if not rules

[supple vortex]

“Low quality posting”

You are weird

Which makes sense

You are anime after all

[coral shale]

You asked a question which couldve been hwk posted word for word

[supple vortex]

Which anime person is that

[coral shale]

and basically fishing for the answer

its not clear how you have expended effort in attenpting it

[supple vortex]

I’m not fishing for an answer I’m fishing for an explanation

[coral shale]

since u didnt say at all which part had u stumped

[supple vortex]

There’s one part

It’s a short question

Just

[rapid junco]

Yeah bro

[supple vortex]

The anime isn’t going to watch itself

[rapid junco]

I just be helping

[coral shale]

You just asked this, and u couldve been stuck on anything from the definition of <g> to what a group is

how are we supposed to know

[supple vortex]

Are you kidding me

[coral shale]

or maybe you hadnt even tried

im saying what it looks like to me.

[supple vortex]

Buddy

[coral shale]

"what is the order of <g> n <h>? I've no idea where to begin"
would've been higher quality

[supple vortex]

How come you get confused and pissy when brayden provided a perfect answer

Why can’t you be like Brayden

Why can’t you just make assumptions

Brayden

[rapid junco]

@supple vortex another thing that's important to note is look at the order of the elements in the groups.

[supple vortex]

When you write <g><h> do you mean like coset multiplication

[rapid junco]

The intersection by definition has to have the same order elements

It's all ab where a is in <g> and b is in <h>.

[supple vortex]

Couldn’t it have less?

[rapid junco]

Wdym less

[supple vortex]

Less elements

[rapid junco]

Well it always has less

[supple vortex]

Sorry

You said order

[rapid junco]

The intersection you mean

[supple vortex]

Yeah

[rapid junco]

Yeah order 1 for identity

[supple vortex]

I misread

Yes

[rapid junco]

But it's cyclic so every element has the same order

[supple vortex]

Is it just lcm(15,16)?

Er wait

[rapid junco]

Think about just integers for now

Integers mod 15 and integers mod 16.

[supple vortex]

Mhm

Which they are isomorphic to respectively

Oh

[rapid junco]

The order of every element divides the size of the group

[supple vortex]

Mhm

[rapid junco]

So orders of 15 are possibly 1, 3, 5, 15.

Orders of 16 are possibly 1, 2, 4, 8, 16.

[supple vortex]

Oh

Wow

I’m stupid

[rapid junco]

Nothing lines up except 1

But the only element of order 1 is the identity

So it has trivial intersection

[supple vortex]

Thanks

We didn’t really need cyclicicty then?

Of <g> and <h>

[rapid junco]

Well 2Z and 3Z. Is another wxample

You could look at, but here its triviallly infinite intersection

[supple vortex]

Is the torsion subgroup a subgroup because

if you take $g_1g_2^{-1}$ in it

[cloud walrus]

splotchvan

[supple vortex]

Then $|g_1g_2^{-1}|=\min{|g_1|,|g_2|}$?

[cloud walrus]

splotchvan

[supple vortex]

Which is finite since |g1|,|g2| finite?

[rapid junco]

Not for non abelian groups

[supple vortex]

Yeah

[rapid junco]

Theres probably some counterexample with matrices but I cant find

[supple vortex]

Just abelian

But does that proof make sense

[rapid junco]

Well you just want to show that the order of g_1g_2^{-1} is not infinite

But the order of g_2^{-1} is precisely the order of g_2.

[supple vortex]

Yes

[rapid junco]

Then order divides the product of the orders

So what does that mean about the product of orders

[supple vortex]

What i just did is write out $\langle g_1g_2^{-1}\rangle$

[cloud walrus]

splotchvan

[rapid junco]

You dont need to do this though

[supple vortex]

But i could

I find that way easier

[rapid junco]

🤷‍♂️ up to u

[supple vortex]

I mean

[rapid junco]

it’s essentially the same thing

[supple vortex]

I even got a formula for |g1g2^{-1}| out of it

Wait

No mines wrong

[round hull]

lol

this entire interaction is -able

[brave agate]

T/F: If G is a group that acts non-trivially on {1,2,3}, and |G| >6, then G is non-simple

What I have so far: let g be the element that acts non-trivially. Then |Orb(g)| is 2 or 3, corresponding to two-cycles or three-cycles. Then |Stab(g)| must be >= 4

[rocky cloak]

Think about how you can frame an action as a type of homomorphism

Like in ||Cayleys theorem|| for example

[brave agate]

$\phi: G --> S_3$ where $\phi(g)$ describes how g acts on ${1,2,3}$

[cloud walrus]

BLONK

[brave agate]

the order of some element g is either 2 or 3, a prime number, thus abelian (not sure if that matters)

my next guess is to show the kernel of this is non-trivial

[coral spindle]

the order of some element g is either 2 or 3, a prime number, thus abelian (not sure if that matters)
This is false.

[brave agate]

whoops

must be the first part that's false

[coral spindle]

S_3 is not an Abelian group.

[brave agate]

that's size 6 tho not 2 or 3

im talking about <g>

[coral spindle]

Well, OK. This was not clear.

I should say that any cyclic group is Abelian – it doesn't matter what order it is.

[rocky cloak]

Showing that the kernel is non-trivial is indeed a good idea, and thinking about the size of S3 should help.

Not sure why you say the order of g must be either 2 or 3, but it's not really relevant anyway.

[brave agate]

yeah i think i got it?

by the first isomorphism theorem:
G/K isomorphic to Im(f)

since at least one element is non-trivial, Im(f) is > 1

Since G > 6, then K > 1 so their orders are equal

so K is a non-trivial normal subgroup, so G is non-simple

[topaz solar]

Is there a nice uniform-ish presentation family for S_n? I don’t touch finite groups much

[delicate orchid]

define presentation family

[topaz solar]

Like literally just presentations for each S_n

None of the words except presentation and S_n are precise

[delicate orchid]

oh lol - they're like the most basic coexter groups imaginable

so just take S_n = <a_i : a_i^2 = 1, (a_ia_j)^3 = 1, j = i+1>

think that works

[topaz solar]

I do not know what a Coxeter group is fr

[rose prism]

its like a symmetric group

[topaz solar]

So true

[rose prism]

ok

[rustic crown]

is a_i = (i, i+1)?

[rose prism]

its like a weyl group

which is like a symmetric group

[delicate orchid]

yus

oh

I forgot the commutators

put the commutators in

[topaz solar]

The fires of Babylon

[coral spindle]

Coxeter group presen—dammit wew

[topaz solar]

That’s uniform enough if it works frfr

[delicate orchid]

you can get a 2-generated presentation with worse relators if you want

using an n-cycle and a tranposition

[topaz solar]

Coxeter looks better

[celest furnace]

@rustic crown Update on my 4 hour test! Crushed it!!!

[white oxide]

is every free module torsion free because if ti weren't, this would imply that the basis is linearly dependent?

[delicate orchid]

yurp

[rustic crown]

yay!!

https://tenor.com/view/anime-pat-gif-19580650

[full quartz]

can someone help me with thiss please

[prisma ibex]

they're both groups of order 2

there is a unique group of order 2 up to isomorphism

[tribal moss]

Most likely the intended solution is to list the elements of each group and show an isomorphism between them.

[supple vortex]

Prove/disprove: every nonabelian group of order divisible by $6$ contains a subgroup of order $6$.

[cloud walrus]

splotchvan

[supple vortex]

how can i do this directly?

[kind niche]

lagrange’s theorem

iirc it says that the order of the subgroup divides the order of the group

[supple vortex]

Yes but it says nothing about existence

of the subgroup

[kind niche]

true

[supple vortex]

Lagrange’s Theorem: If H is a subgroup of G, then |H| divides |G|

any ideas anybody?

[round hull]

you can overkill and use sylow's for 2 and 3

[supple vortex]

we havent learned sylows

[round hull]

but cauchy's theorem is enough

well it's precisely cauchy's theorem you want i guess so maybe that isn't assumed

[supple vortex]

we havent learned cauchys theorem

this is in the Isomorphism chapter of the book

[round hull]

ok

[supple vortex]

http://abstract.ups.edu/aata/isomorph-exercises.html

#21

[round hull]

take the order of all the elements. there must be at least one with order divisible by 2, for otherwise the exponent of the group, which is the lcm of the orders, will be coprime to 2

[narrow wagon]

If a take an arbitary group G and compute the powers of some element in G, then subgroup generated will be cyclic right?

[supple vortex]

Not sure what you mean

I know nothing about the group

[kind niche]

idk if that makes sense but if |H| divides |G| then |H| = |G|n and |G| = 6m, so |H| = 6k, could this lead to something?

[supple vortex]

@round hull ?

“exponent of the group” ?

[round hull]

ignore what i said i thought abelian

[supple vortex]

?

[round hull]

ya you want cauchy's theorem exactly, you might want to search it up

[supple vortex]

but i cant use it

this is in the isomorphism section

this question is asked before cauchys theorem is taught

[round hull]

actually i'm not even sure cauchy's/sylow's theorem would help

maybe this statement is false then

[supple vortex]

i mean it is false

but stackexchange proves it by counterexample

[round hull]

ok cool

wdym but

[supple vortex]

https://math.stackexchange.com/questions/1042702/every-nonabelian-group-of-order-divisible-by-6-contains-a-subgroup-of-order-6

how does he know A4 doesnt have a subgroup of orde 6

[mighty kiln]

Trial and error works

More generally subgroups with index 2 have to be normal

[supple vortex]

is there any easier way to tell A4 doesnt have a subgroup of order 6

[round hull]

well subgroups of order 6 contain a cyclic subgroup of order 3 i think

and one of order 2

so it must be generated by a 3-cycle and a double transposition

[supple vortex]

im lost man

can anyone else help

[round hull]

i'll just leave the computation for anyone else that wants to help. wlog we have (1 2 3) and (1 2)(3 4). if we compose them one way we get (1 3 4); the other way we get (2 4 3). so this subgroup has at least 8 elements, which must be A4 itself

[supple vortex]

how do you conclude it must have 8 elements

[round hull]

it has id, (1 2 3), (1 3 2), (1 2)(3 4), (1 3 4), (1 4 3), (2 4 3), and (2 3 4)

[supple vortex]

i dont even know what subgroup youre talking about

generated by an element?

[mighty kiln]

If it helps notice A4 is the rotation group of a tetrahedron

[round hull]

oh is there a geometric argument Arki

[mighty kiln]

No I just find this easier to visualize

[supple vortex]

Man

I am so lost

Im sorry

[round hull]

cyclic subgroups of order 3 in A4 have to be generated by a 3-cycle

[supple vortex]

Can someone explain it plainly from start to finish

[round hull]

am i excluded

[supple vortex]

No

[tribal moss]

That's not a subgroup: (1 2 3)(2 4 3) = (1 2 4) which is not in your list.

[round hull]

by has i mean at least contains

[tribal moss]

Ah, okay.

[round hull]

i mean i counted 8 elements

so it's never a subgroup anyways

[mighty kiln]

trial and error

[round hull]

Let $A_4$ be the alternating group on 4 elements. We wish to show that there is no subgroup of order 6. For a contradiction say we have found one, $H$. Then $H$ is either isomorphic to $S_3$ or $C_2\times C_3$ by classification; regardless it contains a $C_3$ and $C_2$ subgroup. These must be generated in $A_4$ by elements of order 3 and 2, which must be a 3-cycle and a double transposition respectively. Let us label the set $A_4$ acts on by defining 4 to be the element the 3-cycle keeps still; 3 to be the element the double transposition takes 4 to, 1 to be the element the 3-cycle takes 3 to, and 2 to be the remaining element. Hence our 3-cycle is $(1 2 3)$, and our double transposition is $(1 2)(3 4)$. It remains to show that any subgroup containing these two elements must have size $> 6$.

[cloud walrus]

quasi_semi_group

[tribal moss]

We could say: by conjugating the double transposition repeatedly by the 3-cycle, we can make all three possible double transpositions. But those double transpositions (together with the identity) form a subgroup of order 4, which cannot lie inside a subgroup of order 6.

[round hull]

i hate doing any sort of combinatorics write-up

even if it's easy it just looks like a mess on paper

[supple vortex]

Ok

now how about this one

Prove $U(p)$ is cyclic for $p$ prime

[cloud walrus]

splotchvan

[round hull]

do you have any ideas

hint: you can leverage the field structure using polynomials

[tender mist]

Any finite abelian group has an element of maximum order d under divisibility (any other element has order dividing it). Through polynomials, prove d must be |U(p)|

[supple vortex]

Well really what i have to do is

show it is isomorphic to Z_(p-1)

is that easier?

[warm saffron]

This is a good approach. Showing U(p) is isomorphic to Z_{p-1} will show it's cyclic since u know Z_(p-1) is cyclic

[supple vortex]

yeah but how do i do that

[warm saffron]

U know what isomorphisms are yea?

[supple vortex]

U(p)={1,2,…,p-1}, Z_(p-1)={0,1,…,p-2}

what is the isomorphism? f(x)=x-1?

from U to Z

f(xy)=xy-1

[warm saffron]

that is a bijection for sure

[supple vortex]

nope

[warm saffron]

hang on

[supple vortex]

xy-1 neq x-1+y-1

[warm saffron]

u wanna show then that f(xy)=f(x)+f(y)

[supple vortex]

f(x)+f(y)=x+y-2

thats not the iso

[warm saffron]

hmm it's def a bijection but hmm what if we actually looked at an element in U(p) that has order p-1. If we can explicitly find one, then the cyclic group generated by that must exactly U(p)

[celest furnace]

This approach doesnt work well

[supple vortex]

Z -> U?

[celest furnace]

Something like this works

[supple vortex]

how does that get me an iso

[celest furnace]

Cyclic groups are unique up to order

[warm saffron]

skip the iso part

[celest furnace]

So as long as u show its cyclic you're good

[warm saffron]

look for a generator directly

[supple vortex]

thtats what i asked initially

ok well

if U(p)={1,…p-1}

idk

try 2?

[warm saffron]

does 2 work for all p?

does 2 generate U(p) for all p?

[supple vortex]

<2> = {1,2,4 mod p, …}

idk

im stuck

[warm saffron]

so here's what we're looking for....

You need to find an $x\in U(p)$ for which $x^{p-1}\equiv1\pmod p$ and $x^k\not\equiv1\pmod p$ for any $k\in{1,2,3,\dots,p-2}$. Finding such an $x$, would entail u have found and $x$ with order $p-1$. In which case $\langle x\rangle=U(p)$ for such $x$.

[cloud walrus]

logician

[supple vortex]

Ok so U(p) = {1,2,…p-1}

[warm saffron]

yes. I'm actually still thinking about this problem tbh... But I wrote the modular arithmetic down in case it rings any bells

[supple vortex]

i mean an obvious answer would be x=p right

[warm saffron]

I'm more of a number theory person

[supple vortex]

but p notin U(p)

oh

[warm saffron]

yea so p doesn't matter

look at this part tho

$x^{p-1}\equiv1\pmod p$

[supple vortex]

@celest furnace any ideas

[cloud walrus]

logician

[supple vortex]

mhm

man idk

[warm saffron]

I got an idea

fermat's little theorem

[supple vortex]

how so

[warm saffron]

https://www.google.com/search?q=fermat's+little+theorem&oq=fermats+little+&gs_lcrp=EgZjaHJvbWUqCQgBEAAYChiABDIGCAAQRRg5MgkIARAAGAoYgAQyCQgCEAAYChiABDIJCAMQABgKGIAEMgkIBBAAGAoYgAQyCQgFEAAYChiABDIJCAYQABgKGIAEMggIBxAAGBYYHjIICAgQABgWGB4yCAgJEAAYFhge0gEIMzQ3NmowajeoAgCwAgA&sourceid=chrome&ie=UTF-8#vhid=Vn3iExLnuMgfQM&vssid=l

[cobalt heath]

Idk but existence of principal root could be a bit technical

[supple vortex]

yes we werent taught that

well the question asks me to show U(5) is iso to Z4

then asks me to generalize for any p

so what would be the iso U(5) -> Z4?

would it just be piecewise

no

[warm saffron]

it's all about where u send the generators

for instance Z_4 only has 1,3 as generators

where u send the generators of cyclic groups determines the isomorphisms

[supple vortex]

so what would be an iso from U(5) to Z4

[warm saffron]

define it explicitly

[supple vortex]

what

[warm saffron]

f(1)=....

f(2)=...

[supple vortex]

yeah

[warm saffron]

f(3)=...

[supple vortex]

but what do they equal

[warm saffron]

there isn't necessarily only one isomorphism

[supple vortex]

ik

what could they equal

[warm saffron]

well f(3) could equal a generator of Z_4

since 3 is a generator of U(5)

[supple vortex]

why 3

ok

so then U(5) must have two generators

[warm saffron]

first off, determine all the generators of each of the sets

[supple vortex]

ok

[warm saffron]

yes

[supple vortex]

then map the generators to the generators

and any of the nongenerators to any of the nongenerators?

[warm saffron]

yes

[supple vortex]

how would i show f(ab)=f(a)+f(b) in general for that

[warm saffron]

for this I actually think it's easier to define the isomorphism from Z_4 to U(5). Because then since the operation is addition for Z_4, we could just send 1 to 3 in U(5) and then f(2)=f(1+1)=f(1) x f(1), make sense?

[supple vortex]

ok man ill figure this out tomorrow

[warm saffron]

f(3)=f(1+2)=f(1) x f(2)

etc.

[supple vortex]

but i have another q for you

the division algorithm states

if f(x),g(x) in k[x]

then there exist q(x),r(x) st f(x) = q(x)g(x) + r(x)

right

say we want to do long division of polynomials in Zn

[warm saffron]

I know it in the context of NT. Haven't heard about it in terms of polynomials. Maybe someone else here might know the answer to that

[supple vortex]

well

do you know what i did wrong here?

remainder supposed to be 2x+1

in this case we’re in Z5

i thought for every computation of a coefficient i just do it mod 5

but i got the wrong answer ..?

[warm saffron]

is x^3 - 2x^3 = -2x^3? I don't think so

[supple vortex]

where do you see that

oh

shit

[warm saffron]

ur first line that calculates the first difference

0-2x^3...

because wouldn't that be 4x^3 if this is mod 5

since -1 is congruent to 4 mod 5

[supple vortex]

x3 - 2x3 = -x3

?

[warm saffron]

yes and that coefficient is -1

[supple vortex]

yes

[warm saffron]

which is the same as 4 mod 5

so that should technically be 4x^3 as well, yea?

-x^3=4x^3 mod 5 is what I'm saying

[supple vortex]

uh

[warm saffron]

I mean realistically, I don't think that part should matter

if u switch it or not

[supple vortex]

is that because -1 in {-4,-3,-2,-1,0,1,2,3,4}

[warm saffron]

but yea that was the first line that I saw an error in^

[warm saffron]

no it's because 5| 4-(-1). so 4 and -1 are congruent modulo 5

just like 12 is congruent to 2 mod 5

like u noted

u wrote (3x^2)(4x)=2x^3 since 12 mod 5=2

anyway, I think I'm perhaps confusing u. Just redo the long division. It's tedious but yea silly errors that we're all prone too can yield a wrong answer quite easily

[supple vortex]

i just got confused on the negative

-1 mod 5

[warm saffron]

it's because of this^

[supple vortex]

yeah but

is there any trick i can do

to a negative mod n

[warm saffron]

Because it is so that $-1\equiv 4\pmod 5$, I interpreted $-x^3$ as $4x^3$

[cloud walrus]

logician

[supple vortex]

i guess -n mod m = m - n?

n positive

[warm saffron]

these "tricks" go back to the definition of congruence mod k.

[supple vortex]

i know

but

[warm saffron]

Remember $a\equiv b\pmod k$ iff $k|a-b$. I'm only making use of that difference there

[supple vortex]

is $-n$ mod $m$ = $(m-n)$ mod $m$ ?

[cloud walrus]

logician

splotchvan

[supple vortex]

im so dumb

lol

thats dumb

[warm saffron]

$-n\equiv m-n\pmod m$ is true yes

[supple vortex]

dont read that

[cloud walrus]

logician

[supple vortex]

ok

listen man

[chilly ocean]

Let p>=2 prime number. In the ring Z/pZ we define the sequence (a_{n})n>=0 by a_{n} = n mod p for every n=0,1,...,p-1 and a_{n}=a_{n-1}+a_{n-p} for n>=p

Show a_{n}=a_{n+p^2-1} for every n>=0

Any idea

[sinful kite]

why do you even need mod p here? "a_{n} = n mod p for every n=0,1,...,p-1 " looks fishy

[chilly ocean]

Idk man

This is the problem

[sinful kite]

can you post a screenshot instead? and maybe this belongs in number theory.. idk

[chilly ocean]

maybe beacuse a_{p+1}=0

a_{p+2}=2

And so on

[chilly ocean]

It is in romanian

George Stoica problem from Canada

This is the problem

But I translated it in english here

[rocky cloak]

Have you tried induction?

[chilly ocean]

Yes

For a fixed for every k<=n

But does not work

Or I dont see it

[rocky cloak]

Where do you get stuck?

[novel spruce]

You have a linear recursion of degree p here. If you can prove that the roots of the characteristic polynomial all lie in F_{p^2}, you're done.

[chilly ocean]

Wow nice idea

[chilly ocean]

Induction🥲

[novel spruce]

The product of (x-a) where a runs through F_{p^2} is x^p^2 - x. So you must prove that this is 0 modulo x^p-x^(p-1)-1.

[rustic crown]

Since the polynomial is separable it suffices to check stuff at roots

And if a^p = a^(p-1)+1 then b=1/a satisfies a simpler polynomial

Show that b^(p^2)=b

[wild solar]

Hello, I'm wondering if I could get some help on this item (a)? I'm new to studying modules and I don't really know how I'm supposed to prove this, or how to utilize the condition on A. Any hint would be appreciated🥺

[rustic crown]

Take M_i to be the submodule e_i * M consisting of elements e_i*m as m varies in M. Show these do the job.

[wild solar]

oh oki ty that makes sense

[rustic crown]

[wild solar]

and what about (b)?

[rustic crown]

well e_i*m maps to something in e_iN just by A-linearity

This shouldn't be surprising if you know the corresponding situation for rings. When you have idempotents like that, the ring decomposes as a product of multiple rings, and a module over a product of ring is same as the data of module over each ring

[wild solar]

Wait I'm slightly confused, what is f(N_i)? Do they actually mean N_i?

[rustic crown]

yep

[wild solar]

okok

thank you!!

[rustic crown]

[supple vortex]

Hey @rustic crown

[rustic crown]

[supple vortex]

can you tell me why the elements of order 3 in $\mathbb{Z}_3\times\mathbb Z_4\times\mathbb Z_5$ are (1,0,0) and (2,0,0)

[cloud walrus]

splotchvan

[rustic crown]

if (a, b, c) has order 3, then (3a, 3b, 3c) = 0

[supple vortex]

mhm

[rustic crown]

but you also konw that 4b = 0 and 5c = 0

from this show that b and c are both 0s already

so the element looks like (a, 0, 0), when does this have order 3?

[supple vortex]

why isnt 3a=0

[rustic crown]

it is, but we already knew that

[rustic crown]

.

[supple vortex]

ok but if 3a=0 then a=0

[rustic crown]

nu

that's in Z/3

everything satisfies that

cause modulo 3

[supple vortex]

oh

ok what about in Z6 x Z8

when does (a,b) have order 5

say

sorry

[rustic crown]

what would you think it is