groups-rings-fields

[cyan marten]

Is this an easy way?

[past temple]

wdym

im not rly sure that its easy.. its a ton of computations

thats my current understanding of how character theory is useful atm

is there a simpler way of simply checking if a repn is irreducible or not?

thats what my original question was to begin with

[cyan marten]

Yeah, I was trying to understand what you mean by "easy"

Well, maybe one could somehow prove that it's hard to check that

[past temple]

but character theory isnt applicable for non-algebraically closed fields

and so is there an algorithmic way to tell if a repn over R is irreducible by looking at the matrices?

[brisk spindle]

Im stuck here help

[coral spindle]

character theory is applicable for non-algebraically closed fields

just not all of it works, necessarily

According to this mathoverflow question, an algorithm to determine this kind of question is an area of active research: https://mathoverflow.net/questions/111404/algorithm-to-check-is-representation-irreducible-algorithm-to-decompose-the-r

[past temple]

i see, i dont think my knowledge of representation theory is quite advanced enough to undersatnd that post

but to be more specific about my question

im trying to show that a particular representation of a finite group in R^4 is reducible over R

all of the matrices of the generators of my group have complex eigenvalues, so they obviously cannot have 1-D R-invariant subspaces

this also implies that my representation cant have any 3-D invariant subspaces, because by complete reducibility, that would imply that there's also a 1-D invariant subspace

but im kind of stuck for 2-D

would it be valid to show that the matrix representations of my group elements cannot be similar to a block-upper triangular matrix with 2x2 blocks?

@coral spindle

[coral spindle]

Yes

[vivid tiger]

1 is not in all of them

[white oxide]

nvm i looked up the solution online and i still dont get it

this shit is so frustrating

[tender wharf]

could you post the question here so we can take a look

[white oxide]

and one second let me post the actual question sorry

i was looking at 46

[tender wharf]

right

which part does not make sense

[white oxide]

i still dont get how it shows that the number of generators adds up to n, and that line, thus sigma counts each element of Z_n once and only once as a generator of a subgroup of order d dividing n does not make sense at all to me

ig that would have to follow from every single element in Z_n generating a subgroup

but the problem is there's multiple generators of that subgroup

[tender wharf]

right

but those generators must be coprime

[white oxide]

I think it’s just one of those things that is hiding in plain sight

But I’ve thought too long about this question for my brain to process anything

And I can see that ur right

[tender wharf]

try a few examples

maybe

like say

12

[white oxide]

It’s prolly best to do an example

Yeah

i still dont get it at all

is it like you set up a bijection or something

take Z_4

then by exercise 45 it's subgroups are Z_4 itself, Z_2 and Z_1

phi(4) = 2

phi(2) = 1

phi(1) = 1

sure makes sense

4 generators

Z_4 has 4 elements

[tender wharf]

try looking at U(4) U(2) and U(1)

[white oxide]

so are you supposed to set up a bijection between the generators and n

or something

and ig my question is how do you even prove that

[white oxide]

what does U denote

[tender wharf]

like the set of positive integers coprime to 4

sorry I need to eat

i may be back but if not someone else will be here

[white oxide]

no worries at all bro

i appreciate the help

[main needle]

If $f \in F[X]$ where $F$ is a subfield of some field $E$. If $a \in E$ is a repeated root of $f$, does that force $a \in F $

[cloud walrus]

ru0xffian

[coral spindle]

No. Take F = R, f(x) = (x^2 +1)^2, and a = i.

[main needle]

yeah should've thought about that more. I am trying to show that If $a$ is a repeated root of $f$, then $f, f'$ are not coprime on $F[X]$ but I can't make sure that their divisor must lie on $F[X]$ rather than $E[X]$

[cloud walrus]

ru0xffian

[coral spindle]

The GCD of f and f' in E[X] must divide the gcd in F[X], which is always nonzero.

[main needle]

Oh yeah that makes a lot of sense. Thanks

[boreal inlet]

Prove that if G is a finite group with even number of conjugate classes, then |G| is even.

There's the hint where it says we can use class equation

What I know is

The number of conjugates of an element a in G is the index of Centralizer of a in G

And also, all elements in center of G form a singleton equivalence class by themselves

how should I proceed

[main needle]

Orbit-Stabilizer ?

[boreal inlet]

that says index of Stabillizer is the order of orbit for an element

I'm sorry I'm actually lost

[coral spindle]

I don't see how orbit-stabiliser would be useful.

I think you should look at the class equation.

Remember also that the order of any equivalence class divides the order of the group.

[boreal inlet]

yes, because it's the index of the centralizer of the representative element

[young fiber]

Conjugacy classes partition the group

[coral spindle]

You have all the information necessary to solve this here. I think you should continue trying to solve it, focusing on the aspects that I pointed out

[boreal inlet]

ok thankss I'll try, if I still don't able to i'll ask

[young fiber]

|G| is the sum of orders of conjugacy classes

[coral spindle]

The fact you state is referred to as the class formula, which has been mentioned already.

[boreal inlet]

I'm sorry I got stuck in the equation itself

I assumed k elements were in the center

these form k equivalence classes

after that I tried to use this fact

$i_{G}(C_{G}(g)) = \frac{|G|}{|C_{G}(g)|}$

[cloud walrus]

rikusp2002

[coral spindle]

Don't use that fact, you don't need it.

[boreal inlet]

oh

[coral spindle]

I will say again that you should look at the class equation

and also recall that the size of any conjugacy class divides the order of the group

[boreal inlet]

class equation is about the order of G right?

which is the sum of order of Center and sizes of Conjugacy classes

[coral spindle]

I suggest you go back and look at what it is again

[boreal inlet]

ok wait

https://groupprops.subwiki.org/wiki/Class_equation_of_a_group this says the same as my textbook does

[coral spindle]

Let's put this in different words, then.

The sum of the sizes of conjugacy classes is |G|.

Use this fact now

[boreal inlet]

OH

wait

let's take one conjugacy class

$\frac{|G|}{|C_{G}(g)|} \text{" divides "} |G|$

[cloud walrus]

rikusp2002

[boreal inlet]

which is the size of one conjugacy class

no.

this actually doesn't get me anywhere

[coral spindle]

It does

[boreal inlet]

oh..?

[coral spindle]

I don't know why you're writing out |G|/|C_G(g)| every time

just write |g^G|

As I said, you do not need to use the fact that |g^G| * |C_G(g)| = |G|, only that |g^G| divides |G|.

[boreal inlet]

You know what let's use contradiction

If order of G is odd

|Z(G)| divides |G| and has to be odd

Also all sizes of conjugacy classes would also be odd

[coral spindle]

Go on

[boreal inlet]

How do I show number of conjugacy classes this way is also odd

Wait

Odd minus odd is even

So

Sum of all elements in conjugacy classes represented by non central elements is even

So number of conjugacy classes is odd

I think we're done

[coral spindle]

Simplify this proof, by not considering the central and non-central case

Just consider the number of conjugacy classes

[boreal inlet]

Man it took me forever to do such a simple thing exam preparation pressure sucks

I'll try to shorten this

Thanks a lot though, you helped a lot

[coral spindle]

I'll write out the shorter version

If |G| is odd, then the size of every conjugacy class is odd. If there were an even number of conjugacy classes, then since the sum of an even number of odd numbers is even, |G| would be even, which we assumed was false.

[boreal inlet]

💀

wow.

Yeah now I see what i was missing

i must've appeared dumb

[coral spindle]

It takes time to learn to find proofs quickly. Dw.

[boreal inlet]

[past temple]

if i have a morphism f : U -> V (x) W

are there morphism f1 : U -> V, f2: U -> W such that f = f1 (x) f2

where (x) denotes tensor product

[coral spindle]

This would imply that everything in the image of f is a simple tensor, which is usually false.

E.g. take U = V \otimes W

and of course let V, W be, let's say, vector spaces of dimension at least 2.

[past temple]

i see

[sonic coral]

How do you know if something has index 2?

thats the amount of distinct right (left) cosets right?

[coral spindle]

Yes

[sonic coral]

I was told i needed to use that to show a group of order 2p^n for odd prime p is not simple, but I dont know how to get to the fact that it has index 2

[tribal moss]

Do you have the Sylow theorems available?

[sonic coral]

yes

[tribal moss]

What is the index of a Sylow p-subgroup of the group in question, then?

[sonic coral]

i know it’s gonna be index 2 since you’re asking me, but i don’t really know why

[tribal moss]

Hmm.

[next obsidian]

Do you know what Lagrange’s theorem is

[tribal moss]

What is the order of a Sylow p-subgroup in our group?

[next obsidian]

It gives you a way to compute the index of a subgroup given information about its size

[sonic coral]

yes. the order of the subgroups divide the order of the group

[next obsidian]

No

Lagrange’s theorem says more

[sonic coral]

it gives us the possible orders of the subgroups but not the existence

[next obsidian]

It gives you a specific formula which implies that the order of the subgroup divides the order of the group

At least in the formulation I’ve seen it

[sonic coral]

2 right

[tribal moss]

No, that would be a Sylow 2-subgroup.

[sonic coral]

or 1 i guess

oh p^n

[tribal moss]

So if you have a subgroup of order p^n in a group of order 2p^n, what is its index?

[sonic coral]

the possible divisors of 2?

but it can’t be one so it’s gotta be 2

[tribal moss]

What does "index" mean?

[sonic coral]

.

[tribal moss]

Right. And do you know all of the cosets have the same size?

[sonic coral]

i do now

and they’re disjoint

since they partition right

[tribal moss]

Yes.

So we're partitioning a set of size 2p^n into disjoint subsets of size p^n ...

[sonic coral]

there’s only one way to do that

both have size p^n and are disjoint

[tribal moss]

Which means there are how many of them?

[sonic coral]

2. So an index is the possible ways to partition a group of order n into sets of the same size?

[tribal moss]

No.

You quoted the correct definition yourself.

[sonic coral]

yeah i’m just trying to think about what it is while avoiding the term coset

[tribal moss]

At this point I think you should really take Chmonkey's hint and look at the statement at https://en.wikipedia.org/wiki/Lagrange's_theorem_(group_theory)

(And I'll admit I had also just internalized "Lagrange's theorem" as just meaning "the order of an element divides the order of the group" -- but there's indeed more to it).

[sonic coral]

Let H be the sylow p-subgroup of order p^n

then by lagrange theorem, |G| = 2p^n = [G:H] * |H| = [G:H] * p^n

and so [G:H] has index 2

hence H is normal in G by the index 2 lemma and so G is not simple

[tribal moss]

Right!

[sonic coral]

trivial

thank you for being patient though, as always

[void cosmos]

yo howtf is the last statement true

[quiet pelican]

Given f in C(R)
Take a continuous function, g, such that g(s) = 0, g(r) = f(r)
Then g in ker phi_s, and f - g in ker phi_r

[void cosmos]

yea

so smart

tytyy

[soft ore]

Is anyone familiar with any methods to compute high degree (eg. >1,000) primitive polynomials in finite fields?

[formal ermine]

maybe this can be of interest to you? http://www.seanerikoconnor.freeservers.com/Mathematics/AbstractAlgebra/PrimitivePolynomials/theory.html#AppendixD

[soft ore]

Thanks, I was hoping someone might know of instances where the performance of this algorithm can be exceeded using algebraic tricks.

[soft ore]

https://www.ams.org/journals/mcom/2003-72-243/S0025-5718-02-01478-3/

Something like this would be a pretty good example

[static needle]

I saw this and you immediately know what i mustve been thinking

QUOTIENT GROUP

Ofc im assuming P has total order inwich any $p_i$ and $p_j$ $\in P$ are comparable

[cloud walrus]

messyinterval

[wraith swan]

Moved to #linear-algebra

[static needle]

This deserves to be discussed under #linear-algebra

[coarse musk]

hello

so the question says, describe the elements of z[1/6]

how dyu approach it?

[coral shale]

So you identified earlier the quotient

Z[x]/(6x-1)

right?

[coarse musk]

yep

[coral shale]

Do you understand how this is related

to Z[1/6]

[coarse musk]

as far as I understand, the notation means that we're looking at polynomials with 1/6 as a root

so x=1/6

and generate 6x-1

as that would equate a 0

and generate a homomorphism from there

as far as I understand till now

[coral shale]

Sorry, back

So to be clear - there is an isomorphism between A = Z[x]/(6x-1) and B = Z[1/6]

you map x to 1/6 as you've said.

So you should think about quotients like 'modding' out

[coarse musk]

hmm okay

[coral shale]

(btw I'll also say understanding this quotient is probably not necessary for the questiono)

but its worth it

[coarse musk]

yes

[coral shale]

So, a simpler example

Z[x]/(3)

This should be one you are familiar with

Or even Z/(3)

[coarse musk]

yes, this contains polynomials with coefficients as 0,1 or 2 if i'm not wrong

[coral shale]

yes

integer polynomials mod 3

In the same way, think of Z[x]/(6x-1) as integer polynomials mod '6x - 1'

If I give you a cubic polynomial, you can find an element in its class with lower degree by adding multiples of 6x-1, since that is now 0

6x^3 = 6x^3 - x^2(6x-1) = x^2

for example.

just like modular arithmetic

[coarse musk]

hmm yep i get what you're saying

[coral shale]

So if your ring is also a euclidean domain - then you know for sure you can utilize the euclidean algorithm

to help simplify

if not, it might be a little more tricky

But back to the question - I don't think it wants you to do this at all

Just directly utilize the definition of Z[a], a ring adjoined with an element

[coarse musk]

hmm okay, so then how would I get a general description for the elements of Z[1/6]?

[coral shale]

Yeah, I'm not 100% sure if thats what they want

so you know Z[a] = {n + ma : n, m in Z} ?

[coarse musk]

the division algorithm?

[coral shale]

or Z + aZ (maybe you havent seen this notation, related to modules)

[coarse musk]

haven't

[coral shale]

ok well, id just stick with this

[coral shale]

I mean the actual definition is 'the smallest ring that contains Z and a', but this is more handy to use

[coarse musk]

hmm yes I've seen that

think I get that

I feel like I understand that, but when faced with a problem I stuggle to apply it

for instance, this problem itself

the 1/6 one

[coral shale]

oh i think i mistyped, silly me

[coral shale]

Its polynomials in a with integer coefficients, im clowning

rusty on this

[coarse musk]

yes, I understand this

[coral shale]

,,R[a] = {\sum^n_{i=0}k_ia^n : \text{for some $n$ and $k_i$'s}}

[coral shale]

Ok, in that case, perhaps 'write' some elements out?

[coarse musk]

how would I do that, for z[1/6]? I tried to write some polynomials, see what I get if I divide them by the respective power of 6x-1

but stuck from there

[coral shale]

mmmm

mmm

so just to give an example

2 + 5(1/6) - 3(1/6)^2 is one such polynomial in 1/6

So its a linear combination of non-negative powers of 1/6 with integer coefficients --- another way to view it

Sorry if I'm not being very helpful, the descriptive nature of the Q

[coarse musk]

oh wait....OH DAMN OKAY I SEE SOMETHING NOW

ok wait no

i think i get it

so it would be a/6^n, where a,n belong to Z?

if I'm not wrong

[coral shale]

hmm, I suppose so - it can simplify to this

idk why i didnt think of that

[coarse musk]

oh that's perfect, that's the answer but was struggling to get to that

[coral shale]

Using 6(1/6) = 1, yes you do get to this simplification from here

[coarse musk]

but I had another question, it says show that Z[x]/(2x-1,2)=0

similar, but struggling with this

[coral shale]

0 being the uh

trivial ring im guessing

Ok, so for starters, (2x - 1, 2), do you know what this ideal 'looks' like

[coarse musk]

it says, if u kill a unit, the ring dies, show it using this example

[coarse musk]

I see that it is all polynomials with 0,1 as coefficients, but looking at it as those mod 2x-1

[coral shale]

indeed

you're taking Z[x] and setting both 2 AND 2x-1 to 0

its isomorphic to

Z_2[x]/(2x-1) = Z_2[x]/(1)

[coarse musk]

i odn't get this

[coral shale]

which part

[coarse musk]

how did u come to this conclusion

[coral shale]

how i got to the left?

one sec

[coarse musk]

this i get

[coral shale]

Z[x]/(2x-1, 2) = Z_2[x]/(2x-1) = Z_2[x]/(-1)

So I claim 2 isomorphisms, which don't you get

[coarse musk]

how is part 2 equal to part 3

[coral shale]

because 2x = 0x

[coarse musk]

the second one

[coral shale]

oh sorry, i should write -1

[coarse musk]

ohhhh

[coral shale]

but ofc, thats the same ideal

[coarse musk]

I'm dumb sorree

yea -1 = +1 under mod 2

[coral shale]

no its ok lol

[coarse musk]

oh okay so (1) is a trivial ideal

i get ittttttt

[coral shale]

i dont think trivial is the name

(0) is the trivial ideal

(1) is the ring, or more generally (unit) is the ring (prove this!)

[coarse musk]

oh

wait it says show that the ring dies if u kill off a unit

[coral shale]

thats a bit informal

[elder wave]

[coral shale]

I think it wants you to show R/(unit) = trivial ring

And you can do this by showing (unit) = R

this is defo an exercise worth doing

[elder wave]

Im gonna use that

Rings dying

[coarse musk]

Ok so u can say that 1 maps to 0

Which means every element maps to 0?

[coral shale]

uh

[coarse musk]

As it's an ideal in this situation

[coral shale]

R/(1) case sure

[coarse musk]

Yep

[coral shale]

but your reasoning isnt watertight for a general unit

You need to show an ideal containing a unit is the entire ring

[coarse musk]

How would u do that?

[coral shale]

Thats an exercise for you

if ur stuck, ask for hints

[coarse musk]

Ohh

Okay I'll give a shot!

Thanks a lot for this btw! I get the general idea of how this work

Works

[dreamy chasm]

could anyone help me with this problem? I think the quotient ring is a field, because x^2+x+3 is irreducible. how can I find the inverse of x?

[south patrol]

Well, I think you should check whether it is irreducible again

[quiet pelican]

It’s not irreducible

[south patrol]

Lol

[coral shale]

oops finite ring

[quiet pelican]

Hint: you’re not working over Z

[dreamy chasm]

oh. I thought it wasn't bc it couldn't be factored

[south patrol]

It can

[quiet pelican]

It can’t over Z, but it can be over Z5

[south patrol]

Since it's degree 2, you can just check whether it has any roots

And since you're in a finite field, that is a finite task

[coral shale]

=< deg 3

[south patrol]

Indeed

[quiet pelican]

It’s also a finite task in Z
Size arguments bound the sizes of roots quite easily

[south patrol]

sure I mean

Even the worst possible method is finite

[quiet pelican]

The worst possible reasonable method

You could say, only ever check 0

[dreamy chasm]

well 1 is a root, right? since we're in Z_5

[quiet pelican]

Infinitely many times

[quiet pelican]

Yes

[south patrol]

So x-1 is a factor

Not irred

Also like follow up question for you: why doesn't the question allow for the quotient to be an integral domain but not a field?

[quiet pelican]

5 = 0 in Z5

[dreamy chasm]

whoops. lol

disregard what I said then

[formal ermine]

fwiw pZ is maximal iff p is prime

[south patrol]

Okay, well what I was gonna point out was that any finite integral domain is a field

if R is a finite integral domain and x a nonzero element, then multiplication by x gives you an injective map R -> R

But R is finite, so that map is surjective

That is, there is an element y in R such that xy=1

So x is a unit

[fossil shore]

Hey, don't we need X to be countable here, or am I missing something?

[coral spindle]

It does not need to be countable, no

Hint for showing this: ||indicator functions||

[fossil shore]

I was also using indicator functions but had a hard time showing surjectivity

[coral spindle]

Keep trying

[formal ermine]

let $A$ be an $A$ module, is there any nice condition when $A$ can be decomposed to a direct sum of ideals? i.e. [ A = \bigoplus a_i ] other than when $A$ is a direct product as a ring?

[void cosmos]

ig check out artin-weddeburn theorem

[coral spindle]

We call such rings semisimple, and there are several equivalent conditions.

I realised after I wrote this that you may not have been talking about a decomposition into a direct sum of minimal ideals.

If $R = I \oplus J$, consider the $R$-linear projection map $\pi \colon R \to I$, and in particular consider the image of $1$, which we will call $e = \pi(1) \in I$. Now $e^2 = e\pi(1) = \pi(e) = e$, so $e$ is an idempotent. Now try verifying for yourself that $I = eRe$.

I'll have to keep writing without it then

[lethal dune]

yea decomposing into ideals is same as decomposing 1 into idempotents

[coral spindle]

To finish this thought

$J$ and $I$ are therefore defined by certain idempotents which we'll call $e_I$ and $e_J$. Try verifying that these two idempotents commute.

Furthermore, suppose that you have commuting idempotents that sum to 1. Try showing that this then gives you a decomposition into ideals.

In fact I may have left out that the product of the idempotents should be 0 – perhaps you can check this

[sonic coral]

Can I get a hint for this one? I have shown that H intersect K is a subgroup of G if both H and K are subgroups of G, but thats as far as ive got.

[barren sierra]

What's the definition of N_G(H ∩ K)?

Start there

[sonic coral]

the set of all g in G such that g(H intersect K)g' = H intersect K

[barren sierra]

Ok so start with an element h in H

You want to show that this condition is satisfied

so in particular h(H ∩ K)h^-1 ⊆ H ∩ K right?

[sonic coral]

since h in H is also in G, yes

[barren sierra]

So start trying to show that, and let me know where you get stuck

[sonic coral]

Since H cap K is a subgroup, contained in H, then it is closed and so multiplying by h and h' still gives an element in H cap K?

[barren sierra]

Errrrr

That's close but not right

Closure means that multiplying an element of H ∩ K by an element of H ∩ K gives you an element in H ∩ K

You do not know if h is in H ∩ K, just that it's in h

[sonic coral]

yeah i see that now

[barren sierra]

But you're close

Or at least in the right direction

[sonic coral]

i know H cap K is also contained in K but i dont think that will help me either

[barren sierra]

not just containment, you're given two pieces of information about how H ∩ K is normal in H and K

[sonic coral]

Since H cap K is normal in H, there hk = kh' for some h' in H, right? This is just the definition of normal

[barren sierra]

Yes

So can you rearrange that to maybe be more helpful about proving this?

[sonic coral]

yeha im thinking multiplying on the right by k^-1

so h = kh'k^-1

[barren sierra]

Close. The definition (which is equivalent, check it) of normal that I like is that N is normal in G if for all n in N, g in G, we have gng^-1 = n

[sonic coral]

the n's dont necessarily have to be the same though, right?

[barren sierra]

They are

They don't have to be though cause of closure properties I guess

I think

Oh wait NVM I'm dumb they don't have to be the same

Ok so N is normal in G if for all n in N, g in G, we have gng^-1 = n' for some n' in N

So it's an element in N

[sonic coral]

thats why i like my definition bc it helps me remember that they need not be the same

[barren sierra]

And what do we want to show?

[sonic coral]

h is in H cap K

so in particular we need it in K

h = kh'k^-1 this implies that k is in H?

i think i got that backwards

[barren sierra]

Write out a full proof, as much as you can, in one go. I'm a bit lost myself about what you have shown and what you haven't shown, and also I think you actually have everything you need and you will see that if you try to write it out in one go

[sonic coral]

okay

[barren sierra]

What is k?

[sonic coral]

k is in K

[coral shale]

In particular h = hhh^-1 = ...

[sonic coral]

h(kh'k^-1)h^-1 = (hk)h'(hk)^-1

[coral shale]

also for some h' in H n K surely right?

wait i think i misread

I don't think I follow that . . .

https://i.imgur.com/oXJjDfj.png

k in K, h in H

surely h needs to live in H n K for this to hold

or did i misremember

[sonic coral]

i dont know im lost

[coral shale]

ok ok

N normalsub G
iff
(forall g in G) gHg^-1 = H
iff
(forall g in G)(forall h in H)(exists h' in H) gh'g^-1 = h

I think.

[sonic coral]

x in G?

[coral shale]

And I don't think this quite corresponds to what you've written

or wait

oh oh

ok ok lemme think

[sonic coral]

i think gHg^-1 needs to be contained in H, not necessarily equal to H

[coral shale]

iirc, these turn out to be equivalent

equal is the definition ive nicked from the internet

[sonic coral]

okay, not sure if it matters

[coral shale]

oh right yh poorly written by me - if its equal, i cant quite write that implication

Im not entirely sure if you've swapped it

Let A = H n K. Let h in H, a in A.

hAh^-1 \subseteq A
=>
(exists a' in A) a' = hah^-1
=>
(exists a' in H) a' = hah^-1

==========================
This is what I get

(taking this defn)

[sonic coral]

thats the defn gallian uses also

[coral shale]

im unsure how to get to what you wrote

[sonic coral]

so where did i go wrong

[coral shale]

well its more like how did you get that statement

[sonic coral]

i think i need to take a break from looking at this lol

[coral shale]

I don't see how to get to it from the definitions

I suspect its not necessarily true but uh

yh sure, take a break

[sonic coral]

okay well lets do it the way you did it, let A = H cap K

[coral shale]

I generally start by writing the defns down and then making sure every statement I derive are from those directly

[formal ermine]

I smell cap

[coral shale]

https://i.imgur.com/jJvL2h0.png

Should be all the defns we need

Let h in H is a good start ofc. But its not necessarily clear how you want to use it in the normal subgroup defns

So I would use letters other than h to state those and see what can be done with some manipulation

[sonic coral]

so since H cap K is normal in H, h(H cap K)h^-1 is contained in H cap K for all h in H?

[coral shale]

for all a in H cap K

[sonic coral]

fixed it?

[coral shale]

sorry back. i gtg soon, if you're still stuck later and no one else around ill have a look again

N := N_G(H n K) = {g in G : g(H n K) = (H n K)g}

[sonic coral]

ill look at it tonight or tomorrow, ive been lookin at this for too long

[coral shale]

want to show forall h in H : h(H n K) = (H n K)h

yh gl. seems to be a matter of writing defns down nicely (avoid overload letters)

and letting A = H n K might make it more readable

I think N normal subgroup G := forall g in G : gN = Ng is better normal subgroup defn, from just looking at the above

indeed - it seems you need to show H n K normal subgroup of H

but thats given wtf

ok im confused

[sonic coral]

I think I will just go back to trying to show groups of some order are not simple and ask my professor about this on tuesday

[chilly ocean]

any clever way of doing this im missing? i just tried some stuff till i found an element that isnt in the group that should be

[formal ermine]

91 is a power of 3

[void cosmos]

yea

[formal ermine]

so its multiplicative group is cyclic

therefore all of its subgroups are also cyclic

[coral spindle]

91 = 7 x 13...

[formal ermine]

oh lmao

[south patrol]

Lol

[formal ermine]

I thought 91 = 3^4

[south patrol]

The sum of 9 and 1 is 10

[coral spindle]

This is your Grothendieck moment

[south patrol]

Not even divisible by 3

but dw lol

Tbf I would assume 91 is prime

[chilly ocean]

lol

i also havent gotten to cyclic groups yet

[formal ermine]

cyclic groups are groups generated by one element

qed

[south patrol]

Lol

[coral spindle]

Fun divisibility test for 13: a number 10a + b (where b is the final digit, and a is all but the final digit) is divisible by 13 iff a + 4b is divisible by 13. In the case of 91, we have a = 9 and b=1, so a+4b = 13 is clearly divisible by 13.

This comes from the fact that 4x10 - 3x13 = 1.

[south patrol]

Oh nice that is cool

There isn't really one for 7 that doesn't suck right

[coral spindle]

Well there is a similar one

[chilly radish]

a-2b

[coral spindle]

since 3x7 - 2x10=1, using a similar super secret technique, you get what ShiN just said yeah

so in this case we would get 9 - 2*1 = 7

It's quite fun to try and rederive this, I really suggest you give it a shot

[chilly radish]

Gave this as a hw problem for my students

[coral spindle]

That's a cool hw problem

[south patrol]

nice

i don't like it cause 10 is too arbitrar

y

give me the number in base 7 and then i can tell you if it's divisible by 7

[formal ermine]

bases are stupid. just write everything in base \infty

[coral spindle]

Actually, bases are based

See also cohomology: https://ncatlab.org/nlab/show/carrying#in_terms_of_cohomology

[quiet pelican]

Yes I do really like the number jkłx=ŷ (note this isn’t some form of place value system, this is just a randomly assigned identifier for that number)

[fervent rock]

isomorpyhisms preserve group properties right?

like abelianess and cyclicity

[rustic crown]

yeep

[fervent rock]

bless

[rustic crown]

boytjie uwu

[coral spindle]

Hey det

I was gonna try and be funny and say that this is the definition of a group property, but I decided not to

[rustic crown]

i wanna hear one more story about your kids :p

[coral spindle]

I'll try and adopt some more

[fervent rock]

So its asking me to show that D4, A4, Z_12 and U(21) are not isomorphic to eachother at all. Z12 is cyclic and thus abelian, U(21) is just abelian, D4 and A4 are not abelian. Z12 cant be isomorphic to any of them and neither can U(12). Only thing is to show D4 is not isomorphic to A4 which I can show by listing the order of the group elements?

[coral spindle]

Sure

[fervent rock]

beautifu; thanks bae

[coral spindle]

[upper pivot]

I know this isn’t useful to what you want, but you can formalize what is meant by group property. Any first order sentence in the language of groups is preserved under iso (these are things you can make with basic logical connectives and quantifies, and symbols from the group).

[coral spindle]

We often talk about things that are not first-order sentences though, so this isn't really sufficient imo

[fervent rock]

this is an undergrad algebra course that i just want to pass tbh, this stuff is decent but not my stuff

[upper pivot]

Yeah true

[coral spindle]

e.g. you cannot define what solvability is in the language of groups

and clearly solvability is a group-theoretic property

[rustic crown]

teach det logic >.<

[fervent rock]

i just reckoned that isomorphisms gotta preserve the shits.

[coral spindle]

I guess you'd need some more logic stuff

[fervent rock]

and went from there

[upper pivot]

Right, maybe we need to go second order for those

To quantify over subgroups

[coral spindle]

If Oby were here he would talk about how second order logic is fake

[upper pivot]

But I have 0 exp with second order logic oof

[fervent rock]

nvm its D6 but point still stands

[rustic crown]

lol i read D4 as D6 earlier :p

[fervent rock]

yea i misswrote

[coral spindle]

One way of approaching that might be to count the number of elements of order 3 in those groups btw

I think that should be enough to distinguish them

[fervent rock]

yea

[coral spindle]

I leave the details to you

[formal ermine]

does every nth order logic exist

[fervent rock]

i was just gonna list the orders but makes sense

cause D6 only has the rotations that are order 6 and refelctions that are order 2

while A4 is just bullshit

[coral spindle]

D6 has rotations that are of order dividing 6

[fervent rock]

sure

[coral spindle]

It also has e.g. rotations of order 2 and 3

[rustic crown]

rotation of order 1

[fervent rock]

hmm

oh yea true

but like considering the order of just the sungular rotation by 360/6

[coral spindle]

But anyway my point is, I think you can exclude certain orders in A4

[fervent rock]

something like that

ohhh i get it

a4 have no order 6 but d6 DOes

so its like illegal

[coral spindle]

true you should call the police

[fervent rock]

i will

[coral spindle]

arrest A_4

[fervent rock]

algebra isnt that bad

but like it hurts my brain sometimes

i felt cheacky about this question tho

thxxx

[rustic crown]

[fervent rock]

I have a related question

lets say we have a group on a set of n elements

is there a theorem about the maximum order of the individual elements

or it depends on the operation

[coral spindle]

Well I mean the order of any element is most certainly <= n

is this what you're looking for?

[fervent rock]

yea basically

[coral spindle]

When you see the Theorem of Lagrange you will see more information

[fervent rock]

like the optimization dude?

he was doing this stuff?

[coral spindle]

The same guy, yes

[fervent rock]

crying and shitting rn.

[coral spindle]

Something he can no longer do

[fervent rock]

literally worst news ive heard all semester

[rustic crown]

the lagrange multiplier guy?

[fervent rock]

yea

[rustic crown]

except bernoulli are there many mathematicians with the same name?

cuz i wonder if this lagrange is same as the lagrangian lagrange and the lagrange interpolation lagrange

[fervent rock]

wait arent dihedral groups just permutations on n elements

[coral spindle]

The Neumann couple is a good example

All groups are permutations, under some interpretation

[rustic crown]

yoneda

[fervent rock]

crying so bad rightnow cause i coulda just said its illegal cause d6 is on 6 elements but a4 was on 4

not onto at al

https://tenor.com/view/krill-goodbye-krill-goodbye-krill-world-suicide-lober-gif-26900795

[coral spindle]

no that doesn't work

[fervent rock]

it doesnt

[coral spindle]

it doesn't.

[fervent rock]

i believe oyu

[coral spindle]

That's not a group-theoretic property.

You can define groups in different ways and produce the same results

for example, the dihedral group with 6 elements and the symmetric group S_3 are isomorphic

[fervent rock]

okay so in the end. d6 has r_(60 degrees) which is order 6 but a4 has no element order 6 so boom bam boom

done and dusted

[coral spindle]

According to Wikipedia at least, yes it's all Joseph-Louis Lagrange

[fervent rock]

why cant they stick to one thing

why reinvent the wheel

[south patrol]

gm

Yes Lagrange was a big dude

[formal ermine]

"just use Euler's theorem"
"which one"

[rustic crown]

use Gauss' lemma

[south patrol]

Even when you say like Gauss' lemma for polynomials or whatever you still gotta narrow it down

😿

[formal ermine]

are there only two gauss' lemmas

or more

[south patrol]

I guess two ones that was more focused on in ug curriculum lol

[formal ermine]

in german there's only one lemma

the other one is a theorem

[south patrol]

Lol

Nice

Which one is the theorem

cause to me Gauss' theorem is the divergence thing lol

[formal ermine]

R irred iff frac(R) irred is the lemma

[south patrol]

Lol okay sure

Lol Gauss' lemma in NT feels more like a lemma than the one for polys

[rustic crown]

so the other one is D UFD => D[x] UFD?

[south patrol]

Bruh

[formal ermine]

yes I think so

let me check

[south patrol]

Oh to me Gauss' lemma is like

the one on quadratic residues

[rustic crown]

primitive * primitive = primtive

is this also gauss' lemma?

[south patrol]

that's basically the same as this

I think it's probably fair to call those basically the same lemma

[formal ermine]

ye

[rustic crown]

true true

[south patrol]

Oh okay

[rustic crown]

faktoriell

[south patrol]

genau

[formal ermine]

potato why do you have 100k messages in the german server

[south patrol]

LOL

Why do you bring that up now

or did my saying genau make you check

[formal ermine]

I rember

[south patrol]

Uh

Haven't you asked me this before 😭

[formal ermine]

and like

[south patrol]

But nah I mean I like the language and spent a while on there lmao

[formal ermine]

99k of those are in the math channel

[formal ermine]

I forgor

[south patrol]

I don't think that's true lol

Well

Like in recent years I basically only message that sure

Yeah lol

7k in science channel

[formal ermine]

deutsche sprache schwäre sprache

[south patrol]

and about 117k in total

[south patrol]

lol

okay i haven't seen that joke

deutsche sprache kringelige sprache

jk

[formal ermine]

wat

[south patrol]

like

why did you say schwäre lol

[formal ermine]

pun

[south patrol]

yes

[south patrol]

oh

i thought you were going

"wat" to this message

lol

aalhbgkjahbdlgiukjhgb

[formal ermine]

idk what kringeliege is supposed to mean

[south patrol]

lol

anschnursprache

[fervent rock]

okay so i have one more question

one step test for subgroups, lets say H subset of G, i gotta show that for a,b in G, ab^-1 is in H and H is nonempty

two step, i show closure and inverse seperatly

[formal ermine]

and non emptyness or identity

[fervent rock]

yea true

i just do identity

[primal tusk]

R is a domain. If i have a polynomial in R[X1,.... Xn] that is a form, how would i prove that any factor of a form is also a form

[coral spindle]

What is a 'form'?

[naive wedge]

Multidegree adds when you multiply

[primal tusk]

"We call F homogeneous, or a form, of degree d, if all coefficients a(i) are zero except for monomials of degree d"

[coral spindle]

Suppose f = gh is homogeneous but g is not homogeneous. Inspect the monomials of the product for contradiction

in particular, inspect the multiplication by some highest-degree term in h

[primal tusk]

ah okay thanks thats a good idea

[white oxide]

would it be possible to take m = n for #47 to show that there exists an element a in G such that a^n = e? bc the hint says to use the result of exercise 46 and theorem 10.12 (which is that the order of an element of a finite group divides the order of the group) but why wouldn't this work?

[south patrol]

I don't really get what you mean

[white oxide]

like

[south patrol]

The case m=n always holds

[white oxide]

it feels like cheating to take m = n

[south patrol]

It says for all m

[white oxide]

bc the hint says to use the theorem and the exercise

yeah ik

[south patrol]

so how can you just take m=n

[white oxide]

i know it's because it says for all m

i was just confused as to why they suggested the use of the theorem and the result of the previous exercise

[south patrol]

wait what lol

I mean yes you will need the theorem and previous stuff

[white oxide]

to show that x^n = e?

[south patrol]

Huh

[white oxide]

nah you just use the hypothesis right

nah im just talking about showing x^n = e

where n is the order of the group

[south patrol]

But you don't need to show that

It says element of order n

Just because x^n = e for all x by Lagrange doesn't mean there's an element of order n

Or otherwise all finite groups would be cyclic

[white oxide]

wait im confused

isnt that the definition of an element's order

[coral spindle]

No

[white oxide]

oh wait

the definition of an element's order is the subgroup it generates right

[south patrol]

The order of the subgroup it generates

[coral spindle]

an element x of order 2 also satisfies x^4 = e, x^6 = e, etc.

[white oxide]

would it be a decent idea to use exercise 46 and maybe take d = n

[sonic coral]

it’s the smallest positive integer n such that x^n = e

[white oxide]

wdym

[coral spindle]

Jaxon is defining the order of x, although we're past that now

[formal ermine]

how do you call the mathcal O?

(was it matcal?)

the one used for ring of integers

[rotund aurora]

$\mathcal O$

[coral shale]

why not Z

[coral spindle]

Not that ring of integers

[coral shale]

oh ok.

[coral spindle]

https://en.wikipedia.org/wiki/Ring_of_integers this one

[rotund aurora]

The notation Z_K and Z_L given the field extension K<L is pretty nice too tho

[coral shale]

thought 'algebraic integers' but alright

[formal ermine]

yeh it's mathcal

[south patrol]

Indeed

[formal ermine]

this one

if you read it out loud

do you guys just say

"O"

[south patrol]

Lol

[formal ermine]

or do you go "fancy O"?

[south patrol]

I would say "the ring of integers of..."

[formal ermine]

just like when you see \ell

[coral shale]

[south patrol]

Lol

[rotund aurora]

OOOOOOO

just scream like really loud

[formal ermine]

now try saying that 10 times in a row really fast

[agile burrow]

evocative

[coral shale]

its oh, ohkay?

O_k okay

[south patrol]

Why would you need to do that

😿

[formal ermine]

I have my use cases, ok?

[coral shale]

no u dont

[formal ermine]

yes I do

[coral shale]

ur a highschooler.

no highschooler has a usecase for algebra

prove me wrong

[formal ermine]

did I stutter

here in germany we write a little thesis in 11th grade

gotta prepare for that

[coral shale]

no

[formal ermine]

yes

[white oxide]

would somebody mind explaining the sentence starting with "By the hypothesis..." im confused about how there can only be one subgroup of order d dividing n follows

sorry le tme post the question

[south patrol]

Well there are at most d solutions to x^d = e for each d|n

Now use the first line

[white oxide]

ah i see

thank you~!

[dreamy chasm]

can anyone help clarify what I might be doing wrong here?
x = -4 +_ sqrt(16-b) via the quadratic formula
so we need 16 - b to be a square for the polynomial to have a root
that works for b = 0, 7, 12, 15, 16 (these numbers mod 11)
which are 0,7,1,4,5

[rotund aurora]

you are doing the opposite, you want the polynomial to have no roots

also, for b=2 you also obtain roots

but the main argument is fine (you solved the problem anyway, just make sure b is not any of those values)

[dreamy chasm]

so I should be looking at b =/= 0,7,1,4,5,2?

[rotund aurora]

you want x^2+8x+b to be irreducible

so 16-b=5-b should NOT be a square

there are (p+1)/2 quadratic residues mod p (including zero) and (p-1)/2 quadratic non-residues

[rotund aurora]

yes

[sonic coral]

Let’s say I have 2 Sylow p-subgroups of order p^n, I am not guaranteed that the elements in them both are distinct, right?

[delicate bloom]

they both have the same identity element

[sonic coral]

well yes, how about nontrivial elements

i’m trying to show groups of a certain order are not simple. and i think i’m gong about it the wrong way by not considering the nontrivial intersection between them

[south patrol]

Well, you can often do it in the case n=1, but no it needn't be true in general i guess

[sonic coral]

this counting argument i went with for a group of order 12 and 30 works fine though? I’m just having trouble depicting which orders this argument won’t work and i’ll have work a little harder for.

[static needle]

I got a thing ive been thinking about: How does Z/nZ relate to left cosets?

[knotty frigate]

Z/nZ is a set of cosets

and bc Z is abelian a left coset of a subgroup is also a right coset of a subgroup

[blissful crystal]

does anyone perhaps know what Jacobson means when he says "F(t)[x], t an indeterminate"

[next obsidian]

it just means a variable

[solar glacier]

Indeterminate = variable

[blissful crystal]

i mean yes

but I'm mostly confused as to why he doesn't write F[t][x] or F[t,x] since he used that notation before

[prisma ibex]

F(t) usually means you're adjoining both t and 1/t

so F[t] is polynomials, F(t) is rational functions

[next obsidian]

no this isn't accurate

you're adjoining every possible fraction

adjoining just t and 1/t won't give 1/(t+1) for example

[prisma ibex]

ah yes you're right

[next obsidian]

but the part about rational functions is true

[prisma ibex]

I was probably thinking of F[[t]] versus F((t)) where you can take Laurent series expansions of these rational functions

[remote nymph]

I have to find an example of a latin square that is not the multiplication table of some group and im really struggling with this. I found this random pdf that i feel should be helpful but I dont understand this paragraph

http://web.math.ucsb.edu/~padraic/mathcamp_2012/latin_squares/MC2012_LatinSquares_lecture3.pdf

Why can we assume that f is the identity here? at least thats what i think this paragraph is stating

[solar shore]

a definition from my abstract alg textbook

i thought the set B is referred to as the codomain?

and b is the image

and the subset of B is the image of A

(or range ive also seen that)

is this just me being pedantic?

[barren sierra]

image is used in two contexts here

say phi(a) = b

then b is the image of a

but the image of the whole function is the set {phi(a) | a in A}

[tender wharf]

hello fellow gallian enjoyer

[tender wharf]

do you mean $\func{\phi}{A}{B}$

[tender wharf]

what this generally means (and I believe it also means this in gallian's book) is that phi must be defined on all of A and it has a codomain of B (but not necessarily a range of B)

the image of A by phi is a subset of B yes, or in other words, $\phi(A) \subseteq B$

[solar shore]

right

no i was just

when they said range, i was confused for a second because i usually hear the subset of b is the range

while b itself is the codomain

i think im just being pedantic

[lapis plume]

I am doing part b of the problem. I stuck at the yellow highlighted place, as sth seems to go wrong. Can anyone help me to have a look?

[cunning dust]

It looks like you're trying to prove something you're supposed to disprove. You've overcomplicated a bit too.
In any case m doesn't divide r and k doesn't mean it doesn't divide rk (take for example m=6, r=2, k=3).

[lapis plume]

I see

Umm but what should I do? I can’t find a good reason why $\phi (r+\hat{k})\not=\phi (r)$.

[cloud walrus]

Trenton

[cunning dust]

I think a good way to give you intuition (this also pretty much solves it) is to just notice phi(a) = a (mod m). So you can just use modular arithmetic. And use the hint they gave.

Like consider what phi(1+n-1) is and what phi(1) + phi(n-1) is

[lapis plume]

Got it now, thanks!

[tacit hemlock]

can someone please explain why the cycle notation yields a (135)?

for a composition of permutations

if this is "do p, then q", then wouldn't it go something like:

3 -> 4 (from p), then 4-> 5 (in q)

4 -> 1, 1 -> 4

2 -> 5, 5-> 2

so wouldn't the composition yield (5 4 2) instead?

[coral shale]

3 goes to 5
4 goes to 4

2 goes to 2

how is that expressed in (5 4 2) !?!?!?

[tacit hemlock]

My bad, so would it be (5) (4) (2) and hence not work

[coral shale]

no

3 goes to 5 goes to...?

[tacit hemlock]

2

[coral shale]

u should b chaining these

no it doesnt

5 maps to 1

so 3 5 1

[tacit hemlock]

what

[coral shale]

qp(3)=5

qp(5)=1

qp(1)=3
qp(2)=2, qp(4)=4

(3 5 1)

[tacit hemlock]

yes, but why is qp(3) = 5

Do I do each element one at a time?

[coral shale]

u said it yourself

[tacit hemlock]

i.e., in p, 3 -> 4, then compose by q so that 4-> 5

[coral shale]

well ofc im only talking about the composition

why would what happens in between matter in the final expression

[tacit hemlock]

then qp(5): starting with p, 5-> 2, then map with q so that 2-> 1

[tacit hemlock]

wait, why are you doing it in the order 3,51

as in qp(3), then qp(5), then qp(1)

[coral shale]

let f = qp

f(3) = 5
ff(3) = f(5) = 1
fff(3) = ff(5) = f(1) = 3

[tacit hemlock]

I see! Thanks!

[coral shale]

hence ive just found a factor (351)

cool.

[tacit hemlock]

What is the squiggly arrow notation?

I thought it meant "implies" when used to denote a chain of reasoning, or show the next logical step with laying out your working

[formal ermine]

I think it means t is send to g(f(t))

[summer path]

i think it's the same thing as $\mapsto$

[cloud walrus]

Tubular Cat

[south patrol]

squiggly is cute

[coral shale]

so is det >u<

[rustic crown]

.<

[summer path]

shuwi cute voice :3

[tacit hemlock]

I thought "cardinality" is generally the term used to describe the number of elements in a set?

"The order |S| of a set S"

is this an abuse of terminology?

[tender wharf]

order kind of is only defined for groups?

you wouldn't say order of an arbitrary set

like for example the set of all cats in the world

I guess you could talk about the cardinality of the set of all cats in the world though

but for all intents and purposes you'll get the same value out of it

if G is a finite group then order of G = card G

[tacit hemlock]

Do you guys mostly learn through textbooks or lectures

[tender wharf]

thats more of a #discussion question

[tacit hemlock]

can I make it personal and ask you specifically 🙂

[tender wharf]

textbooks

[summer path]

i mean if it's a class, generally lectures, but also textbook is nice to follow along sometimes; but otherwise textbooks

[tacit hemlock]

What is an intuitive understanding of a kernel of a function?

i.e. the kernel of a homomorphism

[coral spindle]

The definition

Once you see the first isomorphism theorem, you'll get a better idea what it means

[tender wharf]

a normal subgroup

jokes aside every normal subgroup is actually the kernel of a homomorphism, yes

[pastel cliff]

can i get a nudge on this pls

cutie module

[wooden ember]

You’re looking for all polynomials in L that are null

Do you have an example of such a polynomial?

[pastel cliff]

wdym null

like has roots or something

[wooden ember]

As in = 0 matrix

Think in terms of L being any matrix, not the one given

Given a matrix how do I get a polynomial that when evaluated at this matrix is = 0

[pastel cliff]

hmmmm

one sec

[stark cloak]

Hi, there's theorem that says that every permutation can be written as a product of transpositions.
What about a permutation in cyclic form a = (1). What's its decomposition into transpositions?

[pastel cliff]

that's it

[quiet pelican]

You could either do the empty product, or sth like (12)(12)

[pastel cliff]

that's just the identity element

[tender wharf]

i mean thats just the identity permutation lol

wdym transpositions

2 cycles?

[pastel cliff]

this is like saying 1 = .5 * 2

[wooden ember]

It’s explicitly writing the identity as a non empty product of transpositions though so a fair answer

[quiet pelican]

Yes
But it’s still a decomposition

[stark cloak]

good

thanks

Also in cycle notation I believe that writing (a_k, a_k) to denote one-cycle is ill formed, right?

[tender wharf]

thats...

yeah

thats not right

you just ignore it

so for example if f is the permutation that sends 1 to 2 and 2 to 1 and fixes 3

[stark cloak]

Good. That would contradict odd-even theorem.

[tender wharf]

then its just (12)

i guess you could write (12)(3) but its redundant

and implied

[wooden ember]

In general the notation for an n-cycle should only contain distinct elements

[coral spindle]

I have seen people try to write (1 2 3 2) and the like 😦

[pastel cliff]

ok with regards to this i think i need to get clear on what everything is first bc im realizing maybe i dont actually get the question lol

we're treating Q x Q as a Q module

t being an element of the Q that's acting as M if we were talking about an R module M

that means a_i is coefficients in Q that's acting as the ring if it were an R module M

but what is v supposed to be

[coral spindle]

A 2-vector with entries in Q

I.e., an element of Q^2

[pastel cliff]

so asking what Ann(Q) is here is asking what elements of Q we can put into a_i to get the zero matrix right?

perhaps im overthinking but how does the operation come into play?

[coral spindle]

That is what it's saying, yes

after some translation

[feral agate]

Can anyone give me a hint on how to start on this one? m is integer greater than 1 btw

[pastel cliff]

do you have any guesses

thing about gcd and/or lcm

[feral agate]

So dZ where d is any divisor for m expect for 1

[pastel cliff]

try and prove it

[barren sierra]

Why except for 1?

[feral agate]

It is given in the exercise

[feral agate]

I have shown this, but I don't see how I can argue these are all the subgroups that contain mZ

[coral spindle]

Give it a shot

[barren sierra]

Oh so it only wants proper subgroups (since 1 generates Z)? I see

[analog zephyr]

For 2, 8Z<4Z<2Z

[formal ermine]

what?

[summer path]

it's a chain of subgroups, though i don't think that really helps here since it's already been established that those are subgroups containing mZ