groups-rings-fields

[tardy hedge]

thats just the terminology d/f uses for it

you kind of visualize it as those elements "projecting" onto 1

[vagrant zinc]

Can this be considered as a fibres ?

[restive birch]

,rotate ccw

[restive birch]

at the bottom there

confused by the conclusion H has p left cosets -> G/K is isomorphic to a subgroup of S_p

wouldnt this just mean that G/H is isomorphic to a subgroup of S_p?

[restive birch]

wait i think i understand? we take G/K = G/ker pi_H isomorphic to im(pi_H)?

[bitter rover]

You don't take that; that's the actual statement of the first isomorphism theorem!

It's actually an example of what I said earlier to someone else:#advanced-algebra message

The first isomorphism is just that G/ker(f) ≅ im(f) for a group homomorphism f

[restive birch]

right

i forgot K = ker pi H

[bitter rover]

The proof will then conclude that |H:K| = 1, so K=H

But K is normal, so H is too

[restive birch]

okay so the problem is:
-show Q8 is not isomorphic to any subgroup of S_n for n < 8
the hint is:
-show that if A acts on any set of size < 8 then the stabilizer of any point must contain -1
im just not really sure where to go here? maybe its just late and im tired but i cant seem to conclude much about the structure of the induced group action

[rocky cloak]

Use orbit stabilizer

[golden turtle]

"The map sending N to the abelian group homR(M,N) often denoted homR(M,-)"

What is the map?

Also what is a functor?

And covariant or contravariant?

I don't know why my professor just includes these as if they are definitions and does not elaborate at all

I read in Appendix II of the text, but its just category theory

[rocky cloak]

The map that takes in a module N and gives out an abelian group Hom(M, N).

I doubt it's very important what a functor is, if you don't know what a category is, but I guess you can read more about it in the appendix or on Wikipedia.

In essence it's just what is described here. Like a functor is a mapping for example from modules to abelian groups, together with something that takes module homomorphisms to homomorphisms of abelian groups in a compatible way.

So for every homomorphism N -> P you should get a homomorphism Hom(M, N) -> Hom(M, P)

[rocky cloak]

A functor is a "homomorphism of categories", so learning about them is indeed just category theory

[golden turtle]

I have no idea how this map works though because isn't the N in Hom(M,N) arbitrary

[rocky cloak]

Yes, you assign to every module N, the abelian group Hom(M, N).

So that's a map from the set of all modules to set of all abelian groups

(or collection not set whatever)

[golden turtle]

how is this a map at all when its output does not depend on its input

[rocky cloak]

How do you mean?

[golden turtle]

also how can we denote it hom_R(M, -) and not even have this regarding N

[golden turtle]

Like, hom(M,N) is the set of homomorphisms from M to N but N is all arbitrary R modules

so the N doesn't even matter

[rocky cloak]

Like take R=Z and M = Z/4, say.

Then this maps Z/2 to
Hom(Z/4, Z/2) = Z/2

It maps Z/3 to
Hom(Z/4, Z/3) = 0

It maps Z/4 to
Hom(Z/4, Z/4) = Z/4

And so on for every R-module

[golden turtle]

but I'm saying, it doesn't matter what you call the second entry

Hom(Z/4, Z/4)=Hom(Z/4, Z/3)= Hom(Z/4, Z/2) So I dont understand what is going on at all

[rocky cloak]

So if you define a function f(x), then x can take any arbitrary element in the domain of that function.

But if course it matters what x is. For example if f is a function from real numbers to real numbers, f(1) and f(2) might be different

[golden turtle]

but if I say like, let S(y,x)={(y,x) with x in R}
and map n to S(y,n)

then it doesnt matter what n is

and it is like the same here

like either way you get the same thing ...

[rocky cloak]

It's nothing like saying that. It's like saying
f(x) = x^2

[golden turtle]

if homR(M,N) is the set of all homomorphisms from M to N, for N arbitrary

then its just the same as homR(M, X) for any X a module

[rocky cloak]

This is just a badly formed sentence

They want to consider the sets Hom(M, N) for all N. And they want to package them together into this mapping

They don't mean to put them all into a single set

[golden turtle]

So hom(M,N) does depend on N?

[rocky cloak]

Yes

The N in Hom(M, N) represents N

[golden turtle]

but in what way does it do that

[rocky cloak]

Like if N = Z/3, then
Hom(M, N) = Hom(M, Z/3) = the set of homomorphisms from M to Z/3

Which is an abelian group under addition

[golden turtle]

I thought he was defining it so that its all homomorphisms out of M and what theyre homomorphic onto doesnt matter since he said "for all modules N at the same time" 😭 wtf

[rocky cloak]

Yeah, it was worded very confusingly.

The point is that we want to understand all homomorphisms coming out of M, so we parametrize them by this mapping.

Just like if we wanted to understand all square numbers we might study the function f(x) = x^2 as opposed to studying some specific squares

[golden turtle]

Okay, sure

dear god

Now I just have to learn all the other 10 terms he did not define in the first page of our slides

[rocky cloak]

I think I would probably look ahead and see if you actually need those words, or if they're just a fun fact

[tardy hedge]

In my module theory class we proved the result that all modules can be embedded into an injective module and the proof used Hom as an R-module everywhere

I thought it was cool i think its one of the first times ive seen Hom as a module actually being useful

Instead of just being some esoteric thing of oh ok we can consider it as a module, sure

When first seeing that i dont think its that clear on why we would want to do this

At least for me idk man

[dull ginkgo]

That and relating to tensor products

A bilinear form b: X x Y -> Z gives a map from X to Hom(Y,Z) by sending x to b(x, •) and from Y to Hom(X,Z) by y to b(•, y)

[south patrol]

You'll doubtless have used this before: homs of f.d. vector spaces can be identified (after picking a basis) with vector spaces of matrices, and the vector space structure on those is definitely useful

[long geyser]

can someone help me understand why the last sentence is true

sigma is spectrum

[severe linden]

Sanity check: $\mathrm{End} (\mathbb{Q}, 0, +) \cong (\mathbb{Q}, 0, +)$, right?

[cloud walrus]

Yuese

[rocky cloak]

If p(a) - mu is not invertible, that means one of the lambda are in the spectrum of a. So the spectrum of a contains one of the roots of p(z)-mu.

And lambda being a root of p(z)-mu means p(lambda) = mu

[long geyser]

ohh

thanks

[rocky cloak]

In fact it is isomorphic to Q as a ring

[crisp willow]

is abstract algebra worth taking

[mighty kiln]

All of pure math uses algebra

[crisp willow]

lemme try then

[tardy hedge]

Somehow my prof has not introduced tensor products yet in our course

I imagine we will get to it but its halfway thru semester already

We have done the basics, free module stuff, proved the stuff about commutative rings having invariant basis number, and then i think we got to projective injective module stuff ?

[dull ginkgo]

HOW

BOTTOM TEXT

[tardy hedge]

Loool idk man

Maybe she doesnt like tensor products

[dull ginkgo]

Did you get to the horrible PID stuff

[tardy hedge]

Module over pid structure stuff ? No

[dull ginkgo]

Tensor? She hardly even know em

[tardy hedge]

Theres still some good time left in this semester so i imagine we still have a good ampunt to cover

[broken pollen]

if G is an abelian group with ℤ (the integers with addition) as a subgroup, and if G/ℤ ≅ ℤ, must it be the case that G ≅ ℤ × ℤ?

[rustic crown]

yee

(i'm assuming you haven't worked with SES and projective modules.)

[broken pollen]

no i havent

[rustic crown]

pick an element g in G such that under the iso G/ℤ = ℤ, [g] maps to 1

and try to show that the map ℤ x ℤ --> G given by sending (a, b) to a + bg is an isomorphism

[broken pollen]

thanks!

[rustic crown]

[broken pollen]

cant believe that isomorphism evaded me. i was chasing a mayer-vietoris long exact sequence and eventually got that my desired homology group (H₁ of a genus 1 surface) satisfied something of that sort, and was like "the homology just must be ℤ × ℤ now, right? what else would it be?" 💀

but could only conclude that it had to be countably infinite

[rustic crown]

wait looks like you ahve the background ><

[broken pollen]

noh i dont

[tardy hedge]

Wut

[broken pollen]

somehow fumbled my way into an algebraic topology class without seeing projective modules

[rustic crown]

0 --> Z --> G --> Z --> 0 is an SES, and Z is projective so this is a split SES, which means G = Z x Z

[broken pollen]

OH SES stands for short exact sequence?

💀

[rustic crown]

🙈

[tardy hedge]

Bruh Moment! Bruh Moment!

[rustic crown]

det be lazy lazy

[tardy hedge]

Oh Hell Nah!!

[sinful sorrel]

The Sylow theorem states that any two Sylow p-subgroups of G are conjugate. What if G has only one Sylow p-subgroup H, can we conclude that H is normal? I don't think the theorem states that the Sylow p-subgroups are conjugate with them selves as this would mean all of them are normal?

[topaz heart]

yes if it there is one sylow subgroup it's normal

[sinful sorrel]

Right yeah, I misinterpreted the theorem

[topaz heart]

if there is more than one then none can be normal, since by definition they are conjugate in G so their normaliser cannot be the whole of G

sry was still typing

[sinful sorrel]

Thanks though!

[restive birch]

sorry, can you write out explicitly how this works? i think maybe i didnt fully grasp things
i get that the size of the orbit of an element is the index of its stabilizer, but how does that help here?

is it just that the orbit of a point has size at most 7, so that the index of the stabilizer is < 8, so that the stabilizer is nontrivial, which it should be if the permutation was an isomorphism? but this seems wrong because it would seem to work for any group

[bitter rover]

The quaternions Q have order 8.

If Q were isomorphic to a subgroup of S_n for n < 8 then Q would have a faithful action on {1,2,...,n}. Consider n=7.

Orbit stabilizer says |Q| = |Orb(x)| · |Stab(x)| = 8. But the sum of the size of the orbits have to add up to 7.

What goes wrong?

[restive birch]

there has to be a nontrivial stabilizer?

[bitter rover]

Is that a guess?

[restive birch]

well this doesnt seem to use the specific structure of Q8

[bitter rover]

Let A = {1,2,...,7}. For any a in A, |Orb(a)| ≤ |A| < 8 which means |Stab(a)| > 1.

[restive birch]

which makes it seem like this would also apply to D8 which is false

[bitter rover]

You're right, that's not a problem per se.

[restive birch]

alright let me look at it for a few

[bitter rover]

Remember that the kernel of the action is the intersection of the stabilizers.

i.e., the set of elements which act like the identity permutation

So the intersection of the stabilizers is the set of elements that stabilize everything

And also remember that Stab(a) is a subgroup of Q

[restive birch]

ohhh right

because every nontrivial subgroup of Q8 contains -1

the reason this breaks down for D8 is that for instance <r^2> and <s> are (almost) disjoint

[dull ginkgo]

“Trivial intersection” is usually a good way to say it :3

[restive birch]

yea

so more generally, if G is a finite group where the intersection of all the nontrivial subgroups of G is nontrivial, G is not isomorphic to S_n for n < |G|?

[dull ginkgo]

I think there is a name for that

Nvm

[restive birch]

only examples i can think of are like Z_p and Q8

Z_(p^n)

[chilly radish]

What about a nontrivial semidirect product of Z with itself?

oh oops nvm I missed the word abelian

yea in that case there's only 1 option

[rustic crown]

:3

[bitter rover]

Yes.

If G is a finite group acting on a set A with |A| < |G|, then for every a in A you have |Orb(a)| ≤ |A| < |G|, so that |Stab(a)| > 1. So in general, if G has that property then G is not isomorphic to any subgroup of Sym(A) with |A| < |G|.

[restive birch]

okay i know im asking a lot of questions but im just finding it hard to know where to start with these.
-Prove that if H has finite index n then there is a normal subgroup K of G with K ≤ H and |G : K| ≤ n!

i just need a small hint i think

[chilly radish]

Look at the action of G on the set of cosets of H

[restive birch]

okay thanks

[chilly radish]

👍

[restive birch]

okay so heres what im thinking:
let pi_H : G -> S_n be the permutation representation of the action of G on the cosets of H by left multiplication. by first iso thm, G/ker pi_H iso pi_H(G) and |pi_H(G)| <= |S_n| = n! so by lagranges |G : ker pi_H| <= n! and clearly ker pi_H <= H, ker pi_H normal in G

[void cosmos]

yeah

good

[restive birch]

so many thms to remember

[restive birch]

okay another one:
show that any nonabelian group of order 6 has a nonnormal subgroup of order 2

[sonic coral]

can you not just appeal to the only nonabelian group of order six

[restive birch]

no

because i havent proved that that group is unique

thats what this exercise is for

[night tartan]

how much group theory do u know?

[restive birch]

up to D&F chapter 4.2, where this problem is from

[night tartan]

i unfortunately do not know the book

do u know that a group of order 6 has an element of order 2 and order 3?

[restive birch]

this subchapter is on regular representations and cayleys thm

[restive birch]

cauchy's hasnt actually been proven yet i dont think

it was proved for abelian groups

but i think the general proof is later

[night tartan]

hmm interesting

[restive birch]

we're supposed to use this stuff i think

[night tartan]

ngl not too sure how to prove this without cauchy

[night tartan]

i unfortunately do not know what regular representations are

[restive birch]

for H <= G the left regular representation is the map G -> S_n induced by the left action of G on the set of cosets of H by g * H = gH

[night tartan]

oh huh it's that

didn't know it had a name

i'm kinda tired rn so i can't give u an answer involving thinking abt that

[restive birch]

all good

[night tartan]

but normally i'd go groups of order 6 have an element of order 2 and order 3
the order 3 subgroup is always normal, so if the order 2 subgroup is normal, then by DPT it's isomorphic to C6

[restive birch]

DPT?

[night tartan]

direct product thm

[restive birch]

dont think ive seen that one?

[night tartan]

ah k

[restive birch]

idk why the D&F exercises went from being doable to really opaque to me

maybe i stopped reading well

[toxic zephyr]

any hints for (v) to (i)? supposed it was for n,n+1,n+2. was able to show that b commutes with (ab)^(n+1) and a^nab=baa^n, but couldn't get ab=ba

[restive birch]

can someone look over this proof:
statement: if G finite simple with |G| = n > 2, then G is isomorphic to a subgroup of A_n.
proof: take the left regular representation π: G -> S_n and take K = π(G). we show K ≤ A_n by contradiction. If K contains an odd permutation, then we can take the subgroup K cap A_n, which has index 2 in K. Then the preimage π⁻¹(K cap A_n) ≤ G has index 2 in π⁻¹(K) = G, and so is normal. since n > 2, it is nontrivial (and obviously proper) and so G is not simple

[charred iris]

Use the given property to write (ab)^(n+2) = (ab)^(n+1) * ab = (ab)^n*(abab) in three different ways

[night tartan]

nvm i can't read lol

[night tartan]

yeah i think it looks right

[restive birch]

nice thanks

[graceful dome]

Can someone explain to me this stack exchange answer for the proof U(n) is cyclic for n = 2,4, p^k, 2p^k

I understand why they used the chinese remainder theorem and using an isomorphism between these groups to show Z/nZ is cyclic

but this part I dont understand

why for U(2^k) is it equal to Z/2Z x Z/2^(k-2) Z

because then for k = 2 we have U(4) = Z/2Z x Z/Z which is not the prime decomposition for 4, am I misunderstanding something

[unkempt stream]

Z/2^(2-2)Z is trivial

[graceful dome]

Ok but where does Z/2Z x Z/2^(k-2) come from

[bitter rover]

I think the answer is just assuming it as "general background". It's not obvious, but isn't too hard to prove.

[graceful dome]

I see but it's just confusing because they brought up the chinese remainder theorem in the beginning so I was expecting the isomorphisms so continue looking like that

is it that they use the chinese remainder theorems as parts of those "general background" proofs?

[bitter rover]

The CRT directs our attention to prime powers.

In general, the direct product of two cyclic groups isn't cyclic.

The bit about Z/2^kZ being isomorphic to the expression on the RHS is meant to show it's not cyclic for k > 2, but there are more direct ways to prove that if that's all you want to prove.

And the remark that U_2 is trivial is meant to show why n=2p^k is also fine.

(The group of units of Z/2^kZ isn't cyclic for k > 2 because it contains a subgroup isomorphic to the non-cyclic Klein 4-group)

The CRT directs our attention because every subgroup of a cyclic group is cyclic and the group of units has a subgroup isomorphic to each of those factors.

So a necessary condition is that each of the factors is cyclic.

[dull ginkgo]

I forget the specific algebraic reason behind the group of units of Z/2^nZ

But from what I remember it’s due to binomial theorem and powers of 3

[hidden cairn]

if H and H' are isomorphic in a finite group G, is it always possible to find an automorphism of G that sends H to H' ?

[dull ginkgo]

2Z and 3Z in Z are both isomorphic but the only nontrivial automorphism of Z as an additive group is multiplication by -1

Oh wait finite

[thorn jay]

Still no

Most likely

Probably something something symmetric group

Haha

[restive birch]

is there an automorphism of D8 sending r^2 to s?

because if not then thats an example

[bitter rover]

The automorphism group of S_n is all the inner automorphisms, so two isomorphic subgroups that aren't conjugate will do

[restive birch]

okay i was trying this problem:
find all the finite groups with exactly 2 conjugacy classes
heres my thoughts:
if G is abelian finite with |G| > 2, then since every element in the center is in its own conjugacy class, it would have |G| > 2 conjugacy classes. trivially Z2 has 2 conjugacy classes.
if G is nonabelian then take its representating H as a subgroup of a symmetric group. if H has nontrivial cycles of 2 or more cycle types, they are not in the same conjugacy class in Sn and therefore of course not in the same conjugacy class in H, a contradiction. so all the nontrivial elements of H have the same cycle type.
from here i imagine that i want to show that H has nontrivial center, but im not sure precisely how to do that.

[bitter rover]

It needs to be at least S_4

Yeah actually that's fine, there is are non-conjugate copies of V_4

[restive birch]

talking about my 'proof' or still the previous problem

[bitter rover]

Orbit stabilizer

Conjugate classes partition a group and {1} is always a conjugate class.

[restive birch]

and the size of every conjugate class divides the group's size?

[bitter rover]

Yes

[restive birch]

makes sense

so then ofc its just Z2 since 2 is the only number where n - 1 divides n

[bitter rover]

So G\{e} is a conjugacy class of size |G|-1

Yep

[restive birch]

right

is this problem doable for 3?

like find all the finite groups with 3 conjugacy classes

i mean ofc Z3

and S3

its a number theory problem at this point

oh wait no its easy

[bitter rover]

Yes

[restive birch]

nevermind

for n >= 8 it doesnt work so

it is literally just Z3 and S3

can we generalize to saying that any group with exactly n conjugacy classes has order strictly less than 2^n? should hold by induction

[bitter rover]

It gets harder fast

[restive birch]

harder to classify them exactly yea

but like just putting that bound on their size should hold right

i imagine the idea of conjugacy classes is used a lot in the classification of finite simple groups

[grave sedge]

A6 has 7 conjugacy classes and order 360

[rocky cloak]

Appearantly you can do
n^(2^(n-1))

https://math.stackexchange.com/a/1036742/306319

[serene dune]

any map other than the trivial one satisfies this ?

[cloud walrus]

yeshua

$φ :\mathbb N ∪ \{0\} \to \mathbb N ∪ \{0\}$\\
$φ(ab) = φ(a) + φ(b)$,  $\forall a, b \in \mathbb N ∪ \{0\}$

[rocky cloak]

Notice phi(0) = phi(x*0)

[serene dune]

oh yeah

[rocky cloak]

If it was phi from N without 0 to N with, that be more interesting

[serene dune]

i might trip on that

without any arestriction its uncountable right ?

[rocky cloak]

There are uncountably many possible phis yeah. But you can describe all of them quite simply.

[thorn jay]

It's cool how the proof for F[x] being a euclidian domain for a field F essentially just carries out the division algorithm except inductively

[cloud lynx]

Soo I know that I(M)=(xy-1). Now I have to prove the equality. But I have no idea how to prove that I(M) is a subset of (xy-1). Can someone help me?:((

[severe linden]

Am I correct in assuming that the answer is "no" in general? By $1 \cdot x = x$ and distributivity we see that $n \cdot x = nx$, so also $n \cdot (\frac{1}{n} \cdot x) = x$. But this is impossible in for example the Klein group because every element has order $2$, so $2 \cdot (\frac{1}{2} \cdot x) = 1 \neq x$ for each $x \neq 1$

[cloud walrus]

Yuese

[bitter rover]

That proves you can't always extend the structure of M as a ℤ-module to M as a ℚ-module, but maybe there's some other structure that makes M into a (left) ℚ-module.

It's true that M can't be a left ℚ-module, though, but that argument isn't quite it.

Hint 1: ||If M was the trivial group then it could be made into a left-module.||
Hint 2: ||What do modules over a field look like?||

[cloud lynx]

🥲🥲

[severe linden]

But doesn't it directly follow from $1 \cdot x = x$ and distributivity that the structures as a $\mathbb{Q}$-module and a $\mathbb{Z}$-module agree on $\mathbb{Z}$?

[cloud walrus]

Yuese

[severe linden]

It's vague to formulate, hopefully you know what I mean

[bitter rover]

Oh, yes, you're correct.

Well, in any case, the answer isn't just that there are some non-trivial abelian groups that can't be made into a left ℚ-module, but no non-trivial abelian groups can.

What do modules over ℚ look like?

[severe linden]

Yeah I was wondering if that was the question

[severe linden]

Oh wow I'm such a dumbass, $\mathbb{Q}$ is a field so they are just vector spaces

[cloud walrus]

Yuese

[bitter rover]

Yep.

[severe linden]

So they all look like $\mathbb{Q}^n$

[cloud walrus]

Yuese

[bitter rover]

And vector spaces over an infinite field are either trivial or infinite.

So if M were the trivial group then it could be made into a left ℚ-module, namely the trivial vector space.

[severe linden]

Okay not quite because they can be infinite dimensional, but they can't be a finite group

[severe linden]

Yep got it, thanks

[tender wharf]

if F is a field (of characteristic not equal to 2) is there a way to embed F under addition into SL(2, F)?

[rustic crown]

a --> [1, a; 0, 1]

[tender wharf]

thanks

lol how did I not think of that

[rustic crown]

char 2 isn't important, any comm. ring works

[mighty kiln]

Any ring

[rustic crown]

<

[tender wharf]

does SL(2, F) exist for those cases?

[rustic crown]

yee, as long as det : M_n(R) --> R makes sense, so does SL

[tender wharf]

ah okay

[mighty kiln]

Wait is det homo over non-comm rings

[rustic crown]

not sure, prolly not

[serene dune]

if AutG is cyclic

G is abelian

[tender wharf]

yes

[restive birch]

huh, interesting, how do you show that?

[serene dune]

how can i determine the order of G now

[serene dune]

inn being the normal subgroup is cycilc too

which is \cong to G/Z(G)

[restive birch]

ah

right

[serene dune]

thats cycilc

u got it

hmm so

there is this thing that there is some kinda restriction on AutG and perhaps on G with that hypothesis

i dont want any straight answer

but any hint to get further down the road

[bitter rover]

Are you assuming G is finite?

[serene dune]

good point, i havent thought about that at all

but yes for now its fintie ig

[bitter rover]

Well, a finite abelian group G of order n is isomorphic to the direct product of p-groups via the Chinese Remainder Theorem (where the primes are those in the prime decomposition of n).

That tells you what Aut(G) looks like — it'll be the direct product of some other groups related to the factors of G. Because Aut(G) is cyclic, each of its factors must be cyclic.

That puts constraints of what the factors of G can look like.

[toxic sapphire]

can someone help me understand this?

here, D_a is the dihedral group of order a

i.e., D_6 represents symmetries of a 3-gon

shouldn't D_6 be a subset of D_8?

[serene dune]

oh shit i was sleeping on abelian classification, dude that should have been detected by me

[serene dune]

see the orders

more interesting case is splitting into Z2 and a smaller Dn

[toxic sapphire]

ah thanks. so essentially the elements in D_6 (even though we use the same symbols) are different from the ones in D_8?

which is clear because they have different order

[bitter rover]

First, the answer is assuming a slightly pedantic interpretation.

D_6 and D_8 are simply different sets with no elements in common and therefore D_6 by itself can't be a subgroup of D_8.

Just like S_5 isn't a subgroup of S_10, because they're a different set of permutations entirely.

But one could also ask: is there a subgroup of D_8 isomorphic to D_6? That is also false.

(Under this interpretation, no group is a subgroup of another that wasn't defined as a subset to begin with.)

[serene dune]

are your using the fact that Aut is homomorphism

how is that the AutG breaks up nicely as G breaks up into some prime powers

maybe im completely missing the point here

[bitter rover]

Aut(G×H) ≅ Aut(G)×Aut(H) if G and H are finite groups with relatively prime orders.

Write out what G looks like as factors of p-groups. Since they're p-groups of distinct primes their orders are relatively prime, so Aut(G) is a direct product of respective the automorphism groups of the factors of G.

What does a factor of G look like? What does each factor's automorphism group look like?

[restive birch]

just read the proof using the class equation for why groups of order p^2 are abelian

so cool

[golden turtle]

Say I wanted to verify that Hom(D, _) is a covariant functor as claimed
I go to show property (a)
and I immediately get confused because I'm not sure what the objects in C are, if they are R modules or R module homomorphisms

[restive birch]

,rotate ccw

[restive birch]

problem 17 there

having trouble parsing the definition--D is the set of all permutations of A which have a finite number of fixed points?

so like if A = Z then (1 2) (3 4) (5 6).... would be in D but (1 2 3) would not be?

this problem seems odd because this subchapter seems like its almost purely focused on finite grps

[rocky cloak]

The objects are R-modules. R-module homomorphisms are the morphisms of the category (appearing in point b)

[rocky cloak]

No, the opposite. They have a finite number of points that are not fixed points.

[restive birch]

ohh

right

[rocky cloak]

Otherwise the identity would give you trouble

[restive birch]

right

so then i just need to argue that TxT^-1 has finitely many non-fixed points when x has finitely many nonfixed points

but conjugates preserve cycle types, so this is clear? does that work?

[golden turtle]

So to verify (a), I let N be an R-module. And I want to show FN is an object in D which is Hom_R(D,N)

[restive birch]

bit iffy since its not on S_n but should hold since x is finite so has a cycle type

[golden turtle]

FN will be Hom_R(D,N) but I am confused by what the D is coming from

[rocky cloak]

So D should be the category of abelian groups.

So then it's just stating that Hom(M, N) is an abelian group

[golden turtle]

It says that it is a covariant functor from Rmod to Grp

[rocky cloak]

Sure, you can do category of groups

[golden turtle]

so I want to show that F applied to any R-module N is a group

to verify (a)

[rocky cloak]

Yup

[golden turtle]

we have the definition as hom_R(M, _): N -> hom_R(D,N)

I am confused where the D comes from

[rocky cloak]

Yeah, it works the same in S_A as in S_n

[golden turtle]

but if D is any R module then FN is a group so we have (a)

I just don't quite get what the D is

[rocky cloak]

So I don't know what you're copying from, but Hom(M, -) should take N to Hom(M, N)

You wrote Hom(D, -) before, but it's a somewhat unfortunate name collision, as D was already used for the category of groups

[rocky cloak]

Yeah, no M here

Or M=D

[golden turtle]

Oh ok so I wrote it wrong

awesome

(b) is more confusing

we must show for every f in Hom_R(M,N) that F(f) is in Hom_R(FM, FA)

F(f) is defined confusingly, it seems like it would be f'

but then I do not know what f' is

[rocky cloak]

So
f' : Hom(D, M) -> Hom(D, N)
h |-> fh

So you just need to check that this is a group homomorphism

Thats what it means for f' to be in Hom_Grp(FM, FN)

[golden turtle]

FM is Hom_R(M,M) and FN is Hom_R(M,N)

[rocky cloak]

Well Hom(D, M). You get a bit of a name collision if D=M

There's 3 R-modules involved is the point

One used to define F (which is D in the picture)

Then two you have a homomorphism between (M and N)

[golden turtle]

But if we a hommorphism, looking at the picture, isn't it probably in Hom_R(D,N) so its between D and N?

[golden turtle]

looking at this

or I guess actually

its referring to phi

I see

[rocky cloak]

This homomorphism (f in the picture)

[golden turtle]

f in that picture

[rocky cloak]

Should go between modules not related to D

[golden turtle]

is phi in this

right?

[rocky cloak]

It is phi in that picture yes, and it's unrelated to D

[golden turtle]

I understand a little better now

Thank you a lot

[tardy hedge]

freaky font

the freak Hom

[golden turtle]

I want to get freaky with dummit like that one guy

Maybe then I’d understand

[tardy hedge]

fr

[graceful dome]

why is it that in the proof for this corollary that

[chilly radish]

If K<=H then [G:K]>=[G:H]. This is because there is a surjection G/K->G/H given by mapping gK to gH. Check this is a well-defined function (not necessarily a group homomorphism unless K,H are normal)

In fact one can prove the tower law:
[G:H][H:K]=[G:K]

[vagrant zinc]

It is abelian so it is normal, you could get a quotient by the first isomorphy theorem, I would think :c

In abstract algebra think of sets=groups, subsets=subgroup, now think of subsets as “perfect” where it is given by normality, also think of quotients or factors, and natural homomorphisms or canonical homomorphisms, this is the basis of all groups, and rings have more properties, you will notice that the demonstration always talks about these fundamental properties.

[graceful dome]

but <g^n> <= H1
so by what you said [G : H1] >= [G:<g^n>]

but the proof says [G : H1] <= [G:<g^n>]

[vagrant zinc]

Here's to normalcy kid, my colleague just said it.

[graceful dome]

I see, I haven't covered Normalcy yet in my course

I know what you meant by abelian but I haven't covered Normality or Quotients until the next chapter

[chilly radish]

Oops

First inequality is backwards

K<=H

Otherwise [H:K] would also not be defined

[graceful dome]

Ohh okay I found the corresponding thing exercise in my textbook

the surjection you're referring to is just the proof to this right

or rather part of it

[chilly radish]

Kind of. Easiest way to prove it is to note that G is a disjoint union of cosets g1H,...,gnH and H is a disjoint union of cosets h1K,...,hnK by definition

[graceful dome]

So the conclusion here is that in order for the index of H1 to be k = infinity
it must be that H1 is equal to the trivial subgroup, and then obviously H2 = H1

and for finite k, applying the lemma gives us the outcome we want

[chilly radish]

Yes it's asserting that the only subgroup of infinite index of Z (Note that if H_1 is of infinite index the G is infinite cyclic i.e. isomorphic to Z) is the the trivial group

[chilly radish]

Yes

[chilly radish]

It's saying by contradiction that if H was nontrivial, it would have an element of the form g^n, but then its index would be smalelr than the index of <g^n> which is finite, so H_1 must not have nontrivial elements

the wording here is a bit confusing imo

[stone sky]

So I’m reading Gallian’s Contempary Abstract Algebra book, he states if a ring has unity 1 then (-1)a=-a and (-1)(-1)=1 if the unity were say, 2, do i substitute 1 for 2?

or is he using 1 as an identity notation?

[vagrant zinc]

If I is an ideal then it is not necessarily a sub*ring of R?, and if so then I can only be an idea in R.

[delicate bloom]

a subring of of R you mean?

[vagrant zinc]

No, unit means 1, and unitary means +1,-1, so

[vagrant zinc]

Ye,

[stone sky]

but what if the unit is not 1?

[delicate bloom]

does a subring need the multiplicative identity of R?

[vagrant zinc]

Yes!

However, the identity is not necessarily the same as the R-ring, is it?

[delicate bloom]

so if your ideal is a subring, it contains 1. What ideal is generated by 1?

[vagrant zinc]

Must be 1?

[delicate bloom]

I guess it's up to how you define it

[serene dune]

yeah i was really close proving that statement, but i fell asleep
i will start from where i left

[vagrant zinc]

Maybe I'm misinterpreting, I'm not good at translating from English to my language, this is what dummit tells me

[vagrant zinc]

Chinese Residual Theorem

[cloud walrus]

bourbaki

[paper sonnet]

yes, you can see it by imagine 1 is just a placeholder for the element in the ring that satisfy
ba =ab =a (here b = 1)
take for example the set of all nxn matrices, they form a ring but the identity is I_n
Nevertheless, you will be better off by imagine 1 is the identity element, in my example I_n is related to 1 smh, for more examples where you can get an identity not being equal to 1 you can just take a ring that doesnt have the integer 1 as an element

[restive birch]

,rotate ccw

[restive birch]

19 there

ive got that by second iso, |G: HC_G(x)| = |H: C_H(x)| = the size of the conjugacy class around x, but this shows that each conjugacy class of H contained in K has k elements, not that there are k of them

im just super lost on these exercises generally

i think maybe i didnt read the text closely enough

will try again tomorrow

[golden turtle]

I don't understand how it means the first of these starts out being exact

How does Im(f1)=Ker(f2)?

Im(f1) = homR(M,A)

[rocky cloak]

The image of f1 is 0.

The only map from 0 to anything is the zero map, which has image 0

[golden turtle]

I see

then I need to get that ker(f2)={0}

[rocky cloak]

Indeed

Also, there's a missing )

[golden turtle]

yes

I noticed that

but they're not mine

it is my lecture notes

[golden turtle]

so Zm and Zn are supposed to be the free module here

and it states how its well known that their direct sum is isomorphic to Zmn

is that supposed to be like, clearly nonfree for some reason?

I don't understand

[weary frost]

Zm and Zn are not free here but Zmn is

Take for example m=2,n=3

Z6 is a free Z6 module because every element is uniquely written as k*1, where k is in Z6. Z3 is not free as a Z6-module because if you pick any generators you will cover any element twice with different coefficients

[hidden wind]

oooh

[chilly ocean]

Trying to show all groups of order 30 are solvable. Found that there are unique Sylow 5 and 3 subgroups which are obviously abelian and thus normal in G but I can't show that the subquotient of G and either are abelian. I'm guessing that I can probably use the above to show G has a unique normal subgroup of order 15 and thus the quotient would be of order 2 and thus abelian but I don't see any obvious way of doing this. Any ideas/hints?

[serene dune]

i would like to classify the groups at first

as order 15 gives you a unique group upto iso

[rocky cloak]

So you have that the sylow 5 subgroup is normal, call it N, then G/N is a group of order 6. So you just need to prove that any group of order 6 is solvable

[serene dune]

right

$C_{30}, C_{15} \rtimes C_2, D_{10} \times C_2, D_6 \times C_5$

[cloud walrus]

yeshua

[rocky cloak]

As for the technique you're suggesting:

If H and K are normal subgroups, then HK is normal, with order |H|*|K| / |HnK|.

You can try to prove this if you like

[chilly ocean]

Ah that makes sense, thanks

[chilly ocean]

But wait, is it really enough to show a subquotient is solvable? The criteria for solvability is that the subquotient is abelian no?

If this is a property of subnormal series then it hasn't been shown to me in class and I'd have to prove that as well

[hot wadi]

There’s a result that if H=G/N then G is solvable iff both H and N are solvable

Although I would say proving that is a slightly harder exercise that the one you’re faced with

[chilly ocean]

I don't see how this result helps me though

[hot wadi]

If you could prove it, it would help greatly

I don’t think it’s the easiest way to do this problem though

[chilly ocean]

Ah I see

Does this result have a name?

[rocky cloak]

I'm not sure it has a name, but it's an immediate consequences of the isomorphism theorems

[hot wadi]

here’s what I would suggest:

1. Prove G contains an index 2 subgroup (it sounds like you might’ve already done this)
2. Prove that any index 2 subgroup of any group is normal
3. Prove that any group of order 15 is solvable

[chilly ocean]

We have only been using the first isomorphism theorem and skipped the others

[hot wadi]

I wouldn’t say immediate

[rocky cloak]

The (normal) subgroups of G/N correspond to subgroups of G that contain N

[hot wadi]

Oh I see jagr

I guess it is more direct than I remember

Maybe I’m just better at alg than when I learned it

[rocky cloak]

So if you have a series of G/N with abelian subquotients lifts to one for G

[chilly ocean]

Either way this seems outside of the available results I can use

Like I'd have to go out of my way to also prove certain isomorphism theorems from the sound of it

[hot wadi]

Yeah that’s not worth

[chilly ocean]

I haven't been able to show it contains an index 2 normal subgroup, that's exactly what I was trying to do when I asked my original question

[hot wadi]

Have you shown that it contains an index 2 subgroup?

I.e a subgroup of order 15?

[chilly ocean]

I've been able to show this: G either has 10 Sylow 3 subgroups and 1 Sylow 5 subgroup OR G has 1 Sylow 3 subgroup and 6 Sylow 5 subgroups

[hot wadi]

Sorry I didn’t mean to say normal in step 1

[chilly ocean]

No

Although it doesn't seem hard to, since I know it has normal subgroups of orders 3 and 5

[hot wadi]

Well it sounds like you only know it contains one of those

[rocky cloak]

Does this seem like something you think you can do though? @chilly ocean

[chilly ocean]

This seems doable, I'll give it a try for a bit and report back later. My problem that I realize now is that I haven't actually reduced it to 1 Sylow 5 and 1 Sylow 3 subgroups but 10, 1 or 1, 6

I suppose I can treat each case individually but I'm not sure it'll work

I'll come back later if it doesn't, thanks for the help up until now though

[hot wadi]

My suggestion would be to let H be a 3 sylow and K be a 5 sylow and show that because one is normal, the product HK is a proper subgroup of G

And then show from there is has order 15

[chilly ocean]

The problem is that since one is potentially not normal I won't be able to draw that conclusion

[hot wadi]

You know that when EITHER K or H is normal that HK is a subgroup

You can easily bound the order of HK from above and then you can bound it below with Lagrange

[hot wadi]

(Maybe you don’t know this, but I believe it’s an exercise you could solve as a lemma)

[bitter rover]

Do you have access to semidirect products?

[terse crystal]

For an arbitrary group G, denote the subgroup generated by m th power of elements of G by G(m). Given G(m), G(n) both being abelian, how to prove that G(gcd(m,n)) is also abelian? I know I can assume that m and n are coprime, but I don’t know how to proceed.

[chilly ocean]

No

[bitter rover]

You do the only thing you can do, which is expand the definitions of everything involved.

[hot wadi]

Can you construct a surjection from the product G(n) x G(m) onto G(gcd)?

Cufflink’s answer is better

[bitter rover]

Nobody ever does it, though. 😦

They doubt. They think they're being patronized.

I know the definitions, they say. Look, here I am saying it out loud.

[tardy hedge]

wut

Ok tbf i get what u mean

[bitter rover]

Most of these "introductory exercises" boil down to expanding the definitions, connecting matching parts, and remembering some basic facts.

Novices (often) believe that because they can recite the definition in English they understand it.

The doubt that writing down the fully expanded definition on paper and putting it in front of them will give them any insights, so they hear the suggestion and go "Uh huh, I know the definition. Thanks, but I'm going to keep trying other stuff."

It's hard online because it's easier for folks to bounce. In person, I can literally beg them.

[tardy hedge]

Yeah youre definitely right. I get caught up in silly thinking like that too sometimes

[bitter rover]

Please, I am begging you, write it down. Write the whole thing down, literally.

[tardy hedge]

Hahaha

yeah

[bitter rover]

No, don't abbreviate it. Verbatim.

Yes, it will take 10 seconds longer. Do it. Begging you. Please, for me. Your teacher who very much wants you to do well.

They do it and go, "Oh, I see."

The whole time they're like 🙄 OK 🙄 "DAD" 🙄 WHATEVER 🙄 YOU 🙄 SAY 🙄

[terse crystal]

Haven’t figured out… what surjection…

[bitter rover]

See? Always more interested in the tricksy tricks than the direct way.

[hot wadi]

There is a high-brow approach, but it isn’t this. Rather, you need to construct ||an injection of G(gcd) into the product G(n)xG(m)||

[bitter rover]

But they can't see through the fog because they don't have their compass and map in front of them. They're relying on the version in their mind's eye, a novice's mind's eye.

[hot wadi]

Regardless, go with cufflink’s advice rather than my solution

[bitter rover]

@terse crystal Do you know Bezout's lemma?

That's like the #1 thing to reach for any time you're asked about greatest common divisors.

[terse crystal]

I know

[bitter rover]

Second, it suffices to show that any individual elements that are powers of n and m commute. Do you see why?

Or not that they commute...but why it's enough to look at those.

Rather than arbitrary elements of the group generated by those things.

[terse crystal]

You mean a^m b^n=b^n a^m?

[bitter rover]

Yeah

[terse crystal]

I don’t know how to construct the injection in general, but I see now I can do so when n and m are coprime, solving what I needed

[chilly ocean]

So it turns out I did have access to this. We proved it using derived series, I had simply missed the lecture but it was uploaded to yt so I just caught it while going over all this.

Using this the problem became pretty trivial. Thanks for the help!

[terse crystal]

wait I still didn’t get it… like when m and n are coprime is it mapping a to (a^m,a^n)? I can’t see why it’s a homomorphism

[terse crystal]

Showing that is sufficient indeed but I can’t prove it either…. So far I only got any a in G, a^lcm(m,n) commutes with any element

[bitter rover]

Take Bezout, which says that if d = gcd(n,m) then there exist integers p,q with

pm + qn = d

Take an element g. What does g^d look like?

[terse crystal]

g^(pm+qn)

[bitter rover]

Yep. And that equals...

[terse crystal]

(g^m)^p times (g^n)^q

[bitter rover]

Right, so it's a product of two things in G(m) and G(n)

[white yarrow]

i need help with something but i dont know the name of it i have a test tommorow

[terse crystal]

But a^d b^d=(a^m)^p (a^n)^q (b^n)^q (b^m)^p I can only switch middle two

=(a^m)^p (b^n)^q (a^n)^q (b^m)^p, the m- th power terms got separated away

Similarly if I switch the m-th power terms the n-th power terms separated

[white yarrow]

whats that

[terse crystal]

Sorry I was replying to this

So in the end indeed I need to show that a^m and b^n commute, but I still can’t prove it…

[terse crystal]

Yes but what is the mapping from G(d) to G(m) times G(n) though… mapping Πa^d to (Π(a^m)^p, Π(a^n)^q)?, this is a homomorphism but I can’t show it’s injective….

[rocky cloak]

So at least you have
G(gcd(m, n)) = G(m) G(n)
and G(m) and G(n) are both normal subgroups.

So if one could show that their intersection is a direct summand of one of them, maybe that would be a good approach...

Are these finite groups or not necessarily?

[bitter rover]

I wouldn't go that route.

We have pm + qn = d. Diving both sides by d, you get

p(m/d) + q(n/d) = 1

so m/d and n/d are relatively prime. Call those m_0 and n_0, respectively.

Take an element c = (x^d)(y^d) of G(d). What is c^m_0?

[terse crystal]

Not necessarily

[terse crystal]

((x^d)(y^d))^(m/d)

[bitter rover]

Which is (x^m)(y^m), and those commute by hypothesis.

[terse crystal]

How?

[bitter rover]

They're in G(m)

(both being powers of m)

[rocky cloak]

How is it x^m y^m ?

[terse crystal]

Yeah how

[bitter rover]

Oh derp, just a sec

[terse crystal]

And why ((x^d)(y^d))^(m/d) is in G(m)?

[dense root]

hello friends i have a question

for the second part does it suffice to show that R is a subring of the of the set of all Matrcies nxn to prove that R is a Ring than i just prove that it is commutative

since the set of all matrcies is a non-commutative ring it just pass the Ring part and i prove the commutative part by myself

[tropic spade]

You'd want to word it clearly.

Like, it will be a subring of the ring of 2x2 matrices with coefficients in Z.

Subrings of rings are rings.

[golden turtle]

I don't understand how F isomorphic to the direct sum of ker phi and P proves (4)

Why must ker phi and P be free R-modules?

I thought at the beginning of the paragraph it specifically was talking about how P may not be a free module but it must be the quotient of a free module

[golden turtle]

So then this really is not an example of how a direct summand of free modules could fail to be free as it begins by saying?

[weary frost]

it is

[golden turtle]

but Zm and Zn are not free as you said?

[weary frost]

yes

Zmn is

[golden turtle]

so it isn't a direct sum of free modules then

[weary frost]

Zmn is the direct sum, Zm and Zn are the direct summands of Zmn

[golden turtle]

oh I see

summand is the terms in the sum

[weary frost]

yes

LOL

[golden turtle]

I was so confused

[weary frost]

terminology and stuff

[dense sierra]

Can anyone help me with the c part of this question

[dim widget]

Unless I’m missing something obvious this all seems wrong

[dense sierra]

what do you mean

[dim widget]

In part a if x \in X then ab + (a - axa) = 1 which is not in X

Also doesn’t X always have more than one element by assumption?

Also if an inverse of an element in a ring exists it’s easy to show that its unique, indeed the units often form a group

[dense sierra]

it is

[dim widget]

No

Because a*1 = a which is not in general 1

[dim widget]

Ah wait this part I am making a mistake

Part a) makes sense

[dense sierra]

yeah it does

[dim widget]

But part c) is just false in general

Like if R is a field it’s incorrect

Ah wait I see

If it has more than one element

[glad osprey]

two-sided inverses are unique, but an element can have multiple right inverses (only if it has no left inverse)

[dim widget]

Ah okay I was confused

I read “if it has more than one element” as “if it has at least one element”

[glad osprey]

I think the trick for part c) is to show that phi is not surjective

[dense sierra]

If it is not surjective then what?

[dim widget]

Well then the set can’t be finite

Since you have that it’s injective

[dense sierra]

oh yeah makes sense ig

[elfin wraith]

Im not quite sure how to ask this because im just generally confused about how N^n gradings work, like im quite happy with Z or N gradings of modules but what do we mean when were grading by N^n?

The definition given in the book im reading is kinda what youd expect i guess, M = \oplus_ b \in N^n M_b and x^a M_b \subset M_a+b for vectors a,b, but I guess its not clear to me what that means

like what does it mean to index our sum over vectors? What is exponentiation by vectors (im guessing via the exponetial like for matrices but again, not sure here), would appriciate any clarification, thanks!

I suppose the thing about exponentiation is also relevant for modules graded in degree g so I really should understand that

[dim widget]

Why is it unclear what that means? Have you thought about the case when n = 2?

[elfin wraith]

So for n=2 we would just have say M = \oplus M_(0,0) \oplus M_(1,0) \oplus M_(1,1)... i think? I guess I just dont see how this is significantly different to just an N grading, is it to do with ordering or something?

[dim widget]

Is your ring also graded?

[elfin wraith]

Oh i think that may be where my confusion is coming from, im forgetting that in this book S=K[x_1,...,x_n] (I was just thinking of a general ring), so say for n=2, M_(1,1) would be the 1st module of S_1?

I also realise its not entirely clear what im trying to say there, I can probably type up some actual notaition if needed haha

[dim widget]

I think if your ring is ungraded people mean just a family of modules indexed by some discrete category when they talk about a graded module, but if your ring is graded then the grading specifies how multiplication by elements of your ring interacts with the grading

For a polynomial ring in n variables the grading is just by degree (the tuple of degrees of a monomial in each variable)

So this is saying that x_1 induces a morphism M_(k_1, \dots, k_n) \to M_(k_1 + 1, k_2, \dots, k_n)

[elfin wraith]

I dont really know any catagorical stuff so im slightly unsure about the language but I think I get what youre saying, im still a bit confused about the graded in degree thing though, the books definition is "Denote by S(-a) the free module generated in degree a so that S(-a) \cong <x^a>" with x and a in N^n, is this just short hand for (again in the n=2 case), <x_1,x_2>?

[dim widget]

My guess for x^a is that it means x_1^{a_1} \cdot \dots \cdot x_n^{a_n}

[elfin wraith]

Yeah ok that would make sense to me, thanks for that think I got myself a bit lost there!

[terse crystal]

Any idea on this? Still not solved…

[wicked patio]

An arbitrary element in G(gcd(m,n)) looks like g^(am+bn)

[terse crystal]

Product of elements of this form

[wicked patio]

g^(am+bn)h^(cm+dn)

= g^am * g^bn * h^cm * h^dn

= g^am * g^bn * h^dn * h^cm

hmm maybe it's hard to progress further. You can swap around the middle 2 ferms

[terse crystal]

Yeah deadend

[hot wadi]

Wait you’re right it’s not a homomorphism; the group’s not abelian

I was just spitballing without thinking through the fine details. Apologies

[terse crystal]

I didn’t think it’s very hard too, sounds natural but I can’t prove or find a counterexample…

[wicked patio]

Oh I got it lol

Let me think of how to word it

Let k = gcd(m,n) = am+bn

Damn doesn't quite work

[hot wadi]

@terse crystal the problem is false

Nvm

Counterexample doesn’t work sorry for ping

I’m still inclined to believe it’s false though

I feel like you should be able to show the free group on 2 letters mod squares and cubes is not abelian if you try hard enough

[terse crystal]

I don’t know, since it’s not just two relations, like m=2,n=3, it’s not xxyy=yyxx but any word v,w vvww=wwvv

Maybe those infinitely many relations can give you xy=yx somehow

[tough raven]

What's your motivation for this problem BTW?

[terse crystal]

Someone asked me

[tough raven]

Ah.

[grizzled spindle]

if we have a G-set X and g in G, take m_g, an element of Aut(X) sending x->gx. Are we only guaranteed that ord(m_g) | ord(g)?

it can not be equal, but is there any stronger condition

[wicked patio]

I've been looking at the n=2 m=3 case too

[wicked patio]

The free group on 2 elements mod squares should be V_4

[dull ginkgo]

I am not sure off the top of my head but you can probably use the defining map of G into Aut(G)

Since it’s a homomorphism you can just consider what happens to the power lmao

[golden turtle]

I am trying what I thought would be the easy direction of this, assuming that P is free and trying to show that P is projective

I thought that this would have to do wth the universal property of free groups

But I can't see how to make it work

[rocky cloak]

Indeed it does. Maybe try doing it specifically for P=R first.

[golden turtle]

I made the following sketch if it helps

[golden turtle]

I am using F(S) in replacement of P because it helps me understand the universal property more

What do you mean by R?

Like what group is it

[rocky cloak]

The free module on one generator.

So Z if you're just thinking abelian groups

Anyway, you know that a homomorphism F(S) -> M is the same as a function S -> M.

So then the question becomes, given a surjective homomorphism G -> M and a function S -> M, can you lift it to a function S -> G

[golden turtle]

it is confusing me how we are given a set map g:F(S)->M to use in showing that F(S) is projective, when the set map in the universal property is from S to an arbitrary group G, so its not necessarily coming from F(S)

[rocky cloak]

?

The universal property of free modules says that a homomorphism F(S) -> M is determined by where S is mapped and that any function S -> M extends to a homomorphism.

And this holds for any group M yes

[golden turtle]

I understand this to be the universal property

Given any set map phi from S to G

so

like we can G to be any group as you say

but if we want to lift the set map from F(S) to M

it is like, F(S) is not the set its a group

sorry if I'm being confusing

its just the diagrams dont line up when I try drawing them

I find it hard to describe

I think I can ask my question better so I'll retry if that's fine

I look at this, that I'm given from universal property. And then in the question, I'm given pi:G->M and g:F(S)->M and want to lift them to a homomorphism

and so I think I have to replace G in the diagram with M

[rocky cloak]

Maybe you're confused about what projective means...

Like you're given a surjective group homomorphism G -> M and a group homomorphism F(S) -> M
and the goal is to find a homomorphism F(S) -> G that makes the triangle commute

[golden turtle]

because g:F(S)->M and in the diagram F(S) only maps onto G

if I can find f such that this commutes then its projective

[rocky cloak]

Indeed

[golden turtle]

that's my understanding of it basically

[rocky cloak]

And you see how g gives you a function S -> M?

[golden turtle]

g:F(S)->M

but

hmm

can I consider the elements of F(S) like ignoring the group structure

to get a map from S to M

[rocky cloak]

Yes, group homomorphisms are functions, and S is a subset of F(S)

[golden turtle]

Okay cool so then g gives us the phi in this diagram by like a restriction(?)

[rocky cloak]

Well it gives you S -> M at least

[golden turtle]

right we have to replace the G with M

and it gives us that set map

let me do this one sec

so by the universal property we get this diagram and a homomorphism psi:F(S)->M

and we haven't yet used the given pi:G->M

I don't exactly see where to go from here, I know that I probably want to use the given homomorphism pi:G->M but I'm not sure how/where to get G involved in this

[rocky cloak]

So let's think about the case where S = {*} has just one element.

Say G = Z/4 and M = Z/2 and we have the obvious surjective map G -> M.

Now if S -> M sends * to 1. Do you see a way to make a function S -> G that makes the triangle commute?

[golden turtle]

what is the obvious surjective map?

[rocky cloak]

The one that send x mod 4 to x mod 2

0 -> 0, 1 -> 1, 2->0, 3->1

[golden turtle]

Is S={*} the free group on one element

or is it the generating set

[rocky cloak]

S is a set F(S) is the free group generated by the set

[golden turtle]

Ok I'm confused then

you say let G=Z4, M=Z2

and then want to know if I can find f so that the triangle commutes

[rocky cloak]

Yes, that's the eventual goal.

Then we want to use the universal property, so we restrict g to S to get a diagram
G
V
S ->M

[golden turtle]

oh right

I forgot about the restricting

[rocky cloak]

Then the question is can you find a function S -> G

[golden turtle]

You have to have f map 1 and 3 to 1

[rocky cloak]

Yeah, so we can find such a function, great!

Now we know any function S -> G extends uniquely to a group homomorphism F(S) -> G

[golden turtle]

by the universal property

we do i(psi)

[golden turtle]

Wait, how does f map 1 and 3 if its mapping out of S which is only {*}

[rocky cloak]

Oh, guess I misread what you said.
f can map * to either 1 or 3

Because that's how you would get pi(f(*)) = 1

[golden turtle]

okay I understand more now this example

so I labeled some things

we have the bottom most diagram from the universal property

and then still we have to go further somehow from this

[rocky cloak]

So f: S -> Z4 also lifts to a homomorphism F(S) -> Z4

[golden turtle]

Why is that? I thought we went in the previous direction before where we had a homomorphism and restricted it to make a set map

[rocky cloak]

Indeed, so the universal property allows you to translate between homomorphisms from F(S) and functions from S.

Functions are in this case easier to deal with, so to understand homomorphisms from F(S) we first translate to S, find the appropriate functions, then translate back

Like we have the triangle
S -> G
v v
M = M
Which translates to a triangle
FS -> G
v v
M = M
And boom we're done

[dull ginkgo]

Adjunction

[hot wadi]

What I typed deviated horribly from what I meant

[terse crystal]

This is solved: G(mn) is contained in G(m) cap G(n) which is contained in Z(G), any a,b in G, consider x=(a^m)(b^n)(a^-m)(b^-n), x is in G(m) cap G(n). Now (x^m)(b^mn)=(a^m)(b^mn)(a^-m)=b^mn therefore x^m=1, (x^n)(a^-mn)=(b^n)(a^-mn)(b^-n)=a^-mn therefore x^n=1, in conclusion x=1

[rocky cloak]

Nice.

I can see why x^m = 1 and x^n = 1, which then should give x^gcd(m, n) = 1.

How do you get it all the way to x=1?

[terse crystal]

Oh I kept the assumption that m and n are coprime proving G itself is abelian

[rocky cloak]

I see. Makes you wonder if it still holds when m and n are not coprime

[rocky cloak]

This is still pretty cool though

[terse crystal]

It’s general, since G(m)=G(d)(m/d) G(n)=G(d)(n/d) for integers m and n

Oh you mean G(m) G(n) abelian whether G not G(gcd(m,n)) abelian? I guess not…

[regal zodiac]

Hey what is the easiest way to show that the polynomial xz -y^3 is irreducible over C[x,y,z]?

[rocky cloak]

Is this true?

Like obviously G(d)(m/d) contains G(m), but could it not happen that a product of dth powers when raised to m/d is not a product of mth powers?

[terse crystal]

Oh…

[rocky cloak]

Eisenstein I guess

[terse crystal]

d=gcd(m,n), m=du, n=dv. x=(a^m)(b^n)(a^-m)(b^-n) in G(m) cap G(n) contained in C(G(d)), also G(duv) is contained in G(m) cap G(n), (x^u)(b^nu)=(x(b^n))^u=(a^m)(b^nu)(a^-m)=b^nu therefore x^u=1, (a^-mv)(x^v)=((a^-m)x)^v=(b^n)(a^-mv)(b^-n)=a^-mv therefore x^v=1, inclusion x=1

General case

[rocky cloak]

Ah yes of course, same idea just works.

Cool

[terse crystal]

Thanks for pointing out, I thought naively G(m)=G(d)(m/d) when I just saw this problem, and never spotted the error when going further

[dense root]

is there a channel for number theory ?

nvm

im actually blind i found it

[hot wadi]

Asking over C, not Q

[rocky cloak]

You should look at it over C[x, z] actually, but Eisenstein holds over any integral domain

if you pick your prime ideal to be for example (x)

Then y^3 - xz has 1 as leading coefficient, all other coefficients in (x) and constant coefficient not in (x)^2

So is irreducible by Eisenstein

[hot wadi]

Oh I did not know that

[bitter rover]

You can also look at what it'd have to factor into, which is constrained by degree considerations.

That might be tedious here, though.

[rocky cloak]

I guess it's a question of whether "easiest" means quickest or requiring the least machinery.

[vocal ibex]

It feels like it should be the two sphere

[vagrant zinc]

the SO(3) dimension

[rocky cloak]

Well a rotation in SO(3) consists of an axis (basically a point on the 2-sphere) and an angle. So that's 3-dimensional.

It's not actually a 3-sphere though, but SU(2) is

[vocal ibex]

sorry it's been explained in another channel I understand now

thanks though

[vagrant zinc]

xde

[languid trellis]

I'm struggling to see how "it follows easily that..." I can see the identification of an element in T(Z/nZ) = (a0, a1, .... ,an, 0, 0 ,0 ) with the polynomial a0 + a1x + ... + anx^n, where a0 \in Z , ai \in Z/nZ otherwise, but I don't see how nx^2 is in the ideal, for example. And also, what happens to the elements in the tensor algebra that don't have finite support?

[tardy hedge]

Stuck on iv)

[graceful dome]

How is it that (r^i)(fr^j) = fr^(j-i)
same with below

[tardy hedge]

Im running at like 2fps tryna parse this stuff

[graceful dome]

I get rotations commute with each other but how did they get past the flip, what am I missing here

nevermind, is it because Z(Dn) is the center so r^i commutes with fr^j if they are both elements of the center

[rocky cloak]

nx^2 = nx*x
so that's how it is in the iddeal

[languid trellis]

I'm probably being quite silly, but I don't see how x is in the ideal

[rocky cloak]

It's not

It's not supposed to be

[tardy hedge]

Nevermind i think i got it, using the fact that the radical of an ideal is the intersection of all prime ideals containing that ideal

Intuition tho? Ehhh 😅

That will come later hopefully

[rocky cloak]

Like the tensor algebra is
Z (+) Z/n (+) Z/n (+) ...
the first factor corresponds to 1, the second to x, the third to x^2 etc.

So 1 should have infinite order while x^i should have order n

[languid trellis]

Oh right I think I see

[dull ginkgo]

Less so intuition and more defn lmao

[tardy hedge]

Yeah, lol

At least for mow

Now

This stuff is super hard to parse through for me

[dull ginkgo]

The radical is weird

[rocky cloak]

I guess intuitively V(f) captures where f vanishes, while the ideal (f) captures some kind of derivative information about f.

Like for example x and x^2 have the same vanishing set, namely just x=0, but x^2 also has vanishing derivative there.

Taking the radical destroys this derivative information, but keeps the same vanishing.

And the point of schemes is exactly to preserve this derivative information when passing to V(f)

[rocky cloak]

r^i f = f r^-i
is one of the core properties of the dihedral group. "Flips flip rotations"

[dull ginkgo]

What exactly is a scheme?

[rocky cloak]

Basically you package a space together with a cordinate ring for each open set in a nice way.

Less basically it's a sheaf of rings locally isomorphic to the spectrum of ring

[dull ginkgo]

Oh i see

[dull ginkgo]

So the topology is locally the Zariski topology of some ring?

[rocky cloak]

Yup

[graceful dome]

ohh is that because frf = r^-1
so ffrf = fr^-1

[rocky cloak]

That's right

[graceful dome]

just 1 more question, when we define a dihedral group we say its generated by two involutions which are f and fr
Yet I see it defined as G =<r,f> is that just notation

why is it not defined G = <fr,f>

[rocky cloak]

Both (r, f) and (fr, f) would generate the group.

But yeah r is not an involution

So G = <r, f> = <fr, f>

[graceful dome]

I see, okay thank you.

[serene dune]

jagr i think i need your help with something

#linear-algebra message

[tardy hedge]

Trying to show that Spec(A) is compact.

There is a hint in the problem and im tryna figure out why the hint is true.

For a simpler case, if Spec(A) = Xf1 U Xf2, why is it the case that f1 and f2 generate the unit ideal?

Can i get a hint for this?

I know the collection of all prime ideals must capture all non-units

Idk if that is helpful or relevant

[south patrol]

Rephrase in terms of A and prime ideals

[tardy hedge]

All prime ideals of the ring either do not contain f1 or do not contain f2, or do not contain both f1 and f2

Oh ok wait

Then (f1,f2) must not be in any prime ideal, so then it must be the whole ring. (Since any proper ideal is contained in a maximal (prime) ideal, that is the reasoning here right?)

[tardy hedge]

Ok, so I understand why the fi's generate (1), but how can I now say that there is a finite (f1, ... fn) that generate (1)?

im looking at the hint below

I think just because of how we define ideals generated by elements i guess right

ty

[golden turtle]

I am stuck trying to show P projective => P is free

this is my proof for P free => P projective

[rocky cloak]

Do you know that subgroups of free groups are free?

[golden turtle]

Yes

nielsen schrier

So I assume that I should try showing that given P is projective it is the subgroup of some free group

[rocky cloak]

Yeah, that's the proof I have in mind at least

[golden turtle]

Seems fine in concept but I'm also not sure where to begin with that

[weary frost]

surject from a free group to P

[golden turtle]

So say F is a free group

and let g:P->F a surjection

and pi: (a group) -> F a homomorphism

[weary frost]

dont think you can do that

[golden turtle]

why

[weary frost]

F->P maybe

F has no relations

so in general it shouldn't work

[golden turtle]

If P is projective then for any homomorphism pi:G->H and surjection g:P->H one can lift g to G

[rocky cloak]

Think you mixed that up

[weary frost]

I thought start pi: F->P

[golden turtle]

I wrote it stupidly in my notes, I see

yes

pi:G->H is the surjection

g:P->H is the homomorphism

so

let F be a free group and let
pi:F->P be a surjection
and
g:P->P a homomorphism

[weary frost]

nice

maybe take g: P->P as the identity

i think that's all you need actually

[golden turtle]

then we are guaranteed a homomorphism f:P->F such that pi(f)=g

[weary frost]

do you want the solution?

[golden turtle]

Wait I think I'm close

does this work

so

since P is projective

there exists a homomorphism f:P->F such that pi(f)=g
and if g is the identity map then ker(f)=<e>
and by first isomorphism theorem
P/ker(f) iso Im(f)
and Im(f) is a subgroup of the free group F

[weary frost]

works!

basically this

[golden turtle]

😎 😎

tysm

[tardy hedge]

Kinda stuck on vi) now

Ngl i kinda wanna just get thru this question already

[weary frost]

is this functional analysis?

[tardy hedge]

Its an introduction to things in algebraic geometry

Prime spectrum of a ring and zariski topology

[weary frost]

did you ask over here

[tardy hedge]

No but maybe i should i suppose

[rocky cloak]

You can use more or less the same argument as for (v), just replacing 1 by f (or f^n for some n)

[thorn jay]

I've got something which im kinda stuck on
Suppose f(x, y) is a term in the type of groups, in other words, a reduced string of x, y, x^- and y^-.
The condition is that if you add up the number of occurrences of x, and subtract the number of occurrences of x^-, you get 1, and same with y and y^-
(This is the same as saying f(x, e) = f(x, e) = x)
I want to show that
f(f(x, y), z) = f(x, f(y, z)) <=> f(x, y) = xy or f(x, y) = yx
I've got good reason to believe this is true

(I am computing the endomorphisms of Grp in a category ive defined for fun)

[graceful dome]

So I want to show that G is cyclic

I think I know the proof for this btw

I jsut want someone to verify it

typing it out

Proof:
By Langranges since G is finite, any subgroup H <= G must have order dividing G
We then see there are only 2 non trivial subgroups of order p and q.
Once again by lagrange's we can tell these H are cyclic as they have prime order.
So there is an element g in G with order p and an element h of order q.
Then since we know G is abelian, o(pq) = o(p) o(q) = pq
So G has an element of order pq which tells use pq generates G, and G is cyclic
.

if anyone can let me know if I did anything wrong that would be great

I did not follow the instructions of a or use the fact that (p-1) doesnt divide (q-1) or vice versa

[rocky cloak]

I feel like there might be a typo in the problem.

The (p-1) stuff can be used to prove that G is abelian, which feels like what the exercise should be asking

As opposed to just being given that G is abelian from the getgo

[graceful dome]

Q1. Is proving the theorem: If G is a finite group with at most 1 subgroup at each index, then G is cyclic

thats what it's referring to by problem 1

[graceful dome]

how would I use (p-1) doesnt divide (q-1) and vice versa to prove G is abelian

[proven mango]

There are 2 slightly different definitions for normal subgroups. I know that they're both the same idea about the left and right cosets being the same but how could you show that the two definitions are identical?

\newtheorem*{definition*}{Definition}
Prove that the following definitions are equivalent:
\begin{definition*}
A subgroup $H \leq G$ is called \textbf{normal} if its left and right cosets agree: that is, $gH = Hg$ for all $g \in G$.
\end{definition*}
\begin{definition*}
A subgroup $H \leq G$ is called \textbf{normal} if for all $h \in H$ and $g \in G$, $ghg^{-1} \in H$.
\end{definition*}

[rocky cloak]

Do you know the sylow theorems?

[cloud walrus]

Devin ☆

[graceful dome]

No I have either not learnt that yet or I was taught it under a different name

What is it?

[rocky cloak]

If you have a group of order m*p^k (where m doesn't divide p), then the sylow theorems tells you how many subgroups of order p^k there are.

In particular in this case you can use them to prove there are exactly one subgroup of size p and one of size q

[graceful dome]

hmm I dont think I learnt that

What I do know is that given G = <g>, any subgroup H of G with finite index k must be <g^k>

[rocky cloak]

So let gHg^-1 be the set {ghg^-1 | h in H}

Then the second definition is saying
gHg^-1 is a subset of H
gHg^-1 < H
Multiplying by g from the right on both sides gives
gH < Hg
Similarly multiplying by g^-1 from the left gives
Hg^-1 < g^-1 H

Since g is arbitrary we can replace g^-1 with g so Hg < gH.

Then gH is a subset of Hg and Hg is a subset of gH so they're equal

[rocky cloak]

Okay, then maybe it was supposed to say that G is abelian, and they just added the extra stuff to trick you

[graceful dome]

they dont usually try to trick us, the hints usually matter

here is the full problem 1 they were referring to

Which is supposed to be the proof for If G is finite and has at most 1 subgroup of each index then G is cyclic

[rocky cloak]

Well, I guess you can ask however gave you the assignment if it's a typo

[graceful dome]

yea I guess so, though is my proof okay otherwise?

[rocky cloak]

Yes, your proof is good

[graceful dome]

awesome thanks

[thorn jay]

Bump

[rocky cloak]

I feel like you should be able to look at the total number of x^±s in f in reduced form. Then argue that in f(f(x, y), z) you have that amount squared while in f(x, f(y, z)) it's the same amount.

[thorn jay]

Hmm

That's a good one actually

Ok thanks!

I see it

That completes my argument:
$$\text{End}{\mathbf{Eq}}(\Sigma{\text{Grp}}) \cong C_2$$

[cloud walrus]

Z(A) = ∇ <=> A is R-module

[thorn jay]

Yay!

[tardy hedge]

Basic question im forgetting, why if R a field do we have r*x = 0 implying r is 0? x is element of a module over R

x not 0

[next obsidian]

r^-1•r•x = x

[tardy hedge]

Lolllll

Thanks

[next obsidian]

Sorry to be clear

This is for contrapositive

[dim widget]

dont be sorry, it's good to state yourself clearly

[next obsidian]

HELLO MATH DISCORD FINAL GIRL

[tardy hedge]

Dude u beat me to it

MATH DISCORD FINAL GIRL HELLO!!!

[next obsidian]

Chmowned

[tardy hedge]

For M an R module and I an ideal of R, what is the significance of the submodule IM?

I have seen it show up in some proofs and exercises and i cant quite put my finger on what it is

Specifically, ive seen a lot of quotients by this IM kind of submodule

[trail cave]

for a cyclic group G generated by a, can someone explain what the euler phi function tells me about it? e.g. if I take phi(k), is that the number of elements of G of order k?

[next obsidian]

No it’s the size of the automorphism group of the cyclic group of size k

[coral spindle]

Well phi(k) is indeed the number of generators of C_k, i.e. the number of elements in C_k which are of order k. So both are correct.

[next obsidian]

Oh yeah

[coral spindle]

Of course this is really the same thing, since all generators are related by a unique automorphism

[next obsidian]

But like, actually no I still disagree

[coral spindle]

Hahah

[next obsidian]

Because you need to know what k is in relation to the size of the cyclic group

[coral spindle]

Yes indeed

I think our asker did not specify this clearly enough

[trail cave]

thank you

[golden turtle]

I could not find what it is referencing as THM6

I don't really understand what it means by what phi is

but if K=ker(phi)

then doesn't

Im(i)=K=ker(phi) so that is why it is a short exact sequence?

just a little confused by what phi is

[green forge]

the screenshot you sent defines phi
Let f : S -> M be the inclusion map. By the universal property of free modules, there is a unique R-module homomorphism \phi: F(S) -> M such that f = \phi \circ i, where i : S -> F(S) is the inclusion map

[golden turtle]

I see so it is the unique homomorphism from the universal property

thank you

[green forge]

you're welcome!

[dull ginkgo]

call her a short exact sequence the way she splits when she sees my section

[golden turtle]

Facts

[tardy hedge]

Thats crazy 😂

[dull ginkgo]

the malevolent exact functor

[golden turtle]

To verify that this is a homomorphism of short exact sequences, it is clear that alpha, beta, gamma are homomorphisms, and then to show that the diagram commutes I am wondering if there is a standard way of doing this or if you really have to show that each path is equivalent to every other path individually

I've never done one of these yet

[vast stump]

i believe you could show the two ‘squares’ individually commute

[golden turtle]

wow thank you sir

never seen you here befroe

[tardy hedge]

True

Neither have i

Where is wew lads tbh!?!?

[golden turtle]

What does this mean

phi: phi(C') iso C

[tough raven]

It means phi restricted to C' is an isomorphism from C' to C.

[golden turtle]

Oh I see

I'm trying to follow this proof, but I don't see how it follows directly

It seems like it is implying that mu(C) direct sum psi(A) = B