groups-rings-fields

[plucky flicker]

f = X^n - a. Have you tried to find the resultant of f and f'?

If you can compute that, then you can use the formula that connects discriminant and resultant

[tame grove]

In not sure where to start with this one.

What does it mean when it is asking me to prove an isomorphism for a group and its operation? I thought that group isomorphisms had to be between differently defined operations.

[plucky flicker]

You have to construct a bijection on G, such that f(ab) = f(a)*f(b)

[tame grove]

Oh.

Well thats easier than I thought it would be. thanks

[plucky flicker]

like, you can take the identity map on G, which is an isomorphism

[vocal wolf]

i dont think thatll work unless G is abelian

[plucky flicker]

umm yes

I just realized that

[hidden haven]

g maps to g inverse

[viscid pewter]

lol

[plucky flicker]

yepp

[viscid pewter]

i was gonna try and lead them there

[hidden haven]

oh ye

mb 🤡

[viscid pewter]

ah well

because like

(ab)^-1 = b^-1 * a^-1

so it does the flippy thing you want

[hidden haven]

(╯°□°）╯︵ ┻━┻

[plucky flicker]

Do you know about Newton Waring formulas?

I know a solution but that involves NW formulas

Okay, so we denote P_k = x_1^k + ... + x_n^k, and we have two cases

First case: if k <= n, then the formula is the following: P_k - P_k-1 * s_1 + ... + (-1)^i * P_k-i * s_i + ... + (-1)^k-1 * P_1 * s_k-1 + (-1)^k * k * s_k = 0

Second case: if k > n, then the formula is the following: P_k - P_k-1 * s_1 + ... + (-1)^n * P_k-n * s_n = 0

where s_1, ..., s_n are the elementary symmetric polys

Next step: the discriminant of f can be written in terms of P_i, where P_i are polynomials as above, and the variables are the roots of f, and the form is this:

,rotate

[plucky flicker]

And now, using the formula for your polynomial you get a pretty easy determinant to compute, because most of the P_i-s are 0, by NW

I hope it helps

In fact the only P_i which stays in party is P_n because by the Viete formulas you get that for all i from 1 to n-1 P_i = 0 (in the first NW formula all the s_i = 0), and P_n = n*a

And you should check what happens, when k > n: then P_k = 0 for all n < k < 2n-2 (due to the same argument above)

[plucky flicker]

Guys what do you think about this problem here: Let G = <g> a cyclic group of order n. Let us map g^k to the rotation of the (real) plane around the origin by 2pik/n. Show that this is an irreducible representation of G, provided n > 2. So I have to show that if n > 2 then this is an irreducible rep, or if this is an irred rep then n > 2?

[viscid pewter]

you have to show that if n > 2, it's irreducible

[plucky flicker]

Thanks

well basically I have to show that no one dimensional subspace of R^2 is invariant under this action, right?

[lethal cipher]

Hello, I could really use some help on 1). I am not even to sure where to start.

[chilly ocean]

maybe-nuke: fundamental theorem of finite abelian groups

(i don't know)

[lethal cipher]

Actually, if we are using the Fundamental theorem, I think we need to show that if d, d' are divisors, then gcd(d,d')=1

Then we can break it up into a direct product of cyclic groups. Since the order of each is relatively prime, our group is cyclic.

[plucky flicker]

https://math.stackexchange.com/questions/890106/discriminant-of-xn-1

this is a classical solution for x^n - 1

Maybe you can try this method to solve it for your polynomial, it should work

[lethal cipher]

However, I have no idea how to do that...

[lethal cipher]

Tbh, I am not really sure how to apply the Fundamental theorem.

[small bison]

factor it as a product of cyclics and try arguing in each cyclic component

each component is like Z/p^kZ or something like that

[hidden haven]

Does C_2 x C_3 not have that property?

Oh wait

[small bison]

that's C_6

[hidden haven]

Ye

[lethal cipher]

C?

[small bison]

C_n is cyclic of order n

anyway at the end you should be able to show that if Z/p^kZ has that property that k=1

and that should basically give you the proof once you know how to combine the factors back into one big group

[lethal cipher]

Well, small issue with that. The primes dont necessarily need to be distinct when we break G into the direct product of cyclic groups

[small bison]

oh yeah

but you can handle that case anyway

[hidden haven]

Why k=1

Oh wait

[small bison]

cause Z ain't finite

[hidden haven]

Ignore me

[small bison]

think about why something like Z/pZ * Z/pZ isn't allowed

[lethal cipher]

Well, I know why it shouldn't be, because then it wouldnt be cyclic

[small bison]

Like using the condition imposed on G

[lethal cipher]

Ooo. So say you have a group of order d, we will call it D. Then (0,D) and (D,0) would be two subgroups of the same order

But we can't have that, because exactly one subgroup has order d

So why are we looking at cosets and not Z_p itself?

[small bison]

What do you mean by cosets? I’m just giving examples to help you work out the full proof

[lethal cipher]

Isnt Z/pZ the factor group of Z induced by pZ?

[small bison]

Also now that I think about it Z/p^kZ should be okay for k >= 1

Uhh I’m using the fundamental theorem to write G as a sum of cyclical

Cyclics

And I write my cyclic groups as Z/max

Z/nZ

[lethal cipher]

Ah, I see

Okay, so now if it works for k>1 as well, then why can't something like Z_pxZ_p^2 be apart of our product of cyclic groups?

Oh wait, the only subgroup of Z_p is the identity and itself

Then of course {0,p,2p,...,(p-1)p} form a subgroup of order p in Z_p^2

That thought process did the trick. Thank you.

[chilly ocean]

Can someone send me the proof of this theorem?

Or where can I find it

[hidden haven]

Almost any group theory book But it's better to work this out yourself

[chilly ocean]

That's why I'm looking for it

[hidden haven]

Apply both sides to an element x

[chilly ocean]

We had a problem

We didn't do this at class

[hidden haven]

And see how x gets permuted

[chilly ocean]

So

I solved the problem without it

And my friend told me it turned out to be similar to the proof of that theorem

So what book

[hidden haven]

Artin has this I think

[limber vale]

can someone help me with the following?

Let K and L both be field extensions of field F (F is a finite field with q elements) with [K;F]=l,[L;f]=n. where both L and K are contained in and extension M of F.

Proof that [KL;F]=lcm(l,n)
and
[K\capL;F]=gcd(l,m)

my reasoning was since K and L are subfields of KL n and l must both divide [KL;F]

We know that [KL;F]<=ln

but i dont know how to conclude it

for the second with similar reasoning [K\capL;F] must divide both l and n but i also dont know how to continue here

[lethal cipher]

I finished it, but I just wanted to say that this problem is a pain!!

Man, I did not miss the symmetric difference 🤣

[deep nova]

so i have this problem; i managed to show that the function is a homomorphism

for the kernel, im not sure what im supposed to do

[next obsidian]

What’s the identity in GL2(R)?

[deep nova]

identity matrix (1 0
------------------0 1)

[next obsidian]

When does phi(a) equal that matrix?

[deep nova]

a=1

[next obsidian]

You’re done

[deep nova]

thats the kernel?

oh

[next obsidian]

GG

[deep nova]

oh right its supposed to be a one element set right?

[next obsidian]

No

The kernel could be whatever

[next obsidian]

Well, there’s some restrictions

[chilly ocean]

Could someone help me with this one

[deep nova]

ok i see

[next obsidian]

We’re in the middle of something it’s kinda rude to just butt in

This also probably fits better in #discrete-math

[chilly ocean]

Oh am really sorry

And thenks

[next obsidian]

Anyway yes

The kernel could be a singleton, it could be everything

If your set is finite the size needs to divide the size of the ambient set

But in general you don’t really know a priori how big the kernel is

[deep nova]

👍 thanks

[high lava]

is a tuple of complete lattices a complete lattice?

[limber vale]

bump

[civic tapir]

The only definition I can find of defining equations of a group is a system of eqns that completely determines the operation table of the group. What exactly does this mean? My book provides an example and says that (ab)(ab^2)= ... = a^2 b^4. Why is a^2 b^4 "determined" and (ab)(ab^2) isnt, and what does it mean for an entry to be determined?

[viscid pewter]

this sounds a little like generators and relations and presentations

i guess it's just a set of relations that completely defines the group

idk what determined means

[civic tapir]

just got it, sorry for the interruption

[amber stag]

Need to use something special about finite fields. A finite field of q=p^n elements is the set of all roots of x^q-x

In this question we only need the fact that the field of p^n elements is unique (up to isomorphism). If n|m then F_{p^n} is a subfield of F_{p^m}

[tropic spade]

If I have $\Bbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ and a subgroup of the galois group for this field of order 4 that is something like ${e,a,b,c}$, is ${e,a,b,c}'={e,a}'\cap{e,b}'\cap{e,c}'$ where $'$ denotes computing the corresponding fixfield of a subgroup?

[cloud walrus]

DootDooter

[tropic spade]

Now that I think about it idk if fixfield is the word ppl normally use for these things.

My book defines it as something like the field that corresponds to a subgroup of the Galois group for a field under the Galois correspondence.

It seems to me like there should be something sort of like De morgans law kind of deal going on. Where if $B={A_i:i\in I}$ is a set of subgroups of G the Galois group of some field that we should have something like $\bigcap_{i\in I} A_{i}'=(\bigcup_{i\in I} A_i)'$. But I'm not totally sure if that is true.

[cloud walrus]

DootDooter

[tropic spade]

Could Sym(G) be the symmetric group on the set G?

Just a guess.

I have not really seen that notation for that.

Oh I see. I've only ever seen it written with subscripts.

I think you can just do S_X for a set X

But Sym({...}) seems like it could be more convenient if you have a need for the bracket notation.

[next obsidian]

I think it makes sense when you need to emphasize what set it’s bijections on

If you want to say evaluate them on a particular element and use the output for whatever reason

Group theoretically tho it only depends on the size of X

[tropic spade]

So, like if you wanted something like Sym({f,...}) to make it easy to compute something like f(n)?

[next obsidian]

Uh I mean like

Say you have a set of triangle square and circle

And wanted to evaluate on triangle

For some ducking reason

This makes more sense if like, idk you have a subset of an abelian group

And wanted to do like

Sum_f in Sym(S) f(fixed thing in S)

I guess me point is just you want to emphasize these are actual functions, with the same domain and codomain

[tropic spade]

That makes sense.

[vocal wadi]

Could someone help me with the question: how to calculate the direct sum and product of two polynomial rings: k[x,y]/x^2 and k[x,z]/xz ?

now i know they're equal. and i still don't know how to work it out.

x^2 means (x^2), etc

I want to cancel the plus sign

[hidden haven]

Do you mean direct sum and product as k modules?

You don't have direct sums of rings

[robust pollen]

\otimes_{\mathbb{Z}}

[hidden haven]

Do we call coproducts direct sums

[robust pollen]

jk

actually, related question: coproduct in Rng is actually direct sum, innit?

[lethal dune]

so any intuitive way to understand the syllow theorems??

[hidden haven]

How define direct sum of rings?

[lethal dune]

[hidden haven]

Coproduct is tensor

I think direct sum means biproduct

[robust pollen]

no, I mean rngs. So the problem for direct sum with rings is that the obvious inclusions are not ring morphisms, but for rngs, the obvious inclusions are rng morphisms, right?

[robust pollen]

actually yeah that's the abelian way to think about it, you're right

[hidden haven]

I will not be tricked into thinking about rngs

[robust pollen]

dang

what about ?

You know, are like rings, but without any extra structure.

[hidden haven]

[robust pollen]

Some people might denote the category of by Set. But I prefer to call it

[hidden haven]

[vocal wadi]

Sorry, I mean coproduct

[hidden haven]

So you want to prove that for these rings, coproduct = product?

[vocal wadi]

sorry, i can explain the problem as this: Is it possible to cancel the plus sign in "k[x,y]/x2 + k[x,z]/xz" by using another symbol?

where the plus sign means coproduct

[lethal dune]

[vocal wadi]

I probably misunderstood the concept of polynomial rings: I thought the coproduct of polynomial rings can also be polynomial rings. Is it not true?

[robust pollen]

Hm, so if we're just talking about k[x,y] (without quotients), then note that this is the free (commutative unital) algebra on the set {x, y}, which can be seen as the disjoint union of the set {x} with itself.
Call the free algebra functor F. Then k[x,y] = F{x,y} = Fx + Fy = kx + ky in your notatation, since F is a left adjoint and thus commutes with colimits. But I don't see it for k[x, y]/I rn

Is what I write true?

rn = right now, I is an ideal and capital i's look like lowercase L's.

what do you mean?

[vocal wadi]

thanks, i understand what you have written now

[waxen hedge]

Here "+" is the tensor product over k

Since we're working in the category of k-algebra

But I don't know what happens if we forget the k-algebra structure to get vector spaces over k

[robust pollen]

what is this sentence supposed to mean?

https://tenor.com/view/jackie-chan-meme-what-confused-huh-gif-5480485

[winter thorn]

wtf I got confused bw subgroup notation and order notation

today is not a good day

okay, needed to prove that if a is in a group G, then card(a) <= card(G)

The order of a is the largest nonzero positive integer such that a^n = e.

by closure of G, a, a^2, a^3..... a^n are in G.

{a} is a subset of G.

If {a} = G, done.

If not, there exists b in G such that b is not in {a}.

Therefore, card(G) > card(a)

[hidden haven]

Closed subsets are finite or the whole set because a polynomial can have finitely many roots. The only finite irreducible subsets are the singletons

also we have an #algebraic-geometry channel now lul

[winter thorn]

@hidden haven is my solution correct?

[hidden haven]

ye, but you could have stopped at <a> is a subset of G therefore |<a>| <= |G|

card(a) means nothin because a isn't a set. You want to say card(<a>) instead where <a> is the subgroup generated by a

[winter thorn]

oh sorry, it's supposed to be order.

[hidden haven]

I see

You need to also handle the case where a has infinite order

probably

idk what the problem says about that

[winter thorn]

eh, this is the subgroups chapter in a very introductory text, so probably not

[hidden haven]

I see

I am assuming it means closed irreducible subspace

[cloud walrus]

Yeetus

[hidden haven]

right ye

so algebraic sets are exactly the whole space or finite subsets

check this

from each single variable polynomial having at most finitely many roots

and notice that all singletons are algebraic

so finite subsets are algebraic

so only irreducible ones would be the singletons

or empty set 🤡

wait non empty condition

all good

[winter thorn]

how do I even start, no pun intended

[hidden haven]

study the subgroup N of all even numbers

in Z_n

look at intersections

[winter thorn]

okay

hmm basically Z_n for even n is the union of all the even and odd whole numbers less than n

[winter thorn]

no thats true for all of them, there are exactly half evens and half odds in Z_N for even n

[paper flint]

Gallian?

[winter thorn]

yeah

[hidden haven]

manan book collecting

[winter thorn]

He's my role model in that regard.

you know, I got Complex Variables by Kasana today.

[plucky flicker]

Hi! What's the idea behind this construction? In general we can easily construct a 2-dimensional representation of S3, by considering the rotation and reflection matrices. But then the entries would not be integers.

[hidden haven]

S_n has a natural representation (idr what it is actually called) which is n dimensional. You take the free vector space generated by n elements, say x_1 to x_n, and make S_n act by permuting them in the natural way. This is a reducible representation and splits as A direct sum B where A is the set of vectors c(x_1 + ... + x_n) for some c in F, and B is its complementary subspace, of all elements which, when written as a linear combination of the x_i's, have coefficients summing to 0. This B is exactly the one given above

To see that this is the B you are given, take the basis ||(x_1 - x_3, x_1 - x_2)||

[plucky flicker]

by natural representation you mean for every permutation we take the corresponding permutation matrix?

[hidden haven]

yes

[plucky flicker]

and this is a 3d rep, and reducible, and splits into A + B as you have written

[hidden haven]

yes

A is called the trivial rep (it is one dimensional, all elements of the group act on it trivially)

B is called the standard rep

It is irreducible for any n (at least when F = C)

and B has dimension n-1

[plucky flicker]

and it's reducible because for instance a line is not fixed by a 120 degree rotation

i mean the natural repr

[hidden haven]

ye well you can show that A + B is a decomposition

[plucky flicker]

so this is a method for constructing 2d reps for S3 up to equivalence<

?

when I change the basis of B i get another one

but those are equivalent

[hidden haven]

Ye you only really get one representation

so not very useful in terms of constructing representations directly

It is useful because it is an irreducible rep

and for nice fields (characteristic not dividing |G|), any representation of G over that field is a direct sum of irreducible representations

By Maschke's theorem

So characterising the irreps of S_n gives you a lot of info about arbitrary reps

[plucky flicker]

thanks 😄

[barren sierra]

I want to show that $x^2 + 1 \in \mathbb{Z}_3[x]$ is irreducible using Gauss's lemma and Eisenstein's criterion

[cloud walrus]

Spamakin🎷

[barren sierra]

So I am stuck on the Eisenstein's criterion part

(yes I know this is overkill and I can check 0, 1, 2)

[hidden haven]

that's a weird way of showing irreducibility of a quadratic over a field

[barren sierra]

well I just more want to get practice using this theorem

so like

I need some "prime" in Z_3

[hidden haven]

You can't apply it here

[barren sierra]

oh

[hidden haven]

Only prime in Z3 is 0

[barren sierra]

that explains alot why I am struggling

ok so how do you find that out

[hidden haven]

oof

[barren sierra]

[hidden haven]

In fields, all non zero elements are units

So can't be prime

Zero is prime because integral domain

[barren sierra]

I see

ok

makes sense

[naive granite]

I have a question on Abstract Algebra.

[chilly ocean]

semigroups and monoids

[naive granite]

I was able to answer a and b but I have no idea how to answer c

How the hell do I prove if it's a group..

[chilly ocean]

check invertibility of the elements

[naive granite]

oh

So what does set with the operation need to be to be a group

associativity, identity element, closure, invertibility

anything else?

[chilly ocean]

supposedly you've already checked everything but invertibility in parts a and b

[naive granite]

yeah

[chilly ocean]

but yeah you've got it down here

[naive granite]

oooh

okay

sweet

Thank you!

[chilly ocean]

you checked associativity in a and identity in b, so you just gotta check invertibility for c

and closure if you havent already lol

[naive granite]

xd

[plucky flicker]

The Jacobson radical of a division ring is the 0 ring?

[coarse stag]

If you have a center N, if G/N is a subgroup of G, is G/N a normal subgroup?

[hot lake]

how can G/N be a subgroup of G

[next obsidian]

N = {e}

[coarse stag]

If G= N semidirect product with H isn’t G/N=H

[hot lake]

ah yeah but then it has to be a semidirect product

so you're asking if there is a section, is it normal

I think it should be

every element in G ban be written in the form ns with n in N and s in "G/N"

conjugation by anything in N does nothing

and conjugating G/N by anything in G/N gives G/N

so G/N should be normal

[coarse stag]

For groups that can be factored into distinct primes, is G/N not a subgroup?

[viscid pewter]

G/N is never a subgroup

the elements are like

sets

different elements

it might be isomorphic to a subgroup tho

[coarse stag]

Isnt it the cosets tho?

[viscid pewter]

yeah, cosets aren't elements of G

they're sets of (elements of G)

right?

equivalence classes

[deep nova]

im not sure if understand what this means

[stark sigil]

Are you stuck on homomorphisms or the Z_n notation?

[deep nova]

I'm having difficulty understanding how i'd go about solving a problem like this.

I know i have to do something with a kernel and a generator

but im really not sure what it is that i have to do

my book had an example with a similar nature to the problem, but it's a bit vague in how it did it

this is the example given

[fossil shuttle]

Exercise: let $\phi, \psi : \mathbb{Z}_n\to A$ be two homomorphisms. Suppose $\phi(1)=\psi(1)$. Prove $\phi = \psi$.

[cloud walrus]

diligentClerk

[fossil shuttle]

Exercise : Let $a \in A$. Then there exists a homomorphism $\phi : \mathbb{Z}_n\to A$ sending $1$ to $a$ iff $a^n=0$. (By the previous exercise, this homomorphism is necessarily unique if it exists.)

[cloud walrus]

diligentClerk

[fossil shuttle]

If you can solve both of these it should give you all the tools you need to do your problem.

[deep nova]

ill be honest, im not sure how to solve them

[coarse stag]

"if |G| = pq for primes p and q not necessarily distinct, prove that either G is abelian or Z(G)=1"
Where do I even start for this type of question?

[thorn delta]

Well if G is not abelian, then Z(G) is going to be a proper subgroup of G. Maybe start there

[coarse stag]

I reach a point where Z(G) must be a normal subgroup of G with order 1,p, or q, but I dont know where to go from there

[thorn delta]

Maybe note that Z(G) is cyclic and if Z(G) != 1 then G/Z(G) is also cyclic.

[coarse stag]

or atleast im not sure where to make the jump from contains a cyclic group of order p to the entire group is abelian

[small bison]

the center is normal so you get Z(G) -> G -> G/Z(G)

and i think you can do some semidirect product thingy

[coarse stag]

right, but G/Z(G) is not nessesarrily in G right?

[thorn delta]

Well, okay so we want to show G/Z(G) is abelian to get a contradiction right? So if x,y are in G, then xy might not be equal to yx, but we can pass to the quotient and say: xyZ(G) = yxZ(G)
Now you just have to unpack what this means

[coarse stag]

wait how does showing G/Z(G) is abelian arrive at a contradiction?

[thorn delta]

oops, i meant we should show G is abelian, my bad

[small bison]

i think what i wrote works also

0 -> C_p -> G -> C_q -> 0 is short exact

so every action C_q -> C_p gives an isomorphism between G and C_q semidirect product C_p i think

[coarse stag]

wait how do you know C_p is a function image?

[small bison]

G/C_p is order q

q's prime right?

[coarse stag]

yeah

[small bison]

yeah any group of order prime is cyclic

[coarse stag]

but what about Z(G)=1

[small bison]

that's what you want to happen

you're trying to prove that Z(G) = 1

so you would assume Z(G) != 1 so Z(G) = p or q

wlog |Z(G)| = p and run the argument above

[coarse stag]

also I dont get the left hand side of the diagram

0 -> C_p -> G

there are short exact functions from 0 to C_p to G?

[small bison]

do you know what short exxact sequences are?

if not then you should try to prove this a diff way

[coarse stag]

isnt that the image of the previous function is the kernel of the next

[small bison]

yeah

i might've written it backwards

jk it was right the first time

[coarse stag]

whats the image from G/Z(G) to G

[small bison]

yeah it was 0 -> Z(G) -> G -> G/Z(G) -> 0

[coarse stag]

ok I see that, but I dont see what you mean by semidirect product

because the group isnt nessesarily reconstructable by semidirect product of normal subgroup and quotient group right?

[coarse stag]

Are you saying that if f(xy)=f(yx), then we can say xy=yx?

[thorn delta]

yea, you would need to use that G/Z(G) is cyclic

[coarse stag]

wait I dont follow how f(xy)=f(yx) imples xy=yx tho

arent dihedral groups a counterexample?

[small bison]

i think this should be the case

it's a group extension

[coarse stag]

we havent covered extensions yet, is there a line of reasoning that doesnt involve extensions?

[small bison]

hmm in that case maybe you can just sylow it lmao

[coarse stag]

bruh

sylow is hard to get an intuition for

[thorn delta]

yea so lets say cZ(G) generates G/Z(G), and xZ(G) = c^nZ(G) and yZ(G) = c^mZ(G).
then x = c^nk^j and y = c^mk^a for k such that <k> = Z(G).

got it from here?

[coarse stag]

and because k commutes with everything, you can just reorder it to have c on one side and k on the other no matter if its xy or yx?

[thorn delta]

yep

[coarse stag]

Regarding the extension thing, if G is an extension of the quotient by the normal subgroup, then does that imply that you can write G as the semidirect product between the quotient and normal?

because im thinking about say Z_4 and the normal subgroup of 2 and quotient group of 2.

because you get {0}->{0,2}->{0,1,2,3}->{0+{0,2},1+{0,2}}->{0}

which still seems short exact

but you cant write that as a semidirect product

[small bison]

hmm maybe i'm forgetting the details of how it works

[sturdy marsh]

this is not always the case

those are split extensions

there are many non-split extensions

there are uncountably many non-split extensions of Z and Q

[small bison]

yeah ik that but i'm trying to do it with semidirect products

i think you just need in addition a homomorphism from G -> semi direct product to get an iso

[coarse stag]

I think G/H needs to be in G to semidirect product

so you need something from G/H -> G I think?

[small bison]

doesn't it just split in that case?

lmao my group theory is so bad

[sturdy marsh]

oh groups of order pq

yeah all extensions are split

@coarse stag what exactly are you trying to prove

[coarse stag]

one of the discontinued challenge problems

the question I had before was already answered tho

[sturdy marsh]

wait what are you trying to do?

show that 0 --> Z/p ---> G ---> Z/q ---> 0 splits?

if so, this shouldnt be too hard

G has order pq, so it has an element of order q

then we have a homomorphism f: Z/q ---> G

by mapping 1 to that element

[small bison]

Oh yeah that’s easy then

[sturdy marsh]

yeah we just need to show that it is a splitting

call the map G---> Z/q g

for any x in Z/q, we need to check gf(x) = x

i.e. gf(x) -x =0

(im using additive notation because it's hard to type it out multiplicatively)

now, g is surjective, so g(y) = x for some y in G

so we need to check gf(x) - g(y) = 0

i.e. g(f(x) - y) =0

g(f(x)-y) = 0 iff f(x) - y is in Z/p (by exactness)

hmm

y cannot have order pq as we're assuming G is not abelian

so it has order p, q or 1

if it has order p, it cannot map to a nonzero element of Z/q

(assume x \neq 1)

so it must have order q

f(x) has order q

so does f(x)-y have order dividing q

[coarse stag]

f(x) has order q?

[sturdy marsh]

yeah it's in the image of Z/q ---> G

[coarse stag]

oh and by homomorphism

subgroups map to subgroups

so the subgroup generated by x maps to something of order dividing order of x

[sturdy marsh]

the issue is that f(x) and y may not commute

if they do commute, we are done

maybe we can change y by an element z of Z/p so that f(x) and yz commute

[coarse stag]

I believe we assume Z/p to be a center

wait nvm that doesnt help

[coarse stag]

I assume you mean x \neq 0 here btw

wait even if we show that f(x)-y has order not dividing q, how can we show f(x)-y in image of Z_p --> G?

[sturdy marsh]

Z/p is a p-sylow, and it is normal. So it is the only subgroup of order p

yeah this is getting a lot uglier than I thought it would be

the cleanest way is probably showing G/Z(G) cyclic implies G abelian

now im not even sure this works

[sturdy marsh]

ignore what I said

we need to show that it has order dividing p

and q

yeah this shouldnt work

[coarse stag]

so this is a dead end?

[sturdy marsh]

oh wait wtf

Z/p is normal in G

the image of Z/q ---> G has order q

call the image H

Z/p \cap H = 0

as the orders are coprime

Z/p \cdot H is the whole group (as it has order pq)

so G is a semidirect product

[coarse stag]

wait I dont see this?

[sturdy marsh]

if H and K are subgroups of a group G

[coarse stag]

yeah

oh

[sturdy marsh]

then |HK| = |H||K|/|H cap K|

note that HK may not be a group

but the formula still holds

[coarse stag]

x in Z/p and y in H, xy nessesarily must have order |Z/p||H|

wait

is that right?

[sturdy marsh]

no

[sturdy marsh]

but this is

[coarse stag]

what is |H||K|/|H cap K|?

[sturdy marsh]

oh no it is correct

as Z/p is in the center

no it isnt right

as this would imply G is cyclic

but G is cyclic nvm

yeah it works

here's the whole thing

assume Z(G) \neq 1 and G not abelian

then Z(G) = Z/p wlog

choose Z/q ---> G

elements of Z/p commute with everything

so ord(xy) = pq for x in Z/p and y in the image of Z/q ---> G

x and y \neq 0 ofc

so this implies G is cyclic, which is a contradiction

[sturdy marsh]

| something| = order of that thing

[sturdy marsh]

this proof strays too far away from what young_smasher was trying to do though

[coarse stag]

i see

[barren sierra]

I'm trying to show closure for multiplying multiplicative inverses in $\mathbb{Q}\left[ \sqrt[3]{2}~\right]$. Is there something analogous to multiplying by the conjugate for things of the form $a + b\sqrt[3]{2} + c\sqrt[3]{2}^2$??

[cloud walrus]

Spamakin🎷

[barren sierra]

Any ideas?

💀

it's due in 30 minutes and this is all I need lol

just need to show this takes the form $a' + b' \sqrt[3]{2} + c' \sqrt[3]{2}^2$ such that $a', b', c' \in \mathbb{Q}$.

[cloud walrus]

Spamakin🎷

[hidden haven]

I'm not sure I understand what you need

Is p is in this ring, then p inverse is too

Is this what you want to show?

It follows from the fact that Q[cbrt 2] is isomorphic to Q[x]/(x³-2) (using first isomorphism theorem) and (x³-2) is generated by an irreducible in a PID so is maximal so this is a field

@barren sierra

[barren sierra]

eh

what's done is done

[barren sierra]

I needed to show that $\mathbb{Q}\left[ \sqrt[3]{2} \right]$ is a subfield of $\mathbb{R}$

[cloud walrus]

Spamakin🎷

[rustic crown]

if you wanna show by hand, then notice that gcd of x^3-2 and d+ex+fx^2 is 1 assuming not all d, e, f are 0.

[barren sierra]

Haven't learned that 🥲

we haven't done isomorphism theorems idk why

[hidden haven]

Done now

[barren sierra]

it's like we're skating around it

[rustic crown]

so you can find polynomials (x^3-2) * p(x) + (d+ex+fx^2) * q(x) = 1, and now plug x = cbrt(2)

[barren sierra]

oh damn

so a multiplicative inverse must exist

rip

ugh I've used that exactly

on a prior HW

[rustic crown]

oops

[barren sierra]

isomorphism theorems are tomorrow 😭

[rustic crown]

iso thm good

[barren sierra]

🥲

[rustic crown]

midterm not so good

[barren sierra]

[chilly ocean]

do people remember proofs for isomorphism theorems off top of head?

i have intuition for first iso

but i dont for the others as cleanly

[little root]

does anyone have tips for developing better intuition for abstract algebra (e.g. group theory)? I feel like my intuition for analysis and even linear algebra is much stronger because I can visualize things. On the other hand, I still don't feel like I know what a coset or a normal subgroup is, and I actually find that my intuition for group theory is pretty misleading a lot of the time (like something that seems intuitively true to me might only be true for Abelian groups). I think it might be because with abstract algebra, everything is abstract, so there isn't some canonical example I can rely on (unlike in analysis where Euclidean space is very intuitive). I know intuition isn't something I should necessarily rely on, but I do think I am much better at analysis because I have strong intuition for concepts, whereas I don't really have this tool when I'm doing algebra.

[next obsidian]

I can probably reproduce them if I have to

The first iso is definitely just baked into my head tho

[wooden ember]

Imagine replicating the butterfly lemma 😭

It’s so ugly

[lethal dune]

First and the 4th one are actually used the most

don't remember the last time I had to use 2nd or 3rd

[next obsidian]

3rd is used a decent bit

2nd is not very often used

I use 3rd a decent bit with modules

[lethal dune]

lol I use them so less, I can't even recall the 3rd one

[south patrol]

Tfw numbering can be inconsistent anyway

[lethal dune]

For isomorphism, I probably think it's universally understood which number represents which thm

[viscid pewter]

you would be wrong :)

[torpid mesa]

(If this is the wrong channel for this, please feel free to redirect me.) Working on proving the first statement, but I'm pretty stuck. I'm using the definition of the norm (norm-2) in which ||A|| is equal to the max. eigenvalue of A'A but I'm not sure how to use that definition to prove the statement. Is there a different definition of the norm I should be using or some sort of trick I am missing?

I have been assuming it means joining the matrices together (sorry if there's a better term for that I'm missing). The dimensions of the given matrices A_1 and A_2 would allow for it.

[lethal dune]

commutator probably?

[torpid mesa]

Could be, though it looks like commutators usually have a comma in between the matrices like [A_1, A_2]

[lethal dune]

don't think commutator is defined if p,m1 and m2 are unequal so IDK what that means

maybe the joint like sticking them together

as you have mentioned

[torpid mesa]

I think it's just A_2 horizontally appended to A_1, but I asked someone else in the course to confirm

Ya

[lethal dune]

well you can use contradiction for that

say $\norm{A_1} \geq 1$ then you get a vector $x \in \mbb{C}^{m_1}$ s.t. $\norm{x} = 1$ and $\norm{A_1x} \geq 1$. Now you append / pad 0's to x and make it compatible with $[A_1 A_2]$ and you'll get $\norm{[A_1 A_2]} \geq 1$ contradiction

[cloud walrus]

Ryuzaki

[coarse storm]

Might also help to write out [A1 A2]' [A1 A2] as a block matrix, and then consider its norm.

[torpid mesa]

Thanks! Trying the suggestion to write it out as a block matrix, also going through the proof by contradiction that Ryuzaki posted. I may post follow-up questions once I've been able to think about it a bit.

[untold cloud]

Hi, guys, i am confused about how to prove that if p is a prime, b\in Z, and p cannot divide b, then b+p^{n}Z is invertible in Z/pZ for all n>=1.

is there any hint, please?

[plucky flicker]

Hi. So I have this problem here. Implication b => a is trivial, and from a I proved that in R a^1 \neq -1, but how can I get that R is a division ring. Using the Jacobson Chevalley density theorem we know that R is dense in End(S_D), wehere S is a simple, faithful left R-module and D is a division ring, but what's next?

[plucky flicker]

Can we prove that S_D is actually a 1 dimensional vector space?

cuz then S_D is isomorphic to D_D and then R is dense in D, because End(D_D) is isomorphic to D

[plucky flicker]

Solved

[inner acorn]

@rustic crown is a stimpky vee

[rustic crown]

[inner acorn]

[plucky flicker]

I want to be Hilbort in this channel, should I pay for that?

[unreal portal]

wikipedia says that the krull dimension of a ring can be infinite even if the ring is noetherian, how is that possible?

oh nvm it says how

[chilly ocean]

Hi, I'm supposed to prove that there is a right inverse, but all I have is a left identity

Is that even legit?

And there is no right identity

[winter thorn]

all you have is ea = a?

[paper flint]

I don't think you can get inverse just through an identity

[chilly ocean]

Yeah it is bad

I mean

[paper flint]

What kind of structure are you working with?

[chilly ocean]

I'm not interested in getting an inverse

But

If I was

a * a^{-1} = e

Where e is identity

[paper flint]

I don't think just identity suffices, not every semigroup is a group

[chilly ocean]

So I'm not sure how legit it would be to assume that left identity is actually the identity

[paper flint]

Or monoid, I don't know what associative binary operation+identity is

[paper flint]

Hmm, in certain structures left identity automatically becomes the right identity, I'm not sure about the more general case. Again, are you assuming an associative binary operation here?

[chilly ocean]

Yeah I am

[paper flint]

Let's see then

[barren sierra]

I want to find G/H where G = S_4 and H = { I, (1234), (13)(24), (1432)} = < (1234) >. I know there are a total of 24 / 4 = 6 equivalence classes but is there an efficient way to enumerate through these??

[paper flint]

A software package

Jk

[chilly ocean]

But by the definition, a * a^{-1} should be e, for a^{-1} to be a right inverse right

And here, I don't even know if e is the identity

[paper flint]

True

[barren sierra]

I wish

[paper flint]

You do need a both sided inverse

[chilly ocean]

In order to claim that something is a right inverse right?

[paper flint]

I think so, yes. My brain could be buffering though.

In any case you should be able to prove/find a counterexample by gliding through some examples of monoids, semigroups, etc.

[chilly ocean]

But it's like, I'm asked to compute a right inverse without really confirming that I have an identity in the first place :/

I just have a candidate :/

[paper flint]

ae=c, a(ea)=ca, aa=ca Need right cancellation to show a left inverse is also a right inverse

[robust pollen]

but but
G = SymmetricGroup(4)
H = G.subgroup([(1,2,3,4)])
G.cosets(H)

in sage I should add

[barren sierra]

I might do this

cause fuck doing it by hand

just gotta get Sage to work on my laptop

[chilly ocean]

I'm not asked to show that

Just looking at the definition

[barren sierra]

oh wait no I can use a docker container that I have access to

[chilly ocean]

You can't really get a right or left inverse without an identity

[paper flint]

Yeah, then the problem seems ill-posed as such

[robust pollen]

@barren sierra there you go

[barren sierra]

neat

yea idk why sage won't work on my computer

probably fucked over some Path variable or something

I'll figure it out later

[regal carbon]

Hello

[rustic crown]

Hello

https://cdn.discordapp.com/emojis/800205194892804126.gif?size=160

[regal carbon]

Hello I'm preparing prerequisite for abstract algebra

Like referring to proof book. But it's huge I don't know what's important and what's not!

[rustic crown]

I would pretty much say that anyone could start reading group theory... as it's "basic enough"

[regal carbon]

From chapter 4 what can I skip. I mean which topic should I skip

Let me click another photo

[rustic crown]

,rotate

[paper flint]

To be frank, a lot of this can be learnt for the first time within the context of abstract algebra itself

[rustic crown]

yea

[regal carbon]

It's becoming tedious for me for preparing

[rustic crown]

i don't know anyone who took time to learn it separately

[paper flint]

(That's how I started with more abstract math about a year ago)

[robust pollen]

yeah, just dive into it, and make sure you really undrstand the proofs

[regal carbon]

All of these. And since I don't know how this concepts are related to abstract algebra. I am not knowing what to learn and what not to learn

[paper flint]

But you should know some basic stuff about sets, relations, functions (some elementary terminology around them and basic results)

There's a pinned "Introduction to Proofs" in #book-recommendations

I think it should suffice

[regal carbon]

How many pages the book might be?

[paper flint]

Oh, the revised edition with typos fixed is pinned in #proofs-and-logic

[regal carbon]

Sir

[chilly ocean]

Imagine going through this whole book 🤦‍♂️

[paper flint]

One of the users here wrote it, it's just 30-35 pages I think

[regal carbon]

So are those just sufficient for this

I have downloaded the book sir.

[robust pollen]

it's not the size of the boat, it's the motion of the ocean 👌

[regal carbon]

@robust pollen that's the reason I'm doing proofs

Sir.

[chilly ocean]

let f:A to B be a a ring homomorphism

if p is a prime ideal in B

what do we mean by the ideal f^{-1}(p)

do we have to take the ideal generated by the elements f^{-1}(x) where x is in p

is the set of such elements f^{-1}(x) where x is in p automatically an ideal

[robust pollen]

f^-1(p) = { x in A | f(x) in p }

[chilly ocean]

yes but

for example its not obvious to me that if r is in A, x in p, then r f^-1(x)=f^-1(y)

for some y in p

[robust pollen]

no, you have to show: x, y in A with xy in f^-1(p) => x or y in f^-1(p)

use if f(xy) = f(x) f(y) is in p, what does that imply?

[chilly ocean]

if I is an ideal in R, we need that ir is in I for any i in I and r in R

[chilly ocean]

this does not give us that the ideal is closed under multiplication by R

[robust pollen]

you were asking about preimages of prime ideals

preimages of ideals are always ideals

[chilly ocean]

my question is

is the set of the preimage an ideal

or do we need to take the ideal generated by the set

[robust pollen]

if x, y in A and y in f^-1(I), then f(xy) = f(x) f(y) with f(y) in I, and since I is an ideal, f(xy) in I, so that xy is in f^-1(I)

[chilly ocean]

thank you

[robust pollen]

(now understand that preimages of primes ideals are prime)

[simple mulch]

hey guys

I have $V = R^3$ with basis $(e_1,e_2,e_3)$. Let $W \subset V$ be a subspace generated by vectors $v = {e_1+e_2, e_2-e_3}$. What would an element of W be?

[cloud walrus]

mns

[robust pollen]

A linear combination of the vectors in your set v

[viscid pewter]

0

[simple mulch]

uh I see that but

its the fact the way v = {...} that I don't understand 😐

[scarlet estuary]

vectors of the form $ae_1 + (a + b)e_2 - be_3$ for $a, b \in \mathbb{R}$

[cloud walrus]

Namington

[scarlet estuary]

v is just a set

[simple mulch]

oh

[scarlet estuary]

we take linear combinations from that set

thats all

[simple mulch]

I see but how did you came up with the form of the vectors?

[robust pollen]

a (e_1 + e_2) + b (e_2 - e_3)

expand it

[scarlet estuary]

do you know what a linear combination is?

[simple mulch]

Yes?

[scarlet estuary]

by the way, assuming $e_1, e_2, e_3$ are standard basis vectors, this means any element of $W$ has the form [\begin{pmatrix}a\a+b\-b\end{pmatrix}]for $a, b \in \mathbb{R}$

[cloud walrus]

Namington

[scarlet estuary]

then take a linear combination of $e_1 + e_2$ and $e_2 - e_3$ and just rewrite it a bit

[cloud walrus]

Namington

[robust pollen]

@simple mulch write it like this and massage the expression (i.e. distribute and refactor) to arrive at a e_1 + (a+b) e_2 - b e_3

[scarlet estuary]

i just wrote it like that to make it clear what the coefficient of each basis vector is, btw

there isnt, like, a canonical form for this stuff

[simple mulch]

yeah I got that part

[scarlet estuary]

so whats your confusion?

[simple mulch]

the set v

[robust pollen]

it's just notation

[scarlet estuary]

is using the letter v for a set confusing to you? are you expecting it to be a vector since its denoted with a v?

i'll admit that its a bit weird to denote a set with a lower case letter

but v is definitely a set

[simple mulch]

so v = {e_1 + e_2, e_2 - e_3} for all e_1,e_2,e_3 in V ?

[robust pollen]

No, e_1, e_2, e_3 are a basis for V = R^3

they are fixed

[simple mulch]

ok, so every vector in V can be written as a linear combination of e_1,e_2,e_3

[robust pollen]

at hte begining of your statement

[simple mulch]

its exactly the elements of v that are confusing me

I mean, what exactly does the definition of v says?

[robust pollen]

v is just a set of vectors in V

[scarlet estuary]

v is a set that contains two vectors

one of those vectors is e₁ + e₂, the other is e₂ - e₃

W is given by taking linear combinations of the elements of v

i.e. sums of the form a(e₁ + e₂) + b(e₂ - e₃)

[simple mulch]

so for any a,b in R, taking a(e_1+e_2)+b(e_2-e_3) gives a vector in W

[cloud walrus]

expectTheUnexpected

[scarlet estuary]

yes, thats what it means for a set to generate a subspace

[simple mulch]

ok, got the ah! moment, thank you guys

[long obsidian]

Is it possible to have a rng homomorphism between rings where the identities don't match?

I know if it's a ring homomorphism then this should be the case but if we have a rng homomorphism with less structure I'm not sure

[viscid pewter]

imma guess no, unsubstantiatedly

[coarse stag]

What if Z/q —> G doesn’t exist?

[robust pollen]

what about the inclusion R -> R x R? It's not a ring morphism, but should be one of rngs, right?

[sturdy marsh]

it always exists, G has an element of order q

[crude sail]

is there a less cumbersome way of showing that D8 is a subgroup of S4 than showing that every element composed with every other is still in the set?

jus to be clear, D8 contains the 4-cycle (1234) and its inverse (1432), and all possible double transpositions (elements like (12)(34))

[next obsidian]

D8 acts on 4 things faithfully

This gives you a map into S4 which is injective

[crude sail]

is that map a homomorphism?

sorry, i don't get why a map existing between them shows that D8 is a subgroup

i can tell it's a subset but i don't know how to show that it's got closure

[viscid pewter]

it's a group

[chilly ocean]

i have question

with rings with units

units are defined to be invertible elements right

so a*b such that ab=1

group of units is group under multiplication and there is additive group of ring too

is there any meaning to tensoring the two?

uh ig i should be more precise

can you take group of units as R module

and the additive group as a R module over itself

im only thinking about this because I know construction for field of fractions uses direct producting the ring and the multiplicative group, so im wondering if tensoring it had any meanings

[next obsidian]

Yes this is what a group action does. If you don’t know what a group action and the first isomorphism theorem is then I don’t think there’s much of a way to go about it other than just bashing shit manually

[next obsidian]

Not an R module

It’s not closed under multiplication by arbitrary elements of R

As an extreme example, 0 kills everything, and 0 is very much not a unit (unless R is 0)

You could tensor over Z but idk if this has any meaning

[chilly ocean]

group of units is closed under multiplication by units though

thats what i meant

taking group of units being M

and looking at it as R module

so R x M -> R by rm

but that just acts like normal multiplication

oh lol

i see what ur saying

its R x M to M

and 0 not in it

[next obsidian]

Yeah

You can’t define the “action” of R so to speak (the scalar multiplication) on all of M and have it land in M

So like I noted since everything is an abelian group (Z-module) you can tensor over Z

But I have no idea what this does, my gut tells me it isn’t really useful

[chilly ocean]

uh idk lol

i just learned about Tor today

so im just curious about a lot of the properties, and this forces me to understand tensoring R-modules in a way

for R commutative ring

but we phrased it in terms of groups

so like

[next obsidian]

Ahhhhhh

Tor is very cool

I have a suggestion

[chilly ocean]

Tor(G,H) is H_1(F. tensor H) where F. is free resolution of G

[next obsidian]

When I first learned Tor

So you know about the LES Tor gets you?

[chilly ocean]

we also went through ext two weeks earlier

not off top of my head

[next obsidian]

Like how you can extend the exact sequence

Oh it’s just like

[chilly ocean]

wait

loet me try

[next obsidian]

Normally if you have a SES

[chilly ocean]

sont do it

sotp

[next obsidian]

Okay

[chilly ocean]

i got it

ok lol

so

H1(X)tensor G to H1(X;G)-> Tor I forgot lol

something similar to this though

[next obsidian]

If you have a SES

[chilly ocean]

then tensors preserve it

[next obsidian]

0 -> M -> N -> L -> 0

[chilly ocean]

yeah

[next obsidian]

But!

You don’t get a 0 on the left

[chilly ocean]

besides left hand side

yeah

[next obsidian]

Right

[chilly ocean]

its ker something

[next obsidian]

So Tor let’s you extend this

So you get something like

Tor^1(M,K) -> Tor^1(N,K) -> Tor^1(L,K) -> M (x) K -> N (x) K

And like you can keep going even back further

To Tor^2

Tor^3

…

[chilly ocean]

wait

what

[next obsidian]

Yeah

That’s the point of Tor

[chilly ocean]

we didnt have tor with superscripts

we just wrote one

no super script

[next obsidian]

So

Tor was the first homology

Of like

Free resolution (x) K

right?

[chilly ocean]

yuh

[next obsidian]

Tor^2

(Or maybe Tor_2

I honestly forget if super or sub

Is the second homology

[chilly ocean]

probs sub

[next obsidian]

Of Free resolution (x) K

Well

[chilly ocean]

oh

[next obsidian]

It flips

[chilly ocean]

it doesnt matter for abelian groups ig

[next obsidian]

For Ext it’s one

For Tor it’s the other

Nonono

[chilly ocean]

wuh

[next obsidian]

It’s about if the thing is a chain complex

[chilly ocean]

isnt free resolution inly 2

[next obsidian]

Or cochain complex

No

Free resolution might be infinite

Ohhhhhh

Lmfao

[chilly ocean]

not for abelian groups

[next obsidian]

Did you only do it for abelian groups?

Yeah

[chilly ocean]

in general ig

[next obsidian]

Hahaha

[chilly ocean]

ye

lol

[next obsidian]

So yes for Abelian groups

Since Z is a PID

It only goes out to 1

[chilly ocean]

wait

huhh

[next obsidian]

Yeah

It’s true for any PID

That Tor^2 and onwards is 0

Like your free resolution ends at 2

[chilly ocean]

lol idk how many ways im being lied to

[next obsidian]

You aren’t being lied to

[chilly ocean]

ok

wait

[next obsidian]

You just aren’t doing the most general form

It’s different than being lied to

It’s just not opening the door all the way

;)

[chilly ocean]

ive been told its F1->F0->H where the first map is relations and second maps generators

[next obsidian]

Right

[chilly ocean]

but i havent worked out any examples

i just feel its true

[next obsidian]

So this can end here

Because Z is a PID

But in general

You might have relations between relations

And then relations between relations between relations

…

[chilly ocean]

wait

[next obsidian]

So on and so on

[chilly ocean]

why does it correspond to relations exsctly

[next obsidian]

Cuz the relations are the kernel of F0 -> H

So the idea is

[chilly ocean]

like an example given Z_3 x Z_2

[next obsidian]

Take generators of H

Call these h1,…,hn

So let’s assume we have finitely many for ease of writing stuff down

Then the map F0-> H

Is defined by sending ei to hi

Then suppose you have some expression like

Sum aiei in the kernel

This says that Sum ai hi = 0 in H

So that expression Sum ai ei in the kernel is expression some relation among the hi

[next obsidian]

So here

[chilly ocean]

i sort of see it

[next obsidian]

h1 = (1,0)

h2 = (0,1)

[chilly ocean]

ok

[next obsidian]

These generate it

Then there’s basically two relations here

3h1 = 0

And 2h2 = 0

[chilly ocean]

so F0 is Z2?

[next obsidian]

But none between h1 and h2, they’re kinda independent in that sense

Well

Z^2

[chilly ocean]

oh ok

lol

[next obsidian]

Then we have only two relations

[chilly ocean]

one Z for each generator

[next obsidian]

Yuh

[chilly ocean]

lol bruh i forgot they are free groups

[next obsidian]

So now F1 is also Z^2

So the kernel

Is generated by

3e1

And 2e2

Cuz that generates the kernel

So now F1 -> F0

[chilly ocean]

and the map between F1 and F2 is for the relation ship between generators?

[next obsidian]

There is no F2

Yeah?

Oh fugg

[chilly ocean]

lol i meant F1 and F0

[next obsidian]

Sorry yeah I mistyped

So yeah

F1 -> F0

By sending d1 to 3e1

And d2 to 2e2

So this surjective onto the kernel

[chilly ocean]

oh ok

[next obsidian]

Cuz kernel generated by those

[chilly ocean]

so uh

[next obsidian]

And now 3e1 and 2e2 have no relations among them

There’s no way to get

a13e1 + a22e2 = 0

Unless a1 = a2 = 0

[chilly ocean]

(1,1) |-> (2,3) |-> ??

[next obsidian]

Wel

You don’t wanna do it like that

(1,0) -> (3,0)

(0,1) -> (0,2)

[chilly ocean]

oh yeah true

[next obsidian]

But yes

This is the map

And this is injective

Thus your resolution just ends there

[chilly ocean]

there are a couple of things I missed out learning on

[next obsidian]

0 -> F1 -> F0 -> Z3 x Z2 -> 0

We are done

🥳

[chilly ocean]

so i dont remember exactly why but it feels intuitive, elements of Hom(Z,G) for G Abelian are unique determined by where 1 gets sent.
So in guessing similarly f in Hom(Z^n,G) are uniquely determined where (1,…),(0,1,…),(0,0,1,…),… get sent?

[next obsidian]

Yes!

This is cuz Z^n is the free group

Of rank n

The reason is cuz

Linearity

[chilly ocean]

why isnt this true if G not abelian though?

[next obsidian]

Because

The map might not be a group homomorphism

So we define the map by

Say f(1) = k

Then by linearity

You need f(n) = nk

[chilly ocean]

o i see

[next obsidian]

But like

The expression nk

Doesn’t work nice when G isn’t abelian

[chilly ocean]

yeah

because if f(1)=ab

[next obsidian]

Well it’s like

[chilly ocean]

and n(ab)=abab… not equal to a^nb^n

[next obsidian]

nk = k^n if you wrote it multilicstivekt yeah?

Right!

[chilly ocean]

yeah