groups-rings-fields

[ivory dust]

But you seemed to manage

Like when did you learn linear?

[bleak abyss]

I had an REU summer after first year which did some stuff in LA, and we had a linear algebra book, Hoffman-Kunze, in my analysis class

[ivory dust]

Oh cool

[bleak abyss]

Each week were given a bunch of problems and told to read the relevant material in the chapter

[ivory dust]

What are your thoughts on REUs

[solar vessel]

what does REU mean

[ivory dust]

Because I'm thinking of applying to one

[bleak abyss]

This was first quarter second year. And then second quarter patched up certain things which weren't covered first quarter (though in the REU we did them differently)

So depends on the nature of the REU

[ivory dust]

Research Experience for Undergrads

[bleak abyss]

UChicago REU is good

[ivory dust]

Though I'd wager it's more of a "research" experience

[bleak abyss]

Because I don't think undergrads really should be worrying too much about research

The UChicago REU is like

Gather a bunch of kids

Pair them with grad students

[ivory dust]

What I care about is problem solving

And learning how to learn math

[bleak abyss]

Have them learn a math topic and write an expository paper about it

[ivory dust]

Yeah that sounds perfect for me

[bleak abyss]

So rather than like, go out of your way to solve some obscure graph theory problem that's technically "open" and get a publication

[ivory dust]

That's actually kinda what I'm doing over the school year too

[bleak abyss]

You pick a topic you wanna learn, learn about it, get some practice writing. Pretty much a summer reading course with a (small) stipend

[ivory dust]

Yeah that sounds great

Any tips for applying to one?

[bleak abyss]

I did it twice. Summer after my first year I did the apprentice program

You have a 5 week class which used to be linear algebra and basic graph theory (now someone else teaches and it's more geometry)

[ivory dust]

What do you usually need to apply?

Like a cover letter, recommendations, and a resume?

[bleak abyss]

And like it's a class with graded psets. And then you work with your grad student on something, in my case group actions and Sylow theory

Probably not a cover letter

I applied to like, 2 at one point aside from my home school and that's it

The application of Chicago students to Chicago was much simpler lol

[ivory dust]

Oh and you got in? I heard they were quite competitive

[bleak abyss]

If you're not from Chicago it's hard

If you're from Chicago it's easy

[ivory dust]

Makes sense

[bleak abyss]

I got turned down from one of the two that was outside of Chicago, the other one didn't respond in time so I pulled back my app

[ivory dust]

I like the idea of independently learning about some topic and then writing an expository paper on it

[bleak abyss]

(Chicago's deadline for us to confirm whether we were gonna attend or not was super super early)

[ivory dust]

Because it does resemble research in some way

[bleak abyss]

Yeah for undergrads this is far more productive than trying to do original stuff

[ivory dust]

I got a part time job like that

[bleak abyss]

Because you need a lot of background in math to actually hit research level

[ivory dust]

Working for a professor in the cs department in my school

The idea is that I will help present his research in more accessible fashion

Which I really like because it'll let me get involved in actual research in a (hopefully) productive way

Even though I won't be doing any

[bleak abyss]

Makes sense yeah

What's it on?

[ivory dust]

Simple Temporal Networks

Basically a system of linear constraints on a bunch of variables

Like $x-y \leq a$

[cloud walrus]

0.5772156649:

[ivory dust]

And if you think of the variables as events and the value they're assigned as when that event happens, it allows you to represent a scheduling problem

And then if you add more structure on top of that you can account for things like uncertainty

The end goals is to automate scheduling problems

[inner acorn]

I don't really understand this highlighted conclusion

I understand (I think) everything up until this point

Like.. it is my understanding a faithful group action can have some fixed points, however g*x can't equal x, for EVERY x in X unless g is the identity element
however.. here it feels like it's concluding any single fixed point contradicts the faithful property

Looking into their references, it's following Wagon's text

and here the proposition talks about "without nontrivial fixed points"

which is a stronger assumption than being faithful - am I correct?

<@&286206848099549185>

even then .w. I feel like h⁻¹g could totally be the identity element...

[mild laurel]

they assumed that g \neq h so that can't be the case?

[inner acorn]

pretty sure assuming g & h are distinct, is to show g(M) and h(M) are disjoint

I'mma put brackets to emphasize this isn't necessarily a coset

[inner acorn]

yeah spotted that just now

[gentle pendant]

Yep, that was exactly her point.

[inner acorn]

Yeah, pretty sure the first screen shot has the statement wrong.. which was what was causing confusion

[gentle pendant]

They are both there, she asked about an apparent flaw she found in the other source and she is correct.

[inner acorn]

I did, after being dazzled by the undergrad project's flaw, start losing faith in myself... and asking questions I know the answer to ¬w¬

but thank you kitties!

[gentle pendant]

It's not a bad thing to ask yourself dumb questions.

[inner acorn]

your profile picture is a cat

[gentle pendant]

In fact it is one of the best and most productive things.

[inner acorn]

now.. do I ponder about finding a counter example to this undergrad's statement.. or do I correct the statement and move on

[ivory dust]

Same cardinality seems like it would follow easily from the first part

That's the one I'm stuck on

That is g1kk'g2

Obviously those sets of pairs are all in the fiber

But the issue is proving they are the only ones

If I have ab = g1g2

Well my first thought was that for any k, g = ka for some a

Since k is in G so you can multiply by its inverse

But that doesn't really do anything

Next thought was showing some kind of normality

And using that to define a congruence

Wait is this trivial?

Hold on

Well I can turn ab into $g_{1}'k_1 k_2 g_{2}'$

[cloud walrus]

0.5772156649:

0.5772156649:

[ivory dust]

Well for any k we can write a=g_1' k

For any a

For some g_1'

Not neccesarily g_1

I'll write out a formal proof

Wait that does nothing

Really I think the answer is to define a congruence

Ok I'm still stuck

Right

g1^-1 a?

I mean that works

Right

Is it?

I don't really follow

Actually I guess that works if g2b^-1 and a^-1 g1 are both in K

Because you get ab = g1 (g2b') (a'g1) g2

And if the middle two factors are in K we're done

Yeah but we don't need to explicitly show that because they have to be inverses

Since the whole thing also equals g1g2

Actually wait

g2b' is in G2

And a'g1 is in G1

And they're inverses

Does that mean they're both in K?

It does

Because g2b' is the inverse of something in G1 so it's in G1

But it's also the product of things in G2 so it's in G2

Hold on let me write out this proof

Ok cardinality

Well it's trivial if they're the same group

Is it trivial if they're disjoint?

Wait the whole thing might be trivial hold on

Because G1G2 is the set of all products g1g2

Which we just showed had cardinality equal the cardinality of all ordered pairs (g1, g2) divided by the cardinality of K

So the cardinality thing is just a restatement of what we just proved

Yeah

Basically I'm wondering what the "normal" amount of struggling and confusion is when learning new math. Obviously if you never struggle, you're not moving through the material fast enough (which is why I'm pushing myself to take advanced classes) but also if you are struggling too much, it's a sign that you're moving too fast

What is that url

Wait and why should I check it out?

I already have a good source for learning group theory

That sounds cool

But it doesn't answer my original question

[mild laurel]

I think the sign you look for is whether you actually understand the question or not

I think if you understand the question and all the moving parts that come with the question, as well as understanding how to use all the tools that you'd need to do the question, then the pace you're moving at is fine

Even if the question itself is rather tricky

[ivory dust]

That makes sense

[mild laurel]

This is in my opinion at least, but it feels like when I'm getting ahead of myself, I can't really even begin to think about the question

I don't understand the material well enough to even grasp at what the question is really asking

[ivory dust]

But you can confirm that getting confused as a normal part of learning math. Like the way I've always seen it is that you're always confused, the difference is how advanced the thing that's confusing you is

Does that sound right?

Actually I shouldn't say always confused. Frequently confused would probably be a better statement

[mild laurel]

Maybe it's about how confused you are

[ivory dust]

Also @chilly ocean is the napkin project good for undergrads as well? Because I'd like to move away from the "I'm a highschool student trying to get a taste of things" mentality to "I'm an undergrad trying to properly learn things all the way through"

Now that I'm actually in college

I mean I like getting a taste of things before actually learning them

Also

I feel extremely validated

After that conversation last night

[delicate bloom]

a few general milestones that vary from theorem to theorem is: Can I use the theorem? Can I prove the theorem? Do I have examples of the theorem? breaking it down into these parts can help to not overwhelm yourself

[ivory dust]

Ok so Jackbson keeps mentioning the group of ordered pairs of real numbers (a,b) with a not equal to 0 and multiplication defined as (a,b)(c,d) = (ac, ad+b). Does this group have any significance whatsoever or is it just a convinent example

[mild laurel]

Is it supposed to be ad + bc?

[delicate bloom]

this is matrix multiplication of some particular matrices

(a,b) is the first row and (0,1) is the second row

[mild laurel]

Yeah okay maybe that's better, I was going to note that this is close to the complex numbers too

[delicate bloom]

you could relate it to mobius transformations

specifically if you have f(x) = ax+b and g(x)=cx+d then the group operation f*g gives you f(g(x))

is that clear @ivory dust ? you could attempt to try to find a group that corresponds to composition of functions of the form Ax^2+Bx+C or (Ax+B)/(Cx+D) or other ideas you might have to see how far this idea extends

[ivory dust]

Interesting

I don't know mobius transformations but the matrix thing makes sense

[delicate bloom]

don't worry about it, ignore that single line, everything after that stands purely on it's own

it's just composition of linear functions that I describe

compute f(g(x)) to see if you don't see it

[ivory dust]

oh right

that makes sense

f(g(x)) = a(cx+d) + b = acx + ad + b

which is isomorphic to the ordered pair (ac, ad+b)

[delicate bloom]

so if you do the same thing except with quadratics to make a binary operation, will it form a group?

[ivory dust]

I mean I don't see why not

composition of functions is associative

actually will there be an inverse?

too lazy to check that now

[delicate bloom]

think about plugging in stuff to quadratics

[ivory dust]

in what sense

[delicate bloom]

well the fact that f(x)=x^2 gives f(-2)=f(2)=4 is a bad sign

[ivory dust]

yeah

I was starting to think about that

[delicate bloom]

furthermore, f(f(x)) = x^4 is not even a quadratic

so it's not even closed

[ivory dust]

oh right

so you could extend it to just polynomials

and then to rational functions so there are inverses?

or are there no inverses either way

[delicate bloom]

you tell me

well, on the flip side, I gave the other example of (Ax+B)/(Cx+D)

these are called linear fractional transformations, and this does form a group

which when C=0 and D=1 we recover the original group you started with

have you learned about normal subgroups and the center of a group yet?

[ivory dust]

center of a group commutes with every element of the group

normal subgroup satisfies $x^{-1}kx \in K$

[cloud walrus]

0.5772156649:

[ivory dust]

for all k in K and x in G

but I'll think about this later

[strong stone]

Antidisestablishmentarianism

[raw moth]

so in our lectures last term we mentioned briefly that the group of m"obius transformations
\begin{equation*}
\mathcal{M} \cong PSL_2(\mathbb{C}) \cong \frac{SL_2(\mathbb{C})}{{I_2, -I_2}}
\end{equation*}

[cloud walrus]

Denial:

[raw moth]

presumably we have in general $PSL_n(\mathbb{C}) \cong \frac{SL_n(\mathbb{C})}{{I_n, -I_n}}$

[cloud walrus]

Denial:

[bleak abyss]

Negative identity isn't always in SL_n

[raw moth]

oh right, has to be n even

[bleak abyss]

Yeah

It might help to consider PGL_n

That's GL_n/scalar multiples of the identity

[raw moth]

hmm ok

so in general it's just quotient out whatever multiples of the identity are in the group?

I think we did some example of PGL_2/PSL_2 of Z_5

oh wait nah we just did PSL_2(Z_5)

I guess one question I had was like

is say, PSL_4(C) interesting in the same way?

[hot lake]

..... aren't you supposed t oquotient by all multiples of In ?

oh wait, determinant 1

nevermind

[bleak abyss]

This is SL yeah

[raw moth]

like some sort of transformational thing, but it seems like it would be super unwieldy to actually get anything out of it tbh

idk

[bleak abyss]

But yeah so PGL and PSL are related to projective geometry

[raw moth]

I guess I should look into that sort of stuff then

I've been meaning to for a while for other reasons but never got around to it

thanks! :)

[solemn rain]

G and H are groups

|G| = n , |H| = k

show that |G intersects K| = lcm(n,k)

please help lmao

oh yea G and H are subgroups of a cyclic group

sorry ^ thats importnat dk why

[magic owl]

Oh ok

[solemn rain]

yea XD

[magic owl]

I was gonna say lol

[solemn rain]

yea okay

any hints?

ik by fundameental theorem

g intersects k order must dividie both

but why the lowest multiple?

divide*

[magic owl]

Think about generator

S

And their orders

[solemn rain]

generators of G

?

which generators

[magic owl]

G is cyclic

[solemn rain]

yea

[magic owl]

So G=Z/nZ

For some n

Do this explicitly in Z/nZ

[solemn rain]

well ik G is isomorphic to Z/nZ but okay

[magic owl]

My point is that the lcm thing, the order of the group, and the intersection

Should best be thought of in this concrete context

[solemn rain]

yea

yea right

[magic owl]

Play around w some examples

[solemn rain]

okay umm

i am going to think out loud here esorry

so let me rephrase this shit

G is a cyclic group

H_1 , H_2 are subgroups

|H_1| = n , |H_2| = k

so maybe

H_1 = Z_4 , H_2 = Z_3 ?

is that a good example?

H_1 = {1,2,3} , H_2 = {1,2}

[fickle brook]

[solemn rain]

xd ? XXD

what

did i do something wrong

[fickle brook]

can you restate your problem again

[solemn rain]

okay

let G be a cyclic group

H_1 and H_2 be subgroups of order n and k respect

show that |H_1 intersects H_2| = lcm(n,k)

[fickle brook]

uh

that

there's a very easy counterexample to that

[solemn rain]

what really

if G isnt cyclic?

[fickle brook]

no

G = Z/16, H_1 = <8> (order 2), H_2 = <4> (order 4). H_1 intersect H_2 = H_1, and lcm(2,4) is not 2

besides lcm(n,k) ≥ max(n,k) always

while |H_1 \cap H_2| ≤ min(|H_1|, |H_2|)

[static crest]

H_3897498edysh9ojvhkcjgu\p0as9oqAZX8V|ZUA|ZXU2|IUM

[hot lake]

not sure how you're expecting a subgroup of H1 to have a bigger size than H1

[static crest]

its a joke bro

[solemn rain]

lmao okay

i got it

ty all

sorry

[chilly ocean]

How do you use coset enumeration to find the subgroups of a finite group with a presentation?

[chilly ocean]

I guess take H = {e} derp

[keen magnet]

Hello there, I am searching for someone who wants to study a book with me (do a lot of exercises together etc.). If we are several people on the same book it is better but even if we are only two people I think it can be cool.
I was thinking about a book about category and/or a book about algebraic geometry.
If anyone is interested you can send me a message, or just tell me here 🙂

[solemn rain]

xd

[ivory dust]

What does it mean for a group to be normal? Like I know the definition but how should I think of it

Based on the definition and a little that I was reading online it seems to resemble change of basis from linear algebra

Like if you do the transformation in a new basis it's still the same kind of transformation

[potent birch]

for cosets of a subgroup to form a quotient group, it has to be a normal subgroup, so you can think of it like a quotient space

[ivory dust]

Yeah I know that

[potent birch]

besides something that you can evenly split up, i don't know how else you'd think of it

but i don't know much so maybe there's some better way

[delicate bloom]

the way you're describing it sounds more like an inner automorphism than a normal subgroup @ivory dust

[potent birch]

sorry for being unhelpful, i realized i don't really have much intuition for it either, though there was this visual group theory book that had a good diagram for it

[delicate bloom]

what you said earlier was fine, normal subgroups are what allow you to define multiplication between equivalence classes of cosets

[topaz solar]

a kernel of a homomorphism is a normal subgroup

[mild laurel]

normal subgroups are kernels of some homomorphism

[topaz solar]

y e s

[chilly ocean]

Normality is a weaker version of abelian

[mild laurel]

uh what

[willow garden]

I think they may be referring to the difference between a group's centralizer and normalizer?

[jade sky]

All subgroups of an Abelian group are normal, so we can always quotient an Abelian group by any subgroup. Similarly we can quotient a vector space by any of its subspaces

Question for Zopherus only: Is the converse of that statement true? If every subgroup of a group is normal, must the group be Abelian?

[fringe nexus]

oops

[willow garden]

oooh oooh I know the answer to this!

I think

(yep, I was right, @ me if I'm allowed to answer)

[mild laurel]

I know the answer to this is no

But I'm not sure I've ever heard anyone call the condition that all subgroups of a group being normal as normality

[chilly ocean]

I need a good algorithm to find all subgroups of a finite group

[chilly ocean]

Is there a relationship between orbits and subgroups that makes this possible

[mild laurel]

how is the group given to you

[chilly ocean]

Finite list of generators in a matrix space

[chilly ocean]

if G is given by generators g1,..,gn

then you can get the different subgroups by looking at the groups generated by g_i1,...,g_ir

That won't get me all of the subgroups.

[fading wagon]

Hmm, it says that it's a hard problem here:
https://mail.gap-system.org/pipermail/forum/2008/002252.html

but if the order is about 10^5 or less it should be fine

@chilly ocean

[jade sky]

@mild laurel I didn't call it normality. I don't think even @chilly ocean meant that (what I said later) by "normality", but I agree that would be a sensible reading of that statement, given that normality [actually a property of a subgroup qua subgroup] is being compared with Abelianness [a property of the whole group].

[chilly ocean]

Right now I'm looking for a good pen and paper method for iterating subgroups....

[jade sky]

@mild laurel So if at all that property needs a name, I guess "Dedekindness" can be used

@willow garden What's the answer?

[fading wagon]

@chilly ocean Well, you can use Lagrange to say what sort of sizes of subgroups you can get, and then you can try to figure out what sort of groups are of that size, you can do that for pretty small groups of order <= 200 I guess

[chilly ocean]

These are all trial and error

[solemn rain]

if your group is cyclic its easy

ig

u use fundamental theorem

just find a generator

[fading wagon]

choose an element and branch on that, if it is part of the subgroup?

[chilly ocean]

"branch on that"?

What does that mean?

[fading wagon]

sorry, wasn't clear, consider 2 cases, that element being part of the subgroup and that element not being part of the subgroup

then you can repeat this recurively

[solemn rain]

yea lmao

that would be cool

pick an element

assume its in a subgroup

then it must be closed ( multiply by itself) and must have inverse and so on

and identity

cool lmao

[jade sky]

You can begin [this procedure ↑] with the atoms of the subgroup lattice — the minimal subgroups — by computing the cyclic subgroups generated by elements of prime orders

[And then recursively add to these, the elements of the group that are left out]

[chilly ocean]

Thanks Vyn

I know the number of generators grows logarithmically with group size

[jade sky]

@chilly ocean Here's another idea, which may work better for some groups (and is useless for some others [e.g., simple groups]):

1. Find a non-trivial normal subgroup N of G (the smaller the better). Recursively find all subgroups of G/N, and the correspondence theorem gives you all subgroups of G that contain N.
2. Now all that's left are subgroups of G that do not contain N. Somehow find these (lol) (well, you can hunt for other normal subgroups and do Step 1, but even if you do it with all minimal normal subgroups, there may still subgroups that contain none of them)

[chilly ocean]

What are groups of lie type?

[jade sky]

I don't know

[fading wagon]

ping me if someone finds a good introduction for that ^

[clear obsidian]

lie groups?

just groups that have smooth manifold structure also (iirc)

so e.g. Euclidean space as an additive group (vector addition) is clearly a smooth manifold (R^n is for all n in N)

(guessing theres a smooth manifold structure for S^2 then you could define a group addition)

[chilly ocean]

You’ve probably heard of GL_2(R); that’s a Lie group

[woven delta]

S^2 is not a lie group

S^3 is

[bleak abyss]

Groups of Lie type aren't Lie groups I think

It's got something to do with CFSG

[uncut girder]

Why is S2 not a lie group

[bleak abyss]

Easiest way to see it is hairy ball theorem

[uncut girder]

Oh

[bleak abyss]

You can prove that there's an isomorphism of smooth vector bundles TG \cong G\times T_1G

[uncut girder]

S1 is tho right?

[bleak abyss]

But then obv you can find non-vanishing vector fields. S^1 is a Lie group yeah, unit complex numbers

Also S^3, unit quaternions

[solemn rain]

whats a lie algebra

[bleak abyss]

A Lie algebra is a vector space with a bilinear map [-,-]:V\times V->V that basically behaves like a commutator

[willow garden]

@jade sky the answer is no. A counter example is Q_8 which has only normal subgroups, but is not itself abelian. In fact it's the smallest such group, called Hamiltonian groups.

[jade sky]

@willow garden Correct. It's the smallest Hamiltonian group not only in terms of size, but also in the terms of inclusion.

Every Hamiltonian group must contain Q₈ as a subgroup (up to isomorphism)

Better yet: as a direct summand

[willow garden]

🥳

[jade sky]

@clear obsidian No, "groups of Lie type" are not the same as "Lie groups"

[willow garden]

iirc there's not really a single agreed-upon definition for groups of Lie type

I'm not there yet, I just stumbled on that tidbit of information when I discovered on wikipedia something called the Tits group

and concluded that the existence of the Tits group is the sufficient condition for not teaching abstract algebra to high schoolers

bonus points: it's actually pronounced "teats"

[jade sky]

Which isn't a lot better

But definitely don't teach high schoolers about the intersection numbers of sections of elliptic surfaces

[willow garden]

why, what are they called?

[jade sky]

https://wstein.org/misc/sagedays18_papers/cox-zucker-intersection-numbers.pdf

[mild laurel]

Cox actually told me that him and Zucker actually planned to write a paper together for the joke

[jade sky]

That's right

[mild laurel]

which is hilarious that professional mathematicians can still be pretty immature

[jade sky]

Sometimes I feel you can't be a mathematician unless you're still a high schooler at heart

[willow garden]

...ah.

yeah, that pretty much rules out complex analysis too

[jade sky]

When I was in high school I used to try to prove all the big conjectures (Goldbach, Collatz, Riemann…), and believe that it was quite likely I would be able to — and that's the case with many others who later end up working in math. But after learning more math, these attempts got replaced by attempts at answering open questions that were a bit less famous. And now it's reduced to trying to solve only open questions in latest research papers (in a much narrower area)

[willow garden]

that second paper there is missing appendix A

[jade sky]

@willow garden What did you mean in complex analysis?

[willow garden]

no, the mailing list one

edit: oh

cox-zucker talks about elliptic surfaces over the field of complex numbers

[jade sky]

Oh, okay

https://projecteuclid.org/download/pdf_1/euclid.nmj/1118787426

[vale coral]

Hello. I am having troubles with the following general problem. I have a system of congruences and usually if the left hand side of the equations is a lone x (e.g. x = 1 mod 2) one can solve it using the chinese remainder theorem.
However, say I have a system that is as follows:
x^3 + x + 1 ≡ 4 (mod 5)
2x^2 ≡ 7 (mod 11)
x ≡ 1 (mod 3)
From what I understand it should be possible to 'reduce' these equation so that you
end up with only x'es on the left hand side. What I don't understand is how to
perform the reduction so you can apply the chinese remainder theorem on the system. If any kind soul could explain it or point me to relevant resources it would be much appreciated :).

[chilly ocean]

x^3 + x = 3 mod 5
2x^2 = 7 mod 11
x = 1 mod 3

Addition and subtraction work as typical.

And we know these have solutions by CRT

Since these are all coprime moduli.

So with 2x^2=7 mod 11, we have x = 11k + 3 with n any integer

Now, take x^3 + x = 3 mod 5.
Substitute and we get k=5n+1

So x = 11(5n+1) + 3
x=55n+14

Sub into x=1 mod 3

55n+14 = 1 mod 3
55n+13=0 mod 3
n+1=0 mod 3
n=-1 mod 3

Or n=2 mod 3

3c+2 = n

x=55(3c+2)+14
x=165c+124

Thats what I got hm.

,w x^3 + x + 1 = 4 mod 5, 2x^2 = 7 mod 11, x = 1 mod 3

[cloud walrus]

Display results online and refine query

Requested by rudy#0446

[chilly ocean]

Yes, well, that’s misfortunate

[vale coral]

Cheers

[mild laurel]

@vale coral the answer is that since your primes are small here, you can just substitute numbers to find all the possible solutions to your equations

For the mod 5 one, there are only 5 different possibilities to check

If the primes were larger, it turns out that things like the quadratic and cubic formula still work mod p (as long as you avoid the primes 2 and 3)

[fading wagon]

So with 2x^2=7 mod 11, we have x = 11k + 3 with n any integer
no actually +-3

[chilly ocean]

yes :/ @fading wagon

Thats why I said “that’s misfortunate”

I forgot to account for the negative portion

[solemn hollow]

Hi! I need help with the red underlined bit. I've tried considering that I know $H≤C_G(H)$. Also tried some $G$ acting on $H$ by conjugation stuff

[cloud walrus]

EpicGuy4227:

[mild laurel]

I think use the fact that H is normal and look at G/H

If an element commutes with all of H, and since G/H is abelian, it should be true that it must commute with everything in G which is a contradiction

[solemn hollow]

Hm I'm having a hard time showing that this element must commute with everything in G. We are assuming that this element isn't in H right?

@mild laurel

[mild laurel]

Let \varphi : G \to G/H be the natural surjection

and let g be the element, not in H, that we're assuming commutes with everything in H

for any other element f of the group, we have that \varphi(g)\varphi(f) = \varphi(f)\varphi(g) since G/H is abelian

[solemn hollow]

don't we want to show gf = fg?

@mild laurel

[mild laurel]

Hm

this actually might not work

I was thinking that since gfg^{-1}f^{-1} is in the kernel, so is in H

and I was thinking that we could use the fact that g commutes with everything in H to show that g and f commute, but I don't think this works out

[bleak abyss]

So

The point is that if you're not in H, you commute with everything in H but also in the subgroup you generate

So your own centralizer ends up being the whole group because of counting

@mild laurel and @solemn hollow

[mild laurel]

Ah right

[solemn hollow]

ooh ok I got it thanks guys

[urban acorn]

do p-groups in general have a characteristic subgroup of every possible order? i couldn't find anything helpful in google. i have an idea for a proof that this is the case, but there's still a hole in it.

Lemma 1: if $H \mathrel{char} G$, $K \mathrel{char} G/H$, then the preimage of $K$ in the natural projection is characteristic in G. \newline
Proof: Ommited. \newline
Lemma 2: every (non-trivial) $p$-group has a characteristic subgroup of index $p$ \newline
Proof: \newline
let $P$ be of order $p^a$ and proceed by complete induction on $a$. when $a = 1$ the result is (quite literally) trivial. suppose the result holds for all groups of order $p^b$ for every $b < a$, then consider the quotient of $P$ by its center, it's of order $p^b$ for some $b < a$ because the center of a $p$-group is always nontrivial. then the quotient has a characteristic subgroup $K$ of index $p$. Observe that $Z(P) \mathrel{char} P$ and $K \mathrel{char} P/Z(P)$ allowing us to use Lemma 1 to conclude K corresponds to a characteristic subgroup of $P$ of index $p$. Q.E.D. \newline
Theorem: every p-group has a characteristic subgroup of every size dividing its order. \newline
Proof: \newline
let $|P| = p^a$ and let $H_a = P$. We define $H_{i-1}$ recursively with respect to $H_i$ as some characteristic subgroup of $H_i$ of index $p$, the existence of which is given by Lemma 2. That is until we reach $H_0 = 1$. This creates a chain of subgroups $H_0 \mathrel{char} \dots \mathrel{char} H_a$ with each subgroup characteristic in the next and of index $p$ in it. Then let $x$ be any positive integer divisor of $|P| = p^a$, we observe that $x = p^b$ with $0 \leq b \leq a$. We then see that $H_b$ is characteristic in $P$ (because char is transitive) and of order $x$, as desired. Q.E.D.

I still don't have a solid proof for Lemma 1, but it seems true intuitively.

At first I thought that a more general version of it would be a part of the third isomorphism theorem, but then rethinking it, I'm not that sure about it.

I do have an idea for a potential proof for Lemma 1, but I don't think the resulting theorem is likely to be true because I haven't seen it mentioned anymore.

Is there perhaps a mistake in the proof of Lemma 2 or the theorem?

[cloud walrus]

Intel:

[urban acorn]

Nevermind, I figured it out. There's a mistake in the proof of Lemma 2.

It's because the result is not true when a=0, and taking a quotient by the center doesn't guarentee a ≥ 1

[jade sky]

Well, the Klein 4-group is a simple counterexample

[chilly ocean]

Let p be a prime number. Find all units of the ring $$\mathbb Q_p = { \frac{a}{b} : p \nmid b }$$

[cloud walrus]

Godel:

[chilly ocean]

How do I justify that those would be all numbers b/a such that p doesnt divide a

[jade sky]

Why does it render the {? : S

[chilly ocean]

because I put \ before it

it doesnt appear on discord msg

[jade sky]

I see. Thanks

Don't you mean: All the numbers a/b ∈ ℚₚ such that p doesn't divide a?

(So that its reciprocal b/a would also be an element of ℚₚ)

[magic owl]

I mean its kinda just a follow your nose result

Its obvious why those are units

And to show they are the only ones

Just write down what unit means in Q

[chilly ocean]

yeah why is that even a question its fucking obvious

[magic owl]

Huh

You just asked how to do it lol

[chilly ocean]

I mean the answer is clearly what one thinks it is, I guess what you said with showig those are the only ones needs justification

thx

[urban acorn]

i mean i guess you could say that it's a unit precisely when 1/x is also in the ring

which just reverses teh fraction

idk if that makes sense im drunk

[chilly ocean]

why is $\mathbb R \left[i \sqrt 3\right] = \mathbb R \left[i \right]$

[cloud walrus]

Godel:

[chilly ocean]

what function gives isomorphism

when I tried a+bi maps to a +b/sqrt3 i it wasnt multiplicative I think

[bleak abyss]

@chilly ocean not even isomorphic, they're just equal as sets

Like if you consider both to be subsets of C

[wind steeple]

sqrt(3) is in R

[bleak abyss]

Obviously R[i] = C, and (a+bi) = (a + (b/sqrt(3))i\sqrt(3))

And b/sqrt(3) is a real number

[chilly ocean]

OK YEAH THX THATS WHAT IO TOLD MY FRIEND AND HE SAID ITS WRONG

because said that maybe it behaves differently

[wind steeple]

tell your friend he's dumb

[bleak abyss]

Lol

[chilly ocean]

are those two isomorphic?

is gonna be Q[i] and Q[i sqrt3]

and dont know if those are iso

[wind steeple]

nope they aren't

if there were, R would have a square root of -1 and -3, hence a square root of 3, but clearly, sqrt(3) isn't in R

[chilly ocean]

I dont get your idea

Whats R?

[wind steeple]

the field you defined

[chilly ocean]

(i solved it different way, but curious what you mean)

Which one

[wind steeple]

Q(x)/(x²+1)

[chilly ocean]

Oh ok, I see

Thx

Wait actually why would R need to have sqrt - 3

[mild laurel]

If R were isomorphic to S

[cloud walrus]

fields!:

[warped bay]

How to prove this? I don't quite understand the proofs I found online

[mild laurel]

What did you try?

[warped bay]

I read the proof of this in a class I sat-in but it uses corollaries not discussed in my class

I read online that it's easier to prove by contradiction
but it doesn't make sense to consider an n<m such that a^n=e if the order of a is infinite

so I'd rather not prove it using inequalities

[mild laurel]

You can just handle that case separately

If the order of a is infinite, it's not too hard to see that the subgroup it generates must also be infinite

[warped bay]

otherwise, it contradicts a corollary not used in my class lol

and that corollary was never mentioned in my class

so if possible i'd not consider separate cases if i'd have to recall theorems not used in my class

Or is it not possible to not consider separate cases?

[solemn rain]

try contradiction

@warped bay

[maiden ocean]

@warped bay let a be of order r, and then consider an element a^m in <a>

try using the division algorithm for integers and let m = rq + p

and see where this takes u

[ivory dust]

I don't even understand the question

What is |a|

[maiden ocean]

order of a in G

[ivory dust]

The order of the group genrated by a?

[maiden ocean]

the smallest n such that a^n = e

[ivory dust]

But then that's just the definition

[maiden ocean]

(this is equivalent to ur definition)

kinda

[ivory dust]

I mean I guess it isn't quite

[maiden ocean]

basically he just needs to show that the cyclic group <a> where a is of order r

is {1, a, ..., a^(r-1)}

[ivory dust]

I mean that still kinda feels too trivial to prove

[maiden ocean]

its not that trivial

well

its pretty trivial lol

[ivory dust]

Honestly I hate those kind of proofs

[maiden ocean]

u just need to show that every a^m is in the that range

and that they're all distinct

[ivory dust]

Again though that's just definition

[maiden ocean]

so just let a^m, m = rq + p, and then a^m = a^(rq + p) = (a^r)^q * a^p = 1^q * a^p = a^p

[ivory dust]

Like all a^n are in the group

[maiden ocean]

and then since 0 <= p < r

its not definitional

[ivory dust]

How is it not definitional? Every element in <a> is a power of a. That's how generated groups work

[maiden ocean]

its not 100% trivial that every a^n is in the range of 1, a, a^(r-1)

where a is of order r

[ivory dust]

Right that isn't

[maiden ocean]

which is the goal of the proof

[ivory dust]

But it still just follows by divison

[maiden ocean]

i mean yeah its not hard lol

[ivory dust]

Honestly I hate proofs like that

Like I don't want to prove facts that are obvious to me

Honestly it feels like a lot of proofs at the foundational level are like that though

Like the first isomorphim theorem

I saw a joke on reddit about this

That algebra is just
"definition: let a be the set of elements with this nice property
theorem: the elements in a have that property"

[maiden ocean]

how is first iso theorem obvious

the proof is like a solid page

[ivory dust]

Wait what does it say again?

[topaz solar]

You gotta be careful with “this is obvious” stuff though, c.f. Weierstrass function being not nicely differentiable

[ivory dust]

I know

But like still

Wait what is the first iso theorem again?

[maiden ocean]

if u have a homomorphism from G to G' theres a bijection between set of subgroups of G containing the kernel and set of all subgroups of G'

[steep hull]

First iso proof is rather mechanical tho, even if it’s long

[maiden ocean]

to put it simply if H is normal

gH -> phi(g) phi(H)

is an iso of G/H with G'/phi(H)

[ivory dust]

Right

[maiden ocean]

the full statement is lorger slightly

[ivory dust]

I guess not obvious per se

But not super hard either

[maiden ocean]

its an intuitive result

[ivory dust]

Like that might have been a bad example

[maiden ocean]

but the proof is not like

immediate

lol

[ivory dust]

Yeah that's kinda what I mean

Like homomorphims should have that property

[maiden ocean]

ok is sleepy tiem for me

[ivory dust]

And if they somehow didn't then you defined them wrong

Yeah I should go to bed too

[chilly ocean]

Let $S \subset R$ be a multiplicative system, $1 \in S$ and $I \leq R $ ideal such that $I \cap S = \emptyset$. Show there exists a prime ideal $J$ such that $I \subset J, J \cap S = \emptyset$

[cloud walrus]

Godel:

[chilly ocean]

So I showed there exists a maximal J in the set of ideals with that empty S intersection property, but how would I show its prime?

I saw a definition that J is prime iff its set theoretic complement is multiplicatively clsoed, so I think showing S is the complement would work, but don't know how to do it (if that even is true)

[wind steeple]

you can see that if y is in R/J, then there exist s in S sush that s is in y(R/J)

y neq 0

btw your property shows that the intersection of all prime ideals is the set of all nilpotent x)

[chilly ocean]

why is that?

what does y(R/J) mean

[wind steeple]

like aA for a a ring and a in A

{ab, b in A}

[chilly ocean]

like ideal generated by y in R/J right?

[wind steeple]

yeah

[chilly ocean]

ok so if y is some y + J and s in S then why is s in (y+J)

[wind steeple]

bc J is maximal

if not, you could find an ideal K in R s.t. J subset of K and p(K) = (y+J)

[chilly ocean]

J not maximal as an ideal in R

[wind steeple]

and s.t. K\cap S = \emptyset

where p is the canonical projection R -> R/J

[chilly ocean]

by maximal you mean maximal that has empty intersection with S?

[wind steeple]

J is maximal is the set you defined before

[sacred anvil]

isn't it multiset?

[wind steeple]

???

[sacred anvil]

Is Q, defined that way, a set? or multiset?

[wind steeple]

set

[sacred anvil]

but isn't 3/2 and 6/4 the same element?

[mild laurel]

You're right, you have to mod out by an equivalence relation

[sacred anvil]

I just got a little confused 😅

[wind steeple]

???

{1,1} = {1} by definition of sets

there's no problems here

[mild laurel]

I mean, is {6/4, 3/2}= {3/2}?

It's definitely false that {(6,4), (3,2)} = {(3,2)}

Since (6,4) isn't (3,2) but

[chilly ocean]

How do I find all maximal ideals in $\mathbb Z \left[i\right] / \left(7+i\right)$

[cloud walrus]

Godel:

[mild laurel]

Maybe not the right way, but think of the correspondence of ideals in quotients

[chilly ocean]

I thought and it didnt help me

actually wait I have an idea

I have similiar thing in my notes but I don't understand it - (7+i) is (1-i)(3+4i) and I claim that ((1-i) + I ) R/I and ((3+4i) + I ) R/I are maximal but I dont know why

I do get that (1-i) and (3+4i) cointain (7+i) in Z[i]

[mild laurel]

Well, the first ideal is maximal

[chilly ocean]

you mean (1-i)?

[mild laurel]

you might have miswrote the second one, its not even prime

(2+i)^2 = (3 + 4i)

[chilly ocean]

yeah I was just typing that

then I GUESS ((1-i) + I ) R/I and ((2+i) + I ) R/I are the maximal ideals in R/I, but im not sure if I understand the correspondece thing

and since those 2 elements are irreducible then they are the only maximal ideals containing (7-i)

[mild laurel]

you should understand the correspondence thing

its basically the most important theorem in ring theory

[chilly ocean]

I mean I believe it, but like what Im not sure that why if (1-i) is a maximal ideal containing (7-i) then it has to correspond to ((1-i) + I )R/I and not some other element that isn't maximal in R. I'm probably not making any sense but whatever, if you did not understand what I tried to say just ignore it

[mild laurel]

prove it

prove that if J is an ideal containing I, then J is maximal in R iff (J + I)R/I is maximal in R/I

[chilly ocean]

Ok I think I got it, thank you for telling me to do that

(tried to show that if R/J is field then other quotent also is and vice versa)

[mild laurel]

the same thing holds for prime ideals as well

[solemn hollow]

Here's a fun problem: Let $G$ be a group with order Graham's Number. Prove that if $H\triangleleft G$ and $|H|=3,$ then $H≤Z(G).$

[cloud walrus]

EpicGuy4227:

[cloud walrus]

EpicGuy4227:

[solemn hollow]

Just checking if I'm not insane. Looked at some erratas as well, not there.

[mild laurel]

It's possible they just don't consider S_2 as a symmetric group

Lots of people don't really, since its pretty different from the rest of them

But you're right here

[solemn hollow]

Ooh, discrimination :(

I'm trying to find their definition of S_n hm

can't tell much from their section on symmetric groups .-.

thanks Zoph, I'll keep it in mind

[cloud walrus]

fields!:

[fickle brook]

$(F[x])[y]$ and $(F[y])[x]$ are naturally isomorphic to $F[x,y]$ to the point where attempting to make a distinction between the three will in most contexts be too hair-splitting to be worth it, so yes.

[cloud walrus]

Ann:

[coarse storm]

If it is non-commutative, I believe a different notation is used for the extensions.

I just threw away my notes on that last week 😅

So I can't check.

[golden pasture]

well

free algebras usually arejt commutative

until we set it to ge

[warped bay]

ty, got curious

[chilly ocean]

Found this book really useful

[warped bay]

oh? that 2010 version is longer than my 2016 one

by like 100 pages

[ivory dust]

I have zero intuition when multiplying cycles

Like I know how to do it, I just have no idea what the answer will look like until I multiply them

[wind steeple]

that's normal

[ivory dust]

It is?

[wind steeple]

idk, can you imagine (1 4 8 2 3 6 14) ?

[ivory dust]

Basically my question is whether there's an easy well to tell what the partition of the product of permutations will be without computing it

Even one that only works under certain circumstances

[wind steeple]

partition of the product ?

[delicate bloom]

might find this conversation interesting:

https://discordapp.com/channels/268882317391429632/496784958430380033/663952740610736158

[wind steeple]

idk what to say, when you have a product of permutations, you need to compute it, one can't get the result only by looking at the product itself

[jade sky]

@ivory dust I think the usual permutation notation with the first row skipped is called one-line notation, and is commonly written without enclosing parentheses (presumably to distinguish it from cycle notation)

See: https://planetmath.org/onelinenotationforpermutations

[ivory dust]

I don't know how to approach 4

Like I have that it's equivalent to showing there exists a normal K contained in H such that [H:K]|(n-1)!

[wind steeple]

Wut

Take K = {e}

Idk, it works x)

[ivory dust]

I don't think it does

[wind steeple]

Oh the index

[ivory dust]

Since n is the index of H not the order

[wind steeple]

Mb I don't know to read

[ivory dust]

So yeah what do I do from here

[wind steeple]

Ok

Just take the action of G on G/H

[ivory dust]

Yeah that has order n

[wind steeple]

It's the data of a morphism G -> S(G/H)

[ivory dust]

Hm?

[wind steeple]

And now you can only do one thing to exvibit a normal subgroup of G

[ivory dust]

Data or a morphism?

[wind steeple]

That's the definition of an action

of

If you have an action of G on X, then g->(x->gx) is a morphism from G to S(X)

[ivory dust]

Is a morphism just a function?

[wind steeple]

S(X) is the group of bijections of X

[ivory dust]

Or do you mean homomorphism

[wind steeple]

You don't know what's a morphism ?

Yeah it's the same thing

[ivory dust]

Right

Oh

Hold on

[wind steeple]

What kind of general normal subgroup do you know ?

[ivory dust]

I'm not sure what you mean

[wind steeple]

Given a morphism of group A->B, how can you construct a normal subgroup of A ?

[ivory dust]

Oh just the kernel

[wind steeple]

Yep

[ivory dust]

And the kernel is obviously a subgroup of H

And normal

[wind steeple]

Yes

[ivory dust]

And it maps to an element of S_n so its index must divide n!

Right?

[wind steeple]

Nop

[chilly ocean]

Let G,G, H,H, and KK be finitely generated abelian groups. Show that if G×H≅G×K,G×H≅G×K, then H≅K.H≅K. Give a counterexample to show that this cannot be true in general.

I am stuck and need a hint.

[mild laurel]

I feel like you're missing somethings here

But the classification of finite abelian groups should probably do the trick here

[hot lake]

which part are you stuck on ?

[magic owl]

Wait am I losing my mind

Are those supposed to be like

H,H’

G,G’

[chilly ocean]

Youre Just seeing double mate

[ivory dust]

what is the deal with the outer automorphism of S_6

like why does is it the only symmetry group that has one

[solemn hollow]

you could look at a proof

[ivory dust]

I should

[fading wagon]

Consider any automorphism applied to (1 2) and (1 2 3 ... n). Since these 2 generate S_n, it determines the automorphism fully. Any inner automorphism consists of relabelling the elements of the set {1...n}.
(1 2 3 ... n) can only be mapped to one of (n-1)! elements. Upon picking that, consider picking any pair of elements to swap for (1 2).

I'm guessing if they are not adjacent in the cycle structure of (1 2 3 ... n) try to force a contradiction, and then you proven that |Aut(S_n)|<=n!. Since |Inn(S_n)|=n!, we have |Out(S_n)|=1

[potent birch]

what's an example of a group that has elements of both infinite and finite order?

[fading wagon]

Multiplication of nonzero real numbers

@potent birch

any infinite group honestly

because the identity has finite order

[potent birch]

that makes sense but what about something that has more elements than that? like something geometric?

i guess i found some stuff now that i used google, but thanks anyways

[fading wagon]

group of linear transformations?

i.e. just matrices

[potent birch]

yeah that's a good picture, i guess i was just thinking if there's an example of something like a cylinder with infinite length, where the torsion is looping around the cylinder and infinite order going lengthwise

[fading wagon]

yeah you can do that

Symmetry group of that cylinder, preserved on translations along the cylinder and rotations around the cylinder

[solemn rain]

i dont understand why

the condition of normal subgroups would imply that the set of cosets is a group

can some1 verify that for me

sorry basic question

[fading wagon]

What's the condition of normal subgroup as you learnt it?

@solemn rain

gN=Ng?

[solemn rain]

for all h in H and g in G

ghg^-1 is in H

now how i learnt was

[fading wagon]

So the set gHg^-1 is a subset of H

[solemn rain]

okay

[fading wagon]

left multiplying by g^-1 and right multiplying by g, we have

H is subset of g^-1Hg which is subset of H

so the 2 sets are equal

[solemn rain]

g^-1Hg = gHg^-1?

[fading wagon]

H is subset of g^-1Hg

g^-1Hg is subset of H

[solemn rain]

okayy

[fading wagon]

hence g^-1Hg = H

g was arbitrary

So, gHg^-1=H too

[solemn rain]

yea

so

[fading wagon]

Okay now let's show that $g_1Hg_2H=g_1g_2H$

[solemn rain]

the definition we defined on the quotient set

[cloud walrus]

Element118:

[solemn rain]

yes

exactly lmao

thats our operation

[fading wagon]

Well, $g_1Hg_2H=g_1Hg_1^{-1}g_1g_2H$

[cloud walrus]

Element118:

[fading wagon]

$g_1Hg_2H=g_1g_2Hg_2^{-1}g_2H=g_1g_2HH=g_1g_2H$

[cloud walrus]

Element118:

[stone fulcrum]

g'Hg is a subgroup of H, even

[solemn rain]

got it

tysm element

so we showed

the operation is well defined

and i can verify myself that its closed under inverses too

so G/H where H is normal

is a group

correct?

[fading wagon]

yeah

[solemn rain]

tysm

[solemn rain]

another problem sorry

i showed that the canonical projection is a homomorphism

what im stuck with is showing that its kernel is N

f:G---> G/N where N is a normal subgroup to G

f(g) = gN

[topaz solar]

What is gN if g is in N?

@solemn rain

[solemn rain]

thats just N

cuz its closed ig

@topaz solar

[topaz solar]

so that means the kernel contains N

[solemn rain]

wait wait im slow

umm

okay

yea

[topaz solar]

if gN=N=eN

[solemn rain]

the identity is N for htis homomorph

so yea okay

[topaz solar]

So

Assume that gN=N

Can you show a contradiction if you assume g is not in N?

[solemn rain]

okay so assume g is not in N

gN=N

[topaz solar]

gN is the collection of gn, with n in N

Just in case

[solemn rain]

yea yea

if g is not in N then gn cant be in N

right?

so the equality cant hold ig

[topaz solar]

Well an easy example is showing e isn’t in gN

[solemn rain]

yea lmfao

lmao

[topaz solar]

Since if it were, then -g is there

And u h h

[solemn rain]

okay

[topaz solar]

So N < ker < N

[solemn rain]

if gN = N then g must be in N

[topaz solar]

Ye

[solemn rain]

so N is a subset of ker(f)

[topaz solar]

N is in ker because N gets sent to e obviously

ker is in N since any non N element can’t go to the identity

ergo

[solemn rain]

n=k

okay

[topaz solar]

Ker = N

[solemn rain]

got it

yea yea

sorry im abit slow

ty

[topaz solar]

Aight

[solemn rain]

so i am doing this so i can do homomorphism theorem

can you help me prove that?

so its stated like this for me

let phi : G--->H be a surjective homomorphism

then G/ker(phi) is isomorphic to H

with isomorphism phi bar

where phi=phi(bar) o f

f is natural homomorphism

[topaz solar]

So
start from the top

[solemn rain]

what top XD

[topaz solar]

Phi is surjective

[solemn rain]

what

[topaz solar]

So you know it hits everything

[solemn rain]

you mean phi is surjective?

thats new language

for me

okay so assume phi is a surjective homormoprhism

[topaz solar]

And ye too tired

[solemn rain]

so yea every element in H is equal to phi(a) for some a in G

so i tried to just unfold the definition

and to show phi bar is an isomorphism

okay can we call phi bar like gamma

okay

so im given that phi = gamma o f

f is the natural projection

so phi(a) = gamma(f(a))

f(a) = aK where K is ker(f) right

so phi(a) = gamma(aK)

i want to show gamma is isomoprhism

[topaz solar]

showing that phi = phi’ (f) is ez btw since equiv classes js

So

[solemn rain]

okay so

phi(a) = phi'(aK)

i want to show phi' : G/K ---> H is isomoprhism

i should show its well defined first

right?

[topaz solar]

So what you need to do is show that each element a in H has exactly 1 preimage in G/k

[solemn rain]

yea

thats what it means to be well defined ig

right?

[topaz solar]

No

Well defined is independent of representing element

[solemn rain]

oh thats

[topaz solar]

But that parts easy so let’s get to the iso part

[solemn rain]

injectivity

?

[topaz solar]

No

[solemn rain]

what

is this

[topaz solar]

if aK=bK the phi(a)=phi(b) basically

[solemn rain]

i thought this meant well defined

phi' *?

[topaz solar]

no

Anyway moving on

[solemn rain]

okay

so i wanna show this first

so assume aK = bK

phi(a) = phi'(aK)

[topaz solar]

Take k in K

Phi(ak)=phi(a)

Which is easy

For each ak in aK, there is bk’ in bK so bk’=ak

So phi(a)=phi(b)

done

Moving on to the isomorphism

[solemn rain]

okay

[topaz solar]

It’s sufficient to show injectivity

[solemn rain]

we want to show phi'(aK) = phi(a) is an isomoprhism

where phi(a) is a homomorphism

okay

so suppose phi'(aK) = phi'(bK) for a and b in G

wait

i proved something b4

phi' is injective --> ker(phi') = {1}

iff

maybe i can use this lma

[chilly ocean]

Wrong

[topaz solar]

phi(a)=phi(b) so phi(a)phi(-b)=e
so phi(a(-b))=e -> a(-b) in K

[solemn rain]

yes

a(-b) is in K

[chilly ocean]

phi injective => ker(phi) = {0}

[topaz solar]

That’s the same under different notation

[solemn rain]

yea i meant what you mean

N\U

[topaz solar]

Multiplicative stuff can often use 1 as unit

[solemn rain]

yea yea

okay so a(-b) is in K

what else

[chilly ocean]

Ah carry on then

[topaz solar]

So they’re in K

So what does that say about aK and bK

[solemn rain]

they are equal

[topaz solar]

good, since aK(-b)K=a(-b)K = eK=K

So by uniqueness of inverse aK=bK

[solemn rain]

okay

so i think we are done lmao

[topaz solar]

Yep

[solemn rain]

woo

thats cool

i m like mega confused but i think i got it

im still like in a brain fog now

but i understand

[topaz solar]

It’s obv surjective as given, so you showed objective, therefore bijective

And a bijective group homomorphism is iso

Which is quite a nice property

[solemn rain]

okay so

what i showed is

if G and H are homomorphic with a homomorphism f

G/ker(f) is isomorphic to H

right?

[topaz solar]

Not quite

You required that f is surjective

[solemn rain]

yea yea

forgot that

[topaz solar]

But you can just take the image

And do that

[solemn rain]

okay

yea yea so in general

if G and H are homomorphic with a homomorphism f
G/ker(f) is isomorphic to im(H)

[topaz solar]

Yes

[solemn rain]

im(H) = { phi(a) | a is in G }

just to review

right?

[topaz solar]

$f:G\to H$ factors as $G\to im(f)\to H$

[cloud walrus]

Darkrifts:

[topaz solar]

And ye

[solemn rain]

if phi is surjective then im(H) = H

yea yea cool

okay so there are like

alot of fucking

diagrams

and like arrows

wtf is this

[topaz solar]

that’s just how I do shit

Symbols

[solemn rain]

is this useful

?

like later in category theory/

[topaz solar]

The basic ideas are useful for dealing with algebraic structures

And categories sometimes are built on those

But dwai

[solemn rain]

is it okay if im like

confused af

a bit?

[topaz solar]

Yeah, but you’re trying

Also here’s a cool tidbit

$ker(f)\to G\to im(f)$ obviously gives you a trivial map at the end right?

[cloud walrus]

Darkrifts: