groups-rings-fields

[chilly ocean]

yes I said that

[next obsidian]

Then to say ab^-1 is in U(D) means that a/b = u where u is a unit of D

But then a = ub

So they’re associate

You don’t view b^-1 as an element of D since like you said you don’t know it exists, but b^-1 exists in F

[leaden finch]

isnt that similar to one of the ring axioms ?

[next obsidian]

No

A ring axiom does not require multiplication is commutative

Unless you have some weird ass textbook going by weird conventions

[leaden finch]

oh wait you mean this one

[chilly ocean]

like a associated to b doesnt imply the existence of inverse, so this property seems weird

as in, 1/(1-i) isnt even in the ring we are talking about

[next obsidian]

Yes, but it exists in C

Which is the field of fractions

[chilly ocean]

so what

[next obsidian]

This is an equivalent way to state this

??

Look you view ab^-1 in the field of fractions

You have U(D) existing as a subset of the field of fractions

If ab^-1 is in U(D) then you know ab^-1 = u where u is invertible IN D

But then a = ub in D

By how equality is defined in the field of fractions, and since D is an integral domain

[chilly ocean]

k I think I see

thx

[next obsidian]

So the fact 1/(1-i) isn’t in Z[i]

Doesn’t matter since if you know that like uhh

What was the sample?

Example haha

Okay so if you go to C

[chilly ocean]

1+i / 1-i

[next obsidian]

Then you know (1+i)/(1-i) = i in C

[chilly ocean]

I guess its an easy way to find out by just algebra

since we know units

[next obsidian]

You still get that 1 + i = (1 - i)i

[chilly ocean]

of z[i]

[next obsidian]

But now everything exists in Z[i]

[chilly ocean]

ye

[next obsidian]

And this is still an equality in Z[i]

Have you shown Z[i] is a Euclidean domain?

You do this sort of thing in order to do it

[chilly ocean]

like a year ago or so

I remember the argument I think

[next obsidian]

Probably this is how you did it

You look at the fraction in C

[chilly ocean]

no, we made the norm I think

[next obsidian]

Find some element of Z[i] within distance 1 of that

[chilly ocean]

and delta

[next obsidian]

Then multiply through again

Oh

I just showed you can make the Euclidean algorithm work haha

Oh wait no

I still made the norm

But you need to show you can do the Euclidean algorithm with that

In order to do that the easiest way is to go to C, do stuff there, then clear denominators

This gets you back into Z[i]

[chilly ocean]

oh I don't think I've seen this proof youre talking about

[next obsidian]

Ah

Well the idea is just z/w lives in C

[chilly ocean]

but I get the point

[next obsidian]

Yeah

This is a useful thing, I forget when I used it but I used this for something with uhh polynomial rings once

You go to the field of fractions or something and do crap there

Then go back down and everything still works

[chilly ocean]

btw do you know any sources that cover algebraic/trans numbers? (i,.e algebraic independece)

[next obsidian]

I mean...

Any comprehensive algebra textbook haha

[chilly ocean]

nope

[next obsidian]

That’s at least how I learned it

D&F has it

Uhhh besides that

[chilly ocean]

like bare minimum I think

[next obsidian]

If you look into commutative algebra textbooks they cover it sort of

[chilly ocean]

in DF

[next obsidian]

Since you prove Noether Normalization and crap

You definitely are forced to get better at it in order to do commutative algebra stuff

I’m not really too sure though, I used D&F and Aluffi to learn algebra and I know what I know so ¯_(ツ)_/¯

[chilly ocean]

commutative algebra seems too hard, alhtought the name implies its easy since commie - cool

[next obsidian]

Maybe I picked it up randomly from other people along the way too ¯_(ツ)_/¯?

[chilly ocean]

I do have few books that mention algebraic numbers, recently found one that focuses a lot on those, but seems pretty bad

[next obsidian]

I feel like if you’re interested in algebraic numbers that falls more into a number theory place

General thing of transcendence degree and algebraicity of field extensions is algebra though

Once you go to R though I think you run into number theory

[astral galleon]

@next obsidian for giving a counter example of a subgroup is it fine to take the union of two things that don't even live in the same set and claim that as a non-subgroup

[next obsidian]

Uhhhh

Not really

It doesn’t prove your point

Just like, work in Z/2Z x Z/2Z

Take the union of {(0,0),(1,0)} and {(0,0),(0,1)}

[astral galleon]

ok thx

[bronze trench]

Which order if you don't mind sharing?
@next obsidian 56 is giving me trouble

Feit-Thompson - "am I a joke to you"
@sturdy marsh sylow's are simple to prove tho, feit thompson is like a full semester of classes just to prove it

[next obsidian]

Well if you can prove Burnside’s pq-theorem hahahaha

Would you like a general Sylow tip?

[bronze trench]

Character theory isn't on the course and I doubt the professor will like me to just either give references or fully develop character theory up to that point for a homework 😂

[next obsidian]

I’m not certain if you can use this here, but you might not know it??

If you want to do it yourself though I totally respect that

[bronze trench]

hm, tell you what, I want to think about it since I still have time but maybe later I'll ask or even just discuss it with you if you'd like 😄

[next obsidian]

Mkay sounds good 👌. Just @ me whenever

There’s a lot of tricks to these sorts of problems, I think D&F covers them in the text sometimes???

But it’s always great to learn more haha

[bronze trench]

My professor is following rotman's book pretty closely so I'll take a look at that also to see if there's stuff he didn't cover that might be useful, or even some exercises that might help 🙂

[next obsidian]

Sounds good, have fun! I like Sylow’s a ton so I like discussing it with people 😊

[bronze trench]

Awesome, I'll surely discuss it more if my academic integrity isn't on the line (so like after turning the homework in 😂)

I'm more of a rings and algebras guy from what I can tell, but these Sylow things are absolutely fantastic, I'm loving this part of the GT course 😄

[astral galleon]

@next obsidian is this correct so far

Let $G=V$ where $V$ is the famed four-group now let $H= { (1 2 3 4) }$ and $K= {(1 2) }$ taking the union of $H$ and $K$ we see that,

$$H \cup K = {(1 2)(3 4), (1 2 ) }$

^ I think $H \cup K$ is not a subgroup but i'm not sure my reasoning is that there's no identity element

[cloud walrus]

Zophike1:
Compile Error! Click the reaction for details. (You may edit your message)

[next obsidian]

Uhh, H and K arent even sets

Those are just single elements

[astral galleon]

@next obsidian fixed my bad my reasoning is that H union K is not a subgroup since there's no identity element but I'm not sure if my reasoning is correct

[cloud walrus]

Zophike1:
Compile Error! Click the reaction for details. (You may edit your message)

[astral galleon]

@next obsidian you have a look ?

[bronze trench]

Just to be sure, you really want H=(1234) or you want H=(12)(34)? If it's the first option your union isn't even right

[astral galleon]

No H=(12)(34)

[bronze trench]

Also since neither H nor K have the identity you can't claim they are subgroups. Moreover K isn't subset of V (but you may take S_4 instead 😉 )

[astral galleon]

yeah makes sense 😦

[chilly ocean]

@next obsidian how would you use the sylows theorems in that problem about groups of order<=100?

[bronze trench]

but you can tidy up that example (taking those but with the identity, and inside S_4) and you can reach a counterexample, I think you might be on the right track!

[astral galleon]

Ok I see how it works now thank you

[chilly ocean]

oh benny you can answer since you posted the questin

[bronze trench]

how would you use the sylows theorems to that problem about groups of order<=100?
From what I could do so far, Sylow's theorem prove some general results about groups of order pq for p and q primes and also powers of primes

then for some others you can use the result that states that a sylow p-subgroup has r conjugates where r is congruent to 1 mod p and also r divides |G|

[chilly ocean]

ok so for lets say 60 how do you show no group of order 60 is either abelian or simple

(since that was the problem I think)

[bronze trench]

and conclude there's subroups with only one conjugate and you conclude they are normal

that is the only order for which there is one non abelian simple group!

that's the point of the exercise (find it btw, not too hard if you're familiar with the common groups)

[chilly ocean]

im really bad at groups

I dont think I understand how A_n even works

but just wondering if you could show proof of one specific order

[bronze trench]

well A_n is not abelian for n > 3 and simple for n >4 right?

sure, let me find a simple one (no pun intended)

[chilly ocean]

non abelian yeah, I havent studied simple groups so idk

[bronze trench]

simple means no normal subgroups

[chilly ocean]

ok

[bronze trench]

So A_5 isn't abelian and is simple

and |A_5|=5!/2=60 so that's the magic one 🙂

[chilly ocean]

ok fuck I pcked a wrong number then lmao

lets say 35

[bronze trench]

35 is also not great since it's 7*5 and the proof that all groups of order pq are either non simple or abelian isn't trivial

let's say order 40

[astral galleon]

@bronze trench dumb question $S_4$ is just $(1 2 3 4)$ right I think I had that as $K$ in the post I delted

[chilly ocean]

oh I thougt it would be easier for pq

[bronze trench]

40=2^3*5 for example

[chilly ocean]

ok 40 sounds good

[bronze trench]

@bronze trench dumb question $S_4$ is just $(1 2 3 4)$ right I think I had that as $K$ in the post I delted
@astral galleon it's the set of permutations on those 4 numbers yeah

[cloud walrus]

bennycunha97:

[bronze trench]

ok, take 40

from sylow's theorems you know there's one or more 5-subgroups for any group of order 40 right?

So let P be one of those 5-groups. From sylow's theorems you also know that all 5-subgroups are conjugate amongst themselves, and their number is r, where r|40 and r is congruent to 1 mod 5, right?

nothing much, these are just the statements of the theorems. Following so far?

[chilly ocean]

yeye

[bronze trench]

ok now check which are the possibilities for r

[chilly ocean]

1 only?

[bronze trench]

I'll rnu through the numbers that are 1 mod 5. So r=1 is always ok. r=6 doesn't divide 40. Neither does 11...

so on and you get r=1, yes!

[chilly ocean]

and its order has to be 5?

[bronze trench]

So there's only 1 5-subgroup of our group G. Well then it's conjugate to itself

we did pick 5 since it works, for the number itself to apply it just try the prime factos of 40 and see what works

but there's a proper subgroup which is conjugate to itself right?

[chilly ocean]

yeah

[bronze trench]

so...

what does "being a normal subgroup" mean? 🙂

[chilly ocean]

exactly that

gH=Hg

[bronze trench]

I mean sorry, my bad. Every subgroup is conjugate to itself for the identity lol. I mean it's the only subgroup conjugate to itself

exactly that
indeed!

[chilly ocean]

ok so its abelian gotcha

[bronze trench]

so there's a normal subgroup for our group, just from the fact |G|=40, no need to even see what groups there are of order 40

ok so its abelian gotcha
wait no 😛

[chilly ocean]

wait no lmao

[bronze trench]

it might not be, but it's definitely not simple, there's a normal subgroup there

[chilly ocean]

yeyeyey

hmm nice

it works for most orders you said?

[bronze trench]

so yeah you can use that argument plus the pq argument plus the p^n argument to plow through the numbers

and there's like 6 left, 5 of which require an argument a little more sophisticated but not that hard. Then there's 56 which is impervious to my attacks so far 😛

[next obsidian]

Sorry I’m in a call right now for a seminar

[chilly ocean]

,w factor 56

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Godel#2664

[next obsidian]

You can hit huge classes, like pq, pqr, pq^2

And show they can’t be simple

And clearly for p-groups

They’re solvable lol

[bronze trench]

yep but you can't do that for 56 since both 8 and 1 are 1 mod 7 and divide 56 😦

[chilly ocean]

groups are so bad just study rings and fields instead

[next obsidian]

Groups are wonderful

[bronze trench]

I like rings a lot. Fields scare me lol

[next obsidian]

Shaking my head

[chilly ocean]

yeah rings are just numbers and stuff, groups are wierd

[bronze trench]

rings are just numbers
😮

[chilly ocean]

and stuff

stuff is rest of the rings

[bronze trench]

oh yeah then groups are just the trivial group and stuff 😂

[chilly ocean]

and every ring is commutative

[bronze trench]

noooooo

[chilly ocean]

yes

[bronze trench]

the best rings are the non commutative ones

[chilly ocean]

noncommutative stuff scares me

[bronze trench]

it's the best stuff. Endomorphisms are everywhere for example, can't dodge them

I've explored some cool noncommutative rings with a prof of mine, it was fun. You take a polynomial ring over a commutative ring and make the polynomial ring itslef noncommutative 😂

[chilly ocean]

mind asking what year you in?

[bronze trench]

currently between my 1st and 2nd year of my master's

basically I'm on my 2nd year but shit happens and I'll take 3 years instead of 2 so I'm not doing my dissertation yet

[chilly ocean]

is your master's algebra related?

[bronze trench]

It's a "master's in mathematics"

I like algebra so I take all algebra I can in my subjects 😛

[chilly ocean]

but your thesis - what field of math?

[bronze trench]

I do also have a geometry course this semester and a combinatorics one next semester, and also did a topology course, another geometry course and functional analysis

[chilly ocean]

pun intended?

[bronze trench]

oh definitely algebra

not sure what yet exactly but since I'm almost a year from starting anyway I'm taking my time

while still talking to my professors and exploring some random stuff I find interesting

[chilly ocean]

cool

[bronze trench]

but indeed algebra is life and I can't see myself studying geometry or analysis or something for my whole life

[astral galleon]

If $H$ and $K$ are subgroups of a group $G$ and if |H|and |K|are relatively prime, prove that $H∩K={1}$

$\textbf{Proof}$

Since $H \cap K \leq G$ we take $|H \cap K|$ into consideration by Lagrange's Theorem we can make observation that,

$$|H \cap K| = |H| \cap |K|$$

$$, , , , ! ! ! ! ! ! , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , = \big[G: H \big] \cap \big[ G: K \big] = { 1 }$$

Since the $\textbf{gcd}(|H|, |K|) = 1$

^ Is this correct

[cloud walrus]

Zophike1:

[bronze trench]

| in math mode is \mid btw

[astral galleon]

Oh thank you sorry for the typo's I don't know much latex

[bronze trench]

no problem. Also what does it mean the intersection of 2 numbers?

you're intersecting both orders and indexes... 😳

[astral galleon]

no i'm intersecting the number of cosets of H and K respectively

[bronze trench]

it's still a number, you can't intersect numbers... 😛

[astral galleon]

oh true 😦

[bronze trench]

but you're right in guessing Lagrange's theorem will help. Just try and see, H \cap K is a subgroup of what?

[astral galleon]

you can only take the intersection of sets

it's a subgroup of $G$ and $|H \cap K|$ dives $|H|$ and $|K|$ respectively it equals $1$ since gcd(|H|,|K|) = 1

[cloud walrus]

Zophike1:

[bronze trench]

well you're right, the only missing piece is that it divides both |H| and |K| as it's a subgroup of each of them, and then you use Lagrange's theorem

but see you got it!

[astral galleon]

@bronze trench I peaked at the solution to get it

[bronze trench]

ok but the key is that H \cap K is a subgroup of both H and K

[astral galleon]

help me with this next one plz

Let $G$ be a finite group with subgroups $H$ and $K$ if $H \leq K$ prove that,

$$ [G: H] = [G:K][K:H] $$

$\textbf{Proof}$
Using Lagrange's Theorem we have that,

$$[G:K][K:H] = \frac{|G|}{|K|} \frac{|K|}{|H|} = \frac{|G|}{|H|}$$

[cloud walrus]

Zophike1:

[scarlet estuary]

are you familiar with the phrase "coset decomposition"? that's the key idea here

[astral galleon]

Oh my idea is wrong 😦

[scarlet estuary]

well thats the idea i'm familiar with off-hand

[astral galleon]

I used Laragrange's Thoerem to prove it is it wrong to do what I idid

[scarlet estuary]

oh, i read that H was simply a subset of K

if H is a subgroup of a finite group then sure, thats fine

[astral galleon]

all right thanks sorry I need to work on my writing

[scarlet estuary]

nah its my fault for misreading

[bronze trench]

yep that's it!

sorry for only replying now, was having dinner. A tasty lasagna 😄

[astral galleon]

How do you prove that every infinite group contains infinitely many subgroups ?

There was a hint in the book that has me confused

Can an infinite group have only finitely many cylic subgroups which I said no since a finite clycic group has finitely many cylic subgroups

Now i'm not sure how to prove it

[bronze trench]

Well you can "list" infintely many cyclic subgroups of an inifinte group I think

[astral galleon]

After browsing around and looking at a simular problem I have an explanation

Indeed, let G be a group with infinitely many subgroups. Then G has infinitely many cyclic subgroups. An infinite cyclic group has infinitely many subgroups. Finally, G is infinite because it is the union of its cyclic subgroups, which is a infinite union of finite sets.

[bronze trench]

Well there's something wrong there, this is wrong

Therefore, all cyclic subgroups of G are infinite

Take the rationals minus 0 with multiplication

{-1,1} is cyclic and not infinite

[astral galleon]

Oh should it be finite

[bronze trench]

also not finite, Z is cyclic and is a subgroup of Q

the thing is, if you have infintely many cyclic subgroups, this is still valid, no?
Finally, G is infinite because it is the union of its cyclic subgroups, which is a infinite union of finite sets

except remove finite so

Finally, G is infinite because it is the union of its cyclic subgroups, which is a infinite union of sets

[astral galleon]

Ok i'll have to ask about this one in office hours

😦

I'm doing poorly

[bronze trench]

well everyone hits a wall at some point and abstract algebra is indeed... well, abstract

so sometims hard to wrap your head around at the beginning

[astral galleon]

I'm getting some of the problems not all of them like I used ot

[bronze trench]

but you do seem willing to learn and ask and scrutinize answers so I think you should be good at some point 😄

don't worry, math is not easy, it's normal to be stuck sometimes

[astral galleon]

I have to look at a lot of examples and coming up with the examples is sometimes not easy

[bronze trench]

yeah I feel you

[next obsidian]

Did you sort out the infinite subgroups thinf?

[astral galleon]

Yeah I got it with some help

[next obsidian]

Okay

[astral galleon]

Dumb question for the following result: Let $G$ be a finite group and let $a \in G$ Then the order of $a$ is the number of elements in $<a>$

[cloud walrus]

Zophike1:

[astral galleon]

^ does it work for infinite groups

[scarlet estuary]

yes, if we regard Z under + as an infinite cyclic group

(and in fact, its isomorphism class is the only infinite cyclic group)

this is why we often say an element of a group has "infinite order" if there is no m s.t. a^m = e

[astral galleon]

I ask because i'm trying to prove that every infinte group contains infintely many subgroups via another method

In the book at the point where I am defines order of a group $G$ but I don' think it's enough via defintions to prove it

[cloud walrus]

Zophike1:

[next obsidian]

This holds true

Use the exact same proof

Nothing about the infinitude of the ambient group plays a role

[astral galleon]

@next obsidian so just looking at the order of G and it's subgroup is enough

[next obsidian]

What?

By G do you mean g?

The element

[astral galleon]

no I mean the group

[next obsidian]

The order of G literally doesn’t matter

[astral galleon]

oh I can just look at the order of an element to make conclude something about the entire group

[next obsidian]

No

You conclude something abou <g>

[astral galleon]

the group

?

[next obsidian]

That’s a sub group of G

[astral galleon]

ok ok

So just to repeat I can conclude something about <g> by looking at it's order which should be finite ?

@next obsidian ?

[next obsidian]

You don’t need to know |g| is finite

If it is then <g> is finite with the same order

But if |g| is infinite then <g> is infinite too

[astral galleon]

oh I see how the proofs follow since G is infinite it's subgroups are infinte as well

[next obsidian]

That’s not true

G can be infinite and have a finite subgroup

It has to have infinitely many subgroups

[astral galleon]

can you start over now i'm confused

[next obsidian]

G can be an infinite group

And have a subgroup which is finite

[astral galleon]

Oh just take the union of infinitely many finite groups

[next obsidian]

Not the union

[astral galleon]

the intersection

[next obsidian]

But like direct sum or product

[astral galleon]

ahhh ok

[next obsidian]

Anyway yes

Each of those groups in the product appears as a subgroup

And they’re finite

[astral galleon]

ahhh why can't we take the union

@next obsidian consider an infinite group $G$ there exists many subgroups $g_i$ starting at $|g_i|$ it's indeed vaild that $g_i$ is finite hence we take the direct product of $g_i$ for all $i$ infinitely many times to get that $G$ contains infinitely many subgroups $g_i$

[cloud walrus]

Zophike1:

[astral galleon]

Is it correct
@next obsidian

[latent anvil]

@bronze trench [REDACTED]

[next obsidian]

Delete!

[latent anvil]

Oops

Sorry I didn't see the part where you asked people not to spoil 56 😅

[astral galleon]

@next obsidian you didnt answer my previous question

[sturdy marsh]

56 is pretty neat

[golden pasture]

56

@astral galleon if there are only finitely many g_i taking products of them doesnt give infinitely many subgroups

[sturdy marsh]

it's one less than my favorite prime number

[golden pasture]

you may want to consider the case that every element is of finite order and there exists an element of infinite order

[vestal snow]

When computing the trace/norm as the sum/product of galois conjugates, do we allow for repetition?

[woven obsidian]

I'm usually bad at these concrete examples. How do I actually show z is irreducible?

[bronze trench]

@bronze trench [REDACTED]
@latent anvil the notification still showed up with the original message 😂

It's ok, I was probably going to try that too, for now I counted exactly 56 elements lol

There must be some trick or something that I need to get, but I'll persevere 😁

[latent anvil]

sorry lol

I want to say things but I am exercising self control

[bronze trench]

Shit I think I got it. Just tell me yes or no, no details

So, I counted 49 elements of order 7 and 6 of order 2

Plus the identity it makes 56 so not yet over the correct number

But multiply an element of order 2 and one of order 7 and you get one of order 14, hence different from all others already accounted for!

@next obsidian @latent anvil

This assuming it was simple originally, hence it can't be

To get the ones of order 2 and 7 I assume there is more than one 2-subgroup and conclude there are 7 and they all have a trivial intersection so removing the identity from each I get 7 elements of order 2, same for the 8 7-subgroups and get 49 of order 7

The prof showed us this but in his example these steps were enough to go over the order of the group. Now I'm one short but the order 14 element works right?

[vestal snow]

Thanks

[sturdy marsh]

@bronze trench I'm pretty sure you need those elements to commute to conclude that the order is 14

and you probably overcounted the number of elements of order 7

you have too many

not really 'too' many but too many

[bronze trench]

Order at least 14 is good enough

Yeah I overcounted crap. It's 6x8 not 7x7 :(

[sturdy marsh]

yup it's 48

one many

[bronze trench]

Anyway don't tell me more then please

[sturdy marsh]

how did you prove that the order is at least 14?

[bronze trench]

It's a very serious homework, can't get any answers, at most just discuss what I know

[sturdy marsh]

cool

[next obsidian]

My advice, count harder

😊

[jade crag]

Hi, I need help proving that the dimension of the centralisers of a jordan block is = n, where the jordan block is a nxn matrix

https://math.stackexchange.com/questions/3884418/prove-that-the-dimension-of-the-centralizer-of-a-jordan-block-is-n?noredirect=1#comment8011303_3884418

i asked it on stack exchange and got a comment about expressing the N, for J=I+N as well as the centralizer in terms of sums but I dont know how to proceed from there

the comment seems to imply that its clear as to how the sums show the result I want or that its "routine" to check but im completely lost

[latent anvil]

@bronze trench why do the 2 sylows intersect trivially?

[bronze trench]

@bronze trench why do the 2 sylows intersect trivially?
@latent anvil maybe I'm misremembering but I'm pretty sure the prof used that somewhere

[latent anvil]

So you're thinking of the case where the p sylows have order p

In general it can get messier

if the sylows have order p then they're cyclic, and so any nontrivial element generates them, so distinct sylow subgroups must intersect trivially

But if they're like copies of the quaternion group they can intersect in something of order 2 or 4

[bronze trench]

ok so this only works since 5 divides 30 but 5^2 doesn't, is that it?

so every 5-subgroup is of order 5

[latent anvil]

Yup, exactly

[bronze trench]

crap I'm dumb af

[latent anvil]

No you're not this is a really common mistake!!!

[bronze trench]

no not that

[latent anvil]

I make sure to explicitly call it out for my students

Oh okay lol

[bronze trench]

the 2-subgroups have order 8, not 2...

lol

[latent anvil]

Haha yeah there's that too

[bronze trench]

yeah this resolves my issue

maybe

they're not all the same so each contributes with at least 2 "new" elements and then the count indeed exceeds |G|

[latent anvil]

I don't think I agree

Why can't one of them be contained in the union of 2 (or more) of the others?

[bronze trench]

fuck me

[latent anvil]

Haha

[bronze trench]

ok nvm I'll think harder, don't keep telling me stuff, this is academic fraud lol

[latent anvil]

Possible intersections of multiple sylows get very complicated very quickly

Sure

[bronze trench]

ok at least I know how not to proceed xD
thanks and I'll talk later about this when I turn it in

I don't think I should be discussing this further until I hand it over

[next obsidian]

You are very very committed to that haha

Which is admirable

But I think it’s not uncommon for people to get a tiny bit of help or just discuss it to get ideas

Be it with classmates or others

[bronze trench]

homeworks count for 100% of the grade lol
so if I cheat a little on each one it adds up and I don't think it's ethical

[latent anvil]

I mean I think it's different to discuss it with me vs a classmate

[bronze trench]

also this

[latent anvil]

discussing with classmates is good, discussing with someone who knows the proof is not good

[next obsidian]

Oh haha

It explicitly says not to

[latent anvil]

Oh that's dumb

[bronze trench]

yep 😂

[latent anvil]

Banning collaboration with other students makes no sense

[bronze trench]

we're 4 taking this class, and we all have the prof's trust too... 😛

it's only because it's 100% of the grade. I did have subjects in which we could basically do the HW together as long as we wrote up our own solutions, but those were worth 50% or 30% of the grade, not all of it

[sturdy marsh]

there's only 4 people in the class?

woah

[bronze trench]

small-ish math department and most students go for applied courses

so very few in group theory and symplectic geometry for example

but even the compulsory subjects have less than 10 students each year on average, our master's program is kinda small

[sturdy marsh]

huh

[bronze trench]

because of that many subjects don't open every year, only every 2 years or worse

but since we're few it's also easier to talk with the program's director and actually have the subjects we want to open, literally customized to our year if it makes sense

[sturdy marsh]

yeah and small classes are usually more fun too

[bronze trench]

yep definitely

also not many students to have a lot of question at the end of classes, which means I can stay and discuss hopf algebras and augmentation ideals or profinite semigroups with the profs at the end, or some crazy stuff like that

[sturdy marsh]

where do profinite semigroups turn up

[bronze trench]

you mean in math or why did I ask about it? I haven't got a useful answer to the former 😂

[next obsidian]

Why are you talking about a pro finite semi group while learning finite group theory haha

[sturdy marsh]

I would be satisfied with an answer to any of those lol

[bronze trench]

Why are you talking about a pro finite semi group while learning finite group theory haha
the prof does semigroup theory for a living and I searched some of his papers and profinite things turn up there so... xD

also the very end of the syllabus is something like "time permitting, a little introduction to profinite groups"

[sturdy marsh]

well profinite groups definitely come up naturally

[bronze trench]

not that I know that yet 😂

[sturdy marsh]

number theory

[bronze trench]

how? like wtf number theory has it all. étale rings, fucked up fields, also profinite groups wtf?

[sturdy marsh]

Gal(Qbar/Q) is an example

[bronze trench]

I know basically no number theory and tbh I'm not that interested but it keeps bringing in such heavy machinery, it's insane

[sturdy marsh]

I've seen very little class field theory and the wee bit I saw was full of profinite groups

[bronze trench]

unrelated but a bit of a complaint regarding abstract algebra and my education so far

so basically I know 0 category theory

and every time I search for random algebra stuff I encounter it

I'm afraid not learning it with someone who knows about it will be troublesome for me. I mean of course I can read up on it and ask people who know. But that's different from classes and I find it dumb that I don't even have the option to take a class on it in my uni

[carmine fossil]

Aluffi,ig

[sturdy marsh]

Emily Riehl's book is really good

once you know about limits, colimits, adjunctions that should enable you to read a lot of the algebra stuff online

or do an alg top. class

that will have plenty of category theory

or AG

[bronze trench]

I have Tom Leinster's Basic Category Theory I had printed for some reason, people also tell me it's good

[sturdy marsh]

haven't read that

Riehl is great

MacLane is alright

[bronze trench]

or do an alg top. class
I did one, it didn't involve a lot for some reason

or AG
I have PTSD, I had a terrible prof teach me that, I dropped it halfway through, it was awuful

[sturdy marsh]

Peter May's book is great if you want to learn alg top. and category theory

Tom Dieck is also great

[bronze trench]

And I also had a commutative algebra course which had category theory in it but it was assumed I knew it, not taugh. I failed miserably at it lol

I might check it out next semester, this one I'm a bit busy and I don't want to mess it up again like last year :/

[sturdy marsh]

it definitely helps to have some notions discussed in class, our group theory class spent a few weeks doing categories

but again, Riehl's book has a lot of examples and is pretty easy to read

[bronze trench]

my uni has a small department and we have poor undergrad training so we don't do a lot of "hard stuff" and I'm really pissed off at it

people here in Portugal don't have the faintest idea of what math education should be so 40-60 people enter the undergrad program at my uni each year and like 6 go on to even do a master's in pure math or "real" applied math (real applied math as opposed to my uni's master's in "mathematical engineering")

so undergrad education sucks hard

[sturdy marsh]

I have PTSD, I had a terrible prof teach me that, I dropped it halfway through, it was awuful
@bronze trench rip

[bronze trench]

the ring theory course isn't compulsory. In my year we were 5 sitting the exam

what a fucking disgrace. Then in the master's my algebra course so far is stuff I learnt in that ring theory course since a couple of people in my class didn't even take it and yet decided that going to a master's in math was a good idea

sorry for the ran but this pisses me off so much

[sturdy marsh]

the first year grad courses in algebra do tend to repeat a lot of the material from the undergrad courses

but the problems are normally harder

[woven obsidian]

Given a polynomial over a field k in n variables X1,...,Xn is it true that I can bound the number of roots of the polynomials by the maximum n of any X_i^n occuring?

[hot lake]

um

X+Y has many roots I think ?

[sturdy marsh]

X+Y has infinitely many solutions over R

[woven obsidian]

I mean roots in k

[sturdy marsh]

k = R = real numbers

[bronze trench]

but the problems are normally harder
@sturdy marsh my first homework problem for the algebra class was basically to show that if R is the gaussian integers and p is a prime number, R/pR is a field iff p is congruent to 3 mod 4. hard my ass lol

[woven obsidian]

Ah I'm stupid

[bronze trench]

oh and in stages, you had your work cut out for you lol

[sturdy marsh]

well it might get harder as the semester/quarter progresses

[woven obsidian]

So how would you show the last thing?

[sturdy marsh]

are you asking why does a nonzero polynomial over an infinite field nonzero value somwhere?

[woven obsidian]

Yeah, but I think I found the answer

[sturdy marsh]

sorry nonconstant

[woven obsidian]

Just let all but one variable be 1 I guess. Then you get a polynomial in one variable

[sturdy marsh]

not nonzero

[woven obsidian]

Assuming you pick a variable that actually occurs in the polynomial

[sturdy marsh]

Just let all but one variable be 1 I guess. Then you get a polynomial in one variable
@woven obsidian Pretty much. The same argument allows you to extend the result to a polynomial ring over an infinite domain

Is this for the primitive element theorem for infinite fields?

I think it used a similar lemma

[woven obsidian]

No it was for Nullstellensatz, the proof using resultants

It seemed like a really obvious statement but I was honestly not sure why it was true at first

[sturdy marsh]

apparently, there's a proof of the nullstellensatz via model theory lol

(I do not know what model theory is)

[woven obsidian]

Ah really

[sturdy marsh]

yeah wikipedia never lies

[latent anvil]

huh

I would like to see that proof

[next obsidian]

Model theory is set theory shit

Model theory is set theory is logic is model theory is...

[sturdy marsh]

so model theory = logic then

and wikipedia says model theory = universal algebra + logic

so universal algebra = 0

[astral galleon]

Thx @golden pasture

My bad typo :(

[bronze trench]

@latent anvil lol I said I wouldn't discuss it further but here I am
So, I don't need for each 2-subgroup to contribute with a different element duh. So assuming there's more than one of them, 8 come from that, plus one from any other 2-subgroup (which isn't the same so that one element exists). Plus the previous argument that works for 7 since 7^2 doesn't divide 56 gives us 48 elements of order 7 (that aren't on the 2-subgroups of course). So 48+8+1=57 and we already have too many

@next obsidian also you 😂

[next obsidian]

That's exactly true 🤣

[bronze trench]

fuck it wasn't hard, I'm dumb lol

[next obsidian]

i wanted to say like

pls just count

you have it, but messed it up a bit

but didn't want to say anything since you asked us not to

you simply miscounted haha

[bronze trench]

I was trying to see they were all somewhat disjoint and lost a ton of time and I didn't need to at one point as well

[next obsidian]

yeah, you have juuuuust enough space to have 8 Sylow-7 and 1 Sylow-2

[bronze trench]

you simply miscounted haha
not that, at first I thought the 2-subgroups had order 2, not 8 bah

[next obsidian]

but no more than that

OOF

I would say that's miscounting

[bronze trench]

totally forgot the maximal part on the power since the other times I saw this the things used were really just p^1 xD

[next obsidian]

I'm curious why they didn't ask you to show all of them < 120 cannot be simple except A_5

[bronze trench]

like 5 on a group with 30 elements

[next obsidian]

the only cases left are 101 through 119 which I think aren't too bad?

[bronze trench]

dunno the prof thought 100 was a good number lol

[next obsidian]

¯_(ツ)_/¯

I remember when I did this

I just went to wikipedia

there's a list of prime factorizations of integers

[bronze trench]

and there are a ton of exercises, not just this one, maybe he thought a little more time wasted on this wasn't that great 😂

[next obsidian]

since I didn't want to factor them all myself lmfaooo

[bronze trench]

same

[next obsidian]

I'm like nice, this is p

this is pq

this is pq^2

this is pqr...

[bronze trench]

I started to factor them myself but gave up and looked it up 😂

[next obsidian]

Yeah, I think there's no shame in that haha

if you really wanted to, you can say you looked up the prime factorizations lol

I doubt the prof will be mad about that

[bronze trench]

It's not that I don't know how to factor numbers lol

[sturdy marsh]

the only cases left are 101 through 119 which I think aren't too bad?
@next obsidian 101 is prime, 119 is pq

[next obsidian]

yeah but there's still stuff in between

I think most of them fall into p^a, pq,pqr, p^2q and p^2q^2

all of which you can rule out

[bronze trench]

I mean, in an exam with this same prof I factored 57 as 8*7 and the correct answer involved checking if 19 divided 57

[next obsidian]

LOL

[bronze trench]

I did everything right and messed that up lol

[next obsidian]

I once during an exam

had to check some determinant was non-zero to apply the umm

what is that

implicit function theorem?

inverse function theorem

it was one of them

I miscalculated the determinant lol

[bronze trench]

what even is a function lol

[next obsidian]

but all I need is non-zero

and the real answer was non-zero

and my answer was non-zero

so the prof said "this isn't right... but whatever"

[bronze trench]

well that's good at least 😂

[next obsidian]

and gave me full points

But you know 2sham2rock

we were in the same calss

he calculated the determinant wrong as well the first time

BUT HE GOT 0!

[bronze trench]

yeah this little factoring 57 wrong stunt made me not have 17/20 ffs

[next obsidian]

So he was freaking out and recalculated it

[bronze trench]

0! = 1 so that's ok 😛

[next obsidian]

and got the right answer

🤦‍♂️

wdym 57 is prime

[bronze trench]

I make such good jokes lol

[next obsidian]

prime factorization is 57

[bronze trench]

oh right my bad

[next obsidian]

[bronze trench]

nice emoji 😛

[next obsidian]

anyway, I'm glad you got the answer 👍

[bronze trench]

anyway yeah my prof took 0.5 on that exercise for that stunt and I ended up with 16.0

16.5 goes to a final grade of 17 fml

[next obsidian]

Rip

😔

[bronze trench]

yeah btw grades here are up to 20 and 17 is a fucking good grade

depending on the prof of course

but he's one of the tougher ones. I got mad lol

[next obsidian]

That sucks

I once lost points on an exam

[bronze trench]

retook the exam, 16 again. I accepted my fate

[next obsidian]

because I think I said the integral from 0 to infinity of 1/x^1/2 was finite

or something stupid like that

I was trying to bound something and bounded it by infinity 🤣

the prof just wrote in red pen NO!

I don't blame him

[bronze trench]

that was my grade on an exam once. Just NO on the corner of the page

he didn't even add up the score, it was not enough to pass by miles rip

[next obsidian]

Oof :(

feels bad

[bronze trench]

this one was on a scale to 30, got 8/30 in the end lol

pass is at 18 so not even halfway there

rip

I was severely underprepared for that course, did it while in Erasmus in Milan

my shitty undergrad education showed through

[bronze trench]

lol apart from this one excercise today I made negative progress in my homework

that being copying my solutions to latex and realising I had something wrong so I did -1 exercises today

[cinder bone]

so i know some element a is in a group of order 181440

does that tell me anything about what a^2020 is

[bronze trench]

my intuition tells me no, no idea tho

but since the exponent is much less than the order of the group I see no way to say anything

[cinder bone]

yeahhh

[bronze trench]

say G is the cyclic group of that order and a is a generator? then a^2020 is literally just a^2020 and that's it 😂

it's the actual name you'd give to that element and everything

[cinder bone]

ohhhh ok

yeah it would be cyclic

I determined that ō is the product of disjoint cycles of lengths 2,3,4

so it has order 12

but also it’s a member of A_9 since it has an even order

and the order of A_9 is 9!/2

[bronze trench]

oh yeah "in a group of order something" is a much weaker statement than "in A_9", you know what A_9 is and does 😛

[cinder bone]

which is 181440

yeah true

[bronze trench]

but you know the order of sigma so that's enough to see the order of any power of sigma

you say the order of sigma is 12 right? So what's sigma^13 for example?

[cinder bone]

the identity

[bronze trench]

ok not what I wanted but let's go with that. You mean sigma^12 is the identity right?

[cinder bone]

yeah

[bronze trench]

let's call it just s lol
s^13=s^12s=es=s

does it make sense?

since the order of s is 12 you can cancel the s^12 everytime they appear

[cinder bone]

Oh okay

so it’s sigma^4

why do I need to use the A_9 then hmm

[bronze trench]

just because it's some permutations and you know how to compute stuff with them. It could be on S_9 for example

[cinder bone]

would the order of sigma^4 then be 3?

[bronze trench]

maybe it matters for another question

yes it's 3, great!

[cinder bone]

nah, there’s not another question about it, weird

but thanks yeah

[bronze trench]

np 🙂

[cinder bone]

maybe cus there's a chance s in not in S_9?

so i cant really talk about the identity element

??

or is sigma definitely in S_9 lol

[bronze trench]

it is, every element of A_9 is in S_9

but where does it say A_9?

[cinder bone]

that’s theorem 7.9

well I guess theorem 7.9 just says the order of A_n in general is n!/2

but idk why that would be recommended 🤷‍♂️

[bronze trench]

can you check theorem 7.29 and tell me what it says? I'm curious if the professor made a typo or something 😛

because in my mind you'd write them in order so 7.9 before 7.25 😛

[cinder bone]

yeah i think it she meant 7.29

thats what i meant

7.9 dont exist

i have something else i need some tips for if u dont mind

[thorn delta]

Yea just tackle each case one at a time. Every element of {1,2,...., n} falls under exactly one. You know that if i is not in one of the a's, then it is fixed by sigma, and similarly for tau

[cinder bone]

yeah i didnt understand the q at first but now i do

[cinder bone]

still not sure exactly what to show tho

like it seems pretty obvious (a1, a2, ..., i..., ak) (b1, b2, ..., br) = (b1, b2, ..., br) (a1, a2, ..., i..., ak) if they are both disjoint

yeah tha's case 1 i just showed right

[south storm]

Is the automorphism group of a cyclic group necessarily cyclic?

[chilly ocean]

$Aut(\bZ / n\bZ)$ is isomorphic to $(\bZ / n\bZ)^\times$ which need not be cyclic

[cloud walrus]

TTerra:

[south storm]

How does one prove this?

[chilly ocean]

which?

[carmine fossil]

Take (Z/8Z)* as an example

[south storm]

That Aut(Z/nZ) is isomorphic to the (Z/nZ)* under multiplication

[chilly ocean]

look at phi(1)

phi an automorphism of Z/ nZ

i think

[south storm]

Quite sure it isn’t, if you look at Z/Z3 phi would map 3 to 2, 2 to 1, and 1 to 0 while 0 also to 0.

So no element is mapped to 3 and 2 to 0

And even if it were, how does it prove that the 2 groups are isomorphic?

[carmine fossil]

Given a group of order n,an automorphism is of form:
$x \mapsto x^k \text{ and } (k,n)=1$

[cloud walrus]

DrunkenDrake:

[south storm]

Yes

Oh I see now

Thank you

[next obsidian]

This is for the cyclic group

Just to reiterate haha

Intuitively just think of it as “the only automorphisms in Z/nZ are multiplication by m, when m is coprime to n”

Then composing multiplication by m with multiplication by m’ gives multiplication by mm’ which is still coprime to n

So really this is exactly Z/nZ^x

[south storm]

Yup that‘s what I thought

Thanks again

[wispy glen]

Question: $p(x) = x^3 + 9x + 6$. Let $\theta$ be a root of $p(x)$. Find the inverse of $1 + \theta$ in $Q[\theta]$.

My attempt:
The inverse is necessarily linear, say $a+b\theta $

So $a-1+(a+b)\theta +b \theta ^2 =0$

Please give a hint

[cloud walrus]

gelearndeutsch:

[carmine fossil]

The inverse is not linear

[wispy glen]

true 🤦‍♂️

[carmine fossil]

Hint:Try to find a polynomial p such that p*(1+x)=x^3+9x+c

[wispy glen]

I'll try. thanks

got it. dividing by -c gives 1. Thanks

[carmine fossil]

Not exactly -c

[wispy glen]

c-6 or something. and c $\neq$ 6 because p(x) is irreducible

[cloud walrus]

gelearndeutsch:

[carmine fossil]

Yes

[chilly ocean]

this might be too elementary for this channel but... the expression in this image makes no sense, right?

ℕ^n is infinite, i don't think infinite sums make sense for rings in general.

[chilly ocean]

I think you should rather think about ℕ^n as function from n to ℕ, which has n elements

so those are finite sums

by n I guess I mean set {1,2,...,n}, \mu_i would be image of {i}

so its an infinite sum of finite expressions, but it does make sense

if it was finite you couldn't make an infinite ring

if it was finite you couldn't make an infinite ring
that is not true

why

if this was meant to say finite sums of finite products

you could certainly make an infinite ring from that

Any element in Z is a finite sum of 1's and -1's

any element, but not Z

Z = {finite sums of 1's and -1's}

but there are infinitely many finite sums of 1s and -1s

so?

{finite sums of 1's and -1's} is an infinite set

I guess I don't know what youre confused about, but to give an example: take ring of polynomials over some ring R - this can be written as { sum i=1 to n a_ix^i, where i \in N and a_i \in R}

yea

no infinite sums here

nor there are in your example

N^n is finite

n elements

I mean

nope

N^n has same cardinality as N

but like N^n is lets say if n=4 then sth like (1,5,17,2) - you can use this for the polynomial ring definition, where those are the powers of x: a_1 x^1 + a_2 x^5 + a_3 x^17 + a_4 x^2

(1,5,17,2) is a particular element of N^n

yeah and you go over all of those

here \lambda seems to be over all elements of N^n

so it'd be an infinite sum

which is not defined for rings in general

ok yeah and taht's what im saying, if it was finite you wouldnt make all polynomials

can you give an example where it would break? or just where your concern comes from?

the expression itself doesn't make sense afaik

how do you define that.

hmm I guess I understand and you might be right, you can probably just write it in the other way

yea i think so

guess you could just write it as \lambda \in S such that S is a finite subset of N^n

whats lambda and S

S is a finite subset of N^n, and lambda element of S

and you'd run over all possible such S's

I mean yeah, this definition is weird, it's just nothing else but taking all finite sums of elements of X?

finite sums of finite products of elements of X

yeye

oh btw this for commutative rings

i dont think it'd work so simple for noncommutative

yeah all rings are commutative thats a secret you cannot share

xD

but the notation is werid, never seen it being called \mathbbZ[X]

yea they say will explain this notation in later chapter lol

I guess such definition will have use in polynomial rings since thats how they are usually denoted

What book is that btw? Or are those class notes?

they are someone elses class notes lol, i dont even know that person directly

ic

[mint gulch]

That statement requires Axiom of Choice?

The statement that exists a minimal subfield of K containing F and a

Yes, if you define it like that it's ok, but

See the first definitiin

When the author says "The smallest" refers to the intersection?

Yes

I see

Thanks

[cloud walrus]

Zophike1:

[astral galleon]

Dumb question If $i \neq j$ then $a_{i}H \cap a_{j}H = \emptyset $ note that $H$ is all the distinct cosets of $H$ in $G$. Is our statement true since if $i \neq j$ then $h_i \notin a_iH$ as well as $h_j \notin a_jH$ hence taking the intersection results in the empty set ?

yeah

oh there just elements of our cosets @open torrent

a_iH and a_jH

ahhh now I see what's wrong hold on

[cloud walrus]

Zophike1:

[astral galleon]

@open torrent ^

$h_i$ is just elements of $a_iH$ and $h_j$ is just elements of $a_jH$

[cloud walrus]

Zophike1:

[astral galleon]

i'll try this one over again

@open torrent I think I figured it out it seems the way to go is to write out all the distinct cosets of H and compute their respective intersections individually to get the empty set

Our $G$ is a finite group

[cloud walrus]

Zophike1:

[astral galleon]

I could maybe use the fact that $aH \neq bH$ ?

[cloud walrus]

Zophike1:

[astral galleon]

other than that i'm stuck

That if $X$ and $Y$ are subsets of a set $Z$ then their intersection is the set $X \cap Y = {z \in Z: z \in X \textbf{and} z \in Y }$

[cloud walrus]

Zophike1:

[astral galleon]

This means that $g \in a_iH$ and that $g \in a_jH$

hence you can take the intersection

sorry for being dumb

[cloud walrus]

Zophike1:

[astral galleon]

wait didn't I say i t already

Oh ok i'm now that's clear i'm sorry 😦

$a_iH = {a_i * h : h \in H }$

[cloud walrus]

Zophike1:

[astral galleon]

^ That's how you define the set

That means $g$ is within H ?

[cloud walrus]

Zophike1:

[astral galleon]

ahhh ok so now can't we just take the intersection of $a_iH$ and $b_iH$ to get our conclusion

[cloud walrus]

Zophike1:

Viburnum:

[astral galleon]

ahhh okay it seems what we can do from there is also write $g=b_ih$ for some element $h \in H$

[cloud walrus]

Zophike1:

[astral galleon]

then it seems you get $g*h$

[cloud walrus]

Zophike1:

[astral galleon]

can't I just take the intersection at this point otherwise I have no clue where to go form there

yeah sorry it should be $a_j$

[cloud walrus]

Zophike1:

Viburnum:

Viburnum:

[astral galleon]

solve them ?

it seems we can set $a_ih_i = a_jh_j$ for some element $h_i, h_j \in H$

[cloud walrus]

Zophike1:

[astral galleon]

@open torrent

But we can't do it for all $h_i, h_j \in H$

[cloud walrus]

Zophike1:

[astral galleon]

So put $ah_iah_j=ah_iah_j$ ?

[cloud walrus]

Zophike1:

Zophike1:

Viburnum:

[astral galleon]

ahhh okay

[cloud walrus]

Viburnum:

[astral galleon]

so just take the intersection ?

of $h_jh_i^{-1}$ and $a_ia^{-1}_j$

[cloud walrus]

Zophike1:

[astral galleon]

ahhh okay you didn't explicity say that eariler so I having trouble

sorry I got lost

😦

@open torrent just take the intersection of the cosets that should lead us to getting a coset which is a contradiction since there's no element in the coset to begin with

yeah i figured

i'm lost on this proof

It seems like given what we know that a_i a_j^{-1} \in H it seems that we are assuming that this is the element in the intersection of a_iH and b_iH which is a contradiction because a_iH=a_jH for all i=j ?

@open torrent ^

yeah I know

since a_iH and a_jH is a subset of H ?

So just assume that a_i is not in H

I just don't understand how your going about it

I know that but I don't get how your setting up the proof by contradcition i'm having trouble seeing everything can you start over

ahhhh okay now I got it

@open torrent thx for the help sorry for being dumb 😦

[astral galleon]

@open torrent to show the contrapostive would it be correct to show that the cosets or contained within each other

[paper flint]

I have to prove that the centralizer of an element in a group is a subgroup. I have proven associativity, existence of identity and closure. I don't know how to prove that an element x also has its inverse x^(-1) in the centralizer, and would like to get a hint for that.

Nvm I got it.

[eager bobcat]

Given the order of two elements a, b of a finite abelian group what other orders must exist in the group. I understand that we can know the order of ab and that e has an order of 1 but is there anything else we can figure out?

Man pls help my tank is empty

[compact needle]

@eager bobcat if the order of a and the order of b are relatively prime, then ab has order the product of the orders. If you're careful, combining this fact with Viburnum's observation you can make an element of order lcm(order a,order b) in general (but it won't be ab if their oders are not relatively prime).

[carmine fossil]

You can also find elements of order, which divide lcm(order a,order b)

Because this is an abelian group

[nova plank]

@paper flint Still need help? Let's say x is in the centralizer of a. Then $xa = ax$ so $x^{-1}(xa)x^{-1} = x^{-1}(ax)x^{-1}$ and then hopefully you can continue from there

[cloud walrus]

Lunasong:

[paper flint]

Yep, figured that out as I typed it here haha.

I now have to prove that the order of an element and its inverse in a group have the same order.
The finite order case was easy to prove. For the case where the element has infinite order, the inverse cannot have a finite order since that would be tantamount to saying that the element itself has a finite order(I wrote down the expressions to highlight the fact). Is the proof okay?

[nova plank]

Yes, that works, assuming you showed the finite case correctly

[paper flint]

Yepp, if $x^n=e$, then $(x^n)^{-1}=e^{-1}$, which gives me $(x^{-1})^n=e$ and hence ord($x^{-1}$)=$n$ as well.

[cloud walrus]

ted-D:

[paper flint]

(I hope I got it right? 😬)

[carmine fossil]

Well ord(x^-1) could just divide n

[paper flint]

Uh oh

Yeah that just struck my mind

[carmine fossil]

Ofc, That's easily resolved

[paper flint]

I also need to show that's the least possible n

Yepp, I know the trick.

Thanks!

[paper flint]

https://cdn.discordapp.com/attachments/737043741168238673/771728006024396840/697848465819959388.png

I've computed the order of elements, but I cannot deduce any sensible relationship between them.

But |6|=2, |2|=6 and |8|=3?

Am I missing something?

[carmine fossil]

|a+b| divides lcm(|a|,|b|)

[paper flint]

Oh, alright. That makes sense.

Is there some fundamental reason why this happens?

[carmine fossil]

Consider an abelian group G. Choose a,b in G.Let n=lcm(|a|,|b|) ,then |a| divides n and |b| divides n,so (ab)^n=(a^n) (b^n)=1

So order of ab divides n

[wispy glen]

Prove that $x^5-ax - 1 \in Z[x]$ is irreducible unless a = 0, 2 or -1. The first two correspond to linear factors, the third corresponds to the factorization $(x^2 -x + 1)(x^3 + x^2 -1)$.

[cloud walrus]

gelearndeutsch:

[wispy glen]

Eisenstein is not applicable. Plz give a hint what to use. Also how's the last sentence useful?

[carmine fossil]

Let's the polynomial is reducible. Now take cases:
The minumum degree of the irreducible polynomials in the factorisation is
Case 1:1 ,Case 2:2
(Other cases collapes into these 2)

For case 1: it has to be either (x-1) or (x+1)

Giving us a=0 or a =2

[wispy glen]

case 1 is easy. because for big x, x^5 is very big and the rest of the expression is small. so f(x) cannot be zero.
let me see if case ii is easy

[wispy glen]

case ii: assumed two factors. took cases. got the solution. thanks

[carmine fossil]

Btw,Is this algebraic number theory?

[wispy glen]

no. field theory. dummit and foote field extension

[paper flint]

If a nonempty subset H of a group G is closed under the operation of G, and if an element x is not in H then x^(-1) is not in H, is H a subgroup of G?

Since the contrapositive of the second statement is x^(-1) in H implies x in H, I think it should be a subgroup but I'm not entirely sure. Associativity, closure and existence of identity all seem rather trivial.

[carmine fossil]

Yes

[paper flint]

Is this self-evident or do I need to prove that inverses exist for every element in H?

[pallid ember]

should probably show it

but its straightforward

[paper flint]

Thanks!

[bronze trench]

ffs my group theory homework is still giving me a headache

[carmine fossil]

Is it done?

[bronze trench]

Not a chance, I won't turn it in fully done 😂
It's very hard. This prof is known for being an absolute bro correcting sudents' work and giving marks away but on the other hand, gives very very hard exercises on the exams and homework

One of my best friends who is taking the class with me had another class with a similar evaluation method and had an excelent grade, and apparently all of the 4 homework sheets she turned in were missng something somewhere

conclusion: not realistic to expect to have the homework fully done at any point 😂

I think the prof makes it on purpose, in "real" mathematics it's not common to have a problem you want to solve fully solved. Despite these having known answers I think he makes them purposefully a couple of notches above what we're expected to know for us to actually feel what it's like

either that or he's a sadist but I prefer to believe the former xD

[snow cliff]

Likely a bit of a sadist too

For a group G, if G maps to the quotient group G/G, would it be equivalent to say that G maps to the trivial group since G/G isomorphic to the trivial group?

So basically G maps to the identity?

[bronze trench]

dunno what you mean but indeed the function that maps G to the group {1} sending every element to 1 is a homomorphism

I mean it's easy to check it's a homomorphism, if you call it f then f(gh)=1= 1·1 =f(g)f(h)

for any 2 elements g and h on the group

[next obsidian]

no

That means the element has order 2 (or 1 I guess lol)

Take any abelian group which has an element not of order 2

then you get a counterexample

[kindred mist]

started studying dimension of rings now

(chapter 15 of altman klein, so neat)

[bronze trench]

@next obsidian it is done, put to paper, finally

[next obsidian]

Woohoo

[bronze trench]

I want to discuss this whole homework so bad

I'll be spamming after I turn this in lol

[next obsidian]

I really really like finite group theory and Sylow's theorems so I'll be happy to talk about it haha

[bronze trench]

Ngl my semester so far has been pretty boring but stuff like this 2 week long grind for the homework is very fun and challenging

[next obsidian]

If you like these sorts of things maybe after you're done with this class you should try to learn more haha, I'm currently working from a finite group theory focused textbook

part of it includes a no-representation-theory proof of Burnside's pq-theorem

[bronze trench]

That's fun! I know my heart is firmly in the algebra side of things but I think I'll go on to study things with more structure like algebras or rings. But no doubt I'll keep learning group theory stuff, it's also fun and very useful I guess

[next obsidian]

I don't study finite groups for it to be useful, I study it to satisfy my heart

hahaha

[bronze trench]

I can totally relate, that's math for me in general 😂

but what are you doing rn in terms of career? still studying? if so at what level? perhaps Fermat-ing and having a day job yet do math? Maybe I've been talking with a full on math professor? 🤔

this is weird tbh these couple of days I've been randomly talking to people here about GT and yet you could be my grandma and I would not know it 😂

[thorn delta]

don't mean to interrupt, but are you using hungerford? I did that problem last week lol

[bronze trench]

you mean me?

[thorn delta]

yes

[bronze trench]

nah this is a homework question on my class. not just this, this is a sliver of it

the question is this

[thorn delta]

oh okay, i just proved that there are no nonabelian simple groups of order 12, 28, 56, and 200

[bronze trench]

and 56 was the one I was missing btw

[next obsidian]

I'm a third year undergraduate, my main focus is algebraic geometry right now

I'm just studying and preparing for grad school apps

[bronze trench]

are you fucking kidding me rn? You're lower than me on the foodchain, you can't teach me GT 😠

(kidding obviously)

[next obsidian]

XD

[bronze trench]

I guess this shows the piece if shit I was in undergrad 😂

[thorn delta]

ha, wait till you get taught CA by 13 year olds here

[bronze trench]

such is life, and I like it

"if you're the smartest person in the room, you're in the wrong room"

I'm just studying and preparing for grad school apps
@next obsidian what grad schools are you looking into?

[next obsidian]

Uhh

[bronze trench]

also where you at, also a good question xD

[next obsidian]

I want to go to columbia, but IDFK haha

My current list is like
Top Tier/F Me: Columbia, MIT, UMich
Maybe a bit of a reach:I should find some
Safeties: UW-Seattle

There's some others that should be in there too

I guess Stanford in the top one too, basically just schools with strong AG depts lol

[bronze trench]

well you seem pretty prepared lol

[next obsidian]

I hope so ¯_(ツ)_/¯

[bronze trench]

I still don't have the faintest clue on where to do my PhD. I don't even know who in my department I want to be my master's thesis advisor lol

but what's probably going to happen is that I'll hopefully get in the circle of the algebra profs in my department (working on it lol) and maybe find someone whose work really interests me and would take me as a student

not too worried about where tbh, just maybe the person themselves

[sturdy marsh]

part of it includes a no-representation-theory proof of Burnside's pq-theorem
@next obsidian Which book?

[next obsidian]

Isaacs Finite Group Theory

[sturdy marsh]

ah

[next obsidian]

Okay so like

are you a grad student in some Algebraic field?

[sturdy marsh]

me?

[next obsidian]

You know a bunch of algebra

Yeah

[sturdy marsh]

im an undergrad

[next obsidian]

Hmmmmmst

You just know a lot of algebra

So I was like who is this dude

Lmao

Do you just like algebra?

[sturdy marsh]

ye

i'm not doing any algebra this quarter tho

[next obsidian]

Me neither

I’m doing AG tho

[sturdy marsh]

which is why i keep spamming this chat lmao

[next obsidian]

Which is

Eh

Substitute

[sturdy marsh]

yeah I gotta do more AG too

[next obsidian]

Where did you leave off?

[sturdy marsh]

hartshorne 3 stuff ig

we didnt do all of it

the stuff on flat families etc.

but yeah most of chapter 3

[next obsidian]

Rip

[sturdy marsh]

and some of chapter 4

[next obsidian]

I’m like in II.7

I finally made myself finish divisors

My experience with Hartshorne is like

Put myself through hell doing it

Then one particular thm or problem or whatever just breaks me

And then I avoid AG for a mo th

[sturdy marsh]

haha

[next obsidian]

Then I come back and push through lol

[sturdy marsh]

are you using any other books?

vakil?

[next obsidian]

I mean sort of

Yeah

But I like make Hartshorne the main

Which is hard cuz like

[sturdy marsh]

mumford's red book also also not bad

[next obsidian]

A lot of thms seem specific to Hartshorne

I couldn’t find equivalents in Vakil for like divisors stuff

And I supplement with stacks too

[sturdy marsh]

vakil has divisors

[next obsidian]

It’s just a mess overall haha

Yeah, but it like didn’t prove the same theorems I think

Since a lot of Hartshorney stuff was like specific results for varieties