groups-rings-fields

[tough raven]

Try reading https://agag-lassueur.math.rptu.de/~lassueur/en/teaching/COHOMSS18/CGSS18/Kap6.pdf or somewhere around page 800 in Dummit and Foote

[rapid cave]

Please don't send spam

[tulip otter]

well the conclusion i am coming to is that it means X-> X x Y is injective, which doesnt say whether Y is the image of this morphism or no

oh wait i might be confusing myself lmao

[knotty badger]

can you write out the actual equation it needs to satisfy?

[tulip otter]

let p:X x Y->X denote the projection and q: X->X x Y denote the section of p, then poq=id_X

[knotty badger]

mhm

now, $q : X \to X \times Y$ can be expressed componentwise

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[tulip otter]

what does that mean

[knotty badger]

it means any function $q : X \to X \times Y$ is of the form $q(x) = (r(x), s(x))$ for $r : X \to X$ and $s : X \to Y$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[tulip otter]

ah sure

[knotty badger]

if you plug this into this equation, what do you get?

[tulip otter]

poq=r=id

[knotty badger]

Mhm, so you need r(x) = x

So, what’s the form of q?

[tulip otter]

so q(x)=(x,s(x)) and so to each section q of p there is a morphism s:X->Y associated to it

[knotty badger]

mhm

and vice-versa!

[tulip otter]

right

[knotty badger]

given any morphism $s : X \to Y$, you obtain $q : X \to X \times Y$ via $x \mapsto (x, s(x))$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[tulip otter]

so there is a bijection S(X,X x Y)->Mor(X,Y) where S(X,X x Y) denotes the set of sections of p

[knotty badger]

mhm

sometimes it's more helpful to think of functions as sections

very important when you get to things like fibre bundles

[tulip otter]

I see. Actually I didnt really understand the section about categories and functors quite well, ig i will have to return to it soon

(lang has a section about this which is the section I am talking about)

[knotty badger]

i think this is mainly about how sections can be phrased categorically

[tulip otter]

alright tysm i will check it out

I think I didnt understand the things present in that section of the chapter because I was trying to finish it quickly lol

I am currently having this problem which is trying to finish these stuff quickly to get into the other things that look interesting to me like algebraic nt, algebraic geometry etc..

which is a wrong way of thinking lol

but ahh I cant really completely stop myself from thinking like that

tysm pseudo, have a great day/night!

[knotty badger]

[tough raven]

Not necessarily. But I think you can't do both at the same time (in the same subject). Got to pick one strat.

[tulip otter]

right but maybe its bad to try speedrunning a subject which is necessary to study the more interesting stuff if one develops a weak understanding of these fundamental basics as a result of speeding through them

[chilly ocean]

I’m not sure how, but it took you saying this for me to realize it, despite knowing what a product, section, and retract are for quite a while

So thanks

[knotty badger]

oh yay

[chilly ocean]

I never really got why it was called a section but this definitely helps to remember which is the section and which is the retract

[knotty badger]

.

like i imagine taking a cross-section of the product X x Y

[chilly ocean]

Yeah

That definitely helps

[tulip otter]

and then if you decide to read lang he tells you "forget about all this, what you are talking about is not a section"

so ig the standard is what pseudo told me

I will have to keep this in mind and remember that lang defines it in another way if i come across anything related to sections in the book

[fading acorn]

a section is an element of \Gamma(U,F) where F is a presheaf /j

[elfin wraith]

Same here lol

Retraction made sense, never understood why it was called a section until now

[tulip otter]

There always exists a schizophrenic way to define stuff huh

[knotty badger]

wait really

[elfin wraith]

Yeah really, I never made that connection until you pointed it out lol

I just kinda accepted that’s what it it’s called

[knotty badger]

Interesting

[knotty badger]

I guess I’m usually a lot less willing to do that

[tulip otter]

For each x in X, let A_x={(x,y) in C| y in Y}. I claim that C=\sqcup A_x where the disjoint union runs over all x in X

[knotty badger]

oh cool fiber integration

[tribal moss]

Weird terminology; I'd say "relation" rather than "correspondence".

[tulip otter]

First of all, its clear that C=\cup A_x

Now let $x\neq x_1$ be elements of $X$ and suppose that $A_x\cup A_{x_1}\neq\emptyset$, then $\exists (z,y)\in X\times Y$ such that $(z,y)\in A_x$ and $(z,y)\in A_{x_1}$ which then implies that $x=z=x_1$ so that $A_x=A_{x_1}$. Hence, $C=\bigsqcup_{x\in X}A_x$ and $#(C)=\sum_{x\in X}#(A_x)=\sum_{x\in X}\varphi(x)$

ah how to write # in latex

[cloud walrus]

ali yassine

[tulip otter]

ah sorry for the horrible typesetting lol

[tulip otter]

is this correct?

[tulip otter]

wait I meant A_x\cap A_{x_1} instead of \cup at the end of the first line

[tulip otter]

ohhh whats that?

[knotty badger]

it's a generalisation of breaking up an integral over a product into an iterated integral

[tulip otter]

ohhh nice

[knotty badger]

in this case you can think of $#(C) = \sum_{(x, y)} \chi_C(x, y)$, where $\chi_C$ is the indicator function for $C$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

so in a sense, you're "integrating" the indicator function over X x Y

then the formula for 17 just breaks up $\sum_{(x, y) \in X \times Y}$ as $\sum_{x \in X} \sum_{y \in Y}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

which is like an "iterated integral"

by definition, $\varphi(x) = \sum_{y \in Y} \chi_C(x, y)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

it's the result of integrating chi_C over Y while holding x fixed

[tulip otter]

woah interesting.

Actually the idea of partitioning C in this way came to my mind since it reminded me with the proof of $\sum_{d\mid n}\varphi(d)=n$ where $\varphi$ is the euler totient function

[cloud walrus]

ali yassine

[tulip otter]

where you partition the set {1,...,n} into sets A_d={k | (k,n)=d}={k | (k/d,n/d)=1}. Here (k,n) denotes the gcd of k and n

[knotty badger]

oh that's cool

[tulip otter]

tysm

[copper kestrel]

does this make sense? my professor said to find the highest lcm so this was my initial thought

i'll format it correctly and put all of the solution on one page when im done

[marble hinge]

does it mean that if we follow your reasoning the answer will be always just mn? for Z_m x Z_n?

[copper kestrel]

not always

only when m and n are relatively prime guarantees it

[marble hinge]

but 6 and 8 are not

[copper kestrel]

it guarantees it if theyre relatively prime tho

8 and 6 just happened to work

[marble hinge]

so when is it not mn?

[copper kestrel]

no idea

i just compute the lcms and hope for the best

[marble hinge]

there is a line saying LCM(48) = 48

in your reasoning

[rocky cloak]

This seems a lot more work than necessary, and you also got the wrong answer.

The order of an element (x, y) in Z/6 x Z/8 will be the lcm of the order of x and the order of y.

So what you should do is determine the largest lcm of a divisor of 6 and a divisor of 8.

I'm not sure how you got to lcm(16, 3)

[marble hinge]

then we can replace that line with LCM(mn) = mn (which is undoubtedly true) and have mn answer for all 🙂

[copper kestrel]

i thought thats what i was doing but i guess not lol

[rocky cloak]

Well, it's not what you're doing since you're taking the lcm of a lot more than two numbers, and you're talking about partitions. And 16 doesn't divide neither 6 nor 8

[copper kestrel]

my professor said to take the lcm of all the partitions so thats what i thought i did

idk what im supposed to do instead

[rocky cloak]

Well partitions of what exactly? I suspect you may have misunderstood them?

[copper kestrel]

most likely

i was thinking all the partisionf of 6 and 8

so i didi it in terms of primes and stuff like that

[marble hinge]

maybe "partitions" as if in 6 = 3 * 2 = 6 * 1, something like that? I.e. factoring those numbers in various ways...

[rocky cloak]

Well, like I said above.

The order of an element (x, y) is the lcm of the order of x and the order of y.

So what you should do is take the lcm of two things. One being the order of an element in Z/6 and the other the order of an element in Z/8

[livid portal]

Hello, sorry I don't know where to ask so I will ask here, is there an algorithmic way to get matrix representation for generators given a group presentation and generators form?

[marble hinge]

but yeah, I was also somewhat puzzled by this approach and terminology

[copper kestrel]

so i have to do that for all elements in Z/6 and Z/8?

that also seems like it would take too long so im guessing im wrong

[rocky cloak]

Well if your a little clever you might not need to check every element if you know that the order of elements in Z/n are, and are interested in the largest

[marble hinge]

yeah, you just need to look at two elements with orders that have the largest LCM

and there are only a couple of different orders anyway in Z_6

[copper kestrel]

how do i find out which one is the biggest besides brute force

[rocky cloak]

Do you know what the order of elements in Z/6 are?

[copper kestrel]

i know the order is the amount of times you have to add to get 0 mod 6

1 is of order 5

etc

[rocky cloak]

So 1 is not of order 5

If you don't know I can recommend computing the order of all the elements in Z/6

It is quite restrictive what the order of an element in a group can be.

[copper kestrel]

oh wait 1 is of order 6 oh my lord

[rocky cloak]

Indeed. What about the other elements? Does any of them have order 5 for example?

[copper kestrel]

0 is of order 1
1 is of order 6
2 is of order 3
3 is of order 2
4 is of order 3
5 is of order 6

[rocky cloak]

Very good.

Do you see any relationship between these numbers (1, 2, 3, 6)?

A relationship that would be relevant to lcms?

[copper kestrel]

well those are the factors of 6

[rocky cloak]

Indeed. It is in fact true that the orders of elements in Z/n are exactly the factors of n.

Anyway, now which is larger
lcm(2, x) or lcm(6, x)?

[copper kestrel]

i want to say lcm(6,x) but im scared of being wrong

[rocky cloak]

Alright, well maybe you can make an argument for why it should be true?

[copper kestrel]

6 is bigger than 2 LOL

and has more factors

so in the prime factorization it includes 2 and 3 instead of just 2

[rocky cloak]

Yeah, so this is morally what's going on.

Perhaps more formally we could say that any multiple of 6 is also a multiple of 2. So the smallest multiple of 2 should be smaller (or equal) to the smallest multiple of 6

So turning back to the original problem, what does this tell us about the element of largest order in Z/6 x Z/8?

[copper kestrel]

i would think that the smallest multiple of 6 should be smaller or equal to the smallest multiple of 8?

or smth?

if i get this wrong im gonna take a walk

[rocky cloak]

So I think you just need to remind yourself what you're trying to do.

You're looking at x the possible orders of elements in Z/6 and y the possible orders of elements in Z/8 and computing the largest lcm(x, y) can be.

[copper kestrel]

yeah

[rocky cloak]

Now what you've argued so far is that it should be enough to consider x=6

[copper kestrel]

yes

so i would think its similar for y=8

since we're considering the factors

[rocky cloak]

Indeed it is

So then the final answer is?

[copper kestrel]

so then lcm(6,8) is 24 i think

[rocky cloak]

And with this reasoning you can determine the same for Z/n x Z/m for any n and m

[copper kestrel]

so the largest order of any subgroup of a direct product of powers is the lcm(m,n) due to the fact that all subgroups must have orders of factors of m and n, implying that lcm(m,n) is the largest?

i just want to make sure im understanding

also how do people figure this sort of stuff out

bc in my book theres no theorems or examples about finding this stuff

[copper kestrel]

does this make sense at all?

i feel weird about getting an element inside sigma

but i know that each sigma is a permutation of elements so clearly there are elements inside of sigma

[dim wind]

I would think that sigma is a function

So I'm not sure about this

[vapid vale]

there aren't elements inside sigma, its an element itself

[copper kestrel]

yeah i knew i was a bit crazy for that

thanks yall lol

[vapid vale]

also, there are two sets here, the set (which is actually a group) of permutations, and the set of elements which this group is permuting

if you mean x to be an element permuted by sigma, then there is no notion of identity element in this set

[copper kestrel]

rats

[copper kestrel]

okay im still really lost on how to do this, by 2nd best guess is to use the fact that every permutation of a finite set can be expressed as a product of transpositions, and the fact that every permutation can be expressed as a product of disjoint cycles, then pair each orbit of sigma as a product of transpositions or smth?

but then i have to show that n-1 is the maximum and idk how to do that

i was originally thinking like, if we have the n-1, try a product of n transpositions, since we already have every element paired up with something, it must be mapped to itself?

like pidgeonhole principle sort of thing

[vapid vale]

have you tried working out explicit examples

choose a permutation of 1, 2, ..., n and figure out how to get to that permutation by just transpositions

[copper kestrel]

well i know that (1,2,3,4) is (1,2)(1,3)(1,4)

[vapid vale]

and figure out how to minimize that

[copper kestrel]

and i know we can do the same thing for a cycle (1,...,n)

[vapid vale]

right

if you know some statement is true for cycles, what sort of insight does that give you for an arbitrary permutation

[copper kestrel]

that its true since any permutation can be expressed as a product of disjoint cycles

[vapid vale]

i mean "its true" may not be right bc "it" is arbitrary here. but yes, the intuition is that you should be able to make the statement for each of the disjoint cycles of sigma

if a cycle has length k, how many transpositions do you need for it?

[copper kestrel]

at least k-1 since we can pair the left most cycle with all other elements in its orbit

idk how one would write out this proof tho

[vapid vale]

well if you're trying to minimize it you don't really care about the "at least" bit

we can do it in k-1 transpositions

[copper kestrel]

oh

[vapid vale]

and say your permutation is given as a product of disjoint cycles. what can you tell me about the length of all of those cycles

[copper kestrel]

that the length of those is the product of all the lengths since theyre disjoijt

[vapid vale]

what do you mean by the "length of those"

[copper kestrel]

length of the product of disjoint cycles is the product of the lengths of the disjoint cycles

[vapid vale]

i don't know what you mean by length of a product of cycles, but it cant be the order, because (12)(34) has order 2, not 2*2 = 4. it can't be the amount of elements being permuted because that would be the sum, not the product

[copper kestrel]

then idk what im talking about anymore bc i thought thats what was supposed to happen

[vapid vale]

to put it bluntly, for a transposition in S_n, write it as a product of cycles. what do you know is true about the sum of the lengths of each cycle?

[copper kestrel]

i dont know

[vapid vale]

would you agree that it couldn't be greater than n?

[copper kestrel]

yeah

i didn't know if that was the right thing to say though 😭

[vapid vale]

lol

well now you have some cycles of length k_1, k_2, ..., k_m, where k_1 + k_2 + ... + k_m =< n, right?

[copper kestrel]

ye

[vapid vale]

so i claim that you can now use this and you are basically at the solution

[copper kestrel]

i dont see how we can do that but i trust you

[vapid vale]

well how many transpositions do you need for the cycle of length k_1?

[copper kestrel]

less than k_1

i dont know im sorry 😭

[vapid vale]

would you agree with k_1 - 1?

thats what you told me earlier

[copper kestrel]

i would think so

[vapid vale]

right

and so generally for each cycle of length k_i, it can be written with k_i - 1 transpositions

but you know k_1 + k_2 + ... + k_m =< n

[copper kestrel]

i guess that makes sense

[tulip otter]

any hints for part a?

[velvet hull]

what is #(Gt)

[tulip otter]

the cardinality of Gt if thats what you are asking about

[velvet hull]

oh, is Gs the orbit of s?

[tulip otter]

yes

[velvet hull]

in that case ||Gt = Gs||

[tulip otter]

ah right lmao

and then we are done

a question that cant be given a hint without spelling out the answer

tysm

[sacred wharf]

blud when did they add hashtag to math

when was the new version update

math 2.0

we got math 2.0 before gta6

[rapid cave]

you haven't seen music symbols in math yet

$\sharp, \flat, \eighthnote$

[cloud walrus]

ExpertEsquieESQUIE

[sacred wharf]

how do u know i haven't

[rapid cave]

have you?

[sacred wharf]

no i haven't

[white oxide]

I think I'm going crazy how does a k-vector space V having finite dimension imply finite length

[karmic moat]

if you have a basis {v_1, ..., v_n} of V, then your longest chain will be 0 < span(v_1) < span(v_1, v_2) < ... < span(v_1, ..., v_n) = V

show that this chain is actually the longest

actually this should also give you a finite length composition series

[white oxide]

Hmmm

Maybe I could say that each quotient is simple, since the quotient would be isomorphic to a one dimensional subspace

in particular any submodule of this space must have dimension 1 or 0

[south patrol]

Ye

Length = dim by le induction (as both are additive and agree on dim 1)

[white oxide]

hence there cannot be any proper submodules and its simple or something like that

[white oxide]

in reference to this

[south patrol]

Proper nonzero ye

[rocky cloak]

Where does \eighthnote show up?

[rapid cave]

my class uses it for some morphism

the context is sheaves, direct and inverse images

[rocky cloak]

But like... What morphisms?

[rapid cave]

ahhh, ... some morphism in the proof of the adjunction

[rocky cloak]

Was it like you were already using sharp and flat and needed a third symbol?

[tulip otter]

in music theory

[rapid cave]

yeah

weird notation ig

[copper kestrel]

im backkkkkkkk the only thing i can think of is i know that a subgroup H can be the alternating A_n which has all even permutations i think?

or maybe thats the wrong way to go about it

im very confused sorry

[velvet hull]

assume that H has some odd permutation p

otherwise we are done

think about what you could do with it

or what you would want to show in order to prove that it's half even half odd

[copper kestrel]

assume H has an odd permutation p, then we know that it can be expressed as a product of an odd number of transpositions

im not sure what to do from that

hm

[velvet hull]

well, how might we be able to show that H consists of half even half odd elements?

[copper kestrel]

i apologize in advance i am quite dumb

let me check my textbook

maybe something like this will help?

no that doesnt help at all LMAO

im so sorry

[copper kestrel]

i really dont know

[velvet hull]

this is not a group theory question

it's a set theory question

[copper kestrel]

i got a C- in set theory

LMAO

i will need a bit more help because i still dont know

[velvet hull]

you could find some kind of bijection

or find injections / surjections going both ways

[elfin wraith]

Also, how do odd permutations interact? What do you know about subgroups?

[copper kestrel]

i do know about subgroups but now how they interqct

[velvet hull]

i suggest you think about this a little more closely

[copper kestrel]

we could do the sgn function maybe? thats the only function i can think of that really does smth like that

and splits smth into two different categories

[velvet hull]

well, we would want to show that the set of all even permutations in H have the same size as the set of all odd permutations in H

[copper kestrel]

oh

this set theory stuff always comes back to bite me in the butt

[elfin wraith]

I’m pretty sure you can do it this way, but I think this doesn’t really highlight the important understanding of subgroups that you seem to be missing

[copper kestrel]

great :')

[elfin wraith]

Could you tell me what a subgroup is? Not just like the definition from your book, but in your own words

[copper kestrel]

a group inside a group

it has to satisfy different properties tho

closed under inverses closed under the operation and has to have the identity element

[elfin wraith]

Yeah sure, it’s a smaller group, inside your bigger one, under the same operation. So importantly it’s still closed

[copper kestrel]

ye

[elfin wraith]

Now I want you to think about the product of odd permutations, and keep in mind the fact that subgroups are closed under that operation

And by think about, go pick some random odd permutations and actually just multiply them, see what you get

[copper kestrel]

well they work like in nimber theory

odd x odd is even
even x even is even
odd x even is odd

[elfin wraith]

So what, combined with closure, should that tell you about the structure of a subgroup which has an odd permutation

[copper kestrel]

closure?

oh wait being closed under the operation

sorry LOL

[copper kestrel]

it should be even

[elfin wraith]

What should be even?

[copper kestrel]

oh wait sorry i misunderstood

if you multiply by another odd permutation then it will be even

[tribal moss]

Random hint: ||"exactly half of them are even" means "the number of even permutations is the same as the number of odd permutations"|| and ||one can prove the latter if one has a way to pair up each of one kind with one of the other||.

[copper kestrel]

what do you mean by proving the latter?

[tribal moss]

Proving the second of the phrases I put in quotes in the first spoiler block.

[copper kestrel]

like which part is the latter

ohhhh

i dont know how i would do that

[tribal moss]

All the ingredients you need have been explicitly mentioned in the conversation so far -- try rereading it with this in mind.

[elfin wraith]

.

[copper kestrel]

how does that come into play

[elfin wraith]

Keep everything that’s been said so far in mind and see if you can come up with a way to combine it

[copper kestrel]

i'm sorry i must be really missing something

[elfin wraith]

We’ve given you all the ingredients you need

[copper kestrel]

tonight is gonna be a long night

[elfin wraith]

Maybe it’s helpful to explicitly work out all the subgroups of S_3, maybe S_4 too, and see how they behave

As a general piece of advice, if you’re struggling with a generic broad claim, pick a special case you can actually work with and experiment with it

[copper kestrel]

hm

so suppose H has an odd permutation, then we want to show that there is an even permutation that pairs up with it?

[noble nexus]

This is a good way to do it

[elfin wraith]

I think it’s good as a second way to do it, I’m not sure it’s as instructive.

[noble nexus]

Idk I think it's the most natural approach

b/c it's more group theoretical

more approaches is good though

[elfin wraith]

Yes but I think given the issues they’re having this approach is better at making you actually understand how subgroups behave and how permutations interact

I do agree the other proof is a bit “more algebraic” though

My position though is that if you really understand the definitions the proof I’m trying to nudge them towards should be pretty immediate

[copper kestrel]

ruh roh

i don't see any other way than how my approach was

[tribal moss]

You're supposing H has an odd permutation, okay, but you then need to prove that each odd permutation can be paired up with a different even one (such that no evens are left over).

[copper kestrel]

yeyeye

[tribal moss]

(There will generally be many different ways to do that; you just have to show that at least one possibility exists).

[copper kestrel]

or i guess we can say we can H has n odd permutations, then we want to show that there are n even permutations

[copper kestrel]

uh oh

am i on the right track at least

[tribal moss]

This is the track I'm recommending.

[copper kestrel]

okay

maybe we can decompose the odd permutation into an odd and an even?

bc odd x even = odd

[tribal moss]

That could work, with some additional manipulation.

You should probably start giving names to things so you can begin writing symbolic expressions.

Pick a letter to stand for the odd permuation you're assuming exists.

[copper kestrel]

o will denote it

[tribal moss]

Very well. Then you want to construct a function f: (H intersect odds) -> (H intersect evens) that you can later prove is a bijection.

[copper kestrel]

huh??

we havent done anything like that before

[tribal moss]

(Which is to say, start by making one such function that you can merely hope is a bijection).

You haven't worked with functions before?

[copper kestrel]

we have but ive never had to construct one for a proof

i dont think at least

ive been it be done in the textbook

[next obsidian]

Tabbycat

[copper kestrel]

hello :')

i am struggling with very basic algebra

[tribal moss]

To recap, we're trying to steer you towards a proof of the form:

If H consists of only even permutations, then we're done. So assume that H contains an odd permutation o.
Now define the function f : (H intersect odds) -> (H intersect evens) by
f(p) = [gap still needs to be filled out]
Then [gap still needs to be filled out] and therefore f is injective.
On the other hand [gap still needs to be filled out] and therefore f is surjective.
Since f is bijective, (H intersect odds) and (H intersect evens) have the same number of elements, which means that half of the elements of H are even, as desired.

I'm suggesting that the next step will be for you to guess wildly at how we could define f, and then see if your guess is lucky enough that it's possible to fill in the last two gaps.

[copper kestrel]

oh we can multiply by odd permutations to get to the even permutations

[tribal moss]

Yes.

[copper kestrel]

hmm but bijectivity is scaring me a bit

[tribal moss]

You're getting ahead of yourself. First define a candidate function, then try to see if you can prove it bijective.

[copper kestrel]

oh i defined the function to be f(p) = p * odd permutation

[tribal moss]

Which odd permutation?

[copper kestrel]

would any work?

[tribal moss]

You're the one defining the function.

[copper kestrel]

or did you want o specifically

[copper kestrel]

i would imagine it can be any odd permutation

[elfin wraith]

Any would work, but you’ve already given o a name and you know it exists!

[copper kestrel]

true

so then we'll choose o

[elfin wraith]

Good choice

[copper kestrel]

and since H is a subgroup we know o^{-1} exists too right

so then we can easily check injectivity

[tribal moss]

Yes.

(In fact for injectivity in particular you don't strictly need to know o^-1 is in H, but it doesn't hurt).

(And you will need it for surjectivity).

[copper kestrel]

surjectivity wants us to find q in H cap evens such that f(p)=q

[tribal moss]

For surjectivity you assume you're given q, and then you want to find some p that gives f(p)=q.

[copper kestrel]

right! so then we can solve for p in f(p) = q

so clearly p needs to be qo^{-1}

that means we're surjective right?

[tribal moss]

Yes, as long as you're sure that qo^-1 is actually in H intersect odds.

[copper kestrel]

it should

since o^{-1} is in there

[tribal moss]

Indeed it should, but that reasoning is part of your proof.

[copper kestrel]

ahhh i see

well obviously it is since odd times even is odd (i put it much more eloquently in my proof)

does this look okay?

after this i will need more help with the one i was talking about earlier

[tribal moss]

There's a ^-1 missing in the "qoo" line, but otherwise I didn't spot any problems.

[copper kestrel]

thank you!

[kind temple]

you could simplify this by noticing that f is right multiplication by o and f^-1 is right multiplication by o^-1. so you can eliminate injectivity and surjectivity by just arguing that f has an inverse

[tribal moss]

Now I think it would also be fair to write all of this simply as

If H has only even elements, then we're done, so assume it contains an odd permutation o.
Now, by multiplying each odd element of H by o (say, from the right) we get an even element, and this correspondence is reversible by multiplying each even element by o^-1. So we get an exact matching between the odd and the even elements of H, which shows that half of the elements are even.
except in order to hint you towards that instead of just revealing a solution, I had to start describing the thing in terms of explicitly named functions and so forth.

[elfin wraith]

Depending on how introductory this is and what you’ve already shown it could be worth noting why po is even, but I think this looks good!

[copper kestrel]

i like the hinting better and making my own solution with help personally

rather than just heres the solution in a very advanced way of lofic that i dont understand

[elfin wraith]

Of course!

Us just telling you the solution helps no one

[copper kestrel]

yeaaaa

[elfin wraith]

It may be worth exploring the idea you had with the sign function earlier

[kind temple]

there is a solution with the sgn function that might be worth fleshing out if you want to get better with quotients, cosets, and the first isomorphism theorem

[elfin wraith]

I think that also gives a very slick proof but I thought it was important that you understood why this works

[copper kestrel]

i would love to explore it eventually, but that will need to be another week because i need to finish this homework and go to bed to get up early tmr to study even more for my midterms on friday im super stressed about

this is the last problem im working on, i tried getting the n-1 in counting of a cycle but am a bit confused on where to go from here

[tribal moss]

This is one of the (somewhat rare) cases where two-line notation feels more useful than cycle decomposition.

[copper kestrel]

rats

then its back to square one

and now im even more confused LOL

[tribal moss]

I mean, it can also be done with cycles if you already know how to make a k-cycle as a product of k-1 transpositions.

[copper kestrel]

i can

maybe i didnt do it right in the explanation i gave

[tribal moss]

Oh sorry, I must be blind.

[copper kestrel]

its okay!

[vapid vale]

you have the statement for one of the cycles in the transposition

but its true for every cycle in the transposition

consider the permutation (1234)(56)(789)

[copper kestrel]

yep

[vapid vale]

you want to at least show that the theorem is true in this instance

[copper kestrel]

(1,2)(1,3)(1,4)(5,6)(7,8)(7,9)

[vapid vale]

right

[kind temple]

i guess you only have to show that this is true for n-cycles in S_n and the rest follows from strong induction

[vapid vale]

so how do you verify the truth of the theorem in this example?

i mean you can count it (assuming this is in S_9, since there are at least 9 symbols)

[copper kestrel]

it verifies it since its less than n=9

[vapid vale]

right

but more granularly, we broke 9 into 4 + 2 + 3

[copper kestrel]

yes

[vapid vale]

and by your claim about an arbitrary cycle, how many transpositions did we need for the 4, the 2, and the 3?

respectively

[copper kestrel]

3, 1, and 2

[vapid vale]

so no matter how you partition 9 into different summands, youre adding a bunch of things that are less than the summands. this can never be as big as 9

[copper kestrel]

yes

[vapid vale]

this is the essence of the idea – all you have to do to make this a proof rather than an example is let these be arbitrary by choosing names

rather than 9, say n

rather than a 4-cycle, 2-cycle, and 3-cycle, you have a k_1-cycle, a k_2-cycle, etc until k_m

[copper kestrel]

so let sigma be a permutation in S_n and consider sigma in disjoint cycle notation, [talk about how for each cycle c_1 of length k_1 you pair stuff up and make it a product of k_1-1 transpositions], then consider another disjoint cycle multiplied by c_2, and repeat for k_m, add all of them up and show theyre less than n?

but then to show its at most n - 1 we consider an n-cycle, go through the process and get out n-1?

[vapid vale]

considering an n-cycle tells you that you can actually hit the upper bound of n-1. its not necessary to show that, although its a good observation

[copper kestrel]

but is that not the whole point of the problem

like the whole point is to prove that thats the highest

i think??

[kind temple]

i think it is necessary if you are going to formalize it using strong induction

[copper kestrel]

whats strong induction?

[kind temple]

like, if the permutation is already a cycle of length n

[vapid vale]

the whole point is to prove that nothign can be higher than that

[copper kestrel]

ye

[vapid vale]

i would not take "at most" to imply a sharp upper bound. like the statement "every permutation can be written as a product of at most 2^n transpositions" is valid to me, although kind of useless

[copper kestrel]

interesting

[vapid vale]

in any case thats beside the point since the meat of the problem is an arbitrary permutation that you don't know the cycle type of

[copper kestrel]

ye

[copper kestrel]

how should i fix this then?

as much as i want to just give up and turn it in bc ive been working on these last two problems for like over 6 hours

[kind temple]

given a permutation p, decompose it into a product of disjoint cycles c_1 c_2 ... c_m.

each cycle c_i has length k_i

the idea is to apply the theorem inductively to each cycle c_i

[copper kestrel]

ew

[kind temple]

so you know that c_i can be written in at most k_i - 1 transpositions

this is why i said you needed strong induction

[copper kestrel]

yes i can explain how to do that

oh wait is it just basic induction? what makes it strong?

[kind temple]

you need it for every natural number below n

since you could potentially be applying it to a cycle k_i of any length less than n

[copper kestrel]

where did n come from

[kind temple]

n is the n from S_n

[copper kestrel]

oh right

[kind temple]

you also need to take care of one special case

when p is already an n-cycle in S_n

because its cycle decomposition is just p itself

[kind temple]

but this isn't bad, because it already looks like you know how to convert a cycle into a product of transpositions

[copper kestrel]

idk how to explain it well in a proof

[kind temple]

which parts?

[copper kestrel]

is it just "take the left most element in the cycle and pair it with the element directly after it, and put take that as a transposition. repeat this for each element going to the right to get k_i-1 transpositions"

[kind temple]

yea

[copper kestrel]

sick

[kind temple]

that is how to argue that an n-cycle can be written as a product of n - 1 transpositions

so you could even break this out into its own lemma

this is the crux of the proof

[copper kestrel]

id rather not if possible

ive never used my own lemma to prove anything

[kind temple]

i just think of them as helper functions.

but you don't have to do this

[copper kestrel]

i just want to keep it as simple as possible

bc thats all my brain can comprehend

[kind temple]

yea, then i would argue just like you did:

take a permutation p in S_n. decompose it into cycles c_1 ... c_m each of length k_1,...,k_m.
each cycle c_i can be written with at most k_i - 1 transpositions by <insert the process you described>
then the number of transpositions is (k_1 - 1) + ... + (k_m - 1) = n - m <= n - 1

ah, i guess i you don't need induction since you know the theorem directly for a cycle. my bad. you can still do it with induction, but it is not necessary

[copper kestrel]

meh?

[balmy python]

Looking at finitely generated abelian groups. I see that you can write the group as an isomorphism of the product of finitely many cyclic groups. Why is the infinite group part of the product not unique?

[kind temple]

<= at the end, but yea

very nice

[copper kestrel]

yay!

ty yall

[kind temple]

this would be fun to program

like, given a permutation of length n (represented as an array of length n), produce a minimal decomposition into transpositions

[copper kestrel]

i do not know how to program but it does seem fun!

[karmic moat]

macaulay2 does that i think

[kind temple]

is this a youtuber?

[karmic moat]

no, it's a language for commutative algebra

[kind temple]

sweet

[karmic moat]

hmm i can't seem to find it

[vapid vale]

macaulay2 :333

[karmic moat]

https://macaulay2.com/doc/Macaulay2/share/doc/Macaulay2/Permutations/html/_reduced__Words.html

[alpine plank]

It is unique. If two finitely generated abelian groups are isomorphic, they have the same rank

[celest ocean]

what should I do, having exam next week. I didn't study at all. What are easy to go topics in this course that I should not miss. Will do some previous years question to get secured. Sorry doesn't sound ethical. It's my last chance.

[spark veldt]

I'm trying to understand this statement in terms of group actions. Suppose $|G|=p^km$ with $\gcd(p,m)=1$. Let $\Omega$ be the set of all subsets of $G$ with size $p^k$. If $G$ acts on $\Omega$ by conjugation, suppose $P_1$ is a Sylow $p$-subgroup. Then the set of all Sylow $p$-subgroups is the conjugacy class cl$(\Omega)$ such that $P_1\in \text{cl}(\Omega)$. Am I right in saying this?

[cloud walrus]

bluepianist

[cedar vault]

well if you want to understand it in terms of actions i guess it'd be better to say that the orbit of P1 is the set of all sylow p-subgroups or maybe the set of all sylow p subgroups is an orbit

[candid patrol]

The conjugacy class makes sense for an element of Ω, the notation cl(Ω) is incorrect. Moreover, if Ω denotes the set of subgroups of order p^k, then Ω is nothing other than the set of Sylow p-subgroups of G. Sylow’s second theorem tells you that the action of G on Ω is transitive.

[tardy hedge]

Lol

[tribal moss]

What you say is true, but it looks a bit weird that you're applying conjugation to arbitrary subsets of G. There's nothing formally wrong with that, but it's strange that you're ignoring/avoiding the fact that conjugation applied to a subgroup (Sylow or not) always produces another subgroup. It's not clear whether you're doing that deliberately or because you're not aware of that fact, but it ends up sounding vaguely like the theorem is saying something (even) more surprising than it is.

[prime sundial]

What is ur fav theorem in group theory that has very genius proof? (if u dont have just tell me any intresting theorems to learn about that maybe i dont know, bcs there is some theorems i see ppl talk about idk where they found them, i nvr saw them before on the books)

[formal laurel]

I just love burnside theorem (every group which order have only two prime factors is solvable). And the proof uses very simple but clever techniques from the theory of group representations.

[prime sundial]

tysm its intresting

[harsh gale]

I believe basic group theory is pretty easy and intuitive if you go slowly and carefully

[tardy hedge]

i thought bro was NotKnow for a sec

@crystal vale lol

[harsh gale]

Schur–Zassenhaus theorem

pretty long and dense

And maybe construction of free group is a little difficult

[quiet pelican]

just say F_n is defined at the fundamental group of the n-fold wedge product of the circle

(Funnily enough it’s not that hard to prove the universal property using presentation complexes if you use that definition)

[delicate orchid]

yeah so make the definition even HARDER

[quiet pelican]

It kinda just shifts where the mess of the definition is from combi to topology
I wouldn’t really say it’s harder

I would say that definition is harder to work with though

[delicate orchid]

instead consider the forgetful functor Grp -> Set and construct an adjoint Set -> Grp as a Kan extenstion. Now consider the image of a set under this adjoint functor, this is a free group

[quiet pelican]

(Where by “definition” I mean “definition, and proving it’s actually a group”)

[harsh gale]

I prefer hard definitions and easy theorems

[delicate orchid]

topology is combinatorics

[karmic moat]

Definition/exercise:

[quiet pelican]

you say that but I’m definitely doing a lot of combi in my ggt

[karmic moat]

I prefer easy definitions and easy theorems

[proud vigil]

this isn’t really hard to prove but I really like how conjugacy classes partition a group into factors of the group size, it feels so randomly strong of a condition

[delicate orchid]

I say that completely unironically, any tractable topology is just combinatorics

[harsh gale]

that's what we call algebra, bro

[quiet pelican]

Prove CFSG 💜

[karmic moat]

is the proof of CFSG hard because there's so many cases, or hard because there are indvidual cases which are incredibly hard to prove

or is it both

[harsh gale]

it's a few hundreds of different theorems

[delicate orchid]

it's hard because you have to show you have no other cases

[quiet pelican]

Both

[karmic moat]

oh yeah true

[delicate orchid]

that's the difficult part of any classification

[quiet pelican]

Feit-Thompson is scary and that’s basically the starting line

[delicate orchid]

not to say the sporadics were easy to find

[formal laurel]

Feit-Thompson is huge beast on their own

[delicate orchid]

yeah without Feit-Thompson you'd be FUCKED

[karmic moat]

i will never read most of the proof of cfsg 😌

[formal laurel]

But like one is circle goes spining hard and the other is counting stuff which I don't like

[delicate orchid]

I can ONLY count things

[karmic moat]

any of wew's papers

W glaze

[quiet pelican]

I can count
1, 2, 3, 4, 5
That’s all the numbers!

[delicate orchid]

I saw a pretty funny construction where conjugacy classes of subgroups were in 1:1 correspondence with ORBITS OF F_p* acting on the power set of F_p*

[harsh gale]

I can count in any group you like

[formal laurel]

Me reading the proof of Cfsg:

[karmic moat]

yea basically

[quiet pelican]

Every finite simple group is 2-generated

Prove it without CFSG

[delicate orchid]

the humble C_2

unless you allow the same element twice

[quiet pelican]

That’s generated by a and e

[delicate orchid]

the identity as a generator makes me ill

[quiet pelican]

The identity is the best generator

[elfin wraith]

Youre not technically wrong

But i dont like it

[quiet pelican]

🙂

[agile burrow]

a and a^-1

[harsh gale]

as you complain about reading the whole proof of classification of finite simple groups. I still can't recall the list of all simple groups

[rocky cloak]

There you go

[harsh gale]

You are the best

[karmic moat]

dynkin ade 🤤

[rocky cloak]

This one also nice, but less funny

[formal laurel]

Basically ciclic stuff, alternating stuff, sporadic stuff and matrices

[rocky cloak]

lies, tits, monsters and pariahs

[twilit wraith]

what is "the things that really good movies and finite simple groups have in common"?

[copper kestrel]

im a bit confused on how my professor got the (p-1)! possibilities as well as a majority of the solution of a. like why does each subgroup contain p-1 elements of order p? why is the intersection trivial? why is each order of p contained in the subgroup? then we get the (p-2)!

i would think the (p-1)! is a theorem but i cant seem to find it

[karmic moat]

your prof fixes the first entry to be 1, then there are p-1 remaining slots in the permutation, so there are (p-1)! possibilities for choices

[copper kestrel]

ahhhh that makes sense, why can we fix it like that?

[karmic moat]

is this cycle notation? if so you can just move 1 to the front

[copper kestrel]

it is cycle notation

[karmic moat]

okay then yea you can just move the 1 to the front

like i could write (2314) as (1423)

[vocal pebble]

also note that once you fix 1, every permutation of the p-1 remaining things gives you a different permutation

[copper kestrel]

i dont understand why the number of elements is the number of possible elements, i would've thought that the number of elements would be p-1

[karmic moat]

you have p-1 choices for the second entry, then p-2 choices for the third entry, etc etc

[copper kestrel]

but why do we get the choices

i didnt realize that the point of the problem is finding all possibel combinations or order p

well actually i guess that makes sense

[karmic moat]

the intersection is trivial because p is prime and the order of a subgroup must divide p

[copper kestrel]

oh so the intersection of any subgroup must be <= p, right?

[karmic moat]

yeah

[copper kestrel]

like its order?

ah

[karmic moat]

like if i have two subgroups H and K, each of order p, then |H \cap K| has to divide p

by lagrange's theorem

[copper kestrel]

oh i didnt know it extended to intersection

i shall write that down

does it also work for unions?

[karmic moat]

the union of two subgroups isn't always a subgroup

[copper kestrel]

oh

[quiet pelican]

In fact, the union of two subgroups is only a subgroup in the trivial case that one contains the other

[karmic moat]

yeah

why is each order of p contained in the subgroup?

the claim is not that every element of order p is contained in the same subgroup, but that it is contained in some subgroup of order p; this subgroup is actually the subgroup which is generated by this element

[copper kestrel]

i missed that theorem

[copper kestrel]

hm

[karmic moat]

if σ is order p, then it generates a cyclic subgroup {e, σ, σ^2, ..., σ^(p-1)} which has order p

[copper kestrel]

oh wait right since every subgroup's order must divide p

and every subgroup of a prime order is cyclic

so every subgroup of order p is generated by an element of order p

[karmic moat]

yeah

[copper kestrel]

except for the trivial subgroup

[karmic moat]

the identity element has order 1

so it doesn't count

[copper kestrel]

yeye

wait but it still divides p

also question, why cant the intersection of two subgroups equal p? is it because the only element thats in both H and K is 1, so we get H \cap K = (e)?

[tribal moss]

The trival subgroup has order 1 not p.

[karmic moat]

they can, if H = K

[copper kestrel]

but does 1 not divide p

[karmic moat]

but I think it's implicit here that H ≠ K

[copper kestrel]

then i still dont understand lmao

[tribal moss]

It does, but the order is still 1 which does not equal p, and you're being asked about the subgroups that have order p.

[copper kestrel]

ah righy

thank you

[copper kestrel]

wait i just got it oh my god sorry

since its a subgroup it must have the identity

so their intersection must be trivial

[karmic moat]

Ye

If the two are distinct

[copper kestrel]

yeyeye

lord this practice midterm is so difficult

[karmic moat]

Keep working hard and asking questions and you’ll do well

[copper kestrel]

why is each permutation in A_4 be a product of two disjoint transpositions? why cant we have a 4-cycle? is it bc its not in A_4 since it is only expressed as a product of 3 transpositions?

wait no

4-cycle isnt of order 2

duh

[copper kestrel]

are the number of homomorphisms between groups the lcm of the orders of both groups?

[quiet pelican]

No

[copper kestrel]

no wait thats not correct

yeah

sorry LOL

[quiet pelican]

Consider like C_2 and C_3

[copper kestrel]

is there an easy method for finding them?

it cant be gcd i dont think

[quiet pelican]

Not really
Like there are some tricks, but they largely break down outside small finite groups, and some particularly nice infinite ones

[copper kestrel]

rats

how do you even find them then?

[quiet pelican]

There isn’t really a nice formula

[copper kestrel]

is there a strategy?

the solutions in my practice midterm said to compute the kernel

[quiet pelican]

Do you have a specific problem or is this a general thing?

[copper kestrel]

i have a specific problem but im worried its gonna be asked on my midterm tmr

i just dont understand the solution lol

[quiet pelican]

Essentially this is just “the order of the image of an element divides the order of the element” and “the size of the kernel divides the order of the group”
And seeing what we can deduce from those two fscts

[copper kestrel]

i dont understand the deducing parts from that, like i guess it makes sense but idk how to find do it myself

[prime sundial]

Hey, guys I have an idea idk if it's true, that says every group, we can define an order using the group operation, for example the usual order on R we can define it by a is bigger then b iff a-b is in R+

[quiet pelican]

It won’t be transitive on say Z/nZ

[prime sundial]

Ig the condition is it needs to be torsion free

[rocky cloak]

The main issue is what to you pick as your R+?

[prime sundial]

Wdym

[rocky cloak]

If G is an arbitrary group, how are you suggesting the order be defined?

a > b iff a-b is in what?

[prime sundial]

Yes ur right so we need to have a subset of G such that it's closure and it's intersection with the set of it's inverses is the identity and the union is G, right?

[rocky cloak]

Yes, if you have all of that you would get a totally ordered group

And if the group isn't abelian you might need this set to be closed under conjugation as well

[prime sundial]

Ok thanks

[vapid vale]

whats the earliest part that you get confused at

[sly rock]

Wow this is great!

[vapid vale]

have you tried finding a definition or expository article online? i have never heard of a nil clean group ring

although if the topic is so niche they will almost certainly define and introduce the concept

[peak root]

Let $M$ be a $n\times m$ matrix with entries $a_{ij}$. $R_{n,m}$ is the polynomial ring $\mathbb{C}[a_{ij}]$ with $nm$ indeterminates. $I_{n,m}$ is the ideal generated by all 2x2 minors of M. Denote by $S$ the quotient ring $R_{n,m}/I_{n.m}$. I know $S$ is a domain. Consider the ideal $J$ generated by $a_{11},a_{21},\ldots,a_{n1}$ in $S$. $S/J$ is isomorphic to $R_{n,m-1}/I_{n,m-1}$, so it is a domain, and $J$ is a prime ideal in $S$. What is the height of $J$ in $S$? I feel like it should be 1, but I do not know why. Or my guessing is not right?

[cloud walrus]

Dong_Valentino

[peak root]

If n=m=2, it is not very hard to check. Just prove (x,y) in the ring S=C[x,y,z,w]/(xz-yw) has height 1. I check this by hand. Any ideal J' contained in (x,y) will give some zerodivisor in S/J', so (x,y) has height 1. I think this proof should work for n=m=2, but how can I generalize to general n,m?

[tough raven]

Can you tell me why S is a domain? I've wanted to know how to prove this for a while.

[tough raven]

S should be the ring of functions on (the affine cone over) the image of the Segre embedding of ℙ^{n-1} ⨯ ℙ^{m-1}. Hence it should be (n+m-1)-dimensional. As you have remarked, S(n,m)/J is isomorphic to S(n,m-1) and hence (n+m-2)-dimensional. So J has height at most 1 (or S(n,m) would have a larger dimension). Of course it has height at least 1 because 0, J is a chain (geometrically, the affine cone mentioned cannot have an irreducible component of dimension n+m-2 because it is irreducible so it is its unique component and has dimension n+m-1. So this (n+m-2)-dimensional closed subspace is proper and has codimension at least 1).

[boreal tartan]

Guys, I feel like I'm going insane. I had a Modern Algebra exam today and one of the questions was: for the rings $\mathbb{Z}$ and $E={2x|x\in\mathbb{Z}}$, let $f:\mathbb{Z}\rightarrow E$ be defined as $f(x)=2x$. Prove that $f$ is an isomorphism.

[cloud walrus]

supimed

[boreal tartan]

But the thing is that it fails under multiplicative closure!!!!

I brought it up to my professor and he was like "no that's right"

I didn't want to argue with him like during the exam but surely it's not a provable statement right

[noble nexus]

yeah that's really weird, its an abelian group isomorphism but isn't a ring isomorphism no matter how you slice it

Z is unital and E is not so they cannot be isomorphic as rings

[boreal tartan]

Also $f(ab)=2ab$ but $f(a)f(b)=4ab$

[noble nexus]

maybe the question was intended for you to show they are isomorphic as abelian groups?

[cloud walrus]

supimed

[boreal tartan]

We haven't even covered groups lol

[noble nexus]

yeah that's uh pretty dumb

talk to the prof when you get the test back

[boreal tartan]

I'm glad I'm not loud and wrong about this LMAO

[kind temple]

this is silly, but like, if you define 2x * 2y := 2 (x * y), then it works out

so then 2 is the unit

and addition distributes

[peak root]

This might not be what you expect. As you said, the homogeneous ideal generated by 2x2 minors cut out the image of the Segre embedding P^{n-1}xP^{m-1} --> P^{nm-1} in P^{nm-1}. Hartshorne problem I.2.14 showed that the image is a subvariety. So it is irreducible and the corresponding ideal is prime, so S=C[a_{ij}]/(2x2-minors) is a domain. I do not know an purely algebraic proof.

[tough raven]

I know the geometric proof of irreducibility but this (or the version I know) only gives that the ideal has prime radical. How do you show that the image is reduced?

[boreal tartan]

Can you just factor out both 2s as a singular 2?

[kind temple]

no, not with the usual definition

but with this one, that is how i have defined it

[boreal tartan]

I'm not doubting you but it seems sketchy for that to be what my prof expected

I'll talk to him ab it

[kind temple]

if he meant this, he should have specified it during the exam

it's not the student's responsibility to figure that out

[peak root]

Ah, I see. We need to show the map from $S=C[a_{ij}]/I(n,m) ---> C[x_1,..,x_n,y_1,...y_m]$ sending $a_{ij}$ to $x_iy_j$ given from the Segre embedding is an injective map. The element $a_{ij}a_{kl}$ and $a_{il}a_{kj}$ has the same image in S from the relations in I(n,m), so we can freely permute the second indices for any monomial $a_{i_1j_1}...a_{i_rj_r}$ without changing to another element in S. Then for any $x_{i_1}...x_{i_r}y_{j_1}...y_{j_r}$ in the target, we first reordering the $a_{i_1j_1}...a_{i_rj_r}$ to make the first indices agree with the targer, then any ordering of the second indices does not matter as they are in the same class, representing the same element in S. This should prove the map is injective and S is a domain.

[cloud walrus]

Dong_Valentino

[tough raven]

I see. That's elegant. Thanks!

[peak root]

Thanks for the help.

[mint seal]

Consider a free group on two generators F({a,b}). I suspect the word a^2 b^2 is not the square of any other word. Is this true, and how might it be proved?

[quiet pelican]

It suffices to find a group G and an element of the form a^2 b^2 in that group which is not a square

(There’s also a nice way to do this with cyclically reduced words)

[mint seal]

lol, my whole motivation for knowing this about F({a,b}) is to have found an example of such a group G

[quiet pelican]

Ok
I’ll give you two theorems to prove, and that’ll give you your result

Theorem 1: every cyclically reduced word in F_2 which is a square, is the square of some cyclically reduced word

Theorem 2: the square of a cyclically reduced word has twice the length of the original word

Then you just need to show a^2 b^2 is not the square of any word of length 2

[mint seal]

cool, lemme think about those

I see how they give the result I want, and they feel very believable

I like it, thanks

[quiet pelican]

I’d like to find a small finite group which proves it if there is one

I think we can do (12)(3456) in S_6
So (12)(45) = (1524)^2, and (34)(56) = (3645)^2
Their product is (12)(3456)
Now the square of a 2n+1 cycle is a 2n+1 cycle, and the square of a 2n cycle is 2 n-cycles
So (12)(3456) can’t be a square as it has an odd number of 2 cycles

[quiet pelican]

(If you want a small explicit example i got one)

[mint seal]

very nice. I noticed the squares in S^3 do in fact form a subgroup, wasn’t sure how big you’d have to go to find a counterexample

[quiet pelican]

For symmetric groups, you need to go as high as S_6 because you need an even element with an odd number (or 0) of each type of 2n-cycle

So you need 2 different types of 2n cycle in one element, and 2+4 = 6 is minimal

[mint seal]

slick

[kind temple]

introducing any non-trivial relation into a free group will mess things up

[quiet pelican]

It’s not immediate that a^2 b^2 can’t be written as a square

[rocky cloak]

I guess just say w is a reduced word, and consider w^2.

Then w^2 = a^2 b^2 after reducing starts with a and ends with b, hence w must also start with a and end with b, so w^2 is already reduced.

But then w has length 2, so equals ab and abab is not a^2b^2

[kind temple]

my thought was that if there were a word w such that w^2 = a^2 b^2, then F_2 is now apparently given by the presentation <a,b | a^2b^2 = w^2> which is very much not <a,b | nothing>

[quiet pelican]

(This is, not coincidentally, basically the proof of “square root of a cyclically reduced word is cyclically reduced” for this specific case)

[quiet pelican]

If you had say the relation abab = (ab)^2 though, that’s not a contradiction

And it’s not immediate that there is no similar word w for a^2 b^2

[mint seal]

but aba^2 ba = (aba)^2 and that’s fine

[kind temple]

that's why i said non-trivial

[mint seal]

yeah but like, how can you tell if it’s trivial or not?

[kind temple]

i just could lmao

like in general this is undecidable

[rocky cloak]

So the proof goes like:
If a^2 b^2 was a square it would be so in a trivial way.
The problem seems nontrivial, contradiction!

[kind temple]

what are some heuristics for this

like

jagrs is one, analyze reduced words

i think abelianization works in some cases to dismiss potential relations

finding an explicit group where the relation doesn't hold

is it just like, throw all the usual tricks at it and see what makes it break?

[knotty badger]

that's my guess

i don't imagine there's a systematic way to do it

[noble nexus]

I think all of those are kinda special cases of finding an action where the two words act differently

which is I believe the main trick used to study free groups and really geometric group theory in general

[alpine plank]

Can we use covering spaces of S^1 v S^1 to give an argument for this?

[quiet pelican]

I’m not sure you can draw a good choice of covering space any easier than finding an explicit group
You can see the reduced word proof visually, but that’s just a choice of whether you wanna look at the argument visually or combinatorially

[prime sundial]

Can we assume the set of all finite groups such that their order is divisible by a fixed prime p?

[rapid cave]

Its probably not a set

[prime sundial]

I wanna proof that there exist no map F between that set and the union of all groups from that set, such that F(G) belongs to G(p), where G(p) is the set of all elements from G such that their order is p

I guess I did prove it but I assumed that it is a set

[rapid cave]

If it were a set you can use AC

[prime sundial]

Idk how to do that tbh

But does assuming it's a set really a problem?

Bcs I wanna show the proof to see if it's true or not

[rapid cave]

How did you prove this map doesn't exist

[prime sundial]

I did a restriction of the map on the set of finite groups that such that |G(p)|=1 and the intersection of G(p) and Z(G) is empty

if u applied inner automorphism it will give a contradiction since F(G) is in G(p)

[rapid cave]

Can you explain more?

Also you need to tackle the existence of such a group

[prime sundial]

Inner automorphism is phi_g(x)=g^(-1)xg, since F(G) (I will call it x) is not in Z(G), then there exist g such that phi_g(x) is diff then x, but automorphism preserve the order so phi_g(x) needs to be in G(p) but there exist no other element of order p then x, so phi_g(x) equal x and diff then x which is a contradiction

[rapid cave]

.

[tribal moss]

If we only want them up to isomorphism, we can have a set of representatives even without AC, since each of the isomorphism classes has a group where the underlying set is a subset of N.

[prime sundial]

Maybe no such group exists

Bcs if there exist a unique element of order n in general then that element is central, because conjugation of that element by any other element will give the same order but since no other element have the same order then g^(-1)xg =x

But ig we can modify smth in the proof to get a more correct one

[prime sundial]

.

[tribal moss]

What does "such a group" refer to?

[prime sundial]

A group where that has only one element of order p prime, and that element doesn't belong to Z(G)

[tribal moss]

If there's an element of order p, then there are at least p-2 others.

[prime sundial]

Yes based on the cauchy theorem sorry I forgot that

[tribal moss]

Hmm, perhaps you meant only one subgroup of order p?

[prime sundial]

Idk really but let me see if that makes the proof correct

[tough raven]

What you have showed is that there is no group G such that |G(p)| = 1 and the unique element of G(p) is not central.

[tough raven]

Assuming AC (and restricting to isomorphism classes to get a set) such a map does exist. Your argument suggests that you want to prove that there is not a natural choice of element in the sense of category theory. To prove that you just need to find G with order divisible by p such that no element of G(p) is fixed by all of Aut(G).

[rapid cave]

Is Sp an example for the last part?

[tough raven]

Sp?

Oh, S_p.

Uhh, maybe?

[south patrol]

∞-category of spectra

[tidal schooner]

symplectic group /j

[prime sundial]

yess ig so maybe what i am trying to prove is there exist no "canonical" map, right?

by canonical i mean every isomorphism between two groups of that set (lets call it D_p), sends F(G) to F(G')

then we can easily proof the contradiction

by the sam argument i gave before

[tough raven]

This is exactly formalised by the category theory notion of naturality (to be precise, a natural transformation of functors on domain the category of finite groups of order divisible by p with group isomorphisms as morphisms, from the functor which maps every group to {*} to the functor which maps G to G(p).

[prime sundial]

ok tysm

[south patrol]

Then it would be GSp

[tough raven]

Instructions unclear, took type A_n and got GLn(Spn(???)).

[white oxide]

Ok I was reviewing this proof and trying to verify this and I'm a bit stuck. The following diagram commutes:
[
\begin{tikzcd}
A && B \
\
{A_{\mathfrak{p}}} && {B_{\mathfrak{p}}}
\arrow["j", from=1-1, to=1-3]
\arrow["\iota"', from=1-1, to=3-1]
\arrow["{\iota'}", from=1-3, to=3-3]
\arrow["{\overline{j}}"', from=3-1, to=3-3]
\end{tikzcd}
]
I know that we have $\mathfrak{q}^{c^e} \subseteq \mathfrak{q}^{e^c}$. I'm stuck trying to prove the converse however. If $\overline{a} \in \overline{j}^{-1}(\iota'(\mathfrak{q}))$ then $\overline{j}(a) \in \iota'(\mathfrak{q})$ so that $\overline{a} = \iota'(q)$ for some $q \in \mathfrak{q}$. However, I have to show that $q \in j^{-1}(\mathfrak{q})$, which isn't necessarily the case

[cloud walrus]

okeyokay

[sacred wharf]

yo

cn some1 chck my proof on orbit stabilizer theorem ?

I want to prove

IGxI = IGI/IG_xI

Let Gx denote denote the orbit of x in G

Let G_x denote the elements that stabilize x

that is:
g dot x = x

Now let us consider the function that is g dot x --> gG_x (the coset of G_x, with representative g), we will show this is a bijective function

This is trivially surjective.

Injective as well since gG_x = vG_x --> g^{-1}v in G_x, .. (I can show injectivity from here)

Thus we have shown the cardinality of both IGxI and IG/G_xI to be the same, assuming one is finite, the other must be finite

We know that IG/G_xI = IGI/IG_xI

[south patrol]

I am confused where ur map is going from

Oh I guess you have written g.x so it depends on a choice of g and x but sure

Have you shown this is well-defined?

It is easiest to write the map the other way round as it avoids these choices.

Indeed it might remind you of a certain theorem lol

[sacred wharf]

No, I skipped it

I guess it can't be difficult to show

[south patrol]

Just yeah note the inverse is gG_x |-> g.x and it is more standard to show that that is well-defined

[sacred wharf]

suppose we have g dot x = h dot x
We will show gG_x = hG_x
if g dot x = h dot x
then we know that x = g^-1h dot x
g^-1h must be in G_x
and then yeah it becomes simple to show that gG_x belongs to hG_x, and we can use a symmetrical argument to show the other inclusion

is this correct

[sacred wharf]

yea i think this is simpler indeed

[sacred wharf]

but if you do this map, and then show surjectivity, don't you overall do the exact same thing but just switch the order ?

[south patrol]

Sure I just mean like

The map is surjective by definition of Gx

And g.x = h.x iff g^-1h.x = 1 iff g^-1h in G_x iff gG_x = hG_x

So I mean slightly more conceptual

[sacred wharf]

oui

[south patrol]

No u

But no like well done cause u proved it

[sacred wharf]

I would hope i'd be able to since i learnt this proof like 5-6 months ago in my first algebra course, this was simply review

[south patrol]

Ah ok

[sacred wharf]

yo

i be tryna prove no 2

and

this is what i get

fr

Using second iso theorem

KZ(G)/K is isomorphic to Z(G)/Z(G) n K

Suppose Z(G) n K = 1

KZ(G)/K has same number of elements as Z(G)

hmm

no

[proud vigil]

let G act on K by conjugation, which is well-defined since K is normal

you can apply orbit stabilizer to get a class partition of K, and the reasoning is very similar to i

maybe more directly, you can just take each of the classes and intersect them with K

[elfin wraith]

1. yes an isomorphism of groups has to send a generator to a generator. To see that, say we have $\phi:G\to H$ is an isomorphism, and G is generated by $g$. Then every element of $G$ can be written as $g^k$ for some $k$. Since $\phi$ is a homomorphism we have $\phi(g^k)=\phi(g)^k$ and since $\phi$ is a bijection, every element of $H$ can be written as a power of $\phi(g)$. Notice that we do need that $\phi$ is an isomorphism there, because we used the fact that its bijective
   \\
2. A group is cyclic if and only if it’s generated by a single element. From what we’ve said about, we know that if either one of the groups is cyclic, then so is the other one. In another sense you can think of this as the fact that isomorphism for groups means that, algebraically they are “the same”
   \\
   I do have to ask though, what is the group operation here? Because $\mathbb{Z}/2 \mathbb{Z} \times \mathbb{Z}/16 \mathbb{Z}$ isn’t cyclic (it is 3am so I’m possibly being dumb, but I’m 99% sure it isn’t)
   If all you want to show is that $\mathbb{Z} /64 \mathbb{Z} $ is cyclic, that’s straight forward assuming the operation is addition (which is typically assumed with that notation) because you can just generate it by 1.

[cloud walrus]

Nope

[elfin wraith]

Ahh of yes sorry I missed that. So yeah from what we’ve said in question 1) either your map rho isn’t an isomorphism or the operation on the LHS isn’t addition (but I suspect it is). You should convince your self the thing on the left isn’t cyclic (and maybe prove that Z/nZ x Z/mZ is cyclic iff m and n are coprime)

It’s quite late so I can’t be of any more help than that right now unfortunately, but yeah, hopefully that’s helpful, I’d have a think about what you’re trying to show here, and see if that gets you anywhere

Your proof that the product of cyclic groups isn’t cyclic isn’t quite sufficient, but that’s the idea

You can pretty easily prove that Z/nZxZ/mZ is cyclic iff n and m are coprime. Or you can just look try to make a more specific argument that there’s no element of order 32 in that group

By pretty easy I don’t mean that you won’t need to think about it, but it’s like 4 lines and doesn’t use any crazy ideas that you hadn’t seen before

This doesn’t work. Z/2ZxZ/3Z is cyclic of order 6

I do really need to sleep now, but think about the fact I’ve told you about products of cyclic groups. Work out some small examples first, like Z/2Z x Z/3Z and Z/2ZxZ/4Z and see if you can get the general idea

[torn thorn]

"Describe all groups G that contain no proper subgroup", Artin chapter 2 exercise 4.4

How do I approach this? I tried fiddling around with finite cyclic groups, but they don't seem to fit the pattern

[knotty badger]

take a nonidentity element of G (if there is one), and consider the subgroup it generates

[torn thorn]

Interesting 🤔. I'll take another shot at it

[sacred wharf]

Cn some1 explain whats going on here

?

[noble nexus]

imo the way this theorem is explained in most books is overly technical

basically because you are finitely generated, there's a map from R^n onto your module, and the goal is the theorem is to basically show that you can take the kernel of this map to be in a very particular form

[sacred wharf]

Isnt that exactly what is done ?

[noble nexus]

yeah but the matrix terminology makes things a bit confusing imo

The step you underlined is basically just saying that when you multiply the kernel by the chosen invertible matrices on the left and right the quotient remains isomorphic

Informally you should think of the theorem as follows: M is generated by some elements say $g_1,\dots, g_n$ and then there are some relations between them (captured by the kernel K) so we can write out the relations as a bunch of equations satisfied by the generators $r_{11}g_1+\dots+r_{1n}g_n$, $r_{21}g_1+\dots+r_{2n}g_n$ and so on (in fact you can even allow an infinite number of relations if you like, it just gives an infinitely tall matrix).

Then notice performing row and column operations doesn't change (up to isomorphism) the resulting module. If you've done group presentations, this is basically the same as when you change the presentation of a group by tweaking the relations.

[cloud walrus]

Blake

[noble nexus]

Oh I should mention all of the relations should have an "=0" at the end. Because R is a PID, you can row and column reduce into the specific form where all of your relations are just $a_1g_1=0$, $a_2g_2=0$ etc

[cloud walrus]

Blake

[noble nexus]

the theorem as written out in most books just kinda makes all this formal although tbh I think it's really a waste of time

[mental lake]

does anyone know a counterexample so that when you have R*~=S* as groups but not isomorphic as rings

[next obsidian]

Is * units

[mental lake]

just the notation for rings

oh wait no

[next obsidian]

Are you asking about when the additive groups are isomorphic

But not as rings?

[mental lake]

thats the full question

[next obsidian]

Okay so what does the * mean?

[rocky cloak]

Z[x] and Z[x, y] have the same additive group, the same unit group and the same multiplicative monoid, but are not isomorphic rings

It would need to mean unit group for the question to make sense surely

[mental lake]

yea

we havent seen the form Z[x,y] tho

[rocky cloak]

I mean for you exercise Z and Z[x] is enough, since you're just comparing unit groups

[mental lake]

okay ty

[noble nexus]

you can find a lot of examples by looking at multiplicative groups of the rings Z/nZ if you've seen those

[twilit wraith]

The * means additive group here for some reason

[spark veldt]

How is C(G) used here I'm kinda lost for np(G)>1 T.T

[noble nexus]

wtf that's unhinged

[wraith cargo]

centralizer of x

[spark veldt]

yeah right that sorry

[wraith cargo]

lol it's fine

[white oxide]

Anybody know what the dth powers means here

[tribal moss]

The dth powers are { x^d | x in U(Z/pZ) }.

[rocky cloak]

How many elements of order p does G have? How many are not if order p? What does this tell you about how many conjugates C_G(x) can have?

[prime sundial]

What is the full description of R? Does saying R is an ordered field enough?

[kind temple]

no

Q is an ordered field

[prime sundial]

R also is a field and ordered?

[kind temple]

there are a few descriptions of R. i think the algebraic description that you want is that R is the unique (up to unique isomorphism) complete archimedean ordered field

[prime sundial]

Yess

[formal laurel]

The only field R such that [R:R']<infinty where R' is its algebraic closure

[rapid cave]

C

[formal laurel]

ahh yes, 1<[R,R']<infinty

[south patrol]

There is a whole class of such fields

[white oxide]

This is driving me crazy, why is q^c^e contained in q^c^e?

Only field medalists can answer this question it seems

[south patrol]

https://en.wikipedia.org/wiki/Real_closed_field?wprov=sfla1

One good example being the real algebraic numbers I.e. Q bar intersect R.

[south patrol]

Isn't this by the hypothesis that q and q' both contract to p

[formal laurel]

Oh yes, me spreding missinformation twice

[white oxide]

Well I think the contraction here is given by contracting from B_p to A_p which is what I'm confused about

[south patrol]

Ye but I mean like you pass from A,B to these localizations ig

[prime sundial]

Isn't the algebraic closure of R is C?

[south patrol]

Yes

[rapid cave]

it is

[white oxide]

Sorry I still don't understand lol

[prime sundial]

So what does he mean by R' the algebraic closure I didn't understand rly what he meant

[white oxide]

The following diagram commutes.
[
\begin{tikzcd}
A && B \
\
{A_{\mathfrak{p}}} && {B_{\mathfrak{p}}}
\arrow["j", from=1-1, to=1-3]
\arrow["\iota"', from=1-1, to=3-1]
\arrow["{\iota'}", from=1-3, to=3-3]
\arrow["{\overline{j}}"', from=3-1, to=3-3]
\end{tikzcd}
]
By assumption, $\mathfrak{q}^c = \mathfrak{q}'^c = \mathfrak{p}$. Thus $\mathfrak{q}^{c^e} = \mathfrak{q}'^{c^e} = \mathfrak{p}^{e} = \mathfrak{m}$.
We claim that $\mathfrak{q}^{e^c} = \mathfrak{q}^{c^e}$. Indeed, let $\overline{a} \in \iota(j^{-1}(\mathfrak{q})) = \mathfrak{q}^{c^e}$. Then $\overline{a} = \iota(a)$ for some $a \in j^{-1}(\mathfrak{q}) = \mathfrak{p}$. Thus $\overline{j}(\overline{a}) = (\overline{j} \circ \iota)(a) = (\iota' \circ j)(a) \in \iota'(\mathfrak{q})$, since $a \in j^{-1}(\mathfrak{q})$. Therefore $\overline{a} \in \overline{j}^{-1}(\iota'(\mathfrak{q}))$. Conversely, let

[cloud walrus]

okeyokay

[rapid cave]

R there wasn't $\bR$

[cloud walrus]

ExpertEsquieESQUIE

[south patrol]

The statement (though false) is more precisely that R is the unique field F such that 1 < [F':F] < oo where F' is alg closure of F

[prime sundial]

Oh I see

[white oxide]

[
\begin{tikzcd}
A && B \
\
{A_{\mathfrak{p}}} && {B_{\mathfrak{p}}}
\arrow["j", from=1-1, to=1-3]
\arrow["\iota"', from=1-1, to=3-1]
\arrow["{\iota'}", from=1-3, to=3-3]
\arrow["{\overline{j}}"', from=3-1, to=3-3]
\end{tikzcd}
]
By assumption, $\mathfrak{q}^c = \mathfrak{q}'^c = \mathfrak{p}$. Thus $\mathfrak{q}^{c^e} = \mathfrak{q}'^{c^e} = \mathfrak{p}^{e} = \mathfrak{m}$.
We claim that $\mathfrak{q}^{e^c} = \mathfrak{q}^{c^e}$. Indeed, let $\overline{a} \in \iota(j^{-1}(\mathfrak{q})) = \mathfrak{q}^{c^e}$. Then $\overline{a} = \iota(a)$ for some $a \in j^{-1}(\mathfrak{q}) = \mathfrak{p}$. Thus $\overline{j}(\overline{a}) = (\overline{j} \circ \iota)(a) = (\iota' \circ j)(a) \in \iota'(\mathfrak{q})$, since $a \in j^{-1}(\mathfrak{q})$. Therefore $\overline{a} \in \overline{j}^{-1}(\iota'(\mathfrak{q}))$. Conversely, let $a/s \in \overline{j}^{-1}(\iota'(\mathfrak{q}))$. Then $j(a/s) = a/s = q/1$ for some $q \in \mathfrak{q}$. It suffices to show that $q \in A$.

Does anybody know why q is in A?

[cloud walrus]

okeyokay

[prime sundial]

Is this chasing diagram?

[cyan skiff]

Can someone correct if I am wrong. The question being is 2Z isomorphic to 3Z. i believe it is given by the mapping $2k \to 3k$. And a similar argument follows for 2Z being isomorphic to 4Z

[cloud walrus]

Taaha_Tariq

[cyan skiff]

How?

I mean every element in 2Z has a unique representation as 2k

[velvet hull]

wait, sorry, could you clarify what you meant by 2Z here?

I think I may have misunderstood

[cyan skiff]

2Z = {2k for k in Z}

[velvet hull]

okay in that case it works

you're good

[rocky cloak]

They are isomorphic as abelian groups at least

[cyan skiff]

What about as rings?

[quiet pelican]

they’re not rings, rings have identity
No, as 3Z has an element whose square is 3 times it, and 2Z doesnt

(Ie, x^2 = x + x + x has a solution in 3Z but not 2Z)

[cyan skiff]

Oh, then the same can also be said about 4Z and 2Z?

[quiet pelican]

Yup

[lusty marlin]

0²=0+0+0 in 2ℤ

So this should be "has a nonzero element whose square is 3 times it"

[sudden condor]

do there exist subgroups $A, B, C \subseteq \bQ$ such that $A \times B \cong A \times C$ but $B \not \cong C$?

[cloud walrus]

Emily

[rapid cave]

No

[sudden condor]

well uh

why

[rapid cave]

If f is the isomorphism then
f(0, x): B --> C is an isomorphism

[sudden condor]

no?

or like maybe but not a priori

how do you know its surjective?

(to be clear, when theyre subgroups of Q^2, this is false)

I know of the classification theorem of Baer of torsion-free Abelian groups of rank 1

and so ive tried reducing the problem to one where i analyze the sets of primes corresponding to the subgroup

ive shown that at the very least $\tau(B) \Delta \tau(C) \subseteq \tau(A)$, so it feels like i am very close

[cloud walrus]

Emily

[sudden condor]

(where Delta is the symmetric difference)

but im not quite sure how to finish it

[rapid cave]

https://www.jstor.org/stable/24490316?read-now=1&seq=1
Found this

I think it gives a counterexample

[sudden condor]

no it doesn't

$A \subseteq \mathbb Q$ but $B, C \subseteq \mathbb Q^2$

[cloud walrus]

Emily

[rapid cave]

Right

[sudden condor]

(his notation is different from mine—in his notation, its B and D)

[rapid cave]

So I have no idea

[sudden condor]

lol its okay the problem is deceptive

i nearly said "no" without proof for the same reason as you

[sudden condor]

as an aside did u find this paper in a math stackexchange post perchance

because if so i think that may have been me who put it there 😭

[rapid cave]

Are you "seirios"

[glad osprey]

Surely you can't be seirios

[sudden condor]

no, but uhh

[rapid cave]

Oh right

18h ago

Didn't see that

[sudden condor]

LMAO thats so funny

sorry dont mean to call u out

[vapid vale]

lol

[sudden condor]

okay i finished my proof

and then found that Baer also proved this in his paper fml

there exists no such subgroups A, B, C

which is really weird

torsion free abelian groups are so strange

[rapid cave]

Great

[rocky cloak]

I mean there not existing such subgroups is much less weird than if they did exist

[sudden condor]

i mean

in general there's no reason to think they shouldn't exist

as we see in the example that when we replace Q with Q^2 it doesn't hold

so its moreso the fact that like Q^2 but not Q has them that is odd to me

[rocky cloak]

My thinking was any homomorphism between two such A and B are given by multiplication by a rational number, so a homorphism between AxB and AxC is given by a 2x2 matrix.

This put very strong constraints on B embedding in C and C embedding in B and the fact that their rank 1 gives very little room to move around.

You also have a relatively simple classification of all subgroups of Q. Going to Q^2 it scales up massively.

So it's less surprising that everything becomes crazy at Q^2

[rocky cloak]

Q is like way smaller than Q^2 though

[sudden condor]

as a set theorist i take objection to this notion

but i know what you mean