Let H’ be another subgroup of order 7. Consider the size of HH’
#groups-rings-fields
prisma ibex
unique juniper
:o
final pasture
list all groups of order 28
unique juniper
~~list all groups of order 28~~
thats what i did lolol
celest brook
index
prisma ibex
|HH’|=|H||H’|/|H\cap H’| and since H and H’ are assumed to be distinct |H\cap H’|=1
So we’re done
final pasture
how the fact that the subgroup is unique implies the subgroup is normal ?
celest brook
prisma ibex
It’s not
viscid pewter
the intersect is a subgroup, which must divide 7
stupid q, why can you say that there are necessarily two subgroups of order 7 tho
final pasture
how the fact that the subgroup is unique implies the subgroup is normal ?
(I mean without any Sylow)
prisma ibex
You’re assuming there is and then you’re getting a contradiction
unique juniper
yes, sylow p subgroups anyway
final pasture
yes
prisma ibex
Hence it’s unique hence normal (since it must necessarily be closed under conjugation)
final pasture
Hence it’s unique hence normal (since it must necessarily be closed under conjug...
oh lol
viscid pewter
very sweet
prisma ibex
Okay yea this is a nice proof no Sylow
You should still copy the entire proof of Sylow I-III as an embedded lemma in your proof though
unique juniper
lol
prisma ibex
Yes, even the proof of Sylow boils down to this 
unique juniper
i still feel like it can be done differently
viscid pewter
eh, this is nice
prisma ibex
This is a nice enough proof
I don’t see how it can be made much more direct than this
unique juniper
$\varphi : G \rightarrow $ Sym($S$) where S = left cosets of $ \langle a\rangle$ a is of order 7. Its a homomorphism, call the kernal N so $G/N \cong Im \varphi$ i.e N is a divisor of 28 and (28/N) is a divisor of 24.
I got up to here on my own
Is this a dead end?
cloud walrusBOT
Yes
unique juniper
from this N is 2 7 or 14 right?
celest brook
oh using the fact it's characteristic to imply its normal
unique juniper
me neither
celest brook
Just prove contrapositive
unique juniper
@prisma ibex i didnt understand your argument
so we assume that H and H' dont intersect and of order 7
prisma ibex
Mhm
final pasture
no
prisma ibex
No
prisma ibex
Yup
final pasture
which is prime
so well, either the intersection is 7, and then they're not distinct
final pasture
or the intersection is {1}, which is what we want
unique juniper
and then?
final pasture
Then you take HH', you have card(HH') = card(H)card(H')/card(H \cap H')
so here card(HH') = 49
unique juniper
yeah
final pasture
but HH' C G
and card(G) = 28
so absurd
so there is only one subgroup of order 7 in G
unique juniper
how do we know that HH' is a subgroup
hot lake
it's not
final pasture
you don't need it to be a subgroup
hot lake
it's the set of hh' where h in H and h' in H'
unique juniper
hmm
prisma ibex
Err sorry
unique juniper
so theres only one subgroup of order 7
prisma ibex
Yes
final pasture
and now you take a f: x |-> gxg^-1 for some g € G
and f(H) is again a subgroup ('cause f is an automorphism) or order 7
so it must be H
so H is closed under conjugation
and so H is normal
cloud walrusBOT
5G enthusiast
unique juniper
hmm
final pasture
Well so I'll say it anyway since I do not care too:
take x € G with ord(x) != 1, then ord(x) | p so ord(x) = p, and then G = <x> so G is cyclic and a cyclic group is abelian
cloud walrusBOT
5G enthusiast
prisma ibex
Just reprove Sylow at this point
cloud walrusBOT
5G enthusiast
smoky cypress
Let $p$ be an odd prime. A video I watched claimed that to find an integer $a$ such that $a$ is coprime to $p$ and $a$ is not a square in $\bQ_p$, it suffices to find a nonsquare $a$ modulo $p$ by hensel’s lemma. I feel like this doesn’t need hensel’s lemma at all since if you have a square root in the p adics then you can truncate the digits to get a square root in modulo p
cloud walrusBOT
Whoever
oblique river
you have ti backward
you're trying to find a nonsquare in Q_p
not a nonsquare in F_p
if you want to find a nonsquare in Q_p
it suffices to find a nonsquare in F_p
by hensel
smoky cypress
Ah ok
rapid meadow
Hi everyone, I would like to get a nice description of the ring $\frac{k[x,y,t]}{(yt-x^2)}$, has anyone some idea?
cloud walrusBOT
sirfelipe
warped bane
hey everyone
I've been confused recently about the notion of vector spaces
is there, by any chance, a vector space E where E is NOT a ring?
rapid meadow
Well it’s isomorphic to k[u^2,uv,v^2] for starters
Nice, the morphisms is $f(x)=u^2, f(y)=uv$ and $f(z)=v^2$, right?
cloud walrusBOT
sirfelipe
celest brook
if vector space doesn't have cross product? idk
warped bane
and what if it had cross product?
celest brook
well there would be an operation of 'multplication' that would map two vectors to another
which is essentially what multiplication is in a ring
another operation that maps two elements to another element
doesnt necessarily have to be a group under multiplication either so
long beacon
i don't know what the identity is for cross product
warped bane
what confused me most was
celest brook
i mean integers mod n are a ring but they aren't a group under multiplication
in general
celest brook
mm
warped bane
unless n is prime
lapis aurora
its still not very much related to vector spaces
celest brook
ahhh
lapis aurora
how did you get from vector spaces to rings
warped bane
but what I believe is this vector space( field( ring( group)))
lapis aurora
its common to build rings on top of vector spaces but they usually arent rings themselves
warped bane
am I wrong?
lapis aurora
uh
either i misunderstand or that idea is wrong
if you mean "is-a" then no
vector spaces arent fields nor rings
warped bane
i mean from the groups as we add conditions we reach vector space
celest brook
lapis aurora
you only have an additive group in the vector space
lapis aurora
cross product is super useless
its like incredibly specific to one particular vector space
celest brook
i only thought of cross product because it's the only thing I can think of that maps two vectors to another
lapis aurora
the generalization takes more than 2 vectors
old lava
chad exterior product vs virgin cross product
lapis aurora
cross product is kinda unrelated to the vector space concept in general
viscid pewter
cross product is outer product
this doesn't feel right
celest brook
probably because it's the other way around
warped bane
so you guys are saying that E is vector space doesn't imply that (E, +, x) is a ring ?
lapis aurora
there is not even a multiplication operation in the first place
celest brook
it is never a ring
chilly ocean
it is never a ring
R over R
celest brook
mmm
lapis aurora
the vector space definition does not have a multiplication of vectors
it cant be a ring by its own definitions
its just that some rings happen to be vector spaces too
like fields for example
blissful ice
why are you all mentioning cross product, it's not even associative
warped bane
oh that makes sense
viscid pewter
why are you all mentioning cross product, it's not even associative
it was brought up, i replied, who even knows
warped bane
so a field can be a vector space while vector space is not necessarily a field?
celest brook
yes
lapis aurora
yes
warped bane
what a heartwarming news
celest brook
a field over a field over a field
lapis aurora
if you really want your rings, you can fix some vector space and consider its endomorphisms
you can add them and composition serves as multiplication
this will give you a ring
chilly ocean
more generally that works over any abelian group
lapis aurora
those are common rings
celest brook
good to know I am just uninformed and stupid
tight otter
I'm working on a problem regarding the polynomial $x^2+3x+4$ over the field $\mathbb{Z}_5$. I have that the two roots of this polynomial are $\alpha,\beta$ (i don't actually care about what they are) and I'm trying to show that $\mathbb{Z}_5(\alpha)=\mathbb{Z}_5(\beta)$. I've found that if $\beta\in\mathbb{Z}_5(\alpha)$, then $\beta=a+b\alpha$ for some $a,b\in\mathbb{Z}_5$, which gave me $a^2+2ab\alpha+b^2\alpha^2+3a+3b\alpha+4=0$, but I'm not sure how to proceed from here.
cloud walrusBOT
panoramatopia
tight otter
Obvs $\alpha^2+3\alpha+4=0$, and there are some terms in that equivalence that are similar, but I'm not sure how to extract the $b$
cloud walrusBOT
panoramatopia
tight otter
ah, just realized i can pick a and b to get the equation i want 
wait nevermind i can't 
doing so just gives me alpha = beta
golden pasture
nicer way to finish the proof is notice that the polynomial would factor into (x-α)(x-β) then you have α+β=-3
chilly ocean
Yeah, I was gonna say, you can just compute alpha and beta with the good old quadratic formula
chilly ocean
Maybe a nitpick, but "Z5(alpha)=Z5(beta)" makes me wince, if we don't specify a bigger field that everything is happening in
golden pasture
they are indeed equal as fields extensions tho not jus isomorphic
it's kinda like k(x)=k(x+1) thing
chilly ocean
How do you define k(x+1)?
golden pasture
the smallest field extension of k containing x+1 ig
chilly ocean
I would think this is the subfield of k(x) that is generated by k and x+1
chilly ocean
Whereas with k(x) is presumably fractions of polynomials in x
Annoying set theoretic issues 🙂
golden pasture
hence one ignored all of these issues
chilly ocean
I guess so. Also I'm getting a math minor and i have to take one of these two classes
next obsidian
What’s ur major or planned job?
chilly ocean
Computer science
" Re-examination of the calculus of functions of one variable: convergence, continuity, differentiation, the mean-value theorem, and the Riemann integral."
This is the course definiton
next obsidian
Yeah I assume it’s just real analysis
chilly ocean
For advanced calc
next obsidian
I don’t think there’s any real big difference in which would be more useful, so it’s kinda up to you I think
I like algebra more so I’d say to do that, but I think there’s a real argument for that
next obsidian
I assume you’ve seen a lot of similar stuff to the thinfs you’d do in the advanced calc class
Oh lmao
Don’t do abstract algebra then
It’s like linear algebra but cooler, but it’ll resemble that
chilly ocean
Lollll crap
final pasture
Well
next obsidian
I think linear algebra is important for computer science tho
And abstract algebra can be, but it again depends on what u do
chilly ocean
I can appreciate linear algebra , I do not like my grade or teacher 😂
final pasture
abstract algebra ends up being quite useful in a lot of theories in theoretical computer science
next obsidian
If u want to become a software developer at some company it doesn’t really matter much
But as they say if you wanted to do theoretical stuff you’ll need it
chilly ocean
Well out of all my A's in college Theory of Comp Sci and Linear Algebra are 71% C's so
im pretty bad at it
fuuu i dunno
I myself didnt really enjoy linear algebra but really enjoyed abstract algebra
Like ik its important but if i dont pass its gunna really screw up my graduation time
Hm
Its like do I do what i'm supposed to and barley pass , or take the easy way out lol
final pasture
Also "concrete" linear algebra like solving systems, finding ranks of matrixes etc and "abstract" linear algebra gives a really different feeling
and abstract algebra is a lot closer to the second one
So you may end up really enjoying abstract algebra as Godel did even though linear algebra wasn't that fun for you
chilly ocean
So you may end up really enjoying abstract algebra as Godel did even though line...
hmmm
only if i knew
I've done web dev and thats pretty fun, depends on the company tho
Yes actually alot depending on what you do
It could be really simple stupid math , or pretty complex 2d arrays and backend junk
You know, like you need to do 1+2 to know how many widgets to add to the website
final pasture
lol
chilly ocean
You know, like you need to do 1+2 to know how many widgets to add to the website
LOL
You can also incorperate literally anything into web dev, Machine learning, bots , all kinds of stuff
networking
final pasture
may be 2D/3D maths, combinatorics ? A lot of other things too, just giving examples 
chilly ocean
They are trying to make HTML handle VR rn
final pasture
more like webgl (or equivalent) handling VR properly, than HTML handling VR, I think, lol
final pasture
yeah right
unique juniper
guys if $|G| = p^kr$ gcd(r,p) = 1, $H_1 = {a \in G | a ^{p^k} = e}$. why is $H_1$ a sylow p subgroup? i dont see it
chilly ocean
or xml if you're special
But yeah web dev can go from baby poo poo creating a website with "wix " to really complex pretty fast
unique juniper
oops
cloud walrusBOT
Yes
unique juniper
G is also abelian *
unique juniper
not really
final pasture
(my game started, I leave this to someone else sry :p)
final pasture
thx 😄
next obsidian
So Yes. What’s the order of every element in this group?
It won’t be fixed, but it has to divide p^k right?
unique juniper
yes
unique juniper
yes
next obsidian
Okay, so do you agree that H_1 is a p-group in view of Cauchy’s theorem?
unique juniper
yes
unique juniper
hm
unique juniper
a subgroup that is the highest order of a prime that it could be
next obsidian
Right, but there’s an equivalent definition that might make what to do a bit more obvious, but it requires the use of a Sylow theorem
unique juniper
rigt
next obsidian
Do you know that theorem which says any p-subgroup is a subgroup of some Sylow-p subgroup?
unique juniper
yes i do
next obsidian
So this actually says it’s equivalent to say that a Sylow-p is a maximal p-subgroup of G
unique juniper
yes
next obsidian
Because if you aren’t a Sylow-p you properly embed into a Sylow-p
Okay so in view of that, say H_1 isn’t a Sylow-p subgroup
What can we then do?
Shut up simpurk
unique juniper
um find an element that isnt in H_1 but in a sylow p i guess
final pasture
um find an element that isnt in H_1 but in a sylow p i guess
(the p sylow need to contain H_1 though)
next obsidian
It actually wouldn’t
Because the mere existence of this element derives a contradiction
unique juniper
well
final pasture
oh ? how
next obsidian
Well the order of the Sylow-p is p^k
unique juniper
g's order needs to not be a divisor of p^k but its in a sylow p so it must have an order of p^k
next obsidian
Exactly!
unique juniper
yeah i got it
next obsidian
By being in a Sylow-p g^{p^k} = r
But then g is in H_1
So Shika-Blyat this fact alone means it doesn’t matter if the Sylow-p contains H_1 or not
unique juniper
Thanks @next obsidian
next obsidian
I think there’s one thing we havent done tho
unique juniper
ive done it yep
next obsidian
But a priori there’s no reason it is
Okay good
I assume this requires abelianness right?
Since H_1 as a set always exists
unique juniper
yep, G is abelian
next obsidian
It just might be too big
Kk
Just an important point I wanted to make sure you took into consideration
Like... H_1 s a set will always include every Sylow-p
But if G isn’t abelian it might be too big
I think that the condition you need specifically is that there’s only a single Sylow-p
final pasture
oh right, ok i misunderstood what you were trying to do
next obsidian
Then H_1 is a Sylow-p by like size considerations
unique juniper
right
final pasture
sry 
next obsidian
shut halo
I'm trying to get my head around representations. In the following theorem there is a lot of talk about the basis of a representation.
What does it mean for a representation to have a basis? To me a representation is just a group homomorphism and why does the form of the matrix represnetaiton affected by choice of basis?
In particular in the picture above, I don't quite understand point 2.50
next obsidian
It’s basically a change of basis thing
Like, the output R(e^itheta) is this n x n invertible matrix with coefficients in C
But then you can choose different bases on C^n which will mess with that matrix via something like a
P^-1MP
And so the claim is via a change of basis you can put it in that form
long beacon
What does it mean for a representation to have a basis? To me a representation i...
an element of gl_n is abstract-- as long as you turn it into a matrix, you need a basis
shut halo
So the matrices represent the elements of the groups by acting on vectors right?
long beacon
yeah
viscid pewter
random q, any intuition for the fact that any subgroup of index p in finite G is normal if p is the smallest prime divisor of |G|?
hmm
bc it was quite surprising in the book
eh, probs not important
chilly ocean
Bullshit "intuition": a subgroup of prime index is maximal, so there are only 2 choices for it's normalizer. So it's at least 50-50 shot if it's normal.
next obsidian
🤔
Kaisheng the only intuition I can really offer is that it's true when p = 2 by a simple argument that there's not enough space
Namely, one shows that gH = Hg
In the case when an index 2 subgroup exists you have that 2 is the smallest prime divisor of |G| since |G| = 2|H|
So something something it just holds when you replace 2 with the smallest prime divisor of |G| and there's something about the rigidity of the structure of groups (something something symmetry idk) that makes it so you couldn't have a different conjugate
jagged dune
Z[sqrt(3)] = {a + b*sqrt(3)| a, b in Z}
I can't tell if that's a field
I know it's an integral domain
chilly ocean
you might be able to find an explicit formula for the inverse of an element. maybe you can find one that doesn't live in Z[sqrt 3]?
if even one element doesnt have an inverse then it's not a field
chilly ocean
not quite, if by the conjugate of a + b sqrt 3 you mean a - b sqrt 3
jagged dune
yeah that's wrong
hmmm
The reciprocal is obvious but we're in the integers
yeah I'm unsure
chilly ocean
try to write 1 / (a + b sqrt 3) as something of the form c + d sqrt 3
hint: ||multiply the numerator and denominator by the conjugate||
jagged dune
hm
I'm honestly at a loss for some reason
is it its own inverse?
no that doesn't make sense nvm
chilly ocean
you can also look at it in the following way
given $a + b\sqrt{3} \in \bZ[\sqrt{3}]$, you want to try and find $c + d\sqrt{3} \in \bZ[\sqrt {3}]$ such that $$(a + b\sqrt{3})(c + d\sqrt{3}) = 1.$$ if you expand this out, you get $$ (ac + 3bd) + (ad + bc)\sqrt{3} = 1. $$ so you ought to have $ac + 3bd = 1$ and $ad + bc = 0$, right? now you can solve the system of equations to get $c$ and $d$. are they always integers?
cloud walrusBOT
TTerra
chilly ocean
it'll be a field if they're always integers, and it won't be if you can find just one a + b sqrt 3 for which c and d aren't both integers
jagged dune
gotcha
chilly ocean
(i've swept under the rug why it follows that ac + 3bd = 1 and ad + bc = 0. this is essentially because 1 and sqrt 3 are linearly independent over Q. you can try proving this separately, if you want.)
(but it's not really the point)
vestal snow
Did Zoph leave the server?
I wanted to tell him that I figured out the problem I was asking about for 2 weeks
snow flint
hes a weeb loser anyways (no, im not projecting)
chilly ocean
kek
vestal snow
Here's the solution if anyone's interested
Let $\phi(x)=\frac{f(x)}{\prod_i(x-a_i)e^i}$. Then we wanted to show that for infinitely many $a\in k$, $\phi(x)-\phi(a)$ does not have $a$ as a multiroot. This is the same as saying that for infinitely many $a\in k$, $(\phi(x)-\phi(a))'=\phi '(x)$ does not have $a$ as a root. If it did, then $\phi '(x)=0$. We also know that $y^p-y=\phi(x)$ so $-dy=\phi'(x) dx=0 dx$ which would imply $\frac{dy}{dx}=0$ which results in a non-smooth curve.
cloud walrusBOT
Have a Banana, Bitch
vestal snow
I don't think that this is something that is obvious in the paper because the word smooth is mentioned once at the beginning and never again, but I'm glad that I figured it out
latent anvil
Did Zoph leave the server?
Probably. There was a shitshow this morning
He was unhappy with the conduct of the mods
vestal snow
What happened?
oblique river
Oh zoph left too? Damn, two good users down. I guess i almost left as well
vestal snow
Who all left and what exactly happened?
oblique river
From what i know, ultra and zoph
It’s a long story and im on my phone and dobt wanna go throughnit all again..... maybe sham can cover me on this one haha
latent anvil
I am going to do French homework finally
chilly ocean
Will I need to change my server name too?
chilly ocean
please don't
chilly ocean
yes pls don't remove the bitch
vestal snow
I don't think generalizing is a good thing outside of math
vestal snow
yes pls don't remove the bitch
That's what I had to do when I joined my school's discord
But my actual name shows up in the Tex bot lol
Dang since Zoph is gone Chmonkey is the only one I can rely on to verify facts about my research without actually studying the material myself
next obsidian
snow flint
lol ur trash
drowsy quest
"the image of the exponential map lies in the identity component of G" is listed as an elementary property of the exponential map on wikipedia, but i don't see why... can anyone offer assistance 🙂
regarding a lie group G
chilly ocean
the image of the exponential map is connected (exp is continuous and \mathfrak g is connected) and intersects the identity component (exp 0 = e)
chilly ocean
it's a finite dimensional vector space
latent anvil
don't stare people it is rude
chilly ocean
latent anvil
Sean, \mathfrak g is homeomorphic to R^n
latent anvil
it's a finite dimensional vector space <:stare:826920038623346729>
Which is what tterra is saying here
drowsy quest
ahh i see
latent anvil
Yup, just choose a basis
latent anvil
It is very very funny
drowsy quest
thank you both
vestal snow
This proof seems too trivial
Cna someone verify
It asks to prove that $Q\otimes_Q Q\cong Q\otimes_Z Q$.
cloud walrusBOT
Have a Banana, Bitch
vestal snow
I'm worried I might be missing some detail
gritty adder
is the isomorphism supposed to be as abelian groups?
oh as rings
is that supposed to be the universal property of tensor products?
because it looks like you're showing something is a coproduct
sturdy marsh
tensor product is the coproduct in this case
gritty adder
do we know that a priori?
sturdy marsh
I mean tensor over R is the coproduct in the category of R-algebras. They aren't asking banana to prove it.
and looking at the proof, I assume that's what they're using
sturdy marsh
yeah people typically write the structure maps from the base ring
gritty adder
either way i think it's easier just to write down the map
prisma ibex
let's show that both of these are isomorphic to Q. For the first one this is clear. For the second one note that pushouts preserve epimorphisms and monomorphisms
hmm does this finish the proof hold on
A morphism of (commutative) rings which is both an epimorphism and a monomorphism and whose domain is a field should be an isomorphism
small bison
i would just do this by hand
gritty adder
yeah a tensor b -> ab
small bison
my brain is too small to think in terms of categories
ancient night
How can one disambiguate a ring of polynomial functions from a ring of polynomials over some field K? Usually, they are both written as K[x], aren't they?
chilly ocean
I would take k[x] to be the ring of polynomials, I have never seen this to mean the set of polynomial functions. Admittedly I have not seen a notation for the set of polynomial functions
ancient night
It's the notation offered by Wikipedia: https://en.wikipedia.org/wiki/Ring_of_polynomial_functions
visual hare
Hello, I would like to know why isomorphism implies that if AB=C then A'B'=C'
To do this, I have to prove that the isomorphism, call it f, has to satisfy: f(AB) = f(A)f(B), but how do I prove this?
gritty adder
That's the definition of an isomorphism
visual hare
All I know about an isomorphism is that its a bijection from G to G'
Let AB = C in G and let A->A' , B-> B' and C-> C'
now why would C' be equal to A' B' ?
carmine fossil
now why would C' be equal to A' B' ?
A group isomorphism is a bijection with that property
ancient night
"In abstract algebra, a group isomorphism is a function between two groups that sets up a one-to-one correspondence between the elements of the groups in a way that respects the given group operations" (c) Wikipedia
rigid cave
Hello! Is this a valid proof of Cauchys theorem? Let's look at all the $p$-tuples $a_1, \ldots, a_p$ where $a_i \in G$ such that $a_1 a_2 \cdots a_p = e$. There is $n^{p-1}$ such tuples. We can permute all the $p$ tuples in a cyclic fashion $p$ times so we get that $p$ divides $n^{p-1}$, ie. $n^{p-1} = k_1p$ for some $k_1$. If one $p$ tuple only consists of one element then it has order $p$. One such $p$ tuple is $e$. Assume that there exists $q$ such $p$ tuples. Then the remaining $p$ tuples can be permuted cyclically, so $p$ divides $n^{p-1} -q$. We get that $n^{p-1} - q = k_2p$ for some $k_2$. We then get that $q = p(k_1 - k_2)$ which can't be one so there is at least one element of order $p$ that is not $e$.
cloud walrusBOT
older sister
rigid cave
But I haven't used the fact that a prime p divides the order of the group, so I feel like something is wrong
Oh, I forgot to say that the order of G is n in the proof above
carmine fossil
You used p divides n in saying q=p(k_1-k_2)
@rigid cave
If p doesn't divide order of group, you have a nice proof of fermat's little theorem
rigid cave
If p doesn't divide order of group, you have a nice proof of fermat's little the...
Oh wow, look at that! Now it's time to prove fermat's last theorem
Guys, I am about to publish a proof of fermat's last theorem and I want opinions on it. We shall use the proof of contradiction. Assume that it's wrong. Then Andrew Wiles proof is wrong. But he's a genius so we arrive at the desired contradiction. QED.
chilly ocean
that makes no sense
rigid cave
This result is absolutely groundbreaking, you just need to understand it
unique berry
Here's how you prove it. Assume fermat's last theorem has a solution. This would produce a semi stable elliptic curve that is not modular but since every semi stable elliptic curve has an associated modular form it's a contradiction
Boom I just fit the proof into the margins
rigid cave
Naaahhh it's wrong. My proof is better
Here's another one. Proof: left as an exercise to the reader
vestal snow
If $G$ is right adjoint to $F$ and we have a commutative diagram $%
\begin{tikzcd}
& X \arrow[ld, "f"'] \arrow[rd, "g"] & \
G(Y) \arrow[rr, "G(h)"'] & & G(Z)
\end{tikzcd}$. Then is the following diagram also commutative?
cloud walrusBOT
Have a Banana, Bitch
vestal snow
$%
\begin{tikzcd}
& F(X) \arrow[ld, "N(f)"'] \arrow[rd, "N(g)"] & \
Y \arrow[rr, "h"'] & & Z
\end{tikzcd}$
cloud walrusBOT
Have a Banana, Bitch
vestal snow
Where N represents the natural isomophism between the hom sets
shut halo
My physics lecturer keeps talking about compact lie groups but hasn't even bothered defining it.
So what is really a compact Lie Group?
vestal snow
Do you know what a lie group is?
vestal snow
So Lie groups come equipped with a topology right?
shut halo
Ok so it's just a topological space with compact topology?
vestal snow
Yup
To be more precise
You can think of a lie group as a group object in the category of manifolds
shut halo
The problem is that physicists never mention topology but in my notes it is mentioned that compact means "group space is bounded". How do I relate to this notion with the topology of the manifold?
vestal snow
Closed is often the same as bounded
shut halo
So they are talking about topologically closed?
vestal snow
Maybe. You'll need to show us the context
terse crystal
Compactness can be used to define (global)one -parameter group of given vector fields as I understand.
shut halo
Ok here is a theorem in my notes: "If G is compact, then for every representation there is an equivalent unitary representation. This is analogous to Maschle's Theorem for finite groups"
timber grail
can I get help?
vestal snow
Usually if they don't have a prefix then yeah
shut halo
In this case we are basically just working with matrix lie groups so subsets of C^n I believe
I see so their compactness is just closedness I suppose
vestal snow
closed and bounded
timber grail
shut halo
Yeah right
vestal snow
@timber grail Try and think of simple elements of each
final pasture
(what's Z_5[i] ? )
vestal snow
I think it means adjoining a root of x^2+1
timber grail
(what's Z_5[i] ? <:tinkTonk:745526556750774383>)
I think this is it right?
vestal snow
I think this is it right?
Not quite
vestal snow
I don't like this definition
final pasture
more like Z[I]/5Z, even though that ends up being the same thing ig ?
timber grail
i'm lost
vestal snow
i'm lost
Try and work with 2 in each case
and see if you can find inverses
All of them are IDs though
vestal snow
I think we have the different definitions of Z_5[i]
timber grail
I think we have the different definitions of Z_5[i]
old lava
that's Z[i]
vestal snow
According to me, R[i] is "R adjoin a root of x^2+1"
final pasture
ok ? I mean you can see Z[i] as Z[X]/(x^2 + 1) ig, but I don't see your point
timber grail
which is the right one in your opinion
vestal snow
Oh wait my bad
old lava
maybe yall be using some whack shit, Z[i] is polynomials in i as far as I know (which is just elements of the form a + i b)
chilly ocean
maybe yall be using some whack shit, Z[i] is polynomials in i as far as I know (...
and how do you define polynomials in i?
vestal snow
I'm just stupid
old lava
Z[x]/(x^2+1) would be one way to do it
chilly ocean
yep
old lava
or you can just define it directly, Z[i] = {a + i b | a, b in Z}
which seems much easier
to work with (here at least)
chilly ocean
when you quotient by a principal ideal <p> you are basically setting a relation p = 0
old lava
I know
timber grail
and how do you define polynomials in i?
you really like to define things don't you
timber grail
I think he wasn't asking for definitions lol
terse crystal
2^2=-1 in Z/5Z😂😂😂
final pasture
so @timber grail, PorosInMyAshe already pointed out that there's 0 divisors in Z_5[i], so it can't be a field right ? Is it an integral domain ?
Do you think 2 is inversible in Z[i] ?
terse crystal
What do you think that Z/5Z[i] suppose to mean? Z/5Z is already the splitting field of x^2+1
chilly ocean
the great defining of 2021
terse crystal
Like I said it already contains two roots 3 and 2
chilly ocean
tinkywinky is dead.
vestal snow
Is this trivial after we use that every finitely generated module over a PID has a free resolution of length one?
rustic crown
yea but you'll be assuming that Tor doesn't depend on chosen free resolution
timber grail
sad
vestal snow
I'm not exactly sure what i'm supposed to do here
terse crystal
We can assume M, N to be finitely generated because any module is the direct limit of its finitely generated submodules and direct limit commutes with tensor product?
timber grail
i think they just got kicked
not really I just need help Im studying for a test
ya'll seem condescending
rustic crown
I'm not exactly sure what i'm supposed to do here
if M,N are finitely generated and you know the theorem, then very little... just write the resolution for M, tensor with N, because of the tailing 0s the homology is also 0
final pasture
so <@!691554624213090335>, PorosInMyAshe already pointed out that there's 0 divi...
up btw @timber grail
vestal snow
What if I don't know the theorem?
How do I verify it for the PID case
That Tor is not dependent on the resolution?
rustic crown
there was a section about that in aluffi, but i didn't read the properly
probably some judicious use of snake lemma
vestal snow
Wait I'm confused
Does allufi want us to take the result of independence of resolution in good faith and prove it in two lines?
Or does he want us to prove the result of independence of resolution for PID in this problem
timber grail
up btw <@!691554624213090335> <:chino_sip:676473025994031140>
thanks
its the first option
rustic crown
i think you're supposed to take than on faith, he promised the proof in last chapter which i still haven't read
like how else would you define Tor?
final pasture
yeah @timber grail 
timber grail
you da real mvp
rustic crown
like how else would you define Tor?
maybe you can use this kinda of definition
which he has outlined some stuff
:chino_sip:
vestal snow
Thanks.
rustic crown
this was on page 510
the snake lemma gives you a long sequence
that is basically the Tor sequence above
terse crystal
I think that independence of resolutions is because that given two projective resolutions P and Q of M , then 1_M from M to itself can be expanded to f from P to Q. Similarly g from Q to P then it’s easy to verify that gf is homotopic to 1_P and fg is homotopic to 1_Q using the property of projective modules
rustic crown
i left chapter 9 from the start of section on homotopy >.<
terse crystal
f is homotopic to g if there exist s_n from P_n to Q_(n+1) such that g-f = d_Q s + s d_P
magic owl
Direct product
terse crystal
11
magic owl
i.e. direct sum
rustic crown
i think it should be 5?
rustic crown
oh i'm bad
terse crystal
20 elements minus 8 units and the zero
timber grail
i need a tutor in this course
this practice sheet got me feeling stupid
@chilly ocean can you tutor?
timber grail
im having a hard time believing this isn't just some elaborate cheating scheme <...
same
vestal snow
@rustic crown do you think that this is a good enough answer?
timber grail
im having a hard time believing this isn't just some elaborate cheating scheme <...
we're taking the same class together and the instructor is just brain dead
so we need a tutor for a study group
need help with parts 3 and 4
magic owl
u could clear this up by sending picture of the full window
perhaps a title saying practice sheet or something
terse crystal
I believe that fundamental questions like those should be done by yourself...
golden pasture
^this but you'll need to find a reference to learn the topics
terse crystal
Unless you can’t find the definition of those concepts...
golden pasture
(any intro algebra book would have it just poke some and see what you like)
rustic crown
<@!671237809029251103> do you think that this is a good enough answer?
yea looks good to me
vestal snow
It just seems like literally writing out the definition
Hopefully I don't lose too many points
magic owl
wait why would a practice sheet
have an mc selector
why go through that extra effort lol
terse crystal
Could be an exam
rustic crown
Hopefully I don't lose too many points
maybe ask another one of your classmate?
magic owl
yeah i put this at like a
vestal snow
have an mc selector
I've had classes where that has happened
magic owl
10% chance of being legit
golden pasture
Hopefully I don't lose too many points
wait what points could be lost here 
vestal snow
But it is a bit suspicious
vestal snow
It's a homework problem guys
chilly ocean
It's a homework problem guys
you're good
chilly ocean
magic owl
okay combined w that message im going to go ahead an <@&268886789983436800> this is pretty sketchy w @timber grail and mamba
vestal snow
That's for when I take algebraic topology next year
magic owl
if u try to cheat ill alg top i'll cut u
terse crystal
Correct
golden pasture
houp
chilly ocean
golden pasture
chilly ocean
spring 2020 moment
golden pasture
wait when is spring 2020
chilly ocean
vestal snow
if u try to cheat ill alg top i'll cut u
Okay. How about complex analysis?
chilly ocean
Okay. How about complex analysis?
same but replace max with me
golden pasture
wait when is spring 2020
actually im very confused about this
never understood how these seasons things work
vestal snow
vestal snow
Yeah he's taking a test lmao
golden pasture
also wtf is the spacing lmao
tulip barn
okay combined w that message im going to go ahead an <@&268886789983436800> this...
What was the post? I see something was deleted, but not the image itself.
deft spade
The multi choice image
magic owl
So its like a combo of things
first they just posted it, when questioned they said it was a practice sheet
deft spade
The message linked from before
magic owl
which doesn't make a lot of sense w that format
deft spade
mentioning assignment
magic owl
Yeah and in a different channel they claimed it was an assignment
and they wouldn't show more of their screen or anything to demonstrate it wasn't a test
deft spade
Can you post a full screen screenshot of the work @timber grail
golden pasture
we're taking the same class together and the instructor is just brain dead
kindasus
terse crystal
~~That's for when I take algebraic topology next year~~
Algebraic topology has so many details. Gosh.like I doubt that there is any person who can remember how to prove the exactness of Mayer-vietoris sequence two months after reading it.
magic owl
Algebraic topology has so many details. Gosh.like I doubt that there is any pers...
you don't really need to
is the answer
golden pasture
you can kinda spend some time to fill in the fdetails if you realllly need
vestal snow
I think doing Aluffi would be really helpful with AT
vestal snow
Category theory + Homological algebra
magic owl
Ah yeah fair enough
In intro Alg Top courses the hom alg is usually pretty tame
but worth brushing up on
vestal snow
It's a grad course
magic owl
A little less predictable but usually still true
vestal snow
But I was told I didn't need to take undergrad AT before it
vestal snow
Are you into AT?
magic owl
Yes it is probably my PhD topic
magic owl
I'm about to graduate undergrad
magic owl
Well the ideas you learn in a first AT class
are completely necessary
for any algebra-flavored field
esp AG and AT
vestal snow
Gotcha
I'm really interested in taking a class in model theory/universal algebra
but there aren't any classes offered at my uni
terse crystal
But does universal algebra really contain much mathematics context?
Like I thought that it ended up as some kind of computer science...
I mean math at the beginning
vestal snow
There's some neat applications I saw which was something like "if a statement is true in an algebraically closed field of char p for for all primes p, then it is true in C"
There was some condition on the statement, but I don't remember it
terse crystal
I see,interesting
vestal snow
terse crystal
There's some neat applications I saw which was something like "if a statement is...
What is this theorem called?
Sorry I was blind
😂😂
chilly ocean
vestal snow
It's mentioned in the stack exchange post
rose axle
seeing math 490 made me go like :o
Because we have a math 490. Then I realized almost everywhere probably has a math 490
golden pasture
https://math.stackexchange.com/questions/1108667/what-are-some-applications-of-m...
woaawild
magic owl
Logicians care about universal algebra
its not totally niche or obscure
but i would say most mathematicians dont know much about it
bronze talon
hey all. Given two finite posets X and Y, does X x Y inherit gradedness, atomicness, co-atomicness, and/or diamond-ness? I'm having such a hard time keeping track of these definitions rn and would appreciate some guidance
chilly ocean
atomicness (and dually coatomicness) seems trivial to prove that are inherited
cuz if (x1,x2) > (0,0) in XxY then xi>=bi>:0 where bi are atomic so (b1,b2)>:(0,0)
: means > and nothing in between
magic owl
Okay I have been stuck on what I think should be a simple degree computation for awhile and I cannot tell why I am being dumb
So let’s say we are starting with a polynomial algebra P(x_1,x_2,...) where the degree of an x_i is 2^i-1
Let us further equip this polynomial algebra with the following coproduct
$\Delta(x_n)=\sum_{0\leq i\leq n} x^{2^i}_{n-i}\otimes x_i$
cloud walrusBOT
MaxJ
magic owl
Okay now suppose y is an element of this polynomial algebra and you know two things about y
- y has even homogenous degree
- Delta(y) is a sum of even degree terms
Claim: y must live in the subalgebra generated by the squares of the $x_i$, i.e. $P(x_1^2,x_2^2,...)$
cloud walrusBOT
MaxJ
magic owl
I am reading a very terse proof of this fact (in fact this is like two words in a longer terse proof)
And it should be trivial i.e. I think it should be straight from degree concerns
but I keep doing computations with toy examples and I am not convinced it is true
So I was wondering if anyone has input
(For those who do not trust my ability to copy the information here, which is fair, the claim is 3.1.6 on page 61 of Ravenel’s Cobordism and Stable Homotopy)
——————————-
Okay so maybe I will try to find a counterexample. Clearly the degree of any $x_i$ is odd, so any product $x_ix_j$ is a potential choice for $y$ satisfying the first thing we know
cloud walrusBOT
MaxJ
magic owl
Then $\Delta(x_i)$ is a sum of terms $x_{i-n}^{2^i}\otimes x_{n}$. These terms should have degree $2^i(2^{n-i}-1) + 2^{n}-1$
cloud walrusBOT
MaxJ
latent anvil
I'm not sure I agree. I feel like you've swapped i and n
magic owl
Then the product of $\Delta(x_ix_j)=\Delta(x_i)\Delta(x_j)$ should have terms that look something like $x_{i-n}^{2^i}x_{j-m}^{2^m}\otimes x_{n}x_{m}$
cloud walrusBOT
MaxJ
magic owl
I'm not sure I agree. I feel like you've swapped i and n
i think i literally swapped i and n
but thats because im bad at notation
not the math
right?
(my original defn is in terms of x_n)
latent anvil
When we define Δ(xn) we get stuff tensored with xn on the right
latent anvil
magic owl
the definition is wrong the computation is correct
latent anvil
hahahaha
magic owl
okay so let me edit that for posterity
oh texit isn’t updating it it got bored
well $\Delta(x_i)=\sum_{0\leq n \leq i} x_{i-n}^{2^{n}}\otimes x_{n}$ is the correct defn
cloud walrusBOT
MaxJ
magic owl
Then the product of $\Delta(x_ix_j)=\Delta(x_i)\Delta(x_j)$ should have terms th...
Back to this, the left and right terms should both be even here
so I feel like this is a counterexample?
I wonder what im messing up im sure its very dumb
latent anvil
Yeah, I agree
magic owl
(i guess its also possible there is something else I know about y that makes this work but that the book didn’t mention)
okay well at least i am not crazy
I wonder if I can find this claim elsewhere it should be in some paper of milnor
chilly ocean
<a:himb:829995022542962768>
latent anvil
<a:himb:828417606171099158>
magic owl
@sturdy marsh maybe you will recognize this as the dual steenrod algebra and be familiar with the claim idk if you care about these computations
i tried to avoid using \xi but it ended up being more work bc i wrote it out of reflex
viscid pewter
it is??
unique juniper
lol
cloud walrusBOT
Yes
unique juniper
mb
viscid pewter
oh isn't lmao
unique juniper
ok
so what ive done so far is
Sylow 3 has order 3 and there are either 4 or 1 of them.
If there are 4, they trivially intersect
so thats 8 distinct elements that are in sylow 3
but then what?
i cant do something similar with sylow 2
will sylow 2 groups and sylow 3 groups ever intersect?
quaint tree
A problem I've been churning over in my head that I've been unable to solve: Are there any fields ${G,+,\times}$ where the groups ${G,+}$ and ${G^{},\times}$ are isomorphic? (Where $G^{}$ is just $G$ but without 0 so that it can be a group) I haven't been able to think of any, but I also haven't been able to think of a proof that they can't exist.
cloud walrusBOT
Bannanachair Monarch
viscid pewter
F2 
unique juniper
unique juniper
too late
viscid pewter
wait they can't ever be isomorphic because they don't have the same number of elements, surely
or well
the identity of one doesn't map to the identity of the other??
wait no hmmm
cursive temple
It suffices to check the case where the characteristic is 2 right
Because F^* has an element of order 2
And F infinite
chilly ocean
wait they can't ever be isomorphic because they don't have the same number of el...
you are assuming the field is finite?
viscid pewter
yeah i saw the problem right afterwards
cursive temple
Because F^* has an element of order 2
Namely -1
chilly ocean
also i think it needs to be characteristic 0. {G,+} has elements that satisfy px=0, but {G*,x} doesn't
er
fuck
viscid pewter
i feel like this should never work
cursive temple
Ye there arent any
viscid pewter
but i am a complete rube at field theory so
chilly ocean
i think what i said above is right, x^p=x in a field of characteristic p right?
cursive temple
I just know the for 1+1+1+...+1 = 0 where there are p ones
chilly ocean
or was it x^{p^n}=x in a finite field of order p^n?
cursive temple
The case where the field is of char 2 is also pretty easy to figure out
Just note that if F is of char 2 then for every x in F^* you have x^2 = 1
But x^2-1 = (x-1)^2 which clearly only has one root
chilly ocean
(x+y)^(p^n) = x^(p^n) + y^(p^n) in a field of characteristic p
but we are interested in the infinite case
and i believe even simpler, (x+y)^p=x^p+y^p, EDIT: er, maybe just different, not stronger
that's special case of one i said
what is n?
natural
shut halo
What's a nice introductory book for Lie Groups and Representation theory for beginners?
viscid pewter
doing the introductory group action questions and it's weird bc they seem to oscillate between trivial and hard to even start on
probs they'd all be trivial if i really understood them 😔
hot lake
fulton harris
viscid pewter
no
i will just meditate on them and i will find that they are trivial
or else i will just ignore them and move onto more interesting bits of the textbook and come back later maybe
viscid pewter
ok today is not a day for doing exercises
i cannot
today is a day for just skimming the rest of the chapter
sturdy marsh
<@351398829238648832> maybe you will recognize this as the dual steenrod algebra...
I did not read through the computations properly. There might be something of use in https://people.math.rochester.edu/faculty/doug/otherpapers/Milnor-A.pdf
magic owl
i ended up figuring it out
and im now stuck on the next part
which i also cant find a reference for
lavish pike
so the question in the book asks me to "find the order of each group" then lists $Z_{18}, S_{4}, S_{5}, D_{4}, U_{18}$ but i have no clue what S, D, and U represent
cloud walrusBOT
NocuousNick
scarlet estuary
likely the symmetric, dihedral, unitary groups respectively
if youre not familiar with those, check your textbook for a definition
chilly ocean
this is a weird question. i don't know what U is, but the other groups, the order is basically definitional
lavish pike
alright and like finding a specific order for a set how do you do that? we went over like a singular element but not a whole set
scarlet estuary
"order of a group" means "number of elements in the group"
its a different definition than order of an element
(though related)
seems like you missed a couple things, id reread the textbook chapter
oblique river
yes that is the idea
vestal snow
I'm trying to prove that kernels are limits using this definition
Let $\phi:M\rightarrow N$ be morphism with kernel $K$ and the map $i:K\rightarrow M$
cloud walrusBOT
Have a Banana, Bitch
vestal snow
Is the key here to define $I$ as $Obj(I)={M}$ and $Hom_I(M,M)=id_M$?
cloud walrusBOT
Have a Banana, Bitch
vestal snow
And defining C as the category of things like $A \xrightarrow[]{f} M$ where $\phi\circ f=0$ and morphisms as morphisms that force commutativity?
cloud walrusBOT
Have a Banana, Bitch
vestal snow
I haven't worked out the details, but it seems like this is the only thing one can do
The only thing that's bothering me is that $I$ in this proof is not really used at all
chilly ocean
I'm trying to prove that kernels are limits using this definition
May i ask what book is this?
actually nvm i can just search it lol
Algebra Chapter 0
vestal snow
It's a pretty good book
chilly ocean
I'm trying to prove that kernels are limits using this definition
do you know what a pullback is
vestal snow
I think so?
Isn't it like if you have f:A--> B and g: B--> C
then the pullback is gf
chilly ocean
oh, no
i mean the limit of this diagram:
[
\begin{tikzcd}
& A \dar["f"] \ B \rar["g"] & C
\end{tikzcd}
]
cloud walrusBOT
vestal snow
Oh yeah
Isn't that the final object in the category I described above?
Similar to that
Not exactly
But like you consider the category where the objects are diagrams from X into A and B
which commute with f and g
chilly ocean
yes
vestal snow
Yeah
chilly ocean
have you computed pullbacks in Set yet
vestal snow
we haven't given them a name
But I think that's basically what I'm doing right?
We know that the Kernel is the pullback in the category I described earlier
And we're using that to prove that its a direct limit
My question is not very conceptual
vestal snow
My bad
chilly ocean
yeah it's confusing
vestal snow
I'm just making sure that my proof is valid
Here's the detailed argument
I'll send a picture because diagrams are a pain to Tex
chilly ocean
i have to study for my exams now so i can't verify your proof unfortunately, but here is a hint: ||in Set, the limit of the above diagram is {c in C | f(c) = g(c)}||
unique juniper
$G$ is a group where every sylow group is normal. Show G is isomorphic to the direct product of its sylow groups
cloud walrusBOT
Yes
vestal snow
This seems like an induction argument to me
unique juniper
why
vestal snow
It clear that this is true when G = 1
unique juniper
But if we know that G = sylow groups multiplied that does it
since sylow groups dont intersect
and they r normal so they commut with each other
vestal snow
Yes
I guess you wanna somehow show that given a G and a sylow subgroup P normal in it
G/P x P is isomorphic to G
I don't know if this is even true, but it seems like a good starting point
unique juniper
cant we just use the fact that
If H,...,Hn subgroup G and they all commute with each other and trivially intersect and $G = H_1 H_2 \dots H_n$ then $G \cong H_1 \times \dots \times H_n$
cloud walrusBOT
Yes
unique juniper
this would do it right?
rustic crown
you need to be more clear about "trivially intersect"
vestal snow
pairwise i think
unique juniper
intersect with just the identity
unique juniper
like H ^ H_2 = 1
unique juniper
:o
rustic crown
or something similar