it is a bit suspicious though
#groups-rings-fields
rose prism
as it is not even clear how to treat the case when prod ^G_i is not discrete
like in particular, the product of locally compact spaces need not be locally compact
coral spindle
Think about which property of groups they are trying to show here. You will likely realise it if you read the sentence in full
spark veldt
oh i see
another qn
When proving subgroups, do you have to show identity? I feel like some texts say you should, but others not.
rocky cloak
When proving subgroups, do you have to show identity? I feel like some texts say...
You have to prove that the set is nonempty. Once you have that it follows that it contains the identity
spark veldt
It follows that because the inverse exists?
rocky cloak
Yeah, x times it's inverse will be the identity
spark veldt
The reasoning works because I can find $a^{-1}$ so that $aa^{-1}=e$. Here I'm assuming there is $e\in A$. Is this not circular?
cloud walrusBOT
bluepianist
rocky cloak
The reasoning works because I can find $a^{-1}$ so that $aa^{-1}=e$. Here I'm as...
If you have an element a, then you have the inverse a^-1, then you have their product a * a^-1, which is the identity
spark veldt
oh wait i think i kinda get it.. so a^-1 exists because you're in group G. aa^-1 is in A if you've shown closure. then it follows that e is in A?
rocky cloak
Yes, that's right.
So the 3 things needed to be subgroup is
- Be closed under taking inverses
- Be closed under the operation
- Not be the empty set
spark veldt
awesome thankss
frank cosmos
"to conclude each factor is an nth power"
is this because f^n=1^n * ... * f^n and the units in C[t] are nonzero complex numbers?
also, what is wrong with the last part of the first paragraph?
rose prism
is this because f^n=1^n * ... * f^n and the units in C[t] are nonzero complex nu...
what does the first part mean
what is 1^n * ... * f^n
celest furnace
is this because f^n=1^n * ... * f^n and the units in C[t] are nonzero complex nu...
No--f has a factorization, and when you take that factorization to the nth power every factor is a power of n
rose prism
No--f has a factorization, and when you take that factorization to the nth power...
what do you mean
its a factorization of f^n
celest furnace
Like every factor of x^2 for an integer x has power divisible by 2
Like this is all thats saying
rose prism
thats a different statement than what you said
frank cosmos
they say "use unique factorization to conclude that each of the factors is an nth power". my justification of this is that f^n=1 * ... * 1 * f^n. Since C[t] is a UFD each of the (h-zeta g) is either a nonzero constant or f^n*constant
is this incorrect
frank cosmos
wdym
rose prism
how did you conclude that h-zeta g = const. f^n
that clearly isnt true as otherwise it would have degree n^n
frank cosmos
ok, i think the problem is that the 1s are not irreducible
frank cosmos
yes
rose prism
1 is a unit
frank cosmos
we want to factor f^n as a product of nth powers
frank cosmos
as a product of n, nth powers
rose prism
correct
frank cosmos
so how do you do so
rotund aurora
G is a finite group. Does this generalize?
- What happens if you consider representations over bar Q_p or C_p. Must they be definable over p-adic number fields or even better, over their valuation rings?
- What if you consider representations over any algebraically closed field (of char=0, and char>0 too), must it be definable over a finite algebraic extension of its prime field?
rose prism
yeah obviously that factorization need not work
frank cosmos
like, for example take f=(t+1) and n=3
then f^3 is degree 3 and cant factor out as a product of 3 third powers of irreducible factors
rose prism
the h-\zeta^i g are not irreducible
frank cosmos
then how is UFD useful here? ufd says that if you have 2 factorizations into irreducibles, then after a possible reshuffling the factors will be associates to each other (factor times a unit)
rose prism
Like every factor of x^2 for an integer x has power divisible by 2
like this
frank cosmos
power means multiplicity?
rose prism
yes
frank cosmos
ok, so then if f is irreducible how does this work
rose prism
what do you mean
frank cosmos
how do you write f^n as the product of n nth powers
frank cosmos
thx
rose prism
in your setting you can absorb u,v into a',b' by taking nth roots
given a complex number u = re^{ix}, you can check that r^{1/n}e^{ix/n} has nth power u
this works because C is algebraically closed
the equation z^n = u has a solution for any u in C
runic galleon
guys im proving tha this is a ring, but i can't prove that the second operator is distribuitive on the first
pls review the way i solve it
for every a, b , c who belongs to Z : a¤(b+c) = (a ¤b )+ (a¤c)
is this correct ?
narrow wagon
a¤(b+c)=a¤(b+c+1)= a+b+c+1+ab+ac+a= (a*b) +(a¤c)+ab
runic galleon
a¤(b+c)=a¤(b+c+1)= a+b+c+1+ab+ac+a= (a*b) +(a¤c)+ab
i found this
try the other part of the equation
it's there where i find a confusion
narrow wagon
for every a, b , c who belongs to Z : a¤(b+c) = (a ¤b )+ (a¤c)
(a ¤b )+ (a¤c) =a+b+1+a+c+ac
notice ab is missing
runic galleon
(a ¤b )+ (a¤c) =a+b+1+a+c+ac
i made a mistake here
forgot the one
ughhh thought it was a normal addition
i won't have you in my exam tmrw to remind me
narrow wagon
it's normal addition
narrow wagon
it's defined in terms of usual addition
runic galleon
well yeah
runic galleon
wait then
narrow wagon
guys im proving tha this is a ring, but i can't prove that the second operator i...
i think you just need to show distrubtivity holds
i really don't think there is something to prove
runic galleon
(a ¤b )+ (a¤c) =a+b+1+a+c+ac
wait, if you wanna prove that its distribuitive wich formulas would u use
runic galleon
i think you just need to show distrubtivity holds
yeah i know, my english is bad
runic galleon
Dubs i think i figured out my mistake
first operator is * not +
therfore yes i needed to add the one
foggy oasis
given that ζ = exp(2πi/17), how exactly would i go about showing these equalities?
im a bit daunted by the looks. i would appreciate any hints or guide on where to start 😭
delicate orchid
You don’t really have that many realtors of roots of unity to go off of. You know zeta^17 = 1 and zeta+zeta^2+…+zeta^16 = -1, so just start symbol bashing until it works
chilly ocean
In my selfstudy. I am going to study 3 subjects: real analysis (from Abbott's book) , proof writing (from How to prove it a structural approach by velleman). Now from linear algebra (friedberg) and abstract algebra (Farleigh) which one should I pick?
Note : I don't have enough knowledge of number theory. Just some basic like Diophantine equation, divisibility, gcd, primes etc
coral spindle
I would suggest just starting reading
If you enjoy it, keep reading, and if you don't, try something else.
Nobody else can possibly choose for you.
chilly ocean
Nobody else can possibly choose for you.
You are so true. Thank you.
I have some knowledge of abstract algebra (first 4 chapters from Gallian's book).
But now I am using Farleigh. So many things look easy.
(Should I still continue this book? And skip easy problems?)
coral spindle
Fraleigh has a lot of content.
chilly ocean
Fraleigh has a lot of content.
Got it. Thank you
grizzled crane
I’m actually trying to learn more on this, has anyone seen this before?
This is from a number theory i used, Elementary Number Theory by Thomas Koshy
frank cosmos
Why appeal to k(y)[x]? Since k is a field then wouldn't that make k[x,y] a PID which has bezouts?
south patrol
k[x,y] is not a PID
frank cosmos
i see, i thought it would by some induction argument
south patrol
In fact R[x] is a PID iff R is a field
frank cosmos
yeah
south patrol
This proof seems wrong to me though
frank cosmos
oof
south patrol
Well when they say identically 0 etc they seem to be identifying polynomials with polynomial functions
But you can just assume the field is infinite to fix that
Since if the field is finite then this theorem is trivial
Lol
frank cosmos
Well when they say identically 0 etc they seem to be identifying polynomials wit...
by identically 0 i thought they meant the 0 polynoial
frank cosmos
yeah definitely
south patrol
Identically seems more like a thing for functions
frank cosmos
yeah
but if they delete the word identically then i think its right? and no distinctions need to be made for finite/infinite?
rose prism
no
tribal moss
I wonder if there's any context where saying "identically zero" instead of just "zero" actually adds meaning.
rose prism
there are nonzero polynomials which are identically zero as functions
tribal moss
Yes, but I think in that sentence it is "as functions" that conveys the meaning, not the "identically".
frank cosmos
no
why is this though?
(also, couldnt they just say that similarly by doing the same process with k(x)[y] there are finitely many choices for the first component)?
south patrol
Actually is there a funny overkill way to the this
slim kayak
why is this though?
Well, take your finite field F_p of choice then you can construct non-zero function with integer coefficients that's 0 at 0, 1, up to p-1
frank cosmos
i dont think thats whats happening here
youre starting with a polynomial in k[x][y] and plugging in a fixed value for y to get a polynomial in k[x]
rose prism
yes
slim kayak
Switched up msgs, mb
south patrol
Well that's why i said we can just assume k infinite
rose prism
right
south patrol
because if k is finite this is trivial
frank cosmos
got it
since infinite IDs have the property that functions uniquely determine the polynomial
south patrol
Yeah exactly - any polynomial of degree n has <= n roots over an integral domain
Indeed this holds for all polys iff the ring is a domain lol
frank cosmos
"a nonconstant polynomial in x,y with coeffieints from algebraicly closed field has infinite zeroes" - This is a following line in that book. I think its wrong, since f=x+1 for example. I think it should require presnence of both x and y
coral spindle
No, that does have infinitely many zeroes. Remember, it's in x and y already. f(x,y) = x + 1 has zeroes f(-1, 0), f(-1, 1), f(-1, 2) for example.
Remember it says "a nonconstant polynomial in x,y"
Can you explain why you chose to react with a clown emoji, @frank cosmos? I'd like to know what you mean by that
frank cosmos
im a clown for not realizing
coral spindle
OK, that's good. I hope you know that people tend to use 🤡 in a rude way, usually
frank cosmos
sorry mb
coral spindle
Just checking
tulip glacier
Does a group homomorphism preserve the identity element ? I don't think it does.
coral spindle
Exercise: prove that group homomorphisms preserve the identity element
You should try this.
It's one of those proofs that 'everyone should do at least once'
tulip glacier
Exercise: prove that group homomorphisms preserve the identity element
Let G1 and G2 be groups. For all x in G1, we have that f(e) * f(x) = f(e * x) = f(x). This means that f(e) * f(x) = f(x) for all x in G1.
I don't see how this implies that f(e) * x = x for all x in G2.
coral spindle
OK :) You should try proving it
dull marsh
||When you have a group, even one instance of x with ax = x is enough to conclude that a is the identity||
tulip glacier
||When you have a group, even one instance of x with ax = x is enough to conclud...
Hmmm, is it because if a is a "fake" identity for x (in the sense that a * x = x and x * a = x) and e is the real identity, we then have:
a * x * x^(-1) = a * e = a
e * x * x^(-1) = e * e = e
Thus, by transitivity of equality, we have a = e.
I see now. Thanks everybody, and sorry if it sounded like such a silly question.
slim kayak
I think they were aiming for proving the cancellative properties of groups. Namely if ax=bx, then a = axx^-1=bxx^-1 = b
tulip glacier
I see, I see. Thanks!
thorny ferry
So given $z_1 z_2 = z_3$ where $z_n \in \mathbb{C}$ then $|z_1|^2 |z_2|^2 = |z_3|^2$. This isn't very special, but I just saw it mentioned as the "law of moduli" in an old book. Is that really a thing? Why would this be any more notable than $|z_1| |z_2| = |z_3|$ ?
cloud walrusBOT
𝓶𝓸𝓶𝓸𝓶𝓸
coral spindle
Seems to be a strange thing to say, especially considering it is a trivial consequence of the second thing you mention. Probably a typo.
We would just write this as |zw| = |z| |w| btw.
thorny ferry
I'll paste a photo.
Oh wait, I guess I don't have permission to do that.
Or maybe I do.
coral spindle
OK. This is completely equivalent, just take square roots. Idk why I didn't say that before.
thorny ferry
So these are the properties that Hamilton was hoping that his new extension of the complex numbers would have.
I guess the idea is to avoid having to define the modulus.
tribal moss
(4) looks like it promises more than it can keep ...
thorny ferry
Yes, it somehow overlooks e.g. N=0, N'=1.
I think it means to say that inverses are unique.
tribal moss
Oh, wait, it speaks of triplets? Then it's probably building up to saying we can't get all those properties anyway.
slim kayak
I would just assume they meant no particular order there
Either they are equal or sort things around N is non-zero
Nvm
thorny ferry
Oh, wait, it speaks of _triplets_? Then it's probably building up to saying we c...
Yes, this is a history of quaternions.
Which have all of the properties except for commutativity of products.
It's saying that Hamilton was looking for triplets with these properties and failed, but then realised that quadruplets work.
The reason he wanted these properties is that they hold for complex numbers.
(when you set c=0 of course)
low wyvern
Having a bit of trouble with this. First I had to show it was a homomorphism which was straightforward assuming the cycle obeys usual power rules. Now I have to find the kernel and image and I am a little stuck...
coral spindle
Try calculating just a few values of psi e.g. psi(1), 2, 3, etc. See what you find.
low wyvern
so like for n=1 (123).... n=2 (123)(123) etc
coral spindle
Work it out more explicitly, for example what is (123)(123) in disjoint cycles
low wyvern
oh ye lol (132). I guess that sorts the image problem out. I just need to do that for different n ... What actually is the identity of Sn ? Like I haven't got it down anywhere
coral spindle
You should think about that.
low wyvern
I would have said 1 but that doesn't feel right
coral spindle
Remember that (123) is just a representation of an element of S_n. The actual elements are bijections on {1...n}
low wyvern
ah ofc
coral spindle
I would have said 1 but that doesn't feel right
Elements of S_n are not numbers :)
tribal moss
But it's common to use the symbol 1 as a name for the identity anyway.
thorny ferry
In this context, would the kernel be the set of integers which map to the identity permutation under \psi ?
tribal moss
Yes, that's the definition of "kernel".
thorny ferry
Okay, thanks.
low wyvern
Yes, that's the definition of "kernel".
funnily enough, this question made me realise... I spelled my name wrong 
thorny ferry
By "image" I guess you mean the image set, i.e. the set of all permutations that \psi maps to.
low wyvern
update: I have realised that the permutation is of order 3 but I am still no closer to realising the identity of the permutation... That is (123)(123)(123) = (123) idk if this is useful but I was just messing around and spotted it. I guess it is also pretty obvious in the first place lol.
tribal moss
funnily enough, this question made me realise... I spelled my name wrong <:openc...
Oh, I always thought were just a Commodore 64 enthusiast.
low wyvern
Oh, I always thought were just a Commodore 64 enthusiast.
Nope, I am just stupid 🙂
frank cosmos
i have a question on super basic algberaic curves, should i ask here or is a different channel better.
tribal moss
Hmm, immediately I would say either #geometry-and-trigonometry or #algebraic-geometry, depending on how "super basic" it is :-)
frank cosmos
basically definition of rational functions
at that level
(as in algebraic curves with rational parametrization)
ok, i think im going through all this wrong
lol
thorny ferry
update: I have realised that the permutation is of order 3 but I am still no clo...
The identity element of S_3 can be written as (1)(2)(3).
slim kayak
update: I have realised that the permutation is of order 3 but I am still no clo...
Maybe try working it out in different notation
low wyvern
Ye, I got it guys but thanks 🙂 . The answer is any integer multiple of 3 for the kernel.
thorny ferry
Yes, including zero of course (which I guess is a multiple of 3 anyway?).
slim kayak
3*0=0, so very multipleist
prisma bear
I’ve been stuck on this for two days now. Any hints?
toxic zephyr
lmfao
white oxide
why is it that if b is irreducible in a PID R, and a is not divisible by b, then a and b are relatively prime?
rustic crown
what is the ideal (a, b)?
white oxide
what is the ideal (a, b)?
i guess it would have to be R, but i'm trying to think of why
or (1)
celest furnace
i guess it would have to be R, but i'm trying to think of why
Hint: (b) is maximal
Since we are in a PID
white oxide
ye
celest furnace
Hint: (b) is maximal
Does this make sense okey
white oxide
i think so... divisibility has never been a strong suit of mine
celest furnace
So do you know why (b) is maximal?
slim kayak
why is it that if b is irreducible in a PID R, and a is not divisible by b, then...
Well, consider decomposition of a into irreducibles
it either has an associate of b or not
white oxide
will come back lol don't want to get stuck but thanks for the help
unkempt stream
Have you forgotten to take linear algebra
i haven’t taken lin alg yet but I am going off Dummit & Foote :troll:
groups, rings, and fields are more *fun*, bro
slim kayak
Cool
unkempt stream
He is one of the homies (*psychopath*)
I have to take linear alg for compeng at some point and I’m unsure if I know all the content
most of the stuff is a manifestation of more general things afaik (like the Rank-Nullity being a consequence of Splitting Lemma)
slim kayak
Idk you have thing a bit weirdly structured. Usually they are introduced when needed in linalg then be equipped for algebra
Occasionaly you can have more generalized concept proving stuff there, but none of the more interesting proofs really profit from it
unkempt stream
modules are interesting in their own right
white oxide
shouldn't this be unique up to a unit?
slim kayak
modules are interesting in their own right
they have a very different flavor tho, being over a field is just extremely nice and well-behaved
slim kayak
shouldn't this be unique up to a unit?
depends on your "directory of primes", all "up to unit"-ness comes from a prime element times a unit still being a prime element
If your system takes representatives (like e.g. choosing the positive primes in Z), then things are fine
white oxide
hmm ok, thanks
rotund aurora
I’ve been stuck on this for two days now. Any hints?
have you got anything?
rose prism
I have to take linear alg for compeng at some point and I’m unsure if I know all...
i can almost guarantee that you dont lol
rank nullity is a special case of some module thing, yes
its also quite trivial
and proving it is hardly the substance of a linear algebra course
the real heart of linear algebra is absolutely not a special case of module theory
unkempt stream
all i know is that SVD’s are a pain to calculate
pastel condor
In the last paragraph how do you know that H or P \cap H is a p-group? Don't we only know that P is a p-group?
slim kayak
In the last paragraph how do you know that H or P \cap H is a p-group? Don't we ...
Essentially, lagrange theorem
They can only be powers of p, since no other order divides the a prime power order
rustic crown
I’ve been stuck on this for two days now. Any hints?
word problems in general are hard. idk any way which can help you avoid calculations. you know yxy^-1 = x^2, conjugate this equation with z and just do weird calculations until stuff simplifies.
if you had 4 generators with similar cyclic relations, then it turns out that group is infinite.
north parrot
are there any units in C[x,y]/xy^2 which are not complex numbers? Like most polynomials without a constant term are zero divisors so they cant be, but I cannot really see no polynomials are units...
north parrot
you are so right lol
rustic crown
pastel condor
They can only be powers of p, since no other order divides the a prime power ord...
Ohhh just the fact that H is a subgroup of Q, which is a q-subgroup?
celest furnace
Ohhh just the fact that H is a subgroup of Q, which is a q-subgroup?
Yep but H is a p-subgroup not a q-subgroup
naive flare
what are the concepts to prove (ℚ, +, ⋅) is an ordered field?
mighty kiln
Prove field axioms, then prove ordered field axioms
Or any reasonable order actually
E.g.
rotund aurora
I don't understand this, how is the injection x-->(x,1/x) continuous? the projection to the second coordinate is certainly continuous right so this would imply that the map x-->1/x itself is continuous, which contradicts the previous remark
topaz solar
I don't understand this, how is the injection x-->(x,1/x) continuous? the projec...
The injection defines the topology
And it claims R* as topologized as a subset of R x R has a continuous inclusion R* -> R
Which looks like projection to first coordinate?
Inverse map need not be continuous on R* < R, but it is for R* < R x R
Since ya know
Second projection map
rotund aurora
Inverse map need not be continuous on R* < R, but it is for R* < R x R
how
are you saying that the map (x, 1/x)-->(1/x, x) is continuous?
rotund aurora
this is not the inverse map, but ok
doesn't make a lot of sense to talk about inverse map when you have 2 rings, but anyway
rotund aurora
you sent (x,1/x) to 1/x, not x to 1/x
topaz solar
x -> (x, 1/x) is cts by definition
(x, 1/x) -> 1/x is second projection
Therefore
Compose them
QED
This is a different topology on R*
You topologize R* based on x -> (x, 1/x)
rotund aurora
mmh isn't this like declaring for every open U of R that U cap R^x and {1/x : x in U cap R^x} should be open?
in R^x
topaz solar
tf is R^x
rotund aurora
the set of functions form x to R, what do you think
topaz solar
For every open U of R x R, the preimage is open in R*
cloud walrusBOT
Croqueta
rotund aurora
what is a subspace
topaz solar
It’s the subspace topology on the set {(x, 1/x)} < R x R
rotund aurora
ah yeah, this I get it
topaz solar
the set of functions form x to R, what do you think
Aggressive
rocky cloak
are you saying that the map (x, 1/x)-->(1/x, x) is continuous?
Yeah in RxR the map (x, y) -> (y, x) is continuous.
If you restrict this to (x, 1/x) it maps to (1/x, x) which is the inverse. As you can see if you multiply then together
(x * 1/x, 1/x * x) = (1, 1)
The resulting topology will be generated by sets of the form
{x in R^* | x in U, 1/x in V}
for open sets U and V.
steel pulsar
What is the algorithm to write this permutation as a product of cycles? two by two disjoint.
slim kayak
you follow an element on its path
If two element intersect in their orbit, then they are part of the same cycle. So by just seeing what cycle any individual element follows you'll see how it decomposes
dull marsh
Simply track down where the elements go as you apply the permutation several times. (In this case, 1 -> 3 -> 9 -> 11 -> 5 -> 19 -> 1, so one of the cycles in your decomposition will be (1 3 9 11 5 19), do the same on the other elements for the rest of the decomposition)
steel pulsar
ok thank you very much
dull marsh
The preimage of 0
daring nova
f^-1(the identity element) whatever you named it
oh I didn't notice ring, mb
yeah that first one
steel pulsar
But I have a problem
The question is Determine the signature of σ as well as its order. Calculate σ^(59)
I calculated the signature
( 1 3 9 11 5 19 ) ( 2 4 6 7 8 14 18 17 16 10 ) ( 12 13 15 )
(−1) ^(5) (−1)^(9) −1)^(2) = 1
Then for the order we calculate the ppcm
ppcm(10,6,3)
The thing is that I find that this ppcm is equal to 30
But in the correction they say 60
And in fact the 60 makes it easier to resolve the problem because afterwards they say that
Since σ is of order 60 we have σ^(60) = Id. Consequently σ^(59) is equal to σ^(-1)
But I think that 60 is wrong for the ppcm, how to calculate σ^(60)
?
rustic crown
And in fact the 60 makes it easier to resolve the problem because afterwards the...
if σ^30 = id then σ^60 = id, so you can do the last part either way
steel pulsar
if σ^30 = id then σ^60 = id, so you can do the last part either way
why if f σ^30 = id then σ^60 = id ?
rustic crown
yea it should be 30
prime dock
So I was told this result that if F is a field then every finite subgroup of F* is cyclic
But I have a counter example to this like look at U(8)
its not cyclic
and its elements are also in set of real numbers which is field
coral spindle
U(8) is not the group of units of a field
Z/8Z is not a field, and neither does it consist of real numbers
You know this already: in what world do we have 8 = 0 in the real numbers??
prime dock
Oh you are right
hm yeah operation would be different
So its a true result basically
coral spindle
Theorem with correct proof is true! In other news, water wet
prime dock
So its a true result basically
I mean sorry my counter example being wrong doesn't prove that the result is correct
prime dock
Yes, I realised that
steel pulsar
I didn't understand why 10 or 12 are possible orders for an element of S_7
I thought it only went to 7...
Can someone please explain to me?
slim kayak
I didn't understand why 10 or 12 are possible orders for an element of S_7
okay, whats the order of the product of elements with relatively prime oders?
assuming an abelian group for the moment
coral spindle
You are likely confusing S_7 with the group you probably know by the name Z_7
S_7 has 7! = 5040 elements, and Z_7 has 7 elements.
steel pulsar
okay, whats the order of the product of elements with relatively prime oders?
Not understood
my teacher wrote this
prime dock
I may be wrong but order of an element is permutation group is calculated by checking the lcm of length of disjoint cycles
Also order of S_7 is 7!
so to have order of 10 your element will look something like this (5)(2)
which means 5 elements in first cycle and 2 in the 2nd
prime dock
Those are elements in disjoint cycles under type
So for order 2 you have something like (2)(1)(1)(1)(1)(1)
prime dock
Those are elements in disjoint cycles under type
So for order 2 you have somethi...
check lcm here and you get 2
Now tell me how you will create an element of order 12 in S_7
lcm of length of disjoint cycle should be 12 so use that
slim kayak
Not understood
my teacher wrote this
Take an abelian group and let A have order 3 and B have order 4. What order does AB have?
steel pulsar
why is the order of B unknown?
slim kayak
whistling innocently
prime dock
why is the order of B unknown?
Did you understand what I said
is this true - if G is cyclic then Aut(G) is cyclic. Its written in the notes Im reading but take like Z8 we know its cyclic but Aut(Z8) is isomorphic to U(8) which is not cyclic
delicate orchid
Well there’s your counter example lol
prime dock
my teacher smoked before making these notes then
delicate orchid
Double check that’s actually what it’s saying
prime dock
delicate orchid
That second statement is wrong as well lol
prime dock
its supposed to be 2p^k
delicate orchid
Ok then it’s correct
prime dock
hm even prime is already counted
delicate orchid
I assumed that was stated elsewhere because 2 and 4 are separate cases
delicate orchid
? What
unkempt stream
i like how 8 just refuses to have a primitive root
delicate orchid
No there is absolutely a need to mention it’s cyclic for p^k with specifically an odd prime
Because it’s not a true statement otherwise
prime dock
oh you are right
oh wait
yeah my bad p^k will never be a prime regardless
💀
so my reasoning was dumb
unkempt stream
hey chat
prime dock
what
unkempt stream
Z/13Z^X is 12-cyclic so you have the chance to make the world’s most incomprehensible clock
Wait so is 61 LMAO (60-cyclic)
prime dock
12-cyclic meaning cyclic group of order 12?
unkempt stream
12-cyclic meaning cyclic group of order 12?
Yes
generated by one of it’s many primitive roots of unity
log problem means it’d be a pain in the dick to go from that system to normal time lmao
prime dock
Its difficult to memorise these many results in group theory
even if i understand the reasoning and proof i forget it if i don't revise it regularly
unkempt stream
guys we have a chance to make the most annoying fucking time system known to man
delicate orchid
Eventually it gets to the point that you can detect when something is wrong on an elementary level as easily as a spelling error
unkempt stream
7 is prime and is in both
just send hr:min to (7^hr mod 13):(7^min mod 61)
and subtract one if you want it to go from 0-59
mighty kiln
guys we have a chance to make the most annoying fucking time system known to man
Great idea
And instead of starting on different times, timezones have different generators
Which are changed when daylight savings
low wyvern
Eventually it gets to the point that you can detect when something is wrong on a...
Isn't it a shame then that I can't spot spelling errors either -_-
delicate orchid
That’s ok I can’t spell either
cloud walrusBOT
Mike Hawk
untold basalt
Does this work?
unkempt stream
Which are changed when daylight savings
AMAZING
delicate orchid
**Mike Hawk**
Not sure you can use Euclid’s division algorithm in Z_5[x,y] cause it’s not a PID
I’d make note that the ideals (x+3) and (y^2+4) are disjoint
untold basalt
Not sure you can use Euclid’s division algorithm in Z_5[x,y] cause it’s not a PI...
I'm using euclid's division with respect to the variable y
steel pulsar
Hello
Why in this case, for <2> = we have the multiplication and not the addition ?
And another thing, I understand that from 8 we go a little outside the border, because 8 x 2 is greater than 12. But I don't remember what the calculation is that makes us go to 3
coral spindle
Why in this case, for <2> = we have the multiplication and not the addition ?
Context. Context determines it.
steel pulsar
find a generator of the cyclic group (Z/13Z)×
steel pulsar
× is not hte multiplication
coral spindle
I know what the notation means, thanks
steel pulsar
Ok
coral spindle
But the group operation is indeed multiplication – this is the multiplicative group.
If we were looking at some other thing, for example the group Z/13Z, we would use addition when writing e.g. <2>
So as I said, context determines it.
coral spindle
You are aware of what the notation (Z/13Z)× means?
(Z/13Z)× is the multiplicative group of the ring Z/13Z.
Are you familar with this?
steel pulsar
that means Z/13 deprived of zero
coral spindle
In this case it does, yes.
In general it does not, but no matter.
And (Z/13Z)× is a group under multiplication.
Clear now?
steel pulsar
well a group provided with a multiplication generally we note it (Z/13Z, x)
south patrol
Sure, that is not what is going on here
steel pulsar
And another thing, I understand that from 8 we go a little outside the border, b...
someone know for this question please?
coral spindle
well a group provided with a multiplication generally we note it (Z/13Z, x)
No, group theorists generally do not need to do that.
south patrol
also that
coral spindle
And furthermore, (Z/13Z, x) is not a group.
((Z/13Z)*, x) is, though.
It should be very clear to you that (Z/13Z)* is a group under multiplication, and not addition.
steel pulsar
Well you said because there is *
coral spindle
So are we clear?
steel pulsar
yes
coral spindle
Great
steel pulsar
And another thing, I understand that from 8 we go a little outside the border, b...
but not this
coral spindle
8 x 2 = 16 = 3 + 13 = 3 mod 13.
south patrol
||I genuinely computed 8 x 2 as 24||
steel pulsar
why 3 +13 and not 4 + 12 ?
south patrol
it is mod 13
coral spindle
why 3 +13 and not 4 + 12 ?
We are working in Z mod 13, not Z mod 12.
steel pulsar
ah actually it’s clearer like that : 2^(0 to 12) ≡ ... [mod 13]
vivid birch
is it true that if K is a field, then K[x] is a PID?
rocky cloak
is it true that if K is a field, then K[x] is a PID?
That is indeed correct
vivid birch
thanks
rocky cloak
For K a commutative ring, this is in fact if and only if
south patrol
which is a good exercise :)
The hint I would give is to consider how you prove that Z[x] isn't a PID (using the fact Z is not a field)
tardy hedge
poo, pee, and more
south patrol
huh
slim kayak
He's got a point☝️
south patrol
Schemes.
slim kayak
What
south patrol
Unless it's the empty scheme
slim kayak
Yeah those lack points
south patrol
Well it depends on the ideal
delicate orchid
||I genuinely computed 8 x 2 as 24||
rapid junco
how do you prove that -1 commutes with any element under mutliplcation in a ring?
lapis latch
Maybe show that (-1)x=-x=x(-1)
rapid junco
bruh
steel pulsar
Hello
sorry I have last one question
I was told I could write 4 = 2 + 1 + 1 in 12 partitions in S_n
• {1, 2}, {3}, {4}
• {1, 2} {4} , {3}
• {1, 3], {2}, {4}
• {1, 3], {4}, {2}
• {3, 4], {1}, {2}
• {3, 4], {2}, {1}
• {4, 3], {1}, {2}
• {4, 3], {2}, {1}
• {2, 1}, {3}, {4}
• {2, 1}, {4}, {3}
• {3, 1}, {2}, {4}
• {3, 1}, {4}, {2}
here it's 12 partitions
But something missing to me
where is {2, 4}, {1}, {2} ?
or {1, 4}, {2}, {3}
steel pulsar
?
steel pulsar
look at the first 4 and the last 4
yes?
summer path
do you notice anything ?
||they are the same, so yes the ones you mentioned as missing are not listed||
slate forum
#help-1 message
just wanted to try asking this here, as the question in #help-1 doesn't seem to be being answered
rotund aurora
Given n rational points (xi,yi) and another n rational points (zi,wi) can you find polynomials with rational coefficients P(x,y) and Q(x,y) such that P(x,y)=zi and Q(x,y)=wi if and only if x=xi and y=yi for each i?
steel pulsar
do you notice anything ?
ah yes thank you
vagrant zinc
Do you guys know what the 2 refers to?
still ledge
Do you guys know what the 2 refers to?
Probably the variable in polynomial
prime sundial
what 2
prime sundial
the n = 0,2,4,6,...?
vagrant zinc
It would be the even-numbered polynomials
prime sundial
even polynomials yeah. also this could probably be asked in #linear-algebra
vagrant zinc
I have the wrong group, sorry
steel pulsar
There are 6 Sylow subgroups of order 5 in A_5 ? right ?
prisma ibex
yes
n_5 divides |A_5|=2^2*3*5 and n_5=1 mod 5 so either n_5=1 or n_5=6, but you can't have n_5=1 since A_5 is simple
prime dock
I have read that if a homomorphism is an isomorphism then Ker phi = {e} but shouldn't it be like an iff condition instead of if
prime dock
I have read that if a homomorphism is an isomorphism then Ker phi = {e} but shou...
ping me if you answer Im going to turn off discord
kind temple
I have read that if a homomorphism is an isomorphism then Ker phi = {e} but shou...
no. the kernel of a group or ring homorphism is trivial if and only if it is injective
prime dock
no. the kernel of a group or ring homorphism is trivial if and only if it is inj...
Oh, so no need to check onto. But, it is an iff condition instead of if
kind temple
what do you mean?
prime dock
what do you mean?
You said I only have to show injective condition right, and not onto
kind temple
i never said that
these two properties are equivalent for a group or ring homorphism:
- f has trivial kernel
- f is injective
prime dock
Hm oh, so you meant that if f has trivial kernel that doesn't prove that f is isomorphic?
kind temple
right, because it might not surject onto the target
prime dock
Ok thanks
also, Im reading if G/N is cylic then G is abelian
where N is a normal subgroup of Z(G) centre of group
then shouldn't it just be all subgroups of Z(G) cause they would all be normal subgroups
What do you mean
prime dock
Its set of all elements that commute with every element in G
So their element should commute with each other
I think its always abelian it has to be
rocky cloak
So yeah, the word 'normal' is redundant in your statement. But it doesn't hurt to keep it in
prime dock
alright thanks
errant shadow
Are group homomorphisms always linear?
errant shadow
What do you mean by group homomorphisms being linear
Like, the maps?
I was reading up non-linear dynamics
And the additive property of linear systems reminded me of homomorphisms.
So I was wondering if there was some sort of distinction between linear and non-linear homomorphisms or if they were all of the same kind.
dull marsh
You are probably thinking about vector space homomorphisms rather than group homomorphisms
errant shadow
You are probably thinking about vector space homomorphisms rather than group hom...
Ah makes sense
dull marsh
Anyway, the former are defined to be linear in the first place
errant shadow
Anyway, the former are defined to be linear in the first place
Fair
Is there a field of work that involves the intersection of complex systems and group theory?
Like, studying some sort of algebraic complex systems? Can we do that?
icy totem
What is an element of 2Z∩3Z that isn't in 2Z3Z:={elements of Z that are a finite sum of elements of the form i*j, where i is in 2Z and j in 3Z}. I ask this because I know that if I and J are ideals, IJ is not necessarily equal to I∩J, but I can't find counterexamples. Someone told me that with I=2Z and J=3Z should work.
icy totem
Oh I see it now thank you
amber cradle
Are group homomorphisms always linear?
The way you can think about linear transformations is like a stronger version of homomorphism. Homomorphisms only need to protect the group structure, so for a vector space this is vector addition. A linear transformation forms at least a homomorphism between the set of vectors on the operation of addition. The linear map then has the stronger property of commuting with scalar multiplication.
coral spindle
An example of a group homomorphism that isn't linear is the map C → C defined by f(x + iy) = x - iy. This is an automorphism of (C, +) but it is not a linear map on the C-vector space C. It is, however, an R-linear map, so that's cool.
amber cradle
^conjugation
brittle ether
Okay so
A bit of a weird question and I'm not sure if this is the most appropriate channel
But I'm gonna try because I cannot find a better place to put it
I was wondering what addition and multiplication were in its most abstract
I got a bunch of properties, such as that multiplication has to be distributive over addition
And a lot of these ideas make perfect sense, and seem to fit in pretty much every abstraction of addition and multiplication that I can think of
Except one
The top answer here says that addition must be commutative
But ordinal addition is not commutative, but we still call it "addition"
Is there a better way to explain why this is still called "addition" even though it doesn't share the same abstract properties as all of these other abstractions of "addition"?
Or should I resign to say that this is just one of those things where
The abstraction just happens to go in two completely different directions
slim kayak
Both are binary associative operations. One (addition) is always taken to be commutative
brittle ether
People name it however they want to name it
brittle ether
Both are binary associative operations. One (addition) is always taken to be com...
Then why is ordinal addition called addition?
coral spindle
I don't think there is going to be a better answer than simply that people made the name up because they want to think of things like addition.
Same goes for the word 'number,' there is no definition.
brittle ether
Okay, may as well accept that, thanks
slim kayak
Then why is ordinal addition called addition?
Addition usually refers to an algebraic structure
Ordinal addition refers to the adding of two well-ordered sets and equipping it with a new well-order
brittle ether
Sure I get that
But how does that match the other abstractions of additions
Or is it just something totally different
And do we refer to other applications of this well-ordering also as "addition"?
slim kayak
Other applications?
With ordinal arithmetic you are adding together different sets (disjoint union) and then equipping it with other structure
So if you want to insist on some common thread, which shouldn't exist for any good reason, then ordinal addition is about adding ordinal structure onto the addition of sets
brittle ether
Hmmm
Ok seems the consensus is that there isn't a common thread
In which case I can see 3 different ways to interpret add/mult:
- mult is repeated add
- abstract properties like the forementioned distribution and commutation
- with regards to linear transformations/complex numbers, add is translation, mult is rotate/dilation
errant shadow
The way you can think about linear transformations is like a stronger version of...
Ah okay, makes sense, thanks
brittle ether
The third one here I guess is actually number two, as a consequence
But I made it distinct for pedagogical reasons
I guess 4th would be the well-ordering thing
Is there anything else I'm missing?
slim kayak
I mean, for Z/C those happen to coincide in the right spaces
The first 3
Ordinal addition still relates to to the formal addition defined before, like it isn't completely random why it's called addition
brittle ether
Yeah which is why I wasn't sure if it should be it's own bullet
I was thinking making it its own bullet if I made 2 and 3 distinct
To be more consistent
Maybe at this point it's better off in #math-pedagogy
slim kayak
Aight, let's move
chilly ocean
how would I go about this?
dull marsh
What form does k have? What does tell you about its inverse?
jaunty void
Hi, everyone can someone recommend a book about field theory and Galois theory? I would like a book that explains the theories and their implecations(or how to use them)
dire siren
Dummit and Foote
sharp viper
Anyone know NTT? It's related to DFT and FFT
rose prism
WTF
coral spindle
I wish there were more acronyms so I could confuse people on the internet more easily
Anyway, I have no idea what NTT is, but the discrete and fast fourier transforms are a bit out of this channel's wheelhouse. Read channel descriptions, please.
coral spindle
Sure
sharp viper
Anyway, I have no idea what NTT is, but the discrete and fast fourier transforms...
You basically can do multiplications faster in NTT field so you do that then convert it back using Inverse NTT
I'm having trouble regarding it because my code implementation based on FIPS 203 ipd is not working as intended
Idk what channel NTT would be understood. This is the closest one because it has the "fields"
And "rings" that's often described in NTT
wraith cargo
You basically can do multiplications faster in NTT field so you do that then con...
what is a NTT
sharp viper
what is a NTT
Number Theoretic Transform. I purposely omitted the full name because you don't need to ask it if you know NTT
Sorry if that sounds a bit egotistic or something
barren sierra
Idk what channel NTT would be understood. This is the closest one because it has...
do u know what fields are? why the quotes?
sharp viper
do u know what fields are? why the quotes?
I'm just emphasizing the words in the channel name
slim kayak
Okay, shouldnt it just end up being essentially FFT ad verbatim as long you didnt mess up implementing your finite field
sharp viper
Okay, shouldnt it just end up being essentially FFT ad verbatim as long you didn...
I know, but I'm missing some understanding. If my understanding is correct, this should be reversible, right? Idk if it's in my implementation or its in the FIPS 203 ipd
I based it on these algorithms published by NIST(FIPS 203 ipd)
rose prism
what should be reversible
sharp viper
what should be reversible
NTT(x) = y, InverseNTT(y) = x
The value you input into NTT
sharp viper
idk what youre asking though
is the NTT code provided by NIST reversible or did they omit an algorithm step?
slim kayak
Did you copy the code in the pic word for word?
sharp viper
I'm new to NTT and would like to understand it
sharp viper
Did you copy the code in the pic word for word?
I tried my best. It should work, but since it's pseudocode and it's an initial public draft(ipd), I'm not sure if it's correct or I'm wrong
slim kayak
Have you tried using a sample set of smaller values and see if that works
rose prism
what is this variable k
sharp viper
Have you tried using a sample set of smaller values and see if that works
https://www.nayuki.io/page/number-theoretic-transform-integer-dft
This website has a NTT and Inverse NTT calculator of some sort that works
rose prism
they dont seem to use it
sharp viper
what is this variable k
rose prism
nvm they do
sharp viper
https://csrc.nist.gov/pubs/fips/203/ipd
The NIST page for it
Basically, this thing is gonna replace current encryption algorithms, so I have few places where I can search for relevant information
barren sierra
it's not necessarily going to replace
afaik these are all just candidate protocols
sharp viper
afaik these are all just candidate protocols
It's already implemented by Cloudflare to eventually replace all their assymmetric algorithms. Maybe one of us is using it rn without knowing
Cause it's theorized as Quantum-safe
slim kayak
What does it mean for a fourier transform to be quantum safe
barren sierra
What does it mean for a fourier transform to be quantum safe
they're talking about the whole encryption protocol being theoreized as secure against attacks by quantum computers
not just the FFT portion
sharp viper
What does it mean for a fourier transform to be quantum safe
I mean, the whole FIPS 203 ipd is supposed to be quantum-safe where it's also hard for quantum computers to crack the encryption. Current encryptions are vulnerable to Quantum computers
NIST started this drafting competition to protect US security
For the most part
I guess I'll go stare at math for hours. No one here seems to know NTT
prime quail
What are similar elements in group called?
Let G = rubik's cube.
Then every element that is similar to F are {F,R,U,L,B,D,F',R',U',L',B',D'}.
The word I am looking for is not conjugate because F looks different from RFR'.
barren sierra
I know a better server that's more cryptography focused @sharp viper, send me a Discord friend request
barren sierra
What are similar elements in group called?
Let G = rubik's cube.
Then every elem...
what do you mean "similar"
can you give an example in a more standard group?
not necessarily the rubiks cube group (just because I'm not super familiar with it)
sharp viper
I know a better server that's more cryptography focused <@670264081147756564>, s...
Done
rocky cloak
What are similar elements in group called?
Let G = rubik's cube.
Then every elem...
Seems to me you just have the symmetry group of the cube act on the Rubik's cube group.
I don't know if there is a word for two elements in the same orbit, though something like "similar under the action" or something like that seems reasonable.
sterile garden
I guess I'll go stare at math for hours. No one here seems to know NTT
Oooo, NTT!
sterile garden
Basically FFT over a finite field
sharp viper
Basically FFT over a finite field
Yep. You know it? 
sterile garden
Yep. You know it? <:POGGIES:1074551972218286081>
I wrote a toy python implementation of it to understand it a bit a couple of years ago
So sort of?
sharp viper
I wrote a toy python implementation of it to understand it a bit a couple of yea...
Do you have it on Github?
sterile garden
Do you have it on Github?
Nooooo haha, but I'm sure some implementations exist. I probably did it in a Jupyter notebook that's long gone
sharp viper
I've seen and read a lot of implementations of it in code. I'm looking for more to get it to click in my head
sharp viper
Nooooo haha, but I'm sure some implementations exist. I probably did it in a Jup...
Awwww
sterile garden
Do you understand FFT?
sharp viper
Do you understand FFT?
Basic understanding only
Basically isolate frequencies, right?
sterile garden
Yeah, but the NTT doesn't have a physical meaning like that.
It's just exploiting the fact that you can do convolutions (polynomial multiplication) faster
I haven't looked at the thing you posted above and won't have time to read that in detail for a bit
sharp viper
Yep! But my implementation is not reversing/inversing. According to the NIST paper I'm following, it should be inversiblr
sterile garden
Compute it with a small example by hand then compare with your implementation
sharp viper
It uses a lot of optimizations like Montgomery Reduction so everything is clouded in my head
sterile garden
So you can find out if it's your transform or inverse that's giving you problems
sharp viper
Compute it with a small example by hand then compare with your implementation
That's a great idea
Thanks!
sterile garden
I think I used one of those two videos to understand the FFT
This is what I used to understand the NTT
sharp viper
I've been staring at that page for a while, but the algorithm used in the paper I'm following is different
sterile garden
You'll want to know a bit about what a primitive root is. People in this channel or elementary number theory channel should be able to help with that. It's covered in Jones & Jones Elementary Number Theory
Yeah the algorithm is different in your paper as the link I shared is just textbook NTT, but the underlying concept is the same
sharp viper
You'll want to know a bit about what a primitive root is. People in this channel...
I've managed to produce primitive roots with the help of the paper
sterile garden
i.e., they should give the same output for a given input
I have a document my master's supervisor wrote to help explain FFT and NTT a bit I can share with you privately if you'd like
sharp viper
I have a document my master's supervisor wrote to help explain FFT and NTT a bit...
Are you sure?
That sounds like personal stuff
sterile garden
Yeah, it's fine. I have the TeX source so I'll just redact names
sharp viper
Yeah, it's fine. I have the TeX source so I'll just redact names
Thank you! That will be great. Some guy posted in a blog that anyone can implement FIPS 203 without understanding NTT, but my failing code says otherwise lol
icy totem
If I consider A:=Q(√6,√7) and B:=Q(√6-√7), clearly A contains B, and the dimension of B over Q is 4. But I'd say also the dimension of A over Q is four, and when I have two vector spaces with the same dimension, and one contains the other, they are equal. But they are not equal, in this example. What am I getting wrong?
icy totem
Ohhh yea you are right
Thank you everything makes sense now
There is a guy on the internet that explained to me using galois theory (which i havent studied yet) that they were different, so i trusted him
coral spindle
Don't trust anything that guys on the internet say. They're always lying. 
celest furnace
Don't trust anything that guys on the internet say. They're always lying. <:devi...
So true
cloud solar
If (H,•) group where H is included in (0,inf) and has the properties: if x in H then 1/x in H, 2023 is in H and x•y=1/x • 1/y for every x,y in H, show that H is not an interval and (x•x)(x•1/x)=1 for every x in H.
How I can show H is not interval
rose prism
If (H,•) group where H is included in (0,inf) and has the properties: if x in H ...
is the dot just the usual multiplication of real numbers?
cloud solar
is the dot just the usual multiplication of real numbers?
No, it is the operation that form a group with the set H
rose prism
i see
barren sierra
ok first question
is 1/x different than the inverse of x respect to the operation dot
I ask because you say "if x in H then 1/x in H"
after specifying H is a group
cloud solar
Yes 1/x is not the inverse of x in H
rose prism
If (H,•) group where H is included in (0,inf) and has the properties: if x in H ...
in the last part is it the product of real numbers xx and x1/x?
barren sierra
just the actual number 1
rose prism
or the product in the group
rose prism
idc
barren sierra
fair
rose prism
i think its clear what i meant
cloud solar
you don't necessarily mean the identity
Yes 1 is the real number 1.
delicate orchid
what have you tried for this question
cloud solar
in the last part is it the product of real numbers x*x and x*1/x?
It is the usual multiplication between (x dot x) and (x dot 1/x)
If H is an interval then [1/2023,2023] is included in H (I dont think 2023 is relevant for the problem).
delicate orchid
I'm not sure how you're concluding the entire interval is in H
also just out of curiosity, is this a competition maths question
rose prism
If H is an interval then [1/2023,2023] is included in H (I dont think 2023 is re...
can you prove that for 2023 replaced by any integer n? any real number?
slim kayak
also just out of curiosity, is this a competition maths question
did using the "2023" give it away
barren sierra
lmao
cloud solar
I'm not sure how you're concluding the entire interval is in H
2023 is in H and by the property of the group 1/2023 must be in H. Because H is an interval every element between 1/2023 and 2023 must be also in the set H.
rose prism
oh
cloud solar
also just out of curiosity, is this a competition maths question
Idk
rose prism
ignore my question
cloud solar
oh you're assuming for contradiction
Yes
delicate orchid
did using the "2023" give it away
it's the fact that this situation would never actually come up in any real mathematics, but that didn't help either 
barren sierra
that would have been good to state earlier lol
nvm you did state that
I can't read
rose prism
they did
cloud solar
it's the fact that this situation would never actually come up in any real mathe...
I know that. But yeah.
rose prism
1 dot y = 1 dot 1/y for all y
if that interval was in H
then 1 would be in H
hence you can cancel it
so y = 1/y for all y
cloud solar
Oh right.
cloud solar
so y = 1/y for all y
:))
rose prism
using the famous theorem 2023 >= 1
slim kayak
"x•y=1/x • 1/y" every product of distinct elements is its own inverse?
delicate orchid
the 1/ there isn't referring to the . operator, otherwise this would be easy via summoning facts about 2-torsion
rose prism
"x•y=1/x • 1/y" every product of distinct elements is its own inverse?
see the many clarifying questions asked above
slim kayak
fair
rose prism
interesting problem
delicate orchid
I have proven it is commutative 
the problem is I have no idea how the two operators interact
this isn't like a bigroup or whatever
so much like every other comp maths problem there's no reasonable approach one can be expected to take and instead you have to spot some obscure arcane trick
I bet this is a formal group law somehow
cloud solar
so much like every other comp maths problem there's no reasonable approach one c...
Yes. This is from Mathematical Gazette.
slim kayak
I have proven it is commutative <:bleakcat:913340873771716630>
EH to the rescue?
delicate orchid
Yes. This is from Mathematical Gazette.
couldn't have given me an easy one huh
cloud solar
Maybe zassenhaus lemma works 
cloud solar
on which groups lol
Imaginary groups
Idk man
This type of problems are so tricky
Only basic group properties
barren sierra
so y = 1/y for all y
Using this we get that 1 / (x dot y) = (1 / x) dot (1 / y) idk how helpful that is
delicate orchid
uhhh massively? if true
rose prism
what
barren sierra
because 1 / (x dot y) = x dot y right?
rose prism
what i wrote finishes the proof
cloud solar
Using this we get that 1 / (x dot y) = (1 / x) dot (1 / y) idk how helpful that ...
This was a contradiction
barren sierra
oh wait
rose prism
its an immediate contradiction
delicate orchid
yeah I'm focused on the (x.x)(x.1/x) = 1 part
basic rearranging gets me nowhere so we're going to have to start inserting x.x^-1s until it works
cloud solar
If e is the identity of H we have x=1/x dot 1/e = 1/e dot 1/x for every x in H
So e=1/e dot 1/e (doesnt help)
delicate orchid
more than that, we have x = 1/x.1/x I think
barren sierra
what how
rose prism
If e is the identity of H we have x=1/x dot 1/e = 1/e dot 1/x for every x in H
using this
oh wait nvm
delicate orchid
yeah I was wrong, I have something else though
x = 1/x.1/e => x.(1/x)^-1 = e => x^-1 = (1/x)^-1
barren sierra
does that not prove that x = 1/x
rose prism
what
cloud solar
x dot 1/x = x dot x dot 1/e
rose prism
how did 1/e become e
barren sierra
also that
cloud solar
the two operators are really messing with me
Yeah
slim kayak
Maybe agree on using some other notation for the group inverse for now, idk, bars?
delicate orchid
I'm not latexing this
rose prism
oh actually
x = (1/x).(1/e), together with e = (1/e).(1/e), implies that x^{-1} = (1/x)^{-1}
nvm
slim kayak
yeah
delicate orchid
I just keep proving that x and 1/x commute lol
slim kayak
i think i added in the assumption 1/e=e somewhere and can delete some stuff again, how fun
delicate orchid
who actually enjoys these sorts of questions. Seriously.
slim kayak
okay wait,
e = e.e = 1/e . 1/e = (1/e . e). 1/e = e . 1/e . e = 1/e?
cloud solar
Morse code
rotund aurora
What problem are you doing?
delicate orchid
okay wait,
e = e.e = 1/e . 1/e = (1/e . e). 1/e = e . 1/e . e = 1/e?
oh. OH
let him cook chat...
slim kayak
wait, doesnt (1/x)^-1 =1/x^-1 follow from just e=x.x^-1=(1/x).(1/x^-1) too?
rose prism
right
delicate orchid
I don't see it
cloud solar
okay wait,
e = e.e = 1/e . 1/e = (1/e . e). 1/e = e . 1/e . e = 1/e?
Wait now I dont understand that
delicate orchid
then a = b^{-1}
was this directed at me?
rose prism
yes
slim kayak
i think there is a typo
yeah nvm that
delicate orchid
alright buddy if you wanted me to stop contributing you don't have to be so passive aggressive about it in the future
rose prism
what
delicate orchid
I don't see how you cancel the x or the 1/x
the multiplciation does not distrubute in any manner over the .
rose prism
chill out its just a math problem
slim kayak
I don't see it
that e=1/e implies x=1/x?
delicate orchid
wait, doesnt (1/x)^-1 =1/x^-1 follow from just e=x.x^-1=(1/x).(1/x^-1) too?
no, this
rotund aurora
Are you assuming that if x in H then 1/x is also in H?
delicate orchid
I can see that from just the starting relation
rose prism
Are you assuming that if x in H then 1/x is also in H?
yes
rose prism
you have equation a.b = e
you cancel
idk why you got so defensive about it
here a=1/x, b=1/x^{-1}
rose prism
yes ok so we have x.x^-1. (1/x^-1)^-1 = 1/x
yes now cancel the first two factors on LHS
delicate orchid
ok there we go, thank you
cloud solar
I try to show 1/(x.x.1/e)=x.x
rotund aurora
If it were an interval, wouldn't it contain 1 so that y=1/y ?
cloud solar
OH
slim kayak
so what happens with e.y=1^-1 . 1. y = 1^-1 . 1/1 . 1/y = 1^-1 . 1 . 1/y?
cloud solar
I did it I think
delicate orchid
oh yeah 1^-1 is the thing we haven't thought about yet
rose prism
1 cannot be in H
slim kayak
if it was that seemingly forces 1/x=x, a contradiction since then everything must be 1
slim kayak
1^-1 is a bit too cursed to start the week with tbh
slim kayak
you lowkey solved the problem
rose prism
uh
delicate orchid
oh that's clever
slim kayak
sorry but the star notation was the worst choice imagineable :sob
summer path
smol dot or x with a line through it
slim kayak
so what happens with e.y=1^-1 . 1. y = 1^-1 . 1/1 . 1/y = 1^-1 . 1 . 1/y?
For clarity. Let a be the inverse of 1. Then:
x=a.1.x=a.(1/1).(1/x)=a.1.(1/x)=1/x
cloud solar
How do you know 1 is an element of H?
slim kayak
ah, that was for the interval situation
it contains 1/2023 and 2023, so if it was an interval it contains 1
rose prism
how do you get the 4th line (not counting the crossed out line)
cloud solar
how do you get the 4th line (not counting the crossed out line)
I used x.y=1/x.1/y property
slim kayak
how does first step in the first line work
rotund aurora
slim kayak
oh probably add e and then shuffle it around ig
delicate orchid
oh probably add e and then shuffle it around ig
x = x.e = 1/x . 1/e
oh and x . 1/x = 1/x . x
slim kayak
yeah
cloud solar
Now I see that the problem asks to show that there exists a group with such properties
slim kayak
doesnt it happen to work out with giving {2023,1/2023} the structure of Z/2Z?
cloud solar
What operation
slim kayak
call a=2023, then a.a=a, a.1/a=1/a.a=1/a and then 1/a.1/a=a.a=a
(i just put the Z2 structure on there, the law just happened to work out in the last line afaict)
rotund aurora
yeah you can just make 2023 the identity and 1/2023 . 1/2023=2023
slim kayak
wondering about less boring examples
rotund aurora
There are probably many
warped sonnet
why isnt {0,1,5} a subgroup of <Z_6;+>? it satisfies all conditions to be a subgroup (inverse, n.e 0, closed under addition), no?
slim kayak
1/e is always idempotent, so it generated a subgroup iso to Z2. Furthermore we have this neat "reflection" property 1/e . a = 1/a
so kinda feels like they maybe look like G x Z2?
rotund aurora
You just want x.y=f(x).f(y) where f is an involution
And f(x)!=x
Well this last remark is vacuous
So you can ignore it. If f(x)=x for some x then f(y)=y for all y
barren sierra
why isnt {0,1,5} a subgroup of <Z_6;+>? it satisfies all conditions to be a subg...
what is 5 + 5 (mod 6)?
rotund aurora
You just want x.y=f(x).f(y) where f is an involution
Given a group, we can ask if such an f exists
Did someone prove H was abelian?
cobalt heath
You just want x.y=f(x).f(y) where f is an involution
So x^{-1} f(x) is constant
rotund aurora
Did someone prove H was abelian?
It's not
cobalt heath
So x^{-1} f(x) is constant
Nah, I am dumb. But maybe the constant is true, my brain is not functioning
rotund aurora
So you want an element a in the center with a^2=e
And this is the only possibility
slim kayak
Did someone prove H was abelian?
Why are you asking?
celest furnace
Is it true that if $E/F$ is a finite extension then $E = F(\alpha_1, \ldots, \alpha_n)$ for some $\alpha_i$'s and $n \geq 0$?
cloud walrusBOT
rbmuk
rotund aurora
So you want an element a in the center with a^2=e
Does this give a direct product Z/2Z x G ? Probably not idk
slim kayak
Nah
cobalt heath
y . f(y)^{-1} = x^{-1} . f(x)
Ah so it is indeed multiplication by involutive element
rotund aurora
Is it true that if $E/F$ is a finite extension then $E = F(\alpha_1, \ldots, \al...
Yes
rotund aurora
But this is trivial
celest furnace
I wrote a proof but i was suspicious
Just keep adding elements until its the whole field right
Each time you add the degree goes up by a factor of >= 2
So has to end somewhere
rotund aurora
I wrote a proof but i was suspicious
The primitive element theorem is far more interesting