that's easy
#groups-rings-fields
static robin
latent anvil
Given an operation * on a finite set S = {x1, x2,...,xn}, you can make a table where the i×jth element is xi * xj
Make sense?
And the table above does that for S = Z/4Z and the operation +
latent anvil
Okay, well do you believe this table is different for different operations?
static robin
and u got 4 n's
static robin
0,1,2,3
latent anvil
In this case n = 4
static robin
ooh
i'm dumb
yeah gotcha
ohhh so by operation you mean the exact operation not just it's definition
latent anvil
Yes
static robin
gooot it
latent anvil
Definitions are just a thing for humans
The actual structure is what math cares about
An operation doesn't even need to be computable, meaning that you could write a computer program which determines x * y given x and y (however every operation on a finite set will be since you can just use a lookup table like above)
static robin
ok
latent anvil
Anyways, if you have two operations * and • which are different, there must be elements xi and xj of S such that xi * xj ≠ xi • xj, which means the tables for * and • are different (specifically they're different at the i×jth entry)
So different operations give different tables
latent anvil
And given a table T, you can define an operation *_T on S by saying xi *_T xj is the i×jth entry of S
latent anvil
Every table is the multiplication table of some operation, and different operations give different tables
Do you believe that?
latent anvil
So that means the number of tables and the number of operations on S are the same
So suppose you have a blank table. How many ways are there to fill it in with elements of S?
static robin
n^2
static robin
lol ? XD
latent anvil
How many ways are there to fill in the 1×1th entry?
latent anvil
Why do you say that?
1×1 is an index, not an operation
If you prefer, how many ways are there to fill in the top left corner?
static robin
1 way
latent anvil
Why?
latent anvil
What's wrong with the table
0|0|0|0
0|0|0|0
0|0|0|0
0|0|0|0
?
static robin
there is no operation
latent anvil
But we're just counting tables
Also there is an operation, x#y = 0
I proved above that every table comes from an operation, and you agreed
static robin
yeah i did
but this one does have one
i can deduce from the output that it does
but like it's not written like the other one
latent anvil
I'm confused
static robin
oh no :((
static robin
np sorry for confusing u :((
latent anvil
What's wrong with the table
1|0|0|0
0|0|0|0
0|0|0|0
0|0|0|0
?
latent anvil
No, x*y=0 if x≠0 or y≠0 and x*y = 1 if x = y = 0
static robin
but either x or y needs to be different from 0
oh no
wait
i'm confused
yeah ur right
latent anvil
Let's step back
I don't want to think about operations at all
I just want to count n×n tables whose entries come from S
You agreed that there are the same number of tables as there are operations on S
Yeah?
static robin
what do you mean there is only one table
latent anvil
I didn't say that
latent anvil
For a given operation, we can build a table
static robin
yea
latent anvil
But I'm trying to count different operations on S
static robin
ok
latent anvil
So I'm saying we can count the number of tables with entries in S instead
latent anvil
No, I'm saying the number of distinct tables overall. How many ways can you fill in an n×n table using elements from S?
To know that there are exactly as many tables as there are operations, we need to know that different operations have different tables and that every table is the table of some operation
static robin
idk
latent anvil
This is called a "bijection"
static robin
depends on how many operations
latent anvil
That's the whole point!!!!
The point is that tables are easier to count than operations
static robin
yes
latent anvil
So do you agree that the number of tables is exactly equal to the number of operations on S
static robin
YEes
cuz that's the whole point of having a table, you need an operation
i know that
latent anvil
Okay, so let's try and count how many tables there are
Forget about operations
You don't need to think about operations at all
We're just trying to count the number of n by n tables with entries in S
static robin
ok
latent anvil
So how many choices are there for the first entry in the table?
The upper left corner
static robin
4
latent anvil
Yes
static robin
also why is it easier to count tables than operations
latent anvil
And the next entry?
static robin
since it's a bijection u can just see how may operations and go
that's how many tables we have
3
and the 2
and then 1
latent anvil
Why are there three?
static robin
no not three threes
latent anvil
Okay, then why 3 and also why 2 and also why 1?
static robin
for the first one i can put one of the 4 choices so either 0 1 2 3
latent anvil
Yes
static robin
then the second, one is eliminatd (for example 0) and i can go either 1 ,2 or 3
static robin
ok so 4 * 4 * 4 * 4
static robin
yea
latent anvil
How many entries are there?
static robin
4
latent anvil
No
static robin
Well 8
latent anvil
So how many are there overall?
latent anvil
I mean how many tables are there
static robin
oh yeh
latent anvil
You're making 16 choices
static robin
n^(n^2) ?
latent anvil
Yes
latent anvil
And that's the number of operations as well
latent anvil
Tables are easier to count because they're just data layed out
Operations are usually thought of as combining stuff or whatever
But it's really just data
latent anvil
This course is really weird fam. I don't know why you would start out with eckmann hilton and magmas
static robin
IKR !!!!
latent anvil
Is this an undergrad course? Do you mind if I ask where?
static robin
1st Year Uni
static robin
Morocco
Website is this
it's in french thoe
the course it self is from another school as well
lol
again thanks you so much for ur time
oof one more question how of those are commutative ?
i guess then can be all comùutative
so same answer
latent anvil
T A B L E S
what does commutativity of an operation look like for the corresponding table?
They can't all be commutative unless n = 1
static robin
oh yeee
yeah cuz then commutativity is automatic
yeah got it
i need to work on these tables
to get the flow of it
latent anvil
What book are you using?
latent anvil
I would recommend getting a book
Usually I recommend Algebra Chapter 0 by Aluffi but there's a lot of stuff in the first section you probably don't have time for. The review of set theory and functions is good though and I think you'd find it helpful. I would also recommend Dummit and Foote
I've heard Artin is good for a first course too
Piracy is your friend for math textbooks 🙂
latent anvil
Np
hardy barn
latent anvil
I think you want #prealg-and-algebra
static robin
i've got something that goes against the uniquness of an invertible element
fading wagon
??
static robin
i got $x*y=x+y+sin(xy)$ i proved that it's commutative and that e = 0 is the identity
cloud walrusBOT
life > Random:
static robin
but for the 2nd question
2)Show that there are some elements in R(set of reals) that have mutliple inverses
but we know that ,$y=ye=y(xy')=(yx)y'=ey'=y'$
cloud walrusBOT
life > Random:
fading wagon
basically, "uniqueness" of an "invertible" element
fading wagon
hmm
static robin
i can use the plotter to cheat
static robin
and say it's not bijective
but that's cheating
i can also solve x+y+sin(xy)=0 but that's hard
is there a cleaner way to solve this ?
plus i can't use a plotter during an exam
:(((((((((((((
fading wagon
Say we keep xy constant but vary x+y?
static robin
how u do dat ?
fading wagon
and quadratic equations are fine, so we can do that
nah, that might not work...
we just need to say more than 1 inverse exists...
static robin
yee
fading wagon
don't need to find them
(like we just need to prove that they are in this interval and that interval)
static robin
ok how would go about doing that ?
fading wagon
sqrt(pi) looks like a good pick
in fact, an inverse is obviously -sqrt(pi)
do we have a second one?
(by plotting it works)
-2<-sqrt(pi)
Next, check if these are positive or negative
-2+sqrt(pi)+sin(-2sqrt(pi))
-3+sqrt(pi)+sin(-3sqrt(pi))
fading wagon
some plotting
I just plotted f(x)=x+sqrt(pi)+sin(x sqrt(pi))
then the second root can use IVT to isolate
static robin
IVT ?
fading wagon
intermediate value theorem
static robin
let E be a set and * it's associative and commutative binary operation, we have also $x*x=x$
- give examples for such a situation
cloud walrusBOT
life > Random:
static robin
i'm not sure even if i understand the question
i came up with this
E= {1} or for all x in E x is idempotent
does that answer their question ?
static robin
no the operation is $x*x=x$ for all x
cloud walrusBOT
life > Random:
static robin
in E
mild laurel
You have to check that this is commutative and associative
static robin
oh pardon me it is supposed to be commutative and associative.
also why would that play a big role in this ?
mild laurel
Uh because that's what the question asks for?
static robin
no the question asks for a situation that the operation would be verified in
wait, do you mean i gotta check the associativity and commutativity for my examples ,
?
if so we already know that E={1} that * is commutative thanks to the table
and it's obvious that it is associative
mild laurel
Yes that's what I mean
static robin
oh ok
so then my example is correct ?
since i proved that it's associative and commutative ?
dusk lintel
static robin
but the pfp !!! the pfp got me :(((
chilly canyon
I asked myself about something earlier this week. By Intermediate Values Theorem, there exists a real x0 such that exp(x0) is algebraic over Q. Then, I wondered : is there an algebraic x0 such that exp(x0) is algebraic ?
More generally, is there always an m-algebraic number such that its exponential is n-algebraic ? (m and n are integers representing the degree of the minimal polynomial) ? If not, what conditions over m and n ?
chilly canyon
Oh yes I forgot about this guy XD okay that's interesting ^^ Do you have any clue where to find this ?
chilly canyon
Oh thank you very much !
languid moss
What topics will be useful in abstract algebra? Asking myself since I still haven’t taken linear algebra
thorn delta
tacit pebble
Is there any easy way to find group isomorphisms between groups? I've been sort of just writing Calyey tables and trying to guess what the isomorphism is (I know it's a permutation from Calyey's theorem). I haven't been able to find anything on the web so I decided to ask you smarticle particles
golden pasture
ehhhh not much linear alg is really needed
abstract algebra is quite doable without much knowledge about most things tbf
latent anvil
@tacit pebble there's not a general algorithm. You need to figure out why two groups should be isomorphic and then define a map based off of that
tacit pebble
Hmmm, ok I remember in my book it was talking about a bunch of properties which groups share if they are isomorphic (like if two groups are isomorphic then if one is cyclic the other one should be too for example). Would a decent strategy be seeing if these properties are satisfied and if they are try to look for an isomorphism somehow or other?
latent anvil
Using that kind of property is a lot more useful for showing when two groups aren't isomorphic
E.g. If G is cyclic but H isn't, then G and H can't be isomorphic
Checking that the groups share a lot of group theoretic properties is good evidence that they're isomorphic, but it won't tell you what the isomorphism is
tacit pebble
ah ok well that must drive algebraists crazy lol
latent anvil
Nah, not really. It's the same as in any other kind of math. What's the general algorithm for proving convergence of an improper integral?
You pick up strategies and heuristics but they're hard to communicate and best learned by working out problems
tacit pebble
that's true i guess
latent anvil
If there were an exact algorithm, math would be pretty boring
tacit pebble
thanks for your help 🙂
latent anvil
Np. I can probably help more if you have an example of this kind of problem you want help with
Or one you've worked out
Also, I remember feeling pretty much the same way. It's normal for your first course in algebra to feel really slippery and like everything is way too abstract
If that makes sense
tacit pebble
I don't have any specific problem I was just trying to find isomorphisms between groups and I got a few for small groups but when it got to the larger sized groups it was taking a while so I just wanted to see if there was better way of going about it.
And yeah i get what you mean, I'm glad I'm not the only one! I usually try to draw pictures because it helps me with my proofs (that's why diff geo and topology are my favs) but with algebra sometimes it's hard to visualize XD
latent anvil
I'm blessed with being absolutely terrible at visualizing things
It makes me good at algebra
tacit pebble
hahahaha we are opposites then
latent anvil
The more you do group theory, the more tools you get, too
Like, after a first or second course in it you can actually classify all groups of certain orders, up to isomorphism
tacit pebble
wait what do you mean by classify groups?
latent anvil
Let p be a prime number. The only groups of order p^2 (up to isomorphism) are C_p^2 and C_p × C_p
Where C_n is the cyclic group of order n
I don't remember exactly what the classification was, but there are like 4 groups of order 28 and an exercise in week 9/10 of the algebra class I help organize is to figure out what all of them are
tacit pebble
woah that's kind of wacky
latent anvil
By these things called the sylow theorems, you can tell immediately that every group of order 28 has exactly one subgroup of order 7, and it's what's called "normal"
And when I say immediately, I mean that the theorem says that if n is the number of subgroups of order 7, then n is congruent to 1 mod 7 and divides 28/7 = 4
tacit pebble
ohhh yeah i remember that
latent anvil
So it must be 1
tacit pebble
yeah we are just learning about that i'm trying to prove the theorem lol
latent anvil
Oh neat
tacit pebble
but don't tell me i want to figure it out lol
latent anvil
How early in your course are you doing sylow?
tacit pebble
tbh our teacher goes really haphazardly and doesn't follow the book like we learned about permutation groups on the first day and bezout multipliers like 2 weeks in
latent anvil
What are bezout multipliers? I'm not familiar with that terminology
tacit pebble
but it's like the 4th week i think?
latent anvil
Permutations groups are a good place to start
tacit pebble
sorry bezout pair
latent anvil
Still haven't heard of it
tacit pebble
like if a and b are co prime, there exist s and t such that as + bt = 1. These s and t can be computed using the Euclidean algorithim
latent anvil
Oh yeah
tacit pebble
more generally if gcd(a,b)=n as+bt=n
latent anvil
Just the coefficients you get out of bezout's lemma
tacit pebble
yeyeye
latent anvil
The algebra course I'm involved with does basic definitions and examples of groups week 1, group homomorphisms week 2, subgroups and quotients week 3, Lagrange's theorem and the isomorphism theorems week 4, group actions week 5, sylow week 6, composition series and simplicity week 7, in depth look at the symmetric and alternating groups week 8, semidirect products week 9, and the classification of finite abelian groups week 10
It seems like there's a lot of variation in what gets covered when
I've seen some places put off quotients for a while and I just don't understand how you can do group theory without them
Like sure they're hard but also really important
tacit pebble
hm interesting I don't know what quotients are yet lol, but we are using the galliean text but again our prof kinda just goes with whatever which is fine i guess
latent anvil
I find it really questionable to do sylow before quotients lol
The idea of a quotient is that if H is a "normal" subgroup of G, then we can create a new group G/H which looks like G but all the elements of H have been collapsed down to the identity
So as a consequence, if x and y are elements of G such that x = hy for some h in H, then in the quotient group they must become equal, since we want h to become the identity
And this is pretty much the definition. You define two elements x and y of G to be equivalent if x = hy for some h in H, and then partition G into "equivalence classes" by this relation
So if x and y are equivalent, you put them in the same box, and if they're different they get put in a different box
If H is "normal", then you can put a group structure on the set of these equivalence classes/boxes. Given a box B and another box B', you pick an element x of B and y of B', and say B * B' is the box containing the element x * y
Sorry this is not a great explanation. It's useful to be able to study groups by quotienting subgroups out though
tacit pebble
haha thanks for trying though, i will probably understand it better when I read more about it. Also when it's not so late XD
tacit pebble
reason i'm trying to learn group theory is actually cause i'm a physics and math double but i find GR really interesting. But the lecture series i was watching eventually started talking about Lie Groups and I was very confused (more confused than I am usually that is)
latent anvil
can't relate lmao. I got a 2.7 in electricity and magnetism because I can't do physics
I am not good with the actual world around me, that's why I do math
tacit pebble
i feel that EM is annoying lol
latent anvil
I definitely didn't like it
I'm taking differential geometry next year and I guess that'll prepare me to do physics if I ever feel the need
I'm very hyped because I'm taking it with Jack Lee, who wrote the book Introduction to Smooth Manifolds. I've heard the book is really incredible and I'm hoping the class is even better
tacit pebble
woah that's cool I know Intro to smooth manifolds (haven't read but know of it). If you ever want to watch a really math based lecture series on GR i recommend the youtube channel XylyXylyX he does GR from a rigourous maths perspective which if you are a mathemetician you can def appreciate
latent anvil
If you want some fun group theory problems to practice, check out https://math.berkeley.edu/~tb65536/algebra_materials/index.html
I helped write these problem sets and before that I had to do them when in the class. I can vouch that they're pretty great
tacit pebble
oh cool thanks for the resource! I will def check them out
the one thing i really like about abstract so far is game isomorphisms
that stuff is super cool
tacit pebble
i think so
like how the game of 15 is isomorphic to tic tac toe. or peg solitare actually forms a klein 4 group which you can use to find all possible solutions
latent anvil
There's a cool thing in model theory which is sort of the reverse https://en.m.wikipedia.org/wiki/Ehrenfeucht–Fraïssé_game
You use a game to show that two models are "elementarily equivalent", which is a little weaker than isomorphic
tacit pebble
oo i'll check out when i'm more awake there is so much stuff in math lol
tacit pebble
very true
static robin
let E be a set with a associative and commutative binary operation * we suppose that for all $x$ of $E x*x=x$.
i have shown that $xRy \iff xy=y$ is a RAT relation, now i need to show that for all $(x,y)$ in $E^2 sup(x,y)=xy$
cloud walrusBOT
life > Random:
static robin
which i don't know how to show ofc
latent anvil
What's a RAT relation? Reflexive, antisymmetric, transitive?
Also, what does the notation E^2 sup(x, y) mean?
Oh, for all (x, y) in E^2, you want to show that sup(x, y) = x * y
In this case, sup is being taken with respect to the relation R, which we're understanding as an order on E
Do you understand what a supremum is?
@static robin
static robin
sorry i wasn't here let me read it :))
latent anvil
Np, I just wrote it
static robin
yes i do
and yes i thought that there is a something to do with R here
but i don't know where to even start
i know that the sup(A) for example can be the Max(A) if A has a max
latent anvil
That's not true in general
Only if the order on A is a total order
Consider the relation S on {a,b,c} where aSc and bSc
static robin
yeah
latent anvil
Then {a, b} has a sup, namely c, but no maximum
static robin
and that sup(A) is the smallest of all majorants.
latent anvil
Yes
In the problem, a majorant of the set {x, y} is an element z of E such that xRz and yRz
Make sense?
static robin
lol you're giving me so much more info , i didn't know that A has to be in total order to say that
cuz they didn't say that in the notes
latent anvil
Can you quote the part of the notes which talk about orders?
Well okay, I should clarify
latent anvil
If A has a maximum, then sup(A) can be that maximum
static robin
if A has a biggest element then A has a supremum and Sup(A) = Max(A)
static robin
ok
latent anvil
I was thinking about the statement that every finite set A has a maximum
static robin
ohh
latent anvil
This is true iff you have a total order
static robin
ok
latent anvil
Sorry, it's a little late where I am
static robin
oof
latent anvil
Okay, anyways
static robin
thanks anyways :DDD
static robin
yes
latent anvil
Given that a*a = a for all a and that * is associative and commutative
So you need to prove first that x * y is an upper bound for {x, y}
Can you do that?
Upper bound = majorant
static robin
by upper bound u mean prove that it's a majorant ?
oh
no i don't know how to do that eihter
either*
latent anvil
Yeah sorry, upper bound is the standard English terminology
static robin
french hon hon hon
latent anvil
Well when is an element b a majorant of a set A?
static robin
ok so for all x in A, bRx.
static robin
and then that translates to for all x in A xRb <=> x*b=b
latent anvil
Right
So what's the set A we're considering and what element are we trying to prove is an upper bound of A?
latent anvil
So what do we need to do to prove x * y is an upper bound of {x, y}?
static robin
x*y = y so i need to prove that y is the upper bound of {x,y}
and we sorta know that
latent anvil
How do you get that x*y=y?
latent anvil
What?
static robin
before this one
latent anvil
You're trying to prove xR(x*y) and yR(x*y), right?
static robin
well if it wasn't true then for what binary operation are we prooving that sup(x,y)=x*y
latent anvil
We assume there is a set E with a commutative, associative relation * on e
We then define a relation on E by saying, for arbitrary x, y, that xRy iff x * y = y
static robin
no we don't
latent anvil
?
static robin
we suppose that for all x in E x*x=x
static robin
Eyyy :DDDD
latent anvil
We now want to prove that for any x, y in E, sup(x, y) = x*y, where the order for sup is this relation R defined above
It's not immediately true that xRy
static robin
why not ?
latent anvil
xRy is a proposition which is defined to mean x*y=y
static robin
ye
latent anvil
Why would that always be true, for any two x, y?
static robin
cuz x*y is unknown so how can u be sure that it's not defined
oh yae
i see now
yeah yea
static robin
i meant that since u don't know what * is and how it behaves you can't say that it's not true for any x and y
but yeah i get you now
latent anvil
Well sure, if E is a one element set then it'll be true
But in any other case it will fail for some x, y
static robin
yea
latent anvil
So we want to prove x*y is an upper bound of {x, y}
static robin
yes
static robin
we have to to that for x and y xR(xy) and yR(xy)
latent anvil
Yes
static robin
eyy
latent anvil
So what does that mean?
latent anvil
Could you put the character ` around your math?
The formatting makes it hard to read
No I mean like `x * y`
It's a backtick, not an apostrophe
Ty
Can you elaborate on why y*(x*y) = x*y?
latent anvil
Oh I believe that yR(x*y) is equivalent to y*(x*y) = x*y
But you have to prove that this is true
latent anvil
Prove either is true
static robin
how can i do that ?
latent anvil
Try and prove y*(x*y) = x*y using what you know about *
static robin
yeah here it is
(x*y)*(x*y)=x*y => (x*y*y)*(x*y)=x*y => y*(x*y)=x*y
by * being commutative and associative
latent anvil
I don't understand these implications
Why does (x*y)*(x*y)=x*y imply (x*y*y)*(x*y)=x*y?
Are you using the fact that y = y*y?
latent anvil
And then the next step is moving the x over?
static robin
yeah since * is commutative
latent anvil
Wait, but don't you then collapse y*y back down to y?
latent anvil
Yeah, but why expand it in the first place?
static robin
yeah i see it now
latent anvil
Your proof works, though you should probably edit it a little
latent anvil
A shorter one is y*(x*y) = x*(y*y) = x*y
static robin
yeaa
latent anvil
Okay, so we know yR(x*y)
static robin
that's for x
latent anvil
By symmetry, xR(x*y)
static robin
yehs
latent anvil
What do you mean that's for x?
static robin
yeah
static robin
yup
latent anvil
What else do we need to do to prove sup(x, y) = x*y?
static robin
well i can just say that since x*y=y (i got that from R) and since y is in the set {x,y} i can say that Max{x,y}=Sup{x,y}=y=x*y
that's from my notes
latent anvil
fam
static robin
oh no
latent anvil
i thought we already ran into this issue
how can you assume xRy is true for any x, y?
latent anvil
You can't
That argument is bogus because you're mixing up the propositions xRy iff x*y = y and xRy
The first is true by definition, but the second may not be true
latent anvil
How do you prove something is the smallest thing?
You prove it's smaller than anything else satisfying the property you care about
static robin
so for all a in sets of upperbound i need to prove that x*y<a
static robin
ok
let me do it
i'm stuck idk how to prove that
i can't use R for this
can i go like this ?
(x*y)*a=x*(y*a)=x*a=a ?
or did i just make a horrific mistake ?
latent anvil
No mistakes, that's exactly right!
latent anvil
Good job
static robin
thanks
latent anvil
And that proves the result
Okay, cool example time
Let X be any set
Let E be the powerset of X
I.e. the set of subsets of X
And define * to be the intersection of two sets
Make sense?
Then for subsets A, B of X, we have ARB iff A*B = B iff the intersection of A and B is B iff B is contained in A
latent anvil
And you just proved that the supremum of A and B, i.e. their greatest lower bound wrt containment, is their intersection
Also, it turns out that all finite rings whose multiplication satisfies x*x = x are commutative and are powerset rings of some set
When someone showed me that problem, I invented the relation R to solve it!
I wasn't the first ofc but it's cool to see something I came up with independently
static robin
lol
yeah that's what i'm trying to reach
is to invent things to help you in solving problems
function/relations conditions even
but thanks again for the help !!
i know i'm sometimes slow and i really appreciate the help :DDDDDDDDDD
static robin
i have another one : let E be a set with an associative binary operation.
for all a of E we have for all x in E :
we suppose that there is an a in E as f and g are surjectives.
- show that E has an identity for *
now i did this : since f and g are surjective then there must be an x in E so that g(x) = a*x = a and f(x) = x*a = a now what's bugging be is am i right on taking the same x or am i supposed to take 2 different x's and show that they're the same?
2)show that every element in e is invirtible same here i don't know if i have to show that sym'(a)=sym(a)
wind steeple
You have to proove that x are the same
static robin
ok
fading wagon
is E a finite set?
static robin
yes
fading wagon
Ah, so it is bijective too
@static robin
Can you show that?
by f and g, you mean d_a and g_a?
Also, you may want to consider $d_a^{|E|!}$
cloud walrusBOT
Element118:
cloud walrusBOT
life > Random:
static robin
i've never heard of it.
static robin
yes, either i never heard of it, or it's a different terminology
is it this ? $fog$ ?
cloud walrusBOT
life > Random:
static robin
we call this the composition of two function.
i haven't heard of a function composed [insert set here] times.
cloud walrusBOT
Darkrifts:
static robin
i don't think we studied that yet
we just started so maybe along the year we'll see it. but not yet
topaz solar
it's definitely good to know about function composition
static robin
i can deduce from $gof=g(f(x))$ and then say g=f and get it but i mean with a whole set i don't know
cloud walrusBOT
life > Random:
static robin
oh for us we write $CardE$ for example
cloud walrusBOT
life > Random:
topaz solar
that's acceptable but I prefer |E|
static robin
oh well different terminology.
topaz solar
different symbols
static robin
😅 yeah symbols
mild laurel
Is there a finite UFD that is not a field?
or maybe a UFD with non-zero charateristic that is not a field
hot lake
if you have a finite UFD and pick a nonzero x, then x^a = x^b for some a > b >= 1, then you get x^(a-b) = 1, so x has an inverse x^(a-b-1)
tender mist
I think the argument doesn't work, though. x^(b-a) is defined only when x is known to have an inverse, right?
As x^(-a), I mean
hot lake
I managed to mess up the order of a and b
tender mist
Wouldn't it be the same thing, even with the order reversed?
I guess you'd still have to multiply each side by either x^(-a) or x^(-b)
hot lake
x^a = x^b, so (x^(a-b)-1) * x^b = 0
and x^(a-b) is well defined because a-b is a positive integer
now because they are in a domain
either x^(a-b) = 1 or x^b = 0
if x^b = 0 then x= 0 (because domain)
and I picked x nonzero
so we must have x^(a-b) = 1
tender mist
Oh I see, interesting. Thanks, I had never seen this short proof
mild laurel
This is really something you should be telling hs
Why would you ask a question where you don't even know what some of the notation means
languid moss
I'm confused as well
golden pasture
i think its something like the teacher thinks that everyone knows that notation so doesn’t mention, but like it never get mentioned
mental onyx
The integers 5 and 15 are among a collection of 12 int that form a group under multiplication modulo 56. how to find all 12 elements in the group?
I have no idea even to approach it
golden pasture
56=2*2*2*7, and every element in the group must be invertible, so would like 2 be in the group?
mental onyx
I don't know? you saying 2 will be present in the group?
golden pasture
Maybe ill rephrase this way
Since every element must have an inverse, suppose 2 is in the group, does 2 have a inverse?
from this try to like guess what kind of numbers would (not) have an inverse
mental onyx
we need identity element I believe for that!
golden pasture
identity in the multiplicative group mod 56 should be trivial i hope? x*e=x for some e
mental onyx
e = 1
golden pasture
yea
mental onyx
ok so 2*a = 1
golden pasture
So 2*a+k*56=1(in Z) for some a and k
similarly there are a bunch of other numbers that doesn't have a inverse
mental onyx
so how do we find?? 2 doesn't have inverse
golden pasture
yea
mental onyx
so check for 7?
mental onyx
is there a quick way to check that too?
since 1 is identity, I'm adding 56 repeatedly to check if its divisible
unkempt socket
Have you done any number theory before
golden pasture
Suppose a and 56 have a common prime factor
will a have an inverse?
Similarly suppose a and 56 are coprime
does a have a inverse?
unkempt socket
That's right, and what they are talking about is:
Suppose a has a multiplicative inverse mod 56. Can a and 56 have any prime factors in common?
mental onyx
yes
unkempt socket
mental onyx
am I wrong ?
golden pasture
Suppose a and 56 have prime factors in common, let k divide both a and 56
now we can write the fact a has an inverse as
ma+n56=1 for some m and n
However 1=mk(a')+nk(56/k)=k(m(a')+n(56/k)), so 1=k*some integer, which doesn't quite make sense
mental onyx
I thought this would be easier!
mental onyx
is there any easy way?
golden pasture
easy?
unkempt socket
You're working with stuff modulo n, how can it be any easier than invoking some NT 😃
unkempt socket
number theory
mental onyx
any other way to solve it?
unkempt socket
idk, you're almost done at this point.
mental onyx
but I don't understand what you said back there
ma+n56=1 for some m and n
However 1=mk(a')+nk(56/k)=k(m(a')+n(56/k)), so 1=k*some integer, which doesn't quite make sense
There's a thing called bézout's identity, and one of its corollaries is that if a,b are coprime then there are integers m,n such that am + bn = 1
so b = 56 in this case, and we have am + 56n = 1. However, we also said that k divides both a and 56, so multiply and divide by k on the left side:
k( (a/k)m + (56/k)n) = 1
The inner thing is an integer, so we have that k * an integer = 1, which is absurd
unkempt socket
They must be, if a multiplicative inverse for a exists.
mental onyx
you are saying k doesn't exist meaning they must be coprime
unkempt socket
I am saying that k = 1.
golden pasture
consider ax+by=1
Suppose we know x,y and we want to solve for a,b
If x and y has a prime factor k, ax+by can be rewritten as k(ax'+by')=1, but this can't be true unless k is 1, which means x and y must be coprime(then you prob need to proof that a and b exists if x and y are coprime)
golden pasture
rn i cant quite think of some simple proof without directly computing the coefficients
mental onyx
I can't understand this part where you prove x and y are co prime why should we do that?
golden pasture
It basically shows if x,y are not coprime, a,b can't exist
mental onyx
so what do we need to do from that proof?
golden pasture
Suppose x,y are not coprime
there exists a k>1 such that x=kx' and y=ky'
mental onyx
then
mild laurel
What's the question?
mental onyx
I have no idea even to approach it```proof that ax+by=1 has no solutions for (a,b), assuming (x,y) are not coprime with gcd(x,y)=g
mental onyx
wait I think he meant the real question
mild laurel
Doesn't coprime mean that gcd(x,y) = 1?
mental onyx
yea
golden pasture
tl;dr
If gcd(x,y)=1, then there exists (a,b) such that ax+by=1
If gcd(x,y)>1, then there does not exist (a,b) such that ax+by=1
mental onyx
how can I solve my problem with this?
unkempt socket
By eliminating the numbers a where gcd(a,56) =/= 1.
mental onyx
gcd(a,56) =/= 1. ??
mild laurel
gcd(a,56) not equal to 1
I'm not sure what you guys talked about yet
But you should think about what the identity of this group has to be
golden pasture
Let y=56
ax+56b = 1
ax = 1 (mod 56)
there you have an inverse
unkempt socket
We did that zoph
mental onyx
so we need to find all the elements that have gcd(a,56) = 1 right?
unkempt socket
Yes.
mental onyx
but the thing I don't understand is how if a,b are coprime then there are integers m,n such that am + bn = 1 so b = 56 in this case, and we have am + 56n = 1. However, we also said that k divides both a and 56, so multiply and divide by k on the left side: k( (a/k)m + (56/k)n) = 1 The inner thing is an integer, so we have that k * an integer = 1, which is absurd related to the problem?
mild laurel
That's showing why if gcd(a,56) is not equal to 1, then a can't have an inverse
mental onyx
can you explain it please coz I can't clearly get it
mild laurel
What don't you understand
mental onyx
the explanation itself
k( (a/k)m + (56/k)n) = 1
The inner thing is an integer, so we have that k * an integer = 1, which is absurd``` this onewe said a, b are coprime then why we say k divides both a and b?
unkempt socket
Because we are trying to reach a contradiction
mild laurel
Nah this is just pretty poorly written
I think the better way to think about this is that
if there is some k that divides both a and 56
with k > 1, then for every integer you can write as ma + 56n, then this integer will also be divisible by k
So if gcd(a,56) > 1, there doesn't exist an m such that ma = 1 (mod 56) since 1 isn't divisible by k > 1
mental onyx
but we said a, b are co-primes earlier right? are we trying to contradict it?
mental onyx
So if gcd(a,56) > 1, there doesn't exist an m such that ma = 1 (mod 56) since 1 isn't divisible by k > 1
what is ma here?
mild laurel
m is any integer
mental onyx
so if gcd(a,56) > 1
then (ma + 56n) mod 56 =/= 1?
mild laurel
Yes
mental onyx
ok I don't understand why this is true but still I get this as an identity/proof
mild laurel
what don't you understand
mental onyx
why/how is that true part
mild laurel
why what is true
mental onyx
with k > 1, then for every integer you can write as ma + 56n, then this integer will also be divisible by k
mental onyx
ya so we can write it as k( ma' + nb')?
So if gcd(a,56) > 1, there doesn't exist an m such that ma = 1 (mod 56) since 1 isn't divisible by k > 1 wb this part?
mild laurel
Think about it
mental onyx
if k > 1 then it means the same as gcd(a,56) > 1
which means k( ma' + nb') = 1
oh okay then here k , (ma' + nb') are integers so this is wrong
so k = gcd(,) = 1 right?
mild laurel
No you're still not understanding it
mental onyx
mild laurel
Where did I ever say that ma + nb = 1
mild laurel
The whole thing?
I never said that ma + nb = 1, so what you tried to argue is completely wrong
mental onyx
I'm not arguing, how can I argue when you're trying to help me and I don't know any of this?
mild laurel
Arguing as in making a mathematical argument
mild laurel
The whole thing?
I never said that ma + nb = 1, so what you tried to argue is completely wrong
mental onyx
even then I can't understand this there doesn't exist an m such that ma = 1 (mod 56) since 1 isn't divisible by k > 1
mild laurel
Think more
mental onyx
i get ma
mild laurel
Correct
mental onyx
a mod b = a + bn mod b
what next?
so if a, 56 are not co-prime then ma + 56n mod 56 will not be 1 right?
mild laurel
that's the idea
mental onyx
is there any fail proof way to find generators for groups?
mild laurel
depends what you know
mental onyx
atleast the number of generators?
pls say it I'll check
I think for multiplicative group trial and error is the only way is it?
mild laurel
That's true
mild laurel
you can find the number of generators easily
mental onyx
how?
the number of generators of a cyclic group is the number of integers that are co-prime with the order of the group? that's what I did
mild laurel
You have to be careful
Are you sure your group is actually cyclic
Not all multiplicative groups are
mental onyx
cyclic groups have generators right?
mild laurel
That's the definition of a cyclic group correct
mental onyx
the group given to me multiplicative and cyclic
yes not all multiplicative groups are cyclic
mild laurel
Okay then yes if its cyclic, then it will have generators
mental onyx
how to find it?
mild laurel
I told you already, just guessing is as good as you're going to get
mental onyx
at least the number of generators?
mild laurel
That's doable
mental onyx
how?
mild laurel
think about it
Maybe first think about how many generators the additive group Z/nZ has
mental onyx
the number of generators of a cyclic group is the number of integers that are co-prime with the order of the group
is it correct?
mild laurel
Yes
mental onyx
will it work for any cyclic group?
mild laurel
All cyclic groups
mental onyx
thank you @mild laurel @golden pasture @unkempt socket
mental onyx
is there any way to find the number of subgroups?
golden pasture
if it is cyclic, its trivial, simply the number of divisors of the order
else sylow theorem
golden pasture
as in?
mental onyx
I don't know sylow theorem
golden pasture
(its ok you'll learn it eventually)
mental onyx
okay how do I do it?
Since I'm not a math student I will not be taught these things unless I learn them on my own
mild laurel
Finding number of subgroups is hard
mental onyx
the question is find the minimum number of non-trivial sub-groups of a group with 32 elements
also can that group be Abelian?
how do I do this ? by finding the number of subgroups?
mild laurel
There are a lot of groups of 32 elements
mental onyx
meaning?
mild laurel
You're trying to find the one with the minimum number of subgroups
chilly ocean
mild laurel
Read #❓how-to-get-help please
I told you to read #❓how-to-get-help
This is not the right channel for that material
tacit pebble
Can someone give me a hint to find automorphisms of D7 (the dihedral group of order 14)
(r and f are rotation and flip here) I'm thinking that r gets sent to r^k where if n is the order of r, n and k are coprime. This is because we need to send cyclic groups to cyclic groups.
stone fulcrum
Dihedral groups have two generators. We can match generators to generators to find an auto
tacit pebble
Ok dope
tacit pebble
Yeah that's where it gets fishy, for D_2n where n is positive integer we can send f to (r^k)*f but i was getting tripped up with the parity
stone fulcrum
There's no other elements of order 2 so f has to stay fixed
Then, we just pretend we're working in Z7
tacit pebble
oh wait you're right because (r^k)*f only works with evens with odd it's not a generator!
so that means that there are 14 automorphisms?
stone fulcrum
You can move r to any other elements except e or f. Everything else is an order 7 generator right?
Wait, no there's going to be some elements that are order 14
I don't think you can move r to any term with an f
stone fulcrum
There's 7 automorphisms, given all that
tacit pebble
Ok i'm gonna cut out a septogon lol
wait can't we send f to rf? rfrf=rff^-1r=e so order 2 element
tame bear
void marsh
potent lynx
not here
tacit pebble
I don't think this belongs in abstract algebra
barren delta
I need to prove that every cyclic group is abelian. Is my reasoning okay here?
$G$ is cyclic so there is some $a \in G$ such that $\langle a \rangle = G$.
Then for every $a \in G ;\exists ; a^{-1}$ so that $a * a^{-1} = e = a^{-1} * a$
mild laurel
how does this show that a group is abelian
cloud walrusBOT
kickpuncher:
barren delta
oh shit, right because that's only each a with its inverse....
where it needs to be any a with any other....
mild laurel
In any group, every element commutes with its inverse so
tacit pebble
hint: write 2 elements of G in terms of the generator of the cyclic group
barren delta
right that's what I was thinking just now but I'm not sure how exactly...
something like $x=a^n,; y=a^m \implies x*y = a^{m+n}$
?
cloud walrusBOT
kickpuncher:
tacit pebble
yep yep you are almost there
barren delta
heck
tacit pebble
m and n are integers right so what does that mean about their commutation
tacit pebble
lolol i know that feel
barren delta
i've had like 8 cups of coffee so I'm well past my prime today
barren delta
lol
bleak abyss
Alright Silverman time
bleak abyss
Okay time for an example
bleak abyss
As for the other, $\frac{(x+y)^2 + 1}{2(x+y)} = \frac{1}{2}(x+y) + \frac{1}{2}\frac{1}{(x+y)}$, but then $1 = x^2 - y^2 = (x+y)(x-y)$ so we get $\frac{1}{2}((x+y) + (x-y)) = x$
cloud walrusBOT
Dami:
bleak abyss
And similarly for the second coordinate
So these guys want varieties to be irreducible
cobalt pilot
This is more of a commutative algebra question, but I saw no topic for that. Assume that A is a commutative ring with 1. Let $f:A\rightarrow A/I$ be a ring homomorphism and $I \subset A$ be an ideal. Let $J \subset A/I$ be an ideal as well. Apart from $f^(-1)(J)$ being an ideal, what else can se say about inverse images of ideals from $A/I$ in $A$ ?
cloud walrusBOT
Xiphias:
woven delta
There's a correspondence between ideals containing I and ideals of A/I
This is the fourth isomorphism theorem
mild laurel
It's also an inclusion preserving correspondence
cobalt pilot
I'll look that theorem up, thanks!
Care to elaborate your answer @mild laurel ?
(I took commutative algebra 9 years ago and I'm trying to brush up on it to improve my grade)
mild laurel
The correspondence that horsie stated is inclusion preserving
If J \subseteq K as ideals of A, then you also have that J \subseteq K when you reduce them down to ideals of A/I
cobalt pilot
Thanks, I'll try to wrap my head around this. Are there any handy examples to help?
woven delta
Quotient by a maximal ideal is a field
cobalt pilot
Alright. Thanks for your help on this. Probably won't be the last question this year.
blazing hazel
can someone help me with 23 here?, 21 is "regular" and 20 is "effective" I think
latent anvil
Is this an algebraic geometry crossword? lmao
blazing hazel
👀
round niche