groups-rings-fields

[hot folio]

?

[magic owl]

Well we get

$f^{-1}\big(\sum \alpha_i f(b_i)\big)=f^{-1}(0)$

[cloud walrus]

MaxJ:

[magic owl]

$\sum f^{-1}(\alpha_i f(b_i))=0$

[cloud walrus]

MaxJ:

[magic owl]

and then?

[hot folio]

$\sum \alpha_i f f^{-1}((b_i))=0 \implies \sum \alpha_i b_i=0 $

[cloud walrus]

Katydid:

[magic owl]

yep

and what do we know about all the alpha_i then?

[hot folio]

are all equal to zero

[magic owl]

And that's what we needed to show!

[hot folio]

i dont get why $ f^{-1}(0) = 0 $

[cloud walrus]

Katydid:

[magic owl]

if $f$ is linear then $f(0)=0$

[cloud walrus]

MaxJ:

[hot folio]

oh that was the key

[magic owl]

$f(0)=f(0+0)=f(0)+f(0)=0$

[cloud walrus]

MaxJ:

[hot folio]

and since it's bijective only f(0) = 0

[magic owl]

Yeah

Fun fact

The amount of things that f sends to 0 for any linear function f

measures exactly how "not injective" f is

this only works in vector spaces and similar structures, tho

[hot folio]

isnt that the "f is injective if kernel(f) = {0} "

[magic owl]

To see why: if $f(a)=f(b)$ then $f(a)-f(b)=0$ and $f(a-b)=0$. If $a=b$ then $a-b=0$ and this makes sense. But if not, then $f(a-b)=0$ means something extra gets sent to 0$

and yes

[hot folio]

i never thought about it like that

[magic owl]

Also the statement that if $f:A\to B$ then $\text{image } f \cong A/\ker f$

[cloud walrus]

MaxJ:

[magic owl]

where / is a quotient, dunno if you have seen those

[hot folio]

where can i find the proof of that

i mean what should i search for

[magic owl]

It's the second isomorphism theorem

[hot folio]

that will be helpful

this was really really helpful too

i really appreciate the help, many thanks

[magic owl]

np anytime

the intuition there by the way

is that image f should look like A

except sometimes f isn't injective

the kernel deals with that

[tame bear]

im f = A / ker f
is the first isomorphism theorem 🤔

[magic owl]

Wait is it

oh fuck lmao

my b

I've always thought of it as the second for some reason

[fleet igloo]

Allo.

How are you ? Which math can be interresting ?

Document** I want to say.

[languid moss]

Anyone know Bézout’s identity?

[potent lynx]

gcd(a,b) = ax+by ?

for integers x and y

[woven delta]

The number of intersections of 2 curves in CP^n is the product of their degrees, counting multiplicity?

[hazy flint]

Why is S3 a subgroup of S4?

[gentle pendant]

Because Sn is the group of permutations of a 4 element set. Consider those that fix the 4th element of this set.

This subset forms a subgroup and is isomorphic to S3 in a natural way.

[fickle brook]

@hazy flint technically it's not

[faint elm]

Let $V, W, K$ be irreducible representations of $S_n$, $S_m$ and $S_k$. I want to show that: $Ind_{S_{n+m} \times S_k}^{S_{n+m+k}}( Ind_{S_n \times S_m}^{S_{n+m}}(V \otimes W) \otimes K) = Ind_{S_n \times S_{m+k}}^{S_{n+m+k}}(V \otimes Ind_{S_m \times S_k}^{S_{n+k}}(W \otimes K))$. Do you have any hints?

[cloud walrus]

emme:

[simple valley]

S3 can be identified with a subgroup of S4

in a lot of higher math we tend to talk about structures up to isomorphism

[winter vigil]

The number of elements in $N(Z_n)$ is one, which is the element 0, as only $0^k =0mod n$. Is this correct ?

[cloud walrus]

Otoro:

[mild laurel]

Uh, no?

You should read what the question is asking for very slowly @winter vigil

[winter vigil]

Then is it m elements ?

[mild laurel]

Also no @winter vigil

[winter vigil]

Any hints? I think I dont understand what the question is asking, sorry

[mild laurel]

Have you done the proof step?

[woven delta]

Let m=4. Then 2 is in the nilradical

But for example 1 is not

1 will never be in the nilradical for obvious reasons

So no unit is in the nilradical

Think about the nilradical for prime powers

@winter vigil these are just some things to think about

[winter vigil]

Is this proof correct @mild laurel

[mild laurel]

No? How do you get that n divides m?

@winter vigil

2^2 = 0 mod 4, but 4 doesn't divide 2

[woven delta]

Maybe he doesn't know that unique factorization doesn't hold in Z_n for n not prime

[winter vigil]

But then doesn't unique factorization is all primes ?

[mild laurel]

@winter vigil What are you asking?

[winter vigil]

Wait I mean, m is a multiple of primes ??

[mild laurel]

uh yes?

It definitely doesn't say that m is prime anywhere

[winter vigil]

Well if m is a multiple of primes, then one of the prime divides m doesn't it

[mild laurel]

one of what primes

[winter vigil]

Prime in the uniqur factorization

[mild laurel]

unique factorization of what

[winter vigil]

Wait, sorry, is a certain amount of primes from n, divide m, then it is in the nilpotent set ?

But as n is a multiple of primes, then Z_n is the integers mod p right ?

[mild laurel]

what

what's p

[winter vigil]

Prime

[mild laurel]

what prime

[winter vigil]

Wait sorry, mod of the product of the primes

[mild laurel]

The product of what primes

[winter vigil]

The primes of n

[mild laurel]

You really need to be more exact with your language

[winter vigil]

I'm sorry

[mild laurel]

It's still not mod the product of the primes

You have to take the primes to the correct powers

[winter vigil]

Causr I don't really understand what I am doing to be honest

[mild laurel]

Then you should figure out what you're confused about and try figuring that out?

Instead of trying to do a problem where you have no idea what's happening

[winter vigil]

Well, here goes.
So n is a just a number represented by primes. m is a number where m^k=0mod n. As m^k is divisible by n, m^k=jn. Then m^k=j*(p_1^k_1 p_2^k_2 ... p_m^k_m)
Therefore m^(k-1)m = j(p_1^k_1 ... p_m^k_m).
Hence p_k divides m for all k=1,2,...,m.
Is this ok for "only if" ?

[mild laurel]

Yeah this still definitely doesn't work

[winter vigil]

Hmmm where is the problem

@mild laurel

[hot lake]

you should try to not give m two different meanings

also the question has a typo where they have a m instead of j

[mild laurel]

Whoops sorry

My internet went out

How do you figure out that p_k divides m

[winter vigil]

Oh so the amount of primes, j, is an element itself in the nilpotent rings ?

Its alright haha, um its is from where m^k for some k, is divisible by n ?

[mild laurel]

what are you even saying

[winter vigil]

Well he said that there is a typo. So what I'm saying is that the m is replaced by a j, thus j itself as an element is divisible by primes from p1 to pj ??

[fringe nexus]

(Q/Z)/Z = (Q/Z) right?

[mild laurel]

It depends what you mean by Z as a subgroup in Q/Z?

[fringe nexus]

Uhhh

So i'm constructing a homomorphism from Q/Z to the multiplicative group of unity

such that

$\phi(\frac{p}{q} + \mathbb{Z}) = e^{ \frac{i2\pi p}{q}}$

And the kernel of this is Z itself, and I want to show that Q/Z is isomorphic to the multiplicative group of roots of unity in C^x

in this case if i quotient Q/Z by the kernel, is it still Q/Z?

[cloud walrus]

Victoria:

[mild laurel]

The kernel of your homomorphism isn't really Z itself

It's the element 0 + Z of Q/Z

Which is just the 0 element of your group

And obviously if you quotient your group by the trivial subgroup, it'll stay the same

[fringe nexus]

ah

i see

[mild laurel]

So yeah, writing it as (Q/Z)/Z isn't really accurate

[fringe nexus]

its (Q/Z)/(0+Z)

Right?

wait

[mild laurel]

Yeah kinda

The notation is a bit weird here but

Maybe (Q/Z)/({0 + Z}) would be a bit better

[tender mist]

Also since the kernel is {Z}, where Z is the 0 of Q/Z, you could directly say Φ is injective

[fringe nexus]

For this question, the orbit of S_3 on (1,1),(2,2)(3,3) is just (1,1),(2,2),(3,3) and the orbit of S_3 on (1,2) is everything else

Does that mean that the orbit of S_3 on any other element which isnt (1,1)(2,2)(3,3) is equivalent to the orbit of S_3 on (1,2)?

[tender mist]

@fringe nexus yes because the orbits form a partition of the set (and each element is in its own orbit), so if two elements are in the same orbit their orbit is the same

In other words being in the orbit of is an equivalence relation

[fringe nexus]

okie

i dont understand

what it means, find the cycle decomposition

im so confused

[tender mist]

Each element of S_3 represents/is associated through the action to a permutation of the set Ω which has 9 elements

It asks to find for each σ in S_3 the cicle decomposition of the permutation in Sym(Ω) associated to σ

[fringe nexus]

ok say we let a_1 = (1,1), a_2 = (2,2) a_3 = (3,3)

oh wait nvm i dont think i understand this

is this what it should be

[tender mist]

Yes something like that, I don't know the full labelling you used so I can't compare results

[fringe nexus]

ok that makes a lot more sense

Can i have some sort of hint for 2)?

idk why its so small

[chilly ocean]

Stupid quick question, but if I show that this group is commutative , I can assume that it's abelian, right?

[topaz solar]

abelian means it's commutative

[chilly ocean]

Ok

Yeah, I did a big dum where I went "I proved that this group is communtative... how do I prove that this is abelian?"

:P

[winter vigil]

@woven delta Liquid
How does unique factorization not hold for Z_n for n not prime

[fading wagon]

hmm...

So all ideals are principal, because we can easily Euclidean algorithm.
So the ideals are just (a), where a is a positive factor of n.

(p) would form maximal ideals, where p is a prime factor of n.
and this isn't even an integral domain due to all the zero divisors

[mild laurel]

@winter vigil think about it

How could you factor 0

[fading wagon]

You probably exclude 0 from factorisation, right?

but yeah, maybe we can consider the factorisation of a zero divisor...

[winter vigil]

There is no way of factoring zero right ? I mean zero with a nonzero object is still zero

[mild laurel]

That's not what factoring means

[sweet moat]

how do i apply homomorphism decomposition theorem on a homomorphism? what exactly do i have to show?

[mild laurel]

What decomposition theorem are you talking about?

[sweet moat]

thats the thing im not sure what it is it just says apply homomorphism decomposition theorem. from what i see i have to define some sort of diagram

cant find anything online

[mild laurel]

It probably means the first isomorphism theorem

Just look that up

[sweet moat]

will do

[errant drum]

@winter vigil Hint: look at Z_6

[woven delta]

Probably the decomposition theorem is that you can decompose a homomorphism as a injection and a surjection or something

Which is true because of the first isomorphism theorem

You quotient out the kernel to get the image, and then imbed that into the space

So I guess it's a surjection, then an injection

@sweet moat does that help?

What was the question actually?

[smoky cypress]

Is the additive group of Q isomorphic to the multiplicative group of Q^+?

Well Q^*

[tame bear]

is that positive Q or non zero Q

[smoky cypress]

Non zero Q

[tame bear]

non zero Q includes -1
so 1 = (-1)^2

if theres an Iso then theres some b in Q
f(b) = -1
1 = f(b)^2 = f(2b)
but 1 = f(0)
that implies b=0
so f(0) = -1 and f(0) = 1

[smoky cypress]

Ok 👌

[final gulch]

@smoky cypress did you discover anything else?

[smoky cypress]

😅 not really

I figured out that Z can't be isomorphic to the multiplicative group of nonzero Q

[final gulch]

well that's very clear

Z is generated by 1 element

[smoky cypress]

That's true...

[final gulch]

for a rational r, can you solve 2x = r in Q?

[smoky cypress]

Yes

[final gulch]

ok, and if you have an isomorphism with the multiplicative group of positive rationals, what happens to this equation?

[smoky cypress]

f(2)+f(x)=f(r)?

[final gulch]

no, f(x + x) = f(x)*f(x)

[smoky cypress]

Oh

well if an isomorphism exists the inverse is also an isomorphism 🤔

But ok

[final gulch]

yes but I meant to apply the equation 2x = r on the additive side not the multiplicative side 😂

it tells you that the image is closed under taking any kind of roots you want

[smoky cypress]

?

oh wait the right side is always positive

and x+x can take on all the values over rationals

because 2x=r is solvable

so the isomorphism is not surjective

[final gulch]

for example if f is an isomorphism, let f(x) = 2, then f(x/2)^2 = f(x/2 + x/2) = f(x) = 2

[stone fulcrum]

Oh sick we learnin abstract

[final gulch]

which you know is not possible

[smoky cypress]

i see

that's a cool way to do it

[woven delta]

@smoky cypress there is actually a first order sentence which tells apart (Q, +) and (Q^+, *)

Which I guess is what Seoin just said

So in particular they are not elementarily equivalent in terms of model theory

Which implies they are not isomorphic

Whether you can find a first order formula that distinguishes 2 structures is something that is well studied

[tame bear]

w h a t

[woven delta]

@tame bear what?

[final gulch]

@tame bear yeah I don't speak metamath either 😂

[tame bear]

model theory
more like lmaodel theory

[runic mesa]

yay model theory

[mild laurel]

I'm surprised that you can find a first order formula to tell two structures apart

Something like that seems a lot more difficult

[final gulch]

I did some searching and the only other way I found to distinguish (Q,+) and (Q^+,*) was that the latter has a minimal set of generators given by the positive primes, but the former has no minimal set of generators

[woven delta]

You can't always

For example Algebraically closed fields of the same characteristic are all elementarily equivalent

Also lowenheim skolem tells you that there are elementarily equivalent structures of every infinite cardinality

Model theory is useful though

I wouldn't call it just metamath

[mild laurel]

id study it tbh

but my brain is too small

[cerulean current]

So my friend asked me a question that I thought was initially poorly formed, but might have some more interesting implications.
He wanted to know what the average value of the real numbers was.
Now, I told him that this was poorly defined, and that you couldn't really assign a meaning to that.
But then I got to thinking.
Sure, the real numbers or the rationals didn't have a meaningful average, but could you reasonably define an average for other fields? Could only finite fields be assigned a meaningful average, or could something be formulated for fields in general? Could it tell us anything interesting?

[fading wagon]

What does the average even mean for a finite field

it can't be calculated in the field

[final gulch]

I mean
if you add up every element of a finite field... each element is going to cancel with its additive inverse 😂

[fading wagon]

and you still divide by 0 afterwards

[tame bear]

what

you divide by |F|

[cerulean current]

Huh, inverse elements would pose a problem

[fading wagon]

which doesn't make sense inside the field itself?

[final gulch]

you're going to get 0 before you even have to think about division

[golden pasture]

Itll be like 0/|F| where 0 is the 0 of the field cuz every other element have a unique additive inverse

[fading wagon]

but |F| is not in the field

[cerulean current]

I mean, maybe you could do addition in the context of some other field structure? Like for the integers mod 5, sum up all the values as if they were regular numbers, and take the average from there?

[tame bear]

its still what you would divide by

thats how averages work

[fading wagon]

well, problem is poorly defined

[tame bear]

yeah

[cerulean current]

Yeah

It was just a errant thought that seemed kind of interesting

[fading wagon]

I have an idea. We give each element with a weight of 1. Then the average is 1. Yay!

[cerulean current]

xD

[fading wagon]

Or you can take the average of the orders of each element, that might be interesting

sounds like burnside if you ask me

[cerulean current]

Burnside?

[final gulch]

in spite of all this sillyness
I actually do remember a professor using some kind of averaging over elements of a group
it was in the context of representation theory I think
I haven't thought about it for a few years so I've totally forgotten, but I know something like that was used

[fading wagon]

https://en.wikipedia.org/wiki/Burnside's_lemma

[cerulean current]

Burnside's lemma, sometimes also called Burnside's counting theorem, the Cauchy–Frobenius lemma, orbit-counting theorem, or The Lemma that is not Burnside's,

[tame bear]

sum the squares

[stone fulcrum]

Here's an interesting input I have. Let's say you have the average of a set
A = (a + b + c +... n) / |n|

Then you want to add another into the set and know the new average.
(a + b + c +... n + m) / |n + 1|
= (n + 1)A/n + m/|n + 1|

My notation is terrible here but I hope I'm getting the idea across

[fading wagon]

$\frac{\sum_{a\in A}a}{|A|}$

[cloud walrus]

Element118:

[stone fulcrum]

Could we use this to show that, as we add elements a certain way, we can get the average to go anywhere we want?

[fading wagon]

If $b\not\in A$, then $\frac{\sum_{a\in A\cup{b}}a}{|A|+1}=\frac{\sum_{a\in A}a}{|A|}\times\frac{|A|}{|A|+1}+\frac{b}{|A|+1}$

[stone fulcrum]

Oh yeah true, we can't add the same element back in again

Because this looks conditionally convergent to me

[cloud walrus]

Element118:

[fading wagon]

There @stone fulcrum

[stone fulcrum]

I was taking A to be a real number, not a set

My notation was just awful

[fading wagon]

yeah, I know, but I really prefer numbers to be lowercase

[stone fulcrum]

But yeah if we let |A| be the number of elements in the average then that's what I was getting at

It's likely impossible to add the reals like this though, since uncountable and stuff

[fading wagon]

hmm, maybe we need a probability distribution and a mapping to reals

[chilly ocean]

https://www.youtube.com/watch?v=uLSUQYzqXZ0

How do I imitate these vocals

For music 2

[mild laurel]

How about you ask this in #chill

[chilly ocean]

Sure

[tame bear]

[wind steeple]

This is abstract

[dull narwhal]

hey guys me and my friend have a problem. we have a permutation described as: a*xmod(54). This gives back a set {0,...,53}. My friend then asserted that this was even because it could broken down into an even amount of two cycles. However, the textbook im reading asserts that subgroups of the symmetric group can be a product of 2-cycles. This isnt such a subgroup and the presence of 6 inputs that map to themselves(those inputs would be 0,9,18,27,36,45) makes me suspicious of even claiming that this permutation group is a product of two-cycles. If you guys could help me understand what's behind this and if it's even representable as a product of 2 -cycles that would be great.

[mild laurel]

Just because the permutation you have isn't a subgroup doesn't mean it can't be an even permutation?

[dull narwhal]

yeah can you explain that pleas

please

like what's the proof and or test and where can i find it?

aight nvm found the identity stuffs thanks for that though i was very lost in the sauce

[clear obsidian]

so how do I get the conjugacy classes of a group of order n (e.g. n = 6)

[mild laurel]

It depends on which group it is?

[jovial juniper]

Hey guys, the positive reals under * where a*b = sqrt{ab} isn't a group, right?

my reasoning being a inverse would have to be a, b inverse would have to b b, etc

so there isn't one inverse

or am I way off

[mild laurel]

why do you think a inverse would have to be a?

[jovial juniper]

whoops, I'm thinking identity I think

I need an element s.t a * i = a for all a in the positive reals

[mild laurel]

Right

Yeah in that case, your reasoning is correct

But you'd have to show that, under the group axioms, the identity is unique

[jovial juniper]

Since each element is its own identity, does it fail on that alone?

I'm not doing a great job of explaining where I'm coming from I don't think, but I think I can show everything else I'd need to

[woven delta]

So really the nice way to think about this is that sqrt is a group isomorphism

[jovial juniper]

Forgive me if I'm off here, new to algebra, but it seems like you're implying that the positive reals do form a group along with this * yeah?

[woven delta]

Yeah

[jovial juniper]

so it's completely fine to have the identity of a =a, identity of b = b, etc for all a,b in the positive reals

it doesn't have to be one element that does it for all of them?

[woven delta]

What do you mean?

The identity is the same globally

Wait, I'm stupid

I'm wrong

[jovial juniper]

That's what I thought, but I don't see a global identity here

that's why i think it fails the group axioms

the identity is just the element itself (I think)

[woven delta]

Yeah you are right

I thought that because sqrt is an automorphism you can just impose the structure on the group through that

But it doesn't work out

Oof

The induced group structure doesn't agree with the structure you're given

[jovial juniper]

I guess I'm getting caught up in the wording, I just need to figure out a way to say that because the candidate inverse of a would be a, the candidate inverse of b would be b, etc it fails to have a unique inverse

I see what's going on that screws up the group structure

but I cant really manage to phrase it

[woven delta]

Yeah

What you're saying is correct

[jovial juniper]

think it would fly just to word it like that, maybe a bit more formally then?

[woven delta]

Say if it is a group it must have a unique identity globally

But you can find 2 local identities

So you're done

Not even local

Pointwise

[jovial juniper]

There we go! That's a really helpful word here

I can show a inverse = a, b inverse = b and just run from there

thanks man!

[woven delta]

👌

[hoary forge]

I am unable to understand the marked part^

What does it means?

[fading wagon]

it means when you divide d by 4

you get a remainder of 0 or 1

[hoary forge]

oh

[fading wagon]

sort of

more rigourously, $a\equiv b\pmod{c}$ means $c|a-b$.

[cloud walrus]

Element118:

[hoary forge]

(Bmin*Bmin-4*(Amin-1)*(Cmin-1))%4

You know all the values that we're getting has remained -3 or 0

If b is even then remainder is 0 if b is odd then remainder is -3

So, the question looks like if B is odd what values within the given range gives remainder -3

else if B is even what values within the given range gives remainder 0

[hoary forge]

@fading wagon They represent the same values, f(x,y) is the reduced form of other one

[hoary forge]

@fading wagon Optimized more

Got 20 points extra 😄

[clear obsidian]

So I think I saw a couple different definitions for ideal,

I know we're a subset of a ring closed under exterior products

but I saw somewhere say that we need unity to be closed under addition, but isn't that its closed under addition and subtraction anyway?

[hot lake]

an ideal of a ring is a subset closed under addition and under multiplication by a ring element

[clear obsidian]

ok ty

[clear obsidian]

So I found this equation for class number of a group but what is the C(g) (because i dont think its the same thing as Cl(g)) (yes I tried to find it online but I didn't see that notation anywhere else)?

[hot lake]

what's the class number of a group ?

[fading wagon]

This is abstract algebra

@gaunt cipher

Try #prealg-and-algebra

[fading wagon]

what do you think?

[earnest jolt]

No

unless its surjective I dont think it'll be the case

But I dont know how to reason it

[fading wagon]

If no, provide a counterexample

[hoary forge]

Element, do you think a formula exists for that question?

[fading wagon]

which question?

[hoary forge]

One we were solving

[fading wagon]

oh yours?

[hoary forge]

Yep, I have optimized to the best

[fading wagon]

You sure?

[hoary forge]

Yep, I am not iterating* anything in excess

[fading wagon]

Prefix sum included?

[hoary forge]

prefix sum?

[fading wagon]

yeah

[earnest jolt]

element is it true?

[fading wagon]

You can't find a counterexample?

[earnest jolt]

I think subjectivity is a requirement but I can't think of a counterexample

yeah

[fading wagon]

Consider size of X and Y

it's easier to use finite sets

[hoary forge]

I am talking about inequality one

[earnest jolt]

wait what about the size of X and Y?

I am using finite sets at the moment

[fading wagon]

try different cases out

[hoary forge]

What do you mean by prefix sum?

[fading wagon]

There'a a lot of overlapping calculations

between different queries

[hoary forge]

I agree with you

[fading wagon]

you can try some sort of prefix sum to count things

i.e. basically memoising stuff

[hoary forge]

So, I should take 3 parallel arrays, find out the maximum of A,B,C from them. Then, compute the counts for each in one iteration of A,B,C?

[hot lake]

what is this about ?

[clear obsidian]

what happened to my question?

[hoary forge]

@fading wagon Tried that, still TLE.

[clear obsidian]

class number is the number of conjugacy classes of a group @hot lake

can someone answer my question?

[hot lake]

C(g) might be the subgroup of elements who commute with g

[clear obsidian]

alright ty

[earnest jolt]

@fading wagon I can't think of a counterexample and im starting to think its true

can you please correct me if its false?

[tame bear]

look at left and right shifts of infinite sequences

[languid moss]

Bazout identity is by far the most annoying thing to compute

[wind steeple]

Computing is painful

[languid moss]

Is so annoying I get lost

[clear fiber]

normally i see this definition of a Group:

Why is it that the book I’m reading gives a different-looking one?

[woven delta]

They are equivalent

Try to prove that

[clear fiber]

🤔

Is property (ii) the same as saying the set is closed under the operation?

[woven delta]

No, it encompasses a lot more

[clear fiber]

hmm

[woven delta]

This is actually a classical problem

To show these are equivalent

[clear fiber]

the book does go on to prove the existence and uniqueness of the identity element and of inverses using this definition

i see

[woven delta]

Yeah, and it's obvious that if a group satisfies the group axioms, it satisfies those conditions

[clear fiber]

interesting

seems like the book definition is the more big brain one

[woven delta]

By letting x= a^-1 b and y = ba^-1

Eh

[clear fiber]

since there are only 2 requirements

instead of four

[woven delta]

The biggest brain one is a group is pi_1 of a Topological space

Then you just reconstruct group theory using Topology

[clear fiber]

lmao

idk what that means but it sounds super big brain

[topaz solar]

or as a specific example of a lawvere theory (or a model of one, more accurately)

which is like
shit tier big brain

[tame bear]

a group is just Aut of a graph

[smoky cypress]

What about infinite group?

[topaz solar]

[smoky cypress]

🤔 is this a restatement of cayley’s theorem

Or however you spell it xD

[topaz solar]

Cayley is that a finite group embeds into the symmetric group on its generators iirc

or was it on the whole set

i forget

fuck it, going with whole set, since generators is too small

[bleak abyss]

So every group is a subgroup of some symmetric group

Don't even need finite

And yeah you definitely want on the whole set

S_n is generated by two elements

[topaz solar]

yeah realized that a second after

fuckin rart memory

[tame bear]

https://en.wikipedia.org/wiki/Frucht's_theorem
infinite groups are fine

Finally, Johannes de Groot and Sabidussi in 1959/1960 independently proved that any group (dropping the assumption that the group be finite) could be realized as the group of symmetries of an infinite graph.

[topaz solar]

nice

nifty theorem

[tame bear]

they do

and since Aut of any graph is contained in Aut K_n = S_n
where n is the order of the original graph, thats Cayleys theorem

[woven delta]

I like the much easier theorem that every group is the automorphism group of some structure

[tame bear]

very vague

[woven delta]

Nope

Not vague at all

[smoky cypress]

🤔 in Jacobson it is said that every group, finite or infinite, is isomorphic to some group of transformation

[golden pasture]

cayley's theorem?

this jacobson? it says finite

[hoary forge]

@fading wagon https://discuss.codechef.com/t/lapd-editorial/38085

The solution

[smoky cypress]

@golden pasture nowhere does it say finite

Well for first one

Symmetric groups are just finite transformation group

Also I said each group is isomorphic to some group of transformation

Which can be a symmetric group

[golden pasture]

o right

[glad gazelle]

Hi guys, could you tell me about some good resources, books or whatever to abstract algebra in my own? I know basics of group theory, what they are, lagrange's theorem

[mild laurel]

Pick a textbook

Which one depends a lot of you

Things like how comfortable you are with rigorous math, or exactly what material you want to cover

[glad gazelle]

i did olympiad training, i understand most proof methods, and i like proofs

i suppose i want rigorous

cause i'm not interested in learning something i can't prove

what textbook do you recommend?

for example: i do algorithms stuff, and it's quite usual to find algorithms without proof, i always try to find a proof i can understand, i don't like to use things i can't prove (sometimes i need to do it)

[mild laurel]

Algorithmic proofs are much, much different than mathematical proofs

Idk it's hard to say

The usual recommendation is Dummit and Foote

I've heard good things about Fraleigh

I think Pinter is on the easier side of things

[glad gazelle]

i know it's not the same, but i also did math olympiad training and i was really interested on learning more

[mild laurel]

I mean you could look at higher level math competitions

Like the putnam or the IMC

But it depends why you want to do more math olympiad training I guess

That kind of stuff isn't really super applicable anywhere else

[glad gazelle]

yeah but i know about all topics of math olympiad (however i'm not really good solving those hard problems)

but now i want to keep learning math and do it like a mathematician, no like an engineer just taking some formulas and using them without really understanding

[mild laurel]

Then math olympiads aren't really that relevant

[glad gazelle]

i know

it's quite specific, elemental math but hard problems

we never used calculus, linear algebra or group theory

i would like to learn as much as i can in a similar way a mathematician would do

since i can't study math in university at this moment

[mild laurel]

Yeah your two goals aren't really that related

I mean you can definitely just do both

[potent briar]

how do i prove Z_6 is isomorphic to Z_2 x Z_3

[unkempt socket]

Hint: What are the generators of Z_2 x Z_3?

[potent briar]

i dont know that yet

i think they want me to find an explicit isomorphism

[mild laurel]

What have you tried?

@potent briar

[woven delta]

Finding a generator is an explicit isomorphism

[proud ore]

im sooo confused here lol

[mild laurel]

What are you confused about @proud ore

[proud ore]

https://media.discordapp.net/attachments/269573202018041856/623275612051603456/Screen_Shot_2019-09-16_at_2.54.36_PM.png?width=1555&height=1049

some background @mild laurel

having a hard time seeing how C,D,F act on f and g

[gentle pendant]

Write down how they act on x,y and z.

Then compose this with f.

[proud ore]

idk how thats the thing 😂

[queen scarab]

why R ⊆ X x X and x, y ∈ X not R ⊆ X x Y x ∈ X, y ∈ Y? Is it possible for the infix operator to denote this?

[sharp sonnet]

you can also define a relation on the cartesian product X x Y, where X != Y, yes

but that has nothing really to do with abstract algebra

[queen scarab]

hmm so if I define my relation as so I can use the infix operator?

[sharp sonnet]

you can define whatever you want

[queen scarab]

right thanks

wait one more thing doesnt R ⊆ X x X imply that the domain and codomain of a relation is the same?

[sharp sonnet]

i don't think the words domain and codomain are used in that regard, but yes

nvm, they are used

[queen scarab]

I cant imagine it to be rare to find a relation with a differing domain/codomain so why would that example use R ⊆ X x X is it just being formal?

[sharp sonnet]

the graph of a function is a relation on X x Y

where X and Y can differ

your example uses what it uses, because it doesn't need more

[queen scarab]

ah ok thank you

[sharp sonnet]

generally definitions are more or less arbitrary

[queen scarab]

Thanks

[plain sequoia]

How can I show that $\text{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q}) \simeq \mathbb{Q}$?

[cloud walrus]

Markus:

[plain sequoia]

(as rings)

Hm nevermind

[final gulch]

you have to explain what you're talking about

what is the star?

what is "noted multiplicatively"?

what is E?

[static robin]

• is a law of internal composition , noted mutliplicativley is when the law is the $\times$ operator and E is a set

[cloud walrus]

life > Random:

[static robin]

might have different termiology here

[final gulch]

ok I guess I see now

[static robin]

LIC could be also called binary operation

i think

[final gulch]

well you said you can cancel the x's

if p is bigger than q, just cancel q of the x's from both sides

[static robin]

you can't do that

[final gulch]

you said xy = xz implies y = z

right?

[static robin]

yes, that is the definition of the regular element.

[final gulch]

what happens if you take y = xy and z = y

(bad notation because I'm using y in two different ways)

but like, in particular if xy = xz implies y = z then for example xxy = xy implies xy = y

xxy = xy is what your top picture says, and xy = y is your bottom (goal) picture

[static robin]

yeah i thought of that

but they didn't do that

[final gulch]

it's just that, iterated

[static robin]

yeah that's what i'm tryna to understand, how did they manupilate x^p and x*^q to get x^*(p-q)

cuz we do'nt know nothing about * except it's assiciative and being a LIC

[final gulch]

I'm not sure which part isn't clear
as far as I can tell, it's just what I wrote

I wrote it for p = 2 and q = 1 but the idea is the same

just apply that cancellation law you said holds q times

if you start with p copies of x on the LHS, then after you do that you'll have p-q of them left

[static robin]

i'm not sure if i get it yet.

[final gulch]

if you want to prove it formally you can use induction

[static robin]

how can i do that ?

and also i want to understand how they did it

also how did they do this ?

cuz * isn't commutative

assuming that the first one was right

[final gulch]

how can i do that ?
induct over q

also how did they do this ? cuz * isn't commutative
I don't know, we would need more information about what this (E,*) is

[static robin]

all i got is (E,*) being a magma

so * is LIC

and * associative

and E is a finite set

the question is : let * be a assocative LIC over E, E being a finite set and x is a regular element, show that E has an identity

[final gulch]

what is LIC?

[static robin]

• LIC appilication that goes from E X E to E taking (x,y) and outputing x*y

x*y is in E

i think it's called closure

different terminlogy i know :(((((

[final gulch]

I see
so I guess you somehow use the fact that E is finite to find p, q so that x^p y = x^q y

[static robin]

maybe ur familiar with this ?

yeee

so i used this function

f(n) = x*^n

f : N -> E

f being non injective

since E is finite and N is infinite

[final gulch]

I see

then you can multiply both sides of x^p = x^q by any y

[static robin]

so u get this : $x^(p) = x^(q)$

[final gulch]

and assume q > p

[cloud walrus]

life > Random:

[static robin]

to the power ofc

yeaa

and then u mutiply both sides with y

y being in E

[final gulch]

so you can do that from either left or right

and for any y

so you get, for every y
x^p y = x^q y
and
y x^p = y x^q

[static robin]

yes

with the * operator between

[final gulch]

by applying those cancellation laws q times to these two equations you get
x^{q - p} y = y
and
y = y x^{q-p}

so to answer your original question, you don't use any kind of commutativity

[static robin]

what cancellation laws

[final gulch]

you just start from the other equation

[static robin]

??

x being regular ???

[final gulch]

yes

so there you have it x^{q-p} is an identity element

[static robin]

yeah ok but how did u get to x^*(p-q)

[final gulch]

just apply the cancellation law q times like we discussed earlier

[static robin]

it's not just x^(p-q)

ok i'm confused now

[final gulch]

there is only one operation on E

[static robin]

yes

[final gulch]

x^p can not mean anything besides x^*p

[static robin]

when u say that u also assumed that it's commtative

[final gulch]

I did not

[static robin]

how else would u get this ?

https://cdn.discordapp.com/attachments/496784958430380033/623402757700714496/unknown.png

then ?

[final gulch]

we started with yx^p = yx^q

[static robin]

yes

[final gulch]

and apply the cancellation law q times

[static robin]

no no

[final gulch]

that gives yx^{p-q} = y

[static robin]

started with what i copied

we didn't start with yx^p = yx^q

[final gulch]

what did you start with?

[static robin]

https://cdn.discordapp.com/attachments/496784958430380033/623406375883767810/unknown.png

that

[final gulch]

using that one gets you x^{p-q} y = y

[static robin]

ok how did u get that ?

[final gulch]

that is one of the things you need for x^{p-q} to be an identity

but you also know that yx^p = yx^q

[static robin]

did u divid by x^q ?

[final gulch]

and this gives the other part: yx^{p-q} = y

[static robin]

did u divide ?

[final gulch]

I have told you many times
I used the cancellation laws ax = bx implies a = b and xa = xb implies a = b

[static robin]

but you also know this yx^p = yx^q
no

[final gulch]

you assumed that these hold for x at the beginning

yes, you know that
it follows trivially from x^p = x^q

[static robin]

omg yes i'm an idiot yes yes

[final gulch]

lol

[static robin]

but i still don't get this ?

cuz if u divided with x*^(q) then it's impossible

cuz u can't simplify

[final gulch]

this was the very first point we discussed
you know for every a and b, that xa = xb implies a = b

[static robin]

yes that is true

[final gulch]

in this case, take a = x^{p-q}y and b = y

[static robin]

huh ?

i'm slow lol, what about the x ?

[final gulch]

you know that x^py = x^qy
without loss of generality we are assuming that p > q
so we can write this as
x^q x^{p-q} y = x^q y

do you agree so far?

[static robin]

x^p x^{p-q} y = x^q y

u mean this ?

[final gulch]

no

I mean what I wrote

yours has two p's on the left side

[static robin]

urs has two q's

[final gulch]

they cancel

[static robin]

yes

[final gulch]

I am only using the fact that x^p = x^q x^{p-q}

[static robin]

yes i agree so far

but we also have x^q = x^p

[final gulch]

ok so x^q x^{p-q} y = x^q y
now use the cancellation law to get
x^{q-1} x^{p-q} y = x^{q-1} y

ok true what you wrote wasn't wrong then. but still lol

do you agree with the last formula I wrote? I just used the cancellation law to remove one x from the left of each expression

[static robin]

oof i don't even understand it wait XDD

i'm working on it XDD

[final gulch]

xa = xb implies a = b
this says you can remove a factor of x from the LHS of an equation
it's all I'm using

[static robin]

yeah i get that, where did u get the whole x^(q-1) ?

ohhh

[final gulch]

well we had x^q on each side
then we removed one x

[static robin]

yee

i'm stupid

[final gulch]

so do that over and over again

[static robin]

yee

Thanks !

dat's SNEAAKY

that is really sneaky

[final gulch]

✌

[static robin]

GG bois we did it

[final gulch]

[static robin]

first off is i'mm gonna screenshot this whole discussion so even if i forget how to do this i'll comeback to it

[final gulch]

lol

I think the more common setting for this argument is to show that a finite domain contains an identity
there you have a little more structure but from it you automatically get the cancellation property for every nonzero x

[static robin]

OH MA GAAAAAAWD

i get it

like fully

THANK U SO MUCH !

yeah the only thing that i missed or was keeping me back was x^(p-q) * x^(q) = x*(p) which should've been a no brainer

[mild laurel]

This is not the right channel for this material

[languid moss]

OMg! I got beZout identity 😊😊😊

After three days of not understanding it

[mild laurel]

The version of bezout's identity for a general UF?

[languid moss]

🤔

(A,b)=d and get ar+bs=d

[mild laurel]

UFD* sorry

Are you only talking about the integers then

[languid moss]

Yea integers

[mild laurel]

Yeah, so if you know anything about rings, we can actually generalize bezout's

[languid moss]

I’m not there there 😢

[mild laurel]

Someday

It's super cool

[chilly ocean]

y can't you take the product of fields

help

[tame bear]

then (a,0) would be in the product

[chilly ocean]

ah yes thanks

does anyone know how tensor algebras work

I can't understand how we quotient them to get symmetric algebras an exterior algs

[tame bear]

dunno

[woven delta]

Tensor Algebra is product of all the tensor powers

@chilly ocean

[chilly ocean]

yes but how do we quotient by e.g. v tensor v

like T(M) is the tensor algebra

so how do you quotient that by ({v tensor v for all v in V})

since the elements of T(M) look like (r, m, m tensor n, m tensor n tensor o, ...)

@woven delta

[woven delta]

So we have the symmetrization maps

You quotient by the kernel of that

Similarly for exterior Algebra

I'll describe that once I get off the train

[chilly ocean]

ok thanks

[woven delta]

Also there's something your missing

You should think of the Tensor algebra as finite sums of k-tensors

There's a difference between the direct sum construction and the direct product construction

You're thinking of the direct product construction

@chilly ocean

[chilly ocean]

oh yeah right

its infinite

so it looks like m tensor n + m tensor n tensor p ... m tensor m or something

[woven delta]

Yeah, something like that

So you can just define the symmetrization map on the k^th tensor product

[chilly ocean]

by symmetrization map you mean the quotient map onto T^k(M) / A^k(M)

[woven delta]

I mean a certain map into itself

Which takes a tensor and symmetrizes it

[chilly ocean]

oh so v |---> v tensor v

[woven delta]

No

So you take v tensor u and map it to 1/2 (v tensor u + u tensor v)

For example

[languid moss]

So far I’ve done ever single one and this one is confusing. Anyone help? Bezout identity?

[woven delta]

That's the 2 tensor case

More generally you take the sum and divide by n!

[woven delta]

For the n tensor case

@chilly ocean

Does that make sense?

[languid moss]

,rotate -90

[chilly ocean]

yeah I guess

[woven delta]

There's 2 ways to view the kth symmetric tensor power

[languid moss]

,rotate 90

[chilly ocean]

but the way it was defined for us is that you quotient out by some submodule

[woven delta]

Either as a submodule or as a quotient

[chilly ocean]

oh so you say it is the quotient of the symmetrization map

[woven delta]

The submodule you quotient out by is the kernel

Yeah

[chilly ocean]

hmmm

[woven delta]

That works for kth symmettric powers

And you you extend that map

To the direct sum

[chilly ocean]

yeah

[woven delta]

Using the universal property

[chilly ocean]

right

[woven delta]

Well first you imbed that map into the tensor algebra

In the natural way

Then you extend that map from the tensor Algebra to itself

But you get the idea

Then the quotient will be the symmettric Algebra

You can similarly define the exterior algebra

With the antisymmetrization map

Which uses alternating sums instead of regular sums

@chilly ocean is that good?

[chilly ocean]

yeah I'm still not comfortable with how to think about tensor product elements though

is it correct to say that M tensor N = {m tensor n}/~ where (m tensor rn) ~ (rm tensor n)?

i.e. the tensor product consists of all the elements m tensor n but you can move ring elements around

it's not clear to me that all these quotients will work out, i.e. can i say m tensor n tensor p goes to n in the exterior algebra (since we kill m tensor m)?

[woven delta]

You can get some intuition for this from working with vector spaces

Cause they're very nice

[static robin]

@mild laurel where should it be ? cuz it's algebra.

[static robin]

Let E be a set provided with 2 operators
we assume that e is the identity for the * operator and f is the identity for the • operator
Finally, suppose that for all x, y, u, v in E : $(H): (xy)•(uv)=(x$•$u)*(y$•$v)$

1. show that e=f. (my idea was to either prove this directly or show that e and f have the same effect, i failed at both)
2. prove that * and • are identical. (no idea how to do this)

[cloud walrus]

life > Random:

[static robin]

i think i proved 1).

$(ee)$•$(ef)=e\iff e$•$f=e\iff(e$•$f)*(f$•$f)=f$

[cloud walrus]

life > Random:

[chilly ocean]

look up eckman hilton argument online

[static robin]

yeah that helps

but is my proof correct ?

also thanks

also also, for second one i think i can say since we proved that $e=f$ and we know that the identity has to be unique that means that the operators are identical ??

[cloud walrus]

life > Random:

[chilly ocean]

looks good for 1

idk

[static robin]

ok

good

[static robin]

i still don't know how to prove that both operations are identical

well i did prove it but idk if it's right

[static robin]

<@&286206848099549185>

[latent anvil]

I don't understand your argument for why e = f

You stated 3 true things

[static robin]

i used this

i sort of deconstructed e and f

$e=(ee)$ and $f=(ef)$

[cloud walrus]

life > Random:

[hot lake]

what's the missing operation in the left hand side of H ?

[latent anvil]

It's bullet

I'm not sure how what you've posted actually shows e = f

[static robin]

ok

the idea is if i can prove that e and f have the same effect

that means that they are equivalent

so i started from this

$e$•$f=e$

[cloud walrus]

life > Random:

[static robin]

and from (H) i can say this : $e=(ee)$ and $f=ef$

[cloud walrus]

life > Random:

[static robin]

and i also can say this :

[latent anvil]

I'm not sure where H comes in there

Those are true because e is an identity for *

[static robin]

this is where it'll come in

$e$•$f=e$ with H i can say that, that is the same as this : $(e$•$f)*(f$•$f)=f$

wait that's wrong let me fix it up

[cloud walrus]

life > Random:

[latent anvil]

How exactly are you applying H?

[hot lake]

still doesn't involve H

[static robin]

can u hop on mathematics voice ?

pls

[latent anvil]

It would be more productive if you broke down what you're doing in more depth over text

also it's 1:36am where I am so I'm not going to go on voice /shrug

[static robin]

oh shit

i'm starting to see cracks

wait

[latent anvil]

f = f•f = (e*f)•(f*e) = (e•f)*(e•f) = e*e = e

[static robin]

yee mine is wrong

cuz if i used H

i would get $(e$•$e)(ef)$

[cloud walrus]

life > Random:

[static robin]

which doesn't lead to anything

[latent anvil]

Yup

[static robin]

Damn i was so sure

[latent anvil]

What you want is some kind of symmetry

Your approach looked off because you have 3 es and 1 f

[static robin]

yee

thanks

what about 2?

saying that since e=f then the operators have to be the same

[latent anvil]

Nope

[static robin]

ded

can u provide an example where that is wrong ?

[hot lake]

f(x,y) = x+y and g(x,y) = (x+1) * (y+1)

[latent anvil]

Another example where both operations are groups

Let E = { 00, 01, 10, 11 }

[hot lake]

operations can't be groups

[latent anvil]

We can add these as integers mod 4 or add their digits mod 2

00 is the identity in both cases

What do you mean?

Oh you were nitpicking my terminology. Where the operations both induce group structures on the underlying set

[hot lake]

mine were also from groups, but on R and R \ {-1}

[latent anvil]

Ah yeah good point

Well hang on, no. The question is about operations on the same set

[static robin]

ye

[hot lake]

ah yes

[latent anvil]

Anyways the klein four group and the cyclic group of order four can be thought of as two different ways of adding bit strings of length 2

Both with identity 00

For the second bit, we can get x*y = (x•f)*(f•y) = (x*f)•(f*y) = (x*e)•(e*y) = x•y

[static robin]

ok so relying on the identity is not a factor that determines that * is different from •

[latent anvil]

I'm not sure what you mean

Two distinct operations can have the same identity

Does that answer your question?

[static robin]

yeah it does

ok got it thanks !

[latent anvil]

Was this a homework problem?

I'm just wondering where you came across it

[static robin]

yeah homework

[latent anvil]

It has a cool application

[static robin]

yeeeee

[latent anvil]

Do you know anything about homotopy?

[static robin]

Nope Nope Nope

[latent anvil]

Ah well

Basically there's some groups that algebraic topologists care about called homotopy groups

For n > 2, we can combine these elements "vertically" and "horizontally"

And they satisfy this kind of interchange law

So that argument proves that they're really the same operation

And that this operation is commutative

Oh commutativity is x*y = (f•x)*(y•f) = (f*y)•(x*f) = (e*y)•(x*e) = y•x = y*x

[static robin]

gotta go back to lab and look this stuff up

[latent anvil]

I'm a fan of alg. top

[static robin]

i find it so abstract

like

stuff that u can't even imagine

[latent anvil]

Algebra is a lot easier if you're bad at visualizing things lol

I'm mostly joking but it's important to be able to think in an abstract setting without geometry

[static robin]

yeah, i heard a teacher once say that we shouldn't trust proofs that rely on that

geometry*

one more question

if i need to prove that e = f

is it right to think that i need to prove that e and f have the same effects on the set ?

[mild laurel]

That's one way of doing it

[static robin]

ok

thanks

[chilly ocean]

https://math.stackexchange.com/questions/2575511/quotient-of-direct-sum-of-modules

does part 2 in the first answer here for for infinite direct sums?

[signal frost]

Oh hey @latent anvil I've seen your username before XD. Are you in the MathFuN workspace on Slack?

[latent anvil]

@zizek yup, you can apply first iso to the map which projects each factor onto its quotient

@signal frost hi

[chilly ocean]

@latent anvil thanks i wasnt sure if bijecitivity works the same in infinite case

[signal frost]

Hi. I guess that's a yes.

[static robin]

How many binary operations are in a set with n elements?
and how many of those operations are commutative?

umm..

i don't even know where to start

like as many as you want ?

and as many as you want as well..

cuz the human creates the operation

[tame bear]

try some examples, n=1

[static robin]

yeah in that case

there are n many operations ???

even if n = 1 i can define as many operations as i want

say for example $E={1}$ i can say that * in E is for all x,y x*y=1+xy and i can say • is for all x,y x•y=0 so on and so forth

[cloud walrus]

life > Random:

[static robin]

<@&286206848099549185>

[latent anvil]

If x•y = 0, it's not an operation on {1}

Same problem for *

An operation on E is a function E×E -> E

Does that make sense @static robin ?

[static robin]

ye ye ye i'm dumb

but i can define what every operation to just get 1 at the ed

end*

x,y => x*y => x-y+1

[latent anvil]

Those are the same operation

[static robin]

oh

[latent anvil]

When we talk about sets or functions or whatever, equality is "extensional"

[static robin]

so n operation it is ?

[latent anvil]

This means that if two sets have the same elements, they're equal, even if they look different, and if two functions (or operations) give the same output for each input, they're equal

It's not n

For the case n=2, you can find 16 operations

If I solved this problem right

[static robin]

ok

how u do dat ?

[latent anvil]

Have you heard the concept of a multiplication table of an operation?

Also, you should try and solve this yourself. What I said up there is a hint but it won't solve it for you

[static robin]

multiplication table alone yes ,operation alone yes , both no

n^6 ?

[latent anvil]

Are you guessing?

[static robin]

pretty much

[latent anvil]

Do you know about modular arithmetic?

[static robin]

idk, i could know about it under a different terminology

[latent anvil]

Where are you coming across these problems?

[static robin]

Teacher's website

[latent anvil]

Modular arithmetic is usually the first example of an algebraic structure you see. The integers mod 4 is the set {0,1,2,3} where addition wraps around

So 2 + 2 = 0 and 2 + 3 = 1 and 3 + 1 = 0 and so on

[static robin]

nope, he didn't tell us that, first example of an algebraic structure was the magma

[latent anvil]

That's very odd. Modular arithmetic isn't a special kind of structure though, it's a family of examples

[static robin]

we no he didn't tell us anything like this

[latent anvil]

The integers mod n, written Z/nZ is a magma/monoid/group/ring/etc

[static robin]

yee

i know about that

[latent anvil]

Oh okay

[static robin]

but we haven't done the course of arithmetics in N yet

[latent anvil]

Number theory?

[static robin]

he just gave us like a hint or something too look forward too

idk if it's called that let me check

nope i don't think it's that

[latent anvil]

What does it cover?

If it's about arithmetic in N (primes and whatever) then it's what I mean by number theory

Anyways, the "multiplication table" for Z/4Z looks like

Ignore the yellow and red, I didn't make this image

[static robin]

Divisibility, Gauss's theorem, Bézout's theorem, lower/highest common divider

[latent anvil]

Yup

[static robin]

oh