groups-rings-fields

[cloud walrus]

mniip:

[cloud walrus]

creamy shits:

[lyric falcon]

why plus? You had $e^{ix}$.

[cloud walrus]

Gonzo17:

[lyric falcon]

don't think about it as "adding multiples of 2 pi"

How would you express 1 as $e^{ix}$ ?

[cloud walrus]

Gonzo17:

creamy shits:

[lyric falcon]

what's the period of sin and cos ?

and you need $cos(n\pi)=1$

[cloud walrus]

Gonzo17:

[lyric falcon]

yes, precisely because the period of cos is 2pi

you can look at the graph of cos, see where the 1s appear

yes

it has to be even

what's the general form of an even number ?

well you just wrote it

2n

where n is an integer

[cloud walrus]

creamy shits:

[lyric falcon]

so now we want the general form of how a 1 would look when written as $\cos(x)+i\sin(x)$

[cloud walrus]

Gonzo17:

[lyric falcon]

\2 in e^

[cloud walrus]

creamy shits:

[lyric falcon]

Wait tho

The Euler identity says $e^{ix}=\cos x+i\sin x$

[cloud walrus]

Gonzo17:

[lyric falcon]

what's your x here ?

[cloud walrus]

creamy shits:

[lyric falcon]

what value do you have inside cos and sin ?

that's your x

[cloud walrus]

creamy shits:

[lyric falcon]

Yup

[cloud walrus]

creamy shits:

[simple valley]

the "not 0" makes the mmultiplicative group a group

no you first need to figure out which elements are torsion

and in this case which are those?

[lyric falcon]

I advise you not to use n since it's already in use

[cloud walrus]

creamy shits:

[lyric falcon]

is that the condition for z to be in torsion?

cause... it's not

what z with the group operation applied m times?

[cloud walrus]

creamy shits:

[lyric falcon]

which do what?

you want an elemnt z with the operation applied m times to give out identity

for some m

what's the group operation?

what's the easiest way you can represent the condition of z being in torsion?

in 1 equation

but that's just the general representation of 1

say you have any z

and it's order is 5

what would it mean (in 1 equation)?

$z^5=1$ is what it means

[cloud walrus]

Gonzo17:

[lyric falcon]

in general $z$ is in torsion if $z^m=1$ for some m>0

[cloud walrus]

Gonzo17:

[lyric falcon]

and ofc integer m

so we already know z must have radius 1

so we can represent it as $e^{ix}$ for some $x$

[cloud walrus]

Gonzo17:

[lyric falcon]

the trigonometric form, but radius is 1

Let's sum up what we did so far

we went from $z^m=1$ for some $m$ wondering how we can get what z satisfy this

[cloud walrus]

Gonzo17:

[lyric falcon]

you found that z must have radius 1

and then we said we want to use the trigonometric form to represent numbers

we figured out the general form of 1 is $e^{2n\pi i}$

[cloud walrus]

Gonzo17:

[lyric falcon]

and now we want to represent z^m similarly

we know it has radius 1, so it's equal to $e^{ix}$ for some x

[cloud walrus]

Gonzo17:

[lyric falcon]

do you see that?

I mean z, not z^m

$z=e^{ix}$

[cloud walrus]

Gonzo17:

[lyric falcon]

so what would $z^m$ be in this case?

[cloud walrus]

Gonzo17:

[lyric falcon]

$e^{ixm}=e^{2n\pi i}$

[cloud walrus]

Gonzo17:

[lyric falcon]

so now the function $z\to e^z$ isn't injective anymore

[cloud walrus]

Gonzo17:

[lyric falcon]

it is injective up to adding $2\pi i$ (because $e^{2\pi i}=1$)

[cloud walrus]

Gonzo17:

[lyric falcon]

but we already took care of it

by representing 1 in the general form

okay basically you'd do something like this:

$e^{i x m}=e^0$

[cloud walrus]

Gonzo17:

[lyric falcon]

which means $ixm=0+2\pi n$ for some n

[cloud walrus]

Gonzo17:

[lyric falcon]

or also recall that you proved all possible numbers $\theta$ such that $e^{\theta}=1$ are of the form $2\pi n i$

[cloud walrus]

Gonzo17:

[lyric falcon]

not just that $e^{2\pi n i}=1$

[cloud walrus]

Gonzo17:

[lyric falcon]

so if $z^m=1$

[cloud walrus]

Gonzo17:

[lyric falcon]

which means $e^{ixm}=1$

[cloud walrus]

Gonzo17:

[lyric falcon]

then $ixm$ must be of the form $2\pi n i$

[cloud walrus]

Gonzo17:

[lyric falcon]

what we're looking for here is x

m and n are just any integers

we've shown that "z is in torsion" is equivalent to "xm=2 pi n for some n and m"

Yeah it's just saying there exists an m

we've shown that "z is in torsion" is equivalent to "xm=2 pi n for some n and m"

so $x=\frac{2\pi n}{m}$ right ?

[cloud walrus]

Gonzo17:

[lyric falcon]

for some m and n

I just divided by m

which we assumed to be >0 cause it's the supposed order of z

[cloud walrus]

creamy shits:

[lyric falcon]

$z=e^{ix}$

[cloud walrus]

Gonzo17:

[lyric falcon]

and we know what x is

[cloud walrus]

creamy shits:

[lyric falcon]

for some m>0 and n in Z

[stiff scaffold]

[lyric falcon]

btw we can try getting it into the form of $e^{i \pi q}$ for any rational $q$

[cloud walrus]

Gonzo17:

[stiff scaffold]

[lyric falcon]

${\frac{2n}{m}: m>0 ~\text{and} ~n\in\mathbb{N}}=\mathbb{Q}$

[cloud walrus]

Gonzo17:

[lyric falcon]

do you see why?

Yeah but for equality we would also need every rational to be representable like that

${\frac{2n}{m}: m>0 ~\text{and} ~n\in\mathbb{N}}=\mathbb{Q}$

[cloud walrus]

Gonzo17:

creamy shits:

[lyric falcon]

ok to clarify we already found an expression for the torsion:

${z: z=e^{i\frac{2\pi n}m}$ for some $m>0$ and $m, n\in \mathbb{Z}}$

[cloud walrus]

Gonzo17:

[lyric falcon]

I mean, you can make that "nicer"

Yeah this is a totally valid answer

[stiff scaffold]

if we’re raising z^m can’t it be written as e^i2pin

[lyric falcon]

wdym find z?

No that set is a totally valid answer

[stiff scaffold]

👏👏👏👏👏👏👏👏

oh are we not done

[lyric falcon]

It's kinda like as if you got a set ${n+3, n\in \mathbb{Z}}$

[cloud walrus]

Gonzo17:

[lyric falcon]

it's just the set of integers

now we also have a set that can be "simplified", idk if you wanna do it or nah

That set is just ${z: z=e^{\pi i q}, q\in\mathbb{Q}}$

[cloud walrus]

Gonzo17:

[lyric falcon]

no it's the same set

[cloud walrus]

creamy shits:

creamy shits:

[cloud walrus]

creamy shits:

[stiff scaffold]

nice

[sonic current]

why must it be closure?

couldn't it fail with inverses as well?

[raw moth]

it can't be a finite group

because every element of a finite group has finite order

and so the torsion part is the entire group

if it's self-inverse it has order 2

well

technically order 2 unless it's the identity, in which case order 1

the question is to find a non-abelian group G such that the elements of finite order don't form a subgroup?

hmm

I can't think of any off the top of my head

I've not done much with infinite groups

[cloud walrus]

creamy shits:

creamy shits:

[simple valley]

it is abelian

Consider the group $Perm(\mathbb{N})$

[cloud walrus]

mniip:

[simple valley]

with two permutations: $\alpha = (01)(23)(45)\dots(2n\ 2n+1)\dots$, $\beta=(12)(34)\dots(2n+1\ 2n+2)\dots$

[cloud walrus]

mniip:

[chilly ocean]

. Let f : R → R be a continuous function satisfying f (x) = f (x + √
2) =
f (x + √3) for all x. Prove that f is constant.

[simple valley]

@chilly ocean asqrt(2)+bsqrt(3) is dense in R

qed

[chilly ocean]

Show that infinitely many powers of 2 start with the digit 7

[simple valley]

I think they're talking about integer powers of 2

[stone fulcrum]

@chilly ocean
Check out this group:
(a, b)| a² = e, (ab)² = e

a and ab are part of the torsion. But a×ab = b is not.

Sorry, I mean:
Take two elements, a and b.

Form a group with them, such that a² = e, (ab)² = e

Note that b to any power is in this group, so it's infinite

This is actually called the infinite dihedral group, I believe.

Nono, this is a group with infinitely many rotations. Think D∞

Like a corcel

It's impossible to really visualize, but the algebra underneath isn't too bad:
s² = e, (rs)² = e

r^(1408) is part of the group, as is any power of r

So r is not part of the torsion

[thorny slate]

i don't like that presentation

i like <a, b / a^n = e, b^2 = e, ab = ba^-1 >

for Dn

where a is rotation, b reflection

[simple valley]

@stone fulcrum have you seen my Perm(N) example

[stone fulcrum]

Wat be perm(N) tho? I'll look it up

[simple valley]

the group of N<->N bijections

[thorny slate]

@chilly ocean it means generated by a and b

with those relations

ab = ba^-1 because if you rotate and reflect, it's the same as reflecting and then rotating the other way

[stone fulcrum]

So note b can take on any power, b has infinite order

Although I might have gotten the (ab)² = e part wrong, and with @thorny slate's correction, I might have to change the argument

Know how it might go, Jacobian?

[thorny slate]

what argument

@chilly ocean it's just a definition of a group

as a quotient of the group with 2 generators a, b

you can also define it as actual operations in a regular polygon

since Dn is the group of symmetries of the n sided polygon

[stone fulcrum]

We're trying to show that the elements of finite order don't form a group

[thorny slate]

oh what

in which group

[stone fulcrum]

The one you presented, lol

[thorny slate]

oh this is D^infty

the one I presented is finite

it's Dn

your presentation of D^infty is fine

ab = ba^-1 is the same as what you said

[stone fulcrum]

Ok! That makes the argument easy then

[thorny slate]

yeah you can write any element as a^n b^m

using that

and then compute with that

[stone fulcrum]

a and ab are part of the torsion, but a×ab = b is not. So, not a group.

Oh no. We're confusing all the presentations

I mean, for this:
(a, b)| a² = e, (ab)² = e

This is a group presentation. It means "Use elements a, b to form a group, such that a² = e, and (ab)² = e"

Note I didn't give any conditions on b alone, so any power of b is possible

Yus. But a²b = b is not finite order

a × ab = a²b = b

And b has infinite order

Cuz there's no power of b, such that bⁿ = e

No closure at all. This is the kind of thing therapists live for

Yus, b = b
That's always true

Infinite

There is no n, such that
(b²)ⁿ = e

If you remove a, then the torsion is {e}, or the trivial subgroup

I'm not sure what you mean

You probably could, there's likely infinitely many ways to break closure here

I found a × ab = b to be simple, but probs not the only option

You only need one counterexample to show it isn't a group

Yes, we just say "infinite order" for something that will never loop back to the identity

Because the group it generates has infinitely many elements

We've defined the group such that it can't. a² = e, (ab)² = e, but there's no restriction on b alone

Yeah, pretty much "fine picking" what needs to happen such that the torsion isn't closed

[simple valley]

my example isn't a free group with equations

might be easier to understand

@chilly ocean so consider $Perm(\mathbb{N})$

[cloud walrus]

mniip:

[simple valley]

the group of bijections $\mathbb{N} \leftrightarrow \mathbb{N}$

[cloud walrus]

mniip:

[simple valley]

yes

yes

this is clearly a group, under composition and the identity bijection

yes

the set of all bijections is a group yes

[errant drum]

It's quite an important group

[simple valley]

probably not Perm(N) but yeah, Perm(X) for an arbitrary X is an important group yes

[errant drum]

Yeah I meant the family

[simple valley]

do you not see why it's a group?

[errant drum]

Yeah

Permutation group/symmetric group are just different names

Apparently it depends where you get taught

[stone fulcrum]

The symmetric group on infinite elements, another way to think of that, I think?

[simple valley]

the symmetric group is usually for finite sets I thought

rather, for a very specific finite set: {0 ... n-1}

or {1..n} idk

[stone fulcrum]

Fair enough!

[simple valley]

okay now, consider the function alpha that swaps 2k with 2k+1

i.e alpha(2k)=2k+1, alpha(2k+1)=2k

no

Perm(N)

alpha : N -> N

alpha is clearly a bijection, right?

[errant drum]

I think when you say permutation group it's meant to be a subset of the symmetric group. But most people use them interchangeably

[simple valley]

alpha is its own inverse

that means alpha^2 = e

now consider a different function: beta(2k+1) = 2k+2; beta(2k+2)=2k+1; beta(0) = 0

also bijection and self-inverse

alpha and beta are clearly torsion, but look what happens when we compose: beta composition alpha

beta(alpha(2k)) = 2k+2

for all k

(beta composition alpha)^n (2k) = 2k+2n

[cloud walrus]

creamy shits:

[simple valley]

no

reread

[cloud walrus]

creamy shits:

[simple valley]

what

[cloud walrus]

creamy shits:

[simple valley]

no

nonono

$(\beta \circ \alpha)^2 = \beta \circ \alpha \circ \beta \circ \alpha$

[cloud walrus]

mniip:

creamy shits:

[simple valley]

that doesn't matter

what matters is that (beta alpha)^n is never e

identity in Perm(N)

yes

alpha and beta are torsion but beta.alpha is not

[cloud walrus]

creamy shits:

[simple valley]

so that it's a function N -> N

because it has a clause alpha(2k) =2k+1

that covers zero

[cloud walrus]

creamy shits:

[simple valley]

it is here

no

well it doesn't matter but I thought 0 in N was commonplace in the US

that's not a correct use of \mapsto

[cloud walrus]

creamy shits:

[simple valley]

yes

[earnest valley]

$\mathbb{N}\cup{0}$ ?

[cloud walrus]

creamy shits:

[worthy kindle]

0 isn't in N for some people

[full blaze]

I mean that notation is not used just so N subscript 0

[cloud walrus]

creamy shits:

[full blaze]

Ye

That means natural number starting with 0

Yes

Fuck u mean

[worthy kindle]

it's like the "arrays start at 1" vs "arrays start at 0" but in maths

[full blaze]

I mean either way works

Realky

My prof say with 0

If they dont start with 0 then they trash

Continuing your proof by induction

Idk

🖕

[simple valley]

@chilly ocean give me 1 reason naturals should start at 1

natural numbers in peano axiomatization have much nicer inductive definitions for addition and multiplication

when your base is the zero

[simple valley]

so?

variables were invented only a few hundred years ago

admit it you're just too rarted to understand what 0 is

[simple valley]

you mean isomorphic to S3?

"|G| = 6 => G abelian or G = S_3" sounds true

are you able to show that all groups of order less than 6 are abelian?

you have to prove that all groups of order 5 or less are abelian

do you?

well in this task you have a table of order 6

that's not what you're proving here

it's supposed to be obvious that it cannot be both

what you're supposed to prove is that it has to be one or the other

no

can you read?

no it doesn't

what "ways"?

and then what?

no?

you haven't proven what the question asks to prove

no you haven't

no you just can't read

there's no such thing

forall G, (G has order 6) -> (G is commutative) or (G is isomorphic to S_3)

the both part is fucking irrelevant

assume arbitrary G

use its properties

conclude it's either abelian or S_3

[civic linden]

where φ is a homomorphism , φ:D8→D8 where φ(r^is^j)=s^j and 0≤i≤3 , 0≤j≤1 how is φ(r^i*r^j)=e?

[earnest valley]

Because e=s^0.

[civic linden]

G=S3, H=<(12)> find the left cosets of H in G

I got H={e, (12)}

how do i find the others?

left coset is gH right

am i doing (12)H

that gets the one above

how do I find the others?

right so

I have to do the ones that are of 2,1 order

(12)(3)

(23)(1)

because the generator is of that order?

type is probably a better word

oh

eH, (12)H, (13)H , (23)H, (123)H (132)H?

ooooh yes

that makes sense

(123)H = (13)H

[raw moth]

your definition of commutative is wrong

a group is commutative if ab = ba for all a,b in the group

so a group is not commutative if there exists a,b such that ab =/= ba

ya

ya

I don't know for sure

but I would guess that

you should start with a group of order 6, assume it isn't abelian, and show that it is the group S_3

ye

that won't work

because S_3 isn't abelian

so there won't be an isomorphism

all that proves is that S3 isn't abelian

you want to prove that any group of order 6 is abelian or isomorphic to S_3

so take a group of order 6

an arbitrary group of order 6

if it's abelian, then clearly it's abelian or isomorphic to S_3

if it's not abelian, then we need to show that it's isomorphic to S_3

this covers all the possible cases: a group is either abelian or not abelian

so we've covered all possible groups of order 6 and shown that for any particular one of them, either they're abelian or they're isomorphic to S_3

we're splitting the problem into two cases

when the group is abelian, and when it isn't

if it's abelian, then there's nothing to do

because we want to show that either it's abelian or it's isomorphic to S_3

correct

:^)

[cloud walrus]

creamy shits:

[stiff scaffold]

Sup creamy shits

I have to go

[cloud walrus]

creamy shits:

[raw moth]

well, what are the possible orders of elements in a non-abelian group of order 6?

[cloud walrus]

creamy shits:

[raw moth]

the orders must exist in a finite group

but ya each element in the group has order 1,2,3 or 6

if the group is not abelian, can you remove any of these possibilities?

yes

I don't think so

the fact that the element is a generator if it has order 6 shows something more

so what kind of group is G6 if there's an element of order 6?

the entire group is generated by a single element

so it's a cyclic group

because that's the definition of a cyclic group

[errant drum]

For these types of proofs, if it's commutative you don't need to say anything. Just show that it's isomorphic to S3 in the other case

[raw moth]

ya that's what we're doing

a cyclic group is one that has a generator i.e. an element that generates the whole group

ya

exactly

the only element with order 1 is the identity in any group

[errant drum]

Identity elements are always unique

[raw moth]

so we know that all non-identity elements have order 2 or order 3

do there exist any elements of order 3?

you're right :^)

[errant drum]

(123) is an example of something with order 3

Any transposition has order 2

[raw moth]

ok I've got an idea to show there are necessarily elements of order 2 and order 3 in G_6

ya nice

what if they're all order 2?

besides identity

if a,b are non-identity elements

what is the inverse of ab?

generally in a group (ab)^-1 = b^-1 a^-1

so if every non-identity is self-inverse

then b = b^-1 and a = a^-1

so (ab)^-1 = ba

however (ab)^-1 = ab because every element is self-inverse

because ab is itself an element of the group because of closure

so we have ab = ba

wut

[errant drum]

Order 3 is what you have

[raw moth]

how did you get elements of order 2 have no inverses 🤔

I have no idea how you got that from what i said :^)

[errant drum]

Yeah that's what's just been shown

[raw moth]

what we just did showed that if every element is self-inverse then the group is abelian

so there must be an element of order 3

[errant drum]

See if you can get to the end from here

[raw moth]

ya

I don't know if the orders of the elements of a finite group being the same is enough to say that they're isomorphic but it's definitely necessary

ya

[simple valley]

1 self inverse and 4 of rder 3 can be the case

but that'd be an abelian group

[errant drum]

If you had suppose you take 3 elements of order 3

Now a^2 = a^-1

b^2 = b^-1

c^2 = c^-1

[cloud walrus]

creamy shits:

[errant drum]

Yeah I'm not sure it leads anywhere

[raw moth]

well we need to show that an element of order 2 and an element of order 3 commute

lets call the elements in the group something

say

{e,a,b,b^2,c,c^2}

now ca =/= c because a isn't the identity

and ca =/= e because a =/= c^2

also ca =/= c^2 by left-inverses that would mean a = c

if ca = c^2

then c^-1 (ca) = (c^-1) c^2

so a = c

but this isn't true because we've defined them to be different elements

c^2 = cc

c^-1(cc) = (c^-1c)c = ec = c because we have associativity

because they're elements of order 3

ok I'll go through how I defined the set of elements in the group

e is the identity

a is the self-inverse i.e. a^2 = e

b is an element of order 3

thus bb = b^2 is in the group

closure

b operated with itself is always in the group if b is

the closure property is that if you take any two elements of the group (that could be the same but don't have to be) and operate them in either order, then the result is an element of the group

yes

and c =/= e and c^2 =/= e

if c is in the group

then cc is a valid way to choose two elements of the group

ye

so we got to ca = b or ca = b^2

it doesn't really matter which because (b^2)^2 = b^4 = b

it doesn't matter which for reasons you'll see in a minute

similarly we can see that c^2a = b or c^2a = b^2

but we know that ca =/= c^2a

so ca c^2a = bb^2 or ca c^2a = b^2 b

but both are equal to b^3 = e

so (ca)^-1 = c^2a

however (ca)^-1 = a^-1c^-1 = ac^2

so c^2 and a commute

however the same argument goes to show that any element of order 3 commutes with the element of order 2

c^2a = b or c^2 a = b^2

same reason ca = b or ca = b^2

you do the same argument to show that ba = c or ba = c^2

and b^2a = c^2 or b^2a = c

ya the above argument works for any element of order 3 and the element of order 2

but then after that you need to show that b and c commute ye

but that one's not too hard

you just need to show that bc = cb = a

[cloud walrus]

creamy shits:

[raw moth]

bc = b^3

so it can't be an element of order 2 and 4 elements of order 3

[cloud walrus]

creamy shits:

[raw moth]

{e, a, b, c, d, d^2}

[cloud walrus]

creamy shits:

[raw moth]

well in the group S_3, ab = c

[errant drum]

You can define your isomorphism explicitly

It's not elegant but it gets the job done

[raw moth]

do the orders of the elements specify a unique group?

[errant drum]

No but the conjugacy classes do

Which is basically the order iirc

In this case

[raw moth]

don't you need to specify a subgroup for conjugacy

[errant drum]

Conjugates are just two elements where a = kxk^-1

So a and x are conjugates

But its kind of pointless using conjugates now because it would have made the whole question a lot quicker

Since at one point you knew the group had 6 elements. Since it's non abelian you know that the only conjugacy class of size 1 is the identity

Then the only way to partition the group in to conjugacy classes is a class of size 2 and a class of size 3

But I think @raw moth had a better way

[raw moth]

since the size of a conjugate class is the index of the normalizer subgroup of that element hmm

that's nice

[errant drum]

Yeah so it must divide 6

[raw moth]

ya

[errant drum]

But it feels a bit cheaty

[raw moth]

o I guess for individual elements it's the same as the centralizer

the centralizer subgroup of an element consists of the set of elements that commute with the element

I think

[chilly ocean]

Centralizer of an element ya

[errant drum]

I always get centre and centraliser confused too

[chilly ocean]

Centre is just intersection of all centralizers

[errant drum]

I think centralizer = commutes with one element centre = commutes with everyone

[raw moth]

ya

[chilly ocean]

Z(G) boi

We were never introduced to the normalizer kek

[errant drum]

Yeah me too

[chilly ocean]

It'll come up later

They were just like "eh, intersect these centralizers, you'll be fine"

[errant drum]

Pretty much ^^

Although centralisers aren't that special in a group but non trival centres are

[chilly ocean]

They sure have some neat properties you can bounce off of

[errant drum]

And normalisers are basically just the "is pepsi okay?" of groups

Yeah.

Just e -> e_s3
a -> (12) etc

[raw moth]

a morphism from one group to another is just a function from the elements of one group to the elements of the other one

[errant drum]

I'm not sure if the order of every element makes the group unique up to isomorphisms

[raw moth]

a function i.e. a mapping

you're defining a map there

[errant drum]

I think it's more proving it's isomorphic

Since you need to show that f(a * b ) =f(a) • f(b)

You could copy down the proof to Cayley's theorem and adjust it

Because it basically does thst

[errant drum]

Errr it specifies that it's a subgroup of S_n doesn't it?

Well now we've shown it's the whole group S3

So there would still have been work to do

But I assume the purpose of that work wasn't to get you to look up Cayley's theorem and nuke the question

So maybe don't include

It

[raw moth]

cayley's theorem just says that the group is a subgroup of S_6 I thought

[errant drum]

Yeah

But S3 is a subgroup of S6

[raw moth]

right, but then you need to prove that there are no other non-abelian subgroups of order 6

[errant drum]

Yeah

That's a group of 720 things

Yeah

[cloud walrus]

creamy shits:

[errant drum]

Also why \mathfrak{} it's just a regular letter?

I see

Fair enough

Any Greek letter at A Level was pretty high tech NGL

I remember thinking that trig was the 'end' of maths at GCSE

[cloud walrus]

creamy shits:

[errant drum]

And then the hyperbolic bois^^

[raw moth]

love hyperbolics

[errant drum]

Really I felt like hyperbolics were easier

[raw moth]

comes from the definition of the unit hyperbola

[errant drum]

Because you could always break them down in terms of exp(x)

[raw moth]

x^2 - y^2 = 1

and the parametrisation x = cosh(t), y = sinh(t)

[errant drum]

1/x is the true hyperbola

I mean they're the same hyperbola

[chilly ocean]

[errant drum]

Just twist it 45°

[errant drum]

Try and bullet point your arguments

[covert vector]

also no need to copy paste the whole thing every time 🤔

[cloud walrus]

Cosmicrays:

creamy shits:

[errant drum]

Also when you prove something

Do it at the middle

[chilly ocean]

enumerate > itemize

[errant drum]

Same thing ¯_(ツ)_/¯

[chilly ocean]

Not even

[errant drum]

Yeah do that and post the image

$\text{write text here}$
$$a^2 = b$$

[chilly ocean]

😩 👌

I think it meant to center

The parser's just weird

[errant drum]

Yeah

[chilly ocean]

,tex \lipsum[2-4]

[cloud walrus]

Jichael Mackson:
Compile Error! Click the reaction for details. (You may edit your message)

[chilly ocean]

Riperoni

[worthy kindle]

textmode vs mathmode

[chilly ocean]

Oh yeah this is a good point

,tex Howdy hi folks! $$\text{Pretentious tex}$$

[cloud walrus]

Jichael Mackson:

[chilly ocean]

Nah renderer is still borked

,tex ao;sidjfasdhfasdhg;asdhgalsdhgkasdhglkahsdashdlahsd;fajsdfaijdf;ajsdf;ajdf;aosdijfa;doifjas;d $$asdofiajs;dofiaj$$ aosidfjaoi;sdjg;aosdj;aiosdjf;oasdf;ioasd;fioajsdo;fja;doifja;sdojf;aosdjf;aosdjf;oaisdjf;aoisdjf;o

[cloud walrus]

Jichael Mackson:

[chilly ocean]

Hm

That fixed it

[raw moth]

,tex $$\text{\frac{new}{line}}$$

[cloud walrus]

TheFornicatingFalcon:
Compile Error! Click the reaction for details. (You may edit your message)

[chilly ocean]

Uhhh

Mudkip

[raw moth]

:^)

[chilly ocean]

Tf you just do

[raw moth]

put a fraction inside a text

[chilly ocean]

:l

I can see that

Why did it bork so spectacularly?

[raw moth]

¯_(ツ)_/¯

[worthy kindle]

fractions aren't supposed to be used in textmode

[delicate chasm]

Haha grats on discovering that

[chilly ocean]

Nah

[raw moth]

shift and enter

[delicate chasm]

It is actually centered but the extra space is cropped

[cloud walrus]

creamy shits:

[chilly ocean]

,$ 5 \ 3

[cloud walrus]

Jichael Mackson:

[raw moth]

,$5

3

[chilly ocean]

5 \\ 3

[raw moth]

o I guess the tex newline also works

[chilly ocean]

$$5 \newline 3$$

[errant drum]

It is but only the full image

[cloud walrus]

Jichael Mackson:

[chilly ocean]

Eh

You might have to use align environment or something

But cba

[worthy kindle]

I think newline is supposed to be used in textmode

[chilly ocean]

I think so too

$$5 \ 3$$

[cloud walrus]

Jichael Mackson:

[chilly ocean]

Nah gotta use align

Anywho

[errant drum]

That's effort

[chilly ocean]

@yes Get texmaker, you bleb

[delicate chasm]

What the devil you trying to do

[chilly ocean]

Break your dumb bot

Bring back math bot

[delicate chasm]

Lol

[chilly ocean]

Viva la revolucion!

[errant drum]

Trying to align equations to the cropped centre @delicate chasm

[delicate chasm]

Mathbot doesn't even have equation mode

[chilly ocean]

Mathbot is pretty inferior tex renderer tbh

Although

[delicate chasm]

You can use texw if you really really want the full width

[chilly ocean]

I think you should add the lipsum package

For extra spam functionality

[delicate chasm]

Add it to your own preamble if you want it

[chilly ocean]

I can modify my preamble?

[delicate chasm]

Yes

[chilly ocean]

I should just read the docs shouldn't I?

[delicate chasm]

Yes

[worthy kindle]

That's why the bot is fantastic

[chilly ocean]

Wat

Just get texmaker dude

[errant drum]

Where can you read the docs? Help command?

[chilly ocean]

,help

[delicate chasm]

,align
a &= b\
c &= d

[cloud walrus]

Puerøsola:

[chilly ocean]

Lovelehh

[delicate chasm]

[errant drum]

[chilly ocean]

Oml

[worthy kindle]

°ω°

[delicate chasm]

I'm glad you like it btw Tuong :)

[worthy kindle]

( ^_^)

[chilly ocean]

So yeah

Now awaiting approval

Is it downloading the package or something?

Texmaker oof

Which one?

[raw moth]

\noindent

I think

[worthy kindle]

dafak is this software

[delicate chasm]

Lyx?

I hate wysiwyg editors

[worthy kindle]

why are the integrals so small, why no consistent italic, o god why why D:

[delicate chasm]

Please migrate to a real editor sometime

[worthy kindle]

phew

cute

[jagged gate]

the annual income of porn societies in nebraska

that's ur tables?

[oak perch]

er ask and hopefully someone chimes in.

[stone fulcrum]

Yeah! Feel free, here to help

What about it?

The identity in An is the "do nothing" which is an even permutation

[raw moth]

the group operator for the sign is multiplication

I would guess so

just guessing from context tbh

[stone fulcrum]

It's a homomorphism. It takes an element from Sn and returns an element from {-1, 1} under multiplication

Depending on odd/even

The identity of {-1, 1} is 1, obv

Lel

Ker(sgn) = An

[raw moth]

implebian sounds like a group property

[stone fulcrum]

No that's fair, you're getting used to the notation

You're figuring out your implebain

Note the important fact that An is always a normal subgroup of Sn. And seriously, that becomes important.

If you take Galois anyway

Whew

Let's say g generates the group

gⁿ for some n gives any element in the group you may want

φ(gⁿ) = φ(g)ⁿ cuz φ splits over the group operation

[raw moth]

if φ is a homomorphism, then φ(ab) = φ (a) φ(b)

this is what is meant by splits over the group operation

= φ(a) φ(a) φ(a) ... φ(a)

[stone fulcrum]

So φ(gⁿ) = φ(ggg...) = φ(g)φ(g)φ(g)...

φ(gⁿ) = φ(g)ⁿ
So if you know φ(g), you can get φ(gⁿ) for all n, and thus you get φ of any element in the group

Everything's determined

True as well if there's multiple generators, similar proof.

[raw moth]

if your set of generators is {a,b,c} for example

every element is a finite product of a,b,c in some order

[stone fulcrum]

That's exactly the proof we did just now

The group has to be cyclic if it has a single generator

And they define it to be a morphism. I don't know why they're asking you to prove that

Yay

💪

Yeah, that's a mapping

[solar wyvern]

@chilly ocean imagine you mean S_3 not S_6

[sullen flint]

not Cayley's theorem?

[raw moth]

cayley's theorem says every group of order 6 is isomoprhic to a subgroup of S_6

no

it has to be a subgroup of order 6

of which S_3 is one

the cyclic group of order 6 is another

and if there was another non-abelian group of order 6 it would be that as well

so I don't think that it actually helps with the question

unless there's something you can do that I'm missing

but anyway, next part of the question

it is but it doesn't prove it

[cloud walrus]

creamy shits:

[earnest valley]

$\mathbb{Z}_7^\times=\mathbb{Z}_7\setminus{0}$?

[worthy kindle]

Z/7Z without the class of 0 seems fair

[thorny slate]

Its the multiplicative group of units

[worthy kindle]

also happens to be the group of inversibles for ×

ain't they like, the permutations of {1,...,6}?

[solar wyvern]

you wanna map generators to generators

think about what the generators of the multiplicative group are :)

1*1=1 mod 7

[raw moth]

ya that'll work then just from that I think

[solar wyvern]

are there any others

now what can you conclude about 6?

why

it's not an ordered set

also 6=2*3

also consider that 6 =7-1

so it's like -1

meaning?

what's its order :^)

what's the order of -1

👏

there's a lot of tricks here

[spark plank]

6x6 = (-1)x(-1) = 1 mod 7

[solar wyvern]

okay, now try to solve the problem keeping in mind generators of cyclic group

[stiff cargo]

1-1
2-4-8-1-2
3-2-6-4-5-1-3
4-2-1-4
5-4-6-2-3-1-5
6-1-6

[solar wyvern]

you got this boi

[sonic current]

@chilly ocean i'm with you buddy!

[solar wyvern]

oh

um

do you know what the generators of a cyclic group are?

1 is a generator always

since cyclic is just addition

what else though

um

they don't have to be common

you can map one generator to another generator

but you're sorta onto something

a hint is that

you can easily classify the generators of cyclic groups, in general for all natural n

(the multiplicative case is slightly more nuanced)

consider divisibility and stuff

like if m<n and m|n can m be a gen of C_n

why or why not

umm

consider 4 in C_{2^10}

how about like 3 in C_{70}

so you don't want to assume that all groups will work like small order groups like Z_6

3*23=70-1

i guess it doesn't 🤔

um

you'd want 3*71

no, since 3*24=2 mod 70

cyclic group is addition

so the things 3 generates are of the form {3,3+3,...n*3} not {3,...,3^n}

but in any case it can only generate at most 1/3+1/2 = 5/6's of the elements of C_70

no, because it doesn't generate every element 🤷

then 0 would be a generator

if n is odd i think 2 works as gen for C_n

ya

nah, since if m|n you're already boned

um

consider like 3 in C_27

you just would get stuff like {3n \mod 27}

ya ummm

might need a moment to think about this 😄

you can still solve the problem you have with what you're given since C_6 sm0l boi

and since 0,2,3 divide C_6 you know anything divisible by those won't be a generator (hence why you know a priori 4 will not be a gen for C_6)

but you know 1 and n-1 will be generators

um

ya you're right should be 1

yeah, and n-1

(since it's basically -1)

ya

so ya...each has 2 generators i guess

[white turret]

pls sum1 walk me thru dis

[stone fulcrum]

The only proper subgroups are nZ. Does that help?

[white turret]

what is a proper subgroup

[solar wyvern]

a subgroup not the group itself

[stone fulcrum]

And not the trivial group

[white turret]

what is a group

most general math object right

from which rings fields and all are derived

whats the defintion tho

[stone fulcrum]

So a book or a YouTube video may give everything you need here

[white turret]

ur right

[stone fulcrum]

SA was reading from this

https://www.jmilne.org/math/CourseNotes/GT310.pdf

It's advanced, but you should be able to get some of it, then ask questions

[spark plank]

[solar wyvern]

keep reading SA

[spark plank]

what's a Z-module tho

[solar wyvern]

so you know what a vector space is right?

[stone fulcrum]

Think a vector space where the scalars are integers

[spark plank]

gib more reads

t!wiki vector space

[fossil mango]

📖 | ** https://en.wikipedia.org/wiki/Vector_space **

[spark plank]

oh nvm

[solar wyvern]

and a Z-algebra just means you can "multiply" vectors

[covert vector]

Z modules are abelian groups

[spark plank]

kk

[covert vector]

they're like "vectors" over the group elements, with coefficients in Z being the exponent on the elements

you can write them in this fashion cuz it's abelian

[spark plank]

yeah m'kay

so those exponents are basically like scalars for vectors

[solar wyvern]

exactly

[spark plank]

explain last sentence

[covert vector]

torsion elements are the elements of finite order

[solar wyvern]

this is a definition

[covert vector]

yeah it's not a result, it's a definition

[raw moth]

idk the fact that they form a subgroup is a result

[covert vector]

well it's straight forward from commutativity

[spark plank]

define subgroup

[raw moth]

ya

[covert vector]

it's a subset that's also a group

[spark plank]

kk

[raw moth]

subset of elements of a group that form a group with respect to the same group operation

[solar wyvern]

[covert vector]

and in general you can concatenate words, or take out a convenient word to make it sound nice

for general math words

[spark plank]

so basically taking the elements that have finite order form a subgroup

[covert vector]

vector space, subset that's also a vector space, → subspace

yes

note that u need commutativity for this, without that you can construct counterexamples

[spark plank]

kk

define field

oof matrices

🤢

[worthy kindle]

🤢

[covert vector]

commutative addition, subtraction, multiplication, division (by Nonzero), mult distributes like normal

[spark plank]

[covert vector]

any system where you can do the normal arithmetic operations

[solar wyvern]

it's a commutative group in both operations, with your usual distribution

[covert vector]

like R

[spark plank]

kk

[worthy kindle]

some people have their field def have a commutative mult, some other people don't

[stone fulcrum]

This reference is great, I like the speed. I just wish it didn't reference stuff that you need to have passed group theory to know

[covert vector]

which reference ?

[spark plank]

gib basic thing for me to proof about groups

[solar wyvern]

(keep reading)

ummm

[covert vector]

prove that |ab|=|ba|

[spark plank]

|a| = ?

[covert vector]

where a,b are arbitrary group elements and |ab| is finite

[stone fulcrum]

Take an element of a finite group, multiply it by itself until you get the identity. Prove the elements generated this way makes a subgroup.

[covert vector]

|a| is least positive integer n such that aⁿ = e

[spark plank]

oh the order of a

[covert vector]

if it's infinite then it's infinite

I guess it's not a positive integer

you don't really write it down then, you'd just say it's infinite

a better definition is the order of the group generated by it

which is what kaynex wrote

[solar wyvern]

$n \in \mathbf N: a^n = e; \text{moreover for all } m \in \mathbf N: a^m=e, n|m$

[covert vector]

mine is a bit more involved of a proof, but group theory proofs are basically just puzzles

[cloud walrus]

flimflam:

[covert vector]

you don't really need to know anything to do them, just problem solving skills

and good at puzzles

and if u don't have those then lots of patience and persistence

[solar wyvern]

ya, especially if you restrict to finite/countable groups

[covert vector]

either way works

for most questions you'll encounter, it's not like there's some op theorem that'll overkill it

it's just called being a god at puzzles

which is what I like about group theory tbh

XD

[spark plank]

$b(ab)^n=(ba)^nb$ (induction + associativity)\$b(ab)^n=(ba)^m(ba)^{n-m}b$\$be=e(ba)^{n-m}b$\$be=(ba)^{n-m}b$\$be=b(ab)^{n-m}$\$e=(ab)^{n-m}$

[cloud walrus]

Simple_Art:

[spark plank]

I think this proves that $m\ge n$ and we can do the same for $m$

[cloud walrus]

Simple_Art:

[spark plank]

did I do it?

@covert vector @solar wyvern

[solar wyvern]

sec

[spark plank]

ur late

[solar wyvern]

um

[covert vector]

well that shows that |ab| divides |n-m|

[spark plank]

Let n = |ab| and m = |ba|

[covert vector]

ya

I should be vacuuming

[spark plank]

I believe the last line implies we must have n-m ≤ 0

unless m = 0

[covert vector]

it doesn't really matter if it's positive or negative unless u have a double inequality

[solar wyvern]

think it'd be better to show that n|m and m|n

[covert vector]

bounding on both sides

since elements are invertible negative exponents are just powers of the inverse

[spark plank]

but didn't you say that n = |ab| is the least positive integer such that (ab)^n = e?

[covert vector]

ya

[solar wyvern]

i didn't say that

[spark plank]

and m = |ba| is the least positive integer such that (ba)^m = e

[covert vector]

well I did :P

[spark plank]

this means m > 0

[covert vector]

Ya

[spark plank]

If n-m > 0, that means I've proven that n-m = |ab|

[covert vector]

why?

[spark plank]

or rather, n-m is less than n

and satisfies "(ab)^(n-m) = e"

and n-m is a positive integer

Which contradicts the definition of n

[covert vector]

ohoh that I guess is a misunderstanding

you can have (ab)^k = e

where k is negative

[spark plank]

Yes, I know

that's why I said n-m must be ≤ 0

to avoid the contradiction

hence n ≤ m

and you can do my entire argument again to prove n ≥ m

=> n = m

[covert vector]

I don't see how this deduces that n-m ≥ 0

we could have m=2n

[spark plank]

okay

let m = 2n

then we do our proof over again:

[covert vector]

like it probably works but I am slow and skeptic 🤔

[solar wyvern]

i c mupkip

[raw moth]

hi

[spark plank]

$a(ba)^{2n}=(ab)^{2n}a$\$a(ba)^{2n}=(ab)^n(ab)^na$\$ae=e(ab)^na$\$ae=(ab)^na$\$ae=a(ba)^n$\$e=(ba)^n$

this proves that |ba| ≤ n < 2n = m which is a contradiction

wait typos

[raw moth]

I think this one can be nuked

[covert vector]

u can be nuked

[solar wyvern]

no nukes

[covert vector]

ya no nukes tho !!

[raw moth]

hmm

I'll pm and see if you think it's a nuke I guess

[cloud walrus]

Simple_Art:

[covert vector]

ok u can DM me 😄

[solar wyvern]

for abelian isn't $(ab)^n=a^n b^n$

[cloud walrus]

flimflam:

[spark plank]

I never used that

[covert vector]

yes @solar wyvern

[solar wyvern]

you should 🤔

[covert vector]

but it's not abelian

[spark plank]

?

^

[solar wyvern]

rip

i think sa's thing prolly works

[spark plank]

but I think this should prove that |ba| ≤ n < 2n = m

which is a contradiction

meaning we can't have m = 2n

[solar wyvern]

maybe up to something like m=-n

[spark plank]

reminiscent to proving ordinal fixed-points

m and n are positive integers

[solar wyvern]

ok then you're probably good

[covert vector]

Ok!

t!cookie @spark plank

[fossil mango]

Cookie 🍪!

woog has given Simple_Art a cookie!
Easy cookie, easy life.

📥 6 | 📤 4

[spark plank]

🍪

gib 'nother prob

[solar wyvern]

um

you can embed groups into symmetric groups

[spark plank]

gonna go prep lunch

big words

[solar wyvern]

https://en.wikipedia.org/wiki/Cayley's_theorem

[covert vector]

here is a cool problem:

Let G be a finite group, and let p be the least prime number such that p divides the order of G. Prove that if H is a subgroup of G such that |G|/|H| = p, then H is normal in G