groups-rings-fields

[oblique river]

if H is a subgroup of R/Z, if it contains an irrational element, then it's clearly infinite

as nx - mx is never an integer for any n, m in Z if x is irrational

(unless n = m haha)

so you only need to look at subgroups where all elements are rational, i.e., subgroups of Q/Z

[stone fulcrum]

Okay! That makes sense. Let me think

Lol, I don't know if he's still here

[hollow comet]

Does anyone in here know about affine group schemes?

[gusty inlet]

„Does anyone in here know about affine group schemes?“
I know what an affine group scheme is, but I think I can only answer rather elementary questions about them.

The category of affine group schemes is equivalent to the opposite of the category of cogroup objects in the category of rings.

So, up to a category-equivalence, an affine group scheme is a ring A, together with ring homomorphisms μ : A → A⊗A, and ι : A → ℤ, such that some diagrams commute.

while a morphism of affine group schemes, is just a ring homomorphism between the underlying rings going in the opposite direction

what do you have to know about affine group schemes?

[delicate chasm]

Nice question

[cloud walrus]

creamy shits:

creamy shits:

[lyric falcon]

so how could a generator look then

[cloud walrus]

creamy shits:

[lyric falcon]

yeah but in terms of $n_k$ and $m_k$

[cloud walrus]

Gonzo17:

[oblique river]

@hollow comet I know a little bit about affine group schemes but I'm not an expert!

[lyric falcon]

well not necessarily

at least that's not a property we're interested in

we want the generator multiplied by an integer (and then fractional part of it) to be any element H

[oblique river]

1/3 * 2 = 2/3

[lyric falcon]

no, you multiply a fraction by integers

I mean, the group operation applied n times

if the generator is $\frac{1}{4}$ then we want $H\subset {0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, }$

[cloud walrus]

Gonzo17:

[lyric falcon]

if the group is $H={\frac{m_1}{n_1}, \ldots , \frac{m_k}{n_k}}$

[cloud walrus]

Gonzo17:

[lyric falcon]

then what could a viable generator be?

[oblique river]

what if m1/n1 = 1/2 but m2/n2 = 1/3?

[lyric falcon]

what? no you don't need to

[oblique river]

have you ordered these in some way?

[lyric falcon]

like $H={0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, }$

[cloud walrus]

Gonzo17:

[lyric falcon]

1/2 doesn't generate it

ok but working with the smallest element would be really hard

we don't care right now if the generator would even be in H

[oblique river]

what? the generator of a subgroup is always contained in that subgroup

[lyric falcon]

we just want sth that definitely would be a generator if it was in H

yeah I want to show that a specific element $f(m_1, m_2, m_3 ... , m_k, n_1, n_2, ... , n_k)$ that we set up to be a generator is in H from properties of groups

[cloud walrus]

Gonzo17:

[lyric falcon]

so like $\frac{1}{n_1 \cdot n_2 \cdot \ldots \cdot n_k}$

[cloud walrus]

Gonzo17:

[lyric falcon]

this would definitely be a generator if it was in H

[oblique river]

maybe I missed something but I feel like you've already solved the problem. if H is a finite subgroup of Q/Z (or equivalently R/Z), then H is cyclic and generated by the smallest nonzero element of H

and you can prove that statement in that form

[lyric falcon]

yeah it isn't that's the problem

[oblique river]

what isn't?

[lyric falcon]

ok now I look like an idiot

let me just feed you this step

consider $\frac{1}{GCD(n_1, n_2, ... , n_k)}$

[cloud walrus]

Gonzo17:

[oblique river]

@chilly ocean the smallest element of H is obviously contained in H

[lyric falcon]

if we can prove that that's in H then we're done

no this is completely different

[oblique river]

but gonzo why would you do it that way?

you can directly work with the smallest nonzero element of H

[lyric falcon]

I tend to overcomplicate solutions

maybe you can, idk @oblique river

[oblique river]

let K be the subgroup of H generated by the smallest nonzero element x of H. We want to show that K = H. Suppose not. Let y be an element in H but not in K

there is a unique minimal n such that nx > y. Then 0 < (n-1)y < x is still in H, but that contradicts the choice of x.

no nonsense with gcds required

[lyric falcon]

ooo that'snice

my thing would work too though

[oblique river]

and tbh even I overcomplicated it; I never needed to even give K a name. I thought I would need to reference it again but I didn't haha

[lyric falcon]

and it wouldn't be that much longer

[oblique river]

I mean I'm not saying it wouldn't work, but you have to all of a sudden worry about things like reduced forms of fractions, gcds, maybe some kind of euclidean algorithm

[lyric falcon]

yeah that's way easier I can give you that

[oblique river]

like even proving that "if n_1, ..., n_k are denominators of reduced forms of elements of H, then their gcd is, too" is gonna be a little icky

depending on how precise you wanna be

[lyric falcon]

not really

[oblique river]

I guess youre right, you could do it by induction or something

[lyric falcon]

there's always x, y such that $n_i x+n_j y=GCD(n_i, n_j)$

[cloud walrus]

Gonzo17:

[lyric falcon]

ok ignore that

that's for my soln

[oblique river]

because you want to show that the smallest nonzero element of H generates H

by my definition, the smallest nonzero element of H generates K, because K is the subgroup generatd by it

we know that K is contained in H since x is

so to say that H = K (or equivalently that H is contained in K)

is saying that every element of H is some multiple of x

which is what we want

[cloud walrus]

creamy shits:

[lyric falcon]

1/2 + 1/3 is 5/6

which is not in that

[cloud walrus]

creamy shits:

[oblique river]

y is not in K, so by assumption it's not a multiple of x

since the multiples of x are precisely what K is

so there is some smallest multiple of x that is larger than y

or equivalently, some largest multiple of x that is less than y

what do you mean? both are true?

I called the first one n and the second one n-1

what? y is some finite rational number between 0 and 1

so is x

why isnt my message sending >:(

multiply x by 99999999

for some suitably large 99999999

and it's obviously gonna be bigger than y

what

what

what has to be in H

that's not relevant here

what???

that doesn't matter here

forget about K and H

do you agree that given two rational numbers x and y in [0,1), there is a smallest n such that nx > y

you're not forgetting H and K lol

but in any case that diagram isn't accurate

you should think of H as some finite scattering of points in [0,1) and K as some regular scattering of poitns contained in that

start with x, it's smaller than y

look at 2x. maybe it's bigger than y in which case you're done

if not, look at 3x. maybe it's bigger than y in which case youre done

if not, look at 4x

etc

eventually you have to get bigger than y

like if x = a/b, then b*x is larger than 1

but y was less than 1

so clearly bx > y

so somewhere between 1*x and b*x

you had to go from "less than y" to "greater than y"

test what

which set? you're correct that the set {nx : n in N} has no maximum inside Q

yes

true but not relevant

what i'm talking about now is happening inside Q, not inside Q/Z

again I'm working in Q, not in Q/Z

forget about H and K for a moment. Forget about Q/Z. I have two rational numbers x and y with x < y

do you agree that there is some smallest n such that nx > y

as rational numbers

yes but right now I want specifically the smallest one

great, so we can write (n-1)x < y < nx

now subtract (n-1)x from all terms

0 < y - (n-1)x < x

great. now this is true as rational numbers, and everything involved is between 0 and 1

so now the same statement is true in Q/Z

Q/Z is literally just Q but you forget integer parts of things

it doesn't matter if you forget the integer part before you subtract or after you subtract

I mean if you're really concerned about this, just remember that (n-1)x is also less than 1

because it's less than y

uhhhhh I woudln't really go that far

well it could be equal to 1 a priori

but (n-1)x is certainly less than 1

because it's less than y

so the equality 0 < (n-1)x < y < nx \leq 1

is true in [0,1)

now subtract (n-1)x from everything

that's true in [0,1)

and NOW you are allowed to remember H

and you say "wait a minute, x was the smallest thing in H, but y - (n-1)x is in H and smaller than x!"

okay

the order on Q/Z is EXACTLY the same as the order on Q, as long as all of the integer parts of the numbers in question is 0

the integer parts of x and y - (n-1)x are both 0

the fact that there is no maximum in [0,1) really isn't relevant here

maybe you're not thinking of Q/Z in the best way

just think of it as Q intersect [0,1)

except you wrap around back to 0 once you hit 1

oh shit it's 11:30 i have to go

yeah sorry to rush out

here is another related problem that you might have done: Prove that every subgroup of Z is cyclic

how do you prove that? You say "let H be a nontrivial subgroup of Z and let x be the smallest positive element in H"

and then you prove that H msut be generated by x

in exactly the same way we're doing here

try working through that logic

and then see if you can apply that logic to this problem

(that same logic being: if y is in H but not a multiple of x, then there is a smallest n such that nx > y, but then y - (n-1)x is in H but not 0 and also less than x, contradiction)

good luck!

feel free to keep posting your thoughts here, I wont see them for a while but I will respond at some point, and maybe someone else will see them and can help as well

see you!

[errant drum]

I haven't read the question but closure under addition?

So when you combine two things in the group you get something else in the group

Exactly

Remember if y-(n-1)x = 0. Then y = (n-1)x so y is in K

That's why we have the restriction 0<y-(n-1)x

I'm just pointing it out lol

Wait idgi that proof is fine

What part are you stuck on?

Yes

if (n-1)x = y

then y is in K

Because (n-1)x = x + x .... + x n-1 times

It's a proof by contradiction so y-(n-1)x is not in K

No.
If (n-1)x = y then y is generated by x.

[cloud walrus]

creamy shits:

[errant drum]

If y = (n-1)x = x + x.....+x (n-1) times

Correct?

So then y is generated by x. Everything generated by x is in K

Do you agree that y = (n-1)x implies that y is generated by x?

So we cannot have y = (n-1)x

Remember K is the group generated by x. So if we had y=(n-1)x then y is in K

For our proof by contradiction we're selecting some element that is in H but not in K

There's no mention of G

But H is a subgroup and K is the group generated by the smallest thing in H

You're trying to prove H is generated by x

So you're doing it by contradiction

By assuming it's not generated by x

There might be more generators but you're taking the smallest one

It doesn't matter if there's 2. If you have a finite set of numbers one of them will be the smallest

Exactly

Which means the group generated by x = K is the same as H

Your proof doesn't matter if there are 2 generators in K.

Because one of them will be smaller than the other

You're showing that it is cyclic

Wait I just re-read the question K is cyclic

Because it's defined as being generated by x which is the smallest member

That means H is cyclic

If K is generated by x and K = H. Then H = <x>

Yeah but it was a proof by contradiction

You assumed that K = \ = H. And then show that it leads to a problem

The problem is that you can find an element smaller than x

Yeah

No

Let me explain

1. x is the smallest thing in H

2. Let's assume H =/= K. Then by our assumption there must be a y which is in H but not in K

Do you follow so far?

K is the group generated by x

H was a group of rational numbers wasn't it?

Okay and G is a group of what?

So H is some subset of G.

[cloud walrus]

creamy shits:

[errant drum]

And H has a least member

Okay we know that H has a least member

Which is called x

And we also know that we can generate a group with this x

Yes

Now let's assume that they're not equal so K must be smaller than H

And we'll generate K by using x as our generator

Now because H > K there has to be at least one element y that's in H but not K

Yeah

And you know K is a group

Yeah it's easy to prove

You don't need to

Anyway by our earlier argument

(n-1)x <y < nx

Yes?

So if y = (n-1)x

Then y is generated by x

We're taking something that is not in K which is the group generated by x

Now to get from (n-1)x to nx. We add an x

So to go from (n-1)x to y

We have to add something smaller than x

(n-1)x + x = nx yes?

So because (n-1)x < y and because (n-1)x and y are both in H

then y-(n-1)x is also in H

This is the closure property

Yeah because K is a subset of H

Exactly

Now because (n-1)x < y < nx

I'm going to subtract (n-1)x from this inequality

So 0<y-(n-1)x < x

Again because of closure

Everything in this inequality is in H

Let t = y-(n-1)x

Now t<x

And t is inside H

Which means that t is the smallest member of H

But we said that x was the smallest member

So we have a contradiction

There we go ^

Hence K = H

Yeah

[sour plume]

I've been going through the chatlog a bit; you mean finite subgroup, right?

marvellous

(though the order of the two sentences is still a bit wonky)

[solar wyvern]

how about (potential) infinite noncyclic subgroups?

@chilly ocean

it just says any subgroup.

I'm just reading off the problem statement you posted

[cloud walrus]

creamy shits:

[errant drum]

Your assumption was that x was the smallest but it turns out y-(n-1)x is the smallest.

In fact you could say t = y-(n-1)x

And then do the whole thing again

And create something smaller than t

@chilly ocean it means that under the assumption H =/= K. H can't have a smallest member but we know H does so that assumption of H =/= K is wrong

Now either H = K or H = \ =K

Sorry typo

:^)

[cloud walrus]

creamy shits:

[solar wyvern]

u there @chilly ocean

what's "smallest element in H" mean

smallest as in with respect to R?

"set generated by..." should be group?

Omit In addition to H being a subgroup, this is already assumed.

might seem trivial, but u might want to point out that the subgroup <0> is cyclic congruent to C_1

also I'm not sure you can use a minimality argument since G and therefore its subgroups, do not have any order on them

consider $x=\dfrac23,y=\dfrac34$.

$2x = \dfrac13, 3x = 0, 4x=x; \therefore nx<y, \forall n \in \mathbf N$

[cloud walrus]

flimflam:

[oblique river]

yo im back

@solar wyvern as I said a while ago when you're finding n, you're working in Q, not in Q/Z

the proof is correct

what counterexample?

that's not a counterexample?

take n = 2

2*2/3 = 4/3 > 3/4

I dont see the issue...

did you work through the other proof I told you about? that every subgroup of Z is cyclic?

I told you how to do it though

yes

let H be a nontrivial subgroup of Z. let x be the smallest nonzero element of H

we want to prove that H is generated by x. suppose not. Then there is some y in H that is not a multiple of x

let n be the smallest number such that nx > y

then 0 < y - (n-1)x < x

and y - (n-1)x is in H, by closure

and it's nonzero because y is not equal to (n-1)x, because we chose y so that it wasn't a multiple of x

this contradicts our initial choice of x

QED

not sure

:)

the proof you're doing now is literally the same argument

nope

the only place you need the finiteness assumption in the Q/Z case is to guarantee that there is a smallest element of H

because finite sets have smallest elements, whereas infinite subsets of Q need not have a smallest element

however we don't need that assumption with Z because it's just true that every subset of the natural numbers has a least element

(by the well ordering principle)

I have to go again

if you're still struggling I recommend just stepping back for a bit

you've been at this for hours and I know with me, if I stare at the same thing for too long, it gets super twisted in my head and I can't keep things straight anymore

I mean you can work on another question

I have to go again

[solar wyvern]

ya sorry i goofed

[oblique river]

I mean if all you care about is the grade then I've already given you a proof to write down lol

so why not move on to another question and try to understand that one?

maybe a slight change would be beneficial

[bleak abyss]

Grades are for nerds smh

[oblique river]

in any case I have to go now, gl

[solar wyvern]

okay, so how about trying n prime first

so you have an abelian group G

yes.

also since it asks about elements with the order dividing n u can restrict yourself to elements with finite order

🤷

a group of order n generates a subgroup iso to C_n

no just cyclic group of order n 😄

okay u seem like u need brek

[delicate chasm]

Hey crem

How's it going?

Okays

[slow egret]

This is why you get 7k messages a week

I see the secrets

week*

An element?

Hmm

[delicate chasm]

That is the definition

So $a^2 = e$

[cloud walrus]

Puerøsola:

[delicate chasm]

It's also the order of the broup it generates

THe order of a group is just the size

$\langle a \rangle = {e, a, a^2, a^3, \dots} = {e, a, e, a, e, a, \dots } = {e,a }$

[cloud walrus]

Puerøsola:

[delicate chasm]

Yep!

Yep

Lagrange says that every subgroup of your group has order dividing the order of the group

So in particular every element has an even order except the identity, as you stated

So, every element $a \in G$ with $a\neq e$ satisfies $a^{2n} = e$ for some $n > 0$

[cloud walrus]

Puerøsola:

[delicate chasm]

Helps to write things out in symbols sometimes

It does in this case

What you are looking for is an element $b\in G$ with $b \neq e$ such that $b^2 = e$

[cloud walrus]

Puerøsola:

[delicate chasm]

How do you mean split the power?

Write that out, don't just say it

Yeah, but write it down :P

Yep

[cloud walrus]

creamy shits:

[delicate chasm]

Ahh

Not quite

$a^2a^n = a^{2+n}$

[cloud walrus]

Puerøsola:

[delicate chasm]

There is a different power rule you might want to use :)

You nearly have it

[cloud walrus]

creamy shits:

[delicate chasm]

Um

[cloud walrus]

creamy shits:

[delicate chasm]

So close

[cloud walrus]

creamy shits:

[delicate chasm]

http://www.mesacc.edu/~scotz47781/mat120/notes/exponents/review/review.html

Lmao

Do you have it now?

[cloud walrus]

creamy shits:

[delicate chasm]

YES

Then you're done :D

Want me to write the proof as a whole?

Sometimes helpful to see

Sure! And you did most of the work :P

Nah the key point was Lagrange's theorem xD

And seeing that it implied elements had even order

You were so close with $a^na^n$ lol

[cloud walrus]

Puerøsola:

[delicate chasm]

I was like that's a square

Power rules like that should be ingrained in you.. just takes practice, is all :)

You'll be fine

[solar wyvern]

@chilly ocean C3 a subgroup of C12

[cloud walrus]

creamy shits:

[solar wyvern]

@chilly ocean every H doesn't have to be even order (at least not from whatever you've said), but you can show there exist even order subgroups by demonstrating even order elements

like I said

You'll need it, but first you need to be a bit more careful

instead of saying all H are even order, you want to show there exists H even order

try by contradiction that no even order element exist

note this won't immediately net you an order 2 element

Not sure if this is too obvious, but I'll spell it out once:

The order of $a \in G$ equal to the order of the cyclic group $<a> \leq G$

[cloud walrus]

flimflam:

[solar wyvern]

leq meaning subgroup

[oblique river]

what proof are you working on now?

not true

also: what is H?

a random subgroup of G or a specific one?

it's not true that "all subgroups of a group of even order also have even order"

for example, Z/3Z is a subgroup of Z/6Z

I'm not sure if lagrange will help you

[cloud walrus]

creamy shits:

[oblique river]

oh, yeah, that's basically lagrange's theorem haha

so I guess it is helpful!

also wait a minute

nvm thats not true

It would be true if you had $\exists n > 0$ instead of $\forall n > 0$

[cloud walrus]

Buncho Bananas:

[oblique river]

saying an element has order 2n means that a^(2n) = 0

no quantifiers

what IS true is the following statement:

wait a minute nvm im still being dumb

no quantifier on n at all

let me try again hahaha

what is true is the following:

If $G$ is a finite group (which it is in our case) then for all $a \in G$, $a^{|G|} = e$

[cloud walrus]

Buncho Bananas:

[oblique river]

what you wrote at first is not true

because it would imply that every element of G had order 2

which is clearly false in most groups with even order

yeah but then it's trivial because it follows from what I wrote, just taking n to be |G|/2

forget what I said about "change it to exists"

so maybe we should back up a bit: can you go over again what your plan of attack is for this problem?

what is H

H is a subgroup of G

but which one? just a random one?

how does taking a random subgroup of G help you here?

how does taking a random subset of G help you here?

correct

:)

yes

a^2 = e

well that's the whole problem haha

what do you mean by a being even or not?

but G is a random group

what does 2n mean in a random group G

[delicate chasm]

Oof

[oblique river]

but what does it mean for an element of G to be even

I know what it means for the order of G to be even

that's what you're trying to prove

[delicate chasm]

Yah

I can see how to prove it with Lagrange

The standard way is a counting argument

[bleak abyss]

Oh no I think I inadvertently found the nerd chat. But it seems we're talking group theory

[delicate chasm]

Lol

[bleak abyss]

What's the problem?

If you're explaining I won't spoil or smth, just wanna know

[delicate chasm]

It's trivial xD

[oblique river]

@delicate chasm i actually can't see a lagrange proof lol, I can only see the counting argument

[delicate chasm]

Every group of even order has an element of even order

Sorry I didn't check that before crem xD

[oblique river]

@chilly ocean >the order of G to be even just means that the number of elements it has is an even number

yes, the definition of "the order of G" is "the number of elements in G"

[delicate chasm]

element of order 2*

[oblique river]

so if one is even, then the other is too, because they are the same thing

[bleak abyss]

Oh okay I was about to say, that's def a Lagrange argument to just say even order. I mean I guess you could also use Lagrange here as well

[delicate chasm]

That was also my first thought, but...

[oblique river]

is there even a lagrange argument to give even order?

maybe im just being dense but lagrange only tells you that the order divides an even number

[bleak abyss]

Oh right whoops

[oblique river]

but like, for example, what if you had a group of order 10 but all the elements happened to have order 1 or 5

I don't think lagrange is helpful here

[delicate chasm]

Well you can just apply Cauchy and be done with it lol

[bleak abyss]

Lel

[oblique river]

I mean that's obviously not the intended solution

[delicate chasm]

Yeah

[oblique river]

you could say that about any homework problem haha

"just apply the fact that it's true"

:P

[bleak abyss]

Yeah I had a problem like this, I think on my very first algebra pset, far before we did Cauchy

[oblique river]

"if it weren't true then you wouldn't have assigned it, QED"

[delicate chasm]

Lol

[bleak abyss]

With at least one professor here that argument wouldn't hold...

[delicate chasm]

XD

[oblique river]

I can think of several haha

[bleak abyss]

Like with low probabiliyt

[delicate chasm]

Same

[bleak abyss]

The guy who taught analysis last winter would ask students to start the pset early to find inevitable mistakes

[delicate chasm]

Don't google it

I guess you have to go for the standard counting type argument

It's a standard problem

The guy teaching algebraic geometry last semester had several problems on the pset which started with "fix the following problem and solve it"

[bleak abyss]

Yup same problem

[delicate chasm]

I suppose no Lagrange for b?

[bleak abyss]

Nope

[oblique river]

@chilly ocean my honest recommendation is that you should call it a night and tackle it again tomorrow

you've been at this for like 12 hours now

at least haha

don't worry, it's pretty subtle if youve never seen it before!

thats a good goal!

[bleak abyss]

Is this for a pset or are you just kinda doing random problems at your own pace?

Either way it's prob a good plan, just curious in general

Excellent

[patent otter]

i luv u euclid😚

[civic linden]

can anyone explain to me the general idea of orbits/stabilisers and thus burnsides counting theorem?

[chilly ocean]

Do you want hints or the full proof?

[cloud walrus]

creamy shits:

[chilly ocean]

Right

So what are your thoughts

[cloud walrus]

creamy shits:

creamy shits:

creamy shits:

creamy shits:

[hollow comet]

Okay, that is a lot to read

Can you make a summary of where you're at?

Stay strong

Hahaha

Okay, what are you trying to prove?

Okay

I think you are almost there from what I read

Waaaaaaaait

[cloud walrus]

creamy shits:

[hollow comet]

I don't think that a proof by contradiction is the way to go

So for starters the group has even order, so it has at least two elements, right?

So at least you have one element which is not the identity

Sure

Take that element

What does Lagrange theorem tell you about its order?

Sure, you are right

Wait

Wait

Let me start again xD

What does it mean for an element to have order 2?

Yes, but maybe you have to put it another way

[cloud walrus]

creamy shits:

[hollow comet]

This is the proper definition

But what are the important things in groups

Okay

Let us think about inverses

What would be the inverse of an element of order 2?

Nope

It is true

But I mean something more concrete

[cloud walrus]

creamy shits:

[hollow comet]

Well...

I don't mean exactly that

You can literally say which one is the inverse element

Nein

That's it

An element of order 2 is always its own inverse

(how does this work xd)

,$ a a = e

[cloud walrus]

edelopo:

[hollow comet]

Do you see why?

So you agree that an element of order 2 is its own inverse, right?

Is the converse also true?

Exactly

So that is what we are going to look for

To find an element of order 2, we are going to find an element which is its own inverse

Does that sound good?

[lyric falcon]

ehmmm it's not equivalent

[hollow comet]

It is

[lyric falcon]

$e$

[cloud walrus]

Gonzo17:

[hollow comet]

Okay, you are right

[lyric falcon]

is it's own inverse

[hollow comet]

Apart from e

[lyric falcon]

ok then

[hollow comet]

The proof I'm thinking is based on that

No, he is actually right, it is important to point that out

We want a non-identity element which is its own inverse

Cool

So back to the order of the group

We have an even number of elements

And we don't want the identity, so we forget it

We have 2n-1 elements

Yup

Every element has an inverse, right?

Now we go with the contradiction

Let us assume that there are no elements of order 2

Sure

You got it?

[cloud walrus]

creamy shits:

[hollow comet]

?

No, it is not that

Any element has an inverse, and if there are no elements of order 2, this means that there are no elements that are their own inverse

Right?

So we may make pairs of inverses ${g, g^{-1}}$

[cloud walrus]

edelopo:

[hollow comet]

But making pairs always gives an even number of elements

And we have an odd number of elements

So this is not possible

I would suggest trying to word it more carefully, but yes

My personal suggestion is to never say "if all group axioms hold"

Nope

Again

You cannot talk about "the order without the identity"

Order is a reserved word for the number of elements of the group

If you remove the identity you lose the group

The idea is there

But it is not true that you are always left with only one element

Maybe there are more elements of order 2

Maybe all elements are of order 2

It is just the fact that IF THERE WERE NONE, then you would end up with an even number of elements

And you know that there is an odd number of non-identity elements

At this point you are proving more things, which is cool

But it is not necessary

You just say, assume that no element had order 2

Then you could make pairs

But this is not possible

So some element must have order 2

Done

Hahaha

Good luck with that 😃

@chilly ocean ?

[cloud walrus]

creamy shits:

[delicate chasm]

Congrats

[chilly ocean]

[oblique river]

just some advice for when you use TeX. every symbol should be in math mode

so you should say a finite group $G$" and not a finite group G"

[cloud walrus]

Buncho Bananas:

[bleak abyss]

So, given that you'll be checking a lot of "H is a subgroup of G" type statements, it's probably useful to find which criteria you have to check and don't have to check

So here's an exercise

Let G be a group and let H be a non-empty subset which is closed under multiplication and contains the inverse of each element. Then H is a subgroup of G

(In fact, if H is a non-empty subset such that a,b\in H \implies ab^{-1} \in H, you're good, but that's not usually easier to check)

Alright then, go for it. You already saw the most important part of my exercise which is that associativity is free

[errant drum]

We can relate ❤

[bleak abyss]

Rip. Well, whatever your current abilities are or what your assessment of your abilities is, what matters is you learn

[cloud walrus]

creamy shits:

creamy shits:

creamy shits:

creamy shits:

creamy shits:

creamy shits:

creamy shits:

[gloomy furnace]

[chilly ocean]

That's a lot of text

What do you want to prove?

[cloud walrus]

creamy shits:

[ocean marsh]

what in the world is going on in here

[cloud walrus]

creamy shits:

[ocean marsh]

Isn't G an arbitrary abelian group

?

Then why are there coefficients coming from Q?

An element of H is not a multiple of n

n is an integer

we don't know what the elements in our group look like

If a is in H we just know that a+a+...+a (n times) = 0

Well it doesn't really matter, but additive notation is common for abelian groups

you could write it a^n=e if you want

The easiest way to do this is to write down a homomorphism f where H=ker(f)

kernels are automatically subgroups

so you would be done

Do you see what the homomorphism is?

Oh okay

well you can do this problem the long way if you want and just check the group axioms

Yeah. The map will just be from G to G

If you have an element g...what's the obvious thing to do to it (given that you want any element with order dividing n to go to 0)

Not an isomorphism

that would have trivial kernel

we want the kernel to be H

I think you'

re

on the right track

The image does not need to be H

just the kernel

The image does not matter at all

We're just using the fact that the kernel of a homomorphism is a subgroup

Which is easy to prove...and any abstract algebra would go over it or put it on the homework within the first couple of weeks

No

It doesn't have a name

Like it's too obvious to have a name

It's just a basic fact from group theory

It could be way more than the identity

It could be the whole group

The proof is this: you have a map f:G -> K

If a and b are in the kernel, then f(ab)=f(a)f(b)=ee=e, so it's closed under multiplication

If a is in the kernel, the f(a^-1)=f(a^-1)f(a)=f(a^-1 a)=f(e)=e, so it's closed under inverses

that's the whole proof

For this particular problem, yes

No you want a function where the elements that are sent to the identity are exactly the elements of H

Yes

[sick acorn]

kernels are also normal subgrps of the domain grp iirc, right?

sry ignore me

[ocean marsh]

Yes kernels are normal

but this problem is about abelian groups, so there's no need to make a normal/non-normal distinction since every subgroup is trivially normal

You don't need to worry about the normality stuff

Here, I'll write down the map

[sick acorn]

the kernel being a subgroup part?

[ocean marsh]

f: G -> G is given by f(a)=n*a
(n*a is just shorthand for a+a+...+a (n times))

Do you see why this works?

So every element in H has order dividing n

which means if you take a in H and add it to itself n times, you get 0

so f(a)=0 if a is in H

f just takes everything to the nth power (or in additive language, "multiplies" the elements by n)

So the image are the "multiples of n"...but this is not the kernel

No! a subset of the domain

the image is a subset of the codomain

the kernel is a subset of the domain

In our case they might overlap or something

because our domain and codomain are the same

but again, we don't care about the image at all

We just care that we have a homomorphism whose kernel is H

we have to check that it's a homomorphism, but that's basically just f(a+b)=n*(a+b)=n*a+n*b=f(a)+f(b)

where the second equality uses that the group is abelian

(Have you done linear algebra? You might just not be ready for group theory. It will just get way harder from here, and it helps to have a certain degree of mathematical maturity)

An example when n=2: 2*(a+b)=a + b + a + b=a+a+b+b=2*a + 2*b

That's where the swapping happens

remember the multiplication is just notation

so the fact that it distributes over the addition needs to be proved

and the proof will use that the group is abelian

[sick acorn]

you can use an inductive proof, right? for integers, at least

[ocean marsh]

Yeah I think induction is the most rigorous way to prove it

But it's a simple enough fact that most people just mention that the operation commutes and move on

yes

Yes that is the kernel...and you want it to be equal to H

The kernel is a subset of G

H is a subset of G

You want them to be the same subset

Yes

I told you the function

It takes in an element g and spits out g+g+...+g (n times)

or if you like, it spits out n*g

[ocean marsh]

no. ker(f)=H

[sonic current]

@chilly ocean it is not, just be patient and plow through it

[ocean marsh]

The kernel is not a homomorphism...f is a homomorphism

the kernel is a subset of the domain associated with a homomorphism

yes

[sonic current]

what is the question, again?

[ocean marsh]

by associated with I mean for any homomorphism it has a kernel

so it's a subset you get by looking at a homomorphism

[sonic current]

what

[ocean marsh]

Yes

[sonic current]

why morfisms?

[ocean marsh]

It's just the set of elements sent to the identity in the codomain

[sonic current]

sorry, I don't mean to interrupt

[errant drum]

@chilly ocean The identity in G will always map to the identity in G'

[sonic current]

@yes do you have the definition of morphism, etc. in front of you? it should help a lot

[ocean marsh]

the kernel of a homomorphism is a subset of the DOMAIN

[sonic current]

you really must look at the definitions, write them down on your own etc.

i feel you. it will come with time

haha. happens

i have a book I really enjoy

[oblique river]

what is the current problem?

[sonic current]

prove that in an abelian group G the set of elements whose order divides a fixd integer n is a subgroup

apparently

I have no idea how it went from here to morfisms

[errant drum]

Yeah

[sonic current]

unless I'm really missing something

[oblique river]

yeah this is a problem where you just kinda do it, nothing fancy required

[errant drum]

That's just using the properties of a subgroup

[oblique river]

which part are you stuck on

[errant drum]

Show it's closed

[oblique river]

let's start at the beginning. Let G be an abelian group, let n be some positive integer, and let H be the subgroup of elements of G whose order divides n

[errant drum]

Associativity is free. Inverses have the same order.

Identity is free as well

[oblique river]

what is the order of the identity element

does 1 divide n?

[errant drum]

Okay can you show it's closed?

[oblique river]

so therefore the identity is in H. Done.

[errant drum]

1 divides any fixed integer n

[sonic current]

oh I see! you were trying to prove it's a normal subgroup by showing it is a kernel

[oblique river]

don't use kernels for this problem

[sonic current]

yeah, no use

it's overkill and you're clearly not comfortable

[oblique river]

like, literally, just show that H satisfies the subgroup axioms

[errant drum]

It's abelian

There's nothing to prove for normality

[sonic current]

but guys, let's try and not message the channel too much or else it will be more confusing for @chilly ocean

[oblique river]

hold on

what do you mean "multiples of n"

what? if n = 3

1 divides 3 but 1 is not a multiple of 3

[sonic current]

exactly

write it all down, @chilly ocean, and you'll start to see where you are falling off

[oblique river]

and in any case, there is a difference between saying "an element is a multiple of n" and "the order of an element is a multiple of n"

the second statement makes sense but the first one is nonsense if youre working in a random group

[sonic current]

yes you can

[oblique river]

identity is the easiest one

what is the order of the identity element e

[sonic current]

you've done one third!

[oblique river]

does 1 divide n?

okay!

so therefore e is in the subgroup

because the order of e divides n

WAIT

do you know what "divides" means in this context?

"a divides b" means "b is a multiple of a"

not "a divided by b is an integer"

I just told you what it meant lol

"a divides b" means that "b is a multiple of a"

so 2 divides 4 but 2 DOES NOT divide 5

yes, that is the point of the definition

[sonic current]

anyway, let's not lose track!

[oblique river]

what do you mean? we're not losing track, we're making progress lol

@chilly ocean didnt even know what "divides" meant until now lol

not only does it have to be smaller than n, it has to be a factor of n

so let's go back to the identity

what is the order of the identity?

is 1 a factor of n?

okay, so e is in the subgroup

yes

now let's do inverses

[sonic current]

(inverses is so cool in this particular case)

[oblique river]

if the order of x is k, what is the order of x^(-1)?

[you might have to think about this one]

before you try to answer, first you should ask yourself: what would a resaonable answer be

hold up

that doent make any sense

what does it mean to raise something to the 1/k power?

see, that's not resaonable. so let's try agian

the order of an element is a positive integer (or infinity)

yes!

okay, now can you prove that?

WRONG

you can prove it :)

[sonic current]

anyway just to add something I think is important for the sake of organisation,
I'd structure this particular case in this way:

we want to prove for a given x in H, then x^-1 is in H

take some given x. Then x^k = e

so you know the 'form' of any x in the subset.

this problem solves itself as soon as you try to inverse this x

[oblique river]

let's start with our x of order k

what does that mean? it means x^k = e

(and k is the smallest postiive integer that makes that equation hold)

so now we have x^k = e. we want to get some statement about x^(-1)

(namely that its order is also k)

so can we manipulate x^k = e in any way to get a statement about x^(-1)?

how could we try to introduce x^(-1) into that equation?

[sonic current]

@chilly ocean just overdosed on algebra

[oblique river]

they've been having internet problems recently i think

lets think about x^(-1)

[sonic current]

@chilly ocean, you could just assume x^-1 belongs in the subset and prove, by manipulation, it has the same order

[errant drum]

Hint how would you get from x = e to x^-1 = e

[oblique river]

the only think you know about x^(-1) is that when you multiplhy it by x you get e

we have an equation with x^k = e

[errant drum]

Yeah

[sonic current]

that is one way

no

[errant drum]

No

[oblique river]

wahhhhhhhht

we already established that 1/k doesnt make sense as an order

[errant drum]

The order is always an integer

[sonic current]

it's frustrating, i can agree with that

[oblique river]

noooooooo

the order is a postive integer

[sonic current]

orders are positive integers

[errant drum]

So if I asked you x^2 = e. Prove that x^-2 = e

How would you do that?

That's the idea

[sonic current]

@chilly ocean, there are many ways to do this but one very elegant way that highlights the importance of the exponentiation properties is to just take x^-1 and play with it a little

[errant drum]

You have the idea in your head but you haven't written it formally

[oblique river]

(i have to leave again... hopefully you can finish this problem soon! gl!)

[sonic current]

say you assume x^-1 exists, and there's an integer you suspect could be its order

[oblique river]

(what? x^(-1) always exists. there's nothing to assume.)

[errant drum]

^

[sonic current]

yes, but not in the subset.

[cloud walrus]

creamy shits:

[errant drum]

Why h^k•k?

You could but just stick with h^k = e

[cloud walrus]

creamy shits:

[errant drum]

Now you need to sneak in h^-1

How could you do that?

You know h^-1 exists

It's a group

[sonic current]

@chilly ocean because it's in the group

problem is, @chilly ocean, you know x^-1 exists but you must prove it is in the subset

[errant drum]

h^-1 exists but we don't know if it's in our subgroup yet

[sonic current]

one way to prove this is to take x^-1 and prove it belongs in your specific subset

[errant drum]

Okay maybe we'll try with an example

[sonic current]

I think the most elegant solution i've seen to this problem is take x^-1 and raise it to n

[errant drum]

If x = e show that x^-1 = e

Try doing this

[sonic current]

@chilly ocean take x^-1 and raise it to n

you get x^-1^n, right?

is it a fair statement?

now, you know, or should know, a^b^c = a^bc = a^cb = a^c^b
in other words, you can do a lot of gymnastics with exponents

so if you take x^-1 (which we know exists) and raise it to n (just our hypothesis for its order),

x^-1^n = x^-n = x^n^-1 right?

but we know x^n is e

and 1^(-1) = e

so we proved the order of x^-1 is n

which means x^-1 belongs in the subset, for any given x

what part wasn't clear?

well, you raised an element to n and the result was e

no

to be fair, you don't know that is the order -- you know the order of x^-1 is n or a multiple of it

but that is the requirement to belong to your subset

[ocean marsh]

not a multiple, a divisor

[sonic current]

or that

(sorry)

point is:

(x^-1)^n = e

[ocean marsh]

(sorry I'm just sitting here watching this saga unfold)

[sonic current]

hah it's cool

@chilly ocean stay with us

no

but that was not what we did

look closely, I'll write it as if I were writing on my own exercise

We want to prove that, for any given x in H, there exists x^-1 in H:
(also, we know x^-1 exists, because G is a group and x is in G <-- important!)

Take x^-1, the inverse of x, which we know to exist. Let's raise it to the arbitrary power of n.

So, (x^-1)^n.

But, due to the properties of the exponentiation, we know (x^-1)^n = (x^n)^(-1) (this part is were we solve the problem)

And so, since x is in H, and H is the subset of elements whose order divides a fixed integer n, then x^n = e (this is the hypothesis!!)

so, (x^n)^(-1) = e^(-1) = e (any power of the identity is the identity)

and so, x^-1 is also an element, just like any x in H, such that if raised to n, it returns the identity

are you convinced?

no

to be fair, I considered x^-1, and then arbitrarily raised it's power

but that's a legal move, lol

this strategy is very useful when proving subgroups whose defining property relies on a special exponent

i'm sorry if I can't be of any further help, but this is the process I find more elegant

errr

it's basically your way of saying "I think this element, if raised to the n power, is the identity"

and it's fair game!

like, I can obviously see your point that it's somewhat stretching the rules

but you didn't break any rule for the subgroup, so to say

no

this process is like

I have a guess for a power n, and then I used the very strong hypothesis which is that if x is in H, then x^n = e

that is really what you should be considering as the boldest claim

my tip is: every time you prove subgroups, closure and inverse start by assuming there is an x in the subgroup. And you usually must prove something using the property x has to belong to the subgroup

no it isn't, it's fairly different

I know

you're confused and its normal

just be strong and plow through it

eventually it will start making more and more sense

we're here to help

[ocean marsh]

What do you mean when you say the proofs are cop outs?

[sonic current]

@chilly ocean i've been failing algebra for the past three years

this year it is finally clicking

yes

and you proved that it has the same property x has critical to belonging in the subgroup

[ocean marsh]

For the record, protsac didn't define the inverse to be in the subgroup. He wanted to prove that for every x in H, x^-1 is also in H, which is the same as proving (x^-1)^n=e

So he took an arbitrary x in H and proved that (x^-1)^n=e. This is a standard way to prove a "for all" statement

No x will not necessarily equal x^-1

[sonic current]

I really don't see where you take that conclusion, @chilly ocean

lol

[ocean marsh]

No, this is a specific x

[sonic current]

that is wrong

[ocean marsh]

Like there are two fixed objects. x and x^-1

[sonic current]

in the same way 2^2 and 4^2 doesn't imply 2 = 4

[ocean marsh]

you know something about x, you know something about the relationship between the two, and you want to prove something about x^-1

[sonic current]

I really did not in any way shape or form assume x = x^-1 and you can tell that on your own by just stomping your head against the proof

it's more in the vein of

I know x is in the subset, and I know x^-1 must exist. I want to prove x^-1 also belongs in the subgroup

also, it might help you understand if you take for a fact that the order of x and the order of it's inverse are the same, always

[solar wyvern]

you don't have to take it for a fact, it's simple to prove

[sonic current]

well, at this point it's best if you just take this proof, copy it in your notebook and return to it later

and just advance

[solar wyvern]

$({x^{-1})}^{o(x)} = x^{-o(x)} = {(x^{o(x)})}^{-1}$ then show no lower power inverts $x^{-1}$

[sonic current]

were here to help anyway

[cloud walrus]

flimflam:

[solar wyvern]

@chilly ocean ic ur making progress :>

yeah

um

something which divides o(x)

I like to avoid actually using ordering 😄

[delicate chasm]

pops in How's it going?

[solar wyvern]

o: it's pur

yes has recovered

😃

[delicate chasm]

Is that approx 20 questions or the additive inverse of 20 questions?

[sonic current]

lol

[solar wyvern]

avoiding using ordering meaning I'm trying to avoid using something like "the order of x is the smallest integer such that x^(on) = e"

[delicate chasm]

You sound like you're having a blast :D

[solar wyvern]

just by contradiction lol

i think 🤔

that's ur higher power 🙏

[sonic current]

you're using the same variable..

[solar wyvern]

proof isn't hard: assume o(x^-1)|o(x) and differs by nonunit, then plugging back in you'd get a smaller order for o(x) which is a contradiction.

since you want to show the order of the inverse is exactly the order of the element

[sonic current]

it's the order of x^-1

[solar wyvern]

order of x^-1

this is assuming you know o(x) already

which means that for $x^n =e, o(x)|n$

[cloud walrus]

flimflam:

[solar wyvern]

by def

[sonic current]

that is something you want to keep in mind for cyclic subgroups which you'll see later

x raised to powers of the order will yield e, but the converse is not true at all

[solar wyvern]

pretty sure you need to show this--it's clear enough that the order of the element kills off the inverse too, but you want to know it's the legit inverse

@delicate chasm am i doing something unnecessary here

[sonic current]

well

you can do that in the group, can't you

[solar wyvern]

um

repost problem statement exactly?

[sonic current]

any group is closed under operation

so you can raise any element to whichever power you'd like

n is an integer

[solar wyvern]

$n \nin G$ 🤔 ng:

[sonic current]

it's not in the group as an element

because it's a group

[cloud walrus]

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

[sonic current]

of course it will

you really must brush up on group definitions and so on

[delicate chasm]

It's a Z-module

[sonic current]

ok, see it this way

x is in group, yeah?

so x * x is obviously in group

[cloud walrus]

creamy shits:

[sonic current]

yeah

see it this way:

[delicate chasm]

$2x = x+x \in G$

[cloud walrus]

Puerøsola:

[sonic current]

exactly

[solar wyvern]

@delicate chasm thats what i told him

[delicate chasm]

$x^2 = x\cdot x \in G$

[sonic current]

exponentiation is just addition glorified