groups-rings-fields

[upper pivot]

but a and b are different elements (or not necessarily the same)

so a= 13n and b= 13m for some n,m

[half nebula]

Oooooooh okay. Would everything else be the same?

Like f(13m)=17m?

[upper pivot]

yeah

[half nebula]

Okay okay. So it would be 17n+17m on both sides then?

[upper pivot]

yeah and hence its equal

[half nebula]

like phi(13n+13m)=17n+17m

Okay cool!

So that's the end of that? They're isomorphic?

As for 3, I already know that one is false as you can look at groups like Zx8={1,3,5,7} which is non syclic because none of them are generators but all the subgroups are cyclic.

Right?

[upper pivot]

well if you showed its both a bijection and a homo you are done for 2.

yeah

thats right for 3

[half nebula]

Okay cool!

So as for 4, first off like A_n what is that?

[upper pivot]

oh well this needs a bit of explanation

so do you know what S_n is

[half nebula]

Yeah, it's the set of all permutations of whatever n is. Or well that's the best way I can put it into words. I know what it looks like though.

[upper pivot]

yeah so do you know the sign of a permutation

[half nebula]

Like S_3 is e, (12),(13),(23),(123),(321)

The sign?

[upper pivot]

alright so umm theres a few ways to define this but ill use one thats not too involved

so obviously each element of S_n can be thought of as a permutation matrix right

[half nebula]

Yes

[upper pivot]

the matrices than can have determinant 1 or -1 right

this is their "sign"

[half nebula]

Wait but you can only get the determinant of an nxn matrix right?

[upper pivot]

yeah, the elements of S_n are permutation nxn matrices

[half nebula]

Wait never mind.

I was thinking of two line notation.

[upper pivot]

oh so uh do you see what i mean by permutation matrices?

[half nebula]

I don't really know how S_n looks in a matrix no.

Like S_3, what would that look like?

[upper pivot]

hmm might be useful for you to read up a bit on permutation matrices

on like wikipedia rather than me explain it

[hot lake]

you weren't introduced to Sn and An in your lessons somewhere ?

[half nebula]

Let me check actually.

Nope, none that I can tell.

Well S_n yes but not A_n

And even then it's just defined as basically what I put.

Okay I found a good picture that helped me understand permutation matrixes.

Or wait

Oh okay I get it yup.

[hot lake]

so you were one of those who learned determinants before learning about signs of permutations ?

[half nebula]

Yeah I didn't even know what a permutation was until this semester.

[upper pivot]

i mean lol i think explaining using determinants is much easier than sighns

(altho the latter is more useful)

for this problem we only really need determinants so

yeah so basically A_n is the subgroup of permutation whose det= 1

(another way to say it is that its the kernal of the homo det:S_n \to {1,-1})

so do you see how to prove your problem soul?

[half nebula]

Well sort of. Like doesn't it depend on what n is?

[upper pivot]

well the elements do but your problems proof is independent of n

[half nebula]

Okay. so it's true then right?

[upper pivot]

yeah do you see why

[half nebula]

Because like when you square a permutation, either it's going to be the identity which will definitely be in A_n or it will have to have det=1

Or well the identity also has det=1

Right?

[upper pivot]

yeah

because det is +- 1 right

so squaring it will def give 1

[half nebula]

Oh yeah I forgot about that. When you have matrix a multiplied by matrix b, then the det(AB)=det(A)*det(B)

[upper pivot]

yep

(and do you see why det = +-1)

[half nebula]

And since the determinent for the two matrixes are the same, it has to either be 1 times 1 or -1 times -1

which always comes out to be 1.

[upper pivot]

mhm

[half nebula]

Okay sweet!

I do have 5 more problems, that was just the first one. I'll see what I can do on my own and then come back in here. This assignment technically isn't due until tomorrow at midnight but I wanna try and get all of it done today so I can go to a friends house tomorrow before he moves away for college.

If that's all okay I mean.

[upper pivot]

[half nebula]

Coolio ^^

[half nebula]

Okay so for this next one I think I got it, I just want to be sure. What I did was set phi(x):ln(x) so phi(axb)=phi(a^{ln(b)})=ln(a^{ln(b)})=ln(b)*ln(a)=phi(a).phi(b)

[elder valley]

yeah you got it

[half nebula]

For the onto part, I just said that since e is an element in G then you can have phi(e^a)=ln(e^a)=a
And finally for one to one I said that phi(a)=phi(b) implies that ln(a)=ln(b), you can have both be e raised to ln(a) and ln(b) which would mean a=b for a,b in G

And cool ^^

Wait is the onto and one to one correct as well?

[elder valley]

yup

[half nebula]

So for this next one I have a group, G. It has an order of 125. a in G will not have an order of 25. I need to prove that there is an element in G with an order of 125, a generator. There are only 4 possibilities right from the start. An element with order 1, 5, 25, and 125. The order 1 and 125 are what I need, I already know a cannot have an order of 25. As for the a with order 5 though, due to there being no element of order 25, is that enough to say there also cannot be an element of order 5?

[oblique river]

If you have an element of order 125, then you will also have an element of order 25

[half nebula]

What I'm saying is the problem says that there is no element of order 25.

[oblique river]

Then there also can't be an element of order 125

[half nebula]

But then there is no generator.

[oblique river]

correct

[half nebula]

But it says to prove that there is a generator XD

[oblique river]

Maybe you can screenshot the problem?

[half nebula]

I can but I have to restart my pc.

[oblique river]

If g has order 125, then g^5 will have order 25

because (g^5)^(25) = g^(5*25) = g^(125) = e

so if you have a generator, then you necessarily have an element of order 25. So therefore if you have no elements of order 25, then you can't have any generators, either

[half nebula]

Sorry back, family stuff.

PC is good now, give me a second to get it up.

[oblique river]

👍

[oblique river]

oh

so first, do you understand my proof above about why if you have an element of order 125, then you must also have an element of order 25?

[half nebula]

It does make sense yes.

[oblique river]

in general, if you have an element of order n in a group

then you also have an element of order k for every k dividing n

[half nebula]

So if there's an order of 125 then there has to be an element of order 5 and 25

[oblique river]

correct

[half nebula]

Mainly since 125 is only divisible by 5 and 25

and 1 and itself of course

[oblique river]

👍

[half nebula]

So does that mean it isn't cyclic then?

[oblique river]

no, the problem is correct

but you're interpreting it incorrectly

[half nebula]

Oh.

[oblique river]

it says "there exists an element a in G with a^25 \neq e"

which is not the same as "for every element a in G, a^25 \neq e"

(and that's what you asked about earlier)

[half nebula]

Oooooh okay.

[oblique river]

it doesn't say that there are no elements of order 25, it says that there is some element whose 25th power is not e

so there's at least one element whose order is not (a factor of) 25

(if a^5 = e, then a^25 = e, too)

[half nebula]

Oh wait. That must mean that that element has an order of 125. Because there's only 4 possible orders it can have.

[oblique river]

:)

[half nebula]

Wait can't a=e?

[oblique river]

no, because then a^25 = e^25 = e

saying that a^25 \neq e is equivalent to saying that the order of a is not a factor of 25

(you should try to prove that)

[half nebula]

Ooooh okay. I gotcha.

Also I think I got this next one, just wanna be sure.

I have a permutation (17593)(2467)(385) in S_9. I have to find the order of it. After putting in in disjointed cycle form, I got (172465)(389). To get the order, it's lcm(6,3)=6. Is that all right?

[oblique river]

yep

assuming that's the correct disjoint cycle form

which is just to say that I didn't check myself

[half nebula]

I just double checked. It's right.

[oblique river]

nice

[half nebula]

Just being sure as to this, to check if a permutation in S_9 is in A_9, I just have to show that the determinant of the matrix of that permutation is equal to 1 right?

[oblique river]

yep

another way to do it is to write it as a product of transpositions

and count the number

even = A_n, odd = not A_n

[half nebula]

What are transpositions?

[oblique river]

2-cycles

(ab)

[olive mirage]

or there is a formula to do it from the cycle decomposition if you know what that is.

[half nebula]

Ah

I think I do. Not sure. I already got the determinent of these two matrixes I'm just not 100% sure if I'm right or not.

It's asking if the permutations (172465)(389) and (1356)(27489), both in S_9, if either is in A_9

And both ended up having a det=-1

So like I wanna say that my math is right and that neither are in A_9 but I got one of those gut feelings like that was too simple or something. Like I messed up somewhere.

[oblique river]

that looks correct to me

[half nebula]

If nothing else I just needed a second word. Thank you.

[oblique river]

👍

[half nebula]

Okay I think I got this next one but I wanna be sure I understand what it's saying. It says "an element g in G is order 100. List all possible powers of g that order 5." Like I'm just confused as to what that's actually asking.

[oblique river]

it means like

is g^2 order 5? is g^3 order 5? is g^4 order 5?

[half nebula]

I thought that was it. If g is of order 100 then doesn't that mean the only answer is g^20 since 20*5 is 100? Or am I thinking about this wrong?

[oblique river]

that's one answer

but there are more

let's first ask a slightly different qusetion

[half nebula]

Oh wait.

Hmm?

[oblique river]

suppose that a has order 5. what is the order of a^2?

[half nebula]

10?

Or wait

[oblique river]

the order can never go up when you take powers

it can only go down

[half nebula]

2.5 I guess but I don't think you can have an order of 2.5

[oblique river]

no, certainly not. orders are necessarily integers

(positive integers, in fact)

[half nebula]

Okay. So it has to be lower than 5 but greater than 0.

so 1 2 3 or 4

[oblique river]

it can also be 5

[half nebula]

Oh

[oblique river]

there's no reason it has to decrease

[half nebula]

Ah okay.

[oblique river]

why don't you start taking powers of g^2

err, a^2

and keep in mind that hte order of a is 5

so a^k = e if and only if k is divisible by 5

[half nebula]

Well that would mean k has to be 1 or 5

[oblique river]

no

k is divisible by 5

not 5 is divisible by k

[half nebula]

Oh sorry ><

[oblique river]

a^1 is certainly not equal to e

but (assuming a has order 5) then a^5 = a^10 = a^15 = a^20 = ... = e

[half nebula]

So that could be 5x with x in Z >0

[oblique river]

yep

those are the k with a^k = e

[half nebula]

Okay that makes sense.

[oblique river]

so now you can figure out hte order of a^2

[half nebula]

Would it just be 2 then?

Because like a has an order of 5 and a^5 has an order of 5 right?

[oblique river]

what is (a^2)^2 in terms of a

[half nebula]

a^4?

[oblique river]

is a^4 = e?

[half nebula]

No

[oblique river]

then (a^2)^2 isn't equal to e

so the order of a^2 isn't 2

[half nebula]

Oh okay I gotcha.

Oh wait.

a^2 is order 5 because (a^2)^5=a^10

Right?

[oblique river]

yep

so this question is going to sound kind of meaningless but I'm going to ask it anyway

what is special about the pair of numbers "2 and 5"?

like, what property of that pair is what tells us that "if a is order 5 then a^2 also has order 5"

[half nebula]

I'll be honest I have no idea. That 2*5=10?

[oblique river]

let me try to phrase it another way

the order of a^2 is the smallest k such that (a^2)^k = e

i.e. the smallest k for which a^(2k) = e

but we know that a^(2k) = e exactly when 2k is a multiple of 5

so we are looking for the smallest k such that 2k is a multiple of 5

and it turns out that the smallest multiple is k = 5

(because none of 2, 4, 6, or 8 is divisible by 5)

let's try another example: what if a has order 6 -- what is the order of a^4?

(try to work this out now)

[half nebula]

3 because a^12=(a^6)^2

[oblique river]

great

it's certainly true that (a^4)^6 = e, but there happens to be a smaller power in this case

so what's special about 2 and 5 that's not shared by 4 and 6?

[half nebula]

Oh is it the lcm?

Like the lcm(2,5) is 2 times 5 but the lcm (4,6) is 12.

[oblique river]

yeah

basically, gcd(2,5) = 1 but gcd(4,6) > 1

(it's always true that lcm(a,b) = a*b/gcd(a,b) so those two observations are equivalent)

in fact we can use this observation to find a formula

if a has order n, then the order of a^k is equal to [blah]

[half nebula]

So for the one I listed, the possible powers of g would be 20,40,60, and 80 right?

Like g^40^5=g^200

[oblique river]

yes

[half nebula]

Okay but why does it stop at 80 or does it not stop at 80?

[oblique river]

well, in some sense it doesn't

for an example, if a has order 6, then a^8 = a^(6 + 2) = a^6 * a^2 = a^2

so in your case you can talk about powers of g larger than 100

and they're just going to get reduced

so like, g^100 = e, g^101 = g, g^102 = g^2, etc

what does that tell you?

[half nebula]

Ah okay so it just loops. It's cyclic.

[oblique river]

yep

so can you give another power of g which has order 5?

[half nebula]

Well there is already 20,40,60, 80, and now 100.

I'm guessing you just keep going then?

120,140 etc etc.

[oblique river]

what is the order of g^(100)?

[half nebula]

Oh sorry that would be 1.

[oblique river]

👍

see if you can figure out the general formula I said earlier

"if a has order n then the order of a^k is [blah]"

where blah is a formula using n and k. based on the previous observation, it probably has soemthign to do with gcds and/or lcms

I only suggest this because once you think through the general case once, you won't have to do it ever again

[half nebula]

I'll see if I can later, mainly because I want to finish up this assignment. I'm like super tired rn. That happens when you do math for 6 hours straight though I guess <->

Only 5 more problems to go ^^

[oblique river]

good luck!

[half nebula]

Once again, sorry to keep coming in here, I just want to be sure I'm getting this right because it feels too easy.

[oblique river]

you're fine lol

[half nebula]

Hold on my screen snip program isn't working again

[oblique river]

but I need to go for a bit

[half nebula]

It's okay.

[oblique river]

what screensnip program do you use?

[half nebula]

The one built into windows snip and sketch

[oblique river]

ah okay

I use lightshot -- I think it's pretty handy

it basically jsut takes over your printscreen key

alright gl with your work

[half nebula]

Thank ya.

If anyone else can come in, this is the one I'm on right now. I feel like it would just be k= 35+100n where n is all positive integers.

So like 135, 235, 335, etc etc.

[sonic current]

@half nebula that's good but don't forget negatives as well!

it looks like you're working with orders

[half nebula]

Thank ya! @sonic current

[sonic current]

make sure to bang that lagrange theorem as well

[half nebula]

I'm finally on the last problem. Almost done <-> It's a toughy though. At least this first one is. I feel like it uses that homework fact for a but I'm not 100% sure. Also the spacing is off which is annoying me to the point where I'm going to give me professor a low review score at the end of the semester (I won't actually but I want to.)

I think I got the first one. $(a^kb)^2=(a^kb)(a^kb)=(ba^{-k})(a^kb)=b^2=1=e$

[cloud walrus]

Soulgiver831:

[stone fulcrum]

That's good! Can you justify why
aᵏb = ba⁻ᵏ?

Oh wait that's shown nvm

[chilly ocean]

lol

[stone fulcrum]

Then you're good

[chilly ocean]

cute pfp

[stone fulcrum]

My Suki is my favorite

[half nebula]

For the second one, b, I'm a tad bit confused. Would it be {e, a, a^2, a^3, a^4, a^5, ... , a^9, b, ab, a^2b, a^3b, ... , a^9b}?

For the elements that is, not the order of them.

[stone fulcrum]

Yes

[half nebula]

Great! Thank you ^^

[oblique river]

i'm looking back at the problem you posted earlier

the one with 35

does <> mean subgroup?

[half nebula]

I believe so.

[oblique river]

haha you should confirm what it means because it could change the answer :P

[half nebula]

I messaged him and he said what I had was correct. He won't give us hints or help us with our homework but he will tell us if we're right or wrong.

[oblique river]

oh

I'm actually confused because if it means subgroup, then the answer you wrote in here isn't correct

because the subgroup generated by 70 is the same as the subgroup generated by 35

err, maybe not 70, but 105 (which is 3*35)

[half nebula]

I think I may have misunderstood sorry. It's what numbers are equal to 35 in Z100 basically.

Which is addition.

[oblique river]

ohhhh

okay yes

[half nebula]

Yeah ^^

[oblique river]

although what is the purpose of the angle brackets then?

I would have assumed that [35]_100 is the class of 35 in Z100

and that < [35]_100 > is the subgroup generated by [35]_100

[half nebula]

Idk but I don't think we've gone over a question like that plus the notation is similar to a problem we've had before that dealt with cycles.

[oblique river]

hmm, okay

maybe the notation is just nonstandard then

[half nebula]

Maybe.

[oblique river]

I think that < > is pretty common for "subgroup generated by"

like <a> is the subgroup of G generated by a

[olive mirage]

some undergrad texts don't introduce that until painfully late.

[oblique river]

oh, okay

[olive mirage]

or some will introduce <g> but not <g,h>

since in some sense <g,h> is the image of a map from a free group given by a universal property

[oblique river]

https://cdn.discordapp.com/attachments/496784958430380033/720030404824006758/unknown.png

this is the question we're talking about now

[half nebula]

All I know is I only got c left on that last problem then I am good on homework for the next 2 whole days bby ^^ Plus worst case scenario, I lose a few points on the homework that only counts for 10% of my grade.

[oblique river]

and I interpreted that to mean "for which k is the subgroup generated by [k] equal to the subgroup generated by [35]" but I guess that's not how they're using the notation

[olive mirage]

I can't imagine a different interpretation

[oblique river]

apparently soulgiver's prof is using a different one haha

they said they messaged the prof and asked if "35 + 100*k" was correct and the prof said yes

[olive mirage]

Oh I believe you, I just can't look at this and guess

[half nebula]

Honestly this man is one of the weirdest professors I've ever had. We don't have classes so no one ever knows what he wants, he has 1 hour of office time per week, gives us super hard homework (like this one we just got this morning and is due by tomorrow night) but yet he also does things like send us a "happy pride month"email and offers us 10 points on our final just for doing the class survey.

Like I cannot read him at all, he is just crazy.

[oblique river]

weird

well good thing you're almost done with this hw

I gotta go deal with dinner stuff

[half nebula]

I gotcha ^^

[solemn rain]

weird question

all prime ideals in an euclidean domainare maximal

in this proof its really similar ot the cyclic groups subgroup

where it relies heavily on the algorithm

obv we needed to have an euclidean domain

why is it tho that in the cyclic groups subgroups are cyclic we can use the euclidean algorithm directly

why can you do euclidean algorithm in groups easy but not always in rings

[gentle pendant]

It's not "groups" you are doing the euclidean algorithm with @solemn rain , it is the integers. (Exponents of generators).

Eulidean algorithm is specifically defined for rings, it involves both addition and a product operation.

[limpid hawk]

yeah euclidean algorithm is usually used for integers right? which is a ring. idk if there's an analogy for groups

it has + and * in it

[solemn rain]

lmfao yeaaa

got it

ty gomez

[smoky cypress]

it works for euclidean domain, but that's kinda true by definition

the purpose of euclidean domain is to do euclidean algorithm on it

[knotty mason]

you can use the euclidean algorithm on the ring Z[i] though

[delicate bloom]

Z[i] is a euclidean domain

[left monolith]

What do you call the ring $\mathbb{Z}[\sqrt{d}]$ ?

[cloud walrus]

Wass:

[knotty mason]

Z[\sqrt d] is sometimes the ring of integers of Q(\sqrt(d))

[hot lake]

a quadratic ring ?

[knotty mason]

but if d = 1 mod 4 then you want Z[1/2(1 + sqrt(d))]

[left monolith]

I was reading about ring of integers but notation wasnt exactly the same and I was doubting

but if d = 1 mod 4 then you want Z[1/2(1 + sqrt(d))]
@knotty mason
I saw this in the definition but didn't get it. Why is this?

[knotty mason]

It's a bit of a technicality

but to generalize the concept of integer to algebraic integers, we consider roots of monic irreducible polynomials with integer coefficients

e.g. the roots of X^2 + 1 and X^2 + X + 1 are algebraic integers

and the rings of integers containing these are Z[sqrt(2)] and Z[(1/2)(-1 + sqrt(-3))]

in the first case the discriminant b^2-4ac is 2, so the ring is sqrt(2), in the second case the discriminant is -3 so you have got a sqrt(-3) but the half comes in as well in this case

you can prove this will always happen based on D being = 3 or 1 (mod 4)

[left monolith]

Got it :) thanks very much!

[knotty mason]

yw. and the chance to explain this stuff to someone helps me learn it better

[left monolith]

How can I prove 2 is irreducible in $\mathbb{Z}[\sqrt{-10}]$ ?

[cloud walrus]

Wass:

[left monolith]

It should be an easy task but I can't figure it out

[steady axle]

did you check norms

[left monolith]

It's what I'm trying

[steady axle]

why isn't it working

[left monolith]

Got it

I usually get the answer to my problems once I post them 😬

[golden pasture]

type out everything you tried as well and for some reason your brain will click :D

[left monolith]

It usually happens :D

[left monolith]

What are the units of ring D, denoting D as the ring of differentiable functions $f:R\rightarrow R$ ?

[cloud walrus]

Wass:

[left monolith]

It's blowing my mind

[golden pasture]

Start simple what kind of functions R->R are units

[left monolith]

All of those which f(x) =/= 0 for all x?

That's my nearest aproach

Cuz I need them to have a multiplicative inverse so it can't have a 0 value

[elder valley]

Is f(x)=x invertible, and if so what's the inverse?

[left monolith]

f^-1(x)=1/x

[noble saddle]

Is that a differentiable function R -> R ?

[left monolith]

Negative

[noble saddle]

Great, and why not?

[left monolith]

All of those which f(x) =/= 0 for all x?
@left monolith
So this is true?

[noble saddle]

yeah

[left monolith]

Great, and why not?
@noble saddle
Because not differentiable at 0

yeah
@noble saddle
I'm proud of myself right now

And would this ring D have any 0 divisors?

[noble saddle]

Try to build two smooth functions whose product is 0 everywhere

Shit, start with no smooth

and then try to smooth it out

[left monolith]

Smooth you mean continuous? (I'm not an english-based learner)

[noble saddle]

Smooth I meant infinitely differentiable

So like even stronger than just differentiable

[left monolith]

Okk

So that's impossible

[noble saddle]

But yeah, I bet you could easily make a non continuous example

[left monolith]

Cuz either of them has to be 0

[noble saddle]

Is it impossible? If fg =0, then f(x) or g(x) is 0 for all x

but like, they can switch

[left monolith]

Yes, but either has to be 0

At every single point x

[noble saddle]

Right, but the thing about functions is that if f and g are each nonzero at a single point, then that's an example of a zero divisor

Stupid example, because not continuous at all, but let f(1) = 1 and g(0) = 1, and both functions 0 elsewhere

[left monolith]

I get you

So it's enough that at one certain point x, there both non zero so thats a non-zero divisor example

[open mulch]

I've been thinking about a generalization of algebraic structures axiomatized by inequalities a ≤ b (for some partial order ≤) instead of equations (which by a generalized version of the HSP theorem apparently still has varieties, so that's good) - and I'm finding that nearly all standard properties of groups rely on the fact that equations, not inequalities, are used.

Just replacing a single axiom, with a + (-a) ≤ 0 instead of = 0, and leaving the rest as equations, is enough to ruin most elementary group theory facts as far as I can tell

but, I'm not sure how to prove that those facts can't be proved

I can make counterexamples of some of them, however, and it's clear that this enables the existence of idempotents other than 0, makes the left and right inverses not necessarily the same, and makes inverses not necessarily unique.

I don't really have any questions, I just wish there were other people interested in helping me explore my weird theories.

a variant that's even closer to standard groups, but where those things still can't be proven, adds another unary function * where 0 ≤ a + *a.

if -a = *a for all a, the result is a group. otherwise, it's a very slight generalization of a group, but with presumably much more variable properties.

what I'd like is for someone who thinks this is interesting to talk about it with me or ideally be a long-term math friend to explore stuff with.

[noble saddle]

People tend not to care about generalization for the sake of generalizing. Can you give non trivial examples of these? Can you use the theory to answer questions unrelated to your theory?

[open mulch]

I do enjoy generalizing for the sake of generalizing, and no, I can't. I can give examples, easily, but none of them are relevant to anything.

Well - actually, not quite, but it's a stretch.

[noble saddle]

Okay, well that's why other people are unlikely to give a shit about what you're doing

[open mulch]

These are an attempt to axiomatize an idea I had about "what if 1 -1 = z for nonzero z"

where z represents something along the lines of entropy

i.e. irreversibility of the operation of adding 1 - something remains even if you subtract.

obviously this would not actually be standard numbers, but an abstraction of them.

in order to do that though you have to go outside groups into something using a partial order rather than equality.

[upper pivot]

hmm i mean i guess just setwise you can define an equivalence relation on posets by saying a~b if a<=b or b<=a, but idk if any group operation will act nicely under this

[open mulch]

(so that you can say z ≤ 0 but not = 0 - though I prefer using → for the partial order so that it doesn't get confused for the standard real number order)

yeah, that's an interesting thought. but mainly I want to see what is still the case even without equivalence relations. it's highly intriguing to me that nearly every fact people take for granted about groups or rings relies on the use of equality rather than inequality.

another example: in a generalization of a ring but where these "almost-inverses" are used - a + (-a) → 0 for this partial order → (so named to avoid confusion with standard real number order) - then, it is impossible to prove that 0 = 0x for all x, or indeed that there is any necessarily relation between those things, at all.

0 ceases to be absorptive the moment you relax to a partial order.

lots of interesting things here. but because I'm not trying to prove anything in particular - just doing experimental, exploratory thinking - it's difficult to get anyone else interested.

if I had to come up with a specific goal to motivate it though, it would be to model the idea of irreversible operations, which produce some "residue" when you attempt to reverse them.

[dawn kiln]

possible that there might be some type of lattice like structure similar to what you're trying to describe

[errant sierra]

If I have a have group g with finite order n how do I find the order or g raised to some power

[woven delta]

what do you mean raised to some power?

do you mean like direct product?

@errant sierra

[solemn rain]

|g^k| = n/(n,k) where n is the order of g

[chilly ocean]

Write in the form ( x + p ) 2 + q x 2 + 11 x + 1
x2+11x+1
how do i do this?

[golden pasture]

huh?
probably belongs to like #prealg-and-algebra tbh

[knotty mason]

I'm struggling with this problem so much, can anyone help me out

I'm applying the orbit stabilizer theorem and the 2nd isomorphism theorem

I have done (a) and the first part of (b), but i can't get the 2nd part of (b) about r = [H : H G_x]

the 2nd iso gives me the | and || lines equal degree on the diagram above

so [G_x H : G_x] = [H : H_x] and [G_x H : H] = [G_x : H_x]

How can I get the formula for r? My idea was that the |O| * r = |X|

[solemn rain]

yes

[knotty mason]

and then I expressed |X| using orbit stabilizer as |X| = [G : G_x]

[solemn rain]

yes

[knotty mason]

but dividing by |O| I can't get [H : H G_x]

[solemn rain]

use what what you proved

in the b

first part

in b*

[knotty mason]

Wow thanks so much!! It suddenly worked out perfectly

[solemn rain]

cool af no?

[knotty mason]

😄 i don't know how you did that, I spent hours on it

yes its a brilliant result

[solemn rain]

np

gl

[knotty mason]

thank you!!

and I get as a corollary this great fact about an irreducible polynomial splitting into equivalent irreducibles over a galois extension

[solemn rain]

idk galois stuff

[open mulch]

I had an idea for a notion of order on complex numbers. Normally you can't put a sensible order on them since they're 2D, but that's order as a binary relation. A ternary relation, however, could I think work. Call this relation R. Then for three complex numbers a, b, c, say that Rabc iff the three numbers mapped as points on a plane occur in clockwise order around a center point.

This should be conserved by adding the same thing to all three, or by multiplication of all three by any other complex number, I think. (I might be wrong though!)

The main issue of course is that if they all fall on a line together then it's not clear what relation they have. The good thing is almost all ordered triples of points in a plane are non-collinear, so it's still better than a binary order.

[mild laurel]

This vaguely sounds like the study of linear fractional transformations and the definition of the cross ratio

[open mulch]

Not something I'm terribly familiar with tbh.

I'm aware of mobius transformations, though

cross ratio I've heard of but not read much about

[mild laurel]

mobius transformations and linear fractional transformations are the same

[open mulch]

yeah. I mean, I don't know if there's any point to trying to make an order on complex numbers, but if one could be come up with which is compatible with addition and multiplication that would be good, I guess.

ternary "orders" aren't exactly orders, of course.

and I'm not sure exactly what transitivity-like relations would occur here, I'd have to think about it.

[somber bramble]

does the split exact sequence stuff also work for long sequences? i.e. if I have a long exact sequence $\dots \rightarrow A \rightarrow B \rightarrow C \rightarrow \dots$ and a map $C \to B$ that makes the diagram commute, is $B \cong A \oplus C$?

[cloud walrus]

Sascha Baer:

[oblique river]

no

the map A to B need not be injective

consider 0 -> Z -> Z -> 0 -> 0 -> 0 centered on the A = Z, B = 0, C = 0

the maps are all the obvious ones

[somber bramble]

ah, yea, that is a pretty clear counterexample, thanks

ah I just found that in my case, the maps going off to the left and right are the zero maps, in which case I can just replace the stuff outside of A→B→C with zeroes without changing exactness and everything works

[oblique river]

oh, yeah

[woven delta]

Oh btw in the long exact sequence context the condition on the sections is a bit different

[oblique river]

I mean in this case he said the maps ont he end are 0s

so it's all fine

[woven delta]

Sure

[somber bramble]

yea but I only checked that after asking here and then realizing I wasn’t done yet :P

[woven delta]

I remember there is a discussion of splitting for long exact sequences in weibel chapter 1

But I forget what results were proven lmao

But in your case it doesn't matter

[woven delta]

What subgroup of Z/40Z?

@chilly ocean

Lol

Because it is not well defined

So no

[woven delta]

@chilly ocean so in the case you are thinking of there aren't any subrings of Z/40

Because 1 generates it additively

Can you post the question?

Are you sure that maybe it didn't say Z/40 instead of Z/10?

Okay

Oh so I'm thinking of subrings with unity

Did you define subrings to not have unity in it?

Lol

Okay go on

Lol

Okay I was being silly

This seems like Chinese remainder theorem stuff

You can have a unit on part of your ring

That's fine

Do you know the Chinese remainder theorem?

Okay so basically you just need Z/60 isomorphic to Z/12 \times Z/5

Which the Chinese remainder theorem gives you

Lol I shouldn't try to answer shit at 5 am

Each factor is inside the bigger ring

There is a copy of Z/12 and Z/5

Okay so we have an isomorphism between Z/60 and Z/12 \times Z/5

Given by n \mapsto (n mod 5, n mod 12)

Okay so there is a map from Z/12 to Z/60 by going backwards

Where you map n in Z/12 to the unique number k so that k is n mod 12 and 0 mod 5

Okay so you can actually figure out what that number is

Yeah

And anyway that is your ring map

👍

Wow I'm so fucked up rn

Cause it's 5:30 am

Lol I gotta figure out this computability construction

I have to juggle like 10 different conditions for it

[woven delta]

25(5n) = 125(n) = 5n

Mod 60

@chilly ocean

Lol I was being braindead earlier

So don't worry about it

[smoky cypress]

What happened here

[upper pivot]

lmao rip liquid

[woven delta]

Wtf did LHC delete all of his posts?

Or did I hallucinate this entire exchange because I was very tired

@chilly ocean wtf

[soft elm]

[delicate bloom]

what do you think people think of you when you delete all your posts like this, just curious @chilly ocean

[oblique river]

it's the most frustrating thing and for me it usually means I just stop helping them

[delicate bloom]

when I saw it originally I thought this is just a regular person having a normal conversation learning something, but now I just don't know what to think

I guess I just feel sad for them that they feel this self conscious over something so innocuous. I don't really want to call you out on it but it's like, gross

[oblique river]

call who out?

[bleak abyss]

Presumably LHC for the deletion

[delicate bloom]

oh yeah, my internet died when I sent it but guess it didn't go through

[solemn rain]

if M is the abelian group (Z-module) Z/24Z x Z/15Z x Z/50Z

will the ann of M be the set of integers that divide all these 3 numbers? ( 24, 15, 50) ?

my thinking was just to make them 0

would it be the set of all common multiples then?

[solemn rain]

okay and i cant prove that Hom_Z(Z/nZ , Z/mZ) ~ Z/(n,m)Z

[upper pivot]

for the first one it isnt that divide all

for example Z/2z x Z/6Z isnt annihalted by 1,2 or 3

[solemn rain]

okay then

yea

then is it the set of multiplies

multiples* ?

[upper pivot]

yeah

try proving it

[solemn rain]

yea i see why

[upper pivot]

(consider generators)

[solemn rain]

so here it would be

600

right?

[upper pivot]

something like that

[solemn rain]

the set of multlples of 600

atleast

[upper pivot]

,calc lcm(24,15,50)

[cloud walrus]

Result:

600

[solemn rain]

ayy

okay yea cool i get why

now i cant seem to do the second one

i tried a bad homomoprphism

f:Hom --> Z

[upper pivot]

hmm do you want a hint

[solemn rain]

f(phi(a)) = a XD

and tried first iso but no luck

problems are getting hard as fuck and doing even one of them ish ard

[upper pivot]

if you take a homo $\bZ/n\bZ \overset{f}{\longrightarrow} \bZ/m\bZ$ just consider $f([1])$

[cloud walrus]

JohnDoeSmith:

[upper pivot]

(assuming group homomorphisms ofc)

[solemn rain]

Hom is the set of module homomorphisms

but yea

okay so

[upper pivot]

oh i see you specified Z

[solemn rain]

yea

so

is this an isomoprhism

f(1+dm) = d

where d is gcd(n,m)?

[upper pivot]

hmm im not sure what you mean

[solemn rain]

the homomoprhism f between Z/nZ and Z/mZ woud look like this

f(1+nZ) = k+mZ right?

sendding an element mod n to mod m

ops sry meant m

[upper pivot]

right

just say like

[solemn rain]

yea [1[]

[upper pivot]

[1] and [k]

[solemn rain]

okay

[woven delta]

What condition do you have on n and m?

Because obviously this doesn't always work

[upper pivot]

yeah we are figuring out what k has to be

[solemn rain]

what is 'this'

yea

[woven delta]

Well if n divides m then you can always make a map from Z/m to Z/n

[upper pivot]

so anyhow do you see why the gcd must divide k?

[solemn rain]

so if f(1+nZ) = k+mZ

phi(f) = k+(n,m)Z

is this iso

[woven delta]

Are you assuming 1 doesn't have to map to 1?

[upper pivot]

uh we still need to impose conditions on k

this is z-module homomorphisms liquid, not ring

[woven delta]

Okay

Makes more sense now

[solemn rain]

i dont get why

gcd musut divide k

?

[upper pivot]

consider orders

if f([1])=[k]

[solemn rain]

ohh

yea eya forgot about orders

so they must have same order

[uncut girder]

😢 😢

[upper pivot]

not same necessarily

[solemn rain]

if f([1])=[k]

continue

[upper pivot]

oh i might have lied

anyhow figure out the condition

[woven delta]

Can you so some shit with prime factors?

[upper pivot]

with the order idea

[solemn rain]

idk liquid

[woven delta]

I mean prime powers

[uncut girder]

[solemn rain]

xd what did you lie about john

im super confused hahaa

[upper pivot]

forget what i said about gcd lol

[solemn rain]

why is

f(phi(a)) = a wrong

mapping from hom to z

f: Hom --> Z

i dont get the kernel i want right?

[upper pivot]

i mean i dont think that map makes sense?

whats a?

[solemn rain]

some fixed element

phi acts on

a in Z/nZ

[woven delta]

Yeah I think you can do this with the fact you can decompose as the product of prime power cyclic groups and using product and coproduct stuff

[solemn rain]

whats product and coproduct sutff

[upper pivot]

i mean the solution isnt very hard tho

we dont need to complicate it that much?

[woven delta]

Yeah I guess mine is pretty bashy

Although the prime power fact is trivial

[upper pivot]

also it wouldnt be a homo into Z mo2men

try addition

(and the fact n phi=0)

[solemn rain]

yea

okay what do i do now

[upper pivot]

anyhow umm let me go back to what i said before

[solemn rain]

okay so first

let [1] be 1

so if phi(1) = k

how can i define f

[upper pivot]

for each homo from Z/nZ to Z/mZ consider f([1])=[k]

whats f(n[1])

[solemn rain]

fk

nk

[woven delta]

But anyway I think that my solution reduces things to looking at maps between prime power cyclic groups

[solemn rain]

way out of my scope

[upper pivot]

my hint?

[solemn rain]

no what liquidi said

n[k] john

[upper pivot]

yeah and whats n[1]

[solemn rain]

idk

n?

n

[upper pivot]

[n]

[solemn rain]

yeaa

[upper pivot]

which is what in Z/nZ

[woven delta]

Also what does your ~ mean mo2men?

[solemn rain]

isomoprhim

isomoprhic

[upper pivot]

(i assume as Z-modules again)

[solemn rain]

yes

this is lin alg

[upper pivot]

but anyhow whats [n] in Z/nZ

[solemn rain]

0

[upper pivot]

right so f(n[1])=n[k] what does this tell you

(actually ill let u figure out the rest)

[woven delta]

Lol

Yeah John's idea is good

[solemn rain]

phi: f---> f(0) ?

i did something similar when proving Hom(R,R) is R

does this work too here

[upper pivot]

it should but you need to figure out what f(1) looks like, i.e what k is allowed to be

[solemn rain]

do you mean f(0)

[woven delta]

f(0) = 0

Lol

Okay what is the order of an element in Z/m?

[uncut girder]

m/d

[solemn rain]

k has to be mod m

and mod n?

0

0 mod n and mod m?

0 mod (n,m) ig

[woven delta]

Look the point is the image of 1 has a certain order

[solemn rain]

why

[woven delta]

And 1 has order n

[solemn rain]

the idea that phi(1) is not 1

is fucking me bad

[woven delta]

So the order of f(1) has to divide n

[solemn rain]

i dont know how these homs wokr

no

really

why

yea

okay

lmfao

[woven delta]

Lol

[solemn rain]

nvm

[woven delta]

What is the order of f(1)?

[solemn rain]

i hope you mean m

you mean it has to divide m

right?

[woven delta]

No

I mean it has to divide n

[solemn rain]

yea

yea ik why

[woven delta]

Okay great

Now what is the order of an element in Z/m

[solemn rain]

|1^k| = m/gcd(m,k)

m\

[woven delta]

Yeah

[solemn rain]

m sorry

[woven delta]

Also 1^k is bad notation

[solemn rain]

k

yea

[woven delta]

But whatever

Okay so we want to show that the points that 1 can map to are exactly the k in Z/m with m/gcd(m,k) dividing n

And that's entirely obvious

[solemn rain]

yea

thats the order

ig

okay

[woven delta]

Yeah

Okay

Now we also know that the order divides m (obviously)

[solemn rain]

yes

wait

[woven delta]

That's just Lagranges theorem

[solemn rain]

yes

yea

i forgot Z/m is of order m

sry

[woven delta]

Okay so actually I'm not going to be explicit

With the order

So the elements k we care about are the ones with order dividing n in Z/m

What is a generator for that?

Like there is a particular element in Z/m

[uncut girder]

Z/URMom

[woven delta]

Nice

[solemn rain]

k is a multiple of n and multiple of n

m*

right?

divides*

k divides both n and m

[woven delta]

Yeah

Wait no

[solemn rain]

why no

[woven delta]

So k has order m/gcd(m,k)

[solemn rain]

yes

[woven delta]

So let k be (m/gcd(n,m))

[upper pivot]

(there is a bit of a nicer way to look at this btw)

[woven delta]

Then k has order gcd(n,m)

Lol

I'm trying to be super explicit

[solemn rain]

i get it

[upper pivot]

fair enough lol

[solemn rain]

(n,m) generates

the ks

right?

[woven delta]

m/(n,m) does

[solemn rain]

okayy

[woven delta]

Okay now show that this generates

[solemn rain]

okaayt

i mean it does?

what do is how lmfao

i show

(m/(n,m)) are just the multiples no?

[woven delta]

Yeah

[solemn rain]

does each 'k' corrospond

[woven delta]

Hmm maybe there's a good cardinality argument

[solemn rain]

to each multiple

no lmfao ok

can we recap

boys

little lost here

[woven delta]

Okay so I don't think it's hard to show the cardinality has to be at most gcd(n,m). This is because if you take the subgroup, it is cyclic and the order of each element has to divide gcd(n,m)

Furthermore we have a particular element of the subgroup whose order is gcd(n,m)

To recap we are identifing Hom(Z/n, Z/m) with a subgroup of Z/m

Subgroups of cyclic groups are cyclic and the order of each element divides gcd(n,m)

[solemn rain]

yea

[woven delta]

This is what John was saying I guess by mapping f to f(1)

We have an explicit element whose order is gcd(n,m) so we are done

But I guess we really didn't need one

🤷‍♂️

[solemn rain]

HMMMMMMMMMMMMMMMMMMMMMMMM

okay so you were trying to find

the element that divides the orders

we were like

matching orders

right?

so when you do that

you assign this to f(1)

and define phi:f--> f(1)

no?

[woven delta]

Well I think you are thinking about this the wrong way

So we just map f to f(1)

[solemn rain]

okay

[woven delta]

We show this is a homomorphism

[solemn rain]

ok

[woven delta]

Which is fine

Then we show that the order of f(1) must divide n

We have that it divides m already by Lagranges theorem

[solemn rain]

yea

[woven delta]

Therefore it divides gcd(n,m)

[solemn rain]

yea

[woven delta]

Our homomorphism is injective (obviously)

[solemn rain]

so its atleast a subgroup

of what we want to show its iso to

right?

so you were saaying some cardinality

[woven delta]

And the order of each element divides gcd(n,m)

[solemn rain]

argument to show that its gcd(n,m)

right?

[woven delta]

Well I found that the cardinality is at most gcd(n,m)

Since the group is cyclic

And I found an element with order gcd(n,m)

So it is at least cardinality gcd(n,m)

So you are done

[solemn rain]

okay

sorry for taking much of your time boys

@upper pivot

is supposed to be hard

is this

*?

[upper pivot]

uh i mean

[woven delta]

Anyway the way I wanted to do this with products and coproducts

[solemn rain]

i think i got into the math where i have to think about ways to solve problems now

sad

[woven delta]

Decompose Z/n and Z/m as a product of cyclic groups with prime power Cardinality

[solemn rain]

you do this by fundamental theorem right?

[woven delta]

No

This is much weaker

Then you can use the universal property of product and coproduct to make Hom(Z/n, Z/m) into $\prod_{p_i^{a_i} | n, p_j^{b_j} | m} Hom(Z/p_i^{a_i}, Z/p_j^{a_j})$

[cloud walrus]

Liquid:

[solemn rain]

yea i dont know that

[woven delta]

This notation is bad but whatever

[solemn rain]

i see this word alot

'universal property'

[woven delta]

Anyway then if the primes are different you get the trivial group

So you just have to consider if the primes are the same

Suppose you have Hom(Z/p^n, Z/p^m)

We want to show this will be Z/p^{min(n,m)}

Is this actually easier than the general problem?

[solemn rain]

😄 idk

are vector spaces

free modules on their basis?

i wanna see like more links between vec spaces and modules

i am a bit fuzzy on the link between

[upper pivot]

yeah vector spaces are always free

[solemn rain]

when defining F[x]-Modules on a field F and a poly ring R

and a operator on a vector space V

and like sayign that subspaces of V are the T-variant

submodules something like that

im super fuzzy on that

yea

[upper pivot]

well T invariant means TW\subset W

(and same for modules)

[solemn rain]

yea

[upper pivot]

also for modules you will see stronger conditions than this btw

like if $f$ is an endomorphism of $A$-module $M$, then $f(M)\subset aM$ is a nice condition, where $a$ is an ideal

[cloud walrus]

JohnDoeSmith:

[upper pivot]

(and similarly for its submodules and such)

[solemn rain]

what does

'diagram commutes'

mean

[woven delta]

Lol

[solemn rain]

why do you keep saying lol

stupidity makes you luahg

laugh

lol

[woven delta]

So in this case it means that hf= kg

[solemn rain]

im being defined to a homomoprhism of exact sequences

[woven delta]

Those are just maps where the diagram commutes

Lol

[solemn rain]

Lol

[woven delta]

Like you have 2 diagrams

[solemn rain]

yea

and like

2 arrows between them

oe 3

or 3*

[woven delta]

Let's say 0 -> A -> B -> C -> 0 and 0 -> A' -> B' -> C' -> 0

[solemn rain]

ok

[woven delta]

Then a homomorphism will be maps from A to A', B to B', C to C'

So that all of the squares commute

[solemn rain]

yea

got it

whats the use of this?

just a new waay

to look at stuff?

[solemn rain]

yea yea

i see the squares

got it

[woven delta]

Well this is very useful

[solemn rain]

really

i think its just fancy pythagoras

[woven delta]

So you can look at chain complexes as abstract objects

[solemn rain]

idk what those are yet

[woven delta]

Oh lol

I mean you shouldn't care about this sort of thing until you do some algebraic topology

Which you shouldn't be doing

[solemn rain]

yea

i thrink im done with algebra for now

im downloading urdin

rudin

algebra gets hard p fast

for me

like 2 days ago i was solving problems easy

[woven delta]

How much of HK did you do

Dude math is hard work

[solemn rain]

uptill algebras

and polynomials

[woven delta]

Just keep at it

Do some more of HK

[solemn rain]

its literaly harder thaan df

[woven delta]

Lol

Keep at it

[solemn rain]

up till when

it doesnt end

[woven delta]

It does

Eventually

Lol I've been staring at the same 5 pages in soare for days

But you have to keep at it

[solemn rain]

id guess thats much higher

in difficulty

than a linear algebra text

[woven delta]

I mean

It just has like 20 layers of details

[solemn rain]

lmfao

[woven delta]

That you have to keep in mind

[solemn rain]

my issue is with problems

i cant solve problems

[woven delta]

Lol

You just started math

[solemn rain]

its just alwaays some fuckery trick

[woven delta]

You have like less than a years experience

[solemn rain]

some fucking trick makes me mad

[woven delta]

What were you expecting

[solemn rain]

i was exspecting

that

if i finish a section i should be able to solve its problems

thats not the case with df

[woven delta]

Nope

[solemn rain]

for me

wtf

i keep reading on like forums

dont go to another secrtion b4 problems

[woven delta]

Lol

[solemn rain]

if u dont do problems stop

[upper pivot]

you dont need to do all the problems lol

[woven delta]

Yeah

[solemn rain]

i cant do a single one of them

most of the ttime

or not most

there are easy ones

[woven delta]

If you think about them for long enough

[solemn rain]

like proving something is a sub somethign of something or like

[woven delta]

You should be able to

[solemn rain]

proving something is an iso

i didnt know u needed to like

be creative tbh

lmfao

[woven delta]

Lol

Dude

Smh

[solemn rain]

determinants has as ection called modules

in HK

fucking piece of shit

modules fucked me over

hardest section for me yuet

[woven delta]

Lol

Just work harder

[solemn rain]

how harder

[upper pivot]

until you get it

[solemn rain]

problems are just solved

with tricks

i dont find the tricks

what do i do

[upper pivot]

its not like tricks take no time to take

[solemn rain]

damn i never thoguht

i would say that