groups-rings-fields

[lethal dune]

ok can I say something about the intersection of pillow groups? if say |P| = p then they intersect at identity, so we can count the number of elements of order p by n_p(p-1). if |P| = p^k (k>1) however, then is there a way to count elements of all pillow groups

from D6 it looks like we have some common elements

[cyan marten]

There is a stronger congruence for n_p if you can control the maximum cardinality of the intersection of two p-groups: If this is s, then
n_p = 1 (mod p^k / s)

Where of course |G| = p^k * (something not divisible by p)

[lethal dune]

s = p^something?

[cyan marten]

Yes, because it's the cardinality of intersection of two p-subgroups

[lethal dune]

seems interesting

[cyan marten]

This can be used to force s to be small, and then to force the number of elements of order p to "explode" and eventually to conclude that n_p must be 1 after all

[lethal dune]

ngl that looks complicated

[cyan marten]

Well, as you said when we have just p (not p^k for k > 1), we can find out exactly how many elements are there

As we did in the 105 case to conclude that either H or K was normal

If k > 1, I don't think we can really do that in general because of the nontrivial overlaps possible (maybe inclusion-exclusion?)

[lethal dune]

yeah that makes it complicated

[cyan marten]

But suppose we know that, if there are no nontrivial overlaps, n_p must be 1

[lethal dune]

so in general can we say anything about the intersection of the pillow groups?

[cyan marten]

Yeah, we know that the maximum cardinality s makes this work.

There is also a theorem that says if the p-Sylow subgroups are abelian, then there are two of them whose intersection equals the intersection of all of them!

That is, S n T = the intersection of all p-Sylow subgroups, for some p-Sylow subgroups S, T.

[lethal dune]

😣

ok that rabbit hole is too deep

[cyan marten]

Are you looking for something specific?

[lethal dune]

not specific, just looking for some facts that I might have missed/over looked

anyway thanks for you time, I really appreciate it

[cyan marten]

This problem?

[cyan marten]

You are welcome! Also check Isaacs' Finite Group Theory

If you are looking for similar results

[lethal dune]

ok ok

[trim grove]

i have to find a one one homomorphism form cyclic group of order 8 $\mathbb{C}_8$ to $S_5$

but as we know that by isomorphism theorem $\mathbb{C}_8}$ will be isomorphic to some subgroup of $S_5$ , but here C8 has element of order $8$ but S5 does not have element of order $8$ so there exist no one one homomorphism , is my solution correct?

[cloud walrus]

TheStudent
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cyan marten]

Yes

[wooden ember]

So wasn’t sure where to fit this question but on on elf my linear algebra exercises we used some simple algebraic means to get the values for cos(2pi/5), sin(2pi/5), cos(2pi/10) and sin(2pi/10) which gives us the first 5th and 10th root of unity (made use of some polynomials). Now an optional question is to construct the first 5th root of unity USING THIS INFORMATION (not via the standard construction of a regular pentagon) and I’m a bit lost as to how to go about it

The only very unsatisfactory thing I could come up with is to use some generic techniques to construct the rationals and roots I need that make up my sine and cosine

But that seems like more than is needed

[weak oriole]

I don't understand what construct means here

[wooden ember]

compass and ruler construction

by identifying $\mathbb{C}$ with $\mathbb{R}^2$

[cloud walrus]

𝓛ittle ℕarwhal ✓

Hausdorff

[median pawn]

Spoiler/Solution: ||Take u = b^2 + b -1||

[wooden ember]

||u=ba||

[median pawn]

but ||ba may not be a unit, right?|| that's the fun part

[wooden ember]

oh right i didnt read that part lol

[lethal cipher]

Question about rings: is it true that if ac=ab, then c=b? If not, why not?

[hidden haven]

0a = 0b

🙈

[lethal cipher]

Okay fair. And I guess Z_n has some counter examples since they have zero divisors

[hidden haven]

Yes

[lethal cipher]

Welp, it was a fun thought while it lasted

[hidden haven]

Cancellation of a is possible iff a is not a zero divisor

The converse is nice

[lethal cipher]

The converse is trivial...

[hidden haven]

Not a zero divisor → you can cancel a?

Well I guess it kinda is

Thank you dackid, very cool

[lethal cipher]

No, I guess I misunderstood. The converse of my thing was if b=c, then ab=bc. I thought that's what you were referring to, and no duh :p

[hidden haven]

oh lol

[lethal cipher]

Hmm, so why is your claim true. It is not obvious to me

[hidden haven]

Cancellation of a is equivalent to saying that the abelian group homomorphism which is multiplication by a is injective

[wooden ember]

$ab=ac \implies a(b-c)=0$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

since a is not a zero divisor we must have b-c=0 so b=c

[hidden haven]

🙈 👊

[wooden ember]

there now you get react bombed

how does it feel

[hidden haven]

sorry

not that I'd stop it

[lethal cipher]

Oh duh... no zero divisors means the ring is an integral domain 🤦‍♂️

[lethal dune]

[wind locust]

heya

[hidden haven]

We aren't saying no zero divisors though

Only that a isn't one

[wind locust]

the proof for part 2) would've still worked if p weren't prime, right?

sorry for the basic question i've turned my brain into mush today

drank too much coffee

yea i think it works

[chilly ocean]

Yeah I dont think it uses p being prime

[wind locust]

tysm!

also nice cat in the pfp

[hybrid lichen]

can someone help me with this?

I couldnt show that S3 x Z is a solvable group

I just need this part

[wind locust]

well, S_3 and Z are both solvable groups, no? Then S3 x Z, being a finite direct product of solvable groups, must be solvable itself

[hybrid lichen]

yes youre right

I can write a solvable series for S3

but cant write for Z

do you have any examples?

[wind locust]

i think the normal series Z, {0} should work

[hybrid lichen]

thank you so much, that really helped

[wind locust]

no problem, although i only self-studied solvable groups so i might have made a very stupid mistake

[hybrid lichen]

no your statements are totaly true

[wind locust]

nah it should work

yeah

[hybrid lichen]

thank you!

[wind locust]

you are welcome

[wooden ember]

Solvability is weaker than being abelian: in particular every abelian group is solvable (by the trivial normal series)

Nilpotent groups are also solvable if you’ve seen those

[hybrid lichen]

thank you so much

[lethal cipher]

True, I just didnt immediately realize this obvious connection

[broken stirrup]

Hi, why finiteness o I implies that weak direct product and direct products are the same

[next obsidian]

Weak direct product is where only finitely many of the things are not identity

Direct product is when you don’t have this restriction

The two differ for infinite products because a function that is not the identity on anything exists in the direct product but not the weak one

But in the finite case, the condition “finitely many of the things are not identity” is automatically satisfied

[broken stirrup]

thank you

[hidden haven]

Always, assuming that by (R/(a))/(b) you mean (R/(a))/(b+(a)), because b is not an element of R/(a)

This is one of the isomorphism theorems I think, or at least follows easily from the first one

[long obsidian]

I'm reading something demonstrating that there is no such $\mathbb{Z}$ submodule M such that $\mathbb{R}\cong M \bigoplus \mathbb{Z}$.

They start by saying if so then $M\cong \mathbb{R}/ \mathbb{Z}$ by the "five lemma". I looked this up and it had to do with homology. Is there an obvious reason this would be so? I can kinda squint my eyes and believe it but it's not rigorous

[cloud walrus]

fajitas

[woven delta]

Submodule of what?

@long obsidian

[long obsidian]

Of R

[woven delta]

Okay so we have 2 short exact sequences, $$0\to \mathbb{Z}\to M\times \mathbb{Z}\to M\to 0$$ and $$0\to \mathbb{Z}\to \mathbb{R}\to \mathbb{R}/\mathbb{Z}\to 0$$

[cloud walrus]

'quid

[woven delta]

The isomorphism between MxZ and \mathbb{R} induces an isomorphism between the quotients

(you have to pick the map from Z to be the image of 0\times Z)

[long obsidian]

🤔

I think I see what you're saying.

[woven delta]

I don't really see where the five lemma comes in here, but it's not hard to show the isomorphism

[long obsidian]

I didn't know stuff like that could be done with diagrams

[woven delta]

If f is your isomorphism from M\times Z and R, then your map from M to R is m \mapsto [f(m,0)]. Your map from R to M will be [r] \mapsto \pi_1(f^{-1}(r))

You just have to verify that the isomorphism actually makes sense

[r] means the equivalence class of r in R/Z

[long obsidian]

By makes sense does that mean the function is well defined on an equivalence class?

[woven delta]

Sorry I forgot that we are projecting onto the first coordinate

The point is that if you have 2 elements $p$ and $q$ in the same equivalence class you need to show $\pi_1(f^{-1}(p)=\pi_1(f^{-1}(q)$

[cloud walrus]

'quid

[woven delta]

Do you see what I'm saying here?

https://proofwiki.org/wiki/Universal_Property_of_Quotient_Group

This may be helpful

[long obsidian]

Yeah!

Thank you

[long obsidian]

Im seeing the quaternions be represented as $Q_8=<a, b|a^4=e, a^2=b^2, ba=a^{-1} b>$ what do the bars mean?

[cloud walrus]

fajitas

[small bison]

it just separates the generators from the relationship the generators satisfy

[wild sapphire]

Can someone help me with polynomial ideals? I feel like I kinda understand the analogy to group cosets and stuff, but I dont really understand what kind of elements would be in a polynomial factor ring

like R[x]/<x^2+1>

I understand that this is a factor ring "modded out" by the ideal generated by x^2+1, and <x^2+1> I believe is the set of all P(x)*(x^2+1) for all P(x) in R[x]... i think

[upper cape]

Is normal closure the same as galois closure?

[chilly ocean]

it's like R[x] but you pretend x also satisfies x^2 + 1 = 0

[long obsidian]

I can tell that commas separate the generators but I'm not sure how to interpret a^2=b^2 as a generator, it's an equation with two variables

[wild sapphire]

yeah, it like "wraps around" like I'm modding out by x^2 + 1. So if I wanted to find a multiplicative inverse of a function like x+1, I'd need to find what function I could multiply to make it "wrap around" to one?

Like, isn't there a notion of congruence classes or something to be noted? I'm having trouble understanding what would be in a congruence class

[small bison]

that's a relation

not a generator

[wild sapphire]

maybe it's just that I'm not understanding when to use each binary operation, cause it's a ring and all

[long obsidian]

Ohhhh now I see how it reads. I could have preferred (a,b) such that but oh well 😢

Maybe it's just an awkward thing to denote

[barren sierra]

showing the isomorphism

is throwing me for a bender

[next obsidian]

Which one

If it’s the second, you just define a map A -> J by sending (i,j) to j

It’s just projection

[barren sierra]

is it?

hm

yea the second

but then how is I = ker(that map)?

so I have some function f: A -> I x J which I know to be isomorphic

I have some projection function g: I x J - > J

define phi: A -> J such that a maps to g(f(a))

but I need I to be the kernel of phi

to get A/I isomorphic to J

ker(phi) means phi(a) = 0

so then I need f(a) = (0, 0)

[thorn delta]

nope, you just need f(a) to be in the kernel of g

[barren sierra]

hm

ok then I need f(a) = (i, 0)

but how can I guarantee that

do I have some way to know that f(I) = (I, 0)

(scuffed notation but you get the idea)

[long obsidian]

What is an orbit decomposition?

I'm looking at what seems to be a proof that any nontrivial finite p group G has nontrivial center. Say |G|=p^k. They say that the center consistent of all fixed points of the group action of conjugation and write G=C(G)UO_1 U O_2U... U 0_l and call this an "orbit" decomposition. Never heard of such a thing

[thorn delta]

@barren sierra all of these objects are supposed to be rings, right? How are we defining multiplication on I \oplus J?

[barren sierra]

they're rings and I is an idea of R

but we know nothing about J right?

I \oplus J is direct product

so isn't that coordinatewise multiplication?

[thorn delta]

Maybe. Do your rings not need an identity?

[barren sierra]

they don't

so 2Z is a well defined ring

[obtuse kernel]

is the purpose of the primary decomposition theorem to preface other canonical forms like jordan and rational?

[thorn delta]

yea spam idk this is not obvious

[barren sierra]

pain

ok

what about this?

trying to use division theorem

but idk

[thorn delta]

yea, so there are polys r,s such that r(x)q(x) + s(x)d(x) = 1. So you want start something like that

Here's some direction:

Specifying a map $$f : K[x]/(q(x)) \oplus K[x]/(d(x)) \to K[x]/(p(x))$$ is the same as specifying a map $f_1 : K[x]/(q(x))) \to K[x]/(p(x))$ and a map $f_2 : K[x]/(d(x)) \to K[x]/q(x)$ and these two maps are specified by maps
$K[x] \to K[x]/(p(x))$ whose kernels contain $(q(x))$ and $(d(x))$ respectively.

[cloud walrus]

kxrider

[thorn delta]

@barren sierra if what i said above makes no sense I can come up with a different way of explaining maybe

[barren sierra]

Oh hm

I was trying to go the other way

so I have a map like that

it's just that proving that that map is surjective

or rather the reverse

not sure how to use the gcd = 1 part

I have $$\phi: K[x]/(p(x)) \rightarrow K[x]/\langle q(x) \rangle \oplus K[x]/ \langle d(x) \rangle$$

[next obsidian]

This is the chinese remainder theorem

[thorn delta]

im assuming spam doesn't have CRT

[cloud walrus]

Spamakin🎷

[next obsidian]

It’s just the same idea

[barren sierra]

wait no I do

[thorn delta]

lmao oof

[barren sierra]

I just forgot it exists

ok lemme try now

once I prove phi is surjective it's just 1st iso thrm go br

or no

[thorn delta]

well you have to prove the kernel is what you want it to be

and iirc this will use gcd = 1

[barren sierra]

$$\phi: K[x] \rightarrow K[x]/\langle q(x) \rangle \oplus K[x]/ \langle d(x) \rangle$$

[cloud walrus]

Spamakin🎷

[barren sierra]

yea

and then $\text{ker}(\phi) = \langle p9x) \rangle$

[cloud walrus]

Spamakin🎷

[thorn delta]

do you have a candidate map for phi yet?

[barren sierra]

just the quotient maps

for q and d respectively

and then proving my desired kernel is ez

[latent night]

(a) R has no nonzero nilpotent elements
(b) a^2 = 0 implies a = 0
Show that both a and b imply that all idempotents are central

How would I do this? My understanding is that the question is asking me to show that in a ring where either a) or b) holds, any idempotent e, will commute with all other elements in the ring? er = re for all r \in R

Is this correct?

[next obsidian]

Wtf lol

I mean (a) implies (b)

So I don’t get this

But yeah, I guess you just want to use (b) to show what you said

Actually

Aren’t (a) and (b) equal lmfao

[delicate bloom]

yes

[next obsidian]

[latent night]

I know

They are equivalent

That was part 1 of. The question. The second part was the idempotent central thing

[next obsidian]

I guess my initial idea was to look at (er -re)^n

If this is 0 then er-re = 0

But I think it isn’t good enough to just square it, I did the computation i my head

[delicate bloom]

might be useful to consider that e^2=e means we have the zero divisors 0=e(1-e) and (1-e) is also idempotent

in commutative rings it's kind of a fun fact to use $(ae+b(1-e))^n = a^ne+b^n(1-e)$ which might be somewhat helpful to consider

[cloud walrus]

Merosity

[delicate bloom]

trying to draw a connection to the idea of using (er-re)^n here, dunno

[coarse storm]

I am guessing we are supposed to somehow construct a nilpotent element from idempotents?

[next obsidian]

Yeah like if e isn’t 1 or 0 (which are central)

Then what mero said is the main fact I know

But it gives a zero divisor

[latent night]

Wait isn't it really trivial then? e being not 1 or 0 means it's a non zero zero divisor contradicts a)

So the only idempotents in a ring where a) holds are 0 and 1 which obviously are in the centre. Am I missing something?

[delicate bloom]

I think that's conflating zero divisors with nilpotent elements

nilpotent elements are zero divisors but not all zero divisors are nilpotent

[latent night]

Ah, sorry. I forgot the question lol

[delicate bloom]

yeah I made the same mistake earlier when thinking about i(1-i)=0 earlier too for a moment

[delicate bloom]

maybe as a test case of this idea, how would we go about using 3 and 4 in Z/6Z to create a nilpotent element which ends up being exactly 0? Or is the fact that this is already a commutative ring from the start not really useful enough for a test ring

[coarse storm]

Ooh. I think you are supposed to consider e r (1-e).

[delicate bloom]

nice that does it

[coarse storm]

Sssh. Let them work it out!

[delicate bloom]

heh I was one of the fools working it out too

spoilered 😌

[coarse storm]

Yea, thanks. That was an interesting question!

[delicate bloom]

actually just gonna delete it cause it's simple enough, no need to force someone to resist temptation

[coarse storm]

Glad we got there in the end. Would've bothered me the whole day otherwise!

[delicate bloom]

trying to think if there are any rings I like where this has some interesting consequences now lol

[coarse storm]

A potential ring to consider might be generated by

• The upper triangular matrices with all-ones diagonal
• Lower dimensions of the same form embedded as diag(M, 0) for idempotents,
  I think this satisfies the nilpotency restriction.

[long obsidian]

I'm reading about the classification of groups of order 20. take G a group of order 20. By sylows theorem there is a single sylow 5 group, say H. Take a sylow 2-group N. It is clear that N and H are disjoint and that G=NH.

What isn't clear to me is why the solution says that G should be the semi direct product of N and H. How is the semi direct product relevant here?

[hidden haven]

Whenever you have a normal subgroup N of G and another subgroup H with NH = G and N intersection H = {1}, G becomes a semi direct product of N and H. This is because the 2 conditions say that you can write any element of G uniquely as nh for n in N and h in H, so you immediately know G as a set, and just need to know the multiplication. You need to know 3 things: the multiplication in N, the multiplication in H, and how elements of N commute past the elements of H, and then you would know G entirely. And this is the whole point of a semi direct product, so you might be able to deduce the rest from whatever definition of the semi direct product you have

Those 3 things would give you all the information because then if you want to multiply (mg)(nh) = m(gn)h with g,h in H and m,n in N

Then you can commute g and n to get another element n'h' using whatever semi direct equations you have, and then this is again in the canonical nh form

[delicate bloom]

the upper triangular matrices won't work because we have matrices that are nonzero that have A^2=0 with A != 0

[coarse storm]

Hm... May be over a field then? The all-ones diagonal should stop it going to zero.

[delicate bloom]

well it's a ring, so we can subtract one matrix from the other to remove the diagonal

[coarse storm]

Oh right. Forgot about addition. Woops!

[delicate bloom]

yup lol

[cyan marten]

Given groups N and H, the semidirect products (with N normal) all arise from elements of Hom(H, Aut N). But we actually quotient by something here - so when do homeomorphisms f, g : N --> Aut H determine the same group structure?
When constructing CW complexes, two attaching maps that are homotopic give homotopy equivalent complexes. So maybe something similar happens here?
One result in this direction is when H is cyclic and the images of f ans g are conjugate, so in this case semidirect products are in one-to-one correspondence with conjugacy classes of elements of Aut N whose order divides |H|, or something similar.

[long obsidian]

Thank you!! Now it makes sense why it arises

[chilly ocean]

Can anyone with big cyclotomic brain muscles guide me through getting a nice form for the sequence, letting p be prime and $\omega = e^{2\pi i/p}$,
$a_n = 1/p \sum_{m=0}^{p-1} (\omega^m -1)^n$ for $n= 0, 1, 2, \cdots$

[cloud walrus]

zd

[delicate bloom]

https://math.stackexchange.com/questions/1021046/sum-zetap-1-zeta-1n/3924616#3924616
might be worth looking at, I'll think about it more but I doubt it

[chilly ocean]

Hehe the source of this is actually the guy whose book I'm reading, and I used the same trick to show that this is just a simple case (t=1) of that and by his algebraic integer magic he does in the same chapter you can show that this expression has arbitrarily large powers of p as n grows

But finding the exact values for that sequence is a toughie for me rn

[delicate bloom]

idk what you want past that sum representation, but that's already in the question and I rederive a more general version in that answer I came up with there too

which book are you referring to, as far as I knew I came up with that on my own lol

[chilly ocean]

Kurt Mahler, introduction to p-adic numbers and their functions

[delicate bloom]

I have it, where in there is it

that's a pretty fun book imo I don't remember much algebra stuff in there though

[chilly ocean]

Pages 53-58

It's the 1973 version which was randomly in my community college library lol

[delicate bloom]

I don't see anything like that there

[chilly ocean]

And this is for problem 2, the stuff about a_n being null shows that the function is continuous

[delicate bloom]

it's chapter 5 the decomposition of Q_g into p-adic fields for my version

[chilly ocean]

And so I just need to get a nice form for the sequence itself

Oh yeah I think it's a different version

This is in the continuous functions chapter

All the stuff for proving theorem 2

"Cyclotomic extensions of Q" and then "the algebraic integer gamma" basically have that same proof which is really cool that you recreated it yourself lol

Er rather the more general idea where some function is periodic with period a power of p

Btw I love this stuff and I think we chatted ages ago and you inspired me to eventually check p-adic stuff out and here I am finally more than halfway through this book lol

[delicate bloom]

basically I just noticed it by coincidence since to make an indicator function on some open ball in Z_p I realized I could make it two ways, either as its series or as basically a little thing with a p-th root of unity

[chilly ocean]

Oh nice

[delicate bloom]

I'm still not seeing it in my version let me check the year I guess lol

ah this is the second edition 1981

[chilly ocean]

I gotchu

[delicate bloom]

maybe it got removed or something idk, what's the name of the chapter

it's one of the problems at the end of that chapter you're saying?

[chilly ocean]

Yeah

[delicate bloom]

chapter 8 is First properties of continuous g-adic functions

[chilly ocean]

Huh lol there must be a lot more in that version

This version is small, only 87 pages

But dense af of course

[delicate bloom]

then 9 is The interpolation series of a g-adic function

so it's gotta come after that I guess since that's the Mahler series lol

oh well, weird, they sound like fairly different books maybe I can find the first edition by other means

[delicate bloom]

glad to hear it btw 😎

oh I found it

[chilly ocean]

I saw the thing at the end of your se post and I had something similar in my notes for just grabbing the sequence without going into cyclotomic shit

Honestly either one, with or without the roots of unity, I can't really figure out how to get the sequence

[delicate bloom]

actually I think what I proved was more general, it's just I didn't want to over generalize for my answer cause it was long enough

[chilly ocean]

Yeah you did with any power of p I think

[delicate bloom]

but really there's nothing special about p^t Z_p

you can go further and do a+p^t Z_p for an arbitrary ball of center a with radius p^-t

[delicate bloom]

wait which sequence are you talking about

looks like you have it right there a_n = ... already

[chilly ocean]

For the expansion of f(x) = 1 when |x|_p<=1/p and 0 when |x|_p = 1 (f defined on p-adic integers)

The a_n yeah

Oh I was playing around in Wolfram and noticed it seems to be 1,-1,1,-1,...+/- 1, a_p, 1, ... And weird shit happens at a_{kp}

[delicate bloom]

what I do in my answer is take finite differences of it with this representation $$f(x)=\frac{1}{p}\sum_{k=0}^{p-1}\omega^{kx}$$

[cloud walrus]

Merosity

[delicate bloom]

the key is $\nabla (1+a)^x = a (1+a)^x$ so you can just iterate $\nabla^n (1+a)^x = a^n (1+a)^x$ and then plug in $x=0$ to get $a^n$

[cloud walrus]

Merosity

[chilly ocean]

That is a wacky function, what is that?

[delicate bloom]

it's your function

just represented like a fourier transform

[chilly ocean]

The nabla?

[delicate bloom]

oh just means backwards finite difference

[chilly ocean]

Nah I get that I mean the little tringle

Oh ok

[delicate bloom]

$\nabla f(x)=f(x+1)-f(x)$

[cloud walrus]

Merosity

[delicate bloom]

err forwards *

I guess Delta is also common

[delicate bloom]

does this representation make sense or did I pull it from thin air

[chilly ocean]

Oh yeah ok so a_n is originally defined as those successive differences anyway

So if you know how to get those from the function itself then you're good

[delicate bloom]

yeah, exactly

just depends on knowing this representation

maybe worth just saying $f(x)=\omega^x = \omega^{x_0}$ is a continuous function of x that only depends on the first digit, I think in an earlier chapter they say $g(x)=x_0$ is a continuous function too

[cloud walrus]

Merosity

[delicate bloom]

x_0 meaning the first digit

[chilly ocean]

So how would you do that for my f(x), it's not really in that form involving (1+a)^x I don't think is it or am I buggin

Also unfortunately the '70s edition seems kinda sparse so that wasn't in there

[delicate bloom]

I think you're misunderstanding me, I'm saying your f(x) is that and is in that form

w=1+a

I just wrote it that way cause it's the lazier way to write it with (1+a)^x instead of a^x

[chilly ocean]

So we're just doin w = (1+(w-1)) right

[delicate bloom]

the powers just end up being (w-1)^n as the coeffs

yeah

[chilly ocean]

Right ok

[delicate bloom]

here it doesn't really matter but as a bit of trivia, |w-1|<1 if w is a p power root of unity

[chilly ocean]

It seems knowing some discrete calculus really helps for messing with continuous p-adic stuff occasionally

[delicate bloom]

well it kinda matters for other reasons for the divisibility aspect ofc

yeah I suppose so

I just knew from in the past playing around with linear recurrence relations that this kinda thing held true

it's sorta one of those simple cases, the other being knowing that $$\nabla x^{\underline{n}} = n x^{\underline{n-1}}$$

[cloud walrus]

Merosity

[delicate bloom]

here this is the falling factorial

basically it's just like two mimicing cases to the derivative cases

e^{kx} with derivative k e^{kx} was the first and this one is like the power rule

[chilly ocean]

Gaming

[delicate bloom]

the quick way to derive this power rule is to write it as the rule for generating pascal's triangle with binomial coefficients

subtract one to the other side: $$\binom{x+1}{n}-\binom{x}{n} = \binom{x}{n-1}$$

[cloud walrus]

Merosity

[chilly ocean]

Right I think I saw a mathologer vid about this stuff

[delicate bloom]

idk what he covered, but hopefully how to derive stuff like 1+2+...+n = n(n+1)/2 and others by this method

[chilly ocean]

Does the fact there's a kx in there rather than an x change anything? I can see how the 1/p and the sum don't affect anything since taking differences is linear so we can still work inside the sum, but the kx thing seems like we need more

[delicate bloom]

it's a matter of perspective

really it adds nothing if you think of it as (1+a)^k being raised to the x

[chilly ocean]

And the rule still works?

That is wacky

[delicate bloom]

it's not really a rule, you can derive it yourself

h(x)=c*b^x now try to simplify h(x+1)-h(x)

[chilly ocean]

Ok brb

[delicate bloom]

you should have gotten ||nabla h(x) = (b-1) * h(x)|| and so now if b=w^k you have ||nabla h(x) = (w^k-1) h(x)||

[chilly ocean]

Yeah with the k on the inside which just brings me back to the original expression for a_n trollgesad

F

[delicate bloom]

is that bad? lol

should be a good thing

[chilly ocean]

I mean it is cool since it's a much more intuitive way of finding that expression

I just want to get explicit values, it looks like there will be a nice form for this

Potentially

[delicate bloom]

you're confusing me

[delicate bloom]

isn't this the nice form already

what's more explicit than this?

[chilly ocean]

I mean the explicit values of the a_n if you were to write it out as a sequence or piecewise function of n with "simpler" form

[delicate bloom]

it's a sum of integers

I don't see how you think you can get simpler than that

[lethal cipher]

A sum of zeros mayhaps

[delicate bloom]

do you have some other form you know about that you're trying to get it to look like?

[chilly ocean]

Wolfram is giving me interesting results, if you plug n in for some particular p you get a periodic sequence 1,-1,1,-1.. except at n divisible by p, where it's either 0 or spits out some strange integer

Which gives me some confidence that maybe it can be written as a periodic/piecewise thing

I guess you're right that the author was probably fine with the form we had it in, but now I'm just honestly curious haha

Wait that doesn't make any sense I forget what I plugged in actually, it shouldnt be that since a_n needs to get bigger and bigger powers of p

https://tenor.com/view/kirby-i-forgot-i-forgor-gif-22449575

[delicate bloom]

well the indicator function is 0 except when it's 1 every multiple of p, that's part of it

I wouldn't mind playing around further with it I just was making sure I understood what your goal was that's all

[chilly ocean]

Yeah I guess I had it from the start if either of those forms are fine, the one with binomial coeffs or the one with roots of unity
The outline you gave for using differences right away is super nice, that's my big takeaway from this lol

[delicate bloom]

yeah, I guess maybe work out in general when $f(x)=\chi_{a+p^k\bZ_p}(x)$, indicator function on the set $a+p^k\bZ_p$

[cloud walrus]

Merosity

[delicate bloom]

it's actually maybe worth thinking about this in terms of properties of the mahler series itself of how to translate

$$\chi_{a+p^k\bZ_p}(x) = \chi_{p^k\bZ_p}(x-a)$$

[cloud walrus]

Merosity

[chilly ocean]

I figured out the thing btw, it was Wolfram fucking up with the roots of unity one-- the equivalent binomial sum worked fine (checked divisibility for large p^t)

You need to give it specific values of p or else it will do some insane shit

I left p as a variable and plugged n in and my boi Wolfram lost its marbles

[delicate bloom]

heh you hate to see it

[smoky ivy]

Let $\mathbb{K}$ be a field such that $#\mathbb{K} \notin \mathbb{N}$ and $V$ be a vector space over $\mathbb{K}$. Let also $n \in \mathbb{N}$ and $U_1, \hdots, U_n$ be subspaces of $V$ such that $U_i \neq V$ for all $i \in {k \in \mathbb{N} \mid 1 \leq k \leq n}$. Show that
$$\bigcup_{i=1}^{n} U_{i} \neq V$$

[cloud walrus]

lewis

[smoky ivy]

I'm not really sure on what to do, my first intutition is to prove it by contradicition. So assume that $$\bigcup_{i=1}^{n} U_{i} = V$$
holds true. Maybe inducing over $n$ is what I should be aiming for next? But I have no idea on how to approach this.

[cloud walrus]

lewis

[hidden haven]

Reduce to the case where V is finite dimensional. Then subspaces of V are zero sets of some multivariable polynomials over K. Their union will then be the zero set of the product of these polynomials, but over an infinite field, no non zero polynomial can be uniformly zero, so the union can't be everything

There might be a simpler proof

[smoky ivy]

How would I do that

[hidden haven]

Which part

[smoky ivy]

reducing it to the case where V is finite dimensional.

[hidden haven]

Suppose you have an infinite dimensional space which is union of U_1 to U_n which are proper subspaces. Then for each U_i, choose a vector v_i not in U_i. Then W = span {v_i} is finite dimensional but it's still a union of proper subspaces W ∩ U_i

[smoky ivy]

I think I've got this step

But what do I do next

and actually, could you expand on the part why W would be finite dimensional

is it because we're assuming that the union of all proper subspaces is V

[hidden haven]

It is spanned by finitely many vectors

[smoky ivy]

oh, so the span is referring to all "v_i's"

[hidden haven]

The stuff I wrote right after that with polynomials

Yes

[smoky ivy]

so W = span{v_1, ..., v_n}

[hidden haven]

Yes

[smoky ivy]

if we say v_i not in U_i

are we saying it's in V \ U_i

[hidden haven]

Yes

[smoky ivy]

or the union of U_1, ..., U_n \ U_i

[hidden haven]

Same thing

We are assuming V is the union

[smoky ivy]

oh, okay

yeah

what about this part W ∩ U_i

what you wrote

could you expand on this

[hidden haven]

You have those subspaces of W for each i

They are all proper

Because the ith one misses v_i

And there union is W

[smoky ivy]

subspaces of W? isn't the span of v_i a subspace itself

are you maybe referring to the vector space V?

[hidden haven]

Subspaces can have subspaces

[smoky ivy]

yeah

[hidden haven]

Yeah so I mean subspace of W

Which is a subspace of V

[smoky ivy]

im not sure whether i got this step 100%, so we have subspaces of W

[hidden haven]

U_i ∩ W are subspaces of W

[smoky ivy]

Because U_i is a subspace itself?

[hidden haven]

If you haven't seen this then you can try to prove this it's pretty easy

[smoky ivy]

and the intersection of two subspaces

[hidden haven]

U_i is subspace of V

[smoky ivy]

are subspaces?

[hidden haven]

Intersection of 2 subspaces is subspace of both

I need to sleep now. If you have other questions, other people should be able to help

[smoky ivy]

okay...thank you

[covert storm]

I am having difficulties solving a problem, i have to show that if K* is the multiplicative group of a field K, and H = {a^2; a ∈ K*}, then H=K * if the characteristic of K is 2

that's the text of the problem

if H is a subgorup of K* it means that H is contained in K*, so i think i have to show that K * is contained in H but i don't know how

[amber stag]

Use Frobenius map

[wild sapphire]

Can someone tell me what these elements are? I understand that it's like narrowing in all the congruency classes of polynomials F[x] mod p(x) to just those congruent to constant values. I know it's a subfield

But I'm wondering how I would go and add or multiply two of these elements? Like what would the notation look like?

[amber stag]

They are added and multiplied just like polynomials, but they are subject to an additional relation p(x)=0. For example, if p(x)=x^2+1, then whenever you see x^2, replace it by -1

[wild sapphire]

hmm, so addative and multiplicative closure is just really as simple as showing the operations. And like same with multiplicative inverses? Like E is kinda just the field F

?

[amber stag]

addative and multiplicative closure is just really as simple as showing the operations
I think so.
Like E is kinda just the field F
I'd say E is more like the polynomial ring F[x] in your example

[chilly ocean]

looking for guidance. thanks.

[woven delta]

So you need to show the 3 axioms for equivalence relations

$x\sim x$, $x\sim y \implies y\sim x$ and $x\sim y, y\sim z\implies x\sim z$

[cloud walrus]

'quid

[woven delta]

Why is the first one true?

[chilly ocean]

I think we should use this method. but I don't know how

[woven delta]

I'm assuming the equivalence relation is defined by $x\sim y$ if $f(x)=f(y)$

[cloud walrus]

'quid

[chilly ocean]

I was also thinking the same, but it's not the correct solution.

[woven delta]

I'm confused

What is the equivalence relation

Can you post example 1.2.6

[chilly ocean]

are you also from uiuc?

[woven delta]

No, I just read the exercise you posted

Okay my assumption was correct

[chilly ocean]

this is what tutor guided me. I have no idea how to apply it to that.

[barren sierra]

Ok so for the second part of this problem suppose we have $\sigma: A \to I \oplus J$ as our given isomorphism. Also define $\Omega: I \oplus J \to J$ such that $(i, j) \mapsto j$ as a projection function. I want to use the first isomorphism theorem and so have defined a map $\phi: A \to J$ such that $\phi(a) = \Omega(\sigma(a))$. However in proving the kernal of $\phi$ is $I$ I need that $\sigma(I) = { (i, 0) \in I \oplus J }$ which is where I am stuck. Any ideas?

[cloud walrus]

Spamakin🎷

[sturdy marsh]

sigma is an iso

so the kernel of phi is iso to the kernel of omega

@barren sierra

[barren sierra]

hm

well yes

but the kernel of omega is the set of all ${ (i, 0) \in I \oplus J }$

[cloud walrus]

Spamakin🎷

[sturdy marsh]

yup

[barren sierra]

which I already said

[sturdy marsh]

the preimage of that under sigma is I

[barren sierra]

how

that's what I need to prove

[sturdy marsh]

oh it's some arbitrary iso

it's iso to I

[barren sierra]

what

[thorn delta]

You can say $J \simeq A/\sigma(I)$ but a priori you don't have $A/I \simeq A/\sigma(I)$

[cloud walrus]

kxrider

[barren sierra]

yea so I need $\sigma(I) = I$

[cloud walrus]

Spamakin🎷

[sturdy marsh]

you dont

[barren sierra]

or not that

uh

$\sigma(I) = { (i, 0) \in I \oplus J }$

[cloud walrus]

Spamakin🎷

[barren sierra]

this I need this

[sturdy marsh]

you dont

[barren sierra]

wdym by this

i don't?

[sturdy marsh]

youre trying to show that A/I is iso to J right?

[barren sierra]

yea

so I need I = ker(phi)

[sturdy marsh]

A/I is iso to A/sigma(I)

do you agree?

[barren sierra]

I mean I'd like to agree but I don't see how that is

[sturdy marsh]

gimme a sec lemme read the problem again

okay

you have an iso sigma A ---> I+J

[barren sierra]

yea

[thorn delta]

to me this is like saying Z/2Z \simeq Z/Z because 2Z is iso to Z

[sturdy marsh]

let's say the preimage of I x {0} in A is U

U is iso to I

do you agree?

[barren sierra]

yea

cause that second coordinate does nothing

oh

$I \simeq I \times { 0 }$

[cloud walrus]

Spamakin🎷

[barren sierra]

_>

so then $A / I \simeq A / (I \times { 0 } )$

[cloud walrus]

Spamakin🎷

[sturdy marsh]

the difference here is that the iso 2Z --> Z isnt coming from an automorphism of Z

[barren sierra]

@sturdy marsh thanks you're the 🐐

[sturdy marsh]

wait a min

hmm

[barren sierra]

?

o no

[sturdy marsh]

yeah no this doesnt work

you need to argue some more

[barren sierra]

fuck

how tho?

[sturdy marsh]

maybe they mean internal direct sum

[barren sierra]

cause you can easily see ker(sigma) is I x {0}

wdym internal?

direct sum is coordinate wise

[sturdy marsh]

the iso I + J ---> A is induced by the inclusion maps

[woven delta]

Yeah I would hope they mean internal direct sum

[barren sierra]

We never made the specification of internal or external

[woven delta]

I feel like the statement as it's made may be false

Actually if it meant the internal sum that wouldn't really make sense because they use the oplus symbol to mean the regular direct sum previously

[thorn delta]

previously where?

[barren sierra]

the part of "is it true that ..."

first part of the question

[thorn delta]

oh yea

[barren sierra]

I asked my prof on our class forum but idk when they'll answer

[woven delta]

Yeah it may just be a mistake, it seems too hard given the level of the first part of the question

Also is J an ideal or a ring?

Does A have a unit?

[sturdy marsh]

no

[woven delta]

That makes sense

[sturdy marsh]

it could have a unit ig

like R x R

[woven delta]

It can't because you need both factors to have a unit

In order for a product to have a unit

[sturdy marsh]

the first copy of R has a unit (1,0)

[woven delta]

Im saying in I \oplus J, I doesn't have a unit

[barren sierra]

2nd isomorphism theorem doesn't apply right?

[sturdy marsh]

take I = R x {0}, J = {0} x R in RxR

both I and J have identity elements

theyre idempotents in R x R

[woven delta]

Oh I'm being dumb

I totally forgot that you can decompose rings as products of ideals a lot (eg Chinese remainder theorem)

[woven delta]

Okay yeah I think you should just wait for a reply from the professor

[sturdy marsh]

to produce a counterexample, we need a ring which does have many automorphisms

Hmmmmm

[barren sierra]

yea imma just wait

it's due Sunday evening and the final is monday evening fml

[sturdy marsh]

is there a ring automorphism from M_2(k) ---> M_2(k) which takes the ideal of matrices with only 0s in the 2nd column to the ideal with only 0s in the first column

okay

let R be an infinite product of Z s

let I = (0,1,1,1,1,..)

let J = 0

I + J is iso to I which is iso to R

[woven delta]

Gross

[sturdy marsh]

but R/I = Z

[woven delta]

Yeah I think that works

[sturdy marsh]

that should do it

[thorn delta]

is the intuition for this just that if A has enough automorphisms, then there is one that maps an ideal "I" to something which is not "I"

[sturdy marsh]

that's roughly what I was thinking yeah

@barren sierra

there's a counterexample

[woven delta]

I don't like J=0 though

[sturdy marsh]

why

[woven delta]

Idk it feels a bit cheesy

[sturdy marsh]

that was fun

[woven delta]

Fair enough

I guess your example works with J any finite or countable power of Z's

[barren sierra]

interesting

can I ask how you thought of that?

wait is I an ideal?

[sturdy marsh]

yeah

[barren sierra]

oh yea

wait no?

[sturdy marsh]

you can add stuff in it

and it absorbs everything outside

[woven delta]

0\oplus Z^omega

Is the ideal

[barren sierra]

Wouldn't you need {(0, 1, 0, ...) , (0, 0, 1, 0, ...) ...} etc

[woven delta]

Yes

That's what he meant

[barren sierra]

ah

ok then yea

[sturdy marsh]

yeah sorry

everything except for the first factor

[woven delta]

It is funny that I knew what you meant even though you wrote something totally different

Intuition is pretty funny

[barren sierra]

does it have to be infinite btw?

[woven delta]

Yes

[sturdy marsh]

yeah

[barren sierra]

I can't see why it has to be

like infinite product of Zs

why not a product of 5 Zs

[sturdy marsh]

I mean other counterexamples might exist

[woven delta]

Because otherwise I won't be isomorphic to the infinite product

[barren sierra]

oh right

need R iso to I + J

[sturdy marsh]

yeah because it's infinite we have R \simeq I

[barren sierra]

ye

I could have never come up with that wow

[woven delta]

That's a standard counterexample, I've seen it used a bunch of times

[barren sierra]

I'll remember it for sure

but to me it's one of those things now that "will use in future, couldn't have come up with it"

[woven delta]

Also knowing how to use infinity like that comes from seeing it being used before

[barren sierra]

yea just don't have that "seeing it being used before"

[woven delta]

Well now you have

[sturdy marsh]

"find M such that M \simeq M + M" is a pretty common algebra problem

[barren sierra]

ye

now I have seen it before

[sturdy marsh]

:partyblob:

[woven delta]

:partyblob:

[sturdy marsh]

nitro

[barren sierra]

hm I'll have to think on this one

[sturdy marsh]

same idea

[barren sierra]

well 2 is not infinite which makes things harder

[sturdy marsh]

2?

[barren sierra]

M + M has 2 coordinates

vs the chosen counterexample above which has infinitely many coordinates

[woven delta]

Try using the example above as M

It's not too hard to make it work

(M has coordinates indexed by the naturals, M+M has coordinates indexed by pairs of naturals)

[barren sierra]

ah

and so you can make a bijection

neat

[fallow plume]

Could someone explain to me why $N_G(N_G(P)) = N_G(P)$ is a contradiction in this post:

[cloud walrus]

Thomas

[fallow plume]

https://math.stackexchange.com/questions/534686/group-of-order-24-with-no-element-of-order-6-is-isomorphic-to-s-4

Or why $N_G(P)$ is necessarily normal within $G$?

[cloud walrus]

Thomas

[fallow plume]

oh wow I'm dumb

[chilly ocean]

specifically this part

[fallow plume]

If something is in the preimage of a value, then they get mapped to the same value.

[chilly ocean]

I know. but the proof requires this part:

[chilly ocean]

this part

I just don't know how to use this template into that question.

this was my answer but it was rejected

[chilly ocean]

tutor said I should use this method

[fallow plume]

Gotcha, well just to be really thorough, they're both sets so we're going to say we have an object $a\in [x]$.
If $a \in [x]$ you need to show that $a$ is in the preimage of $f(a)$, which is ofc definition of preimage.
This shows that $[x] \subseteq f^{-1}(f(x))$.

Now you assume you have $a \in f^{-1}(f(x))$, which isn't necessarily $a = x$ either.
Show that $a \in [x]$ by the definition of the relation.
This implies $f^{-1}(f(x)) \subseteq [x]$.

[cloud walrus]

Thomas

[fallow plume]

@chilly ocean hope this helps a bit but feel free to followup

[chilly ocean]

yes helped a lot. thanks for sharing your idea.

[chilly ocean]

but anyways, what do you think about this answer?

[fallow plume]

I'm a bit confused because it seems like you're assuming that $[x] = f^{-1}(f(x))$ without proving it. You unfortunately can't just show that $[a] = [b]$.

[cloud walrus]

Thomas

[chilly ocean]

good point. thanks.

[fallow plume]

Np! Good luck

[past temple]

how do i show that additive functors preserve the direct sum?
given the definition of additive functors given here

[thorn delta]

yea it should suffice to show that F(M oplus N) has the universal property of F(M) oplus F(N)

[dense mulch]

I have a super quick question if I can PM someone? Don't want to interrupt this discussion but it's kinda timely lol (it's in the scope of intro to abstract lol)

[thorn delta]

prob just ask, there's not much discussion going on atm

[past temple]

wait is the universal property of F(M) oplus F(N)

the property that like

includions and projections

compose to be the identity

etc

?

[thorn delta]

nah, universal property of the direct sum

[past temple]

like

this one?

[thorn delta]

ye

[past temple]

this thing doesnt use like

projectiosn at all though

its only inclusions

[thorn delta]

right

direct sum is characterized by inclusions into it

[lethal cipher]

@chilly ocean Just a quick addition. The terminology is a bit off.
Equivalence classes are NOT the same as a partition. But, the set of the equivalence classes on a relation form a partition.
It is a small, but notable difference.

[dense mulch]

I want to show
φ: ℤ -> ℤ
x |-> -x is an automorphism.
I’ve shown the bijectivity of φ but need to show that it preserves products. How would I do so? Is the operation just multiplication because we’re in ℤ? I’d assume there’s more to that because of the exact operation I’ve chosen

[lethal cipher]

Are you talking a ring automorphism?

If so, it isn't true

In any ring automorphism, 1 must always map to itself (this is part of the definition)

[tribal moss]

It's a group automorphism of (Z,+).

[lethal cipher]

In (Z,+), then it is a group automorphism and products dont matter

[tribal moss]

Or it could be a module automorphism over Z.

[thorn delta]

(this might make things easier actually, but basically, the idea is that you want to show that the injection/projection maps in the image of F satisfy the same relations that characterize F(M) oplus F(N) up to unique isomorphism)

[lethal cipher]

I am not familiar with what a module automorphism is.

[past temple]

ok yeah i have that

thats what i was talking about

the images of the inclusion and projection maps

satisfy the properties

[tribal moss]

Then that's probably not the interpretation you want to show. :-)

[past temple]

that you would expect

of inclusions and projections

and like

i get the logic that like

the direct sum

is the unique (up to isomorphism) object

such that this should be true

but idk how to show it

@thorn delta

like this definition right?

[lethal cipher]

Well, tbf, it wasn't my question :p

[tribal moss]

Whoops, apologies.

[lethal cipher]

But the way they are presenting their question makes me think they are being introduced to homomorphisms/isomorphisms.
Module homomorphisms seem a bit advanced.

[dense mulch]

Oh yes, I meant group automorphisms

So the operation is addition then?

Is that because x -2x = -x? Because we’re working in Z? Or neither

[thorn delta]

yeah. and if A is some R-module with maps p_k, i_k satisfying all of those things, then A is the biproduct of A1, A2, ..., An

[subtle ivy]

so i usually like to prove all the theorems i come across in texts myself, but i notice that for some theorems in algebra, that is sort of hard to do. for example, i thought the "proof" of the classification of rotation groups in R^3 was sort of unmotivated

im just curious if anyone else feels this way sometimes.

[woven delta]

To expect yourself to be able to do that in general is kind of silly tbh, and can actually be kind of a hindrance to your progress

At least in my experience

[tribal moss]

It's still a good idea to attempt proofs of each claim oneself, if only to get a more vivid feel for what is simple and what is deep.

[tardy scaffold]

i love fields

[lethal cipher]

No, it is because Z is not a group under multiplication?
Is the inverse of 3,4,5,... in Z?

When we talk about Z with respect to groups, it is always referring to addition. This is true for R,Q, and C (0 has no inverse).

[lethal dune]

is R/Q × Q isomorphic to R, +?

it might actually be true if we consider it as a VS over Q

[next obsidian]

Pretty sure as a Q vector space it’s just already iso to R

Something something I think it’s uncountable dimension over Q

[lethal dune]

so isomorphic? can't think of why it should not be but still unsatisfactory

like one of those C \symeq R

[rustic crown]

it's the standard vector space thing that every exact sequence splits
0 --> W --> V --> V/W -->0
we start with a basis of W and extend it to a basis of V, now everything other than the basis of W spans a subspace of V isomorphic to V/W

[lethal dune]

so this gives R/Q ~ R

[rustic crown]

that is true... but it gives V = W (+) V/W

if you forget about it and think how would one show that any two vector spaces are isomorphic, all we have to do is show that they have same dimension

of course this isomorphism will be very whacky

R is isomorphic to R x Q

[lethal dune]

yeah, can't hope to find a good looking one anyway

[rustic crown]

finding a complement is equivalent to giving a one-sided inverse of W --> V. so a map V --> W such that W --> V --> W is identity.

[rustic crown]

there are loads of choices for this, and if there is a nice looking one, then we're heppi

[small karma]

how do i prove or disprove this

[rustic crown]

is D2 a thing?

[lethal dune]

maybe count the elements, idk

[small karma]

idk lol cant find anything in book or even online. this is a question from a last year test at my college .

having trouble with external product of Dihedral group in general

since D group isnt even isomorphic to Z

idk how to proceed in this

[lethal dune]

is there a way to re arrange semi direct prods??

[rustic crown]

i'm assuming D_2 is yet another name for Z_2 x Z_2... but in that case don't both sides have different sizes

left is (3 * 4) * 4 and right is (12 * 2) * 4

[small karma]

yeah

but its a 9 marks question for some reason

so im assuming there should be something more to it

[lethal dune]

I think D2 is just Z2

[hidden haven]

D2 should have 4 elements

Or maybe not given the cardinalities in the problen

[lethal dune]

I can't think of 4 different action on a line

[hidden haven]

I was thinking of defining it by the same relations as Dn

[lethal dune]

[delicate bloom]

yeah it has 4 elements

[lethal dune]

yeah, D2 is something else lol

[chilly ocean]

quick advice question, if there is any algebra book worth memorizing, which one should it be.

[delicate bloom]

rotating 180 degrees and reflecting

[hidden haven]

The polygon symmetry thing is just visualisation sort of

[chilly ocean]

like memorize in entirety

[lethal dune]

but both are same?

[delicate bloom]

no, one reflects keeping vertices fixed in line of reflection

rotation exchanges vertices

[rustic crown]

think of the line having two sides, top side is colored red and bottom is blue

[hidden haven]

But in the problem they are asking if the groups are isomorphic so I'd assume that they'd at least have the same cardinality

[lethal dune]

I was thinking of one edge connecting 2 points

[delicate bloom]

maybe that's the proof that they're not isomorphic

[hidden haven]

Think of Dn as < r, s | s², r^n, rsrs >

[rustic crown]

one could argue that reflection is identity >.<

[delicate bloom]

well the 'top' and 'bottom' side exchange

[lethal dune]

yes that's what I was thinking

[delicate bloom]

like, I realize a digon doesn't exist but that's how you picture it with like a bent sides

[lethal dune]

but generator relation gives 4

[delicate bloom]

yeah

think of D_n

now set n=2

buh-leef

[hidden haven]

Because generator relation thing things of reflection as non identity

[lethal dune]

but you think of them in terms of polygon and suddenly it's Z2

[delicate bloom]

the "polygon" with 2 sides is not a line segment

[hidden haven]

The polygon thing is an action of D_n on the polygon

[hidden haven]

When n ≥ 3 it is faithful, else not

Always surjective

[delicate bloom]

think of D_2 as acting on this shape

https://upload.wikimedia.org/wikipedia/commons/2/2b/Digon_on_circle.png

[hidden haven]

Oh nice you could do this with every Dn

[delicate bloom]

yeah

[hidden haven]

Except D1 😫

[lethal dune]

Thinking Dn analytically

[rustic crown]

D_1 as well >.<

only flips

[small karma]

Express G = Aut(U(81)) as a direct product of cyclic groups. Find the number elements
of order 6 in G.

[delicate bloom]

|D_n| = 2n is the hill I will die on

[rustic crown]

D_0

[lethal dune]

D_-1

[hidden haven]

You'd have to color the different parts differently even though they lie on the same segment

[small karma]

in this ques i gotta turn u into z first then use the aut(z) = u(n) formula rightt

[hidden haven]

Or are you thinking of some other way

[delicate bloom]

a mobius side

[hidden haven]

Wait

I guess it makes sense without colouring

Does it

[small karma]

then turn it again in to cyclic form and find the number of elements

[delicate bloom]

it still is like a coin with a front and back

[lethal dune]

but for Dn n>2 we don't think of edges going forward and backward with a different color, otherwise we'll get another set of rotation of the edges?

[delicate bloom]

so instead of 'reflection' you can still do the flip in R^3

[lethal dune]

like then another xZ2 factor?

[delicate bloom]

but rotation 360 degrees for D_1 is the identity

that's how you can visualize D_1 having actually 2 elements legitimately

[lethal dune]

yeah wiki says the same

[delicate bloom]

like on all odd sided polygons, draw a line through a vertex and the center of the opposite edge, then rotate it 180 degrees in 3D space -> that looks like the reflection in R^2

[lethal dune]

so I'll have believe

[small karma]

one more ques

should i take it as g, + or g , x

its not specified so im confused

isnt it supposed to be specified

whether its for additive mod or multiplicative mod

[paper flint]

Do you have extra conditions on G here?

[small karma]

no this is the whole questioin

[paper flint]

For an arbitrary group, you would just consider its group operation

[small karma]

yeah multiplicative seems easy for this

[paper flint]

Sure

That's just notation

[small karma]

so i was able to transform this into z2 x z18 but since order isnt relatively prime i rearranged it into z4 x z9

and was able to find 2 elements of order 6 after that

am i correct?

[hidden haven]

Z2 x Z18 is Z2 x Z2 x Z9

you can't combine this into a Z4 x Z9

[small karma]

but

then it wont be cyclic?

[hidden haven]

Z2 is cyclic

[small karma]

bcz order of z2 and z2 is not relatively prime

[hidden haven]

It says product of cyclics

not cyclic

[small karma]

oh

right

but then i gotta find 6 order element in terms of lcm (a,b,c)

[hidden haven]

epic pfp btw 😎

[small karma]

considering z2 z2 and z9

[hidden haven]

yeah probably

[small karma]

[small karma]

btw why is that?

[hidden haven]

in general Zm x Zn is isomorphic to Zmn iff m and n are coprime. You can show that
the element with the largest order has order the lcm of m and n

which is mn iff gcd is 1

[lethal dune]

can anyone check these facts for me |U(81)| = φ (81)=3³2 and |aut(U(81))| = ϕ (3³2)=3²2•1?

😣

[small karma]

so 4 and 9 are coprimes. but they will form lets say z36 which has a highest order element as 36 but z4xz9 will also form a highest element of order 36 lcm (4 and 9) are 36

[hidden haven]

yes

Z4 x Z9 and Z36 are isomorphic

[small karma]

yes

oh

z2 and z18 will have a element of highest order as 18

but z9 and z4 will have a highest order of 36

[hidden haven]

yes

[small karma]

so i cant turn it into that

i see

[sly storm]

Hello. If we knew the structure of p-groups completely, then we could classify all finite groups?

[small karma]

Q-29 i know there will be 4 subgroups of order 3 and one of them will be (3,1)

But how do i find others

[rustic crown]

finding subgroups of order 3 is same as finding elements of order 3 as groups of order p are cyclic.

notice that distinct subgroups of order 3 intersect trivially. each contains 2 elements of order 3, so number of elements of order 3 = 2 * (number of subgroups of order 3)

so do you see how you can find elements of order 3?

[small karma]

Basically elements of order 3 = pi(3) * no of subgroups

But i need to find the subgroups

Idk how to find all

[rustic crown]

so let's give it a try... if (a, b) has order 3 what does this tell you about a and b?

[small karma]

One i can find by 9/3 <3> and 3/3 <1> so (3,1) is one subgroup of order 3

[small karma]

They are divisble by 3

?

[rustic crown]

divisible isn't the word.... (3, 1) has order 3 right

[small karma]

Yes

Lcm of a and b is 3

[rustic crown]

not quite...

if (a, b) has order 3, then we know 3 kill it

[small karma]

[rustic crown]

since the components of direct products don't interact with each other

[small karma]

Yes

[rustic crown]

we would get that 3 kills both a and b

[small karma]

They interact with themselves

[rustic crown]

so possibilities of a are {0, 3, 6} and for b are {0, 1, 2}

[small karma]

Yes

[rustic crown]

but both can't be 0 together

[small karma]

Yes

[rustic crown]

else (a, b) has order 1

so this gives you 3 * 3 - 1 = 8 elements of order 3

[small karma]

Yes

[rustic crown]

does this tell you what the other subgroups are?

[small karma]

(3,1) (3,0) (3,2)

And one more subgroup

But idk what

[rustic crown]

we can have 0 in the first slot as well

[small karma]

Oh yes

I see

0,1

Is that correct

[rustic crown]

(3, 0) and (6, 0)
(3, 1) and (6, 2)
(3, 2) and (6, 1)
(0, 1) and (0, 2)
these are the 8 elements

[small karma]

Yes

[rustic crown]

(elements in the same row generate the same subgroup)

[small karma]

Yes

Hence we get 4 subgroups

In col 1

[rustic crown]

yep!

[small karma]

Thanks

[rustic crown]

more precisely subgroup generated by those elements

[small karma]

Makes sense now

I'm ready for my test now

Gtg it's in half hour

[rustic crown]

good luck!

[small karma]

Thanks

[sly storm]

Hello. If we knew the structure of p-groups completely, then we could classify all finite groups?

[lethal dune]

that classification is already done I believe

[rustic crown]

i thought only simples are done

putting the simples together is far murkier

[lethal dune]

yeah I meant simples

[rustic crown]

idk if we can say anything more than like nilpotent groups

which are direct products of their sylow subgroups

[lethal dune]

I though we already know the groups upto isomorphism of order pⁿ

not just abelian but non abelian as well

[rustic crown]

woah really?

[lethal dune]

i thought so

[rustic crown]

i never went past p^3

[weak oriole]

Just 2^n gets very crazy

[weak oriole]

So idt this is true

[lethal dune]

[chilly ocean]

great. thanks for mentioning it.

[hidden haven]

Every partition can be realised as a partition into equivalence classes of some relation though

[next obsidian]

No we don’t

There’s a book on groups of order p^n

That’s is humongous

And it goes up to… n = 6? 7??

[rustic crown]

[lethal dune]

good lord

[next obsidian]

Okay so the book I’m thinking of is 2^n

And this was like pre-computer widespread usage

But it goes up to n = 6

And is like 170 pages

I think there’s some 4 billion groups of order 1024

Or something like that

So we super do not know them

[rustic crown]

didn't expect groups of order 64 to be this much complicated

170 pages!!

[next obsidian]

Lot of variance there

I think even from abelian groups there’s quite a lot that can go on

Something like, number of partitions of the integer 6 is what’s going on

Anyway like for what we know, almost every group is of order some power of 2

I’m sure asymptotically it goes to 100%

Well

I bet

https://math.stackexchange.com/questions/241369/more-than-99-of-groups-of-order-less-than-2000-are-of-order-1024