groups-rings-fields

[bleak abyss]

@prisma ibex look at this bullshit

[brisk granite]

wait, but what does it mean

[bleak abyss]

Yeah he's trying to define A_n

It's something like number of involutions mod 2 I imagine

[brisk granite]

I'm still confused

[prisma ibex]

wow lmao

yea Fraleigh is really not great

[brisk granite]

but what does he mean tho?

by the orbits thing

what is orbit(s) equal to?

[bleak abyss]

Eyy, sorry I had forgotten, so I read through Fraleigh and he might be less braindead than I thought

Okay so

If you give me a permutation, it partitions the set it's acting on into orbits

[marble wagon]

I mean

it's orbits(s)

which makes complete sense

not orbit(s)

which doesn't

[bleak abyss]

See the thing is he's using this to define the alternating group

Which turns out to work but threw me off at first sight

[marble wagon]

yeah that works fine

orbits(s) is fine notation, removing the s turns it into garbage

[bleak abyss]

Since I had mostly seen it as counting inversions mod 2 or really the correct way which is \prod (sigma(i)-sigma(j))/(i-j)

[marble wagon]

embed it into O(n) and pass to SO(n)

[bleak abyss]

Lel

I guess you only really need determinant over R to define sign

And something something product of eigenvalues

I guess over C really

[marble wagon]

over \bar{Q}

[bleak abyss]

One of these days I need to rename the number theory channel to "Representation Theory of \overline{Q}", still feels ever so slightly awkward with Galois theory

[marble wagon]

i was supposed to learn number theory last semester

but nope

it was my weakest class

I didn't learn much

shame

[bleak abyss]

Yeah soon I'd like to get down and do it

For the summer my immediate priority is algebra and probably AT

[marble wagon]

what I need to do is just go balls deep into AG

[bleak abyss]

Once I actually get back to doing some math... My mom's knee is hurting a lot right now and as a result I kinda need to get everything

[marble wagon]

oh

shame

[bleak abyss]

Yeah, meniscus tears and mild arthritis. It really messed her up

Between that, soon I'm gonna have to learn to drive, repair some of the damage that bad sleep and a soda addiction has done... I don't have much time to do things I like :/

[marble wagon]

that's rough

you should try to get some rest before grad school

[full blaze]

Orbits are sets , cycles are permutation

[smoky cypress]

If I have a finite set with a binary operation represented by a cayley table

If I want to show the binary operation is associative, do I have to try out every combination of 3 elements?

[full blaze]

ye looking at a group table and determining it is associative is a pain in the ass

you look at the main diagonal and if there is symmetry on each side then the group is abelian

i would just check associativity of the group its self not by just looking at the table if possible

[prisma ibex]

yea typically you verify associativity directly from the definition of your group operation instead of appealing to the Cayley table

unless your group operation is defined from the Cayley table to begin with, in which case lol

[smoky cypress]

Ugh

In the problem there are 4 tables, each table is a binary operation with a group of order 4

And now I have to check whether the operation is associative

Wait if does it help if the group is abelian?

[prisma ibex]

I mean it halves the number of things you need to check

or something like this

but it's still tedious

[tame bear]

https://en.wikipedia.org/wiki/Light's_associativity_test

honestly i would just openly assume associativity
if its not associative then theres no reason to study it

[smoky cypress]

The problem is determine if the binary operation is a group

I googled it and I found out that, up to isomorphism, there are only two groups of order 4

xD

[full blaze]

in addition every group of order 5 is abelian as well

but ye

[smoky cypress]

?

What do you mean 5?

[full blaze]

every group of order 5 is abelian

[mild laurel]

the fuck are you talking about

he clearly stated that his table was 4 by 4

[tame bear]

i feel like ive seen this exercise in the past but just skipped it

is this from a book

[smoky cypress]

Yeah

[gloomy dirge]

Let $ f \in \mathbb{C}[t]$. Find the Jacobson Radical $\frac{\mathbb{C}[t]}{(f)}$

[cloud walrus]

Raziel:

[full blaze]

all i know is that C[t] is the set of polynomials with complex coefficients of indeterminate t

[mild laurel]

Then why even respond??????

@gloomy dirge what are you confused about

[marble wagon]

it's the intersection of maximal ideals

that's the easiest way

[gloomy dirge]

I cant resolve in quotient ring. I know each definition and proposition, but i cant see a solution in quotient ring. Sorry for my english...

[marble wagon]

what are the maximal ideals of C[t]/f(t)?

i) characterize maximal ideals of A/I in terms of the maximal ideals of A
ii) what are the maximal ideals of C[t]?
iii) use i,ii and the fact that the jacobson radical is the intersection of maximal ideals to find J(C[t]/f)

[chilly ocean]

do you not know what ideals are?

[stone fulcrum]

@smoky cypress
Note that no element should be repeated in any line or column on a cayley table, meaning it's very quick to make the groups of order 4

You just have to check if you have a match

[smoky cypress]

Ok

But the converse is not true right?

Or is it?

[full blaze]

what kaynex said is called the latin square property

[stone fulcrum]

No it isn't. But it is with order 4

@smoky cypress

[smoky cypress]

Oh ok I see

[full blaze]

yes it is

stand corrected buddy

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

so, I think I made a proof for the fact that any permutation can't be a product of an even and odd number of transpositions. Does this look correct? Does it make any sense?

[fickle brook]

a little rough around the edges stylistically but that's actually a nice argument! i hadn't thought of it myself before

[simple valley]

what are you trying to show?

[brisk granite]

How do I make it less rough around the edges?

Oh, should I phrase my answer more like the textbook did?

[rain crescent]

Need some help pls.
Translation: Prove the group (G,*) is abelian if and only if that identity is true for every g and g' in G

[full blaze]

g is abelian under the binary operation if a * b = b * a for all a and b in G

[rain crescent]

dont I need to prove both sides of the implication since it is an if and only if?

[full blaze]

generally in abstract you show a property is 2 sided like inverses, commutativity, identiy

[rain crescent]

so this is my attempt to prove that if the identity holds then the group is abelian, is it correct? I feel like its not valid for some reason

sry for bad quality btw

now I was supposed to prove the opposite ( that if it is abelian the the identity holds)

am I correct in thinking that_

?

[raw moth]

don't think so

you want ghgh = gghh

[prisma ibex]

^

(gh)^2=g^2h^2 iff ghgh=gghh iff hg=gh by left and right cancellation

boom 1 line proof

[brisk granite]

How do you know whether you're supposed to use the tombstone or qed? Does it matter?

[full holly]

Do the filters on a boolean algebra, B correspond to (possibly a subset of) the chain-complexes on the modules over the ring generated by B in any meaningful way?

[chilly ocean]

@brisk granite ?

[stone fulcrum]

Tombstone and QED are generally the same thing

[full holly]

module*

[brisk granite]

I'm finding it rly hard for me to prove the splitting criterion for the conjugacy classes of S_n

any hints or clues on how I might proceed?

[full blaze]

what kind of abstract course are you taking seems quite all over the place

[quartz zodiac]

https://math.stackexchange.com/questions/404656/splitting-of-conjugacy-class-in-alternating-group

the first line of the first answer is a good hint

[brisk granite]

yea, I think I'm just gonna read the whole thing

[royal shadow]

In Hartshorne he says that "we let v :K --> Z be the valuation corresponding to the discrete valuation ring R where K is the quotient field", how is v defined?

p.130

[hot lake]

is the quotient field the field of fractions of R or is it the quotient R/m ? it doesn't really make sense on R/m

[valid umbra]

hartshorne's old so it means the ring of fractions

[clear obsidian]

simple question but im new to abstract algebra, generically speaking, how do I find how many Albean groups of order n (n in N) there are?

[mild laurel]

This is done using the Chinese Remainder theorem

You prime factorize n

[clear obsidian]

ok thankyou

[somber bramble]

Could someone please doublecheck that this argument is rigorous?

We have $d=2$, as $A_8 < S_8$ with index $2$

Now, let $H$ be a subgroup of $S_8$ of index $2$. Then $H$ is a normal subgroup of $S_8$ (known statement). Now, by the second isomorphism theorem, we have $A_8 H / H \cong A_8 / (A_8 \cap H)$. But $A_8$ is simple (known statement) and therefore $A_8 \subset H$. But as $|H| = |A_8|$ by Lagrange, $H = A_8$.

[cloud walrus]

Sascha Baer:

[bleak abyss]

This works yeah

There's actually a nice argument that shows A_n is the only index 2 subgroup of S_n for any n (your argument requires n≥5)

Namely, if you have a homomorphism from a group G to an abelian group A, conjugate elements have to map to the same thing

In particular, if you have a map from S_n to Z/2, transpositions must either all go to 1 (trivial hom) or must all go to -1 (sign hom)

This also uses the fact you used about index 2 subgroups being normal

[fickle brook]

all to -1

til -1 * -1 = -1

[bleak abyss]

Transpositions all go to -1

Product of two transpositions isn't a transposition

[somber bramble]

nice, thanks

could I have any pointers as to where you could even start on b) and c)?

double brackets standing for power series

a) is easy: it’s a UFD because Z is one and there’s that one theorem, I think it was by gauss? that we covered that says R UFD ⇒ R[x] UFD. and I know a counterexample to it being a PID
d) is easy because that’s just ℂ
but the other two I’m lost as to where even start

I don’t remember any theorems about the ring of power series nor about a generic quotient like that (that’s the split-complex numbers isn’t it?)

[hot lake]

well you could try to see who are the invertible elements

[somber bramble]

for the power series, I already know them: all the power series with a₀ ≠ 0

[hot lake]

yes, so what do the ideals there look like ?

[somber bramble]

hm, well, if I have a power series which isn’t a unit, then I can rewrite it as Xⁿ*u, where u is a unit, and so the ideal generated by it would be the same as that generated by Xⁿ, right?
so (p,q) would be the same as (Xⁿ, X^m) = (X^(min{n,m})?

does that work out?

that would make it a PID

and then also a UFD cause every PID is a UFD

[hot lake]

yes

[somber bramble]

aight, I’ll have to think thorugh the details a bit more but I think I get it
which leaves us with R[x]/(x²-1)

[hot lake]

yeah you have to figure out R[x]/(x²-1)'s secret identity

[somber bramble]

I mean it’s the split complex numbers, but I know like, no properties of them

I could just google them but there’s no fun in that

[hot lake]

have you seen the chinese remainder theorem ?

[somber bramble]

I feel like they wouldn’t be a UFD even but I can’t think of an example

I have

[hot lake]

it's not a UFD for a very big reason

[somber bramble]

lemme try to think what the CRT states again

it’s one of those theorems that I lose track of very quickly

wasn’t it just like... let $N = n_1 n_2 \cdots n_m$ be natural numbers and $R$ a sufficiently nice ring (a PID I think?), then there exists an injective ring hom $R^N \to R^{n_1} \oplus \dots \oplus R^{n_m}$

now lemme read up what it actually is

[cloud walrus]

Sascha Baer:

[somber bramble]

isom, not just hom

and nothing with powers of rings

but rather quotients

which makes way more sense

I confused myself with the one theorem from modules over PID

aight, so

$R/(N) \cong R/(n_1) \times \dots \times R/(n_n)$

[cloud walrus]

Sascha Baer:

[somber bramble]

ah, so $\mathbb{R}[X]/(X^2 - 1) \cong \mathbb{R}[X]/(X+1) \times \mathbb{R}[X]/(X-1)$

[cloud walrus]

Sascha Baer:

[hot lake]

yes

[somber bramble]

now I can look at either of those on their own and if either is not a UFD then the product can’t be either

[hot lake]

there is a much more glaring issue when you have a product of rings

[somber bramble]

they’re not even domains, right

[hot lake]

yes

[somber bramble]

cause (0,a) times (a,0) is 0

[hot lake]

yep

also, R[x]/(x+1) and R[x]/(x-1) are both isomorphic to R

if you were wondering

so the ring in c) is isomorphic to R²

if you have a polynomial P you look at P(1) and P(-1), and that gives you an isomorphism

[somber bramble]

so, I know this statement is false, but I need a counterexample and am failing to come up with one (nor can I find one on google): Every group of order n is isomorphic to a subgroup of S_n

the first time it could fail is for n=6 (I’ve checked the n=4 case by hand, and for prime numbers it’s pretty obvious)

but for 6 there’s also only two groups, S3 (clearly a subgroup of S6) and Z/6Z (also clearly one)

actually, I’m not so sure anymore that it is false

A acts transitively on itself. So A is a subset of the symmetries of the set A, which is isomorphic to S_|A|. And since A is a group, it’s actually a subgroup of S_|A|. Is this correct?

[sharp sonnet]

isn't this cayley's theorem?

every group G is ismorphic to a subgroup of the symmetric group acting on G

[oblique river]

yes that is correct

[somber bramble]

but isn’t that “of some symmetric group”?

(not necessarily S_n, when |G|=n)

ah, I see

[oblique river]

I was responding to you when you said "is this correct"

what you wrote is good

[somber bramble]

excuse my sudden surge of quesitons btw, we got the questions catalogue for the algebra exam :P

ideally I should know how to answer all the questions (most are easy, but then some are not) and be familiar enough with the things I use that I could answer some followup questions

(there’s 50 questions on algebra I and another 25 on algebra II, those being “rings, groups and five minutes on modules” and “galois theory”, respectively)

[oblique river]

i mean questions are what this channel is for

[bleak abyss]

Questions catalog for exams?

I imagine they're stricter as a result

But wow

And yeah it's a nifty result for sure

[somber bramble]

I mean it‘s a catalogue of 75 questions (but then the exam is more than a month away so I can actually prep quite well)

in topology they also just released it, there it’s a catalogue of over 150 questions

plus a huge table of properties and whether they are stable under certain operations, and counterexamples of things that don’t hvae them; we’re not supposed to memorize them per se but some are marked as “good exercises” that might be asked (and some are marked as “this is way too hard here’s some references” and one is literally an open question)

this is page 1 of the questions catalogue for algebra

and this from topology

as you can see, the difficulties vary a lot

but I suppose if you get easy questions you’ll have to answer more or get more follow-ups off the catalogue

fuck

[stone fulcrum]

Lol. Do they expect you to know that?

[somber bramble]

apparently

I did not expect it and was not going to learn it, but the prof really really liked it

so I guess I shouldn’t be too surprised

[crimson badger]

note that you don't need to learn the general formula

just the depressed cubic one

cuz you can always put the cubics in depressed form

by telling them that their dad will never love them

[stone fulcrum]

Indeed, all you need to know is how to get the quadratic after the transformation

[somber bramble]

yea I’ll just have to study the thing

it’s not actually that difficult but I would’ve never invested time into it if it wasn’t on the catalogue

[old lava]

where's a good place to learn abstract algebra for a high schooler?

I'm trynna learn algebraic number theory and galois theory. I know the definitions in abstract algebra (like rings, groups, fields, ideals, quotient rings, splitting fields, etc), but I have trouble using them to build upon concepts

[topaz solar]

Well, you might wanna get better at using your normal structures first before touching fancy number theory or galois

uh, do you know linalg?

[old lava]

yea

[topaz solar]

Hm, you might could use Artin's algebra

[old lava]

ok, I'll check that out, thank you!

is it this?

https://www.amazon.com/Algebra-Classic-Classics-Advanced-Mathematics/dp/0134689607?SubscriptionId=AKIAILSHYYTFIVPWUY6Q&tag=duckduckgo-d-20&linkCode=xm2&camp=2025&creative=165953&creativeASIN=0134689607

[stone fulcrum]

I personally used Fraleigh and liked it, there's a number of algebra books to use

https://www.google.com/url?sa=t&source=web&rct=j&url=https://pdfs.semanticscholar.org/1963/b086c84905c94b72f3ef8ff6522b82579f6d.pdf&ved=2ahUKEwidw_-a65TjAhUbU80KHdVwBgQQFjABegQIBRAB&usg=AOvVaw3wq5KN166E1pGTWdhX73RZ&cshid=1562022515293

Free online

[topaz solar]

@old lava that is the one I was thinking of, but kaynex's thing also is nice

[old lava]

ok, thank you for the help!

[bleak abyss]

Use Lang

[prisma ibex]

^

Use Lang or Dummit and Foote

literally anything but Fraleigh unless you're a high schooler

in which case it's a great introduction to abstract algebra

[bleak abyss]

"where's a good place to learn abstract algebra for a high schooler?"

This guy is a high schooler, my Lang suggestion was tongue in cheek

I bought Jacobson actually and it seems good thus far

Dummit and Foote is literally the cure for insomnia

[stone fulcrum]

I am not smart, and I needed something as slow as Fraleigh for an intro.

Not everybody needs it. Just another recommendation

[bleak abyss]

From the epsilon amount of time that I glanced at Fraleigh I don't know if I like it among the slow ones

Mostly the way he constructed A_n made me angry

[stone fulcrum]

:o Any other good examples for gentle intros to abstract?

[bleak abyss]

There was some question someone asked here and I read through a bit of Fraleigh and was tempted to shoot him in the lung

Lol I dunno any good ones, it's more like "This can't be optimal because optimal ought be good and this isn't"

I liked Herstein I guess, though it has less content

"Topics in Algebra"

And also it does function composition as x(f) instead of f(x) which is philosophically the correct answer but is non-standard and thus inconvenient

[stone fulcrum]

I wish composition was left to right but that's just not the way the world works

[tame bear]

Saracino isnt bad

atleast for groups

[bleak abyss]

Never even heard of that

I've heard the names Pinter and Gallian before but I can't testify for either

Really I think Herstein was cute but Artin is probably the correct book with which to learn about algebra, especially if you're not a priori interested in it

Since it apparently situates algebra within other parts of math

Also it does linear algebra if you're rusty on that/haven't had a correct linear algebra class

(Tbh it might even be fine to learn LA from scratch with since it seems to spend non-trivial time on it, but I'm not sure)

[tame bear]

haven't had a correct linear algebra class
yeah that

[uncut girder]

Saracino :D

The one I used

In highschool

Was awesome

[brisk granite]

the coset $aH$ is only subgroup when $a \in H$, right?

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[uncut girder]

Yea

[brisk granite]

k

[uncut girder]

It needs identity

And identity is only in H

[brisk granite]

what?

wdym

[tame bear]

aH needs the identity to be a subgroup

[brisk granite]

yep

[tame bear]

if a isnt in H then a^-1 isnt either
and so aH doesnt have aa^-1 = e

[brisk granite]

oh ok

[full blaze]

its trivial

[uncut girder]

@full blaze you're trivial

[full blaze]

[uncut girder]

[brisk granite]

[sonic current]

guys - would this be appropriate for universal algebra questions?

i'm torn between this and category theory

this should be a very simple question anyway: I want to prove an ideal subset of a lattice is a sublattice

but having a little trouble proving the infimum of x,y in ideal I is in I

using the down-set hypothesis

basically i'm proving the ideal is a sublattice by the definition of ideal

[magic owl]

What is the relationship between inf {x,y} and x?

@sonic current

[sonic current]

@magic owl yeah, inf{x,y} ≤ x

I see your point, I think

if x ⋀ y is x or y, then it will be in the ideal

if x ⋀ y = z, then z ≤ x and z ≤ y and then z = z ⋀ x or z = z ⋀ y; thus, z is in I

[fickle brook]

you don't need that casework

you can just say inf{x,y} ≤ x and x ∈ I, and therefore inf{x,y} ∈ I by (i)

[sonic current]

alright

thanks a lot!

[brisk granite]

Im trying to prove the orbit-stabilizer theorem. does this make sense?

The mapping $f: G/G_x \rightarrow \mathcal{O}(x)$ shall be defined such that $f(gG_x) = gx$. It must hold that that $f$ is surjective because for any $y \in \mathcal{O}(x)$, there exists a $g$ such that $gx = y$, and, since the coset $gG_x$ - or an equivalent coset - is in the coset space, $f(gG_x) = gx = y$. $f$ must also be injective because $f(g_1G_x) = f(g_1G_x) \implies g_1x = g_2x \implies x = g_1^{-1}g_2x$. Hence, $g_1^{-1}g_2 \in G_x$ and $g_1G_x = g_2G_x$.

wut?

what's going on with latex?

$beans$

[oblique river]

not reading #changelog 🤦

[full blaze]

[brisk granite]

can you guys understand my latex code?

[full blaze]

no

[brisk granite]

$pls work$

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

oh nice

The mapping $f: G/G_x \rightarrow \mathcal{O}(x)$ shall be defined such that $f(gG_x) = gx$. It must hold that that $f$ is surjective because for any $y \in \mathcal{O}(x)$, there exists a $g$ such that $gx = y$, and, since the coset $gG_x$ - or an equivalent coset - is in the coset space, $f(gG_x) = gx = y$. $f$ must also be injective because $f(g_1G_x) = f(g_1G_x) \implies g_1x = g_2x \implies x = g_1^{-1}g_2x$. Hence, $g_1^{-1}g_2 \in G_x$ and $g_1G_x = g_2G_x$.

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

Im trying to prove the orbit-stabilizer theorem. does this make sense?

[full blaze]

hmm i didnt learn that in abstract

😢

[brisk granite]

ok

[magic owl]

@brisk granite that looks ok to me

You might want to argue that your mapping is well defined

Ie g’Gx=gGx implies gx=g’x

[full blaze]

valid point max, ye when dealing with mappings with cosets as inputs it could be not well defined

[left copper]

in the strict sense, you must always show that your mapping is well defined

[brisk granite]

Note that the function $f$ is well defined because $gG_x = g^{\prime}G_x \implies G_x = g^{-1}g^{\prime}G_x \implies g^{-1}g^{\prime} \in G_x$. Thus, $g^{-1}g^{\prime}x = x \implies g^{\prime}x = gx$

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

does this look ok?

Also, I've heard statements like two groups are the same "upto isomorphism". what does that mean?

Also, why is it that groups of order 4 must be isomorphic to $\mathbb{Z}/{4Z}$ or $\mathbb{Z}/{2Z} \times \mathbb{Z}/{2Z}$?

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

why not both or neither?

[full blaze]

such that there is a homomorphism between the groups and a bijection

check it

[brisk granite]

ah, so, it just means that there exists an isomorphism

[full blaze]

the gcd of (2,2) is 2

[brisk granite]

ok?

[full blaze]

therefore by theorem that direct product is non cyclic

but is abelian

do you know a non cyclic group of order 4 that is abelian

2 groups one cyclic one non cyclic cannot be ismorphic

[brisk granite]

ah, I see

ok

[full blaze]

so Z_2 x Z_2 is isomorphic to V_4

klein 4 group

[bleak finch]

Which one is correct?

Let H <= G. If [G : H] is a prime power of the smallest factor then H is a normal subgroup of G.

Let H <= G. If [G : H] is the smallest prime factor of |G| then H is a normal subgroup of G.

[bleak finch]

Help

[solemn hollow]

I've unlocked this channel by starting group theory

[full blaze]

attah baby

[uncut girder]

@full blaze call me baby

[full blaze]

its time to learn some lie theory

[uncut girder]

You know diff geo?

[full blaze]

not a lick

im studying for my final in abstract right now, for tomorrow lul

ring theory cuh

[uncut girder]

Oh cool

Summer classes?

[full blaze]

ye

i have a math professor i chat with weekly who specialized in noncommutative geometry

max planck institute fellow

[uncut girder]

That's crazy

[full blaze]

russian mad man

[golden pasture]

wew noncommutative geom

[full blaze]

gonna ask my current abstract prof if he can prep me for grad algebra 1

cuz hes a boss as well

[uncut girder]

Are you gonna be a theoretical physicist

[full blaze]

leaning more to a mathematical physicist

[uncut girder]

Oh I thought they were the same

[full blaze]

sure

[golden pasture]

arent they the same kinda

[full blaze]

[topaz solar]

Well, monic means the coefficient of the highest degree is 1

[bitter mauve]

yup sorry

it said leading coefficirnt 1

so i thought monic meant sth like x^2 + x + 1

all coeffs 1

[topaz solar]

nah fam

[bitter mauve]

so i got confuaed

ok thx0

[topaz solar]

monic just means x^2 + bx + c

(or similarly ofc)

[bitter mauve]

yes

thx

[wind steeple]

@bleak finch the 2nd is correct and called Frobenius theorem. Idk for the first which is a generalization

[oblique river]

@bleak finch the second one is correct. you should be able to easily construct a counterexample to the first statement

(think of groups of order 8 and subgroups of index 4)

[brisk granite]

https://cdn.discordapp.com/attachments/268882317391429632/596835920771284993/unknown.png

What is A(S)?

what does it mean?

[oblique river]

that's not standard notation as far as I know

[brisk granite]

It's from herstein if that helps

[oblique river]

surely that book defines A(S) before this

just look through the previous pages until you find it

[tender mist]

I think A(S) is the symmetric group of S and the homomorphism mentioned is the permutation representation of G on S associated to the (canonical) action of G on the set of right cosets of H

[oblique river]

based on the context that is my guses as well

but it shouldnt be that hard to find where herstein defines that notation

[tender mist]

Out of context alternating group of S would've seemed more likely, maybe

[bleak abyss]

Maybe based on the theorem we can reverse engineer it

Oh

A(S) = Sym(S)

Herstein ples

[brisk granite]

yep, that's what it is

He explicitly states it previously in the book

[fickle brook]

so it's the group of all bijections S -> S then

with composition as your operation obviously...

[brisk granite]

I'd like help starting out on this

I'm not sure where to begin

umm, if I could figure out how to create a subgroup of order p^{k+1} from subgroups of orders p^{1},p^2, p^3, p^4,. . . p^k

then strong induction could work

is that even possible tho?

[frank socket]

Somewhere along the line with k = n + a, where a>0? Idk

[brisk granite]

wdym?

[frank socket]

https://math.stackexchange.com/q/910257/193779

[mild laurel]

Can you first show that the group must have a subgroup of order p?

Giving answers is not teaching

[frank socket]

This may help?

[wind steeple]

Yup you can use induction by quotienting you group with a normal subgroup of order p @brisk granite

[clear obsidian]

Is there any easier way of determining the order of <A> than just computing a bunch of products? Based off my calculations the result seems to be infinity but I’m not sure

[wind steeple]

If A is finite ordre you have n such that A^n=I2

The characteristic polynomial of A is X^2 + X + 1

So if A is finite order, 3 divides the order

Because the minimal polynomial of A muste be X^2 + X + 1 (A is real)

And X^2 + X + 1 divides X^n-1

(X^2 + X + 1 = (X^3-1)/(X-1))

A^3 works I think

Maybe I'm wrong but A is a diagonalizable matrix in C of eigen value j and j^2 where j = e^(2ipi/3)

So A^3 ~ I2

[brisk granite]

I think I figured out an answer! Yay

[brisk granite]

For all p-groups, we know there exists an element with an order of the form $p^k$ where $1 < k < n$. Thus, the cyclic group generated by this element has an order of $p^k$. Call this group $H$. We can use strong induction to prove that there exists a subgroup of $G$ with an order of $p^k$ for all $k$.\

Base Case: We know that there exists a subgroup of order $p^{0} = 1$ (the one containing only the identity).\
Inductive hypothesis: for a subgroup $H$ with an order $p^k$, there exists at least one subgroup, for all $g < k$, with an order of $p^g$.\

From the inductive hypothesis, we've already covered all orders upto $p^k$. Now, consider the Quotient group $G/H$. All subgroups $G/H$ have orders that divide $p^{n-m}$, and, since $n-m < n$, we know that there is a subgroup of $G/H$ with an order of $p^g$ for all $g \leq n-m$ (via the inductive hypothesis). Further, since the subgroup of $G/H$ are in bijection with the subgroups of $G$ containing $H$, there must be some subgroups of $G$ with orders greater than $p^k$. Upon using the bijection $f$ defined in \textbf{Problem Eleven} (for a subgroup $K \leq G/H$ where $K = {g_1H, g_2H, g_3H,\dots, g_rH}$, $f(K) = g_1H \cup g_2H \cup g_3H \cup \dots \cup g_rH$), it's clear that for an arbitrary subgroup $K$ of $G/H$, $\left| f(K) \right| = |H||K| = p^{m+g}$. Since $m \leq m + g \leq n$, we have thus proven the existence of subgroups of $G$ with orders of $p^k$ for all $k \leq n$.

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

This was the answer I made to the question I put up yesterday I think

Does this make sense?

Is my use of strong induction correct?

Here's the question

[oblique river]

I see several issues

what is the point of the group H you define in the first paragraph? all youve done is said "there is a subgroup of some order strictly between p and p^n" which doesn't help you at all (and is also false if n = 2 let's say because there aren't any integers k with 1 < k < 2).

your inductive step is awkward and probably not what you want. I think you are trying to use that "for all p-groups of order less than p^k" or something like that, because when you try to apply it later it seems like you're trying to apply it to G/H

but your inductive hypothesis only says it for subgroups of G, not quotients of G

then in the actual proof, what is H? just something random? is it the H you started with in the first paragraph? What if H is not normal?

[brisk granite]

Thank you

I'll make these corrections

Is the proof itself flawed?

Like, the logic

[oblique river]

it's hard to evaluate because I don't know what your inductive step is supposed to be

[brisk granite]

Ok, I'll repost soon

[oblique river]

also because you haven't guaranteed that H is normal

so G/H might not even be a group

[brisk granite]

Oh yea

[oblique river]

I think what you are trying to do is on the right track though, other than those issues

and also you havent said what f is

oh wait I see

defined in some previous problem

[left copper]

wait, sorry to interupt but does Lagrange theorem has something to do with this question?

[oblique river]

yes but not really

lagrange's theorem is in the wrong direction

but it's at least a related statement

well actually I guess the statement |G| = [G : H]|H| is relevant, I would guess something along those lines is going on in the "bijection from problem 11"

[left copper]

doesn't the theorem guarantees the existance of subgroups of every order that divides groups order?

[oblique river]

no

definitely not

it says that the order of a subgroup divides the order of the group

[left copper]

yes that I know

[oblique river]

also that statement isnt even true for groups of non-prime-power order

the only thing you can say is that if p is a prime factor of the order of the group, then there is an element (and hence a subgroup) of order p

[left copper]

all of that was too much for me back then so i carried that course to this year

now I feel ready : D

[clear obsidian]

Notation question: So do we mean the same thing by G x H if G and H are groups, that we mean by S x V if S and V are just sets in the sense that

if G refers to (G, +) and H refers to (H, +) then G x H means (G x H, +) ?

[chilly ocean]

yeah i suppose

there is a more general definition though

the group operation doesn't have to be the same

the product of $(G, \circ), (H, \star)$ is $(G \times H, )$ with $$ defined as: $$(a, b) * (x,y) = (a \circ x, b \star y)$$ where $a,x \in G$ and $b,y \in H$

[smoky cypress]

It’s like the product of fractions xD

I suppose

[clear obsidian]

Thanks!

[brisk granite]

@oblique river hey, so, I was thinking about the stuff you pointed out, and I tried to improve my proof

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

does this make sense?

Is the inductive step clearer?

I really hope I've grasped strong induction correctly cuz it's my first time using it

Oh, also, the bijection f just take a subgroup of G/Z(G) and sends it to the subgroup of G that is the union of the cosets

[oblique river]

yep this is a much better proof! nice job

TeX comment: every single symbol needs to be in math mode

including all the p's in "p-group"

so you have to type ``$p$-group"

Also, general note, I would say "consider its center Z(G)" not just "its subgroup Z(G)" for maximal clarity

[cloud walrus]

Buncho Bananas:

[brisk granite]

ok, sure

I'll change it rn

[brisk granite]

quick question. Can the reflector in the dihedral group by about any line of symmetry?

[mild laurel]

Yes

As in, reflecting across any line of symmetry will give you a valid symmetry of your polygon

[brisk granite]

no, I mean, you know how the generators of $D_n$ are $\rho$ and $\tau$. Can $\tau$ be any reflection that is a valid symmetry?

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[mild laurel]

Yes that's true as well

[clear obsidian]

Ok so I have 2 questions that I cant find clear answers to elsewhere: (i) So are cyclic groups and cyclic subgroups the same thing? I've seen what seems like 2 different concepts using the exact same notation, i.e. <a> = {e, a, a^2, ... , a^(n-1)} and <a> = {... a^-2, a^-1, e, a, a^2, ... }, & (ii) is Zn = Z/nZ or just isomorphic? because I dont get how they'd be equal if Z/nZ contains a bunch of cosets while Zn contains integers...

[mild laurel]

You can show that whatever rotation and reflection you pick, you generate the whole group of symmetries of your polygon

[brisk granite]

what is Zn?

do you mean nZ?

[mild laurel]

@clear obsidian if a^n = e, then you have that a^(-1) = a^(n-1)

So they really are the same concepts

[tender mist]

Wait Zn is simply a notation for Z/nZ

[brisk granite]

oh, ok

[mild laurel]

Bad notation, but yeah it's commonly used notation

[tender mist]

For the sake of brevity, maybe

[brisk granite]

bad notation because it takes away from the fact that it's a quotient group?

[mild laurel]

Bad notation because Z_p is the p adic integers

[brisk granite]

oh, ok

[topaz solar]

@clear obsidian $C_n \cong \bbZ/n\bbZ$

where Cn is the cyclic of n element groups yes

[cloud walrus]

Darkrifts:

[clear obsidian]

ok so these include Zn ^^^ ?

[topaz solar]

well, Cn ~= Zn

they're both just cyclic groups

Groups you really don't care about a strict equality of the sets and operation, you really only care about isomorphisms

[woven delta]

I'll write Z_n until the day I die

The worst is when you're constructing Z_p using Z_n

[topaz solar]

(Granted, that isomorphism is only for Z/nZ as a group, not to be confused with the version with additional ring structure Z/(n))

kek

[woven delta]

This happened to me recently

[topaz solar]

$Z_p \cong\underset{\leftarrow}\lim ,,\bbZ_{p^n}$ or some shit right

[cloud walrus]

Darkrifts:

[woven delta]

Yeah

[topaz solar]

like who writes that shit

[woven delta]

That looks gross

I would do the opposite

[topaz solar]

tbh

[clear obsidian]

Thanks Darkrifts

[topaz solar]

isomorphism is all you care about with groups

[clear obsidian]

yeah that makes sense, so similar deal with rings then I guess?

[topaz solar]

but uh, some things (i.e. sets) you need stricter equivalencies for

Rings, groups, fields etc you care about isomorphisms

[clear obsidian]

so what does the Z/(n) notation mean?

[topaz solar]

oh that's me being dumb and denoting modding out the ideal by n

literally the same as Z/nZ

[clear obsidian]

ah alright

[topaz solar]

like uh, you know topo?

homeomorphisms are just isomorphisms

[clear obsidian]

ah yeah I got to homeomorphisms recently

[topaz solar]

and you really only care about spaces "up to iso/homeo-morphism"

since you can just go from one to the other without losing anything and all that

[clear obsidian]

mhmm that makes sense, thanks!

[topaz solar]

also that's an informal as hell statement, so uh
be careful with things like those

also don't use Z_n for the cyclic group

[clear obsidian]

so the definition of R as the unique ordered field with Q as a subfield up to isomorphism, for which the supremum property holds up to isomorphism - is an example of this: only caring about iso/home-morphisms thing right?

because I guess theres things isomorphic to R with these properties also

[topaz solar]

well yeah there are things which are isomorphic of course

i.e. multiply everything by 2

(though that's literally the same field afterwards)

note: homomorphisms are not isomorphisms

[clear obsidian]

ah yeah isomorphisms are bijections that preserve the operators instead of just functions right?

[woven delta]

In the group case yeah

Not generally

You also need an inverse that is a morphism generally

[topaz solar]

isomorphisms are maps which allow an inverse such that composing them gets you the identity iirc

[clear obsidian]

I actually havent heard of morphism yet, is that just a relation that perserves the operator that doesnt have to be a function?

[topaz solar]

morphisms are basically generalizing functions in a sense

homomorphisms are your morphisms of Group

That's a shitty explanation on my part, i concede

a morphism is something that morphs A into B

drawn $A \to B$

[cloud walrus]

Darkrifts:

[clear obsidian]

hmm alright, good to know

[topaz solar]

yeah you don't need to worry about them atm, though they're nice to have
isomorphisms require an inverse, normal morphisms don't

[clear obsidian]

Yeah I guess really anything is good to learn for me at this point basically

[topaz solar]

anyway, enough distractions, get back to your groups

hmm, if you're considering structures with a given signature, do the homomorphisms wrt that signature preserve all the operators like $h(x_1 \times_A \dots \times x_n) = h(x_1) \times_B \dots \times_B h(x_n)$ or something?

[cloud walrus]

Darkrifts:

[topaz solar]

(where x_A and x_B are the corresponding function/operators of n arity, i'm horrible at notating this since i'm pleb)

[brisk granite]

Are there any interesting group actions of D_n?

Ones that aren't the obvious ones tho

like, D_n acting on the vertices or sides, etc

[chilly ocean]

@brisk granite seems so:

https://www.maa.org/sites/default/files/pdf/upload_library/22/Hasse/Crans2011.pdf

I don't know music though

[oblique river]

Please don't use Z_n to mean "cyclic group of order n" ever once you leave a first course in group theory. Either use C_n or Z/nZ but be careful cuz they kinda mean different things (which is also why I'm confused; which one do you mean when you write Z_n?)

C_n = a cyclic group of order n, Z/nZ = a cyclic group of order n with a chosen generator

(namely, 1)

I feel like not enough people respect that convention but they should

It's the same as the difference between "a 2-dimensional vector space over R" and "R^2"

[bleak abyss]

I'm fairly guilty of that lmao, I make identifications very lonely

*loosely

[true ingot]

I need to show that U(n) is isomorphic to the semi-direct product of U(1) and SU(n)

But i have no idea how that semi product would look like. I thought that I could take exp(ix) from U(1) and Q from SU(n)

And their product exp(i x)*Q would then be a member of U(n). But that is a direct product then isn't it?

[hot lake]

what are U(n) and SU(n) again ?

technically, a direct product is also a semi-direct product

[magic owl]

Side question: does anyone know what the notation $R\otimes_{\mathbb{Z}} R$ where R and R are abelian groups means? Is it just direct product? What’s the intention of including the Z?

[cloud walrus]

MaxJ:

[tame bear]

probably Tensor product

[hot lake]

I think they view R as a Z-module

then it makes sense

[magic owl]

Oh duh, right

Thanks

[tame bear]

np, i dont even know what the tensor product is tho

[hot lake]

so it's not exactly the same as a direct product

[magic owl]

Yeah I wasn’t thinking when I was looking at it because it was the less-complicated part of the statement haha

[true ingot]

@hot lake U(n) are unitary matrices so QQ^h = I, SU(n) are unitary with determinant 1

The exercise says to show its not isomorphic to a direct product but is isomorphic to the semi direct one

But I dont see how I could construct it

[topaz solar]

direct product is a specific version of semidirect for one

[hot lake]

ah I see

[topaz solar]

But uh, other than that, do you know how to do the semidirect?

[hot lake]

what is your decomposition into a "direct product" ?

And their product exp(i x)*Q would then be a member of U(n). But that is a direct product then isn't it?```

how are you multiplying a size 1 matrix with a size n matrix ?

@true ingot

[true ingot]

Im just treating it as a multiplication of a matrix by a complex scalar exp(i x)

That way I think I get a bijection between U(1)xSU(n) and U(n)

[hot lake]

that won't work

when n > 1 you can multiply a matrix by any nth root of unity and it won't change the determinant

[true ingot]

Oh yea hmm

[whole basalt]

Hey y'all. I'm a bit confused on something

Reading about differences between vector spaces and modules, and how bases work in each

The book I'm reading says to consider Z as a Z-module over itself. Then it says that {2,3} is spanning but can't be reduced to a basis.

[woven delta]

Yes

[whole basalt]

Doesn't Bezout's identity say that since 2 and 3 are coprime, there are integers x and y such that 2x + 3y = 1, and therefore that generates all of Z?

[woven delta]

Yes, that's why they are spanning

[mild laurel]

That's why it's spanning

[woven delta]

Sniped

[whole basalt]

Okay, so then the "basis" part is what I'm not understanding

There's a subtle difference I'm missing.

[woven delta]

So with vector spaces, given a spanning set

[mild laurel]

Basis needs to be minimal

[woven delta]

You can always eliminate some elements to get a basis

[whole basalt]

Oh, and the fact that you can't take some of those elements out and still generate the whole thing is the problem

[woven delta]

Here you cannot do that

[whole basalt]

{2} and {3} do not span all of Z

[woven delta]

Yes

[whole basalt]

Okay, got it.

Whereas with vector spaces, when you have a spanning set that is no longer a spanning set if you remove some elements, you're guaranteed that there isn't an even smaller spanning set that used a different set of elements to start with

Because yay division

[bleak abyss]

I mean you can also think linear independence

Basis has to span + be linearly independent

But {2,3} is a minimal spanning set but it isn't linearly independent

[oblique river]

(just arriving but)

fields are nice and non-fields are non-nice

here is some crazy shit: every vector space has a basis (under axiom of choice). for example, a countable direct product of copies of R, the reals

recall that a basis only allows for finite linear combinations

so a countable direct sum has an "obvious" basis whereas a countable direct product doesn't

however, the countable direct product of Z's does not have a basis as a Z-module so it's not a free Z-module!

it's an example of a torsion-free Z-module that is not free

whereas all finitely generated torsion-free Z-modules are free

[lusty swallow]

Hi, I've been stuck for a few hours on something that I feel should be simple

How can I prove that the intersection of two coprime ideals is the product of the two ideals?

Hoping to get some help on this, I'm really frustrated at myself.

More specifically I'm trying to prove that in a PID, with (x) and (y) coprime, (xy) is equal to the intersection of (x) and (y)..

[oblique river]

write (x) \cap (y) as a principal ideal.

it has some generator z

what can you say about z in relation to x and y?

[fickle brook]

well, one of these inclusions holds trivially...

[lusty swallow]

Alright. I just got it, sorry guys.. I don't know why it was so hard for me. I just realised if (x) and (y) are coprime then there are X in (x) and Y in (y) st X+Y=1. Then some substitution gets me to what I want

Thanks for the help

Getting really frustrated at algebra, feels like I'm not getting any better at solving problems.

[topaz solar]

Well, how do you usually look at them?

[lusty swallow]

I don't think I'm approaching them in a specific manner. I kinda just list things I know about the question and what I have to prove, and start trying to work from there.

[topaz solar]

hmm

[lusty swallow]

It's just annoying because I do enjoy the subject and it's super interesting but I've been struggling with it all semester

[topaz solar]

There's usually more than one way to reach the end goal but almost all are homotopic paths, so don't be afraid of looking an entirely different direction

[lusty swallow]

Which is a little demoralizing I suppose

Sorry, just needed to rant a little, it is getting a little off topic

Thanks anyway!

[topaz solar]

You could have looked at your thing as your principle ideal, but iirc don't you have UFD?

which might've helped if you didn't list I guess (though you didn't really explicitly use it in your proof)

[lusty swallow]

I'm actually not sure how to use the fact that it's a UFD. I think I didn't actually use the fact that I had a PID, but I did assume commutativity.

Thanks for the help anyway. I'm gonna get back to it.

[brisk granite]

what does this mean?

[oblique river]

npg = "no problem, G"

[brisk granite]

nice

do you need context?

[oblique river]

but actually though can you please give context?

[brisk granite]

sure

[oblique river]

is it some sylow theorems nonsense?

or something else

[brisk granite]

Yes, sylow theorem nonsense

[oblique river]

I was right 🤢

[brisk granite]

so, uh, what does it mean

?

[oblique river]

I mean this lemma is proving that np(g) is equal tos omething

but presumably there was some previous definition

[brisk granite]

yea, but I can't seem to find it

[oblique river]

is it the number of p-sylow subgroups of G?

[brisk granite]

yep

that fits

[oblique river]

whelp there you go

[brisk granite]

I'll just work with that assumption

yea, I found where he defined it

so, uh, I missed

it

[clear obsidian]

So is a point with rational coefficients just (r1 r2, ... , rn) with each ri an element of Q ?

[chilly ocean]

yes

[brisk granite]

so, you know how there's an injective homomorphism between $S_{n} \times S_{m}$ and $S_n$

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

And that also means there's an injective homomorphism between $S_{n} \times S_{m}$ and $S_k$ for any $k \geq m + n$

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

But, are there any injective homomorphisms for smaller $k$

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

?

[hollow peak]

between Sn×Sm and Sn?

its impossible

[brisk granite]

yea obv

[hollow peak]

the first one has order nm the second one n

it cant be injective

(if n and m are greater than 1)

[brisk granite]

no, I mean, can there be an injective homomorphism $f: S_n \times S_m \rightarrow S_k$ where $k < n + m$?

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[hollow peak]

no

[brisk granite]

um, why?

I'm confused

[hollow peak]

because if n and m are greater than 1, nm>m+n

[brisk granite]

what is nm and m +n tho?

Those aren't the orders

it's n!m! and (m +n)!

[hollow peak]

oh yes

[brisk granite]

Pardon my stupidity, I'm kinda new to group theory

[hollow peak]

mb

[brisk granite]

ah ok

so, yea, do you think it's possible for a k < n + m?

[hollow peak]

yes

[brisk granite]

example?

[hollow peak]

it could be possible

but you also need n!m! to divide (n+m)!

[brisk granite]

huh, why?

nvm, I see

[hollow peak]

because f(Sn×Sm) is a subgroup of Sn+m if f is a morphism

and then you use Lagarange theorem

[brisk granite]

yea, ok

but that isn't enough

to show that it's possible for some k < n+m

[hollow peak]

no it isnt

you need a subgroup of Sn+m to have the same structure as Sn×Sm

[brisk granite]

I'm not trying to show that there is an injective homomorphism from S_n X S_m to S_{n+m}

[hollow peak]

oh yeah its the same condition on Sk

for all k

[clear obsidian]

Done taking notes on abstract algebra first 30 chapters. Time for Galois theory boys!

[uncut girder]

Why

Which book

@clear obsidian

[clear obsidian]

Because I want the corsera course certificate for my cv, this is the book

[wind steeple]

@hollow peak m parmi m+n ♿ ((m+n)!)/n!m!)

This is a hard problem for general m and n I think @brisk granite

I think you can inject S2xS3 in S4

oh no lel, A4 has not subgroups of order 6

[brisk granite]

Suppose $S_4$ acts on a specific conjugacy class of $S_4$ by conjugation. Specifically, the conjugacy class consisting of elements with one $3$-cycle. This allows us to define a homomorphism $f: S_4 \rightarrow S_3$

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

How do I go about proving f is either surjective or injective

?

[woven delta]

What do you mean consisting of elements with one three cycle?

Can you give me an example of such a thing?

[brisk granite]

Um, if you write the elements of that conjugacy class in cycle notation, it's a 3 cycle

That's what I mean

[woven delta]

Is that a conjugacy class?

[brisk granite]

Wdym?

[woven delta]

Also group actions on a group with n elements give you a map to S_n

[brisk granite]

yes

There are three elements in that conjugacy class

oh shit

nvm

wrong conjugacy class

[woven delta]

🐴

[brisk granite]

the one with two two cycles

that's what I mean't

got confused

but I still have the same question

[woven delta]

So you take the congugacy class of elements of the form (ab)(cd)

[brisk granite]

yep

[woven delta]

And you act on it by conjugation

[brisk granite]

yes

[woven delta]

How big is that congugacy class?

[stone fulcrum]

Three elements

[brisk granite]

the same size

yea

[stone fulcrum]

Which is where we get S3

[brisk granite]

Yea, but I'm trying to show that the map is surjective

I'm not rly sure where to begin

[stone fulcrum]

It's not injective, as that would imply |S3| ≥ |S4|

[woven delta]

🐴

[brisk granite]

um yes

[stone fulcrum]

Oh, yes you know that lel

[brisk granite]

I feel like if the question isn't really simple, then it's hard for me to do a problem about surjectivity. Like, with injectivity, I can always do stuff with kernals but idk what to do here

[stone fulcrum]

Not the easiest way to do it, but you could just find each S3

[brisk granite]

yea, I guess I could do that

It's only 6 elements

[stone fulcrum]

Hopefully something more clever

@brisk granite
There's a very easy homomorphism S4 → S3, simply hold one of the elements. Can you show that they are the same?

I'm not sure if they are but they probably are

[woven delta]

That is not a homomorphism

[stone fulcrum]

It isn't? :O

[woven delta]

Think about why that wouldn't work from S^n -> S^n-1

(that's not the reason it doesn't work, the reason it doesn't work is because it's not a well defined map)

But that should give you intuition why it wouldn't work without going through the work of showing its not well defined

[stone fulcrum]

Oh yeah, duh I can't make that a function lel

What am I thinking of?

S3 is a subgroup of S4, that's what I'm showing

[woven delta]

Yes

You're going the other way

S^3 -> S^4

[stone fulcrum]

Oh true, that works as well

But it doesn't help here :/

[mild laurel]

@quaint wraith this is not the right channel

[brisk granite]

I'm trying to find the number of 2-sylow subgroups of D_n

Suppose $|D_n| = 2^{a}m$ where $m \in (2\mathbb{Z})^{\complement}$. Then, the number of sylow $2$-subgroups can be $m$, but cannot exceed it. \

Now, consider the group of symmetries of an $n$-gon with an order of $2^{a}m$ and a $k$-gon with an order of $2^{a}$ ($D_n = D_{2^{a-1}m}$ and $D_{2^{a-1}}$). There are exactly $m$ $k$-gons that can be inscribed with the $n$-gon (such that the vertices of the polygons meet), and there exists a subgroup of $D_{2^{a-1}m}$ that are the symmetries of a given inscribed $k$-gon. This is one possible $2$-sylow subgroup of $D_n$, and there exactly $m$ of these kinds of subgroups for each of the $k$-gons. Hence, there are $m$ $2$-sylow subgroups in $D_n$

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

does this make sense?

[proud bridge]

I will see

[mild laurel]

That's a terrible way to say an odd number jesus

Don't ever do that

@brisk granite

[brisk granite]

Ok, I'll just write it down next time

but, um, does the rest of it make sense?

[woven delta]

Is your notation that |D_n|=2n or n?

[brisk granite]

2n

[analog oracle]

@brisk granite it makes sense to me but yes don't do that to say odd number o_O I also think it would be less confusing if you used |D_2n|=2n and it would look cleaner since it would get rid of the -1s

but the 2n bit is just my own notation preference, i'm sure plenty of people disagree

[inland bough]

Z/5Z is {0, 1, 2, 3, 4}
is Z/(1+5Z) ={0, 1, 2, 3, 4}?
up to isomorphism

[mild laurel]

No this makes no sense

1 + 5Z is not a subgroup in Z

so you can't take the quotient of it

[inland bough]

oh right

[brisk granite]

it's an element of the quotient group I believe

[woven delta]

Yes

[brisk granite]

so, this isn't really an abstract algebra question, but do you guys also spend more than an hour on a single problem?

Like, it took me a while to come up with the sylow 2-subgroup thing I put above

Am I really slow?

How do I become faster?

Every problem seems so different; so, it feels like I'm climbing a new mountain every time. That's why I'm afraid doing more problem won't help me

[chilly ocean]

It will

You are learning different group techniques and logical steps to follow

[oblique river]

I've spent 6 months on one problem; that's the nature of math

I'm not trying to be a gatekeeper like "you don't know a long time until you've..." i'm just saying that yes, the nature of math is that slow progress isn't always bad

[topaz solar]

big conjectures take the big boyes like years to do

sometimes things take a while to show because 1) you're approaching it the wrong way now and will find a better method by which to attack it or 2) it's just hard to show and you need to build up more and more pieces to finish the final argument

[analog oracle]

I always start my problem sets the day I get them (we get like 2 weeks) because i know I'll have to read it then spend a while understanding the question and then usually several days of sleeping on it and waking up to a new idea which turns out to be another dead end and i feel super slow because most people in my classes and the professors I'm working with can do it all in one go without notes or anything but ultimately if I get there in the end who cares how long it took me (unless it's an exam, then I'm screwed and rarely score above 70%)

And practice and experience help too, obviously every problem is different but you start getting a feel for what kind of things to throw at it

Speaking of what to throw at things! Anyone know any theorems relating to the order of elements in a p-group? I'm working on a problem and every group that meets the conditions we're interested in only has elements of order 1, 2, 4 and 8 (8 only for D16 X C_2^n) so I started listing all possible numbers of elements of each order for 2-groups and I ruled out all but one of order 32 using the fact that the number of elements of order 2 has to be odd, if for example you have an element of order 16 you also have to have an element of order 8 etc and using frobenious' theorem, and i have a lot more left for 64 so I'm wondering what else I can throw at it

The annoying one with order 32 is (n2,n4,n8,n16) = (17,10,4,0). I know from groupprops that there is no such group but I need to know why and since the number of things needing to be ruled out is much longer for 64 (and i don't even want to think about 128) I need some theorem i can apply or something that isn't a case specific proof

So yeah, any ideas welcome! Things that don't apply to the example I gave too

[wind steeple]

(Z/2Z)^n has only order 2 elements

Z/2^nZ has order 2^r elements for 1<=r<=n

So idk if you have a specific propertie for a p-group

[bold coyote]

Can someone help me understand why does this work this way: $$\left[i, a_1, . . . , a_k, j, a_{k+1}, . . . ,a_r\right] \circ \left[i,j\right] = \left[i, a_1, . . . , a_k\right] \circ \left[j, a_{k+1}, . . . ,a_r\right]$$

[cloud walrus]

dog:

[bold coyote]

how do I interpret the composition of cycles kinda, I mean I think I got it once but not sure if I was thinking about it correctly

[uncut girder]

Composition if cycles is just composition of permutations

Do you know how to write permutations with arrows between two layers of numbers?

[bold coyote]

I guess... not sure

like 1 goes to that number 2 goes to that etc.?

[bold coyote]

but not sure how composition works here, like how do you do g(f(x)) if f and g are permutations

[uncut girder]

Do you understand that diagram so far?

[bold coyote]

not really

[uncut girder]

Well on the top row you have the numbers 0 through 4

On the bottom row you also have numbers 0 through 4

The red arrows tell you what the permutation does to each number

[bold coyote]

what is that middle row for

[uncut girder]

A permutation is just rearranging objects, so this is a nice visual

[bold coyote]

Ive seen it in different form

just like a table of domain and codomain

so how would this permutation look like as composition of cycles?

[uncut girder]

In the above table, if we call that permutation p, then p(1)=3

Do you see why?

[bold coyote]

ok sure

but why is there another arrow to 3

oh I see

I get it

[uncut girder]

It's just telling you p(1)=3 and then putting the numbers back in order

The reason we wanna put them back in order is so we can compose easily

[bold coyote]

ok yeah

and how to compose

idk it seems counterintuitive

[bold coyote]

yeah makes sense

[uncut girder]

So you have a blue permutation and you follow it by a red permutation and the end result is the purple permutation

It's really just function composition, if you know how that works

[bold coyote]

ye

[uncut girder]

Permutations are bijections from {1,2,3,...,n} to itself

[bold coyote]

I mean, thats how I imagined it but in the example I shown its not that obvious to me that that transpose kinda cuts the cycle in half

[uncut girder]

Transpose?

[bold coyote]

cycle of length 2 is transpose

right?

[uncut girder]

Yeah

Transposition

[bold coyote]

oh ye my bad

[uncut girder]

You can break a permutation into a sequence of cycles

And the cycles you get are unique

You can break each cycle into transpositions

[bold coyote]

yeah I know that

[uncut girder]

Seems weird to know that but not permutation composition

[bold coyote]

I do know these, you still didnt explain why the transposition cuts the cycle

and its different composition

the one you showed are straight forward, I just didnt know how to interpret composition of n element cycle and 2 element cycle

[uncut girder]

Ok say you have a 4 cycle (1 2 3 4)

How would you break that into a composition of transpositions?

[bold coyote]

[12][23][34] right

[uncut girder]

Yes

[bold coyote]

so like compostitions would be [34] o [23] o [12] but idk

[uncut girder]

Now draw the two layer permutation diagrams to check this works

The composition is actually the way you wrote it first

[1 2] o [23] o [34]

So (3 4) is the first permutation

(2 3) is the second

(1 2) is the third

Remember function composition is right to left

Uhhh

Maybe I got that wrong lol

[bold coyote]

so waht you want to say that the way to explain waht I wanted to understand is to divide the thing I wrote into transpositions or what

[uncut girder]

I gtg

[bold coyote]

cause Im not sure if you actually know why that thing works or jsut saying random things you know

thanks for your time though, you actually helped me understnad other thing I was not sure of

[analog oracle]

@wind steeple re-reading what I wrote that bit is wrong or i wouldn't be seeing the groups I'm trying to classify. Oh well that's what happens when I try to write math up when I'm half asleep

i might be wrong because i only just woke up and my brain is dead but I think $$\left[i, a_1, . . . , ak, j, a{k+1}, . . . ,a_r\right] \circ \left[i,j\right] = \left[j, a_1, . . . , ak\right] \circ \left[i, a{k+1}, . . . ,a_r\right]$$ @bold coyote

[cloud walrus]

Jamie:

[analog oracle]

And how you see that is again if you use the diagrams which I'm on my phone and can't easily make so i guess I can try to explain it verbally

[bold coyote]

I copied what I wrote from the textbook

[analog oracle]

With composition you want to look at the rightmost permutation first, so if you had for example (1 2 3 4)(1 3) the rightmost sends 1 to 3. Then you go to the next one and see what 3 is being mapped to (4 in this case) and the resulting permutation will then map 1 to 4. Then you go back to your rightmost permutation and see what 4 is being mapped to (4 again) and then back to the first which maps 4 to 1 so the resulting one has 4 mapped to 1. Then 2 is fixed by the rightmost one and mapped to 3 by the other one so you end up with (1 4)(2 3)

[bold coyote]

aaah ok I kinda see it, will think about it deeper later but yeah, I guess these small examples are way to go

[analog oracle]

In the one you sent i goes to j and then j goes to $a_{k+1}$ so I goes to $a_{k+1}$. $a_{k+1}$ is mapped to itself by the transposition and $a_{k+2}$ by the cycle so $a_{k+1}$ goes to $a_{k+2}$ in the result and so on. Similarly j goes to i and i goes to $a_{1}$ etc gets you the other bit

[cloud walrus]

Jamie:

[analog oracle]

If you still need help later poke me, hopefully by then I'll be both more awake and not typing on a phone and can be more useful :p

Oh and also! It might be helpful to take your abstract example and turn it concrete

So like for example take (1 2 3)(i j), (1 2 3 4 5)(i j) etc until you get a feel for it (or keep the a_k notation and just set some specific value for r)

Plus also Wolfram alpha should be able to compute permutations like that so you can double check your work

[brisk granite]

https://cdn.discordapp.com/attachments/359052604149465088/601244715912003596/unknown.png

idk where to start

If it helps, I've shown that the number of sylow 2-subgroups in D_n is m when 2n = 2^a \times m.

Also, the notation I am using is |D_n| = 2n

[analog oracle]

@brisk granite what's N? What's its order? What's the order of a sylow 2 subgroup? What about the index? Or look at the index of the intersection of <r> and P in P

I haven't actually done the problem myself but that's where I'd start

[inland bough]

hello, can someone help me with groups?

group G, subgroup H

consider all the cosets of H in G

does there always exist a set of cosets that covers the entire group

*disjoint cosets

[mild laurel]

What have you tried?

[inland bough]

i haven't tried anything

if H is normal

actually idk

[analog oracle]

cosets partition the group so yeah G is the union of disjoint cosets

you don't need H to be normal for that to hold, you only need H to be normal for the cosets to form a group

[inland bough]

ok good lol cuz i proved they were disjoint without using normal

[analog oracle]

if you've already proved they're disjoint try assuming there's some element in G that isn't in any of the cosets and see what happens

[inland bough]

say g in G and not in any of the cosets; but g in gH since 1_G in H; contradiction

[hollow peak]

you can do it directly

the cosets cover G because for all g in G g is in gH

and then you can show that gH et g'H are disjoint or equal for all g and g' in G

[analog oracle]

yeah, I went with contradiction because I was trying to give a hint rathe than straight up give the answer but it's cleaner done directly :p

[brisk granite]

Quick Q. I'm trying to show that the number of sylow p subgroups in any normal subgroup of G must be less than or equal to the number of sylow p subgroups in G. I feel like I'm not really sure where to start again...

Is this an obvious result?

WAit, I think I thought of something

nope nvm

[brisk granite]

Oh, also, I solved that other problem

Suppose $D_n = 2^{a}m$. Then, for some sylow $2$-subgroup $P$,
$$n_{2}(P) = \frac{\left|D_n\right|}{\left|\text{N}{D_n}(P)\right|} \implies m = \frac{2^{a}m}{\left|\text{N}{D_n}(P)\right|} \implies \left|\text{N}{G}(P)\right| = 2^{a}$$
Hence, the relationship $P \trianglelefteq N{D_n}(P)$ becomes an equality: $P = N_{D_n}(P)$

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[analog oracle]

that doesn't look right to me but my brain is beyond fried right now so that's probably why :p

[patent turtle]

Could someone please help me with this notation? G is a group, m is an integer. What does G^m represent?

[mild laurel]

Most likely G x G x ... x G

where x is the direct product

[brisk granite]

@analog oracle why?

I'm just using sylow 3

or technically, I'm not

Is the fact that the number of sylow p subgroups is equal to the index of it's normalizer a part of sylow 3?

[inland bough]

hey guys, i'm confused about how the group operation of the quotient group is defined

its defined as: aN * bN=(ab)N

i don't understand what i couldn't be : aN * bN = (a^-1 b^-1 N)

i don't understand what i couldn't be : aN * bN = (a^-1 b^-1) N

[mild laurel]

Think about it like this

If you take a group G and a normal subgroup N, then you have a group homomorphism from G to G/N

That sends an element a of G to aN

if you describe the group operation of the quotient group like you did, then this would no longer be a group homomorphism

[inland bough]

so the idea is to mimic the structure of G

[mild laurel]

It's true that if you do it like you described, you still get a group, but it's not really related to G or N in any nice way

[inland bough]

thanks that makes a lot of sense

group G, subgroup N

let G/N be the set of all cosets

if there does not exist a homomorphism from G to G/N

then we're just not interested?