groups-rings-fields

[solemn rain]

fuck

you just annihliated me

if you were paying attention to what im saying in #discussion

[bitter mauve]

@mild laurel

sorry for many questions but this is a part of yesterday's question

"and it follows that K=mK"

how

[cloud walrus]

Daminark:

Daminark:

[bitter mauve]

hmm... i might be dumb

i dont get why psi (am) is in psi (M) interset K

sorryy

o wait

i get it

hmm... why is this shit so complicated

u wrote $y=\psi(m)+z$ but shouldnt it be $y = (\psi(m),z)$?

[cloud walrus]

nnutannep:

[bitter mauve]

@bleak abyss

they same?

[bleak abyss]

Well so F is isomorphic to that direct sum via the map (psi(m),z) -> psi(m) + z

[solemn rain]

why are normal subgroups called normal

ping if help

[solemn rain]

How to find the number of elements that have a certain order

specifically in permutation groups

[uncut girder]

Because they let you quotient by them

[valid bridge]

you there algebra champion? @mild laurel

[mild laurel]

lmao I spoke about number theory, I know 0 algebra

[valid bridge]

wtf

[mild laurel]

What exactly are you confused about here

[valid bridge]

what is this stupid ring isomorphic to

[mild laurel]

What would R[x,y]/ <x,y> be isomorphic to

[valid bridge]

R?

[mild laurel]

yeah

[valid bridge]

........

[mild laurel]

And your intuition should be that linear change of coordinates like this shouldn't change anything

[valid bridge]

and its isomorphic to R because

uhh im trying to think of a map

[mild laurel]

uuuuuuuuhhhhhhhhhh

[valid bridge]

LAME

ALGEBRA IS LAME

[mild laurel]

ur just too dumb lmao

dont blame algebra

[valid bridge]

this is why im a gender studies major

[mild laurel]

probably more appropriate

Would you like a hint

[valid bridge]

yea i was just working on part d right now but i'll take a hint for c

[mild laurel]

Basically polynomials are created to play nicely with ring homomorphisms

What this means is that evaluating a polynomial at some value is a ring homomorphism

This is probably too big a hint but

Eh, there's really no hint in between

[valid bridge]

yea too big

the answer is practically

popping off the page

[mild laurel]

Great! Good job! I knew you were so smart!!!

Maybe as an example

We can construct a ring homomorphism from R[x] to R by sending a polynomial f to f(0)

[valid bridge]

u srs

is the answer R^2

[mild laurel]

no

Think about how to extend this to the two dimensional case

[valid bridge]

oh so you mean we send R[x,y] to R by sending f to f(0,0)

[mild laurel]

maybe

see what you get

[valid bridge]

.........

wait a second

holy shit

im an algebra major?????

this is unprecedented.

[mild laurel]

yea

maybe high school algebra

[valid bridge]

wtf........

you see this is why im not taking number theory

i decided to take these DUMB proof based math classes because i thought they'd be useful for physics

[mild laurel]

The algebra used in number theory can be easier than this

idk, learning hard things is hard

you get used to it

[valid bridge]

one of the classes i might take after this is algebraic topology which is taught by the same teacher i had for algebra I

who is godlike

[mild laurel]

booooo

algebraic topology is probably the field that has the least applications in number theory

(also the field I'm worst at, which is not a coincidence)

But nah

It's super cool

The fact that you can use algebra to solve these hard topological problems is magical

[valid bridge]

idk about magic

witchcraft perhaps

[mild laurel]

yea black magic probably

[valid bridge]

wait did you say you were a grad student

[bleak abyss]

Arithmetic topology is a thing actually

I guess that's less about algebraic topology and more geometric

[valid bridge]

say what

[bleak abyss]

Topological modular forms are a thing

[valid bridge]

[insert here] topology is a thing

[mild laurel]

Yeah I probably shouldn't have bet against a field having no applications in number theory

[bleak abyss]

Though tbh there's a non-trivial possibility that tmf is too drugsy to be of irl relevance in number theory

As a friend put it "It's a cohomology theory and I'm not sure if it actually has anything to do with topology or with modular forms"

Probably most of the day to day links are through arithmetic geometry

[mild laurel]

what the fuck does it mean for a field to be drugsy

[bleak abyss]

Like it's approaching so abstract that it lives in the realm of drugs

[valid bridge]

it means for every x in the field there exists a drug dealer in the neighborhood of x

[mild laurel]

So I should start doing acid to understand this field?

[valid bridge]

absolutely you should

right now in fact

[bleak abyss]

I feel like in topology once you reach the term "E_{infinity} ring spectrum" you've gone too far

[mild laurel]

okay let me hit up the drug dealer in my neighborhood

[valid bridge]

wtf is E infinity ring spectrum

[bleak abyss]

No clue tbh

[valid bridge]

lol

[solemn rain]

my question got ignored 😫

how do u find the nunber of elements of certain orders

and why is every group of order 12 has a non trivial normal subgroup

[golden pasture]

what kind of group are we talking about when finding the number of elements order n and what information are you exactly given

if G has order 12 you can proh find a subgroup with index 2

[bleak abyss]

A_4 I think

If a group has order 12 proving it's not simple I think is basically like, okay how many Sylow 3-subgroups are there? Either 1 or 4

If 1 we're done because then it's normal

If there are 4, well they're cyclic so each will be the identity plus two elements

So there are 8 elements of order 3

But then the remaining 4 elements form the unique order 4 subgroup

[solemn rain]

@golden pasture S_9

but in general how

[golden pasture]

eh in general seems kinda painful lol

[solemn rain]

i saw it as a GRE problem on instagram lmal

lmao

i am having very hard times with problems

[mild laurel]

If I give you an element of the permutation group in cycle notation, how can you find it's order

Like if I gave you the permutation

(1 2) (3 4 5)

What's it's order

@solemn rain

[solemn rain]

lcm lengths

idk how to provw that tho xd

[mild laurel]

(12) and (345) commute

This should be a general result you know

That if two elements commute, then the order of their product is just the lcm of their orders

[solemn rain]

yea

olay

okay**

idk hoe to see how cycles commute thst eaasily tho

[mild laurel]

I mean

just check

Another fact you should know, and should be able to prove is that two permutations commute if and only if they're disjoint

[solemn rain]

ugh im so bad with permutations

idk how to proce shit for them

[chilly ocean]

Same mate

[hot lake]

I wouldn't say if and only if

[chilly ocean]

I fucking hate permutations

[hot lake]

I would only say if

and not only if

[solemn rain]

yea

can some1 prove it for me

[mild laurel]

@hot lake oh yeah whoops

@solemn rain You should really prove it for yourself, its not a very hard proof

[solemn rain]

yea there is prob a trick involved

[mild laurel]

There's really no trick at all

if you understand what these words mean, it just falls out

[solemn rain]

okay i got it

lmao

finall

y

okay so anyways

order of a product of disjonint cycles is the lcm of length

[mild laurel]

true

[solemn rain]

so i want lcm(x,y) = 8 ?

xd

[mild laurel]

Uh, not sure what you're trying to do

[solemn rain]

i want to find the number of elements that have order 8

sorry i didnt specify that

let me show you the exact problem

In the permutation group S_6 , the number of elements of order 8 is : ...

multiple choices

@mild laurel

i think this is math csir or whatever

[mild laurel]

No that's not what you want

Because you can multiply 3 disjoint cycles together

Or 4

[solemn rain]

oh fuck

yea

idk

[mild laurel]

What did you try

[solemn rain]

idk how to approach it

[mild laurel]

You have everything you need

[solemn rain]

okay its obv that i need to relearn this stuff

ty

anyways

[coarse stag]

so I understand that a left/right transversal of H is a set that contains exactly one element from each coset of H

if the group G is not a cyclic group, then would every cyclic subgroup be a transversal of some subgroup?

doesnt that still allow for Z/2Z to be a transversal of some subgroup

wait

yeah

because Z/2Z = {1,a} where a^2=1, and xH = yH if and only if x^-1*y in H, then 1*a would always be disjoint of some other subgroup of G if G isnt cyclic

[mild laurel]

Yeah I misunderstood what you were asking

[coarse stag]

oof

also, I'm assuming for each subgroup H of G, there are |H|^|G/H| unique left transversals

[mild laurel]

Hm, I don't think you need the condition that G is not a cyclic group?

[coarse stag]

if G is Z/8Z, wouldnt Z/4Z be the transversal of nothing

[mild laurel]

Yeah right

[coarse stag]

just curious, given two non abelian subgroups H and K of G, is it possible for hk = kh for every h in H and k in K?

[pulsar briar]

G = HxK

[coarse stag]

?

[mild laurel]

If you take nonabelian groups H and K and take G = HxK

[coarse stag]

is that little intersect symbol just latex's in line intersection symbol?

$$H\cap L$$

[cloud walrus]

QuAnTuM:

[coarse stag]

aparently not

[pulsar briar]

it's an old typesetting thing

just use \cap

[solemn rain]

hey

i think i got it

the problem that i haad with finding specific orderrs in groups

so i wanted any1 to confirm if im right with this

problem

' write out the cycle decomposition of each element of order 4 in S_4 ' ( dummit and foote permutations section exercise 6 )

well there is only 1i g

(a_1 a_2 a_3 a_4 )

for the specific porblem i had was

In the permutation group S_6 , the number of elements of order 8 is : ...

okay in S_6 i have

0

0 elements of order 8

hope any1 corrects me

[chilly ocean]

Yes

[solemn rain]

yea ty

1 more problem

Let pi me the m-cycle (1 2 .... m). Show that pi^i is also an m cycle if and only if gcd(m,i) = 1

cool problem but dk how to solve it

[south coral]

Try it out with some values of m and i that make gcd(m,i)≠1 and observe what happens

Then try it with values that do fulfill gcd(m,i)=1 and see what's different

[solemn rain]

okay

ty

[errant drum]

Suppose a group $G$ acts on $X \times X$ pointwise need to show the number of orbits is $$\frac{1}{|G|} \sum_{g \in G} |F(g)|^2$$

[cloud walrus]

Cosmicrayz:

[errant drum]

Not sure where to begin

🤔

[wind steeple]

look at the proof of the burnside formula

[errant drum]

Tried that also the hint on the problem says use burnside's lemma on one of the arguments so I assume there's some fixing an element in the first argument kind of proof involved

[mild laurel]

So F(g) here is the number of elements that g fixes right

[errant drum]

Yes

[mild laurel]

Wait, why is G acting on X x X

[errant drum]

Sorry G acts on X

And you can extend the action to XxX by x,y --> gx,gy

[mild laurel]

Oh hm

[wind steeple]

oh ok

that exactly the burnside formula

but F(g) here is {x st g.x = x}

and in the burside formula you have {(x,y) st g.(x,y) = (x,y)}

that has the same cardinality as F(g)

[errant drum]

Yeah thanks for pointing that out

I thought F(g) was referring to the collection of group elements that fix elements of XxX

Then I was like this can't possibly be true

[wind steeple]

yep

It would be true iff F(g) has 1 elements for all g, hence the action is transitive x)

[simple agate]

how should I show that g is the minimal polynomial?

I've shown that alpha^-1 is a root of g but not quite sure how to show it's irreducible

[wind steeple]

g = X^n f(1/X)/a0

[simple agate]

doesn't that just show that it's a root?

idk I'm kinda confused by what I have to prove. maybe I already did it

[wind steeple]

this shows that if you have any non trivial factorization of g, then you have a non trivial factorization of f

[simple agate]

oh okay, thank you!

[uncut girder]

Problem: Prove that I=(2, 1+sqrt(-3)) is prime in Z[sqrt(-3)]

@mild laurel sorry my first problem statement had a typo

This is the actual problem

[uncut girder]

I got it

Wanna see the solution @mild laurel ?

[mild laurel]

I mean, sure

This is an exercise in ireland rosen right?

[uncut girder]

No it's in Daniel A Marcus Number Fields

Chapter 2 problem 2

[mild laurel]

oh yeah that

I've written down a solution too probably

[uncut girder]

So the key observation is that I is the set of elements with a+bsqrt(-3) such that a=b mod 2

Then you can work out the coefficients of \alpha *\beta in I and use cases to see that \alpha , \beta in I

[mild laurel]

Yeah that's vaguely what I did too

[fierce sinew]

So, I got this 4-element lattice and it is also a heyting algebra

but it has a true complement operator

and both strong and weak de morgan laws hold

I don't know what to make of this structure, is there a corresponding logic to this?

the elements can be described as both, true, false, neither

with the partial order of both < x and x < neither

ok, maybe this is more suited to logic and foundations 😄

[fickle brook]

let p, q ∈ Z[x] such that p(n) divides q(n) for all n ∈ Z. does p necessarily divide q in Z[x]? if not, can i have a counterexample?

[wind steeple]

p = 2; q = X(X+1)

[fickle brook]

[wind steeple]

?

[fickle brook]

i don't know how i didn't think of such a low degree example

[wind steeple]

btw if p is unitary I think it works

you can use euclidean division, q = ap + b

[fickle brook]

unitary?

[wind steeple]

the highest coefficient of p is 1

[fickle brook]

oh monic

[wind steeple]

yep

mb, francisism lel

and you have b(n) = 0 mod p(n) for all n, maybe you can conclude that b = 0

with d°b < d°p

[fickle brook]

d°?

[south coral]

degree?

[fickle brook]

oh lmao

i always just said deg

fuckin baguettes i swear

[wind steeple]

non

In french I've always seen deg

I've seem d° in ian stewart's book, galois theory

[fickle brook]

[wind steeple]

or in Marcus one iirc

[toxic salmon]

WHAT THE FUCK???

@fickle brook FUCK. YOU.

HOW DARE YOU CALL MY PEOPLE BAGUETTES??

NEXT TIME YOU MUNCH DOWN ON A BAGUETTE FOR LUNCH, I WANT YOU TO THINK ABOUT WHAT YOU SAID AND HOW MUCH IT HURT ME.

DISGRACEFUL

[fickle brook]

OH MY GOD WHY ARE YOU YELLING

[toxic salmon]

PROBABLY BECAUSE YOU JSUT FUCKING CALLED MY PEOPLE BAGUETTES??

[fickle brook]

ARE YOU BEING SERIOUS RIGHT NOW OR JUST MEMEING?

[toxic salmon]

OH I'M SERIOUS.

[fickle brook]

TU ES JUSQU'ICI LE SEUL FRANÇAIS OFFENSÉ PAR MON APPELLATION « BAGUETTES » TOTALEMENT HUMORISTIQUE

ok but seriously bruh calm down

[wind steeple]

Baguette > all

[simple agate]

mmm baguette

[toxic salmon]

BIEN DÉSOLÉ QUE J'AI UNE LONGUE HISTOIRE AVEC CE MOT. MA FAMILLE A ÉTÉ PERSÉCUTÉE PENDANT 1500 ANS POUR ÊTRE DES FABRICANTS DE BAGUETTES, CELA ME FAIT VRAIMENT MAL QUAND QUELQU'UN UTILISE CE TERME COMME CELA

[fickle brook]

1500 ans ? waouh

[simple agate]

are there any other big books like dummit&foote covering rings/modules/galois theory/rep theory?

[fickle brook]

je savais pas qu'on vit en fait au XXXIIIème siècle et pas le XXIème !

est-ce que j'étais endormie pendant environ mille ans ?!

[toxic salmon]

Oui

[fickle brook]

srx calme toi

[final gulch]

he is memeing
Link isn't French, he's Hylian

[sullen island]

hi, i need some help with interpreting this question. in the definition of a Lie algebra, the lie bracket is just an abstract symbol which isnt defined to be the commutator (i.e. [a,b] = ab - ba isnt explicitly defined)

is the exercise asking me to prove it using the defined operation ab-ba (with regards to example (3)) ? or is it just asking me to treat it as an abstract symbol

because i wouldnt know how to prove the jacobi identity without knowing what the bracket actually does and just treating it as a symbol

for easy reference, here is the jacobi identity:
[a, [b,c]] + [b, [c,a]] + [c, [a,b]] = 0

sorry if this is a stupid question

[wind steeple]

it's the definition of the 3rd example that you need to use

[sullen island]

oh lol okay

so i got stuck for 30 mins for nothing

thanks ! @wind steeple

[wind steeple]

btw I don't understand why this exercice is so famous lol

[sullen island]

neither do i

its just a tedious computation with no benefit to conceptual understanding of lie algebras whatsoever

xD

[wind steeple]

yes

[golden pasture]

@simple agate jacobson

i think everyobe have seen jacobi and either actually expanded or just by observation this is true

[chilly ocean]

let a be the real cube root of 2. How do I find irreducible polynomial for 1 + a^2 over Q?

[wind steeple]

you can add -1 to get a^2 and take some power to have a rational number

[chilly ocean]

lol wut

I want a polynomial f such that f(a^2+1) = 0

[wind steeple]

les h = a^2+1

h-1 = a^2

[chilly ocean]

idk what then

[wind steeple]

(h-1)^3 = 4

[chilly ocean]

I can do ^3

[wind steeple]

yeah

and you have your polynomial here

[chilly ocean]

and how do I know its irreducible tho

maybe eisenstein works here I guess

[wind steeple]

it's a polynomial of degree 3

check that it has no roots in Q

[chilly ocean]

but I dont get how would the polynomial look like as an f(x). Would it be (h-1)^3 -4 expanded?

[wind steeple]

f(x) = (x-1)^3-4

it's a polynomial

[chilly ocean]

ok holy FUCK im braindead

like, what was stopping me was - how do I know thats the irreducible one? and was too lazy to compute

but eisenstein criterion works here I think

[wind steeple]

maybe but it's too strong

you only need to check that it has no rational roots, and it's obvious bc 4 isn't a cube in Q

[chilly ocean]

How about this one: find irreducible polynomial for $\alpha = \sqrt 3 + \sqrt 5$ over $\mathbb Q \left(\sqrt 5 \right), \mathbb Q \left(\sqrt 10 \right), \mathbb Q \left(\sqrt 15 \right)$

[cloud walrus]

Godel:

[chilly ocean]

So I'm just not sure what does Q(sqrt5) for example mean

I found it for just Q

[wind steeple]

(h-sqrt(5))^2 = 3

[chilly ocean]

but what does Q(sqrt5) mean? Its a smallest field Q + sqrt5 right?

[wind steeple]

yes

[chilly ocean]

ahh ok, so x-sqrt5 is apolynomial there, gotcha, thx

[chilly ocean]

so would x^2 - (8 +2sqrt15) work in Q(sqrt15)?

[wind steeple]

yes

[chilly ocean]

and btw how would I know if its even algebraic in those fields?

oh wait nvm

they're algebraic in Q so they also have to be in their extenstion I guess

hmmm cant quite figure in Q(sqsrt10)

[stoic dirge]

Hey can someone help me out with this problem? I have to prove that $R[x]/{<x^2+x+1>}$ is congruent to the complex field

[cloud walrus]

Benniem88:

[stoic dirge]

I mean isomorphic lol

[chilly ocean]

I guess evaluation at complex root of the polynomial works

[wind steeple]

it's isomorphic to R[j] which is isomorphic to R[i]

where j =e^(2ipi/3)

[chilly ocean]

evaluation at a is a homomorphism such that evaluates polynomial in a f(x) -> f(a). Showing kernel is your polynomial x^2 + x +1 and the image is complex numbers implies they are isomoprhic

thats just first iso theorem

I think its easy to show the image of this homomorphism is C, a bit harder is justifying the kernel is that polynomial I guess

[chilly ocean]

how can I show x^4 -16x^2 +4 is irreducible in Q(sqrt10)

[vagrant imp]

can anyone help me with this?

[chilly ocean]

what is |*|

[chilly ocean]

@chilly ocean The roots are $\pm \sqrt{8 \pm 2 \sqrt{5}}$ so you need only check that $\sqrt{8 + 2 \sqrt{5}}^2 = 8 + 2 \sqrt{5}$ and $\sqrt{8 + 2 \sqrt{5}} \sqrt{8 - 2 \sqrt{5}} = 2 \sqrt{11}$ are not in $\mathbb{Q}(\sqrt{10})$ (i.e. that $\sqrt{5}$ and $\sqrt{11}$ are not in $\mathbb{Q}(\sqrt{10})$, which is easy).

[cloud walrus]

hochs:

[chilly ocean]

k thats what I thought, so that's the 'general way' to determine that? I mean, sometimes it may not be as easy to factorize it right?

this one was as easy as finding roots of quadratic

you can try to pick rational prime $p$ and see if the polynomial is irreducible mod p ("above" in the ring of integers)

[cloud walrus]

hochs:

[chilly ocean]

ohhh yeah I've heard about this one once

Also I guess sometimes checking K[x]/f might be useful

(to see what its isomorphic to)

right, that your polynomial is irreducible over $\mathbb{Q}(\sqrt{10})$ would be equivalent to saying that $\sqrt{10}$ is not in $\mathbb{Q}(\alpha)$ too, where $\alpha$ is a root of ur polynomial (which is from "checking K[x]/f" thing that you mentioned)

[cloud walrus]

hochs:

[chilly ocean]

as $\mathbb{Q}(\sqrt{10})[x]/(x^4 - 16x^2 + 4) \cong \mathbb{Q}[x,y]/(y^2 - 10, x^4 -16x^2 + 4) \cong [\mathbb{Q}[x]/(x^4 - 16x^2 + 4)][y]/(y^2 - 10) \cong \mathbb{Q}(\alpha)[y]/(y^2 - 10)$

[cloud walrus]

hochs:

[chilly ocean]

where $\alpha$ is a root of your polynomial.

[cloud walrus]

hochs:

[chilly ocean]

yeyeye thx

[chilly ocean]

there are other general strategies like taking norm and factoring huge polynomials over Q (and there are known algorithms for factoring polynomials over Z, using lagrange interpolatios, e.g.) or if you're lucky again newton polygons

yeyeye I guess most examples i'll come across won't be that hard

btw, I solved this one, but I'm not sure about one thing

so I assumed the inverse of that element is going to be of degree 2

could it be different degree? Should I assume its of degree 3 and up and while solving it just variables will vanish?

you don't lose any generality by assuming that the inverse will take the form a + b\alpha + c\alpha^2 (since \alpha^n for n >= 3 can be written in terms of lower degrees in \alpha

hmm actually here I figured a^3 is congruent to 2 degree polynomial so I guess that doesnt matter

exactly

ye makes sense

[solemn rain]

let pi be an m-cycle (1 2 3 ... m). show that pi^a is also an m cycle iff m and a are relatively prime

solution:

|a^k| = |a|/gcd(|a|,k)

|pi^a| = m/gcd(m,a)

gcd(m,a) = 1

|pi^a| = m.

the other side is the same asweell

is that riht?

[chilly ocean]

Let $G$ be a group of order $14$. Show $G$ is isomoprhic to either $\mathbb Z_{14}$ or $D_7$.

[cloud walrus]

Godel:

[mild laurel]

This is just a sylow argument

[chilly ocean]

h m m

meaning that its either cyclic (then Z14) or has a 2subgroup and a 7subgroup?

not sure how it implies its exactly D7

and not somethign else

[mild laurel]

I mean, write out the sylow things

There are two possibilities

[chilly ocean]

how

[mild laurel]

Because 7 is 1 mod 2

[chilly ocean]

so there might be 7 2-subgroups

but then there wouldn't me a 7 group no?

[mild laurel]

What

[chilly ocean]

no sorry

does D7 have 7 subgroups of order 2 though?

[mild laurel]

Yes

[chilly ocean]

ohh ok

but what I dont understand is how do I know its not something else than d7 and z14 - does that already imply it has to be d7? Do I need to know all the d7 subgroups and compare their sizes and amount?

[mild laurel]

There's a semidirect product argument you can use here

[chilly ocean]

Probably there aren't many possibilites if I know there are 7 subgroups of order 2, but I would guess its much harder for groups of higher order?

ohh let me check the semiproduct thing I remember reading about it

[mild laurel]

Since the subgroup of order 7 and any subgroup of order 2 will generate the whole group

So you have that the group is the semidirect product of Z/7Z and Z/2Z and there are two ways to form this semidirect product

[chilly ocean]

how do you know it will generate the whole group tho?

[mild laurel]

Think about it

[chilly ocean]

Idk, it seems that if H is of order 7 and K ={1,x} then H and xH will be disjoint but not sure what does it follow from

I guess I see it

but still dont know how to show its either d7 or z14 and not anything else from the fact there might be 7 or just 1 2-subgroup

[mild laurel]

|HK| = |H| |K|/ |H \cap K| is a formula you should know, this is just the third isomorphism theorem

There are only two ways to do the semidirect product, one gives you Z14, the other gives you D7

[chilly ocean]

yes its either Z/14Z or Z/7Z x Z/2Z

[mild laurel]

Uh, those are the same group

[chilly ocean]

how are those the same group

I dont get it

[mild laurel]

Uh

Chinese remainder theorem?

I mean the latter group is cyclic since (1,1) has order 14

[chilly ocean]

D7 isnt cyclic

and I thought its C7xC2

[mild laurel]

Yeah and so your answer is wrong

D7 is not that

[chilly ocean]

like I thought maybe its C2xC2xC2 7 times xC7

but thats too many elements

[mild laurel]

Uh okay

[shrewd halo]

Wait, D7 has 7 subgroups of order 2? That can’t be right, 7 different elements that invert themselves already generate 128 different elements

[mild laurel]

Uh how do you get that number

And if you think of D_7 as the symmetries of a heptagon, each reflection is an element of order 2 and there are seven different reflections

[shrewd halo]

Yeah nevermind

[bitter mauve]

guys

can i get a hint here if possible

so i made this. i think the existence of h and e as h = uc and e = vhd is justified

Ps and Qs are free modules here

so g is justified and then f is justified

i think ill have to show that g or h is surjective and ill have my answer, but it doesn't make sense to have finite presentation as L or N. so maybe another surjective map will come up

i just want some hint

I also just drew i arrow cuz Q2 was projective. wasnt really necessary

[oblique river]

so showing M is finitely generated is easy I think. If you have a surjective A^p --> L and A^q --> N then that second map lifts to a A^q --> M

and composing the first with L-->M you can define an A^{p+q} --> M

that should be easy to shwo is surjective

[bitter mauve]

😮

what is this magic

[oblique river]

that's just how these things tend to work

I'm not convinced your diagram will be particularly useful? the arrows are just kind of going in random directions...

[bitter mauve]

yea, i guess, i just drew every info i had

[oblique river]

but not in a useful way

draw the L -> M -> N vertically

and the "finitely presented" sequences horizontally pointing toward L and N

it's not about just drawing out the info

it needs to be done in a conducive way

like, the only way you could even approach this problem is to try to combine the two presentations of L and N together

but if you draw them on opposite sides of the picture, I don't see how that pictoral representatino would be useful

[bitter mauve]

thanks. i realise that now. next time....... :(

[solemn hollow]

Why did D&F put this exercise in section 5.4 when it's killed by 5.2 FTFGAG?

wait

no

let me think

ye ignore me i think i dum

mit

[solemn hollow]

I have deleted the joke. it was childish and i apologize

WAIT NO KEEP THE WAIT NO IT'S A TYPO

a group can't be a direct product of its subgroups ooof

dwai tho

is it in poor taste say Dumbit and Fool?

im sa

[chilly ocean]

why cant a group be a direct product of its subgroups?

Id say it can

Z24 = Z3 x Z8 ?

[solemn hollow]

I'm taking "is" and "be" to mean "equal to"

[chilly ocean]

what?

[bleak abyss]

@solemn hollow in algebra never make a fuss about things being equal vs isomorphic

[chilly ocean]

how do I show $3 + i \sqrt7 $ isn't prime in $ \mathbb Z \left[i \sqrt 7\right]$

[cloud walrus]

Godel:

[upper pivot]

what have you tried

@chilly ocean

[chilly ocean]

idk

[upper pivot]

well what does it mean to be prime (and what does it mean to be irreducable)

[chilly ocean]

it is irreducible

and prime a if a|bc then a|b or a|c

[upper pivot]

based on that defination of a prime

what do you think should the general outline be of your proof

[chilly ocean]

I know what it should be but idk how

I guess I would need to write this number as some bc

idk

[upper pivot]

well a prime p has the property that p|ab means p|a and p|b right

we want to show your number is not prime

[chilly ocean]

or but yes

thats what I wrote

[upper pivot]

so what should we do? how do you generally show something isnt something

yeah im just helping you do this on your own

[chilly ocean]

idk I tried few things but didnt work

do you see how to do this or are you just trying to work it out with me

[upper pivot]

i know the sol

just answer the question i asked, how do you generally disprove something

you want to disprove 3+isqrt(7) is prime

[chilly ocean]

...

assume its prime

[upper pivot]

ok im just gonna say it

counter example

find a counterexample

to p|ab implies p|a or p|b

[chilly ocean]

yes but I dont know how to find the counter example

[upper pivot]

well whats a number p divides here

a nice number that you could use

[chilly ocean]

idk 16

[upper pivot]

exactly

can you take it from here

[chilly ocean]

yeah I mean thats what I dont get

so my number call it a divides 16

[hot lake]

no

[chilly ocean]

yes

[upper pivot]

ok if p|16, and p was a prime, what must it divide

[hot lake]

you call your number p

[chilly ocean]

same thing

it has to divide p

[upper pivot]

ok we said p|ab means p|a or p|b

if p|16, then what

[chilly ocean]

16 = (3 +i7)(3-i7)

[upper pivot]

ok what are some other divisors of 16 lol

[chilly ocean]

2

[upper pivot]

awesome

[chilly ocean]

but 2 isnt in the ring

[upper pivot]

... yes it is

[chilly ocean]

xd jk

sory

ok so you mean because there are 2 different forms of 16 it would need to work for both

[upper pivot]

yeah

[chilly ocean]

so p would have to divide lets say 2 and 8

[upper pivot]

yeah

[hot lake]

or*

[chilly ocean]

ok

fuCK

[upper pivot]

well in this case and, but in general or yeah

[chilly ocean]

thx

im going insane

[upper pivot]

ok so you should know some general things bc of this

for instance, your ring is not a UFD, so not all irreducables are prime right. Infact this highlights you factorizing 16 in 2 distinct ways

[chilly ocean]

yeah thats what im thinking rn

[upper pivot]

but funnily enough, the quadratic ring of integers Z[ (1+sqrt(-7))/2] is a PID, one of the few quad rings like that

this is an example where taking a ring smaller than an integer ring isnt nice

[chilly ocean]

dude im so pissed for a reason I cant find

btw if R - field, I - ideal of R then R[x]/I[x] field iff I irreduble right?

[oblique river]

taking a ring smaller than an integer ring is basically never nice

[upper pivot]

umm godel

fields have only

2 ideals

lol

do you mean R- ring lol

[chilly ocean]

R field but im looking at R[x]

[upper pivot]

ok but then do you mean I is an ideal of R[x]

[chilly ocean]

yep

[upper pivot]

ok so here the notation I[x] doesnt make sense then

[chilly ocean]

just typed fast

[upper pivot]

alright yeah

[chilly ocean]

but the problem says when is R/I a field

[upper pivot]

R/I is a field iff I is maximal

this is a general theorem for integral domains i believe

[chilly ocean]

yep

[upper pivot]

now what kind of domain is F[x], if F is field

[chilly ocean]

id say euclidean

[upper pivot]

yep

and that also means PID

[chilly ocean]

is maximal iff irreducible?

[upper pivot]

what do the maximal ideals of PIDs look like

for PIDs, maximal ideals are thos generated by irreducable elements

yeah

(unless you mean ideal factorization, which i dont think you do here)

[chilly ocean]

Let $H,F$ be normal subgroups of $G$ and $F\subset H $. Prove the following: $\ \$ a) $H/F $ is a normal subgroup of $G/F$ $\ \ $ b) $ G/H \cong \left(G/F\right)/ \left(H/F\right)$

I think a) I was albe to prove

but got stuck on b)

yes ofc let me edit

[cloud walrus]

Godel:

[chilly ocean]

I guess the usual approach for b) is 1st iso theorem?

but dont know how

[mild laurel]

Yes

[chilly ocean]

We want ker f to be H/F

so f takes the element g + F and spits out elements g'+H

would that kind of identity work?

[mild laurel]

yes check it

[chilly ocean]

I can see the kernel, but how do I check its surjective?

[mild laurel]

by using the definition of surjetivity

[chilly ocean]

nvm

In this one problem they ask to find all irreducible elements of the ring $$R={f/g \in \mathbb R \left(x\right) : g\left(i\right) \neq 0 }$$

Are there any though? Isn't that a field ?

[mild laurel]

Take a picture of the question

[cloud walrus]

Godel:

[chilly ocean]

That's all to it, its actually 'i; and its not said what 'i' is

so I guess a scuffed question

[mild laurel]

nah, just take i to be any real number

and question works well

[chilly ocean]

I mean in that case its a field

I guess just i as in sqrt(-1) and g=/= 0 for any real x is implied right

[mild laurel]

Nah

sorry, g(i) \neq 0 is the right interpretation

yeah i as in sqrt(-1)

and g can be zero for some x

[chilly ocean]

so then irreducible would be f*(x^2+1 ) for non zero f would there be more?

no g cant be zero, f/g is in R(x) (as in field of fractions)

[mild laurel]

think thats it

I mean g can't be the zero polynomial correct

But is it possible that g(x) = 0 for some real x

[chilly ocean]

those are not in the set

[mild laurel]

what set

[chilly ocean]

{f/g in R(x) such that g(i) =/= 0}

[mild laurel]

It is lmao

[chilly ocean]

in R(x) already implies g =/=0 for all real x

[mild laurel]

no

[chilly ocean]

its not R[x]

[mild laurel]

look at your definition of field of fractions again

[chilly ocean]

ok yeah

[mild laurel]

And your idea of not vanishing for any real x isn't closed under addition

[ripe creek]

i need help dont know what formula or how to do dis

[chilly ocean]

lol post it in #prealg-and-algebra

[ripe creek]

ok

[stuck adder]

if i have an equality a²b=c² in unique factorization domain, why does it follow that a divides c?

[mild laurel]

It doesn't

[stuck adder]

well assume b is not zero perhaps

[mild laurel]

In that case yeah, I mean from this you can see that a^2 divides c^2

[stuck adder]

if a² divides c² how is it immediate that a divides c in ufd?

[mild laurel]

you want to show that every prime that divides a also divides c

if some prime p divides c^2, it divides c

[stuck adder]

thanks:)

[valid bridge]

Algebra midterm tomorrow boys

I’m about to spam quadratic equation

[mild laurel]

Your algebra major skills will come in handy

[valid bridge]

No doubt no doubt

[slate brook]

nice algebra

[chilly ocean]

Let $I = \left(4+3i\right) \mathbb Z\left[i\right]$. How many maximal ideals there are in Z[i] that contain I? Prove that $\mathbb Z\left[i\right] /I \cong \mathbb Z_{25}$

[cloud walrus]

Godel:

[chilly ocean]

To the first part - I think its (1+2i) and (2-i) but how do I justify that?

obviously those are maximal because irreducible and they contain I, but how do I know there aren't more? I think it follows from correspondence theorem, but can someone tell me how to write it formally that wouldn't lose any points on exam?

To the second part Idk how to do it without finding an actual formula

[oblique river]

Those two ideals you wrote are the same ideal

Also just fyi when you're using latex, you don't need to do the \left and \right unless the stuff on the inside is oddly-sized. You can just type $\mathbb{Z}[i]$

$\mathbb{Z}[i]$ vs. $\mathbb{Z}\left[i\right]$

[cloud walrus]

Buncho Bananas:

[oblique river]

saves lots of time

but yes, those two ideals you wrote are in fact the same ideal. The right way to do this is the correspondence theorem

which means you need to figure out what Z[i]/I is. Here is a really cool trick that I like:

Z[i] = Z[x] / (x^2 - 1)

therefore by some other isomorphism theorem, Z[i] / (3+4i) = Z[x] / (x^2 + 1, 3+4x)

and then you can quotient by 3+4x first. although unfortunately I guess that method doesn't really work here that well since 3+4x isn't monic

[chilly ocean]

3+4x?

[oblique river]

so Z[x]/(4x+3) is a bit tricky to deal with if you haven't thought about it before

yeah

[chilly ocean]

not like 3x+4?

[oblique river]

oh, I see, yeah I thought it was 3+4i

which is what I also wrote

not 4+3i

yeah it should be 4+3i and 4+3x

but in any case, 4+3x isn't monic which makes it a little tricky

[chilly ocean]

yep

so idk, that doesnt seem much helpful

what I tried to do is like setting 4+3i=0 and try to work with that somehow

25=0

so like, I could construct a homomorphism f(1)=1 f(i) = 7 and check if it works

?

but I think its scuffed

[oblique river]

yeah, there's always that direct route

but it's going to be alittle more work

sry I cant be more help now, I have to run

gl

[chilly ocean]

thx for your time

[sharp sonnet]

what

how is this abstract algebra

[scarlet portal]

Oh ooops

I thought it was in questions

My bad

[chilly ocean]

@chilly ocean $3$ is invertible in the ring $\mathbb{Z}[i]/(4 + 3i)$ so you can invert that first and then use isomorphism theorems: $\mathbb{Z}[i]/(4 + 3i) \cong \mathbb{Z}[1/3][i]/(4 + 3i) \cong \mathbb{Z}[1/3}[x]/(x^2 + 1, 4 + 3x) \cong \mathbb{Z}[1/3][x]/(x^2 + 1, x + 3/4) \cong \mathbb{Z}[1/3] / (25/9) \cong \mathbb{Z}/25$.

[cloud walrus]

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

[chilly ocean]

the last isomorphism since $\mathbb{Z}[1/3]/(25/9) \cong \mathbb{Z}[1/3]/(25) \cong ( \mathbb{Z}/25 )[1/3] \cong \mathbb{Z}/25$ ("localization commutes with modding out")

[cloud walrus]

hochs:

[subtle granite]

Learning a proof for 'A simple abelian group is cyclic of prime order', and I don't really get the argument they're trying to make

so is it trying to say that if there is an element g of infinite order, then g != g^m for any m != 1, which would mean <g^2> is a 'proper' subgroup that is not G or {1}, since g is not contained in that cyclic group?

[chilly ocean]

for any $m \neq 1$.

[cloud walrus]

hochs:

[subtle granite]

yes sorry

[chilly ocean]

then yes

[subtle granite]

Ok thank

:)

[chilly ocean]

neat

[subtle granite]

Simple groups can contain subgroups that are not normal?

[mild laurel]

yes

[subtle granite]

Oh ok

[lime hound]

please explain how they expanded the bracket

[bitter mauve]

why u asking here

[lime hound]

it's an algebra group

[bitter mauve]

just use the formula

[lime hound]

that is algebra

i dont know how

[bitter mauve]

that's nub algebra, ask in #prealg-and-algebra

[lime hound]

okay thanks

[bitter mauve]

$(x+y)^2=x^2+2xy+y^2$

[cloud walrus]

nnutannep:

[lime hound]

it has x to the power of -1

[bitter mauve]

yea y= 1/x in that question

[lime hound]

$(x+x^-1)^2=x^2+2xy+y^2$

$(x+x^-1)^2=

$(x+x^-1)^2

[bitter mauve]

$\left(x+\frac{1}{x}\right)^2=x^2+2x\cdot\frac{1}{x}+\left(\frac{1}{x}\right)^2$

[lime hound]

thank youu!!

[cloud walrus]

nnutannep:

[solemn rain]

Let G be a finite group and let x and y be distinct elements of order 2 in G that generates G. Prove that G is isomorphic to D_(2n) where n = |xy|

help i cant do shit

[wind steeple]

Prove the identity xyx = x(xy)^(n-1)

[rich patrol]

@hushed axle where did you get this question from

[oblique river]

my theory is that you are posting in the wrong channel

[solemn rain]

what is the point of faithful group actions

im still introduced to the notion of gorup actions and i didnt cover much it was just an introduction

what i know that a faithful group action

is a group action where all of the associated permutations induced by the elements

are injective

whats the point of that XD

[fickle brook]

a faithful group action means that no two different elements act the same way on the set

[solemn rain]

cool

thats it

yea got it

[lunar trellis]

De les premier jour ton parfum enivre mon amour

[fickle brook]

what

[faint mica]

Hey people, who can give me a hand with group presentations?

[mild laurel]

I'm here

[faint mica]

Yay! So...

<a, b | ba = a^{-1}b>

I'm having some problems trying to show that every element of this group can be reduced to a "canonical" form (a^m)(b^n) where m and n are integers...

[bleak abyss]

So what does an element of this look like a priori?

[faint mica]

Well, it's just a sequence of a's and b's one next to another without any particular order, no?

[bleak abyss]

Yup, possibly a^{-1} and b^{-1} but yeah

[faint mica]

I actually know how to use the ba = a^{-1}b rule to write them like that, I just don't know how to prove that this is a unique reduced form, that is that (a^m)(b^n) is necessarily different from (a^h)(b^k) whenever the exponents are different.

[bleak abyss]

Ah

Okay here's maybe an idea

This might not work but my instinct is

That relation ba = a^{-1}b keeps the absolute values of orders

Like in a way you can say on both sides "a appears once up to a sign"

Similarly with b

So you might be able to play some game in that regard, just say oh give me a word, take the sum of absolute values of exponents of a, call that O_a, similarly define O_b

And then show that if you use this relation to show that this word is equal to another, then O_a and O_b are not gonna change

Or idk this isn't a well hashed out thing tbh

Zoph what were you gonna say?

@mild laurel

[faint mica]

Oh, this might work, yeah.

Thank you!

[mild laurel]

My idea was to show that the identity is not equivalent to a^n b^m for any integers n,m, but I'm not so sure this is easy

[bleak abyss]

Hmm is this group gonna be π_1(Klein Bottle)?

[faint mica]

Yeah, it's the fundamental group of the Klein bottle.

Also, just checking... that group is not cyclic, right?

[bleak abyss]

It isn't

[faint mica]

Yay, thank you.

And the fact it's not cyclic doesn't directly guarantee that all those (a^m)(b^n) are all distinct from each other, right?

[bleak abyss]

Nah the group <a,b| a^2 = b^3> isn't cyclic I don't think

Since it's a subset of the relations that give you D_3

[faint mica]

Right, so you definitely need that proof you wrote down earlier to show they're all distinct.

[dire shoal]

hey is anyone about

need aid with some uni level pdes

[bleak abyss]

@dire shoal #advanced-analysis lol

[dire shoal]

no one seems active on that room

can i send you the question and see if you can help me with it bro

[bleak abyss]

I'm busy

And I don't know PDE so it's not likely I'll be able to help anyway

[faint mica]

Hey @mild laurel, if you're still online can I ask you one more thing about group presentations?

[chilly ocean]

or just see that there's a semidirect product of Z with Z (1 -> (multiplication by -1)) that is a homomorphic image of ur group, and in this group a^m b^n = 1 iff m = 0 and n = 0.

[chilly ocean]

(for a = (1,0) and b = (0,1) of course)

[leaden finch]

can someone explain to me how he did this proof

[radiant swan]

By the definition of congruence and modulo, you will have [5]*[2a] = [10a], where 10a is a multiple of 10, therefore it’s equivalent to [0] and contradiction

@leaden finch

[leaden finch]

i see many defintions on the congruence and mod

[mild laurel]

and they're all the same

[upbeat juniper]

hi, how does one derive a set of generators and relations from a known group like Q8?

are there some heuristics that I can apply or do I just wing it

[oblique river]

for a group that small, you can just do it

in general, you can take as your generators every element of the group, and for your relations, every possible product in that group

but that's probably not how you wanna do this problem

[fading wagon]

if you remember how the quarterions are formed...

i^2=j^2=k^2=ijk (which is -1) and (-1)^2=1 seems to be good candidates

can you verify it?

[woven obsidian]

Is this statement false: If G is a group of order n and p prime divides n then G has a subgroup of order p

[mild laurel]

What do you think?

[woven obsidian]

Feels false, but then I guess the group would have to be non-abelian and non-cyclic

And for the go-to counterexample Sn it is true so can't use that

[mild laurel]

Hm, one idea is that having a subgroup of order p is the same as having an element of order p

[woven obsidian]

Is it?

For non-cyclic groups?

Ah wait right

[mild laurel]

yeah

[woven obsidian]

Subgroup has to be cyclic

[mild laurel]

yeah okay

I'm not sure there's an easy way to lead you to the solution

so I'll just tell you that this is Cauchy's theorem

[woven obsidian]

Oh really

[mild laurel]

i.e., its always true

[woven obsidian]

Interesting

Not sure if we proved it, can't find it in my notes

[mild laurel]

the proofs aren't too bad

[woven obsidian]

We proved Sylow's thm and wikipedia says it's a generalization of Cauchy's

[mild laurel]

yeah

in particular, Sylow's first is a generalization of Cauchy's

[woven obsidian]

Ah we didn't prove the entire first thm

Only that if order of G=p^k *m where p does not divide m, then G has a subgroup of order p^k

[mild laurel]

Yeah, that's the first theorem

all o fit

You should see how Cauchy's easily follows from that

[woven obsidian]

Well I can write the order as p *p^(k-1)m but then p divides p^(k-1)m

Ah I see now that they left it as an exercise, but you have to reprove a lemma to be more general first

So it doesn't follow from the thm as I stated it

[mild laurel]

Eh, it kind of does

[woven obsidian]

Hmm, how?

The exercise I was doing was to show that every group of order 12 had a subgroup of every possible order except 6, with sylows theorem

So 4,3 followed from my version of Sylow

I showed it had subgroup of order 2 by considering the grouo of the identity and an elt that is its own inverse but wondered if there was a more general thm for primes

So nice to know it works for every prime power

[leaden finch]

can someone help me with this oene

[chilly ocean]

whats definition of prime?

[leaden finch]

p=/= 0 , +-1

its lonly divisots of p are +-1 and +-p

[oblique river]

suppose p satisfies the property you're given and let d be a divisor of p. (Our goal is to prove that $d = \pm 1$ or $\pm p$.)

[cloud walrus]

Buncho Bananas:

[oblique river]

What does it mean that d is a divisor of p?

@leaden finch

[leaden finch]

hmm where did the d come from/

[oblique river]

we're trying to prove that the only divisors of p are +-1 and +-p

so: suppose someone hands you a divisor of p

called d

now you need to show that d is either 1, -1, p, or -p

[leaden finch]

ohhhh

[halcyon siren]

Is anyone here familiar with Lie theory?

[uncut girder]

What's a basis for Z(×)_Z Z

@mild laurel

Wait

Z is not a vector space

I have no intuition for tensorproduct of modules

[oblique river]

it is a good exercise to verify that for any ring R and any R-module M, R \otimes_R M is canonically isomorphic to M

[uncut girder]

Wait that's so weird, the symmetric tensor 1(×)1 in Z(×)_Z Zis not in the image of the symmetrization map

So the symmetrization map doesnt need to leave symmetric tensors invariant

[oblique river]

I'm not sure how that follows from my own statement, but that kinda sounds like an artifact of the fact that you normally would divide by something in a symmetrization map

but you can't divide in Z

[uncut girder]

It doesnt

It's an exercise in DF

Um

I should read the definition of tensor product of modules

Like I get that it's better to use a universal property to define it but that doesnt help you work with the elements of the tensor product

[uncut girder]

So in general, a tensor is a finite sum of simple tensors

[oblique river]

yes

[uncut girder]

Help

Let F be a field where 1=/= -1
Let V be a finite dim vector space over F
Prove there exists a symmetric bilinear map on V that is not alternating.

[wind steeple]

take the trivial one

[uncut girder]

Ie, everything maps to 0?

How is that not alternating ?

[wind steeple]

uh no

the other trivial one xD

tXY

[uncut girder]

tXY?

[wind steeple]

(X,Y) -> tXY

this is a bilinear form

[uncut girder]

I dont understand what tXY means

@wind steeple

[wind steeple]

X and Y are column vectors

[uncut girder]

Ok

[wind steeple]

t is the transposition

[uncut girder]

Ok

Usually it's written XtY

[wind steeple]

Usually, I write it $^tX$

[uncut girder]

Oh interesting

[cloud walrus]

Zak:

[uncut girder]

How is that not alternating tho

When -1=1

Oh sorry

The question is 1=-1

[wind steeple]

idk, take X = Y = (1,0,...,0)

[solemn rain]

Prove that a group G acts faithfully on a set A if and only if the kernel of the action is the set consisting only of the identity.

help

what i tried:

Let G act faithfully on a set A. Define a map A--->A induced by fixed g phi_g(a) = g.a

define a map G--->S_A , f(g) = phi_g

since G acts faithfully , f(g) is injective

consider f(1).

(1 is identity notation wise)

f(1) = phi_1 = 1.a = a for all a

since f is injective

a != g.a for any g other than 1

hence ker of action is only 1

is that correct for the first if statement?

[true dawn]

f(1) = phi_1 = 1.a = a for all a
@solemn rain This should be written f(1)(a)=phi_1(a)=1·a=a for all a in A

Notice that you are defining f() exactly as phi_() (just for your knowledge if you didn't remark)

[latent anvil]

Can someone define the tensor product of two graded modules over a graded ring for me?

Let R be a Γ-graded ring and M, N Γ-graded modules. I want to say that the underlying R-module of M (×)_{R, Γ} N is the tensor product of the underlying modules

I guess because you should be able to factor arbitrary bilinear maps M×N -> module with trivial grading

Not sure if that's correct though

Oh wait hmm, I guess I was thinking that the trivial grading and underlying module functors were adjoint, but I'm having trouble showing that

[solemn rain]

@true dawn yea thank you i got you

[subtle granite]

Back at it again with more groups

oops

I know that I have to use the class equation which says

|G| = |Z(G)| + sum(sizes of possible conjugacy classes in G)

but the thing the result isn't given to us, and I'm not sure how to get to this result from the results that I DO have

[mild laurel]

Use the hint, what are the possible sizes of conjugacy classes in G?

[subtle granite]

since |Cl_G(g)| = |G|/|C_G(g)| by orbit-stabiliser theorem

|G| = p^a, so |C_G(g)| must divide this of course, but I can't think of any other restriction?

[mild laurel]

Nah that's it

So basically it has to be a power of p

[subtle granite]

Oh okay

and if

|C_G(g)| = p^a itself, then Z(G) = C_G(g)?

so we're done in that case

[mild laurel]

uh, yeah

[subtle granite]

yeah?

[mild laurel]

Yeah that's true

It's not super relevant I think

[subtle granite]

hmm okay

okay so

[mild laurel]

What does the center being non-trivial mean?

[subtle granite]

centre being non trivial means it contains more than 1 element

if it was trivial it would just contain identity

right

[mild laurel]

Yep

So |Z(G)| \geq 1

[subtle granite]

yuh

[mild laurel]

Write down the class equation

and think about it mod p

[subtle granite]

ah so we'd have |Z(G)| \equiv 0 (mod p), meaning p | |Z(G)|, so it can't possibly be 1

?

[mild laurel]

yep

[subtle granite]

ohh

that's kinda neat

wtf

I get it

lmfao

[mild laurel]

nice

work

[subtle granite]

thanks a lot !

:)

[mild laurel]

npnp

[subtle granite]

Oof

Need some help proving that there are no simple groups of order 27

So I'm starting with prime factoring 27 into 3^3

but then I'm not sure at all what to do

[upper pivot]

oh yeah this is a classic for prime powers

i will give you a hint

||the center||

[mild laurel]

I mean, use what you just proved

or at least, we just proved here

[subtle granite]

Ohh

Okay so the centre is non trivial

But how does that help us

I was thinking the proof had something to do with the sylow theorems

Which was what I used for a Q asking to prove no simple groups of order 40

[upper pivot]

what can you say about the centre, think about what property it has that is useful to this problem

[subtle granite]

hmm