groups-rings-fields

[delicate orchid]

sure ;3

[crystal turtle]

/s.

[chilly ocean]

How can i show a group of order 312 is not simple

[delicate orchid]

,w prime factors of 312

[cloud walrus]

Results provided by WolframAlpha

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Requested by wewladstbh

[chilly ocean]

Should I look at some sylow subgroup?

[delicate orchid]

yeah

[chilly ocean]

With normalizer index 1

[quiet pelican]

312 = 8*3*13
Look at ||padding 13||
||So there would be 14 sylow 13 groups if not 1 (as it must be less than 8*3, but 14 doesn’t divide 24||

[chilly ocean]

Yeah

Thanks again

[glossy crag]

I really enjoy this image.

[wet zodiac]

@long nebula can i throw 1 into S

[topaz solar]

Yes

[wet zodiac]

[topaz solar]

Since that stays closed

Though, uh, you can get around it as jagr has mentioned

[long nebula]

Lol oops this was an exercise I gave cat bread

Did I make more typos

😭

pain

yes it should contain 1

[delicate orchid]

you didn't claim that the multiplicative set was non-empty and it didn't need to include a 1

but

if it's a stand alone exercise I think it's fine

[topaz solar]

If R doesn’t even have a 1, you might wanna consider other ways to write “1” in Q

[delicate orchid]

the problem with it being in a book is that further sections could use localisations

[long nebula]

R has a 1

[delicate orchid]

and if you weren't able to do the exercise...? you can see my problem now

[wet zodiac]

R is established to be a comm ring

[long nebula]

(with unity)

[topaz solar]

yeah but some authors don’t imply unital 😭

[delicate orchid]

ok lets roll this back then, S has 1

https://cdn.discordapp.com/attachments/496784958430380033/1149436763207647322/Screenshot_2023-09-07_at_9.06.40_AM.png

[topaz solar]

Or, ya know, S u {1} is also closed

[delicate orchid]

I've NEVER seen multiplicatively closed sets defined without 1

is this really that common?

[topaz solar]

I’d believe it if an author didn’t require it

“”Generality””

[delicate orchid]

so see if you can find a map from R into RxS that's a ring homomorphism

[topaz solar]

u(ad-bc) = 0 momento

[wet zodiac]

so now R -> (RxS) where phi(r)=(r,1) is an RH

[delicate orchid]

yur

[topaz solar]

That indeed works when 1 is S

[wet zodiac]

but does this help though

[topaz solar]

Yeah why wouldn’t it

[delicate orchid]

we're literally looking for this map lol

that's the (first half of this) exercise

[wet zodiac]

what

how does this imply that every element in S is a unit

[topaz solar]

And showing that things are units

[wet zodiac]

[topaz solar]

Consider phi(s) for s in s

What might be an inverse

[wet zodiac]

(1,s)

[topaz solar]

Show it

[wet zodiac]

(s,1)(1,s)=(s1,1s)=(1,1)

[topaz solar]

What about injectivity

Swears

D&F hiding nilpotents in exercises

[delicate orchid]

cannot think of a commutative ring with zero divisors that aren't nilpotent

ok thought of one

[topaz solar]

Z[x,y]/(xy) ?

[delicate orchid]

subring of M_2(Z) with everything 0 except either the top or bottom corner

[wet zodiac]

well observe that if there are differing elements s,t in S then (s,1) and (t,1) must be different as (1,s) is clearly an inverse for (s,1) but now if we multiply (1,s) by (t,1) we get t=s if (1,s)(t,1)=1

[quiet pelican]

Z/6Z

[topaz solar]

If S has zero divisors it won’t be injective

[delicate orchid]

which is iso to Z[x,y]/(xy) yeah

[delicate orchid]

well no because if you take 3 and square it you get 3, so if you take it and square it you get 3 and then obviously if you take that and square it y-

idk why I didn't think Z/6Z wouldn't work lmfao

any square free integer works

[quiet pelican]

You’ll never get 0 because your zero divisors (2, 3, 4) don’t have both 2, 3 in their prime factorisation

[delicate orchid]

no no cause if you square 2 you'll get 4, and if you multiply that by 2 you get 2, which when you square you g-

[quiet pelican]

And integer with more than one distinct prime factor works - if p, q divide n, then both p, q are zero divisors that aren’t nilpotent

[delicate orchid]

I think I meant (in my head anyway) a ring where none of the zero divisors are nilpotent but you are correct

[quiet pelican]

(Actually no - no prime works)

[delicate orchid]

anyway I feel like we've gotten a little distracted

[quiet pelican]

I have nothing to be distracted from rn

[delicate orchid]

I meant cat bread's question

the dude seems off doing their own thing anyway so w/e lets talk about uhhhhhh category extenstions or something

[wet zodiac]

what

[delicate orchid]

thank goodness you're back it's tiring filling dead air

[rocky cloak]

Guess the question is, does multiplicatively closed include empty product?

[delicate orchid]

TRUE

[long nebula]

maybe I should actually do the problems before assigning them

usually I do, but I didn't bother with this one

[delicate orchid]

https://tenor.com/view/fnb-get-this-man-a-true-gif-25257480

[wet zodiac]

ok

if S has an element with a zero divisior s such that rs=0

hang on

[delicate orchid]

hanging

[wet zodiac]

i was gonna say (r,1)+(1,s)=(1,s)

[long nebula]

I think if S has zero divisors you also have to be a little careful with defining the equivalence relation

Which I didn't think about at first

[topaz solar]

You didn’t

u(ad-bc)=0 time

[topaz solar]

What’s that sum btw

r/1 + 1/s

[wet zodiac]

(rs+1)/s

[topaz solar]

Ye

[delicate orchid]

I think you can still show non-injectivity even without the u

[topaz solar]

Yeah as cat did

Stuff a lil screwy without id think?

[long nebula]

@wet zodiac imma just go ahead and tell you, consider (a,b) ~ (c,d) iff there exists t in S such that t(ad-bc) = 0

[wet zodiac]

wait a minute

[delicate orchid]

nah it's basically the same you just have extra letters floating around

[topaz solar]

Does t have to be in S

[delicate orchid]

remains to be seen

no it's just a unit

[long nebula]

I copied it from wikipedia

[delicate orchid]

wikipedia LIES

[rocky cloak]

If not just choosing t=0 would make it a little redundant

[wet zodiac]

r where is in R and r maps onto the same element in RxS

[topaz solar]

Ah fair

[wet zodiac]

if thats the case i think

this question is so WACKY...

[topaz solar]

I mean any other outside of S thing that isn’t 0

[delicate orchid]

you've kind of shown that two elements map to the same thing I think

[rocky cloak]

It should still be in S. If you just imagine S={1} you would be saying any zero divisor is 0

[topaz solar]

Ah Yee

[delicate orchid]

you know r/1+1/s = 1/s so what can you conclude about r/1

[wet zodiac]

r/1 must be 0

[delicate orchid]

which is in the image of ur map

[wet zodiac]

so all zero elements will map onto 0

implying noninjectiveness

[delicate orchid]

yur

[wet zodiac]

ok i think going the other way isnt that bad then

[delicate orchid]

and hence we have proven that the set of all zero divisors is an ideal thank u and goodnight

[wet zodiac]

so R is also local

[delicate orchid]

that's non-units

[wet zodiac]

yeah arent zero divisors non-units

[delicate orchid]

there are non-units that aren't zero divisors

[wet zodiac]

oh my god

[long nebula]

xD

[delicate orchid]

society if every ring was local my god

what a world that would be

[rocky cloak]

Every ring is locally local

[delicate orchid]

anyway wasn't there another part to this question

nope nvm guess not

[wet zodiac]

they whipped out the

eric whipped out the fancy 1800s newspaper font in the pset

im scared

[delicate orchid]

At
it's AWAY FROM you FRAUD @long nebula

[delicate orchid]

yeah it's just mathfrak

[rocky cloak]

Nah geometrically it should be at

[delicate orchid]

yeah geometrically cause u zoom in

[agile pine]

https://en.wikipedia.org/wiki/Fraktur

[delicate orchid]

but this ain't geometry!!! I ain't got my compass and set square out for this one!!

[long nebula]

😭 is it actually, I've heard at

[rocky cloak]

So localizing with respect to element 2 is localizing away from 2, localizing at the prime 2 is localizing at 2

[delicate orchid]

it's a convention thing

[wet zodiac]

$\mathfrak{i}$ $\mathfrak{love}$ $\mathfrak{cats}$

[cloud walrus]

weakest eric's pset enjoyer:

[long nebula]

oh does it differ by field

[delicate orchid]

yours is far more common I was just trolling

[long nebula]

ahhh okay ty

[wet zodiac]

@topaz solar

i will

uhh

[delicate orchid]

localise at the prime 0 I dare you I double dog dare you

[long nebula]

that's the next part of the q

[topaz solar]

Q = Z_0

[long nebula]

[delicate orchid]

oh.... bro actually took the double dog dare...

[long nebula]

assuming R is an integral domain

[delicate orchid]

ur puppies will be arriving shortly

yeah obvs or 0 isn't prime

[long nebula]

the next part is to prove that R is integral iff 0 is prime

and then prove the localization is a field

[delicate orchid]

egads!!! I keep spoiling things!!! Gadzooks!!

[topaz solar]

why is that not here yet

[long nebula]

[wet zodiac]

next q is discrete valuation ring

[topaz solar]

I meant the integral iff 0 prime but yes

[wet zodiac]

and then eric talks about nonstandard model of the reals

[delicate orchid]

eric wtf

[wet zodiac]

ultrafilters

[long nebula]

LOLLLLL

[delicate orchid]

bros doing model theory....

[topaz solar]

Clowntown USA

A wild tangent

[long nebula]

good practice for thinking about maximal ideals

[long nebula]

filters are dual to ideals

(ideal in the order theory sense, which is based off the ring theory sense)

[delicate orchid]

mods. Ban this user

[rocky cloak]

Hmm, is there like a universal construction that takes a lattice and constructs a ring with that lattice of ideals

[long nebula]

sounds like stone rep

[delicate orchid]

so u know the errr "free functor"

yeah

it will be easy to generate a ring with that lattice as a sublattice

[summer path]

what if instead of learning about integers in middle school, we only knew about local rings

[wet zodiac]

im curious

how is wew not G+

[rocky cloak]

What is G+?

[summer path]

because he left and rejoine dprobably

[delicate orchid]

because I left the server and rejoined and haven't been bothered to reapply

[crystal turtle]

[delicate orchid]

and it's funnier if I'm just "pending"

[wet zodiac]

https://tenor.com/view/thanos-gif-20009528

[crystal turtle]

Graduate+

[summer path]

it's even funnier if you get denied

[wet zodiac]

This man, G+ Him.

[wet zodiac]

[crystal turtle]

Aka the thing you should be jagr

[delicate orchid]

anyway catbread what's the uhhh local valutation ring exercise

[long nebula]

We're afraid of what he'll do if he gets access to #grad

[delicate orchid]

discrete not local (same thing really)

[summer path]

actually use the vc?

oh

[delicate orchid]

field
coward

Z
coward

[long nebula]

I just copied this problem from my lecturer LOL

I don't know anything about valuations besides what's in this problem tbh

[delicate orchid]

I think I know a little bit but not much

I'll do this exercise though

[crystal turtle]

I mean its enough to cover the important valuations anyways

[wet zodiac]

i know NOTHING

[summer path]

i only know as much as is needed to define the p-adics

[delicate orchid]

it's all just deg(-) really isn't it

oh yeah I suppose you can get a valuation out of a completion

[long nebula]

I added the bit about F[[x]] motivating the definition of local

[delicate orchid]

neat

[long nebula]

But the rest was just an exercise I had to do

[wet zodiac]

eric whats next week

multilinear alg?

[long nebula]

I thought we were on hiatus lol

[wet zodiac]

yeah ik

[delicate orchid]

wait duh it's (can be made into) a graded ring of course you can

[wet zodiac]

after the hiatus

[long nebula]

Maybe we can do some basic module theory, or I could do a pset on properties of polynomial rings

I think what I had scheduled was an intro to field theory but that can wait

[wet zodiac]

will i be able to use multilinear algebra in my linalg class

[long nebula]

Depends how far your class goes

[wet zodiac]

[long nebula]

But the determinant is an important example

[delicate orchid]

hello class this is a matrix and this is the krondecker product and this is the tensor product and this is how the tensor product is used in the compressible euler equations

[long nebula]

So if you learn multilinear stuff you'll better understand the determinant

[long nebula]

YES

[delicate orchid]

stress tensor, or as I like to call it, some garbage + the strain tensor

[summer path]

learn multilinear stuff to better understand , got it

[wet zodiac]

learn multilinear stuff to horrify my prof better understand the determinant

[long nebula]

wew what did you do before getting into rep theory again

[summer path]

wew does rep theory?

[long nebula]

no wew just trolls this server as a job

[wet zodiac]

wew trolled me in the

celeste server

[long nebula]

we hire him to raise the standards for trolls

[wet zodiac]

wew is a hilarious troll

a troll's job is to make everyone laugh while mildly inconveniencing people at worst

if you can't do that you're not a troll

you are a jerk

[delicate orchid]

ahhh I really like question a)

[delicate orchid]

nothing I was just an undergrad but I've done lots of fluid dynamics stuff

[delicate orchid]

I do jacked up wacky rep theory

[long nebula]

ah okay

[long nebula]

I remember being stuck on that for a while

I also changed the order of the parts, so that used to be the last part of the q

[delicate orchid]

I recall reading in a paper that ||The set {a in Frac(R) | \nu(a) > x} for any x in Z is an ideal if you take nu(0) = \infty as you should||
which is also just obvious

[wet zodiac]

i want to learn how to write wicked sick problem sets like this

[long nebula]

this makes sense

[delicate orchid]

oh wait you've only defined it for F^{\times}
annoying

don't care tbh

ok ||now I've shown that R-I_0 = R-{blah blah : blah > 0} = R^\times so that's local done||
wait not quite actually whoops

[white oxide]

Just got fucking destroyed by a quiz

how would one show that $M_p \simeq M/pM$? I did some dumb shit like $m \mapsto m + pM$

[cloud walrus]

okeyokay

[white oxide]

Where $M$ is a module and $M_p = m \in M$ such that $pm = 0$

[cloud walrus]

okeyokay

[white oxide]

And p is a prime

[delicate orchid]

yeah that map seems correct then

the kernel of that map is M_p

oh wait yeah now I see

[white oxide]

wait is that right

[tribal moss]

M_p is the kernel of the map "multiply by p".

[delicate orchid]

nah my line of thinking was backwards

yeah exactly

[south patrol]

That is cursed notation lol

[delicate orchid]

then you get it by first iso

[tribal moss]

Localization of a module :-p

[white oxide]

Eisenbud used the structure theorem after we took the quiz and obtained the isomorphism that way

Either way I got a 10% on that shit

[rocky cloak]

I mean this is not true as stated

[white oxide]

Only think I was able to show was that (1/2, 1/3) was cyclic as a Z-sub module of Q lol

[tribal moss]

But shouldn't it be pM simeq M/M_p?

[south patrol]

i mean take M = Z, p =2, then you are saying 0 = Z/2Z

I guess under certain hypotheses it'd be fine

[tribal moss]

This will make characteristic 2 so much easier to deal with.

[south patrol]

lol

So what is the actual question

[white oxide]

idk, i swear to god it was like if p is a prime and M is a finitely-generated module and R is a PID show that M_p is isomorphic to M/pM

[delicate orchid]

the M = Z as a Z-module in question

[rocky cloak]

It's probably what tropos said

[white oxide]

oh yea maybe i misinterpreted the definition of M_p or something but it was definitely show M_p is isomorphic to something

lol some kid completed the quiz in 10 seconds and i got depressed

other questions involved decomposing such and such into cyclic modules

i forget something of that nature

[young coral]

help me with these two problems please
Q1 Define the sign function from Sn to { 1, -1}. Check that it is well defined and is a group hom. Discuss the kernel of this hom.
Q2 Show that An has no non trivial normal subgroups for n >= 5

[topaz solar]

Hom(R, M_p)?

Or something like Hom(R/p, M)?

[white oxide]

nah definitely not a Hom in the question

idk let me ask a friend. i swear to god it was that

[topaz solar]

Because I know you asked a question like that here

[white oxide]

ye but unfortunately he didn't put any questions on the hw or related to the hw on the quiz

[topaz solar]

Tropo’s sounds right

[white oxide]

well there was that one on jordan canonical form for #7 but i skipped that shit

also M is torsion

idk if that changes things

i'm too lazy to check, i'm doing anal rn

[rocky cloak]

So I'm guessing you know about odd an even permutations, and can guess what the map from Sn -> ±1 should be.

So the main problem is showing that being odd/even is well defined. By multiplying an element with its own inverse you can reduce this to showing that the identity is not odd.

You can do this with a little bit of induction and seeing what happens when you swap the order of transpositions.

For Q2 the usual proof is just checking a bunch of cases, to show that a normal subgroup contains 3 cycles. Then show that An is generated by 3 cycles.

[mossy lintel]

I know if there exists a subgroup of order p, then it must be normal in G and there must also exists a subgroup of order q.

[next obsidian]

Do you have Cauchy’s theorem?

[mossy lintel]

This is an exericise before Cauchy's

[next obsidian]

Do you have Lagrange’s theorem at least?

[mossy lintel]

Yep

[crystal turtle]

why specifc p > q

[next obsidian]

Okay, then if it didn’t have such a subgroup it would only have order q elements and the identity

Maybe from there you can do something?

I don’t know off the top of my head but knowing that each order q element generates an order q subgroup, and all of these intersect trivially you can like write pq - 1 as a product of q-1

This is partitioning up G\{e} into H\{e} where H is a subgroup of order q

And from that I think you might be able to get a contradiction?

I guess if q is 2, then you can’t

Hmmmm

🗿 sorry

[mossy lintel]

I can deal with q=2 separately

[next obsidian]

I don’t know if the numerics line up tho ¯_(ツ)_/¯

[oblique plaza]

help

oh

[mossy lintel]

It seems there's no specific reason, since it also works when O(G) = p^2

[oblique plaza]

guys what’s 0.00633 in scientific notation

[crystal turtle]

not the right channel

[oblique plaza]

HELP IDK THE CHANNELS

[crystal turtle]

Well it's less interesting / easier to show in that case lol. Just that you're showing it has both p and q order subgroups so it's just odd to mention

[next obsidian]

I’m pretty sure this problem is going to go into classification of groups of order pq

Where it matters which is p and which is q

[rocky cloak]

Don't think they do. For example 5*3 - 1 is divisible by 3-1

[mossy lintel]

If q > 3, then pq-1 is not divisible by q-1?

[rocky cloak]

13*5-1 is divisible by 5-1

[chilly ocean]

I just started rings. I have a question about a problem. A finite ring with char(A)=p prime and a in A-{0} nilpotent. Show there is s>=1 s.t. a^p^s =0

I dont understand isn't obvious we can make s large enough?

[topaz solar]

Sure looks like it?

[crystal turtle]

strangely worded problem lol. Should be obvious provided you can prove a * 0 = 0 for all a in A

[chilly ocean]

Yeah..

[topaz solar]

Clown problem

Well ig you also want that p^s gets bigger but

[chilly ocean]

The solution to the problem: let t>=2 with a^t=0. If (t,p^k)=1 then a=0 false. So p divides t so t=p^s × m with m not divisible by p. So (a^p^s)^m=0 and (m,p)=1 and from here we get the only possible solution m=1, otherwise contradiction with a≠0.

I dont understand why this problem exists

[crystal turtle]

how do you conclude tha p divides t

gcd(t, p) can be 1

in fact

if a^n = 0 for a in A, then a^k = 0 for all k >= n

So uhh that conclusion makes 0 sense

actually, none of it makes sense

[topaz solar]

whuh

My child what is the definition of nilpotent

[void cosmos]

can someone give an example

of a primitive ideal that is not maximal

[terse crystal]

I think you can use m^2, for a local ring (R,m)

[void cosmos]

with m being the faithful R/m^2-module?

[ancient ravine]

I've been stuck on this problem for a day and I still have no clue.

What I know is that $\langle A,B,C \rangle$ is isomorphic to $Q_8$, the quaternion group of order 8.

[cloud walrus]

Meno

[ancient ravine]

The induced conjugation action is supposed to be $\operatorname{conj}_g(S):=gSg^{-1}= {gsg^{-1}|s\in S$ where $S \in X$. But it is not technically a group action by $G'$ on $X$.

[cloud walrus]

Meno

[terse crystal]

With some choice we can have R/m^2 itself being simple. Not necessarily local, I shouldn’t have mentioned local. like k[x]/(x^2), it’s a simple k[x] module, generated by x+(x^2) and have (x^2) as its annihilator

[rocky cloak]

k[x]/(x^2) is generated by 1 + (x^2)

[terse crystal]

Sorry my bad, yes , 1+(x^2)

[rocky cloak]

Hence it's not simple

For commutative rings primitive = maximal, since all ideals are two-sided

[terse crystal]

You can translate what this kernel is in a different way, like matrices that are commutative with A,B and C

[terse crystal]

Oh, yes. R/I always have sub module J/I for I contained in J… I only considered choosing a cyclic module completely forgot it should have no proper submodules…

[ancient ravine]

that means kernel is a centralizer of X, but it is not even an homomorphism...

I have no idea how this induced conjugation is a group action

I probably have to think of another kind of mapping

[rocky cloak]

Consider the ring of 2x2 matrices over C. Then C^2 is a simple module.

This has annihilator (0) which is not a maximal (left) ideal. It is maximal two sided though...

[ebon pine]

I am reading Rotman and he defines signum of a permutation in a way I've never seen before. While it works, what's the intuition behind this?

[rocky cloak]

A similar example, let V be an infinite dimensional vector space. And let R be be End(V). Then V is a simple module with annihilator (0).

[ancient ravine]

I think I have some ideas now

the image of S could be isomorphic to X

which I think is likely to be true

[rocky cloak]

I guess you can think of the identity as the product of n "1-cycles". Then multiplying by a transposition will merge two cycles, reducing t and changing the sign

[ancient ravine]

but that means I'm mapping gSg^{-1} to S

[ebon pine]

That makes a lot of sense. Thank you

[tribal moss]

It seems a nicer definition than the usual "count inversions" one, since it doesn't depend on fixing an order of the set we're permuting.

[terse crystal]

I just calculated. If I didn’t calculate wrong , X is the orbit of A under conjugation. I let an element of the group be (a,b;c,d), I checked whether ||a^2+b^2 and c^2+d^2 are 1 and -1 when ac+bd=1 or -1, and a^2+b^2=c^2+d^2=1 or -1 when ac+bd=0||

[rocky cloak]

Why do you say it's not "technically" a group action?

[ancient ravine]

because it supposed to be GxX \to X if it is an action and with some property, but the mapping given above does not map to X but some other set. However, I think that the image is in fact isomorphic to X, so we indeed have a group action.

[terse crystal]

You just need to calculate that they do map X into X, it’s not an obvious fact without any calculation…

[terse crystal]

Like this… need calculation to verify…

[ancient ravine]

I am not sure if that is true? You are saying that the conjugacy class of A := ${gAg^{-1} | g\in GL(2,3)}$ is equal to $X$ ? But X is a set of subsets?

[cloud walrus]

Meno

[terse crystal]

Yes this is what I checked. {gAg^-1: g}=X={+/-A, +/-B, +/-C}

[ancient ravine]

!

[terse crystal]

Now this being said, kernel can be also obtained as intersection of C , uCu^-1, and vCv^-1, where C is the subgroup of matrices commuting A, where u,v are matrices such that B=uAu^-1, C=vAv^-1 . I don’t know which one is more convenient I haven’t calculated the kernel yet…

[rocky cloak]

So you can see that the 6 matrices ±A, ±B, ±C are exactly the matrices with determinant 1 and trace 0.

These are preserved by conjugation, so gAg^- will again be one of these. And then g{±A}g^- = {±gAg^-}

So this does indeed give an action of G on X

[terse crystal]

Oh this is a lot more convenient

[ancient ravine]

After calculation, I think you are right @terse crystal that indeed the orbit of A is the set ${\pm A, \pm B, \pm C}

though $X$ is not that set, $X = {{\pm A}, {\pm B}, {\pm C}}$

[terse crystal]

Yes, though by what jagr said you didn’t actually need this calculation 😂 I mean if you calculate kernel as matrices that commute A,B,C, you didn’t need the fact that there is one orbit actually

[cloud walrus]

Meno

[rocky cloak]

Well you still have to calculate that these are all the matrices with determinant 1 and trace 0. But that calculation is slightly easier

[terse crystal]

A lot easier…

[ancient ravine]

So you mean that A,B,C are the only matrices in GL(2,3) s.t. it has det = 1 and Tr =0?

[rocky cloak]

Yes, those and -A, -B, -C

[terse crystal]

Oh I didn’t notice that X is {{+/-A},…} not {+/-A,… }, then call Y={+/-A,.. }. G’ acts on Y so it acts on X, kernel is bigger than what I said previously… (like C should be {g : gA=+/-Ag})

[ancient ravine]

You are right I have just proved that this is true.

[rocky cloak]

If you want a more clever / non-brute force way to find the kernel

||Q8 has the property that every subgroup is normal||

||±A are the two generators of the subgroup generated by A (and similar for B and C), so Q8 permutes these, hence fixes {±A}||

||So Q8 is in the kernel, but |G'|/|Q8| = 24/8 = 6 = |S3|, so the kernel is Q8||

[ancient ravine]

That is very smart @rocky cloak , I would like to point out a typo in your answer that |G'|=48

[rocky cloak]

Multiplication is hard sometimes

[ancient ravine]

A beautiful result...

[rocky cloak]

It's 8*6 at least, that's all I know

[ancient ravine]

it's indeed 48

[ancient ravine]

@rocky cloak For the last line of your answer, what is your reasoning that |G'|/|Q8| = |S3| indicate that Q8 is the kernel? I know that $GL(2,3)/C_G(Q_8) \simeq Aut(Q_8)/Inn(Q_8)\simeq Out(Q_8) \simeq S_3$, which shows that $Q_8$ is indeed the kernel. But this argument is pretty ugly....

[cloud walrus]

Meno

[ancient ravine]

In fact, I might be wrong.

[rocky cloak]

So assuming you have already shown that the homomorphism is surjective, then you know the kernel has order 8

And Q8 is contained in the kernel, so they must be equal

[ancient ravine]

That's clean

[rocky cloak]

And for surjectivity I guess you can use that two matrices are similar iff they have the same Jordan form

[ancient ravine]

it is surjective, as A , B and C are similar to each other

[ancient ravine]

Thanks for the help! @rocky cloak @terse crystal I've got stuck on this one for too long. 😦

[void cosmos]

yo is the set of all linear operators that have finite column space

is this a tw-sided ideal

?

[rocky cloak]

Yes

[void cosmos]

ty

[white oxide]

how do these two preceding paragraphs imply that G is the disjoint union of the cosets of H? i've already seen this result multiple times, but i can never fully comprehend why it's the case. okay, sure aH, bH not equal implies that they're disjoint, but how do we know that they're not equal in the first place? moreover, how do we know that they make up G?

i've never fully comprehended lagrange ig

[open sluice]

disjoint union = partition = mutually exclusive and exhaustive

these cosets must combine to make up G, because every a is in its own coset aH

for any two cosets aH and bH, there are two possibilities: they are either the same or different

if they are the same then that’s that

if they are different, then one of the cosets contains an element that is not a member of the other

and through groumps theory you can show that distinct cosets are very distinct, in the sense that they are disjoint

[crystal turtle]

Groumps

[white oxide]

hmmm okay i think that helps, but what i'm confused about is how do we know that a_1H, a_2H, \dots, a_nH where a_i \in G are disjoint?

[topaz solar]

See the paragraph right above where you underlined it

If they’re distinct, they’re disjoint

Just apply that to when aH =/= bH

[open sluice]

being a disjoint collection can potentially mean one of two things
either the collection is disjoint as a whole but not pairwise, meaning that the entire collection has empty intersection but two members might overlap
or the collection is pairwise disjoint, meaning that no two members share an object
pairwise disjointedness implies disjointedness as a whole, so we just need to consider two distinct cosets aH, bH and show that they cannot possibly overlap

[delicate orchid]

This is a fun little proof if you haven’t done it before

[crystal turtle]

Alternatively, you can show that they come from an equivalence relation, from which it's clear it's a partition

[white oxide]

ohhh okay so I basically have to show that for any a_1H and a_2H with a_1, a_2 in G that they're disjoint?

[delicate orchid]

Another fun little exercise!

[white oxide]

bruh i should know this by now but the proof of lagrange always confused me lol

[delicate orchid]

Equal or disjoint cause you can have a_1 = a_2

[white oxide]

ah ok

a_1H = a_2H iff a_1 = a_2

?

this is embarassing

wait no that's not true

[delicate orchid]

well no because if a1 =a2h with h in H then clearly

Yeah ok

Lemme write it out clearly

$aH \cap bH = \emptyset$ or $aH$

[white oxide]

i guess i'm confused about is what if aH = bH

ye

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

And if the intersection is aH then clearly aH = bH

[topaz solar]

Then it’s not counted twice in your disjoint union?

[open sluice]

what we're trying to say is that any two cosets are either identical, meaning they contain exactly the same elements, or disjoint, meaning they share no elements
so for any two cosets, it's all or nothing

[white oxide]

OHHHHHH

okay lmaooo

okay that makes sense

okay thank you, sorry for the fiasco

[delicate orchid]

Yeah when we write out a bunch of a_iH it’s almost always the case that the a_is are a transversal of H (in whatever)

[topaz solar]

Truly a classic

Now go reread Lagrange’s theorem bro

[delicate orchid]

I mean you can take the disjoint Union of two equal sets it’s called the coproduct lol

[ember field]

Guys seeing representation theory I dont see its point. Can anyone give example of some group which is difficult but its property becomes clear when seeing at representation. Also is there a strategy on how do I represent a group such that I get good informative representaions. Please give me motivation

[delicate orchid]

It’s a fair point to bring up

[delicate orchid]

You can very easily detect normal subgroups from the irreducible reps of a group by looking at their kernels, for example

The latter question is very vague and has many answers

[white oxide]

incidentally i'm reading about it rn

🤯

[topaz solar]

Corrupted by categories

[delicate orchid]

If you want a general motivation. Linear algebra is solved, group theory is not. Turning group theory into linear algebra solves group theory

[ember field]

Do u have some concrete example or how would you try to find a representation for your group. I am willing to accept heuristics. The question is also vauge.

[delicate orchid]

The classic example is an abelian group, where you decompose it according to the structure theorem and then build the reps back up from the cyclic terms in the decomposition

But uhhh

D_8 is a nice example

[topaz solar]

Jacobian conjecture said:

[delicate orchid]

Who?

[delicate orchid]

Scratch that - S_3 is even nicer

No lifting nonsense required

We have the trivial rep because we always have the trivial rep

And then we have the map sending permutations to their sign (which is one dimensional)

And then we have the two representation given by how it acts on a triangle (S_3 \cong D_3)

These are all of the irreducible reps for S_3 but for a bigger group you’d start off similarly - starting off looking at basic group actions

And then get reps from them in the standard way and then you can then get new reps by taking tensor products of stuff and then decomposing into irreducibles using the inner product and stuff

Or inducing/lifting from subgroups blah blah blah

[rocky cloak]

This is sort of a false dichotomy. Whether or not the representation theory of a group is an easily computed invariant is a completely different question to what the point of representation theory is.

For example Physicists and chemists are not typically interested in groups, but in representations of groups, because in those contexts groups always appear as acting on some space.

Representations are the interesting thing. Why even care about groups if not to study their representations?

[white oxide]

here, would the best approach be to show that (x^2 + 1) is maximal in Q[x]?

should be pretty easy hopefully

[next obsidian]

What do you know about Q[x]

List as many adjectives as possible

Depending on what you know, it makes it easier or harder

[coral shale]

cute

[ashen quartz]

there is a very nice website https://ringtheory.herokuapp.com/

[next obsidian]

[rocky cloak]

The exercise specifically says to use these Q[x] is a PID, what do you know about primes in aPID?

[next obsidian]

It matters what the person themselves know

And is allowed to use

[white oxide]

yea wait let me think about this

[white oxide]

oh wait primes are irreducibles right

[rocky cloak]

That's true

[white oxide]

oh wait

is it true that irreducibl epolynomials generate maximal ideals

i remember that as a theorem

[upper pivot]

oh oh let me try this

[white oxide]

iirc

[upper pivot]

its a.... set?

[coral shale]

adjective

"big"

[white oxide]

so it suffices to show that x^2 + 1 is irreducible in Q[x]

[coral shale]

x - i is a factor sotrue

[coral shale]

prove it when

a bit of suffer but so worth

[white oxide]

nah yea i will this seems to easy

too easy

like to just use that result

but worst comes to worse i'll cite that theorem

😹

[coral shale]

i used that so many times in galois

it might as well have been the only thm

[white oxide]

ok time to prove: primes are irreducibles in PIDs, irreducibles generate maximal ideals in F[x]

now i forgot my irreducible tests

[coral shale]

eisenstein is one

u also got gauss

[next obsidian]

Bro it’s a quadratic

[coral shale]

might be the name

[next obsidian]

Just check if it has a root

[coral shale]

[next obsidian]

A degree <=3 polynomial over a field is irreducible iff it has no roots

[coral shale]

thats a gud one

[white oxide]

OH YEA FACTS LMAOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

[rocky cloak]

So in fact the more general statement that nonzero prime ideals in a PiD are maximal, is true

[white oxide]

i'll try to prove this then

[coral shale]

well the usual statement is

generated by irreducible is maximal in pid

[white oxide]

ye

[coral shale]

So scalar multiplication in modules is a ring action
. : R -> End(M)
If we take M over R/ker ., now this action is injective im wondering if there's anything special about the new module as a result? I don't think its quite free but

This is just rando thoughts

[white oxide]

$x \mapsto i$ and $1 \mapsto 1$ works for the isomorphism right

[cloud walrus]

okeyokay

[coral shale]

yh.

[white oxide]

yh.

ty.

[coral shale]

o ig faithful is the word for injective actions

ok yh found faithful module, cool.

Hmm faithful module seems to be the same as torsion-free to me... maybe im missing something

oh nvm not

[delicate orchid]

Torsion-free modules can still cheat on their spouses

[rocky cloak]

$$\bigoplus_{n=2}^{\infty} \mathbb Z/n$$

Is faithful and torsion

[cloud walrus]

jagr2808

[coral shale]

So in particular, if you have a zero divisor in the kernel

you have non faithful, can be torsion

Uh so like, I think torsion-free implies all elements of kernel of the ring action are zero-divisors. Very tentative thought...

[rocky cloak]

I guess it depends what exactly is meant by "torsion free" if the ring is not a domain

[coral shale]

i think u disregard zero divisors

[white oxide]

here, where do you use the fact that Q[x] is a PID? irreducibles implies maximal ideals, which i'm pretty sure doesn't require that the domain is a PID. maximal ideals implies quotient by maximal ideals is a field, which i'm also pretty sure doesn't use PID hypotheses or whatever

[coral shale]

wait so is there a non-faithful torsion-free module over a domain (which would disprove what i wrote)

[coral spindle]

Let's call this the annihilator of the module rather than the kernel of the ring action. It's true that at least in the definition I know this does ignore zero divisors, but it should be noted this is not sufficient – the module that jagr described has trivial annihilator

[coral shale]

ill have a thonk

[rocky cloak]

Irreducibles need not generate maximal ideals in general

[white oxide]

oh so this theorem requires more context

or like

[coral shale]

shuri learns new terms

[white oxide]

there must have been somea ssumption made earlier in the chapter?

here F is a field as i'm sure ur aware

[rocky cloak]

F[x] is a PID

[coral spindle]

(because F is a field uwu)

[delicate orchid]

Cereal2

[white oxide]

oh so the more general statement would be that an ideal (p) of a PID is maximal iff p is prime over that PID

or you can interchange p being prime with irreducible

since primes are irreducibles in PIDs?

[rocky cloak]

Yeah, so in a UFD we have irreducible implies prime, in a PID we have (nonzero) prime implies maximal. (And PIDs are also UFDs)

[white oxide]

ah okay, that makes sense

thanks

a is irreducible if a = bc implies that either b or c is a unit right

[rocky cloak]

Yes

[white oxide]

ok cool, gonna try to prove prime iff irreducible in PID

altho i do remember reading a proof about it

but like i forget it

[rocky cloak]

(and we exclude units from being irreducible)

[white oxide]

ah ok

[rocky cloak]

So you could frame it as a = bc implies exactly one of b and c is a unit

[white oxide]

ye, if b is a unit then we are done, otherwise suppose b is not a unit, something something a contained in the ideal generated by c

maybe

idk i just need to use the fact that R is a PID

if I can show that (c) = R then i'm done

[rotund aurora]

products in R-mod were the usual cartesian products and coproducts were direct sums (only finitely many entries are nonzero)?

[delicate orchid]

Yeah sure

[crystal turtle]

finite biproduct moment

[dense raven]

is there a phrase for describing when 2 elements of group are the basis for a free subgroup

like "interact freely" or something

[delicate orchid]

Have no non-trivial realtors?

[dense raven]

realtor lmao

[delicate orchid]

We do need a word synonymous with “linearly independent” for groups I agree

[dense raven]

yeah

like when elements are a minimal generating set

which is different from the non-trivial relators thing i guess

[topaz solar]

I think I’ve seen the term “free basis”

[elder lance]

Hey guys, does anybody know how i can prove that if an infinite extension $F \supset k$ has an intermediate field $E$ such that $ F \supset E$ and $E \supset k$ are separable, than $F \supset k$ is separable? In fact, I'm not even sure if thats true. I was only able to prove the result in the case where $F \supset k$ is finite

[cloud walrus]

Maricotinha

[rocky cloak]

Let a be an element of F, and let f(x) = a0 + a1x + ... anx^n be it's minimal polynomial over E.

Then k < k(a0, ..., an) < k(a0, ..., an)(a) are finite seperable extensions.

[elder lance]

ooh cool so same proof as for alebraic extensions

[rocky cloak]

Yeah, the nice thing about algebraic extensions is that you can always reduce to finite ones

[mossy lintel]

Is it true that if $\theta^{-1} (1, 2, 3, 4,..., n) \theta = (2, 1, 3, 4..., n)$ then $\theta$ is an odd permutation?

[cloud walrus]

elgato

[next obsidian]

It’s a general fact that
theta^-1(n1,…,nk)theta = (theta(n1),…,theta(nk)) [or maybe it’s theta^-1 you apply, but it doesn’t matter here]

So what your assumption is saying is that theta simply permutes 1 and 2

So it’s odd

[mossy lintel]

So basically if x is conjugate to y, then there's a unique theta s.t theta^{-1} x theta = y

[next obsidian]

Not necessarily

If you’re in say, S_6

[south patrol]

For instance x and theta could be disjoint

[mossy lintel]

yep (2,1, 3, ..., n) = (n, 2, 1, 3, ..., n-1),

[next obsidian]

And x is like, (123)

And y is (456)

[south patrol]

Lol

Sniped

[next obsidian]

Or uh

Let’s say (12) and (56)

Then (15)(26) works

As does (15)(26)(34)

[next obsidian]

I was assuming that this is in S_n

[mossy lintel]

indeed

[next obsidian]

In that case my argument works because the element is an n-cycle

From that equation you know what every number is sent to. If this was in like S_1000n, then there’s a lot of stuff you don’t know what theta does to

[mossy lintel]

i'm still a little confused, suppose $n = 4$, then $(2, 1, 3, 4) = (4, 2, 1, 3)$, and $(1, 3, 4) (1, 2, 3, 4) (1, 4, 3) = (4, 2, 1, 3)$ , and we found a even $\theta$ satisfies the equation

[cloud walrus]

elgato

[crystal turtle]

I don't think that equation works

Left side maps 3 -> 1 -> 2 -> 2, whereas the right side maps 3 -> 4

[void cosmos]

a commutative semisimple ring is just a finite product of fields right? cuz the matrix rings are isomoprhic to subrings hwich are commutative

like the matrix rings are actually going to be like 1x1

correct?

[south patrol]

Yeah

[void cosmos]

tysm

[south patrol]

Really all you need is that a commutative simple ring is a field I guess and that is clear

Well hm

[void cosmos]

yea

[south patrol]

I think anyway that that is enougg

[void cosmos]

but whenever i see semisimple i just think artin

also

there is a proof that

i can do but there is a detail that jaeger poitned out but i just realiezed i dont get it now

okay so every finite simple ring is just a matrix ring

and the proof goes like

S is artinian so we have a minimal left ideal

so consider like this weird map which i see alot

which is define D = Hom_S(M,M) ( M being the minimal left ideal )

and then S is isomorphic to Hom_D(M,M)

now Hom_D(M,M) is a matrix ring but i dont really see how

like i thought it would be 1-dimensional

but jaeger said it can be any n-dimenisonal

[mossy lintel]

Yep, i was wrong about that one, lol

[mossy lintel]

i thought left side maps 3 to 4 as well ..

i'm applying the mapping from left to right

[crystal turtle]

If you compose them left to right that works lol

real ones composre right to left

[next obsidian]

I go left to right…..

So I was very confused

I am still kinda confused, but it will pass in time

[crystal turtle]

it's function composition

[next obsidian]

Not for me

[crystal turtle]

so right to left

[next obsidian]

I go left to eight

It’s a convention

[crystal turtle]

perhaps

my bad for the confusion

[next obsidian]

I am just confusing myself is all….

[south patrol]

Interesting Chmonkey

I have never met someone outside my uni who does permutations like that

[next obsidian]

I just accidentally did that cuz I did group theory in my second qtr

And was young and noob

And just don’t wanna switch

[south patrol]

Ah okay

[open sluice]

I think I’ve seen a book that does permutations left to right once

I don’t remember which tho

[coral shale]

im used to reading cycles left to right

but really. xfg is clearly the issue here

[crystal turtle]

xfg is immoral. gfx is perfect.

[coral shale]

xcuse your face

[void cosmos]

why does zorn lemma implie axiom of choice

or is this not here

its literally in my algebra past paper tho its so weird

[coral shale]

equivalent.

[void cosmos]

why tho

can someone prove it

[mighty kiln]

Axiom of choice → well-ordering → transfinite induction → Zorn

[coral shale]

why the hell it would be in an algebra paper, though.

[mighty kiln]

Zorn → there is a maximal subset A ⊂ S s.t. there is a function on A sending x to an element of x → choice

(Probably)

(all proofs carried out in ZF)

[topaz solar]

Might be nice to use choice subsets of A = union S, or disjoint union maybe

[coral steeple]

Am I totally off my rocker or is this not H?

[chilly ocean]

methinks it might have a few things possibly not in H

[mighty kiln]

f^-1(f(S)) be called saturation of S with respect to a function f

[coral steeple]

Ah wait if we unroll definitions do we get $\phi^{-1}(\phi(H))={g\in G:\phi(g)=\phi(h)\textrm{ for some }h\in H}={hk:k\in K, h\in H}=HK$?

[cloud walrus]

person2709505

[mighty kiln]

^

[coral steeple]

Groovy

[rocky cloak]

The only matrix ring that is commutative is the 1x1-matrix ring. So if you're assuming commutativity, it will be 1-dimensional.

For a noncommutative example, let S be the 2x2 complex matrix ring, then M = C^2 = S(1,0;0,0) is a minimal left ideal with End_S(M) = C. Here M is 2 dimensional over C.

[rocky cloak]

I mean, Zorn is the one you need if you want to prove stuff in algebra. So it makes some sense you'd want to prove it.

[coral shale]

well i think a lot of courses blackbox

[south patrol]

We just took as an axiom lol

[white oxide]

I'm struggling to see why this is true, I can see the inclusion $f^{-1}\bigl(f(x)\bigl) \subseteq xH$ since $f(g) = f(x) \implies e' = f(xg^{-1}) \implies xg^{-1} \in xH$ (I think at least, I'm not really sure about that either) but since $xH = {xh \mid h \in \text{Ker }f} \implies f(xh) = f(x)f(h) = e' \implies f(x) = f^{-1}(h)$, but this is in terms of $f^{-1}$

[cloud walrus]

okeyokay

[white oxide]

alternatively it's pretty easy to see that xH = Hx, but I just want to see how this argument works

[delicate orchid]

the set of elements that get mapped to x is clearly Hx
so the preimage of f(x) is Hx

[crystal turtle]

well y is in H iff f(y) = e. So we should have z in xH (or Hx) iff z = xy for some y in H, in which case f(z)=f(x)f(y)=f(x)

[delicate orchid]

more generally f^-1(f(A)) = ker(f)A for A \leq G

[crystal turtle]

So z is in f^{-1} ( {f(x)} ); this is the inverse image

[white oxide]

ohhh right bc they become the identity

oops

[white oxide]

ah okay that makes sense

[void cosmos]

oh yeah obviously

yeah

thank you sm

[white oxide]

can somebody help me understand why the kernel of f contains N?

[void cosmos]

N is the intersection of all normal subgroups

the kernel is a normal subgroup that contains H

@white oxide

[white oxide]

OHHHHH

lamo

lmao

i forgot about intersections and stuff

okay thanks

[void cosmos]

np

[rotund aurora]

Is Z[[x]] a PID

Z[x] is not ik

[quiet pelican]

Try the same counterexample

[rotund aurora]

Yeah ok

Idk why I had that written in some notes

Lmao

I think I just meant UFD because I use no PIDness

[south patrol]

If R is Noetherian then dim R[[x]] = dim R + 1

[crystal turtle]

:o

[delicate orchid]

yeah cause u got R and then u got (x)

very wholesome. very chungus

[void cosmos]

what

what does dim mean

in ring

[rotund aurora]

Least integer n st every ideal is generated by n elements

[void cosmos]

hmm

so like

a pid has dimension 1

?

[rotund aurora]

Ye

Very natural concept

[crystal turtle]

AM defines dim A as the supremum over the length of all chains of prime ideals. I imagine these two don't always correlate?

[rotund aurora]

Wait maybe I clowned

I think they dont xd

[white oxide]

yo am i doing this question right??? i did some shit with floor and ceiling functions lmfao, it seems right and i'll post the proof in a minute but like doesn't seem very algebra-y

like

[coral spindle]

It’s a geometric argument

[white oxide]

ah

[coral shale]

im prolly high but istg thats not true. . .

im utterly bamboozled, what.

[crystal turtle]

The norm is not the usual modulus. It's lacking the sqrt

[coral shale]

ah.

[white oxide]

would this argument work?

[crystal turtle]

Not well-defined: 0.49999... = 0.5, but they would be mapped to different numbers

[white oxide]

Wait ok

The tens place doesn’t work

Yeah

[long obsidian]

Can someone verify for me whether a direct limit of groups is an equivalence relation on the disjoint union or the direct sum?

[crystal turtle]

You know that at least one of ceil(x) an floor(x) is less than or equal to 1/2 away. So you always have a choice. Pick one for each x in R. If you oppose using axiom of choice, modify it in a way so that you don't have to make infinitely many choices

[white oxide]

Wait sorry I’m a little confused, why are they equal? Like I know that the sequence that that’s contained in has limit equal to 5, but that term is not specifically equal to 5 right

[void cosmos]

the disjoint union

[crystal turtle]

this is the same as 0.9999... = 1. idk the best short but formal justification, but if you consider the sequences of truncated decimals {0.9, 0.99, 0.999, ...} they clearly converge to the same real number

[white oxide]

ye but like one of those terms is not equal to the number 1 right

like

the sequence (0.9, 0.99, ...) = 1 sure

but like i'm just confused about one of those terms being precisely equal to 1

[crystal turtle]

sure but decimal representation is not perfect. It should be defined (if done with convergence) as a limiting process

[white oxide]

hm ok yea you're prob right

[white oxide]

i'll consider this then

[crystal turtle]

0 . x_1 x_2 x_3 ... := lim{ (0. x_1), (0. x_1 x_2), ... } or something like that

[cloud walrus]

Ryx (Home for flowers)

[crystal turtle]

I think?

idk sounds right

bounded monotone sequence => limit

[white oxide]

hmm ok

[gleaming shell]

Is the following true?

If $G_0 \to G_1\to...\to G_n\to 1$ such that $x\to y \Rightarrow \mathrm{aut}(x)=y$, then $G_0$ is a cyclic group.

[cloud walrus]

Waffles

[crystal turtle]

what are x and y

[white oxide]

here they mean the literal sum of the Rxi right? not the internal direct sum or whatever

moamen this your @void cosmos t

[void cosmos]

yea

[white oxide]

ok thx

[void cosmos]

a direct sum is just the usual sum with that each summand is disjoint from the sum of the rest

its just like a type of sum

[gleaming shell]

they are groups

[void cosmos]

external and internal ithink it means like

which thing u have first

i dont htink they matter tbh

idk much about the difference tbh

but i dont think they matter

[white oxide]

oh wait i thought it was like tuples such that each element in the product was disjoint from the rest

[void cosmos]

they are if the indices are finite

like the direct sum is the same as the direct product

over finite indices

[crystal turtle]

So this is a sequence like $G_0 \to Aut(G_0) \to Aut(Aut(G_0)) \to \cdots \to Aut^n(G_0) = 1$?

[cloud walrus]

Ryx (Home for flowers)

[crystal turtle]

Am I understanding correct?

[gleaming shell]

Yes!

[white oxide]

hmm ok i'll have to review that then

[void cosmos]

just think of the first isomorphism that comes to mind

[gleaming shell]

sorry if my notation is a little weird lol

[crystal turtle]

Huh. Lemma think on this but if I had to guess, I don't(?) think it's true that Aut^n(G) being trivial for some n implies G is cyclic

[white oxide]

oh so internal direct sum each of the Ai is actually a submodule of their sum, external direct sum is the tuples with the disjointness property

and the internal direct sum and external direct sum are isomorphic i believe?

[crystal turtle]

You're good, just wanted to double check!

[void cosmos]

i meant showing that the direct sum is isomoorphic to the direct produc

when the indices are finite

the external and internal sum are more like

say u have two subspaces

then the sum u form is the internal one

( if they have trivial intersection )

[white oxide]

oh ye

[white oxide]

oh yea

[void cosmos]

while the external is more like XD external

[white oxide]

ye that makes sense

[void cosmos]

some shit like that im sure someone more knowledgable can help

but throughout hungerford

it did not use any of that external internal shit

but i think if u managed to do like the exercises of the projective/ionjective modules then u are good enough wiht like

produts and coproducts and their universal properties and stuff

so dont worry about it

[white oxide]

ah okay thanks

[void cosmos]

over infinite indices tho

things change

[white oxide]

right right

but like

[void cosmos]

they are not isomorphic but one is a subset of the other

[white oxide]

ye

[void cosmos]

the direct sum is a subset of th eproduct

cuz with the direct sum the elements are now nonzero for only finitely many indices

[crystal turtle]

See this: https://en.wikipedia.org/wiki/Automorphisms_of_the_symmetric_and_alternating_groups
Aut(S_3) is C_2, the cyclic group on two elements, of which there's only one automorphism. So S_3 satisfies Aut(Aut(S_3)) = 1 but S_3 is not cyclic of course 🙃

[gleaming shell]

isnt Aut(S_3)=S_3?

[crystal turtle]

It's one of the exceptional cases listed

but hmm

you're right

Oh

duh, I misread it

[gleaming shell]

all good

Also |A_3|=3 so its cyclic as well

[rocky cloak]

By Krulls principle ideal theorem, the minimal prime above an ideal generated by n elements has height at most n. So if every ideal can be generated by n elements, the Krull dimension is <= n. But there are 1 dimensional rings that are not principal ideal rings.

[gleaming shell]

Should this question have gone into #advanced-algebra ? I wasn't sure because I'm completely self studied.

[crystal turtle]

If there's a small counterexample, you'd be able to find it here: https://people.maths.bris.ac.uk/~matyd/GroupNames/. Just click on a group's name (on the left) and it tells you Aut(G) (on the right). I can't find any counterexamples though

In fact, it seems as though only small cyclic groups eventually have trivial Aut^n groups. Most others seem to eventually be constant at some other group which is isomorphic to its own automorphism group, say S_n or D_4 or ...

[rocky cloak]

Related paper https://www.jstor.org/stable/119135

[crystal turtle]

ooh thanks jagr

I figured something like this would be true, though I'm surprised it holds for all groups, not just finite. But interesting that they only state that it terminates in finitely many terms for centerless finite groups (or a slight generalization of finite), as opposed to all finite groups.

[rocky cloak]

Shouldn't be too hard to cook up a finite group such that the automorphism groups grow quite rapidly I would think

[white oxide]

can somebody please explain to me why $b_i \in G_{i + 1} - G_i$? I understand that the sequence splitting implies that $G_{i + 1} \simeq G_i \oplus Rb_i$, and here's exercise 15. if we label the inclusion map $f$ as in exercise 15, then we get that the image of $f$ is $G_i$ which makes sense, but I'm having trouble seeing that the kernel of $G_{i + 1} \to G_{i +1}/G_i$ is $Rb_i$

[cloud walrus]

okeyokay

[crystal turtle]

It would necessarily have to be one which never has trivial center though. That's the hard part

[rocky cloak]

Do the groups have to be finite? If not there is a counter example here
https://mathoverflow.net/a/198339/157483

[gleaming shell]

I should of said, but i was only thinking of finite groups

[white oxide]

ok nvm i don't think my reasoning makes sense

[rocky cloak]

Alright, then from what I can infer from the comments on MO it must be cyclic

[white oxide]

ok I understand that G_{i + 1}/G_i isomorphic to Rb_i implies that G_{i + 1} = G_i \oplus Rb_i, but I don't understand why b is in G_{i + 1} - G_i

ohhhh

wait i think i get it now

cuz it's the image of the canonical projection

wait

nvm

cuz that would imply that G_i is the kernel of the inclusion map which is absurd

[rocky cloak]

If b was in G_i, the sum wouldn't be very direct

[white oxide]

lol good point

oops

ok thanks

[crystal turtle]

Okay so to elaborate, I think this might actually be true. Let G be a finite group.
We know that if G is nonabelian, its automorphism group is not cyclic (https://math.stackexchange.com/questions/1898050/if-g-is-non-abelian-show-that-autg-is-not-cyclic). [ As a side note: we also know that Aut(S_3) = S_3 and for other nonabelian groups, their automorphism group must have order at least 8 (https://link.springer.com/content/pdf/10.1007/BF01160361.pdf Theorem 6) ]

We also know that if G is abelian and not cyclic, we can decompose it into A x B for some abelian groups A and B. Then Aut(A x B) has a noncyclic subgroup generated by the automorphisms (a, b) --> (a^{-1}, b) and (a, b) --> (a, b^{-1}) [at least, this should not be cyclic?]. Thus, Aut(A x B) = Aut(G) is not cyclic.

In either case, G noncylic ==> Aut(G) noncylic. By contrapositive, if G is a finite group and Aut(G) is cyclic, then G must be cyclic itself (this fails for infinite groups!). In particular, since the trivial group is cyclic, the only finite groups whose Aut^n(G) is trivial for some n must be cyclic themselves (this should follow by induction I think).

There is also a book/long paper on this: https://sites.math.rutgers.edu/~sthomas/tower.pdf problem in general.

This was a very interesting question!

[gleaming shell]

Thanks so much. I just was scribbling on paper and came up with this conjecture. Also thank @rocky cloak

[dire igloo]

Can anyone explain from p(x bar) =p(x) bar I didn't get that part

And why they have represented p(x bar) as
p(x)(mod p(x))

[eager willow]

So this is really saying in different notation pi•p=p•pi... it's just that canonical homorphic images "image under the quotient map" gets special notation

As for why this is the case, p is a polynomial, so one can write it out term by term and see this property using the definition of pi being a homomorphism

[dense raven]

is "freeness" necesarily preserved by limits?

not quite sure what branch of group theory this is

[dire igloo]

Okay thanks !

[dense raven]

seems like a "groups acting on trees" type of problem

but idk

[dense raven]

nvm, this is just true

i figured it out

[austere escarp]

Is the property of equality implied of an algebraic strucutre? ie if you have some binary operation on a set S, does a = b imply a*c = b*c for some a, b, c in S?

[crystal turtle]

A binary operation is a function, hence is well defined

So this holds for all binary operations

[austere escarp]

can you elaborate on that a bit more

[crystal turtle]

A binary operation on a set S is a function •: S x S --> S. The operation a•b is shorthand for •(a, b). By virtue of being a function, x=y implies f(x) = f(y) for any function f (this is what I mean by being well defined. In essence, it's the statement that each element in the domain is mapped to a unique element in the codomain/range)
Thus, if a = b then a•c = •(a, c) = •(b, c) = b•c

And this works for any c in S, whenever a = b is an element of S

[austere escarp]

Ah, yes that makes sense

x=y implies f(x) = f(y)
helped to think about it like that

[lilac mango]

I'm studying free groups and I'm kind of struggling to understand how to write proofs involving them. Specifically I'm trying to prove that the free group on 2 generators has a subgroup isomorphic to the free group on 3 generators

My problem is that they seem too abstract of a concept to work with

[terse crystal]

I guess you can start with index. Suppose the index of this subgroup is j, 2j-j+1=3 gives you j=2. Now you have index being 2, this will help you find out what kind of generators it has

Square of something , you know…

[lilac mango]

I don't understand what you're saying here tbh

[terse crystal]

I mean now it becomes finding a subgroup G of it, whose index is 2

index 2 gives you hint, you can let G have generator like, for example, a^2

[lilac mango]

How do you know its index is 2?

[terse crystal]

Because index j subgroup G of a free group F of rank n, G is a free group of rank jn-j+1

I saw this in algebraic topology by Rotman in chapter 10. I am sure there exists pure algebraic proof but I didn’t read them…

(F=<a,b>, normal subgroup generated by ||a^2, b^2, abab|| is of index 2)

Oh I should rephrase it. I don’t know its index is 2, but finding a subgroup of index 2 is sufficient

[topaz solar]

This feels like a rather powerful maneuver to use here, ngl, what if you took something like, ab =/= ba for the generators yeah?

so how do ab and a^2 b^2 compare?

[lilac mango]

What do you mean by how they compare?

[topaz solar]

What does <ab, a^2b^2> look like

[lilac mango]

Well the only thing I see is that you can't make arbitrarily large chains of as or bs

So you are not generating the whole free group <a,b>

[lilac mango]

Because taking inverses doesnt give you more as or bs as you get inverses

[topaz solar]

You’ll have a hard time getting a^2 ye

Here’s the thing

This is a free group

What about if it’s 1 or 2 generators

[lilac mango]

Well 1 generator is just a copy of Z

2 generators wasn't what we were working with in the first place?

[topaz solar]

Well, if <ab, a^2b^2> is free on 2 elements, can you see a way to add more?

[void cosmos]

i have

the stupidest question

in existence

why is the quotient of a semisimple module semisimple

cuz quotient preserves artinian , and maximal ideals of the quotient correspond to maximal ideals that contain the ideal so idk how cani say something about the jacobson

[terse crystal]

Semisimple module M is direct sum of some irreducible submodules M_i, I think you just need to show that (M_i +N)/N is irreducible

[chilly ocean]

G1,G2 are subgroups of G
What's G1G2

[rocky cloak]

G1G2 = {gh | g in G1, h in G2}

Sometimes this might also be used for the subgroup generated by this set, if it doesn't already form a group

[mighty kiln]

Surely generated subgroup would be〈G1G2〉

[delicate orchid]

… or G1G2 for short

Exactly the same situation with ideals

[dim widget]

Surely ideals would be spelled <Ideals>

[delicate orchid]

<\Ideals>

[errant shadow]

Is it true that if you remove all the invertible functions from a semi-group of set of functions defined over compositions -- you break closure of the binary operation? Can you give an example?

note - the functions have been defined over positive real numbers as domain and co-domain

[errant shadow]

I have been trying to figure out why we don't get a semi-group here in this case.

[tribal moss]

For example, take functions N -> N defined by: f(n) = n+1; g(n) = max(0,n-1).
Then neither f nor g is invertible, but g o f is the identity which you've just removed.

[errant shadow]

Oh great, thanks a lot!

[errant shadow]

if you take a commutative operation here instead of composition then why does that make it a semigroup necessarily? Is it the non-commutativity of composition that's causing it to break closure?

[mossy lintel]

i'm trying to prove group of order 15 is abelian by showing it's center is of order 15, since the center Z is a subgroup of G, by lagrange, it must have an order of 1, 3, 5 or 15; i have proved o(Z) \neq 3, o(Z) \neq 5 with the class equation, but i can't see why o(Z) can't be 1

[rocky cloak]

Are you familiar with the sylow theorems?

[tribal moss]

If you have two elements whose composition is invertible, that means each of them has at least a one-sided inverse (to different sides for each). If your semigroup is commutative, that is enough to make each of them invertible in themself, so a counterexample cannot exist then.

[mossy lintel]

for n=15; I think the first part of sylow's just states the existence of subgroups of order 3 and 5,

[delicate orchid]

it's Sylow's third that you want here

[rocky cloak]

Yes (and also the second)

[delicate orchid]

depends what you include under each theorem statement I suppose

[mossy lintel]

With the third Sylow's, I get subgroups of order 3 and 5 are unique.

[errant shadow]

Oh cool, that makes sense.

But wait, what if you are defining the set over the commutative operation as a set which as a rule, doesn't contain invertible elements? Not talking about compositions here, just some general operation that gives a Monoid over a commutative operation but I took out all the invertible elements. Why do I get a semigroup, necessarily?

I am confused because not all elements in a monoid have a one-sided inverse right?

[delicate orchid]

yeah so you know that they're normal - see if you can conclude the whole group is abelian from that

[crystal turtle]

(you also know that groups of order 3 and 5 must be cyclic since prime order, and in particular are abelian, which might help)

[delicate orchid]

good point - dunno if you knew that or not

[mossy lintel]

I get it now, something like aba^-1b^-1 = e

[white oxide]

lol is this related to graph theory in any way

sorry i have to veirfy a diagram is commutative rn so i thought i'd go back to this LOL

[mossy lintel]

However this is a problem before the Sylow's are introduced, I think the uniqueness of order 5 can also be derived from Cayley's, I wonder if the uniqueness of order 3 can be argued w/o invoking Sylow's (my ultimate goal is to prove group of order 15 is cyclic btw)

[crystal turtle]

categories moreso

[white oxide]

yea ig that makes sense

arrows go brr

[delicate orchid]

simply remove the arrows in the middle to join the cycles ;3

[crystal turtle]

:3c

[white oxide]

here, lang means the inverse image of the canonical projection right

[coral shale]

:3c

[south patrol]

ye

[white oxide]

why is rX contained in F? bc what if we have some x in X that's outside the generating set S for F that takes us out of F

[rocky cloak]

r is specifically chosen such that for any x not in S, rx is in F. That's what the preceding paragraph is about

[white oxide]

ohh right oops

thanks

[sweet echo]

What justifies the name "antiderivation" here?

[river otter]

hi

[delicate orchid]

It’s a derivation but the operator is going on the opposite term

[sweet echo]

the opposite term as opposed to what?

oh im not commutative

[river otter]

@river otter hi idiot

[delicate orchid]

Oh no I’m silly, there’s a small omega there too

Not too sure then