groups-rings-fields

[rotund aurora]

for G finite I meant

but yeah, I used that if |G| x=0 then x=0

also that the action on Z is trivial ofc

For that part, I did this: consider an exact sequence 0-->A-->B-->C-->0, if B^G-->C^G is not surjective then I can find x in B setminus B^G such that f(x) is in C^G, and gx-x is not zero for some g in G. Then gx-x is in the image of A and since the action of G on A is trivial gx-x=g^2x-gx=...=g^nx-g^(n-1)x, adding all up gives n(gx-x)=g^n x-x

and if n is the order of g, this will give n*y=0 for some y

also I believe the claim that H^2(G, Z)=H^1(G, C^x) is the same as saying that H^2(G, C)=0, considering this sequence?

[south patrol]

Okay sure, but you know that is 0 too

[rotund aurora]

no idk how to prove it

[south patrol]

Oh well from this

The higher cohomology all vanishes

Bur yeah needs more proof

[chilly radish]

This shouldn't be true. I don't think H^2 of a perfect group is always 0

[rotund aurora]

I can quote theorems, but is there a straightforward way of doing this?

[cobalt heath]

Just curious, is there a way to "detect" if certain infinite-dimensional vectorspace is a dual of another vectorspace?

[rotund aurora]

the only restriction is cardinality of a basis

[lethal cipher]

For (3)->(1), what was the reasoning behind hopping in the quotient field ?

[cobalt heath]

Oh, makes sense

[rotund aurora]

I think duals of infinite vector spaces tend to grow, but there is an explicit formula which I don't remember, try rederiving it. Then your question is "which cardinals are of this form", which is set theory

[cobalt heath]

I guess one issue with this is that we don't have canonical morphism from the space V^* to V.

[rotund aurora]

the dual of V is F^k where k is the dimension and F is the base field

[lethal cipher]

Ah, it's because the quotient field is a vector space over R

[cobalt heath]

Nvm about this, I forgot that we don't have canonical V^* -> V for fin dim as well

[rotund aurora]

also if B is the cardinality of a basis for F^k I think you should have BF=F^k. If F^k>F then I think this implies B=F^k, but don't quote me on this one idk for sure

like when F=Z/2Z I believe you are asking which infinite cardinals are of the form 2^k

[lethal cipher]

I could use a little help understanding. In the proof for the corollary, I'm not seeing how a-b, ab in R[a,b] implies they are integral over R from the proposition above it.

Oh! At the very least we have that R[a+b] is contained in R[a,b], which is module finite.
So condition 3 is satisfied, which means a+b is integral over R

Same thing for ab

Whole lotta rubber duck moments today

[real sparrow]

What's the automorphism group of cyclic groups

For prime cyclic groups it's just -1 cyclic group.

Don't know what it would be for cyclic groups of compound n

well for cycling group of nm where n m are coprime, I would guess it's just the product lf the automorphism groups of the cyclic groups pf n m

[cobalt heath]

Cyclic groups are one where operations are like additions, right?

[real sparrow]

yes

[cobalt heath]

Which operation sends addition to addition

That is, distributes over addition

[real sparrow]

multiplication.

[sonic coral]

it is a group of order phi(n)

[real sparrow]

I haven't looked I to group theory recently

What's phi

[sonic coral]

eulers totient function

[real sparrow]

I see

[cobalt heath]

Yeah you can approach this elementary number theoretically

[real sparrow]

Yes

what's the automotphism group of 4 or 9

Powers of a prime

[cobalt heath]

phi(p^k) = p^(k-1) * (p-1)

And since it has primitive roots, the automorphism group is the corresponding cyclic group

[real sparrow]

Interesting

So aut(C(p^n)) = C(p^(k-1)(p-1))

[cobalt heath]

Yes

[real sparrow]

Neat

also any 2 coprime n m
Aut(C(nm)) = Aut(C(n))Aut(C(m))

[cobalt heath]

Indeed

[chilly radish]

Generally the automorphism group of a cyclic group isn't going to be cyclic, though

[real sparrow]

Explain

[cobalt heath]

C12 = C3 * C4, its automorphism group is C2 * C2

C2 * C2 is not C4.

[real sparrow]

Oh and what I was about to say would actually imply that.

👍

[chilly radish]

You were probably thinking you could decompose any integer into its prime power factors then use what you said previously, right?

[real sparrow]

the automorphism group of a cyclic group is a product of cyclic groups. Via that prime power equation you used.

[chilly radish]

Yea

It's just that the orders of Aut(C(p^k)), Aut(C(q^k)) might not be coprime, so you can't use the chinese remainder theorem

[real sparrow]

Finite abelian groups.

What about the automorphism group pf finite abelian groups

C(2)C(2) would have more complexity.

I forget the name of that group but uts automorphism group is S(3) I think.

Permutation group

[real sparrow]

You get interactions from the product of the same groups.

Idk if you get some from product of powers lf the same prime

Like C(2)C(4)

I don't think you get anything from C(4)C(9) for example.

Aut(C(4)C(9)) I would guess is just Aut(C(4))Aut(C(9))

or any coprime n m

[toxic zephyr]

it's been way too long since i've done module theory. im trying to derive the module decomposition of finite dim vector space V into a F[x] module (using an operator T) via p(x)v=p(T)v.
structure theorem gives it as a sum of F[x]/(pi(x))
why must pi(x) divide the min poly of T m(x)? if pi(x)=m(x)q(x)+r(x), then we get r(x)v=0 for all v in the ith module. but im assuming we want to show that r=0. and i'm not sure how to do that.
also, can anyone give the fastest derivation to get to the decomp into F[x]/(x-lambdai)^m for the jordan blocks?

[chilly ocean]

It's a nice exercise to show that a monoid with either right or left cancellativity, and is finite, is a group

[lethal cipher]

Okay, so I now have the tools needed to do (*), but I'm running into a weird issue.
For now, I'll just be looking at k[X]
So the homomorphism mentioned measns L is ring finite, so there is a v in L s.t. L=k[v]. But...what if this v is an element in L that is not a root of any polynomial?

It seems like a silly question, but all (*) says is that k is a subfield of the field L.

[real sparrow]

There are 51 groups of order 32

[lethal cipher]

Any other words, how do we know v has to be integral over k?

[real sparrow]

267 of order 64

2328 or order 128

[sonic coral]

there are a lot of prime power groups

there’s some weird facts i’ve seen with like the number of 2-groups

like 99% of all groups of order at most 1024 are of order 1024 or something

oh actually, up to 2000

[real sparrow]

Seems the number of groups grows by a square root exponential

[lusty marlin]

There's only 1 group of order p for each prime p

[woeful sage]

Yooo guys so I was doing some intro Group Theory exercises in Dummit and Foote

and I noticed something strange\cool

I think the group of n-th roots of unity is isomorphic to Z/nZ?

[lusty marlin]

It is

Because multiplying them is essentially just modular arithmetic

[woeful sage]

noicee right yea

Group theory is very cool already

[clear ingot]

yep

[clear ingot]

ye

Un is isomorphic to Zn is isomorphic to Z/nZ

[woeful sage]

it's denoted by U_n?

[woeful sage]

I suppose more accurately it's (Z/nZ)^+

[fading field]

what is Un here? most things that i know that are denoted U_n aren’t cyclic groups

[clear ingot]

but i searched up and i dont think there is like a uniform notation for the group of nth roots of unity

[fading field]

oh weird, i’ve never seen the group named that way :P

[woeful sage]

we have groups of unity, groups of unreal when?

[woeful sage]

which book is this?

[chilly ocean]

monster group

[summer path]

I've seen \mu_n but not U_n

[south patrol]

U_n if anything means the unitary group or group of units of Z/nZ to me lol

[tardy hedge]

Not sure about last part. “Defining phi r = r phi” makes it an R algebra? Where is this coming from and whats going on here

What is phi r? An action on Hom from R from the right side?

Can we use that to show r . (Phi1phi2) = phi1(r. Phi2) sort of thing?

[rocky cloak]

I'm not entirely sure what they're trying to communicate, but you can just define
(rphi)(x) = r(phi(x))

[tardy hedge]

(4) is proving that HomR(M,M) is an R-algebra

They already defined action of R on Hom and showed that it makes it an r module in a previous part

[rocky cloak]

What is (2) is the relevant question I guess

[tardy hedge]

Full thing

[pliant raptor]

I was doing the following exercise in Jacobson's Basic Algebra 1 (section 1.8, exercise 6): Let $G_1$ and $G_2$ be simple groups. Show that every normal subgroup of $G= G_1 \times G_2, \neq G, \neq 1$, is isomorphic to either $G_1$ or $G_2$. I feel that this is wrong? If $G_1$ and $G_2$ are abelian, the direct product is abelian, so every subgroup is normal? I'm also struggling with the non-abelian case

[cloud walrus]

ImHackingXD

[rocky cloak]

Does there definition of R-algebra include a bimodule or something. I'm not sure why they would need to define a right module structure. But indeed phi r = r phi would follow from it being an R-algebra.

[tardy hedge]

Their definition of A an R-algebra is a ring with 1 that has homomorphism R -> A so that f(R) is in center of A

[rocky cloak]

Have you tried looking at an example. Indeed every subgroup being normal does not contradict every subgroup being isomorphic to either G1 or G2

[rocky cloak]

Airtight, then they should define
R -> A by
r |-> (x |-> rx)

[rocky cloak]

A possible hint is to consider the image of the subgroup under the projections G1xG2 -> G1 and G1xG2 -> G2

[tardy hedge]

Yea this is what ive seen so that makes more sense to me

In (4) they didnt even give any ring homomorphism so i just dont understand what theyre trying to do

[tardy hedge]

For 3, cant i just use the example already given in the book previously, even though it was showing it as an example of a ring hom thats not a mod hom

The example was for F[x] and the map was f(x) -> f(x^2) being a ring hom but not F[x]-mod hom

So … can i just consider F[x] as an abelian group and then just use that same example for this question

But it would also be nice to see another example

[tardy hedge]

Ok I saw an example with C and complex conjugation

Does anyone have an example where the module M is not the ring R

[mighty kiln]

A module is a ring homomorphism R → End(G) for G a group

So take two R-modules M, N with an R-module morphism and pick two wildly different group endomorphisms on M, N

Then you get two R[x]-modules, and the R-module morphism is likely not an R[x]-module morphism

[tardy hedge]

What is End(G) in this case ? Ring of homomorphisms of G im guessing?

[mighty kiln]

Yup, the ring is (f+g)(x) = f(x) + g(x) and (fg)(x) = f(g(x))

[tardy hedge]

Right, ty

I also havent seen that way of defining a module before. Its insightful, thanks

[winter fog]

hello, can you help me her pls quick question #help-4

[topaz gale]

It’s incredibly useful later on too!

[tardy hedge]

Awesome, yea it seems like that formulation could be really useful

[mighty kiln]

An R-module is an additive functor from a pre-additive category with one object to Ab and R-module morphisms are natural transformations

[tardy hedge]

Haha

Not quite there yet but one day

[junior badge]

This is Atiyah-MacDonald Prop. 9.6. Why does the grey part hold? 👀

[agile burrow]

The localization a_m is an ideal of A_m. It's equal to the whole ring iff it contains a unit s. By clearing the denominator, assume that s in a. Then it's a unit in the localization iff s in A - m, hence a is not contained in m

[junior badge]

What do you mean by "clearing the denominator"?

We know a_m contains elements of the form x/s with x in the ideal a.

[agile burrow]

Yeah, so multiply by s to isolate x in a

It'll still be a unit in a_m since the product of two units is a unit

[junior badge]

I see, okay

Oh. Done processing now.

Thanks for the help! 😀

[limpid ferry]

I am finding some popular applications of the group GL(n,F). Also, my teacher said something about it is used in linear transformation

[uncut girder]

If V is a finite vector space, T is in GL(V) and f is a permutation of V such that fTf^-1 is also in GL(V), is it true that f is necessarily also in GL(V)?

[fierce walrus]

Is this a good channel to ask about representation theory?

[coral spindle]

Yes

[rocky cloak]

#advanced-algebra might also be appropriate

[fierce walrus]

It's not really a super difficult thing

In this context I mean

[rocky cloak]

Yeah, it might go in either. Idk, what your question is

[fierce walrus]

I'll post it in a bit

[rocky cloak]

Either is probably fine, lots of "not difficult" questions in #advanced-algebra

[fierce walrus]

I was just trying to look around at where I could post kt

[tropic spade]

For part (c) why can't I just pick the trivial proper subgroup as a counterexample?

[south patrol]

Wdym

Thats just H_0 right

[tropic spade]

H_0 is not defined?

[south patrol]

Well often Z+ includes 0

But eh otherwise sure

[tropic spade]

Ah okay. So here they probably meant to include 0 for the definition of the H_k's you think?

[south patrol]

Yeah I guess

[tropic spade]

That makes sense. Thanks!

[south patrol]

Np

[rotund aurora]

What is the induced representation? H is a subgroup of G

G is finite

I'm thinking that you can take rho in Hom(H, C^x) and consider g-->prod sigma(g) where sigma runs through the characters of G that agree with rho at H

[south patrol]

Well you can always induce reps from subgroups to the whole thing

essentially given a $\mathbf CH$ module $V$ you form $\mathbf CG \otimes_{\mathbf CH}V$

[cloud walrus]

Crystalline Potato

[chilly radish]

You can also describe this explicitly after choosing a set of coset representatives

[rotund aurora]

Could you write it explicitly?

I am familiar with the classical transfer

Ig you can take duals to that

[clear ingot]

any idea on how to solve this?

[lusty marlin]

Any group whose order is a prime power has nontrivial center

This comes from the class equation

[rocky cloak]

You can also kinda brute force it. Like if the group isn't cyclic, then every element has order 3, so now you're looking at a group of order 9 generated by two elements of order 3. There aren't very many things that can happen

[warm badge]

hi! i just started learning about rings and fields. my question is if every integral domain can be embedded in a field, then is it also true that every field is a field of quotients of some integral domain? for instance if i am given the field of real numbers R, can i identify the integral domain for which R is its field of quotients such that R is the smallest field containing it? or is such an existence of integral domain actually necessary or not?

[rocky cloak]

So, yes. Every integral domain embeds into a field and every field is the field of fractions for some integral domain.

But that integral domain is not uniquely determined.

A field is an integral domain, and a field is it's own field of fractions, so you could just pick the field itself.

If we consider for example the field Q, then it's the field of fractions of Z, but also of Q and of Z[1/2] and a ton of other integral domains.

[warm badge]

ohhh i see then what about for R? ive been trying to think about an integral domain D ⊂ R, such that R is its field of quotients, like R = {ab^-1 | a,b in D, b≠0} but i can only think D = R

[void cosmos]

true

field of quotients of a field is the field itself

[warm badge]

like in the case of Q i can find Z ⊂ Q for which Q is the field of quotients for Z but I can't find such a convenient proper subset D of R. is that in any way related to how R has a nontrivial subfield already and hence the integral domain for which R is its field of quotients must be R itself

i am just jumping into conclusions here i have no basis for what i just said

[prisma ibex]

R doesn't have a "ring of integers" like this

[warm badge]

yes that's true

R has Z but Z's already into Q

[rocky cloak]

It's a bit hard to come up with other examples.

One thing that should work is take a trancendence basis {xi} of R over Q, then let D be the integral closure of Q[{xi}] in R.

[rocky cloak]

Another thing I think should work would be take the subrings of R that don't contain 1/2, and pick a maximal one using Zorn's. Pretty sure that should work, but haven't checked

[junior badge]

I'm kinda stuck on this one. Since A is a Dedekind domain, we can write c(fg) = p_1 ... p_k, etc. using factorization into prime ideals. But I don't see why localizing at each maximal ideal would help. Any further hints?

[cobalt heath]

Is ring of integers only defined for either

• Finite extension of Q, or
• Non-archimedean local field?

[warm badge]

oh okie thanks for clearing the confusions i will look them up

[cloud walrus]

Heywood Jablome

[untold basalt]

Is this the standard way to see this or is there a simpler way?

[coral spindle]

Yes there's a simpler way

Q(...) is formed of rational functions of the z_i, \overline z_i

(With nonzero denominator)

The conjugate of such a rational function is another such rational function, QED.

You do not need to choose a basis

[rustic crown]

another similar way is to use that Q(...) is the smallest subfield of C containing z_i's and z_ibar's.

[minor fulcrum]

if we have a finite group $G$ and two subgroups $H,N$ of $G$, can we have $G/N \cong G/H$ but $H \not \cong N$

[cloud walrus]

Y. Beer

[coral spindle]

Yes

[minor fulcrum]

can you give an explicit example, because I suspect as much

[coral spindle]

If you're struggling to find an example, here's a hint: Try ||G = Z_4 x Z_2||

[minor fulcrum]

thank you

[chilly radish]

https://en.m.wikipedia.org/wiki/Induced_representation

See the first construction here

[sly crescent]

It’s defined for any global field

[bitter shard]

when i have a group G with an operation +, is this correct ? for every g in G there are x,y in G such that x+y=g

[lusty marlin]

Yes

[bitter shard]

sorry with the addition that x≠g and y≠g

[lusty marlin]

Then it isn't true

[coral spindle]

Try the simplest group you can think of

(To check if it's true or not)

The most trivial group possible

[bitter shard]

like with only the e?

[lusty marlin]

Yes

[bitter shard]

yeah you are right

[coral spindle]

Sure. Is that a group? And does this hold?

[bitter shard]

what more a group needs to have so the statement to be true ?

[coral spindle]

What about the next simplest group possible?

What's the next smallest group apart from the trivial group

This line of thinking will answer your question for you if you choose to follow it

[bitter shard]

like the Z2?

(0,1)

[coral spindle]

Yeah, so {0, 1}

Not (0, 1), this means something else.

[bitter shard]

oh yeah sorry

[coral spindle]

So does it hold in the group with two elements?

[bitter shard]

it does not, right ? cause 0+1=1 and we cant achieve the 1 with an other way

actually we can with 1+0 but you know what i mean

[coral spindle]

Yeah I agree

OK so it doesn't work for those two groups

Does it work for the group with three elements, Z mod 3?

{0, 1, 2} let's say

Try it with g = 1

[bitter shard]

then 2+2=1 so we can

[coral spindle]

Aha yes indeed

So I'm going to observe something

g = 1. Let's say h = 2 just choosing a name for it

so g = h + (g - h), right?

What do you get from this

[bitter shard]

i am not sure... 😛

[coral spindle]

I am going to leave you to think about what this means.

[bitter shard]

maybe that if h is not the identity, we are good ?

oh yeah so if we have one more element aside from g and e we are good, right ?

because h isnt e so (g - h) cant be e either and h isnt g so (g - h) cant be g either so everything good

[quartz wind]

what is a chain of ideals? is it a chain in the sense of a partial order on R defined by $\subseteq$?

[cloud walrus]

esca (@ with reply)

[coral spindle]

Yes

This is borrowing the terminology of Zorn's lemma

[quartz wind]

ok ty

how is it an equivalence relation? what is there to guarantee that $a-a\in S$, if S is just some arbitrary subset of R?

[cloud walrus]

esca (@ with reply)

[coral spindle]

You are right, indeed you would at least need 0 to be in S.

In fact you would need S to be an additive subgroup in general.

I think it's likely that this was mentioned elsewhere, else this is a rather large mistake.

[quartz wind]

alright thanks. i didnt see it mentioned anywhere but I'll check again

[rotund aurora]

maybe I'm just confused about definitions. But if Z/2Z acts on Z/3Z, then shouldn't you have (1+1)·x=1·x+1·x which forces 1·x=0?

[rotund aurora]

Because 0=0·x=(1+1)·x

[tribal moss]

"Z/2Z acts on Z/3Z" doesn't specify that the action has to be something that's meaningful in terms of arithmetic operations in Z/3Z.

[rotund aurora]

I am talking about a Z/2Z-module structure on Z/3Z (see the last line of the screenshot I provided)

ah okay

yeah I rechecked the definition of G-module

[tribal moss]

I think that's a "module over a group", rather than viewing Z/2Z as a ring.

So you just need a group homomorphism from C2 into Aut(C3).

[rotund aurora]

yeah, thanks. I was thinking that the G-action should also be distributive (as in (g+h)·x=g·x+h·x) but that's not true, you just want g·(a+b)=g·a+g·b

[tribal moss]

The notion makes more sense when G is written multiplicatively so instead of a distributive law to the left there's just the associative g·(h·x) = (gh)·x.

[rotund aurora]

also, it's not into Aut(C3), its just Hom(C3, C3)

(I think)

like you could consider the action g·x=0 for all g and x

[tribal moss]

When a group G acts on X, each particular g·_ always needs to be invertible -- because its inverse must be the action of g^-1.

[rotund aurora]

right, you also need associativity. Good point 👍

[wicked patio]

Let $G$ be a group and let $S$ be a subset of $G$. Suppose $a\sim b$ iff $ab^{-1} \in S$ is an equivalence relation on $G$, show that $S$ is a subgroup.

[cloud walrus]

thewizardofOU

[wicked patio]

kind of funny lol

[quartz wind]

clearly associativity carries over from G. we have

• $a\sim a$ therefore $aa^{-1}=e\in S$, which gives us the identity axiom;

• $a\sim b \implies b\sim a$ therefore $ab^{-1}\in S\implies ba^{-1}\in S$, which gives us the inverse axiom;

• and $a\sim b\land b\sim c\implies a\sim c$ therefore $ab^{-1}\in S\land bc^{-1}\in S\implies ac^{-1}\in S$, which gives us closure

[cloud walrus]

esca (@ with reply)

[wicked patio]

nice

there's a weird correspondence always and I'm not sure whether it's just mathematical coincidence or some crazy category thing I don't understand or what

reflexivity - symmetry - transitivity

[quartz wind]

yeah its like one to one lol

[wicked patio]

identity - inverses - closure

and for metric spaces

d(x,x)=0, symmetry, triangle inequality

positivity is kinda extra

Definitely not the first place I've seen it happen

[coral spindle]

It is indeed a category thing, but it is a little complicated to describe. For example a metric space is a small category enriched over the poset of nonnegative reals

[quartz wind]

damn

[wicked patio]

In category theory, a branch of mathematics, an enriched category generalizes the idea of a category

I think I'll stop there

who decided categories weren't general enough 😭

[quartz wind]

whats the partial order here? order of distance of pairs of points?

[coral spindle]

The partial order on the reals is the one you’re used to

0 <= 1, pi <= 5

Every poset is a category in a natural way

[south patrol]

Enriched categories are actually very common - for example it is common to consider the abelian group of homomorphisms between abelian groups / modules / vector spaces etc

[quartz wind]

as in the total order right? sorry if this is a dumb question lol

[south patrol]

But in that guise it isn't too different from just a normal category; it can be much harder lol

[quartz wind]

is there a reason we're calling it a partial order rather than total

[coral spindle]

Might be a good idea to ask follow-ups in #category-theory in case we go off topic

[south patrol]

It is a total order tbf lol

[coral spindle]

There is no problem

[south patrol]

I guess categorically poset is more natural than toset

[coral spindle]

I would agree

[south patrol]

And being a toset is just a condition than R happens to satisfy

[quartz wind]

i guess im asking if the category thing you described is just due to nonnegative reals inheriting some general property of partial orders

[wicked patio]

What's the product that makes metric spaces a monoidal category?

[south patrol]

The usual one would be the cartesian product

[coral spindle]

The general property is being a category

[wicked patio]

Ah hmm

[quartz wind]

ah okay thanks

[wicked patio]

Specific distance (euclidean vs max vs sum) doesn't matter?

[coral spindle]

Good question, hmm

I think any of those will produce a monoidal category. I’m not sure if there is a standard one. I doubt the category of metric spaces & continuous maps is particularly nice

[wicked patio]

Yeah fair enough

[south patrol]

The product equipped w any two of these are homeomorphic

Since they are equivalent metrices on R^2

[wicked patio]

Ye ik, just not isometric

[south patrol]

Yeah

But yeah I mean having only isometries would be very restrictive ig

Well like every map becomes an embedding I suppose

[coral spindle]

Good point! Perhaps we instead need the category of metric spaces & maps which do not increase distance

[wicked patio]

Well max and sum are isometric in R^2 maybe lol

ah hmm

yeah I didn't think about that

[kind temple]

given G = S_3, H = {(12), (123)}, it follows that the normalizer of H in G does not contain H, correct?

my reasoning is that (12) (123) (12) = (132) != (123) so (12) is not in N(H)

just want a sanity check

[coral spindle]

It took me a second to realise that H is not a subgroup and rather a random set

I must admit I haven’t seen the notion of the normaliser of a non-subgroup very often

But yes, I suppose if we were to extend the definition in a straightforward way that would be correct.

I would advise against using the name H for something that isn’t a subgroup, it’s quite misleading

[kind temple]

im using the notation from Dummit and Foote lol

this is an exercise in there

its more clear with context, but the phrasing still has some issues

this is in section 2.2.

like, obviously if H is not a subgroup of G, then H is not a subgroup of N(H) by transitivity of <=

they meant that H is not a subset of N(H)

[wicked patio]

Well that's a fairly straightforward way of proving that lmao

just get rid of the identity

Oh wait

I flipped the inclusion lol

[kind temple]

what?

[wicked patio]

ignore me I'm crazy

[rotund aurora]

Let A be an abelian category. Must there always exist a ring R (if the answer depends on whether its commutative/unital or not I'd like to know it) and a functor F : A-->R-mod such that F(x) is isomorphic to F(y) if and only if x is isomorphic to y?

[tender wharf]

mitchell embedding theorem gives you a fully faithful exact functor. that might work

[cobalt heath]

Oh, how is ring of integers defined on global field?

[rotund aurora]

That only works for small categories, and I'm not sure how you would use it here

[cobalt heath]

I do wonder why ring of integers are only undefined for Archimedean fields out of number theoretic fields then.

[tender wharf]

locally small could work

you basically "cut out" the bit of the category you want

[rotund aurora]

Locally small means every object is contained in a small subcategory?

[tender wharf]

i think it means Hom(x,y) is a set

(im a bit rusty)

[rotund aurora]

Ah ok

[tender wharf]

if f is an isomorphism from x to y, Ff is an isomorphism of Fx to Fy. conversely if you have an isomorphism of Fx to Fy the fullness of F guarantees there is an isomorphism of x to y

[tender wharf]

yes this is true

[rotund aurora]

And F is?

[tender wharf]

by cut out I mean take a small subcategory

of A

[tender wharf]

the fully faithful exact functor given by the embedding theorem

we're not using the fact that F is exact here so maybe this theorem is overkill but

[rotund aurora]

But the embedding theorem doesn't work in general?

[tender wharf]

what's the context of the question anyway

when I've used the embedding theorem to do proofs it's only being used on a set's worth of objects

otherwise you just need a full functor from A to R-Mod. Not sure if that always exists

[rotund aurora]

I mean reading the proof of the embedding theorem from the Wikipedia article, they use that A is small from the very first line

Namely, they consider the category L of left exact functors A-->Ab

and the ring R is this thing

which is an hom set

if A is not small, there is no reason why Hom_L(I, I) should be a ring

actually all abelian categories are locally small by definition

so idk why you brought that up

[quartz wind]

is $I_1, I_2,\dots$ presumed to be infinite? otherwise this doesnt seem right

[cloud walrus]

esca (@ with reply)

[quartz wind]

also how do i get the slightly fancy looking capital i

[rotund aurora]

$\mathcal I$

[quartz wind]

ah ty

[kind temple]

if it were finite, how could it be strictly increasing?

[quartz wind]

consider $\mathbb{Z}/6\mathbb{Z}, \mathbb{Z}/3\mathbb{Z}$

[cloud walrus]

esca (@ with reply)

[kind temple]

in the sense that you have a sequence of ideals, i.e., a map from N into the powerset of your ring

[quartz wind]

i guess, idk

[kind temple]

like, finitely many distinct terms

its gotta be eventually constant if you want it to be strictly increasing, which is a contradiction

[quartz wind]

is this not a valid sequence

[kind temple]

it would be 6Z, 3Z, 3Z, 3Z,...

[lusty marlin]

They meant infinite strictly increasing sequence in the theorem

The ... was meant to signify that it is infinite

[quartz wind]

ok i see, thanks yall

[quartz wind]

okay yeah this is clear to me now thanks

[tender wharf]

yes sorry I was sleepy

[chilly radish]

I think you were confusing the notion of using FM "locally", i.e. taking the abelian subcat generated by the objects you care about and "locally small" as in the category theoretic term

[chilly ocean]

Im a bit confused on what's meant by "continuously connected" here

[chilly radish]

I think they mean path connected

No idea tho

This is the really a math term afaik (nor does googling it give any results)

[chilly ocean]

Yeah I couldn't really get anything on google either

The group here is the rotation group SO(3) for context

[chilly radish]

SO(3) is path connected so that doesn't make sense, but also SO(3) is defined to be the subgroup of O(3) of matrices with det 1, so it seems like it's asking you to prove the path component of the identity in O(3) is SO(3)

That's my best guess

[chilly ocean]

I see. Thank you

[chilly ocean]

Do multiplication tables help in any way of finding normal subgroups? Just curious.

[coral spindle]

Not really

[mighty kiln]

Do multiplication tables help with anything

[coral spindle]

Cayley tables for groups aren't very helpful in general

[minor fulcrum]

they're a little useful sometimes for making upper bounds on stuff, but yeah not really

[mighty kiln]

Ig if you already have a suspected normal subgroup in mind you can colour each coset with the same colour to verify that the quotient table exists

[coral spindle]

I wonder what the complexity of deciding whether or not a group is cyclic is?
So the decision problem would be: given a cayley table for a finite group, accept iff it's cyclic.

Perhaps assuming it is a group

[mighty kiln]

Surely O(n)

By randomly looking for cyclic subgroups

[coral spindle]

So you choose an element, check what subgroup it generates (which is O(|G|)) and then check if that is the whole group (which is O(|G|)). If not you continue with some element not already checked? Is that the idea?

There are some tricks from then on but I'm too lazy to work them out properly

But my guess is it'd be O(|G|^2) which is indeed O(n) if n is the size of the input (roughly |G|^2)

[mighty kiln]

A = []
while there are elements not in A:
Pick element x ∉ A
For each n:
compute x^n
If x^n ∈ A, next iteration

[coral spindle]

Inefficient asf ew

[mighty kiln]

This is O(n) though

[coral spindle]

I think if you have some tricks you could get it to O(|G| log |G|)

[mighty kiln]

Oh I mean O(|G|)
Wait I underestimated some stuff
Did not underestimate

[coral spindle]

Hmm is it though?

computing x^n is actually O(|G|^2) in the lookup right

Like if we have x^n-1, looking up the product is O(|G|^2)

Well whatever I'm so shite at complexity theory it'd be better just to ask an expert

[mighty kiln]

Oh I assumed multiplication would be constant time and everything else is instant

[coral spindle]

The input is the cayley table so you have to look up the products

[mighty kiln]

Wait why is looking up the product O(|G|^2)

If the elements are just {0, …, n-1} then it's very quick

And if you have to convert between the elements and the index then it's still just additional O(|G|)

[minor fulcrum]

I mean there is no objective answer here, but if a list lookup is O(n) in the length of the list, then the table is a list of |G|^2 elements

[mighty kiln]

Oh the input table is just an unordered list of tuples (g, h, gh)?

In my mind it was like an n×n array of indices in {0, …, n-1}

[chilly ocean]

I ve been stuck on this no 11 excercise of Dummit Foote

Before this there was an excercise that proved many equivalent consequences of a solvable group

Here it is

I am supposed to use this i guess

I was thinking to use iv consequence and consider H intersect N_1 but what if it's empty

[coral spindle]

You are presumably familiar with the isomorphism theorem SN/N iso S/(S n N)? If I'm not mistaken, that is the key.

(Where S is any subgroup of G and N is a normal subgroup of G)

Indeed there is short proof using this

[rocky cloak]

So together with Boytjies hint, another point is that ||H has to intersect one of the N_ts||

[chilly ocean]

Okay this part seems non trivial to me using the second iso thm we consider SN/N iso S/(S intersect N) for each N

Is this the right step

[coral spindle]

What is S and what is N? And what is the common overgroup G? You need to choose appropriate things here

I deliberately didn't tell you what good choices for S and N were

[chilly ocean]

Oh I meant H is for S right?

And N is the one in the iv consequences of 8 no

But the factors of it are not simple

I mean the N_i+1/N_i

Idt I still have the right intuition for this

[coral spindle]

A further hint: consider a minimality condition

[chilly ocean]

What is a minimalty condition

[rocky cloak]

Something being as small as possible

[coral spindle]

Let t be minimal such that (some property)

[chilly ocean]

Okay so as G is finite

This exists

[coral spindle]

What exists

[chilly ocean]

Minimal abelian subgroup?

Wait

No

That's just e

[coral spindle]

I'm going to leave this to you now

[chilly ocean]

🥲👍 ok

Maybe I'll be back for help

What does it actually mean for a group to be solvable tho

Like the book just presented this weird def

Idk what's it supposed to do

[rocky cloak]

The original motivation comes from a polynomial being solvable in radicals iff it has solvable Galois group. So I suppose that's what it's "supposed to do"

But in general it's just a group "built" out of abelian groups, so many properties of abelian groups extend nicely

[chilly ocean]

Hmm okay

[chilly ocean]

They said this in the book but that part is after fields

And I am not even through groups yet

[coral spindle]

C.f. Nilpotent groups, which are super duper built out of Abelian groups

[chilly ocean]

What's C.f

[coral spindle]

conferre, meaning "also look at" or "compare with"

[chilly ocean]

Ok 👍

[chilly ocean]

Also one more thing i would like to ask ..out of the three consequences given here would it be possible to do this no 11 without knowing them

Like not directly involving them at all

[coral spindle]

Idk but it would be less straightforward. I would suggest just thinking about the hints we gave.

[chilly ocean]

Ok thank you for the help so far @coral spindle and @rocky cloak

Oh did I have to consider the minimal normal subgroup of G that is also a subgroup of H

there must exist like such cause If not then just take H

?

[wraith cargo]

Maybe try intersecting the subnormal series with H and see what you get

[chilly ocean]

Wait what is a subnormal series now....I only know composition series

[wraith cargo]

It's just that chain of normal subgroups

[chilly ocean]

Also I think this idea works cause now we know that as G is solvable all it's composition factors are prime

[chilly ocean]

I would get a subnormal series of H

But I don't k ow if the factor conditions still would hold

[coral spindle]

You seem to have forgotten a hint

[chilly ocean]

Minimality?

[chilly ocean]

Also is this right?

[coral spindle]

Try it and see

[chilly ocean]

So using this minimal subgroup with the given property we can form a composition series all of whose factors as prime order

Then uhh

Okay so proving this to be abelian like any other thing i consider the commutator of elements in this minimal subgroup

Okay so this commutator would belong in its undergroup? In the composition series

I guess so

Now consider the conjugation of it's undergroup

As S is normal in G (minimal normal non trivial subgroup of G which is also a subgroup of H)

Then if N normal in S then gNg^-1 is also normal in S? right

Considering that fact that this inner automorphism fixes S

Now

For all such g , commutator is in gNg^-1

So commutator is in the intersection of gNg^-1

Oh this intersection is normal in G and it satisfies the conditions

So it's smaller than S

Which is not possible

So it's trivial right

And the commutator of any two elements this given is trivial

so S is abelian?

@coral spindle is this right?

[coral spindle]

This doesn't make sense to me

[chilly ocean]

Okay 🥲

[coral spindle]

I have a suggestion. Write down the relevant info (the hints & the theorem you found to be relevant) and come back when you think you have a complete proof. Don't bash it out in discord messages – write it up when you think you have a complete proof.

[chilly ocean]

Maybe i messed up

[chilly ocean]

I have taken the time to write what I came up with in latex

[cloud walrus]

Browassup

[chilly ocean]

Is this correct

Should I brute force 4.2?

Like I used the theorem of subgroup having index 2 is normal but I didn't dwell into Z(S_3) or subgroups of order other than 3 yet

S_3 and D_6 are isomorphic

and take your favourite generator of the cyclic 3 group

Now Z_S_3 is trivial

U can just use the dihedral relations rs=sr^{-1}

[chilly ocean]

How? S_3 has 6 elements and D_6 has 12

Oh D_3 i mean

By your notation

Mines a bit different 😅

Also there's a different way

Conjugacy classes in permutations are dependent on their cycle structure

So if you have a single transposition in a normal subgroup

Then you get all the transpositions in that

So it becomes the whole group

@chilly ocean

Oh apologies, I'm not into conjugacy classes yet

[rocky cloak]

You lose me a little on where x and y come from or why prime index is relevant.

But something that would work is to consider [S, S] as a subgroup of S.

This is a lot more complicated then it needs to be though, but that's okay.

[chilly ocean]

Oh I meant g(ab)g^-1 can be any (cd) you like

[woeful sage]

Hello algebruhists I was doing this problem

and I just couldn't prove the 2nd part of the problem

because it was just so obvious I didn't know where to start

like obviously the order of (a, b) is the lcm of |a| and |b|

[chilly ocean]

x,y come from S ... because I need to prove that it's abelian

[x,y] is just xyx^-1y^-1

[rocky cloak]

Yeah, so you're looking at the normal subgroup generated by [x, y] ?

Then that would make sense

[chilly ocean]

No just element wise

I am saying [x,y] is e

When x,y in S

[rocky cloak]

Yes, that's what you want to prove

[chilly ocean]

Then i consider it's undergroup N

In the composition series

[rocky cloak]

So "it's undergroup" means the normal subgroup generated by it?

[chilly ocean]

As S is normal in G we can write a composition series of G including it right?

[rocky cloak]

Yes

[chilly ocean]

So I take that series

N is just the one that's immediately contained in it

It can be {1} too

[rocky cloak]

Okay, so N is an arbitrary maximal subgroup of S

[chilly ocean]

Okay.... 👍

[rocky cloak]

But how do you then know that [x, y] is in N?

[chilly ocean]

Yeah so index of N is prime

S/N is prime order

As G is solvable

It's abelian

[rocky cloak]

Sure, then I follow

[chilly ocean]

So it's okay?

[rocky cloak]

Yes, it checks out

[chilly ocean]

You said there was an easier way

What would that be?

[chilly ocean]

Well firstly (a,1) and (1,b) commute

[woeful sage]

I agree

[rocky cloak]

Here's the proof I had in mind:

Let t be minimal such that H cap N_t is nontrivial.

Then H cap Nt = (H cap Nt) / (H cap N[t-1]) < Nt/N[t-1]

[chilly ocean]

Ah

[coral spindle]

I had the same in mind

[chilly ocean]

Yeah this is great

I don't know, when boytjie said sth minimal the most obvious thing i could think of is that set

Maybe cause I am not very fluent in iso thm results

[coral spindle]

We mentioned the iso theorem & I hinted that you wanted t minimal

[chilly ocean]

Oh so t had to be minimal

I thought I had to find some minimal subgroup

My bad 😅

[chilly ocean]

Now notice that (a,b)^m is just (a^m,1), (1,b^m) = (a^m,b^m)

If it's the identity then both the coordinates are identity

Now I think you can just verify the definition of LCM

And use the fact that order is minimal

[coral spindle]

I.e., use the frikking definitions!!!!!!!!!!

[woeful sage]

that's it?

I guess that's what I did in my head and I just assumed "it's too trivial where do I even begin"

[chilly ocean]

Just to make sure, in dihedral groups, we have to rotate in the way we have numbered the polygon's vertices (if we visualise rotation)?

Because I get two different results if I compose it with flip

[coral spindle]

It doesn't matter; we get the same result up to isomorphism

That is: if you construct the dihedral group one way or the other, you get isomorphic groups anyway.

[tawdry plover]

Is there a way to construct a vector space such that the size of its basis is of any arbitrary cardinality?

**
Also if this problem is not suited here I might aswell move it to wherever you guys would recommend

Wow okay maybe it is

Thanks for the heads up

[coral spindle]

Yes. To elaborate on what Terra was saying, let X be any set and consider the vector space over the field k consisting of functions f : X → k such that f(x) = 0 for all but finitely many x in X. This is a vector space of dimension |X|.

In fact, up to isomorphism of course this is the only vector space of dimension |X|.

[tawdry plover]

Yeah I guess everything is just functions at the end

😅

Or maybe objects and arrows 🤔

[inner needle]

Please help me to show this fact, as I've tried and have gotten stuck.

For a positive integer $n$, let $\mathbb{M}_n$ denote the set of $n \times n$ matrices over $\mathbb{C}$, and let $I_n \in \mathbb{M}_n$ be the identity matrix.

Let $n$ and $k$ be positive integers.

Suppose that $B \in \mathbb{M}_{nk}$ commutes with all elements of the form $I_k \otimes A$ for $A \in \mathbb{M}_n$.

I want to show that $B = C \otimes I_n$ for some $C \in \mathbb{M}_k$.

[cloud walrus]

Transience

[rocky cloak]

Pick a basis
v1, ..., vn for C^n and w1, ..., wk for C^k.

Let Eij be the matrix that maps vi to vj and the other vs to 0.

Commutativity with I (x) Eii means
B = sum Bi (x) Eii
Commutativity with Eij means Bi = Bj

[winter fog]

hello, can someone help me please #help-45

[spice notch]

I needed some clarification. Let V be normal in G, where G/V is p-solvable for an odd prime p, and V is a non-abelian simple group. is it necessary true that V is the socle of G?

[rocky cloak]

Can't you just take G = V x G/V ?

[spice notch]

mm i suppose. what if G acts transitively on V?

[inner needle]

Thank you! Let me work this out.

[rocky cloak]

Acts how?

Like it's impossible for a group to act transitively on a subgroup by conjugation. Are you thinking of some other action related to the setup somehow?

[spice notch]

no, yes i made a mistake, im trying to think of a base case and confused it for a second

[chilly radish]

As opposed to non-free vector spaces

Those are still free

I saw that arki

[mighty kiln]

No you didn't

[chilly radish]

The theory of vector spaces carries over immaculatrly to the division ring case

I was more going for basis-less VS in the absence of choice

[coral spindle]

If you want MY vector spaces ur gonna have to pay up

[winter fog]

hello, can someone help me please #help-45

[quartz wind]

To prove (3), it is clear that if $ra + sb = 1$, $a$ and $b$ are relatively prime.
this isnt at all clear to me, but i think i worked out why? is it because if a and b have a common divisor x, then ra + sb = rxm + sxn = 1, therefore 1 in <x> = R, therefore x must be a unit?

[cloud walrus]

esca (@ with reply)

[coarse kestrel]

yup

[quartz wind]

ok great thanks

[rapid junco]

Is this a well studied algebraic structure?

A group but instead of 1 you have 0

[kind temple]

why is the hint true?

[coral spindle]

Think about the solutions to x^|g| = 1

You know what, that was a bad hint. Let me think some more.

& I'll come up with a better one

I think there may be some missing context. Is there perhaps a lemma above that says that the hint holds for Abelian groups? This may be what you're missing here.

Otherwise this is essentially a helpful lemma

Perhaps you could try proving this first: if $g, h \in G$ are elements of an Abelian group such that $\gen g \cap \gen h = \set e$, then the order of $gh$ is the least common multiple of the orders of $g$ and $h$.

[fading field]

here is a hint that you might like better: show that for an abelian group, if $$ |{x\in G: x^n = 1}| < n$$ for all n, then G is cyclic

[kind temple]

so the hint is true of abelian groups in general?

[cloud walrus]

Boytjie

smaysurable set

[fading field]

hm

and then use the fact that Z/pZ is a field

[kind temple]

ik this

[fading field]

by < i mean leq

[coral spindle]

Fellow newpx appreciator spotted

[coral spindle]

Great. Use it to prove the hint.

[kind temple]

for h in G, if h = g^a, then |g^a| divides |g|. otherwise, <h> cap <g> = {e} so |hg| = lcm(|g|,|h|). but if g has maximal order, then
|g| = lcm(|g|,|h|) = m |h| so |h| divides |g|

since |g| <= lcm(|g|,|h|) <= |g|

[south patrol]

Wdym

[coral spindle]

This isn't quite right

[kind temple]

what’s wrong?

[coral spindle]

for h in G, if h = g^a, then |g^a| divides |g|. otherwise, <h> cap <g> = {e}
For example <h> could contain <g>

But you have the right general approach

[fading field]

sometimes people use this notation for commutative groups

[south patrol]

Should just be a notational difference

[coral spindle]

Here's how I'd do it: suppose |h| does not divide |g|. Then k := h^{gcd(|h|, |g|)} has order coprime to |g|, ergo <k> n <g> = {e}...

[kind temple]

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

so |g^a| divides |g|

[coral spindle]

Ah wait in fact I made a mistake, my reasoning isn't correct

It may not have order coprime to |g|.

[kind temple]

right

[coral spindle]

The problem isn't that, it's the otherwise part

[kind temple]

oh

[coral spindle]

These are not the only two options

OK. The lemma is still correct in general, but it's easier if we bring in some extra machinery.

[kind temple]

i think this fixes it tho: if <h> cap <g> is non-trivial, then h = g^a for some integer a.

[coral spindle]

That's still not true

But in any case, here's a way to repair it

[fading field]

i am missing context are y’all trying to make statements about abelian groups?

[coral spindle]

Think about the factorisation of the polynomial x^|g| - 1 in Z/pZ

[coral spindle]

Indeed but I think it's a better idea here to bring in the specifics of this situation

The hint is true in any finite abelian group, as one can see via the classification, but it's harder to prove

[kind temple]

it splits into p-1 linear factors

oh

[coral spindle]

And what are those factors?

[kind temple]

um

i don’t

actually know that

[coral spindle]

OK

[kind temple]

i thought this was x^{p-1} - 1

[coral spindle]

The question has a better setup

x^d = 1 has at most d solutions

So now suppose that g is of maximal order

[kind temple]

yea, Z/pZ is a field

[coral spindle]

What are the solutions to x^|g| = 1?

[kind temple]

<g>

[coral spindle]

That's exactly right. So if x has order dividing |g|, then it is in <g>

That's precisely what this polynomial having the solutions <g> means

So you can use this to make proving the hint less fiddly

[kind temple]

i can see how <g> satisfies the equation x^|g| = 1

but if x^|g| = 1, then |x| divides |g|

[coral spindle]

That's right

[kind temple]

but how do you show that x is a power of g

[coral spindle]

How many elements are in <g>? Now recall that there are at most |g| solutions to x^|g| = 1

So...

[kind temple]

|<g>| = |g| oh

lol

[coral spindle]

So in fact <g> is a complete list of solutions to x^|g| = 1

Any solution x is a member of <g>

[kind temple]

alr sweet

where did we use the fact that |g| is maximal

[coral spindle]

We haven't used that yet

The maximality is going to help us prove that |h| divides |g|, which is going to tell us (by what we just proved) that h is in <g>

[kind temple]

can you restate what we just showed for clarity?

[coral spindle]

We proved that if |h| divides |g| then h is in <g>, put simply. (For elements h, g of (Z/pZ)*)

[kind temple]

this doesn’t hold for g,h in a more general abelian group?

[coral spindle]

Most certainly not.

[kind temple]

okay

[coral spindle]

Z/2Z x Z/2Z is a counterexample

[kind temple]

this is where we use the fact that Z/pZ is a field

alr cool

[coral spindle]

That's right

[kind temple]

so, the solutions of x^|g| = 1 are <g> for g in Z/pZ

how do we show that |h| divides |g| for all h in Z/pZ?

[coral spindle]

So now things get a little bit fiddly and we need to use maximality. You're essentially going to do a proof by contradiction, assuming that |g| is maximal and |h| does not divide |g| to get an element of greater order than |g|

[rapid junco]

No im talking about a zero element

Not just a rebranding of 1

[fading field]

oh it won’t have anything that behaves like multiplication by 0

[rapid junco]

Maybe include 1 but say there is a zero element too, but bo afdition.

[fading field]

you can prove such a thing

[rapid junco]

You can prove a group can never have a zero element?

[fading field]

if the size of the group is more than 1

[kind temple]

if |g| is maximal and |h| does not divide |g|, then for g^a = h^b, we have (h^b)^|g| = 1, so |h| divides b |g|, and in turn, |h| divides b so g^a = 1 = h^b, and <g>, <h> have trivial intersection

[fading field]

if |G| = 1 then the one element of G behaves like 0

of course

but other than that it’s not possible

[rapid junco]

How come?

[coral spindle]

and in turn, |h| divides b
This doesn't follow, you've assumed primality of |h| essentially. But the good news is, you may be able to reduce to this case.

[rapid junco]

Oh if g g = g

Then g = 1

I see

[kind temple]

but we assume that |h| does not divide |g|

oh

they aren’t coprime

[coral spindle]

They may not be coprime, yes

[kind temple]

dude i miss topology honestly

[fading field]

me too

[coral spindle]

Here is how I might approach this, in two cases:

• |h| has a prime factor p that is coprime to |g|. Think about h^{|h|/p}.
• |h| has some power p^n not dividing |g| such that p^(n-1) divides |g|. This is a slightly tricker case.

[kind temple]

is this at least in the right direction? we want to show that <g> and <h> have trivial intersection in this case

[coral spindle]

No unfortunately that doesn't follow. You could have, for instance, that <h> contains <g> entirely.

[fading field]

the line that lets you conclude that g = 1 is just that there has to be some other element x such that gx=1,

[kind temple]

this is kind of painful

[coral spindle]

Yes... it is a techical lemma

[fading field]

but gx is also g

[coral spindle]

Previously I alluded to the fact that this follows fairly immediately from the classification of finitely generated Abelian groups, so if you're happy with using that you can. But it works out.

[kind temple]

i was trying to avoid it because i wanted to see how it worked without appealing to it

it seems like an elementary approach

[coral spindle]

Regarding the 2nd case: if p^m divides |h| then h^(|h|/p^m) has order p^m and g^(p^n-1) has order |g|/p^(n-1) which is not divisible by p.

I've kinda laid it all out here... not doing a great job of hinting.

My bad.

It's honestly just one of these relatively brute force things.

[kind temple]

no, ur all good

this feels like a tough one

[coral spindle]

You kinda have to fiddle with the orders and make it work. The good news is, this is a big step towards classifying finite Abelian groups

[kind temple]

okay, so saving the proof of the hint, the rest of the proof would go as follows:

if d <= p-1 is a maximal order of some element g in (Z/pZ)*, then for each x in Z/pZ, |x| divides d, so x^d = 1 has exactly p - 1 solutions, so d = p - 1 and we are done.

does that sound right?

[coral spindle]

I would add just:

...has exactly p - 1 solutions, namely the elements of <g>, so d = p - 1....

But yes you have the logic right

[kind temple]

thanks

i’ll try and fiddle with the orders as you say

topology would never do me like this

[kind temple]

is this any easier?

[fading field]

if you want to use the fundamental theorem of finitely generated abelian groups then it’s easier

(finite abelian groups )

[south patrol]

FTOFGAG

Classic

[fading field]

if you don’t then maybe you’ll just have to do what you’ve been doing

though i’ve only barely been paying attention to the conversation

since i’m at the lake rn

[kind temple]

aluffi gives this as a series of exercises in chapter 1 lol

[fading field]

is that notes from the underground

[kind temple]

idk what that is

[fading field]

or algebra chapter 0

[fading field]

which book

[kind temple]

oh, chapter 0

[south patrol]

Dostoyevsky

Lmao

Well that is notes from underground but ye

[fading field]

bruh

[spare hound]

is this the correct place to ask for a module theory proof verification?

The problem is if every finitely generated R-module is free show R is a field

solution is as follows

Assume for contradiction R is not a domain, then let x in R be a zero divisor, the ideal (x) is a torsion R module, whereas R^n contains 1_n which implies R^n is necessarily not torsion, hence (x) is not free so R must be a domain

Now let x != 0 be an element of R, it follows that R/(x) is free by assumption, but xf(1+(x)) = f((x)) = 0 implies by the domain property that f(1) = 0, hence R/(x) is isomorphic to R^0 = 0, which implies that (x) = R, i.e. x is invertible
since x was arbitrary we are done

is this correct?

here I write n, but this is abuse of notation it really may be any cardinal (under direct sum not product)

[void cosmos]

whats f

[spare hound]

its the isomorphism

[void cosmos]

what isomoprhism

[spare hound]

R/(x) to R^m as modules

sorry the existence is implicit in R/(x) being free

[void cosmos]

xf(1+(x)) =f((x)) i dont understand

maybe it's j ust me

[spare hound]

this is writing as cosets

where we take the ideal generated by x

and use linearity of module homomorphisms

so that xf(1) = f(x) = f(0)

since (x) is the zero coset

[void cosmos]

i see

yeah it's gucci ig if you just slightly explain why f(1)= 0 implies that it has no basis

or ig it's whateber

[spare hound]

yea its an isomorphism

so you get R/(x) isomorphic to 0

[void cosmos]

yeah yeah

ig one thing though is ur assumsing ur ring is commutative

[spare hound]

oh yea

this is in the book

[void cosmos]

and has identity ofc

yeah sure

[spare hound]

I forgot to say

ur right

[void cosmos]

you kinda implied it by saying "field"

if it weren't then it would just be a division ring

bla bla whatever

[tardy hedge]

Why is the module Z2 x Z2 over Z can be written as an internal direct sum?

Something in the book said that its not a free module over Z but it can be written as internal direct sum

Z2+Z2 does not have uniqueness tho? 0 = 0+0=1+1

So Z2xZ2 is not isomorphic to Z2+Z2?

[coral spindle]

Z2 x Z2 is the internal direct sum of the Z-submodules Z2 x {0} and {0} x Z2

In fact the cartesian product of modules is the (external) direct sum of modules.

It's unclear to me what you mean by Z2+Z2.

[coral spindle]

@tardy hedge did you look at this

[tardy hedge]

Not really yet im at work

[tardy hedge]

I meant elements n1+n2 such that n1 and n2 are in Z2

[coral spindle]

So Z2 + Z2 = Z2.

So I don't see how that's relevant

[south patrol]

I guess you mean Z2 (+) Z2, and the confusino is because by 1+1 you should say (1,0)+(0,1)

[spare hound]

I don’t see how this isn’t free maybe I’m being dumb

2Z isomorphic to Z

Oh

You mean it’s not free as a 2Z module?

[rocky cloak]

Z2 means Z/2Z

[spare hound]

Oh yea u right

I didn’t read this as Z_2

Okay yea definitely not free

[coral spindle]

Zoo wee mama

[coral spindle]

I'm not sure kiand knows this notation considering he's writing Z2 x Z2

But if and when he sees this, let it be known that this is the more standard notation.

[spare hound]

When life gives you finitely generated modules over a PID

You make a direct sum of cyclic modules

[next obsidian]

True af

U a real one for that

[south patrol]

and if it is not fg you just write it as a filtered colimit and wlog fg

[sly frost]

this is a stupid question, but what's the shortest way here to get from a ∈ bH to a⁻¹b ∈ H, it's kind of confusing to me

I know that a ∈ bH means a = bh for some h ∈ H, but this just means b⁻¹a = h, which is not what we're proving

do I have to reverse it here to show a⁻¹b = h⁻¹ ∈ H because H is a subgroup?

I know it's true, but the way it's presented here is confusing, seems like too much of a leap

[coral spindle]

Yes

But this is straightforward because (b⁻¹a)⁻¹ = a⁻¹b

[sly frost]

I guess you could also show that a = bh, so ah⁻¹ = b, so a⁻¹b = h⁻¹ ∈ H

the only reason I'm asking is that the implication seems a little too non-obvious to me, like I'm not getting something

but maybe it's a stupid thing to be worried about

I'd say instead that aH = bH, so b = be = ah for some h ∈ H, so so a⁻¹b ∈ H

the way this is presented here seems confusing

[coral spindle]

The point is it's completely equivalent.

[sly frost]

well of course it is

my point was just that the presentation here was a bit confusing

[tardy hedge]

@sly frost Honestly tbh that stuff tripped me up too for the same reasons

All the equivalent statements for coset stuf

I had the same feeling of not “getting” something , even though i understood what was happening. Kind of a weird feeling

I think when u work more with it, you start to more easily see how all the statements are equivalent and you wont feel as confused

[tardy hedge]

modules over a field F are vector spaces over F. So, now that our module has its ring multiplication defined by a field, we go back to our old linear algebra language like now theyre called "scalars", now we have "dimension" etc.

What is it about the field F in particular that allows us to interpret linear algebra (vector spaces over F) and their homomorphisms (linear transformations) as geometric, like we are "scaling" and rotating and such ?

Is it because ax = 0 iff x or a is 0? And so, any ax value can be interpreted as a scaling and rotation of the element x? And this is an IFF implication for F a field being the ring over the module?

Forgive if this is obvious, I'm high right now

I just dont think ive thought about this before

When I was first learning linear algebra

[leaden heart]

You can do linear algebra over a finite field, such as Z_p, and, well you can kind of interpret the linear maps as "scaling" and "rotating" if you draw a picture, but its definitedly not in the same way the reals do it

[tardy hedge]

having a hard time visualizing how a finite field is a vector space

need to think about that again

[leaden heart]

The important difference is the fact all non zero scalars are invertible, which lets you do gram schmidt, so all vector spaces over F (that are spanned by finite many things, at least) have a basis

over just a ring, modules can have torsion, but over a field, vector spaces are isomorphic iff their dimension is the same

you can think about (Z_5)^2 as like, a 5 x 5 grid of dots

and the linear transformations shear it, rotate it, and stuff, they just wrap around

[tardy hedge]

Wow this is so cool to me rn

[leaden heart]

its very fascinating to me that the linear algebra machinery still works over finite fields, yes

for instance, you calculate the determinant in the same way, and its still true that a matrix is invertible iff the determinant isnt 0

even though it could be zero because you calculate, like, 1 + 4 = 0 in mod 5 land

[tardy hedge]

Interesting

I think I should go back and study linear algebra in a deeper way now

Because I dont think I appreciated the connection between the algebra and the geometric visualization

I sort of knew it was there but didnt really think about it too much and just accepted those visualizations of the stretching and rotating as just being the way it is

[topaz gale]

For a cool reference, Module Theory by Blyth does an excellent job of connecting linear algebra to modules to a level I think you might really much enjoy. He starts on an elementary level and then link all of it to linear algebra in the second part of the book.

I really liked how he first motivated the study of modules by expressing how we already, when calculating eigenvalues, extract it from a matrix whose determinant comes from a polynomial ring. Hence, why not start from this general setting? It’s brillant.

[tardy hedge]

thank you!

[chilly ocean]

I would like to verify that a group $G$ with prime order $p$ (which implies it's cyclic) is a subgroup of the center $Z(H)$ for some group $H$, $G \leq H$

This is because take any $a^m \in G$, $m < p$, then

$(a^m)^p = 1$ (by Lagrange)

$\implies za^m = a^mz = (a^{-m})^{p-1}$, $z \in H$

[cloud walrus]

iceball

[tender wharf]

do you mean proper subgroup? because if you take H to be G then Z(H) = G

perhaps the original question might help

[chilly ocean]

Apologies for the improper wording

[cloud walrus]

iceball

[dense forum]

does anyone know any book or any book (or any other resource) in order to get introduced to noetherian rings and algebras?

i would like to learn about it in order to start reading about algebraic geometry (and any other subject that could interest me in the future), but in the courses that my university offers there is no mention (at least, at the moment) about it

[sly frost]

so any book on commutative algebra? you could try the one by Atiyah or Clark

[dense forum]

okay i’ll go check, thanks!!

[chilly ocean]

Ignore this, idek why did I even ask this wrong question. Apologies.

[hidden mist]

This might be a stupid question, but I'm a bit confused about the part I underlined. Firstly, \bar{y_i} is an element in G/H, but 0 is an element in G. It seems that it should be \bar{y_i} = \bar{0} here. However, if \bar{y_i} = \bar{0}, it seems to contradict the condition y_i \in \bar{x_i} because \bar{x_i} is a coset of H.

[barren ingot]

regarding Abstract Algebra could someone tell me the:

1. Prerequisites
2. Related subjects that can be explored parallely
3. Resources
4. Topics that are a natural extension

[small bramble]

Prereq is really only an intro background in proofs for intro to groups, but some linear algebra is needed for rings and fields

[sly frost]

for groups too, since matrix groups are some of the basic examples of groups

although a good introduction to abstract algebra should teach all that

[barren ingot]

I've covered linear algebra, also think I'm fairly ok with proofs

[barren ingot]

I see, and what are related topics to abstract algebra

Which can be covered parallely

[sly frost]

in general I would also get acquainted with calculus before learning it

not sure what are related subjects, maybe set theory, number theory and topology?

[barren ingot]

Oh I see

And uhh, any good resources?

[coral spindle]

Just find a book on abstract algebra & start reading. I liked Fraleigh when I was starting out.

[barren ingot]

Fair enough, thanks a lot,

[sly frost]

I like Artin (more basic) and Aluffi (more advanced, but also introduces you to category theory which is helpful to many other topics)

[barren ingot]

Oh yes! I wanted to proceed to category theory as well, is it a natural follow up to a typical first course in abstract algebra?

[small bramble]

Ik Dummit and Foote is semi-controversial, but I started that out w/o knowing linear algebra and with only a crash course in proofs for a group theory class (whose lectures I never attended so it works on its own) and things worked out

[sly frost]

in theory category theory can be taught without any abstract algebra, although it would probably seem very abstract and confusing, so it's better to know algebra first to know what it applies to better

[rocky cloak]

Cryptography, number theory, algebraic topology and algebraic geometry are all subjects closely related with algebra. Though I guess for most of them you'd want to cover the basics of algebra before you jump in

[small bramble]

It did have stuff on matrix groups iirc but I think it explained that fairly well

[barren ingot]

Thanks a lot, I think I've a fair idea about how to proceed now

💌

[sly frost]

yes, algebraic topology follows from abstract algebra, but you also need to know topology to understand it

[sly frost]

other topics that follow from abstract algebra are commutative algebra, which precedes algebraic geometry and algebraic number theory, also you might be interested in topological groups after learning topology, and set theory would prepare you for mathematical logic and model theory

[barren ingot]

Thx a lot

[chilly radish]

Also noncommutative algebra which broadly studies the nontrivial cases where commutative algebra makes things trivial

Both require about the same level of mathematical maturity

[chilly ocean]

,rotate

[chilly ocean]

I am stuck on 3

[rocky cloak]

Here's a hint.

Think about what you need about s and t to describe how s^-1 t s acts on an element. Then use that the group is abelian and transitive

An I guess you want to frame it as t(a) = a, and then prove that t is the identity

[chilly ocean]

well firstly if they commute then it's just how t acts on the element but in general uhh

We need to know how s and t act individually on A ?

[rocky cloak]

So let's say t(a) = a for some given a

Can you choose s wisely to learn more about t?

[chilly ocean]

And this is where G is transitive comes into play?

uhh

For any a' in A there is a f such that f(a)=a' and now f(t(a))=a'

Wait I need to learn about t

Oh wait is sts^-1 (a) just t(s(a)) or sth

[rocky cloak]

It's not quite that, but you have the right idea

[chilly ocean]

So what does sts^-1 actually do

It's just t acting on a permuted set of A?

[rocky cloak]

Well, think about what you know.

s maps a to a', so s^-1 maps a' to a. And you know t maps a to a. So how can you put this information together into something useful?

[chilly ocean]

This is also fixing a

sts^-1 I mean

Or wait

[rocky cloak]

Well almost

[chilly ocean]

oh it fixes a'

;-;

[rocky cloak]

Corectomundo

[chilly ocean]

So basically we can get a sts^-1 that fixes some a' in A

As G is transitive

Right?

Now G is also abelian

So that means t fixes everything?

So t is actually in the intersection of no 2

So it's the identity?

I think it works

[rocky cloak]

Indeed, t fixes everything, hence is the identity permutation

[chilly ocean]

Okay thanks I got done with most of it

Now I just need to prove the cardinality part

It seems kind of orbit stabilizer except A is not necessarily finite

[rocky cloak]

Orbit stabilizer also works for infinite sets.

But you can't divide by stuff so you'll have to phrase it as
size of orbit time size of stabilizer equals size of group

[chilly ocean]

Hmmm

[rocky cloak]

Or you can just say the size of the orbit is the index of the stabilizer

[chilly ocean]

Index is defined here?

[rocky cloak]

Number of cosets

[chilly ocean]

Cardinality of cosests?

[rocky cloak]

Yeah, cardinality of the set of cosets

[chilly ocean]

Hmm, but how do I write it

Like I want a bijection from G to A

Oh it's the usual Lagrange theorem?

But wait

[rocky cloak]

I mean if you fix an element a then
g |-> ga
is a bijection

[chilly ocean]

🤯

Okay that's it i guess

Too bad I couldn't see that

Maybe it needs time to "see" these stuff

Thanks @rocky cloak for all the help 🙏

[chilly ocean]

,rotate

[chilly ocean]

So some observations I made are the (i,i) elements form a orbit and them only

No other ones