groups-rings-fields

[long geyser]

what will not be isomorphic?

to what?

[ornate tiger]

nvm that's stupid

but imagine i took 1/2q so we get 2qx = 1 so qx =1/2

nvm this is also stupid

[coral spindle]

Hint: the rationals have a nice ordering on them.

[tribal moss]

You want a proof of this rough shape:

Suppose Q is cyclic, then it has a generator p/q. The subgroup generated by p/q consists exactly of (.....). However, the number (...) is not one of those, because (...). This contradicts the assumption that p/q generates all of Q.

[next obsidian]

Generates all of

Z

[coral spindle]

Well that still works as a contradiction of course, since Z is not a subset of Q

This is what comedians call a "callback" sweaty

[languid trellis]

Alternative idea: prove FTOA holds in Q, primes 1/p and p, then use a divisibility argument (:

[coral spindle]

So true!!!!!!

[south patrol]

should've used my argument with automorphisms smh

[languid trellis]

Haha 'Q is not iso to Z as group qed'

I guess we'd also need to say it's not iso to any subgroup of Z

Then we've knocked off all the cyclic groups

[chilly radish]

all subgroups of Z are Z

you want quotients

[dull ginkgo]

Z into Q is epimorphic and monomorphic

fuck rings

[coral spindle]

No thanks

[languid trellis]

is 2Z not a subgroup?

[chilly radish]

Yea, but it's isomorphic to Z

[languid trellis]

Oh, I see what you're saying

yeah fairs

[dull ginkgo]

me when cyclic

[chilly radish]

In general whenever an algebraist says two things are the same, it's worth adding (up to some natural/coherent notion of isomorphism) in your head

[dull ginkgo]

It’s worse when they don’t specify some structure

[dull marsh]

Z and Q are isomorphic (sets)

[dull ginkgo]

Or when the actual like

Isomorphism in question is vital

Like V and V* for finite dim spaces

Each is to some extent a special bilinear form

[dull ginkgo]

I don’t know much about canonicity or naturally of isomorphisms outside of like the specific case of adjoint functors

[round jay]

Does anyone know have tips for showing that the normal subgroup of a set of relations in a free group is the kernel of some homomorphism?

It seems like generated normal subgroups in free groups are too complicated to work with

For example, right now im trying to show that the kernel of the canonical homomorphism from the free group on a group G to the group G is the normal subgroup generated by the multiplication table of G

[tardy hedge]

There is something kind of boring about dummit and foote

[round jay]

Trying to show that every finite group is finitely presented basically

[tardy hedge]

Maybe cause it just feels really serious

[coral spindle]

You are aware of the first isomorphism theorem I hope...

There is a natural map phi : G → G/N. What's its kernel?

[round jay]

Yes I meant a specific homomorphism, I elaborated in my second message

[coral spindle]

There you go, it's the natural map G → G/N. Specific.

[round jay]

The canonical homomorphism from the free group on G to G

[south patrol]

Well this is like tautological and doesn't need the free group

[round jay]

In Aluffi's Algebra chapter 0 finitely generated uses free groups

[south patrol]

Oof

But yeah i mean you don't need the kernel though for that wither

Just the fact that the canonical map is a surjection

Which is clear

[round jay]

You still need to prove that the kernel is finitely generated

[south patrol]

That's finitely presented not finitely generated

[round jay]

we want to show G is finitely presented by showing that the kernel of the canonical homomorphism is generated bya finite set of relations

[south patrol]

Sure, you just didn't say finitely presented

Then I agree

[round jay]

my bad

I mispoke earlier

[south patrol]

Dw

[round jay]

Yeah I just dont know how to show that cuz free groups still seem complicated to me

[south patrol]

I guess one thing is that if G has cardinality n then consider the relations g^n for each g

Once you mod out by those you only have finitely many words left anyway

[round jay]

Not sure I follow

[south patrol]

Nvm it was slightly wrong because of non commutativity

[round jay]

The kernel should be generated by the conjugates of the multiplication table of G but idk how to show that

[tender wharf]

well the usual example of naturality is the isomorphism between V and its double dual

unless you're looking for an intuitive explanation of naturality?

the adjoint functor thing is the same concept of naturality really

so if you have intuition for that it should carry over

[patent girder]

If I have a vector space over a field with char=0 $V$ such that End$(V)\cong\mathbb{C}[G],$ can I use properties of the group to determine decompositions of $V$ into irreducible subspaces? Particularly, how many there will be. Can I make a statement like: under suitable hypothesis, it will have |G|/#{generators}? What would suitable conditions be?

[cloud walrus]

Zander

[rocky cloak]

Take G your finite group and let F be the free group on |G| generators {f_g}. Show that the kernel of the canonical map F ->> G is generated by f_g f_h (f_gh)^-1 (which is a set of cardinality |G|^2)

[rocky cloak]

I feel like you're missing some context here. Does V have more structure than just being a vector space? If not, what do you mean by irreducible subspaces?

[patent girder]

pi,V can be a representation of some other group. I am thinking about using the fourier transform, but struggling to set it up

I was being a bit vague to see if theres some nice general statement, which there never is

[limpid ferry]

Is it correct to say:"A group G has no nontrivial proper subgroup if and only if |G| is prime"?

[daring nova]

some may say nontrivial and proper is a repetition that is not necessary (depending on your definitions) but it's definitely correct math speak anyway

[coral spindle]

Yes although perhaps the trivial group should be mentioned as a boring exception :)

[daring nova]

what is bangbang

[dull ginkgo]

Tldr Sylow stuff yeah

[coral spindle]

No Sylow stuff necessary.

[dull ginkgo]

I wonder if there is an easier way to show G has no proper subgroups -> prime order

Wait

[coral spindle]

It’s very easy

[dull ginkgo]

Chose random element

Consider order

[coral spindle]

:)

[dull ginkgo]

it must generate the whole group if it’s not the identity

Ergo cyclic

[fading field]

no

[dull ginkgo]

If G has no proper subgroups then <x> must either be trivial or the whole group

If it’s trivial, then x is the identity

[fading field]

lol you didn’t make it clear that this was your assumption about G lol

[dull ginkgo]

Oh lmao

[dull ginkgo]

I was following this lmao

To show it’s prime just use an argument on if there was a nontrivial divisor d of the order, then <x^d> would have to explicitly be a subgroup and otherwise would contradict minimality of the order of x

Does anyone find the standard proof of the Sylow theorems through like, striating the normalizers of p-subgroups

Like really neat

[old hollow]

If it’s cyclic of order n, it’s isomorphic to Z/nZ, then you can say. Suppose n = ab is composite, then <a> has order b < n, so we have a proper subgroup, contradiction.

[errant wedge]

Can someone explain how this is please?

[vapid vale]

theres a number of ways you can show this but since it says corollary, maybe you can send what its a corollary of to see how this follows from that

[errant wedge]

Right sorry

[vapid vale]

try to derive this fact from 2.23(3)

[errant wedge]

Thank you

Sorry, could I have a hint

nvm got it

[dull ginkgo]

Just a small review

So H_A stabilizes the set A, so H_A acts on A

[prisma ibex]

tfw orbit stabilizer theorem is POWERFUL

[dull ginkgo]

Need to ponder for a second

[cobalt heath]

H * A = A

So one has H -> End(A)

[cobalt heath]

This yields H × A -> H * A = A

[dull ginkgo]

Since H_A stabilizes the set A, then we can restrict H_A to act on A.

Now left multiplication by any element is a bijection on G. Thus the restricted action is the same for A.

[minor fulcrum]

very much so

[dull ginkgo]

Okay this is fucking with me

So for any a in A, we have the orbit H_Aa in A, which is in bijection with H_A by left cancellativity. As these orbits are cosets, we have that they partition A (since cosets are disjoint or equal), so the order of H_A divides the size of A = p^k so H_A has prime power cardinality less

[next obsidian]

Cancellativity

[errant wedge]

How would u get smth like ba in <ab, a^2b>

o wait nvm

[minor fulcrum]

are you good with this

[dull ginkgo]

I don’t like how this is stated

[errant wedge]

Yee

i remembered inverses exist

Also quick sanity check, 2Q works right

[long geyser]

failed the sanity check I'm afraid

[minor fulcrum]

big if true

[errant wedge]

Wait its not even closed whoops

[long geyser]

? it is closed

issue is it is not proper

it is just Q

[errant wedge]

OH whoops

[long geyser]

p/q = 2 * p/(2q)

[errant wedge]

Ill go review fractions </3

[minor fulcrum]

do you want an answer

||the fractions of the form a/(2^n)||

[errant wedge]

ill open that if im still stuck in like 15 but tysm

[errant wedge]

I did this by showing $\langle\frac{p_1}{q_2},\ldots,\frac{p_n}{q_n}\rangle\leq\langle\frac{1}{\mathrm{lcm}(q_1,\ldots,q_n)}\rangle$ and we know all subgroups of a cyclic group are cyclic, but is there a (not horrible) constructive proof?

[cloud walrus]

Sara

[dull ginkgo]

I assume this is multiplicative

[errant wedge]

Additive

[minor fulcrum]

Clear denominators

[errant wedge]

wdym

[minor fulcrum]

Multiply every element of your subgroup by the product of all the denominators

This makes it a subgroup of Z

[errant wedge]

Oh W and this is equal to the finitely gen subgroup?

[minor fulcrum]

Yeah, isomorphic to it

[errant wedge]

cool ty

[cobalt heath]

Sometimes I cannot believe that subgroup of a cyclic group is cyclic.

[errant wedge]

for (a), I set $X=F_k$ so $F_5X^3=F_5F_k^3=F_5F_k=R_{5-k}$ and $X^9F_{13}=F_k^9F_{13}=F_kF_{13}=R_{k-13}$. Solving $5-k\equiv k-13$ (mod 28) lends $k\equiv 9$ (mod 28). But verifiably, $5-23=-18\equiv 10$ (mod 28) $\equiv 10 = 23-13$, but $23\not\equiv 9$ (mod 28)? What am I doing wrong

[cloud walrus]

Sara

[errant wedge]

Not a field. am dumb

[errant wedge]

is this true for additive groups also

[lapis latch]

As in the additive group mod n?

[errant wedge]

yes

[lapis latch]

yeah, the ring isomorphism restricts to a group isomorphism

[slim kayak]

Ring homomorphisms are always also additive group homomorphism

[errant wedge]

Fantastic thanks

[desert stone]

what all does serge lang algebra cover

[winged void]

hey i have a small question

is there an easy way to determine all rotation of a platonic solid that is possible

without having one in your hand

for example a cube

its 24 in total but is there a way to determine all of them without having a cube in your hand

and i mean more like not the total number of rotational symmetry

but symmetry per axis or plane

like for example with x-axis you can do 3 rotation not taking to account the identity element or rotation in that case

[slim kayak]

Wdym, read the table of contents it's free

[ornate tiger]

in (G,+) , G={(0,0),(1,0),(0,1),(1,1)}

Why it's not cyclic

do we need like to get <(0,0)> and from it get all G ?

[cold dust]

Not <(0,0)>, that'll always just be trivial. Cyclic means there exists some element a in G such that <a>=G. Here, however, for any nontrivial a, |<a>|=2, not 4, so the group isn't cyclic

[ornate tiger]

so we need to get <(1,0)> and it should generate all G

[cold dust]

I mean, it doesn't need to be (1,0) specifically. It's cyclic if any element generates all of G. Here, though, no element generates all of G, so G isn't cyclic

[ornate tiger]

oh ok so can we test <(1,0)> and see what we will get ?

[cold dust]

Sure. If you test all of the elements and never get G, it isn't cyclic

[ornate tiger]

ok i have an answer for <(1,1)> = {(0,0),(1,1)}

how did we reach there

[cold dust]

Well, using the + operation, <a>={a,2a,3a,...}. If a=(1,1), then 2a=(1,1)+(1,1)=(0,0). 3a=2a+a=(0,0)+(1,1)=(1,1), and we're back to where we started, so <(1,1)>={(0,0),(1,1)}

[ornate tiger]

are we in modlo 2 ?

[cold dust]

Yes

[ornate tiger]

because we have a vector from 2 Coordinate ?

[cold dust]

Ehh the reason depends on how you define $G$. The way you defined it, as an explicitly written set, it's kinda just a fact that doesn't have a reason other than "that's what works". If you take the definition of $G$ to be $G\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, then the mod 2 comes from the product of the $\mathbb{Z}/2\mathbb{Z}$ groups, each of which is, by definition, in mod 2

[cloud walrus]

evelyn

[ornate tiger]

if we had Z3 * Z3 it will be mod 3 ?

[cold dust]

Yes

[ornate tiger]

OK THANKS , i have another question

[cold dust]

Okay

[ornate tiger]

i have a prove or it's kind of prove like it's short that every cyclic group is abelian
h in <g>
there is n such that h = g^n
h_1 = g^n1
h_2 = g^n2
this say h_1h_2 = h_2h_1

like i don't understand why h = g^n ?

how to try with the TeXit bot

[cold dust]

Surround math text in dollar signs to get the tex bot to make it look nice

[cold dust]

If you have a group $G=\langle g\rangle$, then for every element $h\in G$, there exists some integer $n$ such that $h=g^n$. That's the definition of a cyclic group

[cloud walrus]

evelyn

[ornate tiger]

oh so h is always g^n

[cold dust]

In a cyclic group, yes

[ornate tiger]

is this related with the order of elements ?

[cold dust]

kinda! The order of an element $g\in G$ is the smallest positive integer $k$ such that $g^k=e$, where $e$ is the identity of $G$.

[cloud walrus]

evelyn

[cold dust]

This is for any group, not just cyclic groups

Another equivalent def of a (finite) cyclic group is a group where there exists an element whose order is equal to the size of the group

Oh, other definition of the order of an element $g\in G$ is $|\langle g\rangle|$

[cloud walrus]

evelyn

[ornate tiger]

yeah but h here is not e right

[cold dust]

The definition of the order uses e, the identity

[ornate tiger]

the definition for cyclic groups doesn't use e

[cold dust]

No, but the first definition of the order of an element above does

[ornate tiger]

ok cool thanks

[cold dust]

Another way to make sense of this is that a cyclic group $\langle g \rangle$ has the set ${g,g^2,g^3,\ldots,g^{n-1},g^n}$, where $n$ is the order of $g$. In this definition, it is clear that any element of that set is some power of $g$

[cloud walrus]

evelyn

[ornate tiger]

oh yeah this is a good one thanks

[cold dust]

yup!

[slim kayak]

What is your definition anyways?

[regal star]

Quick question.
I am confused about how to find the dim of SO(n), O(n) and SU(n) rigorously.
for example, for n=3 the dim of O(3) is 3 and the constraint condition doesn't reduce the dimension further. How to justify this rigorously using an argument from real analysis, perhaps? I know that since the det function is not a constant on O(3) it has disconnected components SO(3) and the other part.

Also, the dim of U(2) is 4 and the det constraint reduces the dim to 3. How to justify this as well? Here the det function is continously maps from U(2) to e^i theta but I am not able to justify it

[kind temple]

the dimension of these groups as lie groups?

[regal star]

yes.

[kind temple]

you can use the regular level set theorem or compute the dimension of the tangent space at the identity based on the defining maps from Mat(n,R) to itself. what do you mean by, "the determinant doesn't reduce the dimension"?

[regal star]

U(2) has the dimension of 4 right. but for SU(2) we need det = 1 which is another constraint and this makes the dimension of SU(2) to be 3.

[regal star]

Is it possible to prove it without using the dim of tangent space or any other ideas of differential geometry?

[kind temple]

i mean, the tools of differential topology/geometry are tailored for tackling problems about lie groups/algebras

im not sure if there is any way to do this purely without differential topology since the definition of their dimension is in terms of the dimension of the manifold they form

[south patrol]

this is really smth i'd normally put in #diff-geo-diff-top btw

but yeah hm

[regal star]

Okay. makes sense

[south patrol]

one thing you can do if you just care about topological manifolds is note that like it suffices to consider a neighbourhood of the identity and then you can check that that is homeomorphic to R^k for appropriate k

but it is best imo just to use regular values to do this cleanly and show they're smooth at once

[kind temple]

how would one go about choosing charts?

like, if all i have is QQ^T = I, where do i start making the correspondence

[south patrol]

for topological ones you don't need charts usually though right

[kind temple]

oh shoot

my b

well

you need local homeos
(i call charts any homeo from an open subset of the manifold into R^n that is appropriately C^k or D^k for whatever type of structure you want)

[south patrol]

though i think if you pick smooth charts in one neighbourhood you can get charts everywhere by post composing with stuff in the group

[kind temple]

sure, but how are you going to get a local homeo near I?

[south patrol]

true idk how to do that bit any easier lol, seems to basically come down to the same thing

[kind temple]

thats what i was thinking too

[frozen dust]

suppose we have a commutative group (G, +), and 2 subgroups H, H', an element a of H.
if nor H is a subset of H', nor H' is a subset of H, is it possible for the coset a+H' to be a subset of H?

i think that the only elements for which a + a', with a' from H', could be an element of H, would be if a' were the identity element or if a' were the inverse of a. If a' were the identity element, and the only element in H', then a+a' could be indeed be a subset of H but if H' only contains the identity element it is a subset of H so this does not work. I thought the only other possibility could occur if a' is the inverse of a. But if the inverse of a' is in H', the inverse of a', a, must also be in H' or it would not be a group.

But I have no idea whether my reasoning is correct and/or there are other such examples for which this could still happen

[south patrol]

If a+H' is a subset of H then H' is too, because a is in H

like H' = (a+H')-a subset H-a = H

@frozen dust

Oh I guess you want the general case hm

if a is in H, a' in H' and a + a' is in H then a' is also in H, so a' is in both

and yeah you can always add elements of the intersection to an element of H to get smth in H

[frozen dust]

but does this also apply for the a' that are not in the intersection of H' and H?

in order for a+H' to be a subset of H, does it not need to apply for every a' in H'?

[ornate tiger]

Hi i have a question in (Z_6 , + ) or (G, +) i don't know how to write it anyways the order of <2> is 3 does that mean the subgroup <2> is not cyclic ? and the order of <5> is 6 does that mean it's cyclic subgroup ?

[fading field]

they’re not cyclic for the reason that you said, there’s no order-checking needed here. the subgroups that you mentioned are both cyclic, but that’s because every subgroup of a cyclic group is cyclic

also, subgroups that are generated by a single element are cyclic

by definition

[ornate tiger]

why <2> is cyclic we don't get G from it

but <5> we get G

[fading field]

what does cyclic mean to you?

[ornate tiger]

nothing it's just something in math

[fading field]

well if cyclic doesn’t mean anything to you how could you assert that anything is cyclic or not cyclic?

it makes it harder to talk about a concept when you don’t have a clear definition for it

[ornate tiger]

well there's g such that <g> = G

<5> = G that's correct

<2> is not

[fading field]

but that would mean G is cyclic

but 2 generates <2>

so <2> is cyclic

[ornate tiger]

what's inside <2> ? just 2 ? or 0,1,2 ?

[fading field]

in the group (Z_6,+) which you mentioned before, the group <2> would be all the multiples of 2 in that group, so {0,2,4}

[ornate tiger]

yeah true

but how does that say it's cyclic

[fading field]

well, a group is cyclic if there is an element that generates it

[ornate tiger]

ok wait can we say <5> is a cyclic group not a cyclic subgroup

[fading field]

but <2> is generated by 2

[fading field]

a cyclic subgroup is a subgroup that is cyclic itself, you don’t have to look at the structure of the big group to determine whether the small group is cyclic

it’s not about the relationship between the subgroup and the big group at all

[ornate tiger]

ok <2> is {0,2,4} why it's cyclic

[fading field]

because it is generated by 2

[frozen dust]

can you write all elements of {0, 2, 4} in terms of one of it's elements?

[ornate tiger]

what does this mean give me an example english is not my first language

[frozen dust]

if i have {0, 1, 2, 3, 4, 5} i could say 0 = 0 * 1, 1 = 1 * 1, 2 = 2 * 1, 3 = 3 * 1, 4 = 4 * 1, 5 = 5 * 1, so i can write every element as a multiple of 1

[ornate tiger]

so if we say <3> it is {0,1} ig right

does this mean it is subgroup

no wait it is {0,3}

right ?

[fading field]

that’s right, and the group would be generated by 3, since 3*1=3 and 3*2=0

[ornate tiger]

is this what u talk about ?

what if we had some <a> = {e} is it also cyclic

[fading field]

yes it would be generated by a

all of these <a> groups are cyclic by definition

[ornate tiger]

when do we say a group is not cyclic

[fading field]

if it’s not generated by a single element

[ornate tiger]

do u have an example ?

[fading field]

yeah, (Z_3 x Z_2,+) with addition coordinatewise, like (a,b)+(c,d) = (a+c,b+d)

that’s not cyclic

because there is no element that generates the whole group

[ornate tiger]

why it's not a+c , b+d ?

[fading field]

lol yes

ur right i am sick and not thinking straight sorry

but this group isn’t cyclic

[ornate tiger]

is this possible like i know Z_2 x Z_2 but idon't remember soemting like this (Z_3 x Z_2,

[fading field]

Z_2 x Z_2 is also not cyclic

you can take the product of any two groups

like this

[ornate tiger]

ok i have a qustion about Z_2 * Z_3 how do we write the vectors there like Z_2 * Z*2 are like this (a,b) right

[fading field]

yeah, and if you have GxH then the elements will be of the form (g,h) in general

[ornate tiger]

if we had Z_3 * Z_3 * Z_3 does that mean (a,b,c) ?

[fading field]

yes

[ornate tiger]

ok thanks

[frozen dust]

are there finite abelian non cyclic groups that are not a direct product of cyclic groups?

[fading field]

see the fundamental theorem of finitely generated abelian geoups

[slim kayak]

You may later learn a theorem that any finite group embeds into the symmetric group of the corresponding order. So they are a place where often want to check for counterexamples

[frozen dust]

im gonna bump my question if any1 could help me with this

[slim kayak]

Can you phrase your question again?

Like, the original question or a new one?

[frozen dust]

the original question was this:
suppose we have a commutative group (G, +), and 2 subgroups H, H', an element a of H.
if nor H is a subset of H', nor H' is a subset of H, is it possible for the coset a+H' to be a subset of H?

my attempt was this: i think that the only elements for which a + a', with a' from H', could be an element of H, would be if a' were the identity element or if a' were the inverse of a. If a' were the identity element, and the only element in H', then a+a' could be indeed be a subset of H but if H' only contains the identity element it is a subset of H so this does not work. I thought the only other possibility could occur if a' is the inverse of a. But if the inverse of a' is in H', the inverse of a', a, must also be in H' or it would not be a group.

after getting a reply i think i might misunderstand the question though. do i have to show that for every a' from H', when added to an element a of H, it ends up in H in order for a + H' to be a subset of H

[slim kayak]

Lets replace H' with K. If a+K is a subset of H, then in particular a is contained in H. Since any element of a+K is of the form a+k, and all a+k are in H, then all -a+a+k are in H. So all k are in H, hence K is a subset of H

[frozen dust]

Got it. So as your conclusion is that K is a subset of H, and a given is that K is not a subset of H, can we conclude that it is not possible ?

i'm a bit confused about this step though " and all a+k are in H, then all -a+a+k are in H."

[slim kayak]

Sure, but thats kinda disappointing

[slim kayak]

yeah? If a is in H, so is -a cuz its a group

[frozen dust]

ohhh okay i was confused about the notation then

[frozen dust]

why is that?

[slim kayak]

A more interesting related result:
If K is a subgroup of H, then H contains all cosets h+K where h is an element of H.

[frozen dust]

there were 3 questions, but i found the first 2, and one of those questions was to show that

i notice both you and the previous person who helped me started from the assumption that a+K is a subset of H

to then arrive at the conclusion

how did you know you should begin your proof with that?

[slim kayak]

wdym?

[frozen dust]

because i was trying to prove "it is possible", but you assumed it was possible and then made some conclusions by which you concluded it's not

is it some intuition u build through repeated exercise? because i feel like i lack that intuition

[slim kayak]

Uh, well. If you are asking if it possible it makes sense to see what follows if a+H' subset of H was the case

[frozen dust]

ok

thanks for your help!

[slim kayak]

I am still thinking hold on

[frozen dust]

oh i thought that you were telling me 'it makes sense to see ..' i misunderstood

[south patrol]

Z_2 isn't cyclic but 1 is a topological generator

[slim kayak]

It's mostly about putting the questions into a more "standard form". Your original question couldve been into this form:
Does for all (list your "ingredients", here a random element a, subgroups H' and H) a+H' subset of H imply H' subset of H.

So you either perform a general proof to show that it is always true, or you find a counterexample - in which case finding an example where what you showed was true.

In either case its usually good to try to find a (constructive) proof of the positive, general statement and just "feel out" the situation

[frozen dust]

I got it, that clarifies it

thank you

[glad osprey]

Isn't Z_3 x Z_2 generated by (1, 1)? Since gcd(3, 2) = 1

[fading field]

oh fuck, honestly I’m just gonna put all my mistakes on me being sick for now, i for sure am not thinking right

yes you’re right lol

[spice whale]

smay remembers the structure theorem

[glad osprey]

I know the feeling, get well soon ❤️

[limpid ferry]

Given the statement: "If group G is cyclic with order n and generator g, then for each positive integer d that divides n, there exists a unique subgroup of order d, which is H=<g^(n/d)>." I need some hints to prove the unique part of this.

[winged void]

can somoeone please answer my question

regarding if there is an easy way to determine all rotation of platonic solid

if there is an easy way

[rocky cloak]

The nice thing about platonic solids is that the symmetry group acts transitively on faces/edges/vertecies.

So you can describe each rotation by, for example, which face is "on top" and how that face is rotated.

So for example a dodecahedron has 12 pentagonal faces, so a rotation is determined by placing any of them on top, and at one of five orientations. So 12*5 = 60 rotations in total

[winged void]

the easiest one is about the faces

this thing i can do it like in my mind

i was solving a question

[rocky cloak]

What was the question?

[winged void]

this is the question

i do not want to look at the answer

but what i did

well first of all we can draw an axis between the upper one and lower one

then we see that the upper face and the lowe face since they do not change they can have a diffrent color and all other rectangular face

must have the same color

thus 2^3

we can do the same thing for the rectangular face then we will see that they are going to have 2^5

and we should multiply that offcours with 3

because there are 6 faces in total

so 6 / 2

= 3

but now for the vertex rotation and edge rotation is difficult idk where to begin

and yes offcour the one doing nothing

2^8

this is identity rotation

[ornate tiger]

I have this question which i don't understand :
there is m,n . when does nZ ⊆ mZ
the answer is when m|n
let's say m =12 , n=60
12Z ⊇ 60Z
12.5 = 60
60∈ 12Z
Therefore 12Z ⊇ 60Z
i don't understand what is this

[rocky cloak]

mZ is the set of multiples of m.

So this is saying every multiple of 60 is also a multiple of 12, which should make sense, since 60 is a multiple of 12

[ornate tiger]

yeah it is but is this the only thing the question talk about ?

is it 60|12 or 12|60 like i can see i said m=12 but is n =12 ?

[vapid vale]

yes

the fact that m divides n means every multiple of n is a multiple of m

[ornate tiger]

so n =12 and m =60

if it was m|n

[rocky cloak]

12|60 means 12 divides 60, so it should be m=12 and n=60 just like you said first

[ornate tiger]

oh ok it does complicate me a little bit

thx

both

[winged void]

Yes @rocky cloak did you look at it

[rocky cloak]

Yeah, is a little hard to follow what you're saying. But the symmetries of the prism are the 6 rotations in the horizontal plane, and the 6 ways to "flip it over".

[ornate tiger]

b here is not equal to e right

[noble lynx]

found this on stack exchange, why cant this 3/4 bound not be increased to 1/2?

[barren sierra]

what is e?

[noble lynx]

we know that S^2 = G and everything in S commutes with everything else in S and so s'(s_1s_2) = (s_1s_2)s' which means S is contained in the center of G which means the center has to be the entire group by lagrange? Don't know what I am missing

[ornate tiger]

identity

[barren sierra]

I think this is well defined if b = e

because then what is the order of b?

1

And then k = n in this case

so then d = gcd(n, n) = n

and order of b = n/n = 1

or you could say k = 0

in which case d = gcd(0, n) = n and it sitll works

[ornate tiger]

mmm

can we build an example ?

like if we in Z_3

what is the order of this group 3 ?

[barren sierra]

order of a group != order of an element

do you know the definition of the order of an element?

[ornate tiger]

yeah a little bit

[barren sierra]

tell it to me

[ornate tiger]

when we reach the identity then it's the order of the element i mean the power is the order

like if g^n = e

[barren sierra]

it's the least such power

[ornate tiger]

then the order is n

[barren sierra]

least such n

not just any n

[ornate tiger]

oh yeah

[barren sierra]

but yea you seem to have the idea

ok lets work with a more interesting example than Z_3 since non-identity element of Z_3 has order 3 (check this yourself)

lets try Z_4

ok so Z_4 is integers mod 4 with addition as the operation

what's a generator of Z_4?

[ornate tiger]

wait let me try

omg i did one of these questions like before 15 mins and now i forgot

[barren sierra]

well there are only 4 elements to check

0, 1, 2, 3

[ornate tiger]

yeah like we do <0>

then 1*0

[barren sierra]

but does 0 generate Z_4?

[ornate tiger]

no it's just 0 right

[barren sierra]

yea, the subgroup generated by 0 is just 0

check the other 3 elements

[ornate tiger]

oh it's 1

right ?

[barren sierra]

1 is a generator

[ornate tiger]

yes

[barren sierra]

ok so lets take b = 2 = 2*1 so our k = 2 agreed?

[ornate tiger]

(1) * 0
(1) * 1
(1) * 2
(1) * 3
when u wrote 2*1 is the 1 u wrote is the one i put between ()

[barren sierra]

right so 1 is a generator but I'm just trying to demonstrate the statement of the problem

[barren sierra]

do you see what I wrote here and does it make sense?

[ornate tiger]

k =2 because b is 2 ? like if we took 3*1 does that make k = b = 3 ?

[barren sierra]

well our group is written additively

so instead of writing 1^2 = 2 like in the problem

I'm writing 2*1 = 1 + 1 = 2

it's a coincidence here that k = b

it's just that if I want to express the element 2 in terms of the generator 1, I would write 1 + 1 = 2

2*1 is shorthand for 1 + 1 just like how a^3 is shorthand for a * a * a if I were working in a group generated by an element a

so since 2 = 2 * 1 = 1 + 1 we have that k = 2

[ornate tiger]

ok

[barren sierra]

so do you agree that if b = 2, then k = 2?

there's alot of notation going on here so I want to make sure you follow

[ornate tiger]

yeah i can see this for now but i think we nearly finished right ? we are in Z_4 so |Z_4| is 4! which is 24 then we do 24\2 ?

[barren sierra]

no, |Z_4| is not 4!

Z_4 only has 4 elements

and that's not what the question is asking

[barren sierra]

the question is asking about orders of elements, not the order of the group

so we have that Z_4 is a cyclic group of order 4, so n = 4

[ornate tiger]

oh

[barren sierra]

and our generator is a = 1

[ornate tiger]

true

[barren sierra]

and we are considering b = 2*1 = 2 (so k = 2)

ok what is the order of 2 in Z_4?

[ornate tiger]

ok well 1

[vapid vale]

quaternion group

[ornate tiger]

no wait 2

[barren sierra]

how did you get that the order of 2 is 1 in Z_4

yea it's order is 2

[vapid vale]

i -> -i, j -> -j sends 6/8 elements to their inverses

[barren sierra]

ok so n = 4, k = 2 so d = gcd(2, 4) = 2 and n / 2 = 4 / 2 = 2

so the question aligns with what you just found that the order of 2 in Z_4 is

so that's what the question is talking about. So try the question in general now

[ornate tiger]

wait the order of b is 2 ?

i forgot the quesstion

yeah the order of b is 2

ok i have an example that i want to follow what u said and see if i will succeed
a = 13 , G = <a> and it asks what is the order of 39

i feel like there's something missing

[noble lynx]

whats wrong with my reasoning?

[rocky cloak]

Why do you say everything in S commutes with everything else in S?

And how do you know S^2 = G?

[noble lynx]

oh I think I assumed S is multp. closed

but S^2 = G comes fro this problem

[rocky cloak]

I see. Yeah, S will not necessarily be multiplicatively closed, so the elements don't necessarily commute with each other.

[noble lynx]

makes sense thanks again

[dull ginkgo]

JACOBSON?

[dull ginkgo]

Oh shit this is the one where also if |S| = 3/4 |G| then S is like a normal subgroup right

Or something like that

oh yeah fuck I forgot

[noble lynx]

yeah lol tryna get through most of the problems

[dull ginkgo]

I took a small hiatus but I’m starting the PID section

[dull ginkgo]

The problems are hell

[noble lynx]

honestly, I like they dont put too many

it's like the minimum to feel confident you understood it

but yeah some problems are like wtf, there's no way anyone would come up with that

@dull ginkgo are you also working through jacobson?

[cobalt heath]

O I thought you finished the book, just how long is this dam book

[dull ginkgo]

Yes

[dull ginkgo]

400 pages

[cobalt heath]

Huh, that's shorter than I thought

Maybe it is full of exercises

[dull ginkgo]

I am a bit slow because I want to finish Abbott analysis but I will come back and go from the PID section

[dull ginkgo]

No, not many, but they tend to be very difficult

[noble lynx]

oh thats awesome

[cobalt heath]

Am talking based on the number of exercises you've shown here

There are just tons of 'em

[dull ginkgo]

I tend to try to work through them in chat

So I realize that I’m a dumbass

I need to start doing that in the thread again

[noble lynx]

I tried to read vol 2 then went to the module chapter of vol 1 and then the galois chapter and now Im revisiting groups, since I realized I dont know anything about groups

[cobalt heath]

You can't say that when you managed that damn Ore localization

Even nlab said it is difficult 💀

[dull ginkgo]

I generalized the fuck out of it

I did a vast majority of the ring section ones

Pretty much every one that isn’t rote showing properties that are obvious pretty much

By far the group theory part though was way harder

[cobalt heath]

Yeah group theory is difficult tbh

[dull ginkgo]

It gets easier after Sylow but still

[cobalt heath]

It's just that group (somehow) has less defining properties

[dull ginkgo]

Like some problems are annoying by themselves

Like proving <u,v> is dihedral where u and v are involute and uv is torsion is a pain in the ass

Actually frankly anything with presentations is a pain in the ass

[cobalt heath]

Ahhh

Did not learn much but I think rep theory could help there

[dull ginkgo]

Volume 2 covers rep theory stuff

[dull ginkgo]

Like it’s not unintuitive it’s just annoying

[cobalt heath]

Indeed

Computation galore

(That said, it seems research level is also full of computations )

[dull ginkgo]

Like most of the computation is to show the presentation and any other relations would violate the minimality of the order of uv

[cobalt heath]

Yeah

I guess Jacobson loves these annoying things

[ornate tiger]

a = 13 , G = <a>
what is the order of 39 ?
the answer is 40

[delicate bloom]

not enough information to answer this

also what have you tried so far

[ornate tiger]

well i thought to generate <13> but i don't know what is our group

what is written after the question mark is this i don't know why and how did he know :
k = 3 , n =120
gcd(k,n) = 3
therefore the order of 39 is 40

@delicate bloom

[crystal vale]

Can you share a screenshot of the problem?

[ornate tiger]

it's not in english

but i said what is wrriten all of it

[crystal vale]

There is one result that if a has order n then for a positive integer k, a^k has order n/gcd(n,k)

[crystal vale]

I think they work in Z_m where 13 has order 120

[ornate tiger]

what is Z_m

[crystal vale]

Z/mZ now I am looking for m such that 13 has order 120 in that group

m = 13×120

[ornate tiger]

sorry i didn't see your message it didn't appear

[ornate tiger]

so n is for the order

can u reply to me when u answer because if u didn't i will not see that u sent anything

[crystal vale]

Since 39 = 13 ×3

[crystal vale]

Now using this one

[ornate tiger]

wait is 13 or 39 are one of the the k or n ?

we can use this

we are using it

[crystal vale]

Yes

[ornate tiger]

ok so 13^k = b

this is the only thing i can seee

[crystal vale]

So according to it let k = 3 and b = 39

[ornate tiger]

why

because 13 *3 = 39 ?

[crystal vale]

Yes and here * is addition modulo 13× 120 operation

[ornate tiger]

so it's like n*13

3*13

but why they write it as power

[crystal vale]

I am so confused with this n,k

[ornate tiger]

like 13^3 = 13+13+13 ?

[crystal vale]

Yes

[ornate tiger]

but why they write with powers

[crystal vale]

It just multiplicative notation

[ornate tiger]

ok so k=3 a =13

b=39 ?

[crystal vale]

Yes

[ornate tiger]

ok we will find the order of b with n/gcd(k,n)

[true ingot]

Any hint for b)? I got that one orbit consists of all invertible matrices, but how do I divide the other in n orbits?

[ornate tiger]

what is our n ?

n is the order of the cyclic group

how do we know it

[crystal vale]

I just assume here the group is Z/mZ, where m =13×120 because you give that n = 120 so I let 120 is the order of 13

[ornate tiger]

i don't know how did he find that n =120

[crystal vale]

Is there any other information?

[ornate tiger]

no

[prisma ibex]

the orbits correspond to ranks

your invertible matrices have rank n. The remaining n orbits are those with ranks 0 through n-1

[ornate tiger]

the answer is 40 so yes it is 120/gcd(3,120) but how i don't know

[terse crystal]

Rank of a

[chilly ocean]

Are the groups here Z6 for abelian and S3 for non-abelian?

[coral spindle]

Yes, those are the only groups of order 6 up to isomorphism.

[chilly ocean]

Thanks

[minor fulcrum]

what is the asymptotic growth rate for groups of order n

like obviously it can't grow faster than n^2

Unrelated: are all lie groups the quotient of two matrix lie groups?

[amber prism]

If i am asked to state what is normal subgroup then is it ok to just say ah in H and ha in H

[mighty kiln]

<@&268886789983436800>

[amber prism]

wowow <@&268886789983436800>

[obsidian sleet]

why is it not deleting pls

discord...

i manually did it

sadness

ok good

[wispy urchin]

[coral spindle]

No, this isn't sufficient explanation. In fact if you are indeed saying what I think you're saying (it's not clear) then this is totally incorrect.

[tough onyx]

gdi I accidentally misclicked and joined the server

[mighty kiln]

Number of groups up to isomorphism?

[coral spindle]

I think anything else might be an easier counting problem...

Ah yes, 1 -> a class-sized object, 2 -> a class-sized object...

[lusty marlin]

Why not?

[rocky cloak]

I guess you mean it can't grow faster than n^n^2 ?

Anyway, there is some results here
https://math.stackexchange.com/a/422133/306319

In short n^(c log(n)^2) for some c, seems to be about right

[amber prism]

so what would i meed to explain more

[lusty marlin]

Yeah n^n^2 makes more sense

[coral spindle]

Just state the definition.

[amber prism]

👍

[wraith cargo]

Do you mean it can't grow faster than n!?

[ornate tiger]

the order of 5 in Z_12 is 12 or 2 ?

because i'm watching a video now that it says 2 and i was like wtf because i thought it's 12

[mighty kiln]

12

[ornate tiger]

like 15 = 5
25 = 10 = 10
35 = 15 = 3
45 = 20 = 8
55 = 25 = 1
65 = 30 = 6
75 = 35 = 11
85 = 40 = 4
9*5 = 45 = 9
10 * 5 = 50 = 2
11 * 5 = 55 = 7
12 * 5 = 60 = 0

right

[ornate tiger]

yeah now i'm confused

because she's solving a question i want

[tribal moss]

The multiplicative order is 2.

[ornate tiger]

but we are in Z_12 aren't we with +

[tribal moss]

The additive order is 12.

[ornate tiger]

ok this make things more clear but i thought in Z_n we only use +

[tribal moss]

Z_12 often means the ring of integers modulo 12, so both addition and multiplication are fair game to consider.

[coral spindle]

You are right that we are no longer working in the group Z_12 but rather the group (Z_12)^x also know as U(12), the group of units of Z_12.

[ornate tiger]

ok cool guys thx

[tribal moss]

If you want to make it more unambiguous that you're talking about the additive group, the notation C_12 (ie the cyclic group of order 12) is unambiguous.

[ornate tiger]

C for complex numbers ?

[tribal moss]

No, C for cyclic.

[minor fulcrum]

yes I meant this, and that's much slower than I'd expect

[minor fulcrum]

no, I mean't n^(nxn)

[minor fulcrum]

update on this: I asked my mentor who works with lie groups and he said "I don't know"

[mighty kiln]

By Ado's theorem every connected Lie group is a quotient of the universal cover of a matrix group right

Are universal covers of matrix groups matrix groups

[minor fulcrum]

Not at all

[sly crescent]

Sometimes but not always

[rose prism]

i think the universal cover of SL(2,R) will be a counterexample

but not sure

i want to say that if M := universal cover of SL(2,R) is the quotient G/H of a matrix group G, then G should be isomorphic to M which would be a contradiction

[old sun]

What's the degree and basis of Q(sqrt3 + sqrt5) over Q

[rose prism]

oh i guess someone was hinting at this above

[long geyser]

how would you show this

[south patrol]

you can sorta just write out what k is explicitly

like what do the elements of k(A) look like

alternatively you can show that like

$\bigcup_{A' \subseteq A \text{,finite}} k(A')$ is a field containing $A$

[cloud walrus]

PD potato

[south patrol]

so it is k(A)

[long geyser]

rational functions in A?

[long geyser]

hmm, I guess this leads to the same thing?

because when A' finite it is rational functions in A'

and if you multiply two rational functions in A' and A'' respectively you get one in A' union A''

[rocky cloak]

Well, this way you can do it without knowing anything about what k(A') looks like.

You would just have u in k(A') for some finite set A', and then you're done

[long geyser]

I think it is trivial either way, whether you show k(A) is rational functions in A or that huge union

because if u is a rational function in A, there are only finitely many of the guys in A showing up in the rational function

important step is being able to describe k(A) in one of those forms

[south patrol]

yes i mean the second was just to avoid knowing what it lookslikeas jagr said

but yeah useful anyway to know it's rational functions in elements of A

[long geyser]

I see now

you don't even need the rational function description for that union

you get closure for free

and all field axioms hold cuz u belong to one guy in the union

[barren sierra]

Is $R / ( \overline{x}, \overline{y} )$ not isomorphic to $k[z] / (z^2)$ so then how is $R / \mathfrak{p}$ a domain?

[cloud walrus]

Spamakin🎷

[rotund aurora]

this is poorly written, k[x,y,y] ?

you could interpret it as k[x,y,z^2], then it's true

[barren sierra]

I think it's meant as k[x, y, z]

if it's k[x, y, z^2] that's an insane typo >_>

[rotund aurora]

k[x,y,z]/(x,y,xy-z^2)=k[z]/(z^2) as you suggested, which is not a domain

[barren sierra]

yea

[rotund aurora]

it could also be that they meant xy-z instead of xy-z^2

[barren sierra]

ok cool just wanted to make sure I wasn't doing something dumb

yea idk this wouldn't be the first HW assignment from this course that has some weird typos / questions that aren't well founded

[rotund aurora]

but I wouldn't waste time on this, it's your professors fault, they should know better lol

[barren sierra]

well I'm not in a course rn

just doing all past HWs from this class to prep for quals

but yea not wasting my time

[long geyser]

in this case, Gal(E/Q) has order 27 and the other 2 have order 9, correct?

(Z_3)^3 for E/Q, (Z_3)^2 for the other two

hmm, actually no

that can't be

the other two gotta have order 6

[cobalt heath]

Where does your order 27 come from?

[long geyser]

and I hope Gal(E/Q) has order 18

[cobalt heath]

I don't think it has order 27

[long geyser]

no it doesn't

I mean, it doesn't have order 27

I hope it has order 18

[cobalt heath]

Yeah, it does

[long geyser]

I was first thinking you have to add omega and 2^(1/3), and both of those have irreducible poly of third degree

but that doesn't account the fact that they are linked

[cobalt heath]

You can consider how roots are exchanged.
At least there is
Gal(B/Q) -> Gal(E/Q)
G(C/Q) -> Gal(E/Q)

[cobalt heath]

The secret is that omega has less degree irreducible polynomial!

[long geyser]

wait, x^3 - 1 isn't even irreducible

fml

ye it is like x^2 + x + 1 right

cyclotomic guy

[cobalt heath]

Yep.

[limpid ferry]

I just want to confirm what is the notation φ^-1, is it the inverse mapping of φ or something else?

[lusty marlin]

ϕ^-1(H') is the set of elements of G that map to H' under ϕ

[cobalt heath]

What is the fundamental reason behind [C : R] = 2?

Also can one extend exponentation to quaternions? I wonder if noncommutativity poses issue.

[cobalt heath]

I guess you can just take lie algebra for quaternions instead

[delicate bloom]

I'd imagine you have to mess with campbell-baker-hausdorf if you want to do it

[delicate bloom]

I just figured this was a consequence of being the field for the prime at infinity somehow made it boring by comparison to the other Q_p. There's probably someone on this server that can give a real answer to that though, but it ain't me

[cobalt heath]

Wow.

Hmm, did not think about that. That said, Infinity being simpler makes sense.

[limpid ferry]

I just managed to prove the part 4) Is it true vice versa tho? That is, if phi is a homomorphism from G to G', and H is a normal subgroup of G, then phi(H) is a normal subgroup of G'.

[fading field]

no, you can take G itself, normal in G and map it to a subgroup that’s not normal in G’

like if H < G then the inclusion H -> G is a group homomorphism

no matter whether H is normal or not (but H is, in fact, normal in H)

[long geyser]

even more extreme, you can map everything to 1 in any group

[fading field]

{1} is normal in every group

though

[next obsidian]

Hi smay

[long geyser]

oh wait

idk what I was thinking lol

[fading field]

real it’s all good

[fading field]

hi chmonkey

[next obsidian]

Chmonkey tired

I gonna sleep

[limpid ferry]

I am not sure I am understanding, G is a normal subgroup of itself, which maps to G', also a subgroup of itself, that does not contradict my statement

[fading field]

group homomorphisms don’t have to be surjective

like phi(G) may very well not be G’

the statement is fine if you additionally require phi to be surjective actually

[mighty kiln]

A degree n extension of R induces a continuous injection RP^(n-1) -²→ S^(n-1), which must be a homeomorphism

Though that doesn't explain why it's finite

[cobalt heath]

Hmmm

Where does the map come from?

[mighty kiln]

Squaring

[cobalt heath]

Ah

[tardy hedge]

GUYS I KEEP ON FORGETTING WHY COSETS PARTITION THE GROUP

the fricken cosets, quotients, etc etc stuff i always learn then forget

it does NOT stick in my brain! Damn!!

[mighty kiln]

g^-1h ∈ H is an equivalence relation

I.e. gH = hH is an equivalence relation

[tardy hedge]

ik personally that doesnt do much more for me besides just memorizing that cosets partition group

at this point i remember a lot of those facts, but i keep on forgetting the understanding of why it makes sense

[mighty kiln]

Are you able to derive this fact from scratch

If so that's probably good enough

[rotund aurora]

imo the observation that cosets partition the group is an important observation, but it is not deep or anything, in fact it is trivial to prove. Maybe you just lack experience/examples, idk

[tardy hedge]

Yeah im not sure why i always get hung up on it honestly

im just wanting it to be like "yeah obviously" when i recall that fact, instead of me being like "wait why is that true again"

yk what i mean?

[rotund aurora]

You said the word "forget". You don't need to remember the proof, you just need to be able to prove it. In fact, when you "forget" such basic notions I'd suggest that you don't try to "remember", but instead try to prove them as if you had never heard of them before

[tardy hedge]

hm i see

[coral spindle]

Is it obvious that:

1. For any surjection f : A → B, the preimages f^-1(b) (where b in B) partition A.
2. The cosets of H are the preimages of the map G → G/H.
   If you don't know 2. you should tattoo it on your forehead

[rotund aurora]

bro was a group theorist but ended up in CECOT ☠️

[coral spindle]

I think face tats should be more popular amongst mathematicians

[mighty kiln]

Does this count https://youtube.com/shorts/hd2J1f3YeQ4?feature=shared

[coral spindle]

No this is just cringe

[mighty kiln]

Bro has a webpage about his tattoos

[glad osprey]

This assumes you have already proved that G/H is a partition of G, right? Since it doesn't use the group axioms for H

[south patrol]

All you need is that G/H exists (which requires the group ofc)

But from then on this is just soome statements about equivalence classes partioning the set

[glad osprey]

Yeah, just seems a bit roundabout to think about preimages of the map G -> G/H, when G/H is itself a partition of G

[coral spindle]

Well that's because one should really think abstractly via the first isomorphism theorem.

What's really happening is that you think of G/H — at least in the case where we have a normal subgroup H — as being some other abstract group A and then looking at preimages of the map f : G → A

Then this is really just saying that quotienting by H is the same as quotienting by the equivalence relation induced by f

In some way the pedagogy is all wrong here. The point first and foremost is that we're taking classes under a nice equivalence relation, it just happens to be that the classes are cosets.

[glad osprey]

Hmm, it's tricky to understand how you would think about G/H without talking about cosets, but is the gist that you start by looking at some map f : G -> G/H, without assigning any meaning to G/H? Then you assert that f is a surjective homomorphism and ker(f) = H, and from there you can conclude that G/H is (isomorphic to) a partition of G by cosets?

[hidden wind]

hot

[tribal moss]

As often as not, we don't start with a normal subgroup and take its quotient. Instead we start with having a surjective homomorphism G -> K where K is some other group that already exists. Then we let H be the kernel of the homomorphism (which we know will be a normal subgroup) and then we know K is isomorphic to G/H. We can then use our theorems about quotients to learn about K.
So what we Boytjie's property 2 actually tells is is that the preimages of any group homomorphism are either empty or cosets of its kernel.

[tribal moss]

Can we not? And also I think I'm just gonna delete that image.

[hidden wind]

[glad osprey]

I see, that makes sense 👍 this way of thinking is still somewhat unfamiliar to me, since most introductory textbooks are very focused on concrete sets. I have started reading Aluffi tho, which I think will help

[limpid ferry]

G = Z_12, H = <4>. Is there a group that has the same Cayley table as the quotient group G/H?

[coral spindle]

This is a funny thing to say

Yes, because G/H is a group

What you mean to say is probably: what is the group G/H isomorphic to amongst a list of common names for groups

To see what it is: note that the image of a cyclic group is cyclic, and compute the order of G/H.

[limpid ferry]

Yes I can see this, only if the mapping is an isomorphism. I think I am not understanding what that means by two groups with the same group tables, is it equivalent to two groups that are the same?

[coral spindle]

Me neither. Post the question.

[limpid ferry]

Here it is. The last question

[fading field]

,rotateccw

[fading field]

by "same group table" they just mean "find a group that is more familiar to you that's isomorphic to this group"

there are only two groups of order 4 up to isomorphism, so just see which one works!

[coral spindle]

n.b. they say familiar group

[coral spindle]

Which means what I was saying here makes sense.

[vapid vale]

WHOA

[round hull]

reading my past work and embarrassingly can't understand what i wrote

i'm supposed to show that a group of order 40 contains a normal p-sylow subgroup and then i've written

If the 5-Sylow group P is not normal, then there are 6 conjugates of P.
how does that follow

[rocky cloak]

The sylow theorems tell you (among other things), that the number of p-sylow subgroups is 1 modulo p, and divides the order of the group. And the sylow subgroup is normal iff it is unique.

[round hull]

oh lmao right

i see

[rocky cloak]

Now 6 is 1 modulo 5, but it doesn't divide 40

So can't be 6 either

[round hull]

and 11 is too large

[rocky cloak]

Indeed

[long geyser]

is this guy just k[r]?

[rocky cloak]

k[r] usually means the subring generated by k and r yes

[long geyser]

well

I mean

polynomials with coefficients in k and "indeterminate" r

[rocky cloak]

The ring will look like polynomials evaluated at r yes

[long geyser]

right

and this guy is k(r)

it must include all polynomials in r (subring generated by k and r)

must include their inverses

so rational functions

so every division ring including k and r must include k(r), but k(r) is a division ring itself

[coral spindle]

No, not all rational functions. For example Q(i) does not consist of all rational functions evaluated at i, because for example 1/(1+x^2) cannot be evaluated at i.

You need the localisation. In the case of Q(i), for example, you need to consider Q[x]_(x^2+1).

[sly rain]

When we want to show Hom_R-mod(R,M) /cong M as R-Modules where R is a Ring and M an R-Module.
Do we assume R commutative? Because idk much module stuff and the (standard (?)) action that makes Hom(R,M) into an R-mod needs commutativity i think

[long geyser]

ah, hmm

[long geyser]

so like, I'd say all f(r)/g(r) such that g(r) =\= 0? does that not work?

[coral spindle]

This is the localisation

[long geyser]

oh, I don't know the term

[rocky cloak]

No, you don't need R to be commutative.

The action is defined by
(rf)(x) = f(xr)

[coral spindle]

D:

Please accept my apologies for the wrong information...

[sly rain]

No worries lol

[rocky cloak]

This works because R is a bimodule over itself

[coral spindle]

Oh of course

Right right the sidedness swaps. Nice.

[shrewd sandal]

I'm looking for an example where I have a group G, a subgroup H, and a normal subgroup N of H such that there doesn't exist a normal subgroup K of G such that N=KnH.
Any help would be appreciated, thanks!

I feel like there should be an easy example but I'm terrible at finding counterexamples in groups

[coral spindle]

... G such that K=NnH.
But N <= H so K = N. So you're looking for a normal subgroup N of H such that N is not a normal subgroup of G.

Are you aware of the normal subgroups of the symmetric group S_n for n >= 5?

This is one way to find an example

[dull ginkgo]

Is there other normal subgroups of S_n, n >= 5?

[coral spindle]

...uh

That may be the point

[dull ginkgo]

Oh fair

I was brain melted for a second

[coral spindle]

This is squeral's exercise tho

[dull ginkgo]

You made me gaslight myself and my group knowledge

;-;

[shrewd sandal]

It's not really an exercise I'm trying to prove something in universal algebra and I wanna make sure I'm not wasting my time lol

[shrewd sandal]

Sorry! I meant N=KnH

[coral spindle]

Have you seen the wreath product before?

[shrewd sandal]

I can't say I have

[coral spindle]

Well actually now that I think of it, I can just do the simplest example

[shrewd sandal]

Ah so there is an example?

[coral spindle]

Yes.

[shrewd sandal]

Well shit that sucks

[coral spindle]

Consider the dihedral group of order 8, Dih(8) = <r, s | r^4 = s^2 = 1, srs = r^-1>

Now let me see

Right

s and rsr^-1 are both involutions but are these the ones I want...

Right I think this works, I'm terrible at Coxeter stuff but here you go.

So there is a subgroup H of Dih(8) given by H = <s, r^2sr^-2> isomorphic to C_2 x C_2. (I think I corrected an error here)

Now then N = <s> is normal in H

But if we have some normal subgroup K of Dih(8) containing N, then rsr^-1 is also in K, so srsr^-1 = r^-2 is in K, so H is in K.

So such a thing is impossible.

In my head I'm just describing G = C_2 wr S_2, the subgroup H = C_2 x C_2, and the normal subgroup N = {1} x C_2.

@shrewd sandal That's a summary.

[shrewd sandal]

Thank you so much!

[coral spindle]

Have you encountered semidirect products before? They are a good way to construct a large number of groups. The above was an example.

[long geyser]

I'm trying to classify all the left ideals in this ring
am I right in observing that any left ideal is of the form (n) in the (1,1) spot, Q in the (2,1) spot and either 0 or Q in the (2,2) spot?

except the 0 ideal

[shrewd sandal]

I have but a while ago

[rocky cloak]

That's correct, except there's also the one that's 0 in (1,1) and (2,1), then Q in (2,2)

[long geyser]

oh, that's right

[rocky cloak]

Actually there are a few more missing

Namely those generated by
[0, 0; x, y]

[long geyser]

hmm, I see

because the (2,2) entry in the left matrix hits both

[rocky cloak]

Yeah

[long geyser]

now trying to think of a right ideal that is not finitely generated... would [0, 0; dyadics, 0] be finitely generated? that kind of stuff is the only thing that comes to mind

there's no way it can

and I believe it is a right ideal too... so I hope it works

[rocky cloak]

It works, but I guess [0, 0; Q, 0] would have been simpler

[long geyser]

ah lol, didn't even think of that

just assumed it would be finitely generated... now I realize it is not due to the same reason

thanks for all the help

[lethal cipher]

Okay, I'm struggling a little bit. I need to show that $I=(X_1-a_1,\dots X_n-a_n)\subseteq k[X_1,\dots, X_n]$ is a maximal ideal.
I'm struggling to figure out what I can do for this one.

[cloud walrus]

dackid

[cobalt heath]

Where did you get stuck?

[lethal cipher]

Oh, I should mention, k is a commutative ring with a unit, and a_1,...,a_n are in k

[lethal cipher]

Well, I'm doing Willian Fullton's book on Algebraic Curves. So my attempt may not be made for this channel, but I'll at least tell you what I tried

My idea was to take an ideal that is properly contains I, we'll call it I'.
However, $I\subset I' \Rightarrow V(I)\supset V(I')$ (V(I) is the set of all points in $k$ that make all polynomias in $I$ to be 0).
$V(I)$ is just the singleton $(a_1,\dots, a_n)$, so that must mean $V(I')$ would have to be the empty set. And so if we look at the ideal of this set, we have $I(V(I'))=k[X_1,\dots, X_n]$. Which...would be great if we knew $I'$ was an ideal of a set in $A^n(k)$, but that's not guaranteed.

[cloud walrus]

dackid

[lethal cipher]

So I don't think this idea works.

As for doing it directly in k[x_1,...,x_n], I'm not entirely sure I have an idea of where to start for that.

I know it would be sufficient to show 1 is in I'

[cobalt heath]

I think you might be overcomplicating things

[lethal cipher]

I would agree

[cobalt heath]

I'd like to think of quotienting by maximal ideal, which would give a field.

[lethal cipher]

Are you implying that if the quotient group is a field, then it's a maximal ideal?

[cobalt heath]

Yep

Like in this case, you cannot have bigger proper ideal

[lethal cipher]

I dunno if it changes anything, but I forgot k is a field.
Though, when dealing with poly's, I dunno if that's that significant of a difference

Okay, hmm. So in order for it to be a field, we need the quotient group to have mult. Inverses

[cobalt heath]

Why wouldn't it change anything though?

Can you compute the quotient, like, what the elements would look like?

[lethal cipher]

Yea, so $$k[X_1,\dots, X_n]/I=G+\sum_{i=1}^n (X_i-a_i)F_i,$$
where$G$ and each $F_i$ are in $k[X_1,\dots, X_n]$.

[cloud walrus]

dackid

[lethal cipher]

That sum on the right is what the elements in the ideal look like.

[cobalt heath]

Let's consider simpler cases.

[cobalt heath]

What happens for n = 1?

[lethal cipher]

Okay sure, then the sum is just (X-a_1)F for some F in k[X]

[cobalt heath]

Hmm wait

[cobalt heath]

Wdym?

Why F_i?

[lethal cipher]

Yea, that should just be I. My mistake

G+I

[slim kayak]

You know some characterizations of being a maximal ideal?

[lethal cipher]

Just the base definition

[cobalt heath]

Welp I do not want to spoil..

[slim kayak]

I mean, its pretty essential (comm) alg knowledge IMO

[cobalt heath]

But, can you iterate all elements of k[x]/(x) @lethal cipher

[slim kayak]

Okay here is a fact, maybe it rings a bell, an ideal I is prime iff R/I is an integral domain

A similar fact holds for maximal ideals, try to figure out what it can be

[lethal cipher]

That's the one absta wants me to get to know (show R/I is a field)

[slim kayak]

In that case, it might even be easier to do the general proof

[lethal cipher]

Okay, If I have done these before, it was a fair bit ago. Imma see if I can prove these first. I'll be back in a bit

[slim kayak]

good thing to remember is that x is either in I, or x+I contains 1 when maximal

[lethal cipher]

Oh? I didn't know that.

[slim kayak]

well

it is maximal after all, and adding (x) to any ideal still gets u an ideal

[lethal cipher]

Ah, I see, yeah.

[cobalt heath]

Huh. For a ring Z and I = 3Z, does 2 + 3Z contain 1?

[lethal cipher]

I need a brain restart. I'll come back to this once I'm ready (and once I try those two problems).
Thank you for the help so far

[slim kayak]

Rumor has it that 2 isnt an ideal

I meant the principal ideal of x ofc

[cobalt heath]

Ah, do you mean (x) + I?

[slim kayak]

yup

[south patrol]

Take the quotient

to show it is $k$ note you equivalently want to create a surjective (well, that is automatic) map of $k$-algebras $k[x_1,\dots,x_n] \to k$ with kernel $(x_1-a_1,\dots,x_{n}-a_n)$.

[cloud walrus]

PD potato

[lethal cipher]

Okay, so this one's actually easy to prove. As for the maximal ideal one, is the goal just to show 1+I is in R/I?

[south patrol]

and you don't have much choice

[south patrol]

Well it always is

[lethal cipher]

Yea, I said the wrong thing. We want a field, so I need to show every non-zero element in R/I has an inverse

[south patrol]

Sure

Though these things are really fundamental to commutative algebra so I'd make sure you get used to them

[lethal cipher]

From what I understand, that's what I am aiming for

[south patrol]

and the correspondence theorem

(which implies this)

[cobalt heath]

Oh no, I tried hard to not spit this fact

[south patrol]

Oh lol well it is just repackaging what it means to be iso to k but yeah

oopies

[lethal cipher]

Don't worry. That's the last part of the problem anyways

[cobalt heath]

Just saying, your approach can be different

[lethal cipher]

Okay anyways. I'm currently just trying to show that if I is a maximal ideal, then R/I is a field.
I'm having a hard time seeing how they connect here

The prime ideal one was basically free tbh

This one feels less so

Just to be clear...1+I should be the mult. Identity of R/I right?

[barren sierra]

Yes

[lethal cipher]

Okay, I'm a bit stuck. So if we let a+I in R/I, then the goal is the show there is a b+I s.t. (a+I)(b+I)=1+I.

[barren sierra]

yea

well, specifically nonzero a + I

[lethal cipher]

I guess this is also equivalent to saying (ab-1)+I=I

[barren sierra]

What does it mean for a + I to be nonzero in R/I? What is the relation between a and I?

[lethal cipher]

True, that's important

[lethal cipher]

It can't be in I itself

[barren sierra]

yea

by assumption I is maximal

so now you have an element a not in I, I is maximal

go from there

[lethal cipher]

Maximal just means there's no proper ideal containing I.

[barren sierra]

ya

[lethal cipher]

So a is not in an ideal (that is not R itself)

[barren sierra]

sure, but we can do better / be more specific

construct an ideal in terms of a and I that strictly contains I

there are some operations we can do to combine two ideals to form a new ideal, I'm sort of assuming you know these operations

[lethal cipher]

Well, (a) is an ideal

[barren sierra]

true. But can we say that (a) strictly contains I?

[lethal cipher]

Correct me if I'm wrong, but is (a) the sum of the powers of a? (I remember the group generators, not the ring one that well).

[barren sierra]

For a finite set ${a_1, \ldots, a_n}$, the finitely generated ideal is $(a_1, \ldots, a_n) = {r_1 a_1 + \cdots + r_n a_n \ |\ a_i \in R }$

[cloud walrus]

Spamakin🎷

[barren sierra]

so for a single element a, (a) is just elements of the form ra

easy to check that this is an ideal

[lethal cipher]

Oh! So it'd just be ra. That makes way more sense

Okay, so you said we wanted to include I, so why not (a)+I?

[barren sierra]

you know the sum of two ideals is an ideal right? So yea that's the right ideal to consider

[lethal cipher]

Yea, this makes perfect sense

Okay, so this has to be R since it is an ideal that is strictly larger than I itself

[barren sierra]

yup

[barren sierra]

if you haven't proven that the sum of two ideals is an ideal, and that the intersection of two ideals is an ideal, do that after this. It's not hard and comes up a ton, very useful fact

[lethal cipher]

The sum one I can see automatically. That really isn't that hard

[barren sierra]

do it later, lets not get side tracked from this R/I field iff I maximal proof

[lethal cipher]

Intersection makes sense too.

[barren sierra]

ok so we have that (a) + I = R

[lethal cipher]

It just comes from the fact we have distribution on our side

[barren sierra]

and remember our goal is to find an element b such that in R/I, (a + I)(b + I) = 1 + I

[lethal cipher]

Yep, and since (a)+I=R, there is an r s.t. ar+I=1

[barren sierra]

you mean ar + i = 1 for some r in R, i in I?

[lethal cipher]

Yes, true true.

Which means... (ar-1)=-i, so ar-1 is in I

Which is what we want

[barren sierra]

yea, ar - 1 in I means that ar = 1 R/I

ok so we've shown that if I is maximal R/I is a field

do the other direction now

[lethal cipher]

Cool!

[cobalt heath]

Ah sorry, were you just beginning commutative algebra

[lethal cipher]

No, I just need to freshen up on my ring knowledge.

[cobalt heath]

Oh. Then sorry for assuming again..

[lethal cipher]

Hmm, okay. Suppose R/I is a field, then for a in I, there is a c in I s.t. (ac-1)+I=I, so ac-1 is in I

So far just treading backwards at the moment

[barren sierra]

you want to show that I is maximal

so let I be an ideal and consider an ideal J strictly containing I

[lethal cipher]

Ah, okay.

[barren sierra]

hint: if you can show that J has a unit, then you are done (why?)

[lethal cipher]

Because any idea that contains a unit has to be the ring itself

[barren sierra]

yea

[barren sierra]

treading backwards is fine but you have to keep in mind what the goal is

[barren sierra]

hence your treading backwards should have started with this

[lethal cipher]

True, true.

[barren sierra]

because in fact c only exists if a + I is nonzero in R/I