groups-rings-fields

[tribal moss]

I.e. for each corner there's a unique rotation or reflection that takes that corner to each of the other corners.

[coral spindle]

Assuming that we have some kind of 'general position' constraint — which is to say that all the side lengths are different so there aren't any strange rotations that occur — this should just be the n-fold direct product of the Klein 4-group, aka C_2 x C_2.

[tribal moss]

Except, hmm, the one symmetry that interchanges each corner with the directly opposite is not really a "reflection", so it's special from a geometric point of view.

[coral spindle]

Ah, or not quite

[tribal moss]

So (classifying symmetries of a general rectangular box), there's a) the identity, b) three rotations by 180° around a perpendicular line through the center of a pair of faces, c) three reflections in a mirror plane parallel to a pair of faces, d) the single symmetry that inverts all three coordinates.

[primal tusk]

that last symmetry is weird. What makes something a symmetry even?

[tribal moss]

In this discussion we're implicitly defining "symmetry" as "a distance-preserving transformation of space that takes the shape in question to itself".

There are more abstract concepts of symmetry that generalizes this one. At the most abstract extreme we can simply choose some set and some group action on it, and declare that the action of each group element is to be called a "symmetry" for the purpose of whatever we're doing.

[primal tusk]

ohh okay that makes a lot of sense actually thank you

[late snow]

How exactly would I go about doing this problem? It feels weird to me because each homomorphism is a group, which means this would be the number of non empty sub-groups

[void cosmos]

think about generators

[coral spindle]

Hint: any homomorphism out of Z_10 (or in fact Z_n in general) is determined by where it sends 1. Think about which ones are surjective from then on. You may want to think about something relating to modular arithmetic.

[south patrol]

"Each homomorphism is a group" wdym?

A homomorphism is a map

[late snow]

thanks for the hints, will get back to this.

[karmic moat]

Hom has a group structure right

just curious

im pretty sure the answer is yes

[coral spindle]

It does not, typically.

[late snow]

yeah homomorphisms are group structures

[coral spindle]

No.

[karmic moat]

aww man

[late snow]

wait does that mean something else than just "a group"?

[coral spindle]

In the case of Abelian groups (and modules!) Hom(A,B) does have the structure of an Abelian group (or module) but not for general groups.

Statsbro I have no idea what you're referring to.

Typically the set of homomorphisms between two groups does not have a natural group structure.

There is a slim chance that you mean, in the style of universal algebra, that the graph of a group homomorphism is a group. This is true, but I doubt this is what you mean.

[random sail]

is this for rings or groups

[karmic moat]

true. i was thinking because it has a module structure it has a group structure

but i forgot that it's specifically abelian groups

tbh i forgot groups arent always commutative lol

im too used to working over rings and modules

[coral spindle]

I'm used to working over rings and modules too. Group rings, and reps of groups

[random sail]

what is a group ring

[coral spindle]

A ring derived from a group, whose modules are exactly representations.

Google it for more details on the construction.

[south patrol]

I like how the group ring is usually thought of as an algebra

Slightly silly nomenclature to me

[rocky cloak]

I usually use 'group ring' to refer to ZG and 'group algebra' to refer to kG.

[coral spindle]

Tbh it is quite useful to think of kG as an algebra specifically. Without that we wouldn't get a lot of results relating representations over non-splitting fields to Galois cohomology.

(I could've mentioned Schur indices, but I didn't. Are you proud of me?)

[late snow]

This is making some serious sense. Considering f: Z_10 -> Z_10, f(1) = 1, f(2) = f(1 + 1) = 1 + , etc. Seems like this means that there would be 10 homomorphisms. In terms of the isomorphisms, I think I need to find all the generators of Z10?

[coral spindle]

Yup, you're almost there.

[late snow]

Excellent. I was slightly unsure about whether f(0) counts as an isomorphism. I think the trick for generators of Z_n is to just take all the relatively prime numbers, so that would be 1, 3, 7, 9?

[coral spindle]

I was slightly unsure about whether f(0) counts as an isomorphism.
f(0) is an element of Z_10, not even a function. What do you mean when you say it may or may not ‘count’ as an isomorphism.

Apart from that, yes those are the numbers coprime to 10 in Z_10.

[dire igloo]

I am trying to find total number of group homomorphism between s_3 and z/6z .
Since s_3 is generated by transposition and 3 cycle so order of image transposition must divide 2 and order of image of 3 cycle must divide 3 so there 3 such elements in z/6z so according me there are 4 homomorphism I know it's wrong I can't find my own mistakes can anyone point it out thanks in advance

[dire igloo]

<@&286206848099549185>

[barren sierra]

How did you get 3 such elements => 4 homomorphisms @dire igloo

[dire igloo]

0,2,3

[barren sierra]

What does that even mean

Anyways if you came up with 4 homomorphisms can you write them down?

And check that they actually are homomorphisms

[dire igloo]

Order of these 3 elements divide divide 2 and 3

[barren sierra]

Ok sure, the order of 2 is 3 and the order of 3 is 2

But you haven't described to me any morphisms from S3 to Z/6Z

Anyways here's my hint

[dire igloo]

f(x)=0

[barren sierra]

Kernel is a normal subgroup right?

[dire igloo]

Yep

[barren sierra]

What are the possibilities of kernels of a morphism from S3 to Z / 6Z

Investigate each of these

[dire igloo]

{e} , A_3 , S_3

[barren sierra]

Your mistake here is that while you're right that the image of the transposition must be an element of order 2 or order 0 and the image of the 3 cycle must be an element of order 3 or order 0, you haven't actually checked anything.

[barren sierra]

Cool, so use Isomorphism Theorems and some casework for each case

2 cases should be easy, 1 case isn't as easy but not too bad

[dire igloo]

Oh okay

[dire igloo]

Okay thanks !

[barren sierra]

You've given an upper bound, there are at most 4 homomorphisms

That's what your previous idea showed

[dire igloo]

Yeah

[icy bear]

I made a cute little proof for that
a is the smallest positive integer
z is some positive integer
z - ka = a for some k, since a is the smallest integer, if for all k z - ka was either negative or positive different than a, a wouldn't be the smallest, since the smallest positive in this case would be different and a is the smallest step you can take
so z = (k+1)a, a multiple of a

I mean, it's a more intuitive proof

[verbal tulip]

Quick question, why is the commutator subgroup a subgroup when we know that the product of commutators is not necessarily a commutator?

[south patrol]

The commutator subgroup is defined as the subgroup generated by the commutators

for that reason!

[late snow]

Isn't this sort of obviously true, given if they were both zero then all non-zero numbers would be factors of both - making the gcd != 1

[verbal tulip]

I'm not sure I'm getting it so let me put some definitions. The commutator subgroup of some group G is defined as the set {a b a^-1 b^-1} where a and b are members of G. Since it is a subgroup, then for every a, b, c and d of G the element a b a^-1 b^-1 c d c^-1 d^-1 must belong to the commutator subgroup. In other words, [a,b][c,d]=[e,f] for some e and f. But we know that the product of commutators is not always a commutator. What am I missing?

[barren sierra]

It's not that set

It's the group generated by that set

[verbal tulip]

I will think about it again. Thank you.

[rocky cloak]

I guess it becomes obvious once you show that
f(a) = 0 implies (x-a) divides f

[teal vessel]

this is quite the challenge tbh. Why are the questions with the shortest texts always the hardest?

[barren sierra]

Wait that's not too bad

Nvm

Has to be the same operation

[teal vessel]

yeah, my first thought was something like the reciprocals of all primes greater than 2 or something, but that's a multiplicative group.

the operation on Q is assumed to be addition, btw

I've previously proven that any finitely generated subgroup of Q is necessarily cyclic, so I need an infinite generating set.

maybe the set of all fractions with even numerators?

yeah, wait, that would work

because adding and multiplying even numbers preserves parity.

so there's no way to get the number 1/odd

no, wait

[delicate orchid]

the main point is that subgroup you just mentioned cannot be cyclic, I don't know what this odd business is

[teal vessel]

I have to show it's a proper subgroup

[delicate orchid]

I mean that's obvious

[teal vessel]

even numerators isn't sufficient, because reduction is a thing

[delicate orchid]

I presumed you meant denominators

say it's generated by 1/2^n, then there's no way for you to get 1/2^{n+1} so it cannot be generated by 1/2^n, repeat for all n

[teal vessel]

even denominators would work, though, yeah.

also all fractions of the form 2m/(2n+1)

it would be impossible to reduce that to every fraction

[tribal moss]

But 1/3 is also 2/6, for example, so you still get all of Q that way.

How about odd denominators instead?

Or power-of-two denominators rather than merely even?

[rocky cloak]

I think any condition such as 'these numbers don't divide the denominator' should work.

[teal vessel]

hence even numbers over odd numbers, the the product of odd numbers is still odd, while a mixed product is even, and the sum of even numbers is still even, therefore it prevents reduction to odd numbered numerators, thus excluding 1 (and all odd integers, technically)

[rocky cloak]

Or I guess the numbers in question has to be prime powers.

So odd = not divisible by 2, power of 2 = not divisible by odd prime and so on

[late snow]

Honestly, I'm having a lot of trouble with why we're using "x-a" with these. I don't 'get' why that works

[rocky cloak]

Because evaluating x-a at a equals 0. Or what are you asking about 'works'?

[late snow]

So if I have any a in F, that is specifically a constant? So x would also be a constant, not an f(a) like x^2?

sorry for the stupid question im really tired

[rocky cloak]

I'm not sure what you're asking, but a is a constant yes. And x is the variable

Also not sure what you mean by comparing f(a) and x^2.

f(a) is a number (it's the value of f at a) while x^2 is a polynomial

[teal vessel]

according to common convention: letters like "a,b,c" typically denote indeterminate/arbitrary constants meaning specifically some member of whatever structure you're working with (in this case a field) while "w,x,y,z" typically denote variables which are used to describe the structure of functions.

[south patrol]

though here we know what x is and a is as elements of F[x] and F

thinking about constants/variables ig isn't really needned

[prime sundial]

what is this a^+ notation? A is a commutative algebra, P and Q are A-modules

[rocky cloak]

Seems to be just the definition.

So Hom(P, Q) can get a structure of an A-module in two different ways, and for one of them they denote the action by a^+.

More generally if A isn't commutative this actually makes Hom(P, Q) into a bimodule, where the a^+ thing is the right action.

[untold basalt]

yeah that's what I meant

[prime sundial]

ah i see

thank you

[alpine island]

man my homework literally has a question just saying "classify all simple groups of order <100"

This is doable but I feel like it will take a long time

[rustic crown]

not really, play with various prime factorizations that will reduce it to a very tiny set of orders that you need to look separately

[alpine island]

yeah prime powers are easy

and abelian is trivial from factorization

so I just need to care about non-abelian non-prime powers

[rustic crown]

sylow theorems give you existence of a normal sylow subgroup in certain conditions

you can also use counting methods to show that number of conjugates can't be too large, e.g. show that groups of order pqr (p < q < r) can't be simple.

there are also some fun actions you can consider with non-trivial kernel, thereby giving a normal subgroup

[elder wave]

here's a fun hint for this: G isn't simple if there's a subgroup H of index k such that |G| does not divide k!

[rustic crown]

(me was gonna type that next :p)

[elder wave]

yeah i figured after the message involving actions :D

[alpine island]

Interesting

[rustic crown]

(interesting that we both described this thing as "fun" )

[barren sierra]

Fun = problem I've previously solved

[elder wave]

The problem definitely wasn’t fun

Ended up writing down a whole table of prime decomps to see what orders can’t be simple and whatnot

[languid trellis]

Exercise: Does every group whose order is a power of a prime p contain an element of order p?

My attempt so far: Consider G whose order is p^m for some natural m. My main idea is to essentially try create a chain of proper subgroups. By lagrange's theorem and the fact that these are proper inclusions, we will eventually arrive at a group of order p. By the fact that each group of order a prime p is cyclic, it follows that there's an element of order p. So, I considered looking at cyclic subgroups generated by elements of G. If there is no proper subgroup then each of these cyclic subgroups must generate G and I feel like there's a contradiction coming but I can't quite finish this argument off. Can someone give a hint or affirm/deny if i'm even going in the right direction? thanks 🙂

[rocky cloak]

So a hint could be if x is an element in G, what's the order of x^p?

[languid trellis]

I'm sorry for whatever reason I can't figure this out

Certainly <x^p> \subset <x> so the order of x^p divides the order of x

but that doesn't say much in itself

If x^q = 1 (order of <x> is q) Then the order of x^p would be the smallest natural a such that ap = some multiple of q

[rocky cloak]

So maybe first think about what q could possibly be

What are the possible orders of an element in G?

[languid trellis]

p^1 , . . ., p^{m-1}

[rocky cloak]

Indeed, so say x has order p^k

What is then the order of x^p

[languid trellis]

k

[rocky cloak]

Not quite

So (x^p)^k equals x^pk

Which is not x^(p^k)

[rocky cloak]

So it's just like you said here you want to find an a such that
ap = p^k

[languid trellis]

oh right, then a = p^{k-1}

[tardy hedge]

Bruh im randomly hitting a wall while trying to read through ideas and factor rings section

Its like so many abstract concepts being piled on to each other i need to take a step back

[rocky cloak]

Indeed, so do you see how you could do something similar to get an element of order p?

[tardy hedge]

I feel myself losing a bit of a foundation

[languid trellis]

consider the order of x^(p^{k-1}). Then we're looking for an a such that a(p^{k-1}) = p^k Then a = p

[languid trellis]

I gotta go to bed now bit I appreciate your help jagr 🙂

[tardy hedge]

Of course, 0 is always in the ideal right

Its an additive subgroup so i guess obviously it needs the 0 from the ring

Then i suppose if the ideal has 1 or not is an important fact

Cuz if it does then its just all of the ring

[white oxide]

indeed

and this leads to the fact that a field has no proper nontrivial ideals

so you can make a whole lot of arguments with that fact

[tardy hedge]

Yeah

[white oxide]

i'm blanking right now

what's an example of an epimorphism in Grp which fails to have a right inverse

[next obsidian]

There are none

Or do u mean a right inverse in Grp?

[white oxide]

yeah

[next obsidian]

Errr

Oh

Take Z -> Z/2Z

If it had a right inverse it would give an embedding Z/2Z -> Z giving a torsion element in Z

[white oxide]

oh nice

yeah makes sense

ty

[south patrol]

Remember as well that uh at least in abelian groups if A -> B is epic and admits a right inverse (section) then B is a direct summand of A

[delicate orchid]

You’re epic

[south patrol]

So you can easily construct examples like this where B is not a direct summand of A

Thank you, no u

[wraith oak]

is "epic" an actual term used in algebra

[coral spindle]

Yes

Though I do see people more often call them epimorphisms rather than say a morphism is epic.

[teal vessel]

can't I just say that because the order of the rotation subgroup is n/2 and D_n is finite, the rotation subgroup must be maximal? it seems that any maximal subgroup of a finite group with order n would be of order n/p where p is a prime divisor of n (since by lagrange's theorem, any supergroup must have an order that is a multiple of the subgroup's order, and a proper finite supergroup must be a non-unit multiple).

except the edge case where the order of the group itself is prime, which would have no maximal subgroup

[coral spindle]

Yes

[south patrol]

Yeah it is very quick

[south patrol]

Well, the trivial subgroup

I guess

So it still works lol

[teal vessel]

oh, duh, if the order is prime then there is only the trivial subgroup as a proper subgroup, so no edge case at all

[south patrol]

Ye

[teal vessel]

this is what I get for typing while thinking

tangential question: is there a term for the complement of a prime factor? we talk about primes so much, but it's interesting this motivation seemed to employ specifically excluding those primes.

is the specification of the order of G here merely convention? I'm not sure I can think of a group of order less than 1

[coral spindle]

It is unnecessary.

[next obsidian]

Groups have at least one element because they need an identity

I believe the point is just that n is a number so it’s written (in a weird way) to say the group is finite, and to introduce n. The >= 1 part is just superfluous

[teal vessel]

I don't think so, because it doesn't specify a finite group.

[next obsidian]

…

[coral spindle]

for some prime p dividing n

Right. Sure.

[next obsidian]

I mean the result is true if you say any prime divides infinity

[teal vessel]

but I suppose this is technically asking for me to restate the notion I gave earlier so yeh

[next obsidian]

But infinity >= 1 and p | infinity is an… interesting statement

[teal vessel]

the first part of that conjunction seems rather non-contentious

[next obsidian]

When someone says “a group of order n” they almost without a doubt mean that n is finite

[teal vessel]

the second.... depends on the definition of divisibility. If you assume that $p|n \Leftrightarrow \exists q \in \mathbb{Z} (pq=n)$ then there are no divisors of infinity.

[cloud walrus]

GoldenPhoenix

[coral spindle]

Ofc it depends on the definition, what else would it depend on??

[teal vessel]

almost every other example in the textbook has specified n<infty when specifying a finite group, so the break in style threw me for a loop

[coral spindle]

Extending the definition here unfortunately just isn't feasible if you'd like it to work in all circumstances. I will underline that you should simply move on, with the understanding that the group is finite.

[next obsidian]

I mean

In this case it’s actually true if you do let n be infinity and say p | infinity always

[coral spindle]

In this particularly nice case, yes

[teal vessel]

anywho, this particular question is a direct corollary to the statement I made about maximals being of order n/p where is a prime divisor of n

[bitter olive]

Definition 1.15. Let $R \subset S$ be an inclusion of rings, and let T be a subset of S.
Then we define the subring R[T ] of S generated by T to be the intersection
of all subrings of S that contain both R and T .
Conversely, if T $\subset$ S is a subset with S = R[T ], then we say that T generates
S over R, or that T is a generating set of S over R. If the ring S admits a
finite generating set T over R, then it is called finitely generated over R.

How can this be written in more symbolic notation like

$R[T] = \bigcap_{ \begin{matrix} S \ subring \ of \ ? \ R \subseteq S \ T \subseteq S \end{matrix}} S$

[cloud walrus]

Tobi

[delicate orchid]

Strange thing is that it actually does still work for the infinite cyclic group

If you are ok with “every prime dividing infinity” then the maximal additive subgroups of Z are pZ for primes p

[delicate orchid]

$R[T] = \bigcap_{X \text{ subring of } S, R \cup T \subseteq X} X$

[cloud walrus]

Wew The Lads Tbh

[delicate orchid]

Alternatively you can just think of it as polynomials with variables in T and coefficients in R

[bitter olive]

So our highest structure is S and S is a ring, right?

Yeah it is xd. Thx I think I got it. R for Ring, S for Subring but here it is actually in opposite way. +1 on confusion

[rocky cloak]

It's more like R for ring, S for next letter in the alphabet

[civic ferry]

does anyone have a hint on proving that all fields either have characteristic 0 or prime?

[rocky cloak]

A hint could be that fields don't have zero-divisors

[civic ferry]

so uh
i could distribute (1+1+... a times)(1+1+... b times)=0 to get (1+1+... ab times)=0
so one of the factors has to be zero?

that sounds right afaict

[rocky cloak]

That works

[civic ferry]

is there a neater way to do it? the ... is annoying me

[teal vessel]

sum from 1 to a of 1

[coral spindle]

Or just

a.1

[teal vessel]

yeah

if you wanna be boring about it

[civic ferry]

but a is an integer and 1 is an element of the field

[coral spindle]

Yes

So you define a new product

This is totally standard.

[civic ferry]

oh huh

[topaz solar]

“Yeah Z/nZ is a subring as <1> for n = char F”

[coral spindle]

||every ring is a Z-algebra uwu owo||

[teal vessel]

"let A times n denote the repeated application of field addition on the object n A times"

[topaz solar]

You know you can do f: Z -> R with f(1) = 1 yes?

[coral spindle]

I have never seen anyone denote it \times, no

[teal vessel]

didn't denote it as \times

just didn't bother writing out \cdot because I'm lazy

[topaz solar]

Why would you not just write

ax

[civic ferry]

no i have not come across that yet

[topaz solar]

[topaz solar]

Well, show that’s a (unique!) ring homomorphism

[teal vessel]

because I arbitrarily chose to write it as I please and didn't feel to put much thought into it.

[coral spindle]

If you're interested, you should see where this leads (as Sharp also suggests). As it turns out this gives you a nice relationship between the kernel of f and the characteristic of the ring. See if you can find out what it is!

[teal vessel]

I'm busy using the axiom of choice to prove that all nontrivial finitely generated groups have maximal subgroups.

[coral spindle]

OK that's nice.

[chilly radish]

If you wanna be truly pedantic, the product is actually defined recursively on N and then extended to integers by multiplying by -1

[civic ferry]

is the kernel defined the same as in linear transformations? how does that work?

[topaz solar]

Kernel is when f(x)=0

[coral spindle]

I'm surprised you haven't seen this! The kernel of any map between groups is what gets sent to the identity. In the case of rings, we look specifically at what gets sent to the 0 in the additive group.

[civic ferry]

im in first year ok😔

[coral spindle]

I'm not dissing you, just surprised

[topaz solar]

So, when would this be 0

[coral spindle]

I suppose you could see characteristic of a ring before kernels

[untold basalt]

Field extensions question

[civic ferry]

would this be when char(R) divides n?

[topaz solar]

Prove that

Show that it can be turned into a unique ring Hom too ofc

Then imagine throwing first isomorphism theorem at it

[civic ferry]

||whats the first isomorphism theorem again||

[topaz solar]

im f = X/ker f

That one

[civic ferry]

is X/ker f like a quotient group?

[next obsidian]

Yes

Unless X is a ring in which case it’s a quotient ring

Etc etc

[coral spindle]

You will likely see soon that you can define quotients of rings in terms of something comparable to a normal subgroup, known as an ideal

And the familiar properties of quotients hold.

[civic ferry]

ok so
i can start by defining the map f
eg say it must satisfy f(1)=1 and f(x+1)=f(x)+1
then it follows that f(0)=0
and also f(a+2)=f(a+1)+1=f(a)+f(2), and inductively f(a+b)=f(a)+f(b) for b>0
setting a=-b it must also be the case that f(-x)=-f(x)
then ig f(2a)=f(a)+f(a)=f(a)(f(1)+f(1))=f(2)f(a) and inductively can generalise to f(ab)=f(a)f(b)

so thats a ring homomorphism
then for uniqueness, suppose f and g are both homomorphisms Z->R
for all a in Z, we have f(a)-f(a)=0=g(a)-f(a)
so g(a)=f(a)

[topaz solar]

The uniqueness part lookin a lil scuffed

But basically just induct again

[civic ferry]

oh i meant suppose f(a)=g(a) as a premise

oh i can see how that doesnt work

oops

oh right so itll be like f(1)=g(1)=1
if f(a)=g(a) then f(a+1)=f(a)+1=g(a)+1
and if they agree on positive integers then they have to agree on the negatives

[topaz solar]

Basically yeah

Induction time

Now, if n = char R obviously f(n)=0

Can you show what you said earlier about divisibility?

Can we have f(n)=0 for n where char R does not divide n?

[civic ferry]

ok and to prove that if char(F)=n, ker f=nZ im thinking another induction?
f(0)=0
also f(n) is zero
and if x>0 and f(x)=0 then f(x+n)=f(x)+f(n)=f(x)+0

[teal vessel]

question about set chains: the definition I have is given as follows:

C is called a chain iff for any X and Y in C, either X⊆Y or Y⊆X
this "in" here seems mildly ambiguous whether X and Y are subsets or elements. Contextually it appears to be subsets, but I just want to confirm I'm reading this right.

[civic ferry]

did not read this oops

[civic ferry]

wait this isnt quite what i was trying to prove hold up

[topaz solar]

That last line you might need to justify

[topaz solar]

No no I think it is

[civic ferry]

but ive shown if n|k then f(k) = 0
not that if f(k) is zero then n|k

[topaz solar]

Ye

Well, if f(k) = 0, what’s k mod n

We can subtract qn for any integer q since that goes to 0 ye?

What’s the definition of char

[civic ferry]

ohh its the smallest such n

[topaz solar]

Yep

[civic ferry]

right that makes sense

[topaz solar]

And we can reduce via this mod operation to 0 <= k <= n

Hmmm

[civic ferry]

time to try this now i suppose?

[teal vessel]

my brain: The union of two finitely generated subgroups is the group generated by the union of their generating sets
me: sounds about right, let me write out a whole proof based on this fact.

my brain: what if ab isn't in group A or group B?

I love psyching myself out, time to go on a tangent.

[topaz solar]

Take free group, union <x1> and <x2> for generators x1, x2

[teal vessel]

dunno what a free group is yet

[topaz solar]

Basically the answer is no

[teal vessel]

I'm fairly confident in that now, yeah, just frustrated

[topaz solar]

Unless you have some degeneracy conditions like A < B

[teal vessel]

there's chains involved that could change that, but the wording is weirding me out

[topaz solar]

Take a, b which are exclusive to each

[teal vessel]

proving part A right now. This wording is throwing me off regarding what objects are where

the beginning on the previous page is just "let G be a finitely generated group"

I know for a fact that the union across S is necessarily a subgroup of G, and i'm fairly sure that any union of parts of S are likewise, but idk how to PROVE it.

[topaz solar]

It’s a chain

So each step satisfies the A < B thing I mentioned

[teal vessel]

yeah. the definition I have of chains is a bit ambiguous regarding subset vs membership so it's been a ride so far

[teal vessel]

.

[topaz solar]

What

[topaz solar]

They’re elements obviously

[teal vessel]

then I'm just stupid, sorry.

[topaz solar]

Why would they be subsets

These are all subsets but ya know

[civic ferry]

hmm ive been thinking about this for a while and im not quite sure what im supposed to be turning into Hom

[topaz solar]

You did, for f

The Z -> R thing

[civic ferry]

huh

[topaz solar]

Subgroups of G but elements of C

[topaz solar]

You showed it can be turned into a ring homomorphism uniquely

[civic ferry]

oh wait was hom short for homomorphism
i thought you wanted me to come up with some kind analogue of the vector space hom(U,V)

[topaz solar]

No

[topaz solar]

Just homomorphism

[chilly radish]

Let let $k$ be a field of characteristic $p$, and let $f(x) = \sum_{i=n}^m a_ix^{p^i}$ be an additive polynomial, where $a_n,a_m\neq 0$. The book claims that this polynomial has $p^{m-n}$ distinct roots in the algebraic closure, but I haven't been able to understand why.

[cloud walrus]

Vile Differential

[next obsidian]

@chilly radish what is an “additive” polynomial

[chilly radish]

So if $n=0$ then this polynomial is separable and this is clear. In the case that $0$ is a repeated root i'm having a harder time seeing why this is true.

[cloud walrus]

Vile Differential

[chilly radish]

this polynomial is additive as a function over k because it's a polynomial in powers of p

that's all

[next obsidian]

Ah okay okay

[chilly radish]

like I tried just factoring out the repeated roots of 0 to end up with a separable polynomial but i'm not getting the correct number

[next obsidian]

The issue with factoring is that you end up with something not additive

So you can’t apply induction, yeah?

@chilly radish I think I have a solution

I haven’t totally fleshed out the last bit, but basically it goes like this

When looking at roots we can look at this defined over any finite extension of k right?

[chilly radish]

sure

[next obsidian]

So let’s say we have n > 0, we can look at k^1/p^n and basically take p^n-th roots and we end up with a polynomial of the same form where n = 0

And basically we transpose between the two using Frobenius

[chilly radish]

I don't understand what you mean by k^1/p^n

[next obsidian]

Adjoin all p^n-th roots

[chilly radish]

oh

[next obsidian]

Actually do we even need that?

I think not

A solution a to p(x)

Gives a solution a^{p^n} to p(x^1/p^n)

That latter expression actually makes sense because we assumed n is minimal in that expression

And a solution a to p(x^1/p^n) gives a solution a^1/p^n to p(x)

It’s okay to take p-th roots because the a live in k-bar

Something like this is the idea I think

[chilly radish]

I think you're right

you can do a substitution of x^1/p^n to reduce to the n=0 case

nice

[next obsidian]

Swag

Always Frobenius

[chilly radish]

that's kind of what I was trying to do but by factoring instead

ye

[next obsidian]

That’s what I tried to do at first

But I realized subtracting is not good

[chilly radish]

which doesn't play well with the powers

exactly

[next obsidian]

Yup

That’s why I thought of Frobenius

[chilly radish]

I'm still wondering why this approach is wrong: Factor out x^(p^n-1), then you get a separable polynomial of degree x^(p^m-p^n+1). Might not be additive anymore, but it doesn't matter, then this has p^m-p^n+1 distinct roots (inc. 0)

like these 2 expressions in general should not be equal, so what's the mistake in my argument

[next obsidian]

I’m not sure the polynomial is separable (so we don’t know about the # of distinct roots)

When you take a derivative on the additive polynomial with n = 0

A bunch of p^k come down and kill stuff so the derivative is just a_n

But when you factor out x^{p^n-1} when you take the derivative a bunch of numbers come down which aren’t divisible by p

So the derivative is an actual polynomial so it has roots which might coincide with the original polynomial’s roots

[chilly radish]

right, this is the problem

when you factor it doesn't kill the higher terms in the derivative

only when you substitute

[next obsidian]

Right, so that’s why your proof falls apart

[chilly radish]

indeed

thanks ch

chm

[next obsidian]

[chilly radish]

[tardy hedge]

In a ring, can multiplication be viewed as repeated addition only if the ring is a principle ideal domain

Tbh im just saying this without really thinking about it

So forgive if its stupid

Well, it surely needs to be an integral domain for multiplication to be viewed as repeated addition

[charred iris]

I wouldn't say multiplication in F[X] is repeated addition (for F a field)

[tardy hedge]

Is there some sort of homomorphism from Z going on here

Yeah true

[tardy hedge]

Can multiplication be viewed as repeated addition only if there is some ring homomorphism out of Z into the ring

Im really just spitballin here

[charred iris]

I have messed up the direction of my logic aka I missed the 'only'

[charred iris]

But I do think this is correct (edit: misread as 'onto the ring')

And I think in order to even say what 'multiplication is repeated addition' is you need that

I'm interepreting it as for each $a,b\in R$ you can write $ab$ as some $a+\dots+a$

[cloud walrus]

Edward II

[charred iris]

And then taking $a=1$ gives you the homomorphism $\bZ \to R$ being surjective

[cloud walrus]

Edward II

[charred iris]

oh I suppose this is when rings have identity

[rocky cloak]

You always have a ring homomorphism from Z, given by mapping n to 1 + 1 + ... + 1 (n times).

Then multiplication by n is repeated addition, but of course every element isn't necessarily an integer. For example multiplication by pi does not look like repeated addition of real numbers.

[chilly radish]

There is always a homomorphism from Z to your ring

It's the one sending n to n*1

Oh jagr already said this

[coral spindle]

If the homomorphism is surjective it works

But this is what jagr said, indirectly

[rustic crown]

multiplication by -2 is adding -2 times

[median rock]

sry to bother, is this proof valid?

[hidden haven]

Does G/H have to be finite?

[median rock]

nope

only that G is not finite

[hidden haven]

Then the first line is wrong

[median rock]

the |G/H|=..

right?

[hidden haven]

Yes

[median rock]

yeah i thought it could be wrong

thanks m8

it G/H was finite then is it right?

[sonic coral]

I’m trying to show that a group of order 288 is not simple. I have shown that the order of the normalizer of the intersection of two Sylow-3 subgroups is 36

i’m not sure how this helps me but I am stuck

[hidden haven]

No, you seem to be assuming that each aH is a group and is finite (when you say that a' ∈ aH implies ord a divides | ord aH |)

[median rock]

ok thanks for the help

[celest furnace]

Is this always true?

Ah i figured out where i was going wrong right as i posted this lol

[sinful holly]

Hmm ... it's a nice question. You can prove it with a bit of representation theory (Burnside's theorem does it), but it's a nice question to do it with the Sylow theorems

[celest furnace]

Can anyone help me understand the line "It follows directly from the lattice iso theorem that..."

Really not seeing it i thought that N_G(H) was the union of all subgroups of G containing H such that H is normal in that subgroup but i also dont think this is true

[sonic coral]

i mean yeah i know burnsides theorem would knock it out since it’s prime factorization only has two distinct primes but i can’t use that

i want to do it using sylow theorems and am stuck

i know that a group of order 36 either has a normal sylow 2 or a normal sylow 3 i guess

the issue is that the sylow 2s can intersect in multiple different ways so i can’t pin down enough about them to know anything useful

[white oxide]

i stg all the things i'm having trouble verifying nowadays come in the form "it is immediate that..." "it is obvious that..." 😭

how exactly do we know that {1, a} spans E? i understand that they're linearly independent

is this a linear algebra theorem that i forgot

[celest furnace]

Well you have a linearly independent set with 2 elements

So it generates a 2 dimensional subspace

[white oxide]

oh right

[celest furnace]

The only one in K is just K

[white oxide]

has to be K

thanks

[white oxide]

also how do we know that there's a relation of the given form

why sully lol

nvm i'm dumb

[swift python]

anyone about?

[white oxide]

if $[F: K] = m$ would the isomorphism $F \simeq K^m$ just be given by $k_1a_1 + k_2a_2 + \dots + k_ma_m \mapsto (k_1, k_2, \dots, k_m)$ where ${a_1, \dots, a_m}$ are a basis for $F$ over $K$

[cloud walrus]

okeyokay

[mighty kiln]

Yes

[white oxide]

what's the purpose of the vector space column? (wikipedia's article on K-algebras)

[karmic moat]

it says it in the first line of the article

"In mathematics, an algebra over a field (often simply called an algebra) is a vector space equipped with a bilinear product."

[long obsidian]

Yeah so it's just highlighting the underlying vector space structures

I could talk about R4 and not think about the quaternions.

[white oxide]

ah in the sense of isomorphism

i guess?

yeah

cool ok thanks

[long obsidian]

Apparently there are no homomorphism from the rationals Q->S3 into the symmetric group on 3 labels.

Is this just because all elements of S3 have finite order but most of Q doesn't. I can't exactly remember why it follows from this if so.

[white oxide]

so is the only difference between a K-algebra A and a ring A that A is considered a vector space over a field K?

what if K isn't a field and is instead a commutative ring, like Z

are they equivalent then

[long obsidian]

I think some people call K algebras modules with a bilinear product. I can't remember. There has to be a name for that

[charred iris]

That's not the reason it doesn't exist, e.g you have homomorphisms from Z to cyclic groups

[long obsidian]

Damn I knew that wasn't the reason

[charred iris]

It doesn't exist because Q is Abelian but S3 is not

[long obsidian]

Oh okay thanks

[long obsidian]

Why couldn't there be a homomorphism into the center of S3?

[charred iris]

What is the center of S3

I assume you mean non trivial homomorphism btw

Because there always is one that sends everything to the identity

[long obsidian]

Oh okay I see. There is no nontrivial center of S3. I guess if there was one you could pick another permutation that moved something that thing in the center moved and those wouldn't commute. Okay thank you

[mighty kiln]

The extra structure in the K-algebra is the ring homomorphism K → Z(A) from K to the center of A

[white oxide]

does anyone know what the notation (a1, a2, ..., an) represents?

[white oxide]

oh interesting

[mighty kiln]

A vector in F^n

[white oxide]

wait rlly?

here's the context

[charred iris]

I have made a huge mistake

S3 does have cyclic subgroups

So in theory you could have maps into those

[mighty kiln]

Ye A^n

Or you can just think of a collection of n elements of A

[charred iris]

I think you can get something from f(1/n) = f(1)^-n for each integer n

(And each homorphism from Q is deternined by f(1))

Oh even better: f(q) = f(6q/6) = f(q/6)^6

(I initially got confused by a recent example I saw where Abelianity was the deciding factor and monkey brain go brrr)

[long obsidian]

Ah I see my issue. I mean there are no group homomorphisms from Q to S3

[topaz solar]

Well, 2 * 1/2 = 1

[charred iris]

You might be right, and this is a sign I should go to bed. However, this bit works regardless

[long obsidian]

The rationals are not generated by their additive identity

[white oxide]

i'm a bit confused, how does R being finite-dimensional over K show that it's surjective? say {a1, .., an} is a basis of R over K. how do we get the element k1a1 + ... + knan using h if we don't necessarily know that a is invertible?

[next obsidian]

Pigeonhole principle

[white oxide]

oh goodness

[next obsidian]

Except you call it rank-nullity

[white oxide]

i don't remember that

but i'll look it up!

oh dim R = ker h + im h

there we go

[topaz solar]

They are not, but if f(1) = x, then 2 f(1/2) = x

And similarly for any other 1/n

Nothing saying f(1/2) needs to be uniquely determined but that’s a pretty strict condition

[long obsidian]

My bad. I was referring to a group homomorphism from Q to S3. I don't think S3 is a ring

[topaz solar]

It isn’t

But 2 f(1/2) still is a thing

f(1/2)^2 rather

[long obsidian]

How is the ring structure relevant to the group structure. I was only concerned about the group homomorphisms from Q->S3. Did I miss something?

[topaz solar]

It’s literally not the ring structure homie

1/2 + 1/2 = 1

f(1/2) * f(1/2) = f(1)

It’s a group homomorphism

[long obsidian]

Oh

[celest furnace]

Every element of Q has infinite order except for 0

Every finitely generated subgroup of Q is cyclic, while S_3 itself is not

Theres many reasons why

[random sail]

1. Y'all should be careful with wording: there is the trivial homomorphism from Q to S3
2. Those statements do not imply the nonexistence of a nontrivial homomorphism. As a counterexample, Z also satisfies the above conditions, but there are nontrivial homomorphisms from Z to S3

[celest furnace]

Ah I read it as S3 to Q

Also what is an example of a nontrivial homomorphism from Z to S3?

Like just send 1 to (123) ah

[halcyon star]

How do I proof $\forall n>2 S_{n}$, all permutations can be decomposed into elements of $S_{2}$?

[cloud walrus]

Living Hydrogen Cyanide

[halcyon star]

Also if $\sigma = \tau_{1} \tau_{2}\dotsi \tau_{r} = \kappa_{1} \dotsi \kappa_{p}$ (i.e. sigma can be decomposed in two different ways, $\tau$ and $\kappa$, then $(-1)^{r} = (-1)^{p}$?

[cloud walrus]

Living Hydrogen Cyanide

[prisma ibex]

yes

[halcyon star]

I have no idea how to proof this

I havent done abs alg in so long

[prisma ibex]

look at the kernel of the morphism sign:S_n->{+1,-1}

[halcyon star]

I believe the proof isn't that hard I just lost the skill to proof it

[prisma ibex]

you can't be both in the kernel and not in the kernel

[halcyon star]

so $\phi : S_{n} \rightarrow {+1,-1}, ker $\phi$ is locked to only have one Im?

[cloud walrus]

Living Hydrogen Cyanide
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[prisma ibex]

yeah

[halcyon star]

Sweet

[prisma ibex]

those in the kernel are those consisting of an even number of transpositions

[halcyon star]

Yeah

Thanks ig

Cleared up my mind a bit

[prisma ibex]

to show that any permutation can be written as a product of transpositions it's enough to show that any cycle can be written as a product of transpositions

since any permutation can be written as a product of disjoint cycles

and then the case of cycles is not so hard, you can try it for some small cycles and figure out what the general pattern is

it's enough to do this for a basic n-cycle (1 2 ... n) since you can just relabel everything

[dull marsh]

"Show that a finite group of even order has an odd number of elements
of order 2"

Any hints on this?

I thought of considering the subgroup generated by elements of order 2, but it doesn't look like the order of that subgroup can be determined

[rocky cloak]

The map g |-> g^-1 is a bijection on G. How many fixed points does it have?

[dull marsh]

As many as the number of elements of order 2

+1

Because of identity

Is there some relevant theorem regarding fixed points? I don't know any

[dull marsh]

Wait

[dull marsh]

The number of fixed points = the number of the non-fixed points, right?

Hmm no this is wrong

[rocky cloak]

So how many elements are not fixed points? Is that an even or odd number?

[dull marsh]

Ah wait I think I see

[dull marsh]

Do we pair every not fixed point with its inverse and get that the number of not fixed points is even?

I realized that the map g |-> g^-1 is some kind of permutation of the elements with order > 2

And realized this must be some product of disjoint transpositions

I.e., even number of elements were permuted

[rocky cloak]

Exactly, so there is an even number of 'not fixed points'. So is there an even or odd number of fixed points?

[dull marsh]

So let's say n is the number of elements in our group (even), k is the number of elements with order 2, then there are k+1 fixed points (the extra 1 is due to the identity element) and, if m is the number of not fixed points (even as we have just discovered), then we should have k + 1 + m = n and k = n - m - 1

The latter is odd

Looks like this can be generalized to n of any parity

So k and n will always be of different parities

Thank you

[rocky cloak]

Indeed, the argument also works when n is odd, but then it's a little boring since k=0

[dull marsh]

Ah wait you are right, the order of the group being odd forces elements to have odd order

Can something similar be derived for elements of order 3?

[novel spruce]

For every prime number p dividing the order of the group, there exists an element of order p (Cauchy's theorem).

The way this is typically proved is as follows: consider all sequences (a_1, ..., a_p) whose product is the identity e. There are #G^(p-1) of those: the first p-1 elements can be freely chosen, and this fixes a_p uniquely. The cyclic permutation (a_p, a_1, a_2, ..., a_{p-1}) also has product e: this follows from the fact that a_p is the inverse of a_1...a_{p-1}, and inverses work on both sides. Cycling p times gives either p distinct sequences or p sequence that are the same, the latter being the case if all elements are the same (and thus leading to an element of order dividing p).

Since e is an element of order dividing p, the number of elements of order p has to be -1 modulo p, in order for the total number of sequences to add up to a multiple of p, which #G^(p-1) is.

[south patrol]

Tbf, it's also easy to do directly right

I remember it being funny - Maynard in some lecture notes commented how a baby could do this naturally (rearrange objects by swapping two at a time)

I would basically just say like given σ in S_n, compose it w a transposition swapping σ(n) and n

then σ fixes n and we're done by induction

[elder wave]

Proof by baby that rearranges

[south patrol]

i mean the baby could probably do the same thing right lol

put the first one in the right place then go to the others

[novel spruce]

A baby would probably only swap two objects right next to each other, and this type of transposition is actually sufficient to get all possible permutations as well.

[elfin wraith]

Orbit stabiliser is out, rearranging babies are in

[coral shale]

a baby would eat the blocks 😬

[south patrol]

True, just need to swap adjacent blocks to do the transposition I said

[untold basalt]

Find the Galois group of the field extension $\mathbb{F}_2(x) / \mathbb{F}_2(x^2)$

[cloud walrus]

Seagull

[untold basalt]

How do i approach this?

I can't figure out the degree og the extenion

[south patrol]

can you write down the minimal polynomial for x over F_2(x^2)

[untold basalt]

I'd say there is none and x is trascendental on F_2(x^2)

[knotty stag]

Hey folks

[tardy hedge]

Hello friend

[knotty stag]

Hey

[novel spruce]

The field extension is finite but inseparable, so there is no Galois group.

[ember field]

y^2-x^2 =0 is minimal satisfied by x. But yeah It is inseparable so only one element in Aut group I guess.

[tardy hedge]

The theorem that says for any nonconstant polynomial over F there is an extension field E that contains a root of the polynomial

Is the idea basically that you factor the polynomial over F into irreducibles and then you basically create a field in which one of the irreducible factors are the “0” element in the new field?

[coral spindle]

Yes

[tardy hedge]

Thanks

[untold basalt]

Show that a field extension $E/F$ is separable iff $E$ contains the splitting field of the minimal polynomial of an element $a \in E$.

[cloud walrus]

Seagull

[untold basalt]

If the the extension is normal for any a the minimal polynomial is irreducible so it's splitting field is contained in E

But If I know there's this a how do I show that all irreducible polynomials split in E[x]?

[chilly radish]

This is not true as stated. Take F_p(x)=F and adjoin a p-th root of x, so E=F_p(x^1/p), then E/F is inseparable but is also the splitting field of t^p-x

[south patrol]

indeed according to this every extension is separable since you can take a = 0

[untold basalt]

ok nvm my professor's notes are trash

doesn't know how to write

[viral halo]

if two groups have the same orders, and the same amount of elements of each order, are they isomorphic?

[barren sierra]

No

Consider

From here

[viral halo]

ty

[karmic moat]

is this proof ok

im just trying to figure out why the isomorphic homs implies ismorphic modules

is it just by universal property of tensor product

[rocky cloak]

Seems like it's proving something different to the stated aim. Though you could prove it in a similar way, by

Hom( A(x)S(x)F, X) =
Hom_S( S(x)F, Hom(A, X)) =
Hom_R( F, Hom_S( S, Hom(A, X))) =
Hom_R( F, Hom(A, X)) =
Hom( A(x)F, X)

Then yonedas lemma.

But this is just a consequence of the fact that tensor products are associative, so it would make more sense to just prove that.

[karmic moat]

yeah, im just not sure how to extend the fact that tensor products are associative to the case of tensoring with two rings

[rocky cloak]

I mean, the tensor product is associative doesn't depend on the rings you tensor over being the same.

[karmic moat]

yeah, i'm just trying to prove that

oh nvm i found a math se post

https://math.stackexchange.com/questions/1485199/associativity-of-extension-of-scalar-tensor-product

its a shame i cant use yonedas, its a fun proof

[cloud solar]

Let f in R[x] with f=x^2019-x^2017+a_{2016} x ^2016 + a_{2015} x^2015 +...+ a_{0}. Knowing that all the roots of f are in R show that the lenght of the minimal interval where the roots is in [sqrt(2019/509545), 2].

The sum of roots is 0 and the product -a_{0}.

Now idk how to catch that lenght of the interval

I need some inequalities

For abs(max(roots)-min(roots))?

[sonic coral]

i think looking at the normalizer of a sylow-2 of the group of order 36 would work

[dire siren]

@cloud solar wrong channel;
actually I had that problem in the contest, but I didn't solve it;
for the l<=2 part you can use the fact that sum for 1<=i<j<=2019 of (x_i-x_j)^2 is 4.

[cloud solar]

La Gheorghe lazar?

[dire siren]

da

[cloud solar]

Super. Un sfat pt onm clasa a 12a? În general problemele de polinoame ca alea sunt mai dure.

[dire siren]

@cloud solar n-am prea vazut polinoame la ONM;
in orice caz; pentru algebra recomand sa parcurgi tot ce e semnat de Marian Andronache - el are cele mai bune probleme de algebra
daca ai gazeta electronica da un search dupa "Andronache"; iarasi Ion Savu si Ioan Baetu au probleme bune de algebra
daca nu ai gazeta electronica, poti sa-mi dai pm si iti trimit arhiva; e o resursa pe care o consider esentiala

[cloud solar]

Nu am gazeta electronica. Îți dau dm?

[dire siren]

da

[sinful holly]

Oh yeah, that's very cool! If the Sylow 2-subgroup is normal in the group of order 36, then the normalizer definitely has to have order >=72 and then you can easily show that the group is not simple. If it's not normal, it's less clear to me, but this seems like the right path.

[sonic coral]

well a group of order 36 either has a normal sylow 2 or sylow 3

so i supposed you’d just do something similar with that

i’m not sure it’s clear to me why the normalizer of a sylow 2 which is normal in a group of order 36 has order atleast 72 though

a normalizer of a normal subgroup is just the whole group which it is contained in right

which would be 36

[icy bear]

what does it mean to "classify every group"?

is it classifying in isomorphism classes?

cus at least when I first heard of it, "classifying every group" just seemed like something that would end when people were in the mood

doesn't seem like an objective goal

but isomorphism classes are objective ig

[wispy torrent]

sorry but this there a way to remove myself from the helper role I think I accidentally put myself on it. I get pinged every 5 secs its a bit distracting

[icy bear]

(I mean, every simple group)

[white oxide]

i think so, yeah

given a group G and a normal subgroup N, does there always exist an endomorphism G --> G such that im N is not contained in N?

[charred iris]

N = 0

[white oxide]

oh i meant for like any arbitrary subgroup sorry

my thought process is that this should be true

take any element in G - N and consider left multiplication

surely there must be some element that takes us out of N

otherwise G would be all of N

i think that works?

i think that works, but i just realized it's irrelevant to the result i'm trying to establish LOL

[sinful holly]

It's normal, because the normalizer contains a subgroup of order 9 (of course), but the normal subgroup of order 4 is contained in a sylow 2-subgroup and normalizers in p-groups always get bigger, so there's also a normalizing 2-subgroup of size >=8, giving us the subgroup of size >=72.

[sonic coral]

H (normal sylow 2 in group of order 36) is contained in P (sylow 2 of group of order 288).

[sonic coral]

which subgroup are you considering the normalizer of after this?

[white oxide]

could I get a hint for this question (in the case that $M$ has index 2): I approached it by letting $\varphi: G \to H$, and considering two homomorphisms $f_1: H \to H \to H/M$ and $f_2: H \to H \to H/M$, and am trying to show that $f_1 \circ \varphi = f_2 \circ \varphi$, but $f_1 \neq f_2$. I know that I couldn't consider any homomorphisms $f_i: H \to H/M$ since they would just end up being the trivial ones and hence being equal, so I need to factor through something. the problem i've encountered after realizing this is the condition that $f_1 \neq f_2$ but $f_1 \circ \varphi = f_2 \circ \varphi$, since all homomorphisms that I end up constructing either end up being the trivial ones or end up wtih $f_1 \circ \varphi \neq f_2 \circ \varphi$

[cloud walrus]

okeyokay

[south patrol]

Woah interesting

[white oxide]

ikr this problem is fun

[sinful holly]

None. Once you have a group of order 72 you have a subgroup of index 4, which gives you a non-trivial homomorphism to S_4, so the kernel of that map is your normal subgroup

[sonic coral]

no i understand this argument. i’m just not sure where the group of order 72 is coming from to get to this point

like how are you constructing the sub group of order 72, it’s not quite clear to me

[white oxide]

but anyways, if say $k_1: H \to H$ with domain $M$ sends elements of $M$ to elements of $H - M$, then we have $G \xrightarrow{\varphi} H \xrightarrow{k_1} H \xrightarrow{\pi} H/M$ nontrivial. on the other hand, if $k_2: H \to H$ with domain $M$ with $\text{im } k_2 \subseteq M$, then $G \xrightarrow{\varphi} H \xrightarrow{k_2} H \xrightarrow{\pi} H/M$ is trivial.

[cloud walrus]

okeyokay

[sinful holly]

It requires that the 4-subgroup is normal in the 36 group. Just consider its normalizer. It contains a subgroup of order 9, but the 4-group also lives in a genuine sylow 2-subgroup of order 32. The normalizer in that group has size >=8, so the normalizer group has size at least 72.

[white oxide]

so this leads us to consider homomorphisms with domain M into H either with range M or not M separately

idk maybe i'm babbling

[sonic coral]

gotcha okay

[supple vortex]

If I,J are prime ideals of a comm. ring R, is I intersect J a prime ideal of R?

[white oxide]

prove it

wait

[supple vortex]

Are you sure

I don’t think its true

[white oxide]

wait nvm i'm dumb

[supple vortex]

But I’m not sure how to proceed

[white oxide]

uhhh

i think considering ideals of Z works

something about prime integers

[supple vortex]

Wym

Are ideals of Z prime ideals?

[white oxide]

well the prime ideals of Z are of the form (p) for a prime

[supple vortex]

Oh

Do you mean <p>

Generated by p

[white oxide]

yeah

[supple vortex]

So wait

$np_1=mp_2$

[cloud walrus]

splotchvan

[supple vortex]

$p_1=m\cdot p_2/n$

[cloud walrus]

splotchvan

[supple vortex]

And p2/n impossible?

If pi prime

[tawny dune]

i think you can do it in a more simple way

you just need to give one counter example

[supple vortex]

Yeah but the more general the better

[tawny dune]

so pick two specific prime ideals in Z and show their intersection is not prime

[supple vortex]

I like this way

[tawny dune]

its sometimes true though

that the intersection of two prime ideals is prime

[supple vortex]

Oh wait

Is <gcd(p1,p2)> the intersection

[tawny dune]

no lcm

[supple vortex]

Yes sry

Are Z and Z[i] isomorphic?

Something tells me no way but

[tawny dune]

no

[supple vortex]

How should i show

What’s the best way

[tawny dune]

show that there is only one ring homomorphism between Z and Z[i]

[supple vortex]

Hmm

[tawny dune]

Hint: || What does 1 in Z have to map to in Z[i]? Does this determine the rest of the map? ||

[supple vortex]

f(1)=1

[tawny dune]

Yes, so what does that say about f(n)?

[supple vortex]

f(n)=n?

[tawny dune]

Yes

[supple vortex]

Wait how

[tawny dune]

f(n) = f(1 + 1 + … + 1) = n • f(1) = n • 1 = n

[supple vortex]

Yeah

Ok

And then f wouldn’t be surjective

[tawny dune]

Yep

[supple vortex]

Thx

[supple vortex]

Any online calculator for polynomial division over Zn[x]

Want to check

I’m on mobile so i can’t code it

[sonic coral]

would this work if i had a normal sylow-3 in the group of order 36

the normalizer of that would contain a subgroup of order 4

i don’t think it quite works though since the normal 3 subgroup of order 9 in the group of order 36 is still a sylow 3 in G

[supple vortex]

Nvm

how can i use the first iso theorem to show Z[x]/I is isomorphic to Z2?

Where I is the ideal of Z[x] consisting of all polynomials with all even coefficients

Wouldn’t i need a homomorphism f : Z[x] -> Z[x]/I with f(Z[x]) = Z2?

[random sail]

you need a homomorphism f: Z[X] -> Z2

where ker f = i and I'm f = Z2

[random sail]

Z2 is not a subset of Z[X]/I

[supple vortex]

Yeah that’s why i was confused

[tawny dune]

Aren’t those not isomorphic tho

[white oxide]

can somebody explain to me how exactly part a) follows from F10? I understand that L1 is a subset of E, and so we can consider L2(L1)/L2 as in F10, but doesn't this require the hypothesis that L1/L2 is algebraic? and we don't necessarily know that

did they mean if L1/L2 is algebraic?

[supple vortex]

Does I have identity?

I don’t think so?

Since all coefficients have to be even that means a_0 also

In a polynomial

So 1 notin I?

[topaz solar]

L2 is algebraic over L2, and L1L2 = L1(L2)

And L1 is gonna be K(L1)

Hmmm

[topaz solar]

But no since in that case L1 L2 = L1, since it already contains L2

[white oxide]

ya but in order to apply the theorem don't we need L1 = L2

[topaz solar]

No?

Yeah no duh that’s algebraic in that case

L2 contains K

And L1 algebraic over K

[white oxide]

oh wait we can just say that L1(L2)/K is algebraic

[topaz solar]

L2(L1)/L2

[white oxide]

and algebraic elements remain algebraic over lifting

so L1(L2)/L2 is algebraic

math prodigy?

nvm

[tawny dune]

wdym by this

But anyway Z[x]/I is isomorphic to Z/2Z[x]

[supple vortex]

Yeah

But i also need to show I is a maximal ideal of Z[x]

[supple vortex]

Wait

Are these even isomorphic

Z[x]/I and Z2

I don’t think they’re homomorphivc by the indicator homomorphism phi(f(x)) = 0 if f in I, 1 if f notin I

But phi has ker phi = I and im phi = Z2…

What other homomorphism is there???

[supple vortex]

Because consider two polynomials in Z[x] of same degree and all odd coefficients. The sum of all respective coefficients will be even and thus the sum polynomial will be in I

So phi(f(x)+g(x)) = 1 neq 0 = phi(f) + phi(g)

Am i crazy???

Im a fucking idiot

[random sail]

f + g \in I, so phi(f + g) = 0

[supple vortex]

How can i prove that I is a maximal ideal of Z[x]?

[random sail]

uhh

suppose J is an ideal such that I < J \leq Z[x]. Then let j in J \setminus I.

uhhhh

this isnt going to work

[random sail]

Z[X]/I is a field, so I must be maximal

[supple vortex]

How do you know it’sa field

[random sail]

because Z[X]/I is isomorphic to Z2 which is afield

[supple vortex]

Oh

Lol

Ofc

[random sail]

wait the indicator homomorphism isnt a homomorphism

phi(x + 1) = 1
phi(x) + phi(1) = 1 + 1 = 0

try phi : f |-> f(0) mod 2

[supple vortex]

Oh

Ofc

Then a0

[random sail]

wait im stupid

omfg

the kernel isnt I

what am i doing lol

rip

[supple vortex]

Oh

Yea

Well wtf

[random sail]

$\mathbb{Z}[X]/I$ is definitely not $\mathbb{Z}_2$. For instance note that $1 + I \neq x + I$, since $1 - x \not\in I$, so there are at least 3 elements

[supple vortex]

You’re saying they aren’t iso?

[cloud walrus]

wrench

[random sail]

...yeah

since there are at least 3 elements: 1 + I, x + I, and 0 + I, but Z2 only has 2 elements

Oh I figured it out. Consider $\phi : \mathbb{Z}[X] \to \mathbb{Z}_2[X]$, which sends $a_0 + a_1x + ... a_nx^n$ to $\overline{a_0} + \overline{a_1}x + ... + \overline{a_n}x^n$. Clearly $\ker \phi = I$, and $\phi(\mathbb{Z}[X]) = \mathbb{Z}_2[X]$, so there you go

oop

[cloud walrus]

wrench

[sonic coral]

@sinful holly oh. i think since we are assuming G is not simple, we know that there are more than one sylow 3 subgroups of order 9. in particular, the subgroup of order 36 has more than one sylow 3 subgroup, so it can’t be normal

so the only case is that it has a normal sylow 2 in which we’re done

[supple vortex]

What is the bar

[random sail]

means equivalence class mod 2

[supple vortex]

Yeye

But

[supple vortex]

You used Z2[x]

Its supposed to be just Z2

[random sail]

It's not though!

It's Z_2[X]

by the first isomorphism theorem

[supple vortex]

Are you saying this is a typo

Or am i missing something

[random sail]

its definitely wrong, just think about it

[supple vortex]

So it’s a typo

[random sail]

0 + I, 1 + I, x + I, x^2 + I, etc... are all distinct elements in Z[X]/I

[random sail]

uhhh are you sure I is the set of polynomials with only even coefficients?

[supple vortex]

Oh fuck

Only a0 is even

[random sail]

LOL

[supple vortex]

My god

Oh my god

So what’s the isomorphism then

Er homomorphism

[random sail]

Well, you want to send everything where a0 is even to zero, and where a0 is odd not to zero

Can you think of a natural way to do that

[supple vortex]

a0 mod 2

[random sail]

(Note that it might be easier to show a0 mod 2 is a homomorphism if you use f(0) = a0 (in Z))

[supple vortex]

Ok

Final question for you

Let I be the ideal of Z[x] consisting of all polynomials a_2x^2 + … + a_nx^n for any n

So all polynomials at least degree 2

(Union 0)

Is Z[x]/I an integral domain

@random sail

?

[tawny dune]

It’s not, since Z[x]/I is an integral domain if and only if I is prime, but x • x is in I but x is not so I is not a prime ideal

[supple vortex]

Thanks all for your help

[random sail]

You really should know the theorems for a commutative ring R

• R/I is an integral domain iff I is prime
• R/I is a field iff R is maximal
  pretty well

[cobalt heath]

Can be similar said about the noncommutative ring?

[rustic crown]

ig you can say for a left ideal I, the left R-module R/I is simple if and only if I is a maximal left-ideal. dunno how one talks about prime ideals in non-commutative setting

[cobalt heath]

Hmmm

Seems I forgot how prime ideals require commutativity

[rustic crown]

you could always make up a weird definition, but i haven't run into those in the wild so far

[cobalt heath]

I see.

[ember field]

If i subset of J or J subset of I this is true. However in other cases I contains a element x not in J and J contains a element y not in I. Then try observing xy

[rocky cloak]

The definition often given is that a (right/left/two sided) ideal P is prime if for any (right/left/two sided) ideals I, J if IJ < P then either I or J is in P. Or equivalently for any two elements a and b if arb is in P for all r, then either a or b is in P.

In these cases R/P is called a prime ring, but I don't think the concept comes to very often.

[cobalt heath]

Oh, that sounds quite natural

[sinful holly]

Ahh very nice. I do understand. You constructed the group of size 36 by looking at the normalizer of an intersection of Sylow 3-subgroups, so the normalizer contains at least those two groups of order 9. OK, that's cool---I never would have thought to try to prove that a group of order 32*9 is not simple using Sylow theory.

[icy bear]

doing that thing halmos said of trying to prove everything in a section before reading it is really worth it I think

[tardy hedge]

it probably is worth it

i dont think im patient enough for that

[icy bear]

proving everything in artin's second chapter from homomorphisms turn inverses into inverses to the first isomorphism theorem was definitely an experience

[tardy hedge]

@icy bear you are still in high school?

[icy bear]

ye

[tardy hedge]

how old are you

[icy bear]

15

[tardy hedge]

Nice

[icy bear]

why

[tardy hedge]

ur thing says pre university so i was curious

[icy bear]

oh

[tardy hedge]

its also cool youre learning this so early though

[icy bear]

ye, I like higher math a lot

[tardy hedge]

at 15 i was just doing like my grade 10 curriculum math which was extremely basic stuff lol

you are planning to study it in uni i presume?

[icy bear]

yes

[tardy hedge]

nice

future mathematician

[icy bear]

but I really enjoy studying those things so I just do it

[icy bear]

I hope so

[tardy hedge]

ya i do too. its sort of addicting isnt it

[icy bear]

yeah

I even have to limit myself so I don't burn out lol

[tardy hedge]

hahaha

[icy bear]

are you in uni?

[tardy hedge]

i graduated my bachelors degree last may but i am taking an abstract algebra course rn at a random school for fun, cuz beleive it or not i never did it during my degree lol

i could have but i wanted to maintain a high gpa

the degree was in math but the courses i took was just a lot of random stuff

probability, complex analysis, dynamical systems

more applied courses

[icy bear]

what's a gpa

[tardy hedge]

grade point average

[icy bear]

interesting

[tardy hedge]

i just wanted to maintain highest grades i could

so i shyed away from the hard stuff

[icy bear]

oh

I hear a lot of stories like that

[tardy hedge]

but i want to learn the hard stuff now

yeah, tbh i think it was a good decision

it allowed me to for eg apply to medical school too cause i have a very high grade average

[icy bear]

getting away from difficult stuff in uni to mantain a high gpa

[mighty kiln]

[tardy hedge]

and i can always learn the hard stuff later, like i am doing rn

[icy bear]

what's gpa for?

[tardy hedge]

u dont like that dont u

[icy bear]

like, why do people want to mantain it high

[tardy hedge]

well it helps for grad school stuff

law school, med school etc

if u end up deciding those routes

[icy bear]

hum

[rocky cloak]

Grades are the deciding factor for getting into certain schools / study programs many places.

[icy bear]

so sad Sn has order n! because lagrange's theorem means nothing to it 😭

[woven obsidian]

Well you still know there are no subgroups of S_n whose order is divided by a prime > n.

[icy bear]

oh yeah

didn't think about that

and that the greatest cycle would be the n-cycle (which is obvious but interesting that you can use lagrange to prove that)

[rocky cloak]

Not sure what lagrange tells you about cycles. For example S5 has an element of order 6, and I don't see that you can abstractly distinguish cycles from other elements.

[icy bear]

it's cus the cycle has order n

[icy bear]

how can S5 have an element of order 6 if 6 doesn't divide 5!?

[rocky cloak]

6 divides 120...

[icy bear]

ohhh

I forgot that logic only works for primes

[cloud walrus]

Leno

[sonic coral]

i went all the way from 1-300 haha

[rocky cloak]

Looks good to me.

[tardy hedge]

When proving rings, groups etc are isomorphic in a theorem does one usually employ an isomorphism theorem?

It seems like thats the strategy so far

Do something with the kernel that makes stuff work out

[void cosmos]

yeah

[tardy hedge]

Thanks

[south patrol]

Well not really

They give you one way to construct isomorphisms

From surjections

[tardy hedge]

Ive just seen it in a few different places thats why im wondering. Right now i was reading thru proof of quotient field of D has subring isomoprhic to D

[south patrol]

But there are other ways

[void cosmos]

there are ofc other ways

if u open up an undergrad textbook in algeb ra

check the exercisies

most of the time its constructing a surjective homomoprhism isntead

and then applying first iso

[cloud solar]

G finite abelian group. We say a in G has property (A) if there is a subgroup H<=G s.t. the product of elements of H is equal to a. Show that the set of elements with property (A) form a subgroup of G

Let a, b in G with property (A). So there is H_a,H_b<=G s.t. prod(H_a)=a, prod(H_b)=b. Because G is finite abelian prod(H_a)=prod(order 2 elements in H_a) and same for H_b

[delicate orchid]

yeah product of all elements seems a tad sus lol

[cloud solar]

Since G abelian then H_aH_b is a subgroup of G

Now I want to find somehow a minimal subgroup of H_aH_b

[delicate orchid]

h contains h^-1 for all H in h, so the product of all elements in H is the identity

[cloud solar]

No

[delicate orchid]

what you mean no

[cloud solar]

The product is the order 2 elements

[delicate orchid]

yeah true carry on

no wait I still don't like it. Let a, b in H order 2, then ab is in H, so the expression ab(ab) appears in the product of all elements

what's gone wrong here

H could be cyclic of order 2, right ok

so if H has a unique involution it doesn't multiply to 0

now I'm happy

[cloud solar]

If I look qt the set of elements with order <=2 in H_a this set is a subgroup cuz H_a abelian.

Let s call it S_a

Same for S_b

If I look at the set S_aS_b this is a subgroup

Maybe this is the subgroup with the the product of all elements =ab

[delicate orchid]

this situation is actually very cool. Let $V = {v_1, ..., v_m}$ be the group of involutions, none of the products of involutions are equal to each other, then we get a nice cancellation as the product of all elements is $\prod_{k=1}^{|V|}\prod_{v \in K \subset V, |K| = k} v = \prod_{i=1}^m v_i^{m_i}$ with $m_i$ even unless $m = 1$,

[cloud walrus]

Wew The Lads Tbh

[delicate orchid]

so since all of the m_is are even, the product is trivial

[rocky cloak]

Another solution could be to notice that such an element ||has order 2|| and that ||all elements of order 2 have property (A)||

[delicate orchid]

I have a hunch there's a cohomological interpritation to this, the nerve of an exponent 2-group should be the full n-simplex with n = the rank of the group

[delicate orchid]

oh i don't care about the original problem I'm just sludge posting

[cloud solar]

Yes i noticed it

[delicate orchid]

yeah and you also noticed that all elements of order 2 form a subgroup in an abelian group

[tardy hedge]

“Are Z9 and Z3 + Z3 isomorphic as rings?”

Answer: no?

I saw that Z9 has 8 units and Z3+Z3 has 4

Isomorphism preserves units

[delicate orchid]

Z9* = {1,2,4,5,7,8} which is 6 not 8

[tardy hedge]

Lol oops i meant 6

[delicate orchid]

now lemme think about Z3

[tardy hedge]

Z3+Z3 i mean to write direct product

I think thats what it called

[delicate orchid]

uhhhhhh taking groups of units is an adjoint so yeah we just have Z_2 x Z_2

I buy it

[tardy hedge]

Haha thats cool how ur doing the same question but just from a more abstract pov

I was just doing it nuts and bolts way

Like idk wtf adjoint shit means