groups-rings-fields

[chilly ocean]

Do you know what a vector space is

Direct product is how you give product of groups a structure of a group

Direct sum of groups is kinda tricky because it's not actually the coproduct

For abelian groups, coproduct is the direct sum

But like you said, direct product of groups is a group in which all of the groups don't sit freely but they must commute

For me it's just a leftover from the category of abelian groups

Defined in more generosity because, why not?

the pdf is from here: https://canvas.harvard.edu/courses/90105/files/12964322?verifier=QP2B5WOaM0eXd9csuVTxy0DqGdOgUDOS1Si0RbO3&wrap=1

hum... the pdf says "given infinitely many groups there are two different notions (of product group)"

[hollow mica]

this is purely a definition, but it has some higher level significance (see the categorical stuff Blitz mentions above)

if you are coming across these constructs for the first time, it should be okay to accept these definitions at face value

In fact I asked the same exact question about a month ago, so I am just relaying the advice I was given 😄

[trail stump]

are simple groups always a normal subgroup of a bigger group?

[chilly ocean]

In S_3 this should be false for a copy of Z_2

Z_2 is simple

[trail stump]

Z_2 do you mean like {e, (1,2)}?

[chilly ocean]

Yeah

It's clearly not normal

Easy to see from t(1, 2)t^-1 = (t(1), t(2))

[trail stump]

yeah

damn i feel so stupid for asking such question

[tribal moss]

Here's one way to think about it. If you have a family of infinitely many groups G1, G2, G3, ... , then each of the original groups can be thought of as a subgroup of the original group, if we identify the element g in G3 with (e,e,g,e,e,...) in the product group, and so forth.
The direct sum of all the groups is then the smallest subgroup of the product that contains all of the original groups.
This sometimes makes it useful if you don't need the additional elements that are only in the product for your application.
In particular, if you have a homomorphism from the direct sum you can reconstruct the entire homomorphism just by knowing what it does to elements of the original groups; this is not the case for a homomorphism from an (infinite) direct product.

As Blitz points out, it is even more useful for abelian groups. If you have a bunch of homomorphisms G1->H, G2->H, G3->H, ... where H is abelian, then the entire family of maps always corresponds to a single combined map from the direct sum of the Gi's. This is not necessarily true if H is not abelian -- but it is still the case that IF you can represent all of them by a combined map, THEN the map is unique, which is sometimes enough to be useful.

[ember field]

it's supposed to work for any h' st pi(h')=h

[next obsidian]

Yeah so you have to show that this action doesn’t depend on which h’ you chose, it’ll be the same if you pick any h’ with pi(h’) = h

Once you do that, e is a valid choice of e’

And then it’s clear that e’a(e’)^-1 = a

Since you can choose e’ = e

[wise nimbus]

,tex
Hello. If $\alpha$ is a root of $f(x)=x^3-7x+7$ How do I determine $Gal(\mathbb{Q}(a)/\mathbb{Q}$ ?

[cloud walrus]

chedug

[wise nimbus]

By Eisenstein criterion, f(x) is irreducible over rationals

So f(x) has either 3 real roots or 1 real root and 2 complex roots

Is it even important to compute the roots explicitly in order to answer this question?

[chilly radish]

You need to determine if Q(a) is even Galois first

Appending one root doesn't necessarily split f

[wise nimbus]

So I need to be sure that all of my three roots are in Q(a)

[chilly radish]

If you can show they are, then by the fundamental theorem Gal(Q(a)/Q) is if order 3 (since Q(a) is a degree 3 extension), and there's only 1 group of order 3

[wise nimbus]

yes

Z_3?

[chilly radish]

Yes

There's a classification of the galois groups of degree 3 polynomials, but I'm assuming you don't know it

[wise nimbus]

Yes I don't know it. My professor thinks that we should work out these problems ourselves but as soon as I start one, I am lost

[chilly radish]

Depending on the tools in your disposal this may be difficult

[wise nimbus]

Thank you for the insight

[chilly radish]

Np

[rotund aurora]

Lattices from order theory and lattices used in group theory (like Z^2) have "nothing" do with each other right? In the sense that the fact that they share the same name is just a coincidence (?)

[formal ermine]

let $H$ be a subgroup of $G$, why if $g H g\inv \subseteq H$ and $g\inv H g \subseteq H$ for all $g \in G$ then $H \trianglelefteq G$ i.e. why is it that if those conditions apply then $g H g\inv = H$

[cloud walrus]

illuminator3 (#eric4honorable)

[rotund aurora]

conjugation is a bijection

[coral spindle]

What's the size of gHg^-1? Think about this

[formal ermine]

right, so if $K \subseteq H$ and $|K| = |H| \implies K = H$?

[cloud walrus]

illuminator3 (#eric4honorable)

[coral spindle]

This is insufficient for infinite sets but it's fine for finite sets.

Now prove that gHg^-1 = H for infinite sets.

[flint niche]

When I try to solve it, I get that there is no such p. Could you help, please?

[formal ermine]

no idea how I would do that. does it have something to do with $g\inv H g \subseteq H$?

[cloud walrus]

illuminator3 (#eric4honorable)

[coral spindle]

It requires that, yes.

n.b. for all g in G

Think about what Croq said.

[formal ermine]

what does n.b. mean?

[coral spindle]

google it, I can't remember the latin phrase

[tribal moss]

Surely what you want there is |gHg^-1|=|H|.

[coral spindle]

No.

I mean if gHg^-1 is a subset of H for all g in G, then gHg^-1 = H.

(for all g in G, also)

[tribal moss]

Sorry, I missed some context. Go on, don't mind me.

[coral spindle]

No worries, it comes with the discord format

[coral spindle]

Perhaps you can use the rational root test?

I also am inclined to believe that there is no such p but am unsure

[formal ermine]

doesn't gHg^-1 \subseteq H imply g^-1Hg \subseteq H

[coral spindle]

This is the same statement for g^-1, yes

I don't think the statement for g implies the statement for g^-1, but we are given that it holds for all elements of G so they are both true.

[formal ermine]

I'm still not getting it

so we have phi : H -> gHg^-1, h -> ghg^-1 which is a bijection, and ghg^-1 is always in H (because gHg^-1 is a subset of H)

but how does the bijection help us here

but even in the non finite case H must have as many elements as gHg^-1 because there exists a bijection, so H = gHg^-1, no?

[chilly ocean]

you want H -> H sending h to ghg^{-1} to be a bijection

maybe it's more clear if you see it written like that

[formal ermine]

yeah that's a bijection

and if they're subseteq then they must be equal because there's a bijection?

[chilly ocean]

or you can skip all this bijection nonsense and just go from H = (gg^{-1})H(gg^{-1})

(though this step in one form or another is what you'd be doing to prove the thing i just wrote is actually a bijection)

[chilly ocean]

i suggest you carefully write out what it means for the map i just wrote to be a bijection (in particular, surjective...)

[formal ermine]

,, h = \varphi(g^{-1} h g), \forall h \in H

[cloud walrus]

illuminator3 (#eric4honorable)

[formal ermine]

oh wait

nvm

I'll go get some food and think about this again when I'm no longer hungry

thanks for the help so far though

[lethal dune]

@formal ermine you want to show that conjugation is surjective right?

injective is obvious

[formal ermine]

no

[formal ermine]

this is the original question

[rotund aurora]

So it was proven by Keremedis that the assertion that every vector space over Z/2Z has a basis implies the axiom of choice. Given a field k, the question of whether the assertion that every vector space over k implies AC is an open problem. What do most people think about this problem? Is it conceivable that there exists a field k such that "every vector space over k has a basis" does not imply AC? If so, would this field k have some special properties relevant to algebra?

I assume the answer to the last question is no, since all of this is very "theoretical", but just curious.

[lethal dune]

@formal ermine gHg' ⊆ H and g'Hg ⊆ H implies H ⊆ gHg'. combine to get gHg'=H

[toxic zephyr]

i'm trying to find what $\bZ[x]/\langle 2x+4\rangle$ actually looks like by seeing what it is isomorphic to. i think i have a solution using the FHT with the map
$$\vfunc{f}{\bZ[x]}{\bZ_2[x]\times\bZ}{p(x)}{([p(x)]_2,p(-2))}$$
i'm still working on finding the preimage in $\bZ[x]/\langle 2x+4\rangle$ for a given element $(\overline{p(x)},n)\in\bZ_2[x]\times\bZ$, but can anyone let me know if i'm on the right track? not 100% sure that $\ker f=\langle 2x+4\rangle$, and i don't want to waste my time if i got the wrong ring.

[cloud walrus]

nix

[formal ermine]

how do they imply it?

[lethal dune]

g'Hg ⊆ H
gg'Hgg' ⊆ gHg'
H ⊆ gHg'

H ⊆ gHg' ⊆ H

[chilly ocean]

please

[formal ermine]

ahhhh thank you!

that makes sense

[coral spindle]

It's OK Terra I noticed you, you are heard 🫂

It was a good suggestion, but apparently that doesn't cut it...

[chilly ocean]

i'm used to it by now

[chilly ocean]

🫂

[toxic zephyr]

oh err idk if it's surjective actually. (1,2) doesn't have a preimage i don't think

[lethal dune]

is that a fish inside an aquarium?

[south patrol]

Lol

[cloud walrus]

randomspades

If $A$ is a matrix, then what does $(A_{\mathcal{B}_1}^{\mathcal{B}_2})_{i,j}$ denote?

[burnt flower]

is it just the entries?

[coral spindle]

A_{B_1}^{B_2} is the matrix but from the basis B_1 to the basis B_2

then i,j just tells you the entry number, yes

[burnt flower]

okay thanks so much

[coral spindle]

no worries

[celest cairn]

Hi, how do I simplify this? I need help understanding it.

[burnt flower]

wouldn't u just find the 7th cyclotomic polynomial

and multiply

what's the question?

[celest cairn]

I need to find the minimal polynomial of Zeta(5) + Zeta(7).

[hollow mica]

what is an example of two R-modules A and C, such that A is a submodule of C, where there exists no submodule B such that A + B = C

one example I'm pretty sure that works is R = A = integers and C = rationals, but the proof is a bit involved

are there any nicer examples

[tribal moss]

I think that is only possible if it is "no submodule $B$ such that $A \oplus B = C$" rather than "no submodule $B$ such that $A+B=C$".

[cloud walrus]

Troposphere

[hollow mica]

oh I thought + also meant direct sum

what does + mean

[tribal moss]

It might sometimes mean direct sum, but most often it would be understood as simply the set { a + b | a in A, b in B }.

[hollow mica]

ah

yeah I was thinking some more

the only submodules of Z (over Z) are nZ

[tribal moss]

A simple example would be $R=C=\bZ$ and $A=2\bZ$.

[cloud walrus]

Troposphere

[hollow mica]

but mZ and nZ always intersect at an infinite number of places

yeah

two submodules must intersect at only the identity for there to be a direct sum right?

[tribal moss]

Right.

If they have another shared element, that element would be a sum of something-from-A and something-from-B in at least two different ways.

[hollow mica]

wait I'm confused

some sources define the direct sum of two modules as their cartesian product

there's no way that's equivalent, right?

we're talking about a different direct sum

[rustic crown]

there are two versions of a direct sum. the cartesian product thingy is called external direct sum

[formal ermine]

$|G/H| = |G\backslash H|$?

[coral spindle]

No.

|G/H| = |G| / |H|.

[formal ermine]

jeuss christ latex

[coral spindle]

you want \setminus

remember that \ means something in latex.

[cloud walrus]

illuminator3 (#eric4honorable)

[formal ermine]

yeah

[coral spindle]

I have already answered the question in any case.

[formal ermine]

and \ escaped is a newline

[agile burrow]

is that a setminus or the number of right cosets?

[hollow mica]

ah

[formal ermine]

index of right cosets

[coral spindle]

LOL

OK

Yes

[agile burrow]

Then it's true

[coral spindle]

There are the same number of left and right cosets.

Try proving this

Hint: there is a map that sends left cosets to right cosets and vice versa

you will be very familiar with it...

[rustic crown]

yea, if you have some module M and two sub-modules A and B which intersect trivially, then you could show that A + B = {a + b | a in A, b in B} is isomorphic to the external direct sum of A and B

[formal ermine]

for abelian it's trivial

ohhhh

nice

color change

[hollow mica]

makes sense

[formal ermine]

lagrange's theorem is true for both index of right cosets and left cosets, no? because it's pretty much the same procedure

and then $|G|= |H| \cdot |G/H| = |H| \cdot |G\backslash H| \implies |G/H| = |G\backslash H|$

[cloud walrus]

illuminator3 (#eric4honorable)

[coral spindle]

I consider this cheating

[rustic crown]

True, but the theorem is still true even when the index or the group is infinite

[coral spindle]

^

This too lol

Try proving it directly.

[rustic crown]

in that case, we are asserting that the two sets have same cardinality, i.e. there is a bijection between them

[coral spindle]

See here

You will be very familiar with this map

[formal ermine]

phi : left closet -> h^-1 left closet h for the h that 'generated'(?) the left closet?

[coral spindle]

Prove it

[burnt flower]

yeah I think that's a difficult question to straight up get. The coeficcients are quite large in the minimal polynomial.

this depends what you mean by zeta(5) and zeta(7)

If it's a fixed root of unity and you're asking for a minimal polynomial of $\zeta^5+\zeta^7$ then that's one problem but if you're looking for what I assumed was the minimal polynomial primitive 5th root of unity plus the primitive 7th root of unity, denoted by $\zeta_5 + \zeta_7$

[formal ermine]

$\varphi : G/H \to G\backslash H, l_g \mapsto g\inv l_g g$ the rhs is a right closet because $g\inv l_g g = g\inv (gH) g = Hg = r_g$

[cloud walrus]

illuminator3 (#eric4honorable)

[rustic crown]

this map isn't well defined 😦

[formal ermine]

and $x \to g\inv x g$ is bijective for all $g \in G$

[cloud walrus]

illuminator3 (#eric4honorable)

[formal ermine]

why

[rustic crown]

because l_g is sent to g^-1 l_g g

and l_g doesn't uniquely determine g

[formal ermine]

oh right

[cloud walrus]

randomspades

[coral spindle]

Here's another hint, illum

You want a map that turns something like gH into Hg' for some other g'

So it's like it reverses things

What map do you know that reverses the operation in a group?

[formal ermine]

inverse?

[formal ermine]

yea but idk the g that was used to generate the left coset when mapping

[coral spindle]

That doesn't matter.

It's coset, not closet btw

[formal ermine]

coset, yes

[coral spindle]

If you think that might work, try it.

[formal ermine]

that's what I tried here

cuz for every coset I have to apply the inverse of the element that was used to generate the coset, no?

[coral spindle]

That's not what you're doing there, no.

Try again.

[formal ermine]

hm

[coral spindle]

What you're doing there is conjugation.

[formal ermine]

is it not just literally $gH \mapsto Hg\inv$?

[cloud walrus]

illuminator3 (#eric4honorable)

[coral spindle]

You haven't proved this is well-defined, so I have no reason to believe it works

[formal ermine]

right

sec

[coral spindle]

Similarly you haven't proved it's a bijection. Stop guessing and start proving!

[formal ermine]

being well defined means that iff $a = b$, $f(a) = f(b)$ right?

[cloud walrus]

illuminator3 (#eric4honorable)

[coral spindle]

Yes.

[formal ermine]

ok

[glass grail]

just if

only if is injective

[formal ermine]

right

[coral spindle]

Oh, I didn't see the iff.

It's just if.

[formal ermine]

okay so, let $\varphi : gH \mapsto Hg\inv$. $\varphi$ is well defined because if $aH = bH$ (firstly it follows that $b\inv a H = H \implies b\inv a = e$) then
$$\varphi(aH) = \varphi(bH)$$
$$Ha\inv = Hb\inv$$
$$H = Hb\inv a \text{ (using the fact from above)}$$
$$H = H$$

$\varphi$ is injective:
$$a = \varphi(aH)a = \varphi(bH)a = b \text{ (also using the fact from above)}$$

$\varphi$ is surjective:
$$H = \varphi(e \circ H)$$

[cloud walrus]

illuminator3 (#eric4honorable)

[coral spindle]

The proof of well-definedness is invalid

You cannot end with H = H and say that this finishes the proof

"Proof that the eiffel tower is upside-down: Consider 2, and 1+1. Clearly 1+1 = 2, so 2 = 2. So we're done"

Start with the assumption that aH = bH and conclude that Ha^-1 = Hb^-1. The latter should be your last line, not something already known to be true such as H = H.

The proof that phi is injective does not make sense to me, as a cannot be equal to phi(aH)a: the latter is a set but the former is an element.

The proof of surjectivity is simply wrong.

It does not prove surjectivity at all.

[formal ermine]

ah

[coral spindle]

Try again. Write things out on paper before typing them up in LaTeX. I know getting TeXit to make your equations look nice is a lot of fun, but it is distracting you, I think.

[formal ermine]

I did write them out on paper before typing them up here

for injective I first thought that it follows from my well defindness 'proof' tho

[formal ermine]

right

I should probably go to bed

it's already really late here

[old hollow]

I don't understand how they're allowed to do this

they're talking about the product of two cosets

how can you just regroup the parentheses? we're not multiplying elements here

[coral spindle]

Well idk how to say this more than it just works

Like the definition is just

AB = {ab | a in A, b in B}

and you can verify for yourself that this is associative

[old hollow]

hm alright

yeah I proved it myself but I did a bunch of set builder stuff

so does a(Nb)N mean {a * n1 * b * n2 | n1, n2 in N}

?

[coral spindle]

Yes

If you want to shoehorn it into the notation above, just replace a with {a} and b with {b}.

[old hollow]

ah I see

[coral spindle]

This is what's meant, really. In some ways this notation would be clearer than the standard

But we tend to know what's meant

[old hollow]

yeah

I mean the lowercase letter is that distinction I guess

[coral spindle]

Yeah typically

[old hollow]

aight

thanks

[formal ermine]

if $aH = bH$ then $a = b$ as $H = a\inv b H \implies a\inv b = e$,
$$aH = bH$$
$$aHa\inv = bHa\inv$
$$a\inv aHa\inv = a\inv b Hb\inv$
$$Ha\inv = Hb\inv$$
which shows that $\varphi$ is well defined.
injectivity:
$$Ha\inv = Hb\inv$$
$$H = H(b\inv a)$$
$$\implies b\inv a = e \implies a = b$$$
surjectivity: $gH = \varphi(g (gHg))$

[hollow mica]

I'm so close to proving this, can anyone help me out on the final step?

[solar glacier]

small question

oh theres a question at play

shux

[formal ermine]

two

[hollow mica]

If V is a finite-dimensional vector space over a field k, B: V x V -> k is a non-degenerate bilinear form, and S is a subspace of V that is killed by B, i.e., B(a, b) = 0 for all a, b in S, show that dim(S) <= 1/2 * dim(V)

[solar glacier]

so there realls isnt room for a third one then

😦

[formal ermine]

open a thread with your question

top right thread icon with the little 2 next to it

that way you won't disturb any ongoing convos

@solar glacier

[solar glacier]

yes?

i think i solved my own question lol

[hollow mica]

my current proof:
Let s_1, ..., s_k be a basis for S. By the definition of S, B(s_i, s_j) = 0 for any i, j. In particular, if we have B(s_i, -) = 0 where - is taken to be any s_j. Since B is nondegenerate, we must have B(s_i, t_i) ≠ 0 for some t_i because the form B(s_i, -) cannot be the 0-map. Now each t_i is linearly independent from all of the s_i, so now it suffices to show there exists a choice of t_1, ..., t_k such that they are all linearly independent, then the result follows.

thing is I don't know how to ensure the t_i are linearly independent

[coral spindle]

if $aH = bH$ then $a = b$
This is false.

[hollow mica]

It would be great if I could say "choose t_i such that B(s_i, t_i) = 0 and t_i is linearly independent of t_1, ..., t_{i - 1}"

[coral spindle]

Well good news, you can! It's just induction on the dimension of V

Alternatively, I think you can use a Gram-Schmidt on a basis

Well, am I sure about that... no

But anyway, there is another problem with your proof that I have

It's not at all clear to me why there should be $k$ such vectors

oh god I shouldn't have invoked the bot

Anyway

Yeah this is a big gap in your proof

[hollow mica]

oh I forgot to say let dim(S) = k

is that what you're referring to?

[coral spindle]

Well tbh I just read it again and realised I misunderstood yet again

I totally agree with the problem you're having. I also don't see any way to guarantee that the t_i be linearly independent, and I suspect in general you can't actually guarantee this with a similar setup. I think there's information we need to use.

I'm going to think about this

[hollow mica]

ok

in the meanwhile, I will try induction like you suggested

just induction from the very beginning?

[coral spindle]

The induction method doesn't work.

You need more information.

[hollow mica]

ah rip

[sage spruce]

I guess it could be good to use the rank theorem

[hollow mica]

One thought I had was, just choose the t_i ahead of time to be the basis completion of the s_i.

[coral spindle]

I'm thinking about how that could be used indeed.

[hollow mica]

And then you might run out of t_i to pair off to the s_i

[sage spruce]

Just take the left or right application

[coral spindle]

But a single t_i could be a witness for all of the s_i, so that doesn't necessarily work.

[sage spruce]

Like $B(x, \cdot)$ or $B(\cdot, y)$

[coral spindle]

Well I leave it to you then

[hollow mica]

yeah you would have to prove that the span of a witness isn't in the support of the other guys

which is hard

[hollow mica]

do you mean the rank nullity theorem

[sage spruce]

Yes that might be called this in english

dim(Ker)+rk=dim E

[hollow mica]

yeah

hmm

you're saying to use it on the B(s_i, -) linear forms?

[sage spruce]

Yup

That might come in handy

[hollow mica]

ok if I use rank nullity on a_i := B(s_i., -), I get

dim ker(a_i) + dim im(a_i) = k
but more specifically
dim ker(a_i) ≠ k so
dim ker(a_i) < k and dim im(a_i) > 0

this is all you can say right

[coral spindle]

Well specifically

dim im(a_i) = 1

by nondegeneracy

[hollow mica]

oh shit true

[coral spindle]

so dim ker = k-1

[hollow mica]

yeah

[coral spindle]

Which is pretty significant

you know that there's exactly one vector (up to scaling) which s_i doesn't annihilate

[hollow mica]

dim ker(a_i) = k - 1 and dim im(a_i) = 1

yeah

[coral spindle]

q?

[hollow mica]

typo

[coral spindle]

Right cool

yeah

[hollow mica]

sooo

[coral spindle]

Oh and

this holds for any vector, not just the s_i

[hollow mica]

yeah true

[coral spindle]

so if B(a, v) =/= 0 and B(b, v) =/= 0, then a = lambda b for some nonzero lambda

for any nonzero v

I think this is what we needed.

[hollow mica]

oh so now can I use this

to finish my original proof

[coral spindle]

Your original proof should work now, yeah

Neat

[hollow mica]

This is awesome

[coral spindle]

It's very nice

Thanks couscous for pointing out the right way

[hollow mica]

what's not awesome

is this problem

Let V be an inner product space, and let T be a self-adjoint operator where T ○ T = I and dim ker(T + I) = 1. Show that T can be expressed as T(x) = x - 2<x, v> v, where |v| = 1 and <a, b> is the inner product of the vectors a and b.

this makes sense geometrically in real space

[coral spindle]

Sorry, I don't know how adjoints work but hopefully someone else does

[hollow mica]

but I don't know how to exploit the self-adjointness

ah

self adjoint just means <v, Tw> = <Tv, w> for any vectors v, w

the only property I know of self-adjoint operators is that if W is an invariant subspace under T, then so is the orthogonal complement of W

[chilly ocean]

it might be useful to note that v must be in ker(T + I)

[sage spruce]

T is an orthogonal symetry

Is it a finite or infinite dimension space?

[hollow mica]

finite

[sage spruce]

Okay

[hollow mica]

you're saying that T(v) has to equal -v ?

[chilly ocean]

yes

[hollow mica]

oh

yeah you plug it in

nice

[sage spruce]

So you know Ker(T+I)+Ker(T-I)=V

[hollow mica]

I never knew this

I guess they're disjoint

and uhh

[sage spruce]

It comes from the kernel lemma

[chilly ocean]

since you assumed that ker(T + I) has dimension 1, you might as well try to pick v from there. it might also be worth noting that the map x -> <x, v>v would then be precisely the orthogonal projection onto ker(T + I)

[sage spruce]

Exactly

[chilly ocean]

just some thoughts. no particular solution in mind

[sage spruce]

In that case Ker(T+I) and Ker(T-I) are orthogonal

So checkmate eh

[hollow mica]

wait all the kernel lemma says is that V = ker(T) ⊕ im(T) when they are disjoint

[chilly ocean]

https://fr.wikipedia.org/wiki/Lemme_des_noyaux

[sage spruce]

No it's about a polynomial of T that cancels T

[hollow mica]

oh

hm

[sage spruce]

Here (T+I)(T-I)= 0

So you have your result

[hollow mica]

that means they're coprime?

oh because they are inverses of each other

[sage spruce]

Like X+1 is prime with X-1 in R[X]

[hollow mica]

mhm I see

sorry I have never worked with polynomial rings before, this is why I'm struggling

[chilly ocean]

linear algebra is a great place to get started doing so

[sage spruce]

Don't be sorry

[chilly ocean]

most of your favorite decomposition theorems about linear maps and finite dimensional vector spaces can be seen as special cases for nice modules over nice polynomial rings (in one indeterminate)

[hollow mica]

will have to read more then

[hollow mica]

I am trying to understand from the beginning because I didn't understand all the steps.
Why does Ker(T+I) ⊕ Ker(T-I)=V imply that Ker(T+I) and Ker(T-I) are orthogonal complements?

[sage spruce]

It does not but here it is self adjoint

[hollow mica]

oh I see

[sage spruce]

I'll try to write it

[hollow mica]

you're using this basically

right?

[sage spruce]

$\langle T(x) + x, T(y)-y \rangle =\langle T(x + T(x)), T(y - T(y)) \rangle = \langle T(x) + x, y-T(y) \rangle$

[chilly ocean]

bot is dead

[sage spruce]

So it's equal to 0

[sage spruce]

Bloody hell

The only time I can't use a darn pencil

[hollow mica]

hmm

let me process this

[sage spruce]

Thx

[hollow mica]

is x in Ker(T + I) and y in Ker(T - I)?

[sage spruce]

Yes

[hollow mica]

but then isn’t T(x)+x just 0

and same for y

I thought we need to show <x, y> = 0, because obviously <0, 0> = 0

[sage spruce]

Yes I made a mistake

But y=T(y)

[rotund aurora]

If in a vector space/module we remove the restriction that 1\cdot v=v for all v, do we get something interesting? It is of course clear that probably (almost) all basic results of vector spaces/modules are lost, but idk if something nice can be said in those cases. It is weird.

It is weird because $1\cdot(1\cdot v)=(1\cdot 1)\cdot v=1\cdot v$, so some vectors will be fixed

[tribal moss]

You could end up with every scalar product being the zero vector.

[rotund aurora]

Sure, but that's not the only possibility

Like for example, there are only two possible scalar products on the module Z over Z (dropping the "identity axiom" I mean)

So in that case it is reasonably behaved, but if you consider a module over Z of rank 2, I think then there are just too many uninteresting possibilities

[tribal moss]

I think the vectors that satisfy 1·v=v would form an honest module, and they'd have a direct complement that's killed by all scalar products.

If that is true, then what we'd be looking at would always decompose into a cartesian product between an actual module and an abelian group, with the two parts not interacting.

[rotund aurora]

kind of

oh yes

Multiplication by 1 is a homomorphism of the abelian group. Then if V=M+F (this decomposition is "as" an abelian group so far) where M is the actual module, F is closed under fake scalar multiplication, so its kinda nice.

I'm so sleepy xd, but this is fun

Then idk what can you say about F

sure for every v in F, 1*v=0, but if you have things that are far from the rationals in your ring, then idk if you will also get zero. I think that need not be the case

oh wait nah, it will also be zero

so this is done

[celest cairn]

Yes I’m trying to find the minimal polynomial of $\zeta_5 + \zeta_7$

[burnt flower]

Oh i cheated and used wolfram alpha

but like

there's probably an interesting answer to this

https://math.stackexchange.com/questions/155122/how-to-prove-that-the-sum-and-product-of-two-algebraic-numbers-is-algebraic/155153#155153

hope this helps u

here's an example of how they'd actually calculate this on the computer

Let me see if I can find the result in mathematica for u by this algorithm

[sage spruce]

The complexity is extreme

[burnt flower]

yeah you're gonna get a gigantic matrix

[hollow mica]

Is Hom(Z^m, Z^n) ≅ Z^{mn} where everything (including the set of homomorphisms) taken to be a Z-module?

my intuition was that each basis element can be sent anywhere, and this uniquely and sufficiently specifies a homomorphism, so there are (Z^n)^m = Z^{mn} total ways to do this

not sure how to formalize though

[sage spruce]

And an n! complexity

[next obsidian]

This is true

Your logic also works

Just prove in genera that Hom(Z^n,A) ≈ A^n

Basically doing what you said

[hollow mica]

nice

The natural homomorphism f: M -> (M*)* between a module and it's double dual is defined by f(x) = (g -> g(x)), right?

the same way as it is for vector spaces

[agile burrow]

That's right

[next obsidian]

This is not always an isomorphism though

[hollow mica]

yeah

[next obsidian]

If the map is injective M is said to be torsionless (not torsionfree!)

[hollow mica]

I'm trying to come up with a counterexample

[next obsidian]

And if it’s an iso it’s called reflexive

I think every module of the form M^* is reflexive tho

At least with some amount of hypotheses

[hollow mica]

I don't see why the proof why f is injective when we are dealing with vector spaces doesn't also apply here

let me run through the proof why f is injective for vector spaces

if f(a) = f(b), then for any form g in M* we have that g(a) = g(b)

[next obsidian]

You need the following:

[hollow mica]

but then g(a - b) = 0, so a - b is in the kernel of everything

this only happens when a = b

[next obsidian]

For any x,y in M, there exists a map M -> R such that f(x) ≠ f(y)

If M for example doesn’t admit many maps to R then this can be an issue

Not a problem for vector spaces tho because there’s always tons of maps

[agile burrow]

Maybe it's worth pointing out that it's not an isomorphism if the vector space is infinite dimensional

[next obsidian]

Okay true

Very true

[hollow mica]

wait do we at least know that f is surjective?

for modules

[next obsidian]

Nah

If so then every torsionless module is reflexive

[hollow mica]

why is it surjective for vector spaces tho

[next obsidian]

DIMENSION!

It’s injevtive because you can find maps which like, distinguish x and y

[hollow mica]

oh

in finite dimensions

[next obsidian]

And M and M* have the same dimension

By just making an iso

[hollow mica]

injectivity is enough

right

hmm

so then f doesn't have to be surjective or injective for modules?

[agile burrow]

right

[hollow mica]

I don't see anywhere in the proof of injectiveness that doesn't work for modules is the thing

[next obsidian]

It’s what I said

You’re asking for what x is the map
(f -> f(x)) the zero function (this is the kernel of M -> M**)

Ranging over all maps M -> R

Consider eg Z/2Z over Z

There’s only a single map Z/2Z -> Z

[agile burrow]

yeah, you're relying on the existence of some very specific functionals

[next obsidian]

The zero map

[agile burrow]

functionals that exist because of the structure of vector spaces...

[next obsidian]

So the image of 1 in Z/2Z under this map goes to zero

[hollow mica]

wait are you invoking the definition of f is injective iff (f(x) = 0 -> x = 0)

[next obsidian]

Yes

And the map is sending x to the function (f -> f(x))

So when is that function 0 is asking when does x map to 0 under every map f:M -> R

And I gave an example

In fact, for any n, the only map Z/nZ -> Z is 0

So actually the kernel of this map to the double dual is everything

[hollow mica]

if we let g(x) = (f -> f(x))

and we work in the module Z/2Z

you've shown that g(1) = 0 = g(0)

right?

and that's enough to show that g is not injective hence not an isomorphism

[next obsidian]

Yeah

I mean (Z/nZ)* is 0

So the double dual is also 0

So g sure as hell better not be an isomorphism

This actually works for any torsion group

G* will be 0 in that case

[hollow mica]

(Z/nZ)* is 0 means the only form is the 0-map

so (Z/nZ)** will send the 0:= 0-map to somewhere in Z

[next obsidian]

To 0

[hollow mica]

i'm trying to see this

[agile burrow]

homomorphism

[next obsidian]

It’s a group homomorphism

[hollow mica]

ah 2 * k = 0 -> k = 0, where k is the image of the 0-map

[next obsidian]

No like

Any group homomorphism sends 0 to 0

[hollow mica]

oh

[next obsidian]

I mean I guess your proof

[hollow mica]

i guess i just reproved that

lol

[next obsidian]

Does tell you this

[hollow mica]

has anyone ever seen this quantity before

T is an invertible linear operator on a real inner product space V (of finite dimension), and v is an element of V

apparently C(T)^2 is the ratio between the largest and smallest eigenvalues of T* T

no idea how to show this

someone told me spectral theorem but that confuses me because T is not assumed to be self-adjoint

[chilly ocean]

the numerator is the largest eigenvalue of T*T and the denominator is the smallest. a fun way to prove this is using lagrange multipliers, assuming you're working in R^n with the usual norm (which you can do by picking an orthonormal basis)

you can prove the spectral theorem in R^n using lagrange multipliers btw

i'll write out the details if you'd like. it's not really algebraic

[hollow mica]

Sure I'd love to see

[chilly ocean]

actually wait i may be baiting you

i think i made a few errors, oops. sorry for the early ping

i'll stare at it and see if i can fix

i know lagrange multipliers will give you something

[hollow mica]

yeah I am reading this right now and there seems to be a big connection https://journeyinmath.wordpress.com/2018/08/18/lagrangemultipliers/

[chilly ocean]

mm

|Tx| = <Tx, Tx> = <x, T*Tx>, so try taking A = T*T

[hollow mica]

how do we know <Tx, Tx> = <x, T*Tx>

[chilly ocean]

T* is the adjoint of T

[hollow mica]

oh adjoint not matrix mult

looking back at the original problem, we're trying to maximize <Tv, Tv> on the domain <v, v> = 1 (that's what the numerator says at least)

but <Tv, Tv> = <v, T*T v>

and <v, v> = 1 is the unit sphere

so the link does the same thing

hmm

gradient of inner product is confusing

[chilly ocean]

the numerator of C(T)^2

ignore my initial message, i was quite wrong lmao

[hollow mica]

yeah

how do they compute the gradient of f(x) = <x, Ax> to be 2Ax

oh we're working in R^n so this is the dot product and not some general inner product

I get that part nvm

what is the last line saying

I think they mean x_0 to mean a local min/max

but all x_0 in S_{n - 1} satisfy ||x_0|| = 1 by definition, so g(x_0) = 0

[chilly ocean]

i think they meant \nabla g(x_0) \neq 0

which you need to know to apply lagrange multipliers

[hollow mica]

is their statement wrong then?

[chilly ocean]

no it's correct

the one you just copy pasted, at least

they meant to write \nabla g(x_0) \neq 0 (which holds for all x_0 in S^{n - 1}), which rules out the first case and ensures the second holds (which gives you the eigenvalue of A = T*T)

[hollow mica]

oh I see

wait

I feel like the jump from the supremum to the maximization is hasty

because the supremum doesn't have to exist in the image of ||Tv|| right

If the domain were finite then this wouldn't matter

but there are an infinite number of vectors satisfying ||v||=1

[chilly ocean]

but a maximum exists

as does a minimum

you're looking at a continuous function on a compact set

[hollow mica]

ah I see

well I don't exactly see

but I don't know much analysis yet so I'll take the fact "continuous functions on compact sets have minimums and maximums" as granted

[chilly ocean]

this is the extreme value theorem

[hollow mica]

oh lol

[toxic zephyr]

any ideas for constructing an onto ring homomorphism

$\func{f}{\bZ[x]}{?}$ such that $\ker f=\langle 2x+4\rangle$?

i tried $\vfunc{g_1}{\bZ[x]}{\bZ_2[x]\times\bZ}{p(x)}{([p(x)]_2,p(-2))}$ but it is not onto because $p(0)\equiv p(-2)\pmod{2}$ so $(\bar{1},2)$ has no preimage.

[cloud walrus]

nix

[lethal dune]

$Z[x] \to \frac{Z[x]}{\ip{2x+4}}$ should do the job

[toxic zephyr]

haha that is true... but i was looking for something else so that i could find what that quotient ring is isomorphic to lol 😆

[next obsidian]

This is Z[1/2]

Because x is 1/2

Or well, x is actually -1/2 lol

Wait I’m an idiot x = -2

Hurbbbb

This is Z

Errr

Okay everything I’ve said is wrong, I shouldn’t do math at 2 am lol

[lethal dune]

[next obsidian]

Okay it’s F_2 I think…

So there’s a map Z[x] -> Z given by setting x = -2

This is obviously surjevtive and the kernel contains (2x + 4)

So write I for the kernel, then quotient Z[x]/(2x + 4) is the same as

[lethal dune]

kernel also contains (x+2) which is not an element of (2x+4)

[next obsidian]

Yeah I know I’m getting there

Wait

Okay I seriously should not do math

I was trying to say the quotient is the quotient of Z by the image of (2x + 4)

But this is also not true

Hurb

Thinking

So the quotient surjevts onto Z…

This quotient fucking sucks

I’m too chleepy

💤

[lethal dune]

maybe $Z_2[x] \times (\text{what you said})$?

[next obsidian]

Hmm

I mean the other part is just Z

[lethal dune]

yes I was talking abt the map

[next obsidian]

I would guess the thing is F_2 x Z

[lethal dune]

(reduce modules 2, your map)

[next obsidian]

Oh no

F_2[x] yeah

This is what CRT predicts

Except you can’t apply it

[lethal dune]

yes

[next obsidian]

Yeah so you should take the product of the maps

f(x) -> (f(x) mod 2, f(-2))

The kernel is the intersection of kernels

The kernel of the map to Z is (x + 2)

The kernel of reducing mod 2 is well… things which are doubles

So the kernel is (2x + 4) yeah

And then showing it’s surjevtive is annoying

[lethal dune]

nah

wait it is

[next obsidian]

It’s true I think

I ran a shitty argument in my head that I think works

But basically like

Okay, so

[lethal dune]

maybe take binary decomp of n \in Z

and do shit

[next obsidian]

Say you want to map to (g(x), n)

[next obsidian]

Yup

Like

g tells you the parity of the numbers you have to use

For the thing mapping to g

And then you have to modify these by 2 to guarantee this maps to n

So like

Okay let g(x) = Sum a_ix^i

Where a_i is in {0,1}

Set f(x) = Sum a_i x^i, this maps to g

Now f(-2) is off of n by some number k

Hmmmm

Actually this seems bad

No, you get 2Z

Ah!

Or uh…

Wtf

This is fucked up

The image is like

(g(x),even or odd numbers depending on if the constant of g is 0 or 1)

[lethal dune]

how do you get the elem (x, 1)?

[next obsidian]

Like if the constant of g is odd, then any f mapping to g has to necessarily map to an odd number

Yeah that’s my point

If g’s constant germ is 0

Then any f mapping to g has to have f(-2) even

[lethal dune]

can you get it? LHS says f(-2) be even

[next obsidian]

Yeah

[lethal dune]

lol

[next obsidian]

So it’s like

[lethal dune]

I don't think you can actually get the elem

[next obsidian]

{(a_0 + higher order terms,n) | n is odd if a_0 is 1, even if a_0 is 0}

And this is closed under sum and product

[lethal dune]

so why not just take the set itself to be the image set of the map

[next obsidian]

Idk

This ring sucks

[lethal dune]

oh that's what you were trying

[next obsidian]

I think this is correct

[lethal dune]

I happy with im(f)

[next obsidian]

I am not

Because I wrote it explicitly

I think

Okay so

[lethal dune]

[next obsidian]

Suppose you have (g(x),n)

Of the form I claimed

Let f(x) = g(x) where you write g’s coefficients as 0 or 1

Then f(-2) and n necessarily differ by an even number

Because of the condition I specified on n

So write their difference as 2k

Now write k in the binary expansion

Modify f’s coefficients by that

QED

[lethal dune]

2x^2 + 4x gets mapped to 0 tho

lol not injective

[next obsidian]

But this is in 2x + 4

[lethal dune]

oh lol

[next obsidian]

Remember I quotientes by that already

I computed the kernel as (2x + 4)

[lethal dune]

yeah I wasn't looking at the quotient

[next obsidian]

Yeah I mean it doesn’t matter too much

[lethal dune]

I believe it works now

[next obsidian]

Swag

[lethal dune]

go to sleep

[next obsidian]

This ring is shit

TRUE

[hollow mica]

I am reading that if V is a vector space over C with basis b_1, ..., b_n, then b_1, ..., b_n, ib_1, ..., ib_n form the basis of an analogous vector space over R

I don't see how this makes sense, because if one of the b_j are not pure imaginary thehn ib_j won't be an element of R

[chilly ocean]

but they don't need to be

they are elements of V

not R

not scalars

[hollow mica]

ohh

what does it mean for a map f: V x V -> C to be a "hermitian pairing"

[lethal dune]

means f is a Hermitian form

[hollow mica]

oh

[toxic zephyr]

aw shoot nobody pinged me and i missed everything. i am not understanding what @next obsidian was saying here
#groups-rings-fields message

I'm not sure what the codomain would even be

[lethal dune]

tried contacting the image explicitly

[toxic zephyr]

is it just Z?

[lethal dune]

constructing*

[toxic zephyr]

I'm not sure what it's saying in the first place 😦
it seems like it loses the property of the kernel requiring f(-2)=0

or wait is it like...

$f(p(x))=
([p(x)]_2,kp(-2))$ where $k\in{1,2}$ has the same parity as $p(0)$?

[cloud walrus]

nix

[toxic zephyr]

no that doesnt make sense

[wooden ember]

$R[[x]]\cong\prod_{i=0}^\infty R$ as modules right?

[cloud walrus]

𝓛ittle ℕarwhal ✓

[coral spindle]

Yeah

[wooden ember]

coolio

would there be a slightly less ugly example than Z[[x]] when looking for an infinitely generated Z module not isomorphic to some direct sum of cyclic Z modules?

my group theory is rusty so i cant think of a nice abelian group of this form

[coral spindle]

Not really

None that come to mind, at least

[wooden ember]

alright ig it's too much to ask since it's asking for something ugly anyways

[coral spindle]

Oh actually I thought of a much better example, specific to Z

Just Q

Can't believe this didn't come to me immediately

[wooden ember]

oh lmao fair enough

[coral spindle]

It's pretty easy to show that this isn't free

I leave it to you

[wooden ember]

yeah i think i can do that

[coral spindle]

You definitely can

Just a little bit of creativity needed

[wooden ember]

is it smth like

oh wait im a moron

if p_1/q_1 and p_2/q_2 are two elements of Q then q_1p_2(p_1/q_1)-q_2p_1(p_2/q_2) = 0 so it has no basis of size > 1 but clearly Q is not isomorphic to Z

[coral spindle]

Yup

There are no linearly independent elements in Q

hence it cannot be free, since it is not generated by a single element

good stuff

[chilly ocean]

Can't you just do direct sum of Z/Z2

[wooden ember]

if im looking at the cokernel of a linear map whose matrix is not equivalent to a diagonal matrix (so over some non PID ring) can i say that the cokernel is not isomorphic to something of the form given by the structure theorem?

[wooden ember]

well no the point was to not have something that's a direct sum of cyclic Z modules

[chilly ocean]

Oh dear. My bad

[gusty thistle]

I am doing an exercise about a fourth degree irreducible polynomial $f(x)$ over $\mathbb{Z}$ with a Galois group isomorphic to $A_4$. I don't understand how this is possible. Because $f(x)$ is irreducible and thus seperable, from which follows, with $L$ the splitting of $f(x)$, that $|\text{Aut}(L/\mathbb{Q}| = [L:\mathbb{Q}]$. Now $f(x)$ is seperable thus $f(x)$ has four distinct roots, thus $|\text{Aut}(L/\mathbb{Q}| = [L:\mathbb{Q}] = 4$. But $A_4$ has 12 elements. Does anyone know where I am wrong?

[coral spindle]

You have to ask texit explicitly I think, like so

,tex Tex here

[gusty thistle]

thanks

[coral spindle]

Oh, I guess it's just down again

Nvm

[wooden ember]

maybe im dumb but why can you say that f is separable?

nvm im dumb

[gusty thistle]

Aah i see know that [L:Q] can be bigger than four

[wooden ember]

lmao

i was literally writing that

[gusty thistle]

we are both smart now 😎

[wooden ember]

took me way too long to notice tho

[gusty thistle]

me too 😭

[hearty charm]

hmm

[rotund aurora]

Given a field K and an abelian group V, when can we give V a vector space structure over K?

[chilly ocean]

If V isn't trivial then it has to have at least cardinality |K| for us to be able to give any sort of vector space structure to V

[rotund aurora]

sure

[chilly ocean]

So K = R and V = Z works as a counter-example

[rotund aurora]

I said "when", not that you always can

[chilly ocean]

there's a characterization of such groups when K = Q

they are the torsion free, infinite divisible abelian groups

If V is a vector space over K, then it must be in particular a vector space over Q, so it must satisfy above

assuming K has characteristic 0

[rotund aurora]

That's if K is of char 0

yes

I don't think there's anything close to a nice answer in general though, because the infinite divisible condition almost seems redundant

[chilly ocean]

If you take it as an answer, $V$ needs to be isomorphic to $\bigoplus_{i\in I} K$

[cloud walrus]

Blitz

[chilly ocean]

for some set I

[rotund aurora]

yeah right

[chilly ocean]

taken as a direct sum of abelian groups

why I don't think there's a nice answer is because K isn't a fixed field

[rotund aurora]

I was also thinking about the following: Given an abelian group V and a field K, we can turn V into a vector space over K iff we can define a scalar product $K\times V\to V$ satisfying the vector space axioms (definition). Given that we have one such scalar product $\cdot$, then if $\varphi$ is an automorphism of $K$ we can define a new scalar product by the rule $k*v=\varphi(k)\cdot v$.

[cloud walrus]

Croqueta

[rotund aurora]

Then I was wondering if, given that V has a scalar product over K, are all other possible scalar products obtained in this way?

In the case of Q there is only one possible scalar product. I was considering the case of R, since it has no non-trivial automorphisms. Q is dense in R, but idk if that tells you anything

okay so I think definitely not. You can give R a vector space structure over Q(x) by defining f(x)·r =f(pi)·r for all real numbers r and where pi is some transcendental number

But given a vector space V over R, idk if the scalar product is unique

[lethal dune]

@rotund aurora Assuming you have 2 products on V over K, call • and ×, then since (V, •) and (V,×) VS over same field of same dim, they are isomorphism. There is f: (V, •) → (V,×) s.t.
f(k•x)=k×f(x).
f is definitely an automorphism of K (as abelian groups) but also K equivariant which is even stronger

[rotund aurora]

I'm not seeing how f induces an automorphism of K

[formal ermine]

let $H$ be a subgroup of $G$. for $a, b \in G$, why doesn't $aH = bH$ imply $b = a$? my line of reasoning was that if you move b to the lhs you get $b\inv a H = H \implies b\inv a = e$

[cloud walrus]

illuminator3 (#eric4honorable)

[formal ermine]

what's wrong here

[next obsidian]

You can’t cancel H

Try an example

[chilly ocean]

H is not an element of G

[south patrol]

aH =H just means a is in H

[formal ermine]

mhm

ok wait

[formal ermine]

I'm having a hard time coming up with an example

[next obsidian]

Think harder

[formal ermine]

question

if I give you a set like { e, a, a^-1, b, b^-1 }

how do you actually define a meaningful operation on it

[cloud walrus]

MyMathYourMath

MyMathYourMath

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

[next obsidian]

There’s only 2 groups of order 6, one is abelian and the other isn’t

Compute whether or not the holomorph is abelian

[solar glacier]

Thats what I thought too! But how do I show if this semi direct product is abelian or not? Im new to semi direct products.

do I just take two elements and show wether or not they commute? under the given binary operation

[cloud walrus]

MyMathYourMath

[next obsidian]

Yeah

I mean that’s how I’d do it

I think because this is a semi direct product of Z2 and Z3 it might be forced to be the direct product, but this is all just vibes

Or uh

Nah, I take that back

[solar glacier]

lol

[formal ermine]

if G = { a, a^2, a^3, a^4, a^5 = e } and H = G then aH = { a, a^2, a^3, a^4, e } = H but a \neq e

is that a correct counter example

[next obsidian]

Sure that works

[wooden ember]

i doubt that's true because Aut(Z_3) acts non trivially on Z_3

[formal ermine]

epic

[wooden ember]

but i feel like you might be able to show it's not a direct product in that way and then it's necessarily S_3

[next obsidian]

Yeah that’s what I was saying later

I was misremembering my thing

[wooden ember]

oh oop

i shouldve read the whole thing

[next obsidian]

For the classification of groups of order pq

When p | q-1 there’s a thing where all the maps which are nontrivial differ by an automorphism

So all the nontrivial semi direct products are isomorphic

I was mixing that up with the statement all semi direct products are isomorphic

[formal ermine]

if I want to show that $\varphi : gH \mapsto Hg\inv$ is a well defined homomorphism don't I only have to show that $\varphi(aH \circ bH) = \varphi(aH)\varphi(bH)$? or do I have to show that if $aH = bH \implies \varphi(aH) = \varphi(bH)$ too?

[wooden ember]

right yeah

[cloud walrus]

illuminator3 (#eric4honorable)

[wooden ember]

you also have to show that second part yes

[solar glacier]

can someone help me with this

and show their produtcs are distinct when comptuing using the binary operation on semi direct product

[cloud walrus]

MyMathYourMath

MyMathYourMath

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

[solar glacier]

does phi*** map

[formal ermine]

no idea how to do the second part, been trying to figure it out for the last hour, any hint is appreciated

[wooden ember]

It’s what I said earlier, the automorphism group of Z_3 acts non trivially on Z_3 so you can’t get the direct product Z_3 x Z_2. The only other group of order 6 is then S_3

You could also build it explicitly from an internal direct product in S_3

Namely take the normal subgroup generated by (123) and the subgroup generated by (12) and see that action by conjugation of the second on the first is the only possible non trivial action, which is the same as given by the group of automorphisms of the first group

So you get an isomorphism with that internal direct product which has to be equal to S_3 since it’s a subgroup of order 6 of S_3

[solar glacier]

that makes sense

[wooden ember]

In general Holomorph isnt a direct product unless on Z/2Z I’m pretty sure

[lapis trail]

I need help finding the number of conjugacy classes of group homomorphisms from S_n - > D_m and from D_m -> S_n

[ember field]

is the set of split extensions in bijection with the set of possible semidirect products of H and G ?

[coral spindle]

Yeah that's right

Split SESes of groups correspond to semidirect products

Well actually I don't think the bijection is quite perfect

Extensions being isomorphic is a little stronger than the semidirect products themselves being isomorphic

However, although I would need to check this, I think that they are the same up to extension isomorphism if you equip H \rtimes G with the appropriate inclusion/projection maps

[hollow mica]

Wait I was thinking about this some more and I don't see why this is sufficient

If B(s_1, t_1) \neq 0 and B(s_2, t_2) \neq 0, nothing goes wrong if t_2 = λt_1 .

Each B(s_i, -) could be nonzero on the same one-dimensional subspace, is another way to express what I'm saying

Holy shit

I just figured it out

and it's a very satisfying argument

[chilly ocean]

maybe a dumb question but how come for a ring homomorphism f we construct ker(f) = { a | f(a) = 0} but have no construction for the set { a | f(a) = 1}

like this seems to be a multiplicative group i guess

[hollow mica]

multiplicative inverses need not exist in a ring

[chilly ocean]

Ahhh

[hollow mica]

wait actually i misinterpreted your question

i'm not sure what you're asking

[chilly ocean]

i was wondering why associated to a ring homomorphism we dont construct the set of elements which get sent to the multiplicative identity but what you said answers it

it doesnt form a group under multiplication because multiplicative inverses dont necessarily exist in the original ring (although when they exist theyre in the set)

so the set doesnt have any structure which would explain why that construction doesnt seem to appear anywhere

[hollow mica]

oh yeah

inverse of the additive identity will be an ideal

inverse of the multiplicative identity will be a monoid at best

preimage I should say

[chilly ocean]

i wonder if this is fixed by taking the elements of the units which get sent to the multiplicative identity

then it looks to form a multiplicative group

like f: R -> R' ring homomorphism and let A = { a \in R^x | f(a) = 1}

[hollow mica]

sure

you're guaranteed to have multiplicative inverses if you just grab the units

[chilly ocean]

a little odd that ideals have so many useful properties but this which looks very similar probably does not

[hollow mica]

yeah

If G, H are two subgroups of a group K, what is meant by GH and G x H ?

I came across the statement "If G ∩ H = {1} and G and H are normal, then GH is isomorphic to G x H"

[chilly ocean]

GH is the set of products {gh} with g in G and h in H

[hollow mica]

so it has no structure?

[chilly ocean]

not necessarily

for it to be a group one of G or H must be normal

however the direct product G x H = { (g , h) | g in G , h in H} is always a group

[hollow mica]

ah I see

[chilly ocean]

where you multiply componentwise, (g,h)(g',h') = (gg', hh')

[hollow mica]

so if A and B are normal, then you can think of AB ≅ A x B as their direct sum

direct sum in the vector space sense

[chilly ocean]

well they have to have trivial intersection

[hollow mica]

oh yeah and that too

[chilly ocean]

but yeah this is how i think about it as like an analogue to direct sums of linearly independent subspaces

[chilly ocean]

only one being normal is good enough

[hollow mica]

I see

Right now I'm trying to prove that if a group G has two normal subgroups A, B of orders 3, 5, respectively, then G contains an element of order 15

I thought the construction AB may be useful since |AB| <= 15

[chilly ocean]

this theorem is the way to go i think

since 3 and 5 are prime, and all prime order groups are cyclic A and B are cyclic. So if you can show the subgroup AB is isomorphic to A x B then since A and B are cyclic so is A x B and it is of order 15 so it has an element of order 15 ...

wait nvm

direct product of cyclic groups is not cyclic

[hollow mica]

this is relevant I think

[chilly ocean]

direct product of cyclic groups is cyclic iff their orders are coprime so looks like the argument works actually

[hollow mica]

oh nice

wait @chilly ocean

I found a proof of why A, B normal means that AB ≅ A x B

but I don't see where they used normality

[chilly ocean]

For AB to be a group

In order for it to be closed under multiplication

[hollow mica]

ohh

[chilly ocean]

Like for aba’b’ to be in AB you’ve gotta write as a product of A times product of B

And closure under conjugating can do this

[hollow mica]

I see

Wait don't we need A and B to have trivial intersection to have AB ≅ A x B

[hollow mica]

This proof relies on it, at least

@chilly ocean

[chilly ocean]

yes

[hollow mica]

oh, of course their intersection is trivial

because of orders

[chilly ocean]

i mean otherwise A = AA is clearly not isomorphic to A x A

yeah if A is finite then |AA| = |A| but |Ax A| = |A|^2

[hollow mica]

well I was just saying

every element in A has order 3 (except identity)

same with B except 5

so done

[chilly ocean]

so youre saying if x in A has order 3 and y in B has order 5 then xy in AB has order 15?

[hollow mica]

no I was arguing why A and B have trivial intersection

[chilly ocean]

ahh yes

[hollow mica]

wait could I also just say that C_3 and C_5 have trivial intersection

or does intersection not carry over isomorphisms @chilly ocean

[chilly ocean]

yes! in fact it carries over for all bijections: f(A intersect B) = f(A) intersect f(B)

[hollow mica]

wait I'm tripping

[chilly ocean]

injectivity is even good enough

[hollow mica]

C_3 ∩ C_5 = C_3, no?

if you view them as subgroups of C_n for n > 5

wait no that's not true

yeah i don't think it makes sense to talk about C_3 ∩ C_5 if you don't establish what they are subgroups of

[chilly ocean]

yeah, it still remains that if you have two subgroups of order 3 and 5 of a group that they intersect trivially. Let x be an element of the intersection, then since it is an element of a group of order 3 then the order of x divides 3. similarly the order of x divides 5. but then the order of x is one, so x = e

[hollow mica]

No what I'm saying is the object A ∩ B doesn't make sense if A and B are not subgroups of some other group G

[chilly ocean]

thats true

because it otherwise would have no reason to be a group

[hollow mica]

like sure, if you want to show A and B have trivial intersection, and you find two subgroups C_1 and C_2 of the same group C such that A ≅ C_1 and B ≅ C_2, then showing they have trivial intersection is equivalent

but idt it applies here

[chilly ocean]

i think the most direct way of showing that A with |A| = 3 and B with |B| = 5 intersect intersect trivially is by the divisibility argument above

[hollow mica]

yeah

[hollow mica]

@chilly ocean apparently there's a hint I didn't know about to that one sup/inf eigenvalue problem we worked through earlier

If V is a finite-dimensional vector space over the reals, and T: V -> V is a self-adjoint map, and v in V is a vector satisfying |v| = 1 and <Tv, v> \geq <Tw, w> for every w in V satisfying |w| = 1, then v is an eigenvector of T.

I'm not sure how to prove this hint though

This time we can use the spectral theorem though

Okay there's a hint to prove this hint lol:

---

Let f(w) = 1/||w||^2 <Tw, w> and consider f(v + tu) for <u, v> = 0 and t in R.

[hollow mica]

well from the spectral theorem i can say that such u, v are guaranteed to exist

ok I calculated
let k_t = 1/||v+tu||^2 for fixed v, u
then
f(v + tu) = k_t <T(v + tu), v + tu> = k_t <Tv + tT(u), v + tu> = k_t (<Tv, v> + 2t<Tv, u> + t^2<Tu, u>)

[south patrol]

This is actually a way to prove the spectral theorem which is cute

[formal ermine]

what does $x \equiv^l y \pmod{H}$ mean

[cloud walrus]

illuminator3 👻(#eric4honorable)

[hollow mica]

How so

[south patrol]

You can put a topology on V using your inner product and show that the map v -> <Tv,v> attains a max

[hollow mica]

oh how to prove the spectral theorem

the problem itself is pretty hard

wait can we even claim the existence of some nonzero u, v such that <u, v> = 0 without the spectral theorem?

[chilly radish]

Yes, as long as your space is of dim >1

For example, just apply gram-schmidt to 2 linearly independent vectors

[hollow mica]

I see

[hollow mica]

If V is a finite-dimensional vector space over a field k, and B: V x V -> k is a symmetric bilinear form, then there exists a nonzero v in V such that B(v, v) = 0

any hints

[coral spindle]

Rank-nullity theorem

That’s my guess at least

[formal ermine]

bump

[coral spindle]

No idea illum.

[formal ermine]

rip

[hollow mica]

yesterday rank-nullity came in handy because the bilinear form was non-degenerate