groups-rings-fields

[chilly ocean]

okayy

that's true

we only need one root

but uhm

we'll need to show that the extansion contains all roots

Right?

Or am I missing something?

[oblique river]

Didnt we just do that?

^

[chilly ocean]

then what about this

[oblique river]

As a linear combination of 1 and zeta

[chilly ocean]

yes

[oblique river]

I was explaining why the detree wasnt 2

[chilly ocean]

hmm

[oblique river]

{1, zeta} is not a basis

[chilly ocean]

right right

yes understood

[oblique river]

But {1, zeta, zeta^2, zeta^3} is

[chilly ocean]

what is zeta

[oblique river]

You tell me

You used it before i did haha

[chilly ocean]

this is zeta

Q[zeta]

[oblique river]

Yea

[chilly ocean]

okay okay

[chilly ocean]

we can all roots from this

Right?

[oblique river]

Yes

[chilly ocean]

Understood

[oblique river]

For any irred poly of degree d with root alpha, the set {1, alpha, alpha^2, …, alpha^(d-1)} is a basis for Q(alpha)

Regardless if that field contains the other roots of the poly or not

[chilly ocean]

hm

but It should contain other roots to be splitting field

[oblique river]

Yes. In most cases, adjoining a single root isnt enough to get all of the roots

But in this case, with x^4 + 1, it is

Because the other three roots are just powers of the first one

[chilly ocean]

no

are they?

[oblique river]

Yes

The roots are zeta, zeta^3, zeta^5, and zeta^7

[chilly ocean]

Right

[oblique river]

There are two separate facts here which have separate and unrelated proofs. The first is that Q(zeta) has degree 4. The second is that Q(zeta) contains all the other roots of x^4 + 1

There are multiple ways to prove the second fact, several of which have been given here already.

[chilly ocean]

okay

[chilly ocean]

and how to get this

I mean how do you know that?

[oblique river]

Thats how roots of unity work

If zeta is an nth root of unity then so are all powers of zeta

Or just do direct computation

Zeta satisfies the poly x^4 + 1 = 0. Show that those three powers also satisfy that polynomial, and then show that they are all distinct

Or even more direct computation

[chilly ocean]

hmm okay

[oblique river]

You already know that the four roots are (1/sqrt2)(+- 1 +- i)

So pick one and start raising it to@powers

And youll get the rest

[chilly ocean]

yes

i've done it

i do get it

then i think all four

have same argument

[oblique river]

Yes, all are related to roots of unity.

[chilly ocean]

okay

Thank You

[oblique river]

Np and gl

[chilly ocean]

but wait

but zeta^2 will be in Q[zeta]

as zeta into zeta should be in field

[oblique river]

Yes

Is that a problem?

[chilly ocean]

so what will be the splitting field

[oblique river]

Q(zeta)

[chilly ocean]

uhmm okay okay

sorry that's right

but it will requre 4 basis elements

[oblique river]

Yes, it has degree 4

[chilly ocean]

Okay now it is more clear

😅

[boreal egret]

Can anyone help me solve this question? I think I am halfway there

[lethal dune]

Remember that $\tau (12)(34) \tau^{-1}=(\tau(1) \tau(2))(\tau(3)\tau(4))$

[chilly ocean]

need help with 1)

I don't know how to show closure

I know I have to show closure, commutativity, associativity, existence of inverses and identity.

[oblique river]

what does it mean for it to be closed

[chilly ocean]

a, b in G => ab in G

[oblique river]

ok, so what does that mean in this context

what does being in G entail

[chilly ocean]

being in G entails being an an element that is coprime to n in Z_n

so the "product" of two coprime elements to n is coprime to n (?)

[oblique river]

yes, that's what you need to show -- not sure why you have scare quotes around the word "product"

if a and b are each coprime to n, show that ab is coprime to n

[chilly ocean]

should I use bezout's lemma

gcd$(a,b) = ax + by$?

[cloud walrus]

n!

[chilly ocean]

I have an idea

but I'm not sure if it makes sense

[oblique river]

well, it does say right below the question

to use that gcd(m,n) = 1 iff you can write 1 = mx + ny

[chilly ocean]

okay so

suppose we have $\bZ_n$

[cloud walrus]

n!

[chilly ocean]

and $a, b \in \mbb{U}_n$ such that gcd$(a, n) = 1$ and gcd$(b, n) = 1$. That is, $$1 = ax + ny, 1 = au + nw$$

[cloud walrus]

n!

[chilly ocean]

uh

what do I do from now on

[oblique river]

what was your idea?

[chilly ocean]

well

my idea was to multiply

[lapis plume]

I am stuck at part b of the problem. I guess I can use part a when I can find distinct m and n, but idk why there exists such distinct m,n satisfying a^m=a^n.

[chilly ocean]

these two equations

but I think that's stupid

[oblique river]

well, if you can't think of anything else to do, you might as well try that

like, at least try something before you give up all hope

[chilly ocean]

$1 = (ax+ny)(au + nw) = a^2 x u + axnw + nyau + n^2 yw$

[cloud walrus]

n!

[lapis plume]

Not sure is it due to G is finite. This is the only property I haven’t used.

[chilly ocean]

no that's not gonna work

[oblique river]

you're missing a b somewhere

i think you mean (bu + nw)

the first factor comes from (a,n) = 1, so your second factor should come from (b,n) = 1

[chilly ocean]

gcd(a,n) = 1 and gcd(b, n) = 1

so

ax + ny = 1

bz + nw = 1

(ax + ny)(bz + nw) = 1

axbz + axnw + nybz + n^2yw = 1

hm

[oblique river]

is that of the form abX + nY = 1?

[chilly ocean]

it is

a(xbz + xnw) + n(ybz + nyw) = 1

[oblique river]

that's... not what we want

[chilly ocean]

pain

[oblique river]

that just shows that gcd(a,n) = 1

again

[oblique river]

this is what we need

we need to show gcd(ab, n) = 1

[chilly ocean]

oh

OH

ab(xz) + n(axw + ybz + nyw) = 1

ez

[oblique river]

sometimes you just have to do something to solve a problem. you can't just stare blankly at it

its ok if you dont know how the proof is going to end when you get started

if you have an idea, just try it

[chilly ocean]

does this make sense

[oblique river]

yes

[chilly ocean]

W rizz

thank you sir

so as for the other axioms

commutativity and associativity are present due to the fact that this is multiplication

what about the identity and inverses

[oblique river]

what about them? you tell me

[chilly ocean]

uh

so

not sure how to show their existence

[oblique river]

what is the identity for multiplication

[chilly ocean]

1

[oblique river]

okay

is that in U_n

[chilly ocean]

gcd(1, n) = 1

so yes

it is

right?

[oblique river]

yes

[chilly ocean]

ok

inverses

uh

thats the tricky part

[oblique river]

break down what you need to prove: if gcd(a,n) = 1 then there exists b such that ab = 1 mod n (and gcd(b,n) is also 1)

[chilly ocean]

(ab - 1)/n = k

ab = nk + 1

[oblique river]

well, you want to prove that such a number b exists

[chilly ocean]

can I do it in a different way

I haven't got a clue how to show it like this

[oblique river]

like what?

you have to prove that if a in U_n then there exists b in U_n such that ab = 1 (mod n)

that is, if gcd(a,n) = 1, you need to find some number b which also has gcd(b,n) = 1, and for which ab = 1 mod n

[chilly ocean]

ok so if a in U_n then gcd(a, n) = 1

[oblique river]

you can't assume that you have b

[chilly ocean]

cant i just multiply

hm

[oblique river]

multiply what

[chilly ocean]

why do I need to show this

[oblique river]

that's what it means for a to have an inverse

like, that's the definition of inverse

[chilly ocean]

hmm

ok

I will try

thank you @oblique river

[oblique river]

np and gl

[south patrol]

Galois correspondence probably easiest

[rustic crown]

you could use some gauss sums, you have some g in Q(zeta_p) such that g^2 = (-1/p) * p

(that parenthesis is the legendre symbol)

from that you easily see that sqrt(-3) and sqrt(5) lie in your field

[south patrol]

If this is like my course you probably won't have that technology at hand lol

[oblique river]

"you can solve this problem by simply knowing how to find the answer ahead of time"

[south patrol]

Lol

[rustic crown]

maybe working separately with Q(zeta_3) and Q(zeta_5) would be a little less messier

[oblique river]

yes i do agree with det's suggestion

to work with zeta_3 and zeta_5 separately

even weaker than that is that Q(z15) contains both Q(z3) and Q(z5)

and each of those has a unique quadratic subfield Q(sqrt(a)) and Q(sqrt(b)) (by galois theory)

so Q(z15) contains both of those and therefore also Q(sqrt(ab))

you don't actually need that Q(z15) = Q(z3, z5), just that Q(z15) contains the right side

just work with them one at a time

instead of looking for quadratic subfields of Q(z15)

look for quadratic subfields of Q(z5)

and Q(z3) is itself already quadratic

[pastel cliff]

det

[oblique river]

tha'ts not waht quadratic subfield means

it means quadratic over the base field

[south patrol]

Oop

[rustic crown]

oopsie

[south patrol]

Gg

[oblique river]

you want index 2 subgroups

yeah, there should be 3

well, again, like det suggested

it would be a little less messy to start by looking for quadratic subfields of Q(z5)

because Q(z5) \subseteq Q(z15), so a quadratic subfield of Q(z5) is also a quadratic subfield of Q(z15)

well, with any luck, we wont have to work with those subgroups at all

we know by galois theory that there are definitely three quadratic subfields. so, if we can find them by any means, we konw that they have to be all of them

that's right

to be clear -- you could go further and actually use the subgroups as you suggested

and tried to figure out what theyw ere directly by calculating their fixed fields

and that would lead you to the correct answers as well

but, in this case, it's just easier to work with simpler fields

that's right

just because like, writing down something like z15 + z15^4 + z15^7 + z15^13

(which is fixed under your second subgroup)

and trying to figure out what that's equal to

looks like it could get a little messy

ok det you can take over again cuz i gotta go

:^)

there is only one

but once you have Q(sqrt(a)) and Q(sqrt(b))

you also have Q(sqrt(ab))

[pastel cliff]

posting here bc #category-theory seems to get way less attention

this counts as #groups-rings-fields bc i need to show products exist in Ring

[next obsidian]

I bet you’d see how to do it if u deleted your messages in Chmonkey thread

[barren sierra]

it was answered in #category-theory 😎

[formal ermine]

why does the existence of one G invariant subspace imply the existence of another one

[next obsidian]

Look up Maschke’s theorem

[formal ermine]

oh lmao

maschke's comes right after this

so irreducible reps are like prime elements basically?

[delicate orchid]

Uhhh cause as one move other stay still so all stay still but in half

[pastel cliff]

chmonkey no

it doesnt have to be like this

[formal ermine]

I don't understand

[next obsidian]

It does

[pastel cliff]

https://tenor.com/view/crying-crying-meme-gif-26327776

[delicate orchid]

Maschke’s theorem

[formal ermine]

.

[delicate orchid]

Yup

Or as I like to call it, baby’s first blocks

[formal ermine]

rep theory is nice but sometimes it's too much linear algebra for my taste

[next obsidian]

I love doing rep theory for F_p^n and thinking about representations of p-groups

How does that make u feel wew

Do you enjoy that?

Does this bring you joy?

[delicate orchid]

Quite

[next obsidian]

You enjoy not being able to write down 1/|G|?

You like not having Maschke’s theorem?

[prisma ibex]

I am literally suffering this pain right now

[next obsidian]

Shouldn’t have done langlands

[delicate orchid]

Oh wait no it doesn’t because it turns out unlike the brauer ring the F-stable character ring isn’t free so I am in for a world of pain

[prisma ibex]

THIS IS D-MODULES STUFF

I HATE POSITIVE CHARACTERISTIC

[next obsidian]

Didn’t ur mother ever tell you to stay away from the scary men with their representations of Galois groups?

[pastel cliff]

[next obsidian]

I hate you

[delicate orchid]

Omgggg literally just like… compute it pmggggg

[delicate orchid]

Galois groups
I don’t care

[pastel cliff]

the second i delete my message i become a chmonkey dick rider

[next obsidian]

Me when I’m in char p and do ^p

[next obsidian]

Hate you

[next obsidian]

Literally every finite group is a Galois group

[delicate orchid]

Don’t caare

[next obsidian]

It’s not proven yet

[delicate orchid]

unaware

Awww ruin the fun

[formal ermine]

it's easy to show over any arbitrary field

but over Q it's unknown

[delicate orchid]

I know

[rotund aurora]

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

[formal ermine]

like if you take any field K and L = K(x_1, ..., x_n) then Gal(L/L^S_n) = S_n

then you just apply cayley

[delicate orchid]

So u can just apply cayley

Becomes more and more surprising it’s unsolved to me

[rotund aurora]

what are you trying to say? xd

the fixed field will usually be stupid as fuck

in F<F(x1,...,xn)

and its likely to contain algebraic elements not be purely transcendental

[formal ermine]

any G --> S_n so if Gal(L/K) = S_n then Gal(L/L^G) = G

[oblique river]

that's just how fields work. If Q(sqrt(a)) is a subfield of K and Q(sqrt(b)) is another subfield of K, that means K contains sqrt(a) and sqrt(b). therefore it also contains their product, whcih is sqrt(ab)

and hence it contains the quadratic subfield Q(sqrt(ab))

not quite, no

notice that in what I wrote, i specifically took generators of the quadratic fields of the form sqrt(a)

z3 isn't of the form sqrt(a) for an integer a

so therefore the third one would be Q(sqrt(-15))

pop quiz for you:

we've now shown that Q(z15) contains both Q(sqrt(5)) and Q(sqrt(-15)). So, by the reasoning above, shouldn't it also contain Q(sqrt(5 * -15)) = Q(sqrt(-75))?

but isn't that too many subfields?

(what error have i made in my reasoning)

@willow geyser

that is related to the answer, but that in and of itself isn't the issue

meaning, the error in the reasoning is not "the sqrt(a) , sqrt(b) ==> sqrt(ab) fact only works for coprime a, b"

(it's true for any a and b)

[pliant forge]

For b \in B where B is an ordered multiplicative basis, and v_i's are in a set of orthogonal idempotents such that 1 = \sum_{i=1}^n v_i.
Why is it that only one v_j from the orthogonal set multiplied from the right gives b?

[oblique river]

it is true that Q(z15) does have Q(sqrt(-75)) as a subfield

yes, that's right!

but which one

not quite, no

the "Q(sqrt(a)), Q(sqrt(b)) => Q(sqrt(ab))" process will always give you a new field

(assuming that sqrt(a) and sqrt(b) aren't in Q to begin with)

so in fact Q(sqrt(-75)) = Q(sqrt(-3))

indeed, sqrt(-75) = sqrt(-3 * 25) = 5sqrt(-3)

so adjoining sqrt(-75) is algebraically the same thing as adjoining sqrt(-3)

;P

so that's kind of where the "not coprime" thing comes into play

both 5 and -15 have a factor of 5

so their product has a factor of 5^2

[pliant forge]

I suppose since any basis element can be written as a product of two other basis elements, which is the case here.

[somber sleet]

I wanted to thank you for the nice explanation, I got it now. Sorry for not answering earlier 😦

[formal ermine]

"degree 6 over Q"

[oblique river]

you can also answer this from context. the problem gives you an example -- Q(z7). that field does indeed have degree 6 over Q, but since [Q(z21):Q] = 12, it means that [Q(z21):Q(z7)] = 2

so by context we can assume that it is indeed asking for subfields of degree 6 over Q, not subfields of relative degree 6

(but even without the context, illuminator3 is right -- the degree of a subfield means the degree of teh subfield over the base field. otherwise, i think you would say "relative degree" or something like that)

[fringe heath]

could i get some help with this please 😓
my friend tried to help me before but i didnt rly understand and i felt bad asking her repeatedly
i know i probably have to use tower law in some way but its not rly clicking

sorry to interrupt; its just that the channel seemed to be busy for the past few hours

[oblique river]

yes

index 6 subgroups will give subfields of degree 6

[oblique river]

let M = K[x]/f(x) = K(alpha) where alpha is a root of f

What can you say about [M : K]? what about [LM : K]?

think about [LM : K] in two different ways, once using the tower K < L < LM and once using the tower K < M < LM

[fringe heath]

oki ill give it a shot and come back

[rustic crown]

do you still have this question? sorry i was away >.<

[rustic crown]

[rustic crown]

okie :3

[void cosmos]

Sf is symmetric proof: Sf = sum(f) over all permutations on S_k (k is dim) , phi(S(f)) = S(phi(f)) = sum(f_phi) over all permutations on S_k = sum(f) over all permutations on S_k as just this phi will be composed with itself giving out another permutation as S_k is a group
is that correct

[rustic crown]

looks correct

[void cosmos]

ty ty det

i am going to have to ask about determinants

soon

[rustic crown]

only weird part is that you need to verify that this gives a left action

[void cosmos]

this is given

in the text

[rustic crown]

sometimes this could be weird since you're defining it in terms of indices

[void cosmos]

yea i get u

good catch

but arent i going ot have 2 * S(f)

or no not 2 *S (f)

i meant like one will bee repeeating

[rustic crown]

$\phi \cdot S(f) = \phi\cdot \sum_{\sigma} \sigma \cdot f = \sum_{\sigma} (\phi\cdot \sigma) \cdot f = S(f)$

[cloud walrus]

det

[void cosmos]

yea thats what i meant

but i mean like

[rustic crown]

so what's repeating?

[void cosmos]

phi . sigma is some other permutation

call it tao

but tao is in this sum

right?

[rustic crown]

right, but everything gets permuted, this tau goes to phi . tau

[void cosmos]

lmfao yea

im so stupid

yea yea

got it

tysm

[rustic crown]

[past temple]

this is an incredibly basic and kind of stupid questin but

if G is a group and g is in G

is gG = G?

[next obsidian]

Yes

Try to prove it

[past temple]

ok if thats the case im running into a major conceptual roadblock

because if you have some representation V of a finite group G

then u can take any vector v

consider its orbit under G

add all of those up

then the span of that forms a G-invariant subspace, making V reducible

[coral spindle]

What if that subspace is just 0

(which it typically is, btw)

[past temple]

o h

i see i see

[coral spindle]

funny, this same thing has happened twice today

[past temple]

ppl asking this same qusetion?

[coral spindle]

Someone else forgetting that this same sum can be 0

[past temple]

💀

generally speaking though, is taking sums of the elements in the orbit of a particular vector by a subgroup a standard trick for producing subrepresentations?

[coral spindle]

Considering it will only produce the trivial representation, no.

That's super boring.

[past temple]

by a subgroup

not the whole group

[coral spindle]

Right. I don't think I've seen it but it would work, I guess

[past temple]

particularly if this subgroup is cyclic

[toxic zephyr]

for an automorphism $\func{\phi}{A}{A}$ on an abelian group $A$, does it make sense to write something like $p(\phi)$ where $p(x)\in\bZ[x]$? and $p(\phi)$ will still be a homomorphism, but won't necessarily be an automorphism itself, correct? or am i completely wrong...

[cloud walrus]

nilpotent nix

[chilly ocean]

makes sense to me

if p(x) is zero then you definitely won't get an automorphism

[coral spindle]

I'd understand this notation, as long as you had a little explanation beforehand

[toxic zephyr]

great thank you both sm! @chilly ocean @coral spindle

[coral spindle]

Should be noted that you've discovered that endomorphisms on Abelian groups determine Z[x]-modules :)

[delicate bloom]

idk, throwing ideas out there: check when T=0 that it's true, then prove the derivative is 0, maybe that's easier. then take care of the special case when 0 is a root or something.

another idea: factor out f(T) and integrate, get something like log lol use log rules --> ???

[toxic zephyr]

can someone please explain how these definitions are equivalent? it's not obvious to me, and i can't find anything about metabelian groups in the textbooks i have... is it possibly under a different name? or does anyone know of any textbooks that do cover it?

[tribal moss]

=> should be obvious: take A to be the commutator subgroup.

For <=: If G/A is abelian, then A must contain every commutator -- and so the commutator subgroup is a subgroup of A and is therefore abelian.

[toxic zephyr]

sorry could you explain why G/A being abelian implies that A must have every commutator?

[tribal moss]

If f is the projection from G to G/A and G/A is abelian, then f(xyx'y') = f(x)f(y)f(x')f(y') = 1, and therefore xyx'y' is in the kernel of f -- that is, A.

[toxic zephyr]

ah that makes sense

thank you @tribal moss!

[agile burrow]

oh yeah, so I never got an answer to this question even after asking people irl. But my best guess is that PSL(2, R) is perfect, so we can consider it's universal central extension, and B_3 arises as the group which surjects onto PSL(2, Z) under this extension

[white oxide]

does associativity only hold for triples in G? for example is it not always true that (a * b) * (a * b) = (a * b) * (b * a)

[tender wharf]

you just commuted it

groups must be associative by definition

[tender wharf]

this will hold if G is abelian ofc

associative means (ab)c = a(bc)

[white oxide]

so say we had a group G with identity e such that every element x in G is its own inverse, i.e. x * x = e for all x in G. then x = a * b for some arbitrary a, b in G. would the following line of logic be valid:

x * (a * b) = e, so x * a * b = e --> a * x * b --> a * a * b * b = e = a * b * a * b, so following the left and right cancellation laws, we see that a * b = b * a so G is abelian?

[next obsidian]

What’s supposed to be in the middle of your chain of implication

[tender wharf]

x * a * b = e --> a * x * b this I am not convinced by

[next obsidian]

You only wrote axb

It isn’t a statement, it’s just an element

[tender wharf]

even if xab=e => axb=e still not convinced tbh

[next obsidian]

There’s no =

[white oxide]

oh wait im actually stupid i didnt use associativity

[next obsidian]

It’s just a single element

[tender wharf]

Yeah even if it was stating it was =

[next obsidian]

You’ve been using it by just not writing parentheses

[white oxide]

yeah just not explicitly is what i meant

[next obsidian]

Anyway

Your last thing

Your chain of 2 equalities

Is all you need, I don’t know what the point of the middle thing in your chain of implications is

you know abab=e because x is its own invwrse

[white oxide]

yeah ur right

[next obsidian]

you also know e = aabb

[white oxide]

oh right

because every element is it's own inverse

[next obsidian]

Well I mean

[white oxide]

and ee = e

[next obsidian]

You wrote it down even

[next obsidian]

Like in this block of text you wrote it haha

[white oxide]

but would you able to get rid of the parentheses to get aabb = abab

[white oxide]

yea i know g

[next obsidian]

I mean both are e

There’s nothing about parentheses, you’re just using transitivity of =

[white oxide]

what does that mean

[next obsidian]

If a = b and b = c then a = c

[white oxide]

sorry im confused how does that apply here?

[next obsidian]

You know that aabb = e

And e = abab

So you get aabb = abab

There’s nothing about parentheses here

[white oxide]

oh so ur saying that it's okay to go from (a * b) * (a * b) to a * b * a * b

?

[next obsidian]

I mean if you want to explicitly write down associativity you have to worry about parentheses or whatever

abab is meaningless if your operation isn’t associative

But it is, so it’s just equal to anything you can possibly write down with any parentheses

[white oxide]

oh okay so it just follows from associativity

[next obsidian]

I mean sure

If you want to go write all the parentheses you can work out where you swap them

But it’ll work out

I refuse to think about it because it’s annoying lol

[white oxide]

so associativity is not only for triples

because in the definition or rather associativity they state that a(bc) = (ab)c

but that's only with three elements

[next obsidian]

I mean, you can just iteratively do this

If you are confused by this maybe it’s best for you to write the parentheses down and figure it out

[white oxide]

so say i'm trying to prove the following:

Show that if $(a * b)^2 = a^2 * b^2$ for $a$ and $b$ in a group $G$, then $a * b = b * a$

[cloud walrus]

okeyokay

[white oxide]

would this be an acceptable proof?:

Since $(a * b)^2 = a^2 * b^2$, it follows that $a * b * a * b = a * a * b * b$, so the left and right cancellation laws imply that $a * b = b * a$

[cloud walrus]

okeyokay

[plain bronze]

hi, any suggested readings to better understand how to factor a polynomial over pretty small (non-prime) fields?

[delicate bloom]

basically but I would probably not say cancellation laws, I'd say multiply on the left by a^-1 and on the right by b^-1

[white oxide]

okay they just call them cancellation laws in the book

but that makes sense

[plain bronze]

i see some things on factoring x^n - x (n is a power of a prime) over a field but not sure how to generalize this to other polynomials

[delicate bloom]

sounds redundant since a group has inverses in it by definition, how do they describe the group axioms?

[white oxide]

yea ur right

one sec lemme pull it up

[delicate bloom]

hmm looks pretty usual I guess

when I think of cancelling I think more like an integral domain or something

idk doesn't matter I guess

[white oxide]

it is a first course in abstract algebra type textbook so it is pretty gentle

the problems are kind of sloppy though in my opinion

for instance, when checking the existence of an inverse in this problem you must assume that given any a in G that a is nonzero for its inverse is 1/a

unless i'm mistaken

but that's not mentioned in the problem

[tender wharf]

hold on

isn't the notation specifically a^-1

to avoid this issue

[white oxide]

no clue

i thought it would be not enough to state that the inverse of any a in G is a^-1

but even still that's equal to 1/a

literally speaking of course

in the definition of a^-1 it's never specified that a is nonzero

or any inverse for that matter

[delicate bloom]

depends on what you mean by zero

[white oxide]

wdym

[delicate bloom]

if the group is the set {0} with the binary operation + then the inverse of 0 is 0

if you're saying 0 is something that has the property that x*0=0 for all x, then it forces it to be a group with one element

because in a group you always have inverses

[white oxide]

im so confused then

[white oxide]

so what would be an inverse for any a in G here?

you can't use the fact that G is a group because we haven't proven that yet

[delicate bloom]

it's a group because they say it is in the first line

[white oxide]

well they say show that it's a group

so we're proving that it's a group right

otherwise the proof would just be like

<G, *> is a group because it's stated that it's a group

[delicate bloom]

ok we were being ambiguous

I was talking about <G, .> not <G, *>

[white oxide]

oh sorry

i meant <G, *>

ah

so by definition it has an inverse

[delicate bloom]

well there's some facts about all homomorphisms

they always send identity to identity and inverses to inverses

so probably worth proving those if you haven't already

[white oxide]

oh yeah haven't proven those yet

or covered them yet

[delicate bloom]

although idk gotta scroll up and see if we know it's a homomorphism or not

[white oxide]

so the identity for <G, *> is 1 right

so we need to produce an element b such that a * b = 1

or in other words ba = 1

[delicate bloom]

if you're calling the identity of <G, .> 1, sure

[white oxide]

wdym

i was talking about <G, *>

[delicate bloom]

the identity is arbirary notation, sometimes people use e or 0 or 1

depending on the operation

[white oxide]

wait so im so confused

it would be sufficient to just say the identity is e?

[delicate bloom]

maybe I'm actually thinking you meant more than you did here

[white oxide]

because to prove <G, *> is a group i proved associativity

now i need to produce an identity and 1 works

because a * 1 = 1(a) = a

for all a in G

[delicate bloom]

you have a*b = ba so if you plug in b=1 you get that a maps to a for all a, so the identity will trickle through as itself as well

a*1 = a

that's the same 1 that gives us a.1 = a

[white oxide]

oh okay

wait im confused

are you trying to get to the existence of an inverse

[delicate bloom]

I gotta go to sleep, keep playing around, someone else might help, good luck

[tender wharf]

okeyokay

what are you confused about specifically

[white oxide]

about an inverse to produce

[tender wharf]

right

[white oxide]

we are proving an existence statement

in other words that an inverse exists

[tender wharf]

okay

[white oxide]

so it suffices to produce an inverse

for a * b = ba

a physical inverse

[tender wharf]

right

[white oxide]

not some vague element such as e

because nobody's gonna take that seriously in the proof

like

idgi

[tender wharf]

okay

what is your group

[white oxide]

"To prove the existence of an inverse, take e."

<G, *> correct?

[tender wharf]

okay

[white oxide]

maybe this is where i'm strugglig

[tender wharf]

you know by definition of a group

[white oxide]

struggling

[tender wharf]

there must exist an inverse always?

okay so here's the notion

I noticed you were talking about a being nonzero

[white oxide]

right but we're trying to prove that it's a group

we don't know it's a group

[tender wharf]

right

[white oxide]

ye

[tender wharf]

so you've assumed that ab = ba

for all a, b in G

and of course I'm guessing that operation is associative

[white oxide]

yeah i proved that

[tender wharf]

have you found an identity?

[white oxide]

wait how did i assume that it was abelian

yes

[tender wharf]

okay

[white oxide]

1

[tender wharf]

can you send the question

[white oxide]

yeah

[tender wharf]

alright

yeah okay so G under . is already a group

[white oxide]

right

[tender wharf]

how did you find e?

[white oxide]

oh wait

wait wait

so we know that ba = a * b

[tender wharf]

right

[white oxide]

so maybe if i switch the equations around it'll make more sense?

[tender wharf]

let's see it

[white oxide]

so are you saying that we have ba = a * b, so since <G, .> is a group then by definition it has an inverse for each x in G, so that means * must have an inverse for each a in <G, *>?

or something along the lines of that?

*since brain is not working

[tender wharf]

it should have an inverse, but it doesn't immediately follow

let's consider any element in G

say a

[white oxide]

ok

[tender wharf]

under . it must have an inverse, yes?

[white oxide]

right

[tender wharf]

so let this inverse be say y

[white oxide]

okay

[tender wharf]

so ya=e

now

consider *

[white oxide]

so ya = e = a * y

[tender wharf]

awesome

[white oxide]

OHHHHH

OKAY

THANKS SO MUCH

[tender wharf]

[white oxide]

that makes a lot of sense ur the goat

thank you

[tender wharf]

do you know how to make the isomorphism in the next bit?

well they already produced the map so you just need to prove it

[white oxide]

right i already started trying to prove it but injectivity seems a bit difficult but i haven't really thought that much about it

phi(a) = phi(b)

[tender wharf]

do you know how to show something is injective

[white oxide]

so the inverse of a = the inverse of b

sorry

yeah

so i'm trying to show that a = b

from the inverse of a = the inverse of b

[tender wharf]

right

recall one special property of inverses

can we have more than one inverse?

[white oxide]

nope, uniqueness of inverses comes to mind

ok wait

wait

i see where you're going

wait let me think about hthis

so the inverse of a is equal to the inverse of b, so more precisely let a' be the inverse of a which is the same as the inverse of b. then a'b = e = a'a, which is a contradiction since each element in a group G has only one unique inverse so a = b?

[tender wharf]

what are we trying to prove here

[white oxide]

that a = b

oh wait

[tender wharf]

yeah you don't need the contradiction

[white oxide]

right

we can just use the cancellation laws

or probably the fact that a' has an inverse

[tender wharf]

you literally did it lmao

[white oxide]

yeah

lmao

[tender wharf]

okay surjectivity is trivial

[white oxide]

right

[tender wharf]

you know how to show the last property?

[white oxide]

every element has an inverse

yeah let me try to show it

it's called the homomorphism property correct?

[tender wharf]

yup

although some authors just call it "preserving the group operation"

[white oxide]

oh okay interesting

ig that makes sense

okay

so

for the homomorphism property i basically have to show that the inverse of the product of two elements is the same as the product of the inverses of the two elements

in other words (ba)' = a'b'

by plugging it into the homomorphism equation or whatever you call it

[tender wharf]

phi(ab)=phi(a)phi(b)

yeah

notice how you are actually already done

[white oxide]

yeah i feel like there's something obvious hidden in plain sight

ok so phi(ba) is equal to (ba)'

[tender wharf]

ok let me give you a massive hint: you actually already said the correct answer

[white oxide]

this?

[tender wharf]

hmm

not quite

the hint here is that one of the lines

contains the thing you said

[white oxide]

i don't feel like this would suffice right

[tender wharf]

that thing is already true

but this is already true though

[white oxide]

because that's what we're trying to prove

[tender wharf]

(ab)^{-1} = b^-1 a^-1

[white oxide]

oh i did not know that

i mean i knew that from linear algebra

[tender wharf]

to see this

notice (ab)(ab)^-1 = ab b^-1 a^-1 = e

[tender wharf]

funny thing about linear algebra is that a lot of abstract algebra came from there

[white oxide]

ohhhhh

okay so ab(ab)^-1 clearly equals e

[tender wharf]

gallian calls it the "socks-shoes property"

basically if you put on your socks and then your shoes

[white oxide]

which is equal to ab(b^-1)(a^-1)

[tender wharf]

you must remove your shoes first before your socks

[white oxide]

so by the cancellation property or whatever you can get to (ab)^-1 = b^-1a^-1

[tender wharf]

yep

[white oxide]

oh interesting lmao

[tender wharf]

here we left-cancel off (ab)

so yeah when you said that line you've already proven it

[hollow mica]

If G is a group, X is a set, and a: X x G -> X is a (right) group action on X that is free, then
a(x, g) = x implies g = e (the identity). This is the definition of a free action.

But doesn't the inverse of g fix a(x, g) for any x and g? (the calculation being a(a(x, g), g^-1) = a(x, gg^-1) = x) So by freeness we must have g inverse be equal to the identity, but this clearly isn't true for all elements of the group

I don't believe my definition is wrong, so I went wrong somewhere in my logic

[tender wharf]

what is the calculation of a(x, g)

is this xg or gx

[hollow mica]

xg

[tender wharf]

mathworlf suggests a different definition for free action

[hollow mica]

wikipedia

[tender wharf]

this says gx

[hollow mica]

whether or not the action of left or right isn't relevant to my question though right

like the point still stands

[tender wharf]

let's see

I am not convinced the calculation is correct

from a(a(x, g), g^-1) to a(x, gg^-1)

[hollow mica]

oh is that not just associativity

definition of a group action

[tender wharf]

xg is not a group action

letting * denote

x * g = a(x, g)

and let . denote the group action

[hollow mica]

wait you just defined * to be the group action?

[tender wharf]

nope

I let * denote a(x, g)

and . denote the group action

[hollow mica]

oh

for specific elements x, g

wait i'm confused

[tender wharf]

ok let's look at

[hollow mica]

a(x, g) is literally the group action

[tender wharf]

uh

now I am confused too

[hollow mica]

If G is a group, X is a set, and a: X x G -> X is a (right) group action on X that is free, then
a(x, g) = x implies g = e (the identity). This is the definition of a free action.

[tender wharf]

but if thats the case your action isn't associative no?

because (xg)(g^-1) should be the same as x(gg^-1)

idk

[hollow mica]

associativity is a part of the definition of an action

I just wrote down the definition of a free action above

[tender wharf]

honestly I'm confused too sorry

I'm popping off now, someone else will probably be over to explain

[hollow mica]

no problem

[lapis plume]

I stuck at part c of the problem. I have tried the method of contradiction but I think I can’t say since ab not equal to ba then (ab)^2 not equal to (ba)^2. Is there any better method to show?

[tender wharf]

expand it out

this hopefully is not an exam or a test

[tender wharf]

contradiction is overkill

[lapis plume]

It is hw

[tender wharf]

alright

[lapis plume]

You mean my negation is false?

[tender wharf]

overkill meaning we can use a direct proof here

also what you wrote looks like a contrapositive rather than contradiction

[lapis plume]

Ok but I am not sure how to prevent square root

[tender wharf]

this is abstract algebra

[lapis plume]

Umm

[tender wharf]

we don't do square roots like that

[lapis plume]

I see

[tender wharf]

look at the left side

get rid of the ^2 and write it out

[lapis plume]

OH

(ab)^2=(ab)(ab)

Ummm

[tender wharf]

close

[lapis plume]

It is associative

[tender wharf]

which means (what can we do here?)

[lapis plume]

So I can write a(ba)b

[tender wharf]

I think you got it

[lapis plume]

Oh wait I cant assume it is commute

[tender wharf]

err

what?

we dont need to assume that

we're literally supposed to prove that

[tender wharf]

compare to the RHS

[lapis plume]

a(ba)b=a^2b^2

But why? It is not commute

[tender wharf]

expand on RHS

[lapis plume]

Oh

[tender wharf]

we aren't assuming it is abelian at all

[lapis plume]

abab=aabb

[tender wharf]

the result follows directly

[lapis plume]

But why we can cancel out the left a and right b?

[tender wharf]

left multiply by inverse

[lapis plume]

Oh yes

Nice

I got it

[tender wharf]

are you lacking sleep by any chance

[lapis plume]

Lol perhaps

[tender wharf]

get some sleep

[lapis plume]

I slept at 3am last night

[tender wharf]

you need sleep

[lapis plume]

Yeah

Thank you so much

[toxic zephyr]

say $N$ is a normal subgroup of $G$, and $S={s_1,s_2,\ldots}$ is a left transversal of $N$. i believe conjugation of elements of $N$ by $s_k$ ($\phi_k(n)=s_k\inv ns_k$) is a homomorphism on $N$. will the elements of $S$ give all such homomorphisms?

[cloud walrus]

nilpotent nix

[chilly ocean]

How to show this??

f(x) and g(x) are polynomials

[tender wharf]

#calculus

[plain bronze]

perhaps we can find an n such that the polynomial we wish to factor divides x^n - x, then we use the factors of x^n - x as factors for our polynomial? is this the right thinking?

[chilly ocean]

need help with b)

[rustic crown]

what are you thoughts?

[chilly ocean]

I think $a = \pm 1$ works

[cloud walrus]

isomorphism

[rustic crown]

yep!

and what about the right inverse of some element a?

say e is a left identity element

[chilly ocean]

not sure

do I set |a|b = e

then |a| = b^{-1}

[rustic crown]

yep, because definition of the right inverse will depend on the left identity you chose

[rustic crown]

ah this has a little error

e could be both +1 or -1

[chilly ocean]

oh

true

so

|a|b = 1

and |a|b = -1

[rustic crown]

yep, and you wanna solve for b

[chilly ocean]

so $|a| = \pm b^{-1}$

[cloud walrus]

isomorphism

[chilly ocean]

I think?

[rustic crown]

sort of true? >.<

[chilly ocean]

did I make a mistake

we have to find the right inverse element

uh

[rustic crown]

not really, but it's a little confusing the way you wrote

because |a| can't be something negative right

[chilly ocean]

so oh

a = %

a = pm b^-1

[rustic crown]

in any case, we want b and not a

a is given to you

and you want to find a right inverse

[chilly ocean]

im just confused with the notion of right inverses

hmmmmm

[rustic crown]

you just want to find all b's such that a * b = e

[chilly ocean]

right

oh

OH

[rustic crown]

which like you say is |a|b = e

and so b = ?

[chilly ocean]

|a|b = \pm 1

b = |a|^-1?

[rustic crown]

yep, if e was 1.

b = e/|a| in general

[chilly ocean]

b = $\pm \frac{1}{|a|}$

[cloud walrus]

isomorphism

[rustic crown]

won't write +- here

[chilly ocean]

how come

[rustic crown]

like i said, the definition of right inverse depends on the choice of e

[chilly ocean]

right

we have two choices for $e$

[cloud walrus]

isomorphism

[rustic crown]

so once you fix e, there is a single element b which gives you a right inverse of a

[chilly ocean]

okay

lets fix e to be 1

[rustic crown]

yea that all is good

but i think the point of the exercise is to show you that the identity element may not be unique in genearl

[chilly ocean]

but isnt the identity always unique

or do you mean one-sided identities

[rustic crown]

yea, my bad... meant the left identity

if you change the axioms very slightly

[chilly ocean]

I see, that makes sense - I was surprised we had two choices for the left identity

[rustic crown]

say

• is associative,
  there is a left identity
  and for each element there is a left inverse

then this would automatically imply the left identity is unique and each element also has a right inverse

basically in the axiom of groups, you could only ask for left identities instead of two-sided and same for inverses without really changing what they are

[chilly ocean]

ohh

that makes sense

[rustic crown]

but like you saw, left identity and right inverse don't imply all the group axioms

because we had an example with two distinct left identities

[chilly ocean]

thank you @rustic crown

I got it

[rustic crown]

[chilly ocean]

for this problem

the order of x^2 is 3

so x^4 cannot be the identity right

is that sufficient reasoning?

[rustic crown]

well you don't know what |x| is

and hence also not |x^2|

[chilly ocean]

hmm

if $x^6 = e$ then $x^5 = x^{-1}$

[cloud walrus]

isomorphism

[rustic crown]

yee

[chilly ocean]

so

x^2 \neq e implies that x \neq e right

which must mean that

if x \neq e

then x^{-1} neq e

[rustic crown]

yep

[chilly ocean]

hence x^5 neq e

[rustic crown]

yee :3

[chilly ocean]

thank u

[chilly ocean]

is this logic consistent though

regarding the order of x^2

[rustic crown]

if you know order of x is 6, then yep

[chilly ocean]

so that proves that x^4 cant be e

[rustic crown]

even if order of x was 3, that works

but how do show that this will be the case

[chilly ocean]

i see

thank oyu

that makes sense

[rustic crown]

that's why the problems asks you to show x^4 and x^5 are not e

that way, only options for order are 3 and 6

[chilly ocean]

x^5 = x^-1 implies that x^4 = x^-2and x^2 neq e which means x^-2 neq e which means x^4 neq e

[rustic crown]

yee :3

[chilly ocean]

yay

thank u sm

[rustic crown]

you could also argue via contradiction, that's sometimes a little easier to follow in head

x^4 = 1 would mean x^2 = x^6/x^4 = 1

[chilly ocean]

that's smart asf

holy shit

[rustic crown]

and similarly x^5 = 1 would mean x = x^6/x^5 = 1 which means x^2 = 1

[chilly ocean]

then you can solve it in one step

that's actually big brain

[rustic crown]

(oh btw, when i write x^6/x^4, i actually mean x^6 * (x^4)^-1 etc)

[chilly ocean]

right

yeah

[rustic crown]

(i get to be lazy because <x> is cyclic)

it doesn't matter whether you write (x^4)^-1 * x^6 or the thing above

[chilly ocean]

yeah

I have one more question that I kinda don't know how to approach because the dihedral group is not abelian

I think I have to use associativity

but i'm not sure

I thought I can leverage $a^{-1} b^{-1} = (ba)^{-1}$ somehow

[cloud walrus]

isomorphism

[rustic crown]

you need to use the relations a and b satisfy in D_n

[chilly ocean]

what's that

[rustic crown]

bab^-1 can be written in terms of a

what is it?

[chilly ocean]

uh

e?

no wait

I'm dumb

why do we care about bab^-1

[rustic crown]

because if bab^-1 = a^k then, ba = a^k b

so you can use this relations to push all the b's to the extreme right

[chilly ocean]

how do we know bab^-1 = a^k

[rustic crown]

that's what i'm asking :p

in D_n this is true

do you recall the relations >.<

[chilly ocean]

I don't

[rustic crown]

(like what's your definition of D_n?)

[chilly ocean]

well

D_n is the group of symmetries of an n-gon

right?

of order 2n

[rustic crown]

hmm okie

so in that case notice that if b is a reflection and a is a rotation, then ba would again be a reflection.

[chilly ocean]

I think I have to try this out with an example

like D_3

so I can see the relation

[rustic crown]

yee give it a try

[chilly ocean]

thank u again for helping me

[rustic crown]

[chilly ocean]

I'll get back to you once I'm done

[trim grove]

Can someone please help mw with this, i m not getting how they assuming function g?

[rustic crown]

,rotate

[delicate orchid]

It seems to be the same g that appears in the previous proof - or at least a specific example of such a g

[trim grove]

Is there any reason they are defining g like this?

[rustic crown]

you're forced to define that because you want f = g o alpha

[trim grove]

Thats the thing how they get f=ga even they have not defined a 🥲

[rustic crown]

ah

alpha is the map X --> F

[delicate orchid]

They have defined a

Two lines up

[rustic crown]

yea they did

[pliant forge]

Let $F \subseteq E \subseteq K$, a field extension such that $F \subseteq E$ is algebraic. Then if $a \in K$ is algebraic over E how can we show it is also algebraic over F?

[trim grove]

Let me try

[delicate orchid]

What are you trying?

[pliant forge]

ye

[trim grove]

Trying to show f= ga

[delicate orchid]

Good luck

[cloud walrus]

Eso

[rustic crown]

so say a satisfies some polynomial x^n + c1 x^n-1 + ... + c_n = 0 where c_i are in E. what can you say about a over F(c1, ..., c_n)?

from that what can you say about F(c1, ..., c_n, a)/F

[pliant forge]

hmm then it would be a root

[blazing viper]

when can an upper triangular matrix commute with a down triangular matrix

[rustic crown]

right, and the polynomial also lives over the field L = F(c1, .., c_n). so not only a is algebraic over E, it's also algebraic over this (possibly) much smaller subfield L

[blazing viper]

when one of the matrices is diag?

[pliant forge]

ye i kinda was here but idk how to show there is a poly in F[x] which has this root as well

[rustic crown]

yeap, so instead of looking for an explicit polynomial, just try to argue using degree. recall a is algebraic over k if and only if [k(a):k] is finite

you know that a is algebraic over L and L is f.g. and algebraic over F

[pliant forge]

thx ❤️

[rustic crown]

[pliant forge]

i totally overlooked the thm

[trim grove]

Is This ok?

[formal ermine]

how do I show that [L(alpha) : K(alpha)] = [L : K] for some alpha in the algebraic closure of K if [K(alpha) : K] and [L : K] are coprime

[rustic crown]

hint: both [L:K] and [K(alpha):K] divide [L(alpha):K]

[rustic crown]

not quite. you could only apply g = f o alpha to an element of X. and alpha(x) is 1.x in the copy R_x

there won't be any summations there

[trim grove]

Okk, let me check this again

[trim grove]

But 1.x is element of F now, so we can write this in summation form.

[rustic crown]

have you seen that Exercise 2.2 which they talk about?

the element x in F actually means the tuple (0, 0, .... 1, 0, ... 0) wehre the only non-zero element is 1 which is at the x^th spot in the direct sum of Rs

[trim grove]

Thankyou for the help, i will try

[frank cosmos]

How can I use induction to show that the direct product of a family of disjoint subgroups of a group is isomorphic to the group itself?

showing that it is a homomprhism and surjective is trivial, im having difficulties showing the kernel is trivial

commutative subgroups*