groups-rings-fields

[rustic crown]

what's really the question here?

[lethal dune]

lol

[sacred pier]

Was reading the article about orthogonal matrices. The definition is weird, but now knows its meaning now lol Thank you

[lethal dune]

weird? in what sense?

[sacred pier]

If Q mean orthogonal matrix, I don't know why we need to mention Q "is in general linear group"

The author of another article just define Q = orthogonal matrix

And I read this definition in another book, he didn't define Q = orthogonal matrix

[rustic crown]

your english is cute

[lethal dune]

to emphasize it's a subset. Also there's more set theoretical reasons

[sacred pier]

how to tell O(n) is subset of Q in this equation (if we already told the readers Q is orthogonal matrices)?
By his equation, its more like telling O(n) "is orthogonal matrix", more than "is the subset of orthogonal matrix".

[vagrant zinc]

guys a question for this exercise 32
My idea is this
let a,b belong to the group G we know that (ab)^2=(ab)(ab) so if a=a^-1 and b^-1 then we have
(ab)^2=(ab)(ab)
=(a^-1b^-1)(a^-1b^-1)
=(ab)^-1(ab)^-1
=(ba)^-1(ba)^-1
=(b^-1a^-1)(b^-1a^-1)
=(ba)(ba)
=(ba)^2

[lethal dune]

I don't really understand what you mean

[sacred pier]

how to tell O(n) is subset of Q in this equation (if we already told the readers Q is orthogonal matrices)?
By his equation, its more like telling O(n) "is orthogonal matrix", more than "is the subset of orthogonal matrix".

[lethal dune]

Q is not a "set of orthogonal matrices" it's just a placeholder

[sacred pier]

Yeah Q is a placeholder, and I think the whole expression inside the bracket is representing"orthogonal matrice"

[lethal dune]

33 or 32?

[vagrant zinc]

32 jeje

[lethal dune]

why is a'=a?

[rustic crown]

see either you define orthogonal matrices by saying those invertible matrices whose inverse is the transpose. or you expand this out more, i.e. matrices Q such that Q * Q^t = Q^t * Q = I. since for orthogonal matrices Q, the inverse is given by transpose, they're a subset of GL(n, k)

[vagrant zinc]

in order to prove commutativity

[lethal dune]

ahh no, you can't just assume a and b are of order 2

only thing given is (ab)²=a²b²

hint: || abab=aabb now cancel||

[sacred pier]

yes. I know these 2 facts"what orthogonal matrix is", and also "ortho. matrix is the subset of GL(n,k)".

What I don't understand is, Its text description said "orthogonal group is the subset of orthogonal matrix". Which I cannot see O(n) is subset of it in its equation. The right hand side of the equation is more like defining what is orthogonal matrix itself

[rustic crown]

ah i see your question

they define orthogonal group to be "the" subset of orthogonal matrices

[lethal dune]

looks like typo

[sacred pier]

?

[rustic crown]

as in they're refering to the subset which contains all the orthogonal matrices

and not defining it as "a" further subset of the subset of orthogonal matrices

[lethal dune]

or not

ngl even I fell for it

[rustic crown]

whatever, english is weird. ignore it

[sacred pier]

Can I write as such for this sentence ? O(n) x O(n) -> O(n)

So far I can't tell the difference between orthogonal group and the set of orthogonal matrices

[rustic crown]

at the level of sets there is no difference

[chilly ocean]

you just wrote the only difference

[rustic crown]

when you say orthogonal group, it contains the data of both the underlying set of orthogonal matrices, along with the multiplication operation

[vagrant zinc]

[rustic crown]

[sacred pier]

why we highlight it is a group then?

[chilly ocean]

dan

[lethal dune]

because it is

[chilly ocean]

because... we care about groups?

[lethal dune]

[rustic crown]

because more structure makes mathematicians happy

[sacred pier]

oh ok. A set is just a bag of items. A group is a data structure that supports binary operations.

[rustic crown]

yee

[sacred pier]

Oh... thank you

[rustic crown]

that's sounds too cs for my comfort >.<

[lethal dune]

yeah

[sacred pier]

It sounds easier with cs words

[rustic crown]

btw, the language might be a little weird here... when you say "support", it shouldn't mean that there exists a group structure on it. it should mean that the group structure is part of the data >.<
just a tiny point i wanted to say >.<

for any (non-empty) set X, you can give it a group structure by defining some weird operation

if X is finite, use the structure of Z/nZ

if X is infinite then |X| = |F(X)|, so transport this structure

[white oxide]

how would we show that H intersect K satisfies the existence of an inverse x^-1 for every x in H intersect K? that would basically amount to showing that if x in H and x in K, then x^-1 in H and x^-1 in K. but the problem is that we don't know that x in H and x in K from the information we're given (or at least you guys might, i'm struggling on figuring out how to prove it). We know that their intersection is nonempty (e in H intersect K) but other than that i'm stuck

[barren sierra]

I think you have some of the directions of the proof wrong

[cloud walrus]

Spamakin🎷

[barren sierra]

@white oxide

[white oxide]

oh ok thanks

[barren sierra]

So you're starting with that fact

[white oxide]

but how do you know x in H intersect K?

[barren sierra]

by assumption

[white oxide]

don't you have to prove existence first?

[barren sierra]

no

[white oxide]

that there is something other than the identity in H intersect K

[barren sierra]

for all you know, the identity may be the only thing in there

you want to show that if anything is in there

so is it's inverse

that's all

[long geyser]

yes you are supposed to show a subgroup is non-empty, but this is trivial because identity is in both H and in K

[barren sierra]

yea you already showed that so I didn't mention that

[white oxide]

ah okay

so the element may just be x = e

[barren sierra]

but as far as any other element, doesn't matter. Note that no where did I say that x was a non-identity element

[long geyser]

sure it may

[barren sierra]

yea or it may be something else (for example if H = K = G and all the groups are non-trivial)

[white oxide]

right

okay thanks that helps a lot!

[barren sierra]

similar logic should be used to show closure under the group operation

[white oxide]

right i think i got it

gracais

gracias

[barren sierra]

why would someone make someone do this

[vagrant zinc]

guys I just made this cayley chart of Z_3xZ_3
two questions
the first one is that the table if it is ok or not ?
at some point I found
2+2=4=1mod3
no problem for this ?
now i would like to know how can i get the subgroups out of this table ?

[vagrant zinc]

this exercise appears on youtube in a class at UNAM

[barren sierra]

the table looks fine to me

[vagrant zinc]

heehe nice exercise

[sonic coral]

was literally on my test today

[vagrant zinc]

what question?

[sonic coral]

highlighted one

[vagrant zinc]

how did you do it?

[lethal dune]

are you still stuck?

[vagrant zinc]

to find the subgroups of Z3xZ3 if any idea ?

[tender wharf]

there arent many

[sonic coral]

On my test today, one of the questions was to prove that the cyclic group G is abelian

so i did something like for any x,y in G, x=g^a and y=g^b for some integers a,b respectively

then xy = (g^a)(g^b)=g^(a+b)=(g^b)(g^a)=yx

would that be right or am i completely off

[lavish nexus]

It’s correct

[tender wharf]

yeah that's fine

[white oxide]

ohhh ok thank you so much that makes complete sense!!!

[formal ermine]

do you know lagrange's theorem

[lusty marlin]

Any subgroup will have order 1,3 or 9. (Why?)
Order 1 implies singleton identity, order 9 implies the whole group

Now find the subgroups of order 3

[lusty marlin]

Identity must belong to each of these

[somber sleet]

is it right to define Z[i/25]= {(a+bi)/25^k| a,b in Z and k in N}

[somber sleet]

following question, what is a nice/fast way to show that two generated rings are the same?

[chilly ocean]

how do I prove these two groups aren't isomorphic?

[south patrol]

Think about some properties of the groups involved

[lusty marlin]

What do S,C and D mean? Do they have their usual meanings?

[somber sleet]

why is it enough to show that the generators are contained?

[chilly ocean]

is there any general method to show groups are not isomorphic/are isomorphic?

[chilly ocean]

S_3 is the symmetric group of order 3, i.e. the group of bijections from {1, 2, 3} to itself

and C_6 is the cyclic group {0, 1, 2, 3, 4, 5}

D is the dihedral group for uh.... not sure

[lusty marlin]

To show that two groups are isomorphic, construct/demonstrate the existence of an isomorphism.
To show that they are not isomorphic, either assume an isomorphism and prove a contradiction or examine properties of the elements such as their order or something, or examine subgroups.

[lusty marlin]

What is the group operation here?

[chilly ocean]

addition

modulo 6

[lusty marlin]

Ah ok

S_3 is not a cyclic group, C_6 is

Hence they are not isomorphic

Why is S_3 not cyclic? It cannot be generated by a single element (Prove this)

[chilly ocean]

wait, so if a group is cyclic but another isn't then we cannot construct an isomorphism between them?

(why?)

[lusty marlin]

Right

[lusty marlin]

Left as an exercise to the reader

[chilly ocean]

but how come?

[lusty marlin]

Prove it, not that tough

[chilly ocean]

no I mean, why is it that two groups must both be cyclic in order for an isomorphism to be constructed between them

or not both cyclic

[lusty marlin]

Let G and H be two isomorphic groups, such that G is cyclic. Now consider a generator of G, and an isomorphism ϕ from G to H.

[lusty marlin]

That is what I suggested you prove

[lusty marlin]

This is a hint, although I would call it basically the whole answer

[chilly ocean]

oh

[lusty marlin]

Even easier, all cyclic groups are abelian. So it is sufficient to demonstrate that S_3 is not abelian, which is trivial, to show that it is not cyclic.

[indigo nymph]

How can I answer part b of this question?

[rustic crown]

i get scared when i see [4] on the right >.<

[lusty marlin]

If H is a subgroup of index 2 of a group G, then H is a normal subgroup of G.

(Prove it)

[formal ermine]

S3 doesn't have order 3

how many ways are there to arrange 3 objects?

[indigo nymph]

Thank you!

[chilly radish]

3

[rotund aurora]

Sanity check: Just send (m+n)(S+T) to (mS,nT). You can check that this is well defined, since if m+n=m'+n' modulo S+T then m=m' mod S and n=n' mod T, as S and T are disjoint. This is clearly surjective and to see that it is injective, suppose (mS,nT)=(m'S,n'T). Then there exist s in S and t in T such that m-m'=s and n-n'=t. Adding both expressions, we get (m+n)-(m'+n')=s+t in S+T

Another way is to just consider the "canonical" mapping M+N --> M/S + N/T, defined by m+n --> (mS,nT) and see that it is surjective and that the kernel is S+T, and apply first iso

[warm wyvern]

le epic universal property

[chilly radish]

I am confirming your sanity

[rustic crown]

mS is a weird notation >.<

the operation is + so use m+S >.<

[rotund aurora]

I didnt want to write m+S

so many + signs otherwise

xD

[rustic crown]

[glossy crag]

This works for arbitrary index sets, too.

[rustic crown]

this is the only way

other way doesn't exist 🙈

[rotund aurora]

yeah

both ways are clearly the same sure, but in one you have to invoke the first iso, in the other you kinda do it in the fly I guess

[rotund aurora]

what do you mean?

[rustic crown]

like arbitrary direct sums and not just 2

[rotund aurora]

👍

I was aware of that long ago, but looking through some old exercises I realized I did not write down a proof of that, so wanted to make sure

[somber sleet]

how can you show that R[X1,...,Xn] is isomorphi to R[X1,..., Xm][Xm+1,...,Xn]
I know that it has to be done with the universal property, but I can't visualise the commutative diagram

[glossy crag]

I too require a sanity check: if $A$ is integrally closed, $K$ its fraction field, $L$ finite extension of $K$ and $B$ integral closure of $A$ in $L$, then for $a\in B$, $N_{L/K}(a)\in A^\implies a\in B^$ (I don't like the standard approach with embeddings):
$N(a)$ is a power of $b_n$ (up to $\pm1$), where $x^n+b_1x^{n-1}+\cdots+b_n$ is the minimal polynomial of $a$ (which has coefficients in $A$), so if $N(a)$ is a unit, so is $b_n\implies$ multiplying the identity $a^n+b_1a^{n-1}+\cdots+b_n=0$ by the inverse of $b_n$, you get $1=a(\cdots)\in B\cdot B\implies a\in B^*$.

[rustic crown]

show abstractly for any ring R, you have a natural iso R[x,y] = R[x][y], then rest would follow from induction

[cloud walrus]

Ocean Man

[somber sleet]

I can't really visualize the definition of the second set

[rustic crown]

so they're polynomials in the variable y and coefficients in R[x]

something like
(4 + x +x^2) + (3x - x^7)y + (x^99 - x^999) y^2

[somber sleet]

pff haha omg, thanks

[rustic crown]

and the isomorphism would think of this polynomial as a bi-variate poly with coefficients in R

[somber sleet]

this makes sense

[rustic crown]

btw what's your definition of R[x,y]?

[somber sleet]

you don't want to see it

[rustic crown]

i realize we might have to prove R[x1, ..., xn, y] = R[x1, ..., xn][y] depending on the way you define it

[somber sleet]

Rn mentioned above

[rustic crown]

ah that

[somber sleet]

yup

but they show it with the universal property

I just don't see how you get your Isomorphism

[rustic crown]

yee, that's the best approach

[somber sleet]

do you just define your homo from R[X] to any set S

[rustic crown]

you never would want to define a map with elements and then check it preserves + and *, that's too much work

[south patrol]

Big up universal properties

[glossy crag]

hehe, homo

[rustic crown]

okie let's just do it directly

we wanna define map both ways R[X1, .., Xn] --> R[X1, .., Xm][Xm+1, ..., Xn]

recall that any polynomial ring comes with a natural map S --> S[Y1, ..., Yk]

R --> R[X1, .., Xm] --> R[X1, .., Xm][Xm+1, ..., Xn]
so use this map

and the universal property to extend it to R[X1, .., Xn]

similarly define the other direction and then we need to check that the composites are just identities

[rotund aurora]

if |G|=n then for all g in G we have g^n=1 correct?

[south patrol]

Yes

[coral spindle]

You can prove this, croq.

Use Lagrange

[rustic crown]

to check that, it would suffice to check it only on the generators X1, ..., Xn by uniqueness in the universal property.

another idea is to verify that R[X1, .., Xm][Xm+1, ..., Xn] satisfies the universal property of R[X1, .., Xn]

[rotund aurora]

ah right

[rustic crown]

which maybe a little nicer now i think about it :p

[south patrol]

Oh interesting that the non-lagrange proof breaks down in non abelian case

Well if G is abelian you can consider the product of all elements of G, then the product of all elements of gG (which is also G ofc)

[rustic crown]

(you're sane uwu)

[south patrol]

and then you get like the second is the first times g^n

So g^n=1

But ofc only works for abelian, sad

[rustic crown]

(not any "set", it should be a commutative ring together with a map from R)

[west sinew]

Will a simple ring always be a simple module over itself? Even if it's not unital, commutative, etc.?

[rustic crown]

nope

iirc for simple rings, you require no non-trivial two-sided ideals

for if you think of it as a (left) module over itself, then the definition would only look at no non-trivial left-ideals

i don't wanna think about non-unital :p

[west sinew]

You could have it as a two-sided module over itself tho, no?

[rustic crown]

yea but usually people don't do that ig

like it becomes an R-R bimodule

which is same as R⊗R^op left module

[west sinew]

Okok I see what you mean

[rustic crown]

[west sinew]

thank

[somber sleet]

yes you're right, oopsie

thanks for the explanation though

I worked it through

I have another question though which is confusing me rn

why is the unit Ring of R contained in the one of R[X], it's logic, I know, I understand it too

but somehow my definition doesn't make it work

[rustic crown]

unit ring?

do you mean the unit group?

[somber sleet]

the group of unit of R(?)

[somber sleet]

I'd say so

[rustic crown]

yee that's a group

they actually have the same unit groups

because the stuff you add to your ring can't be inverted because of degree reasons

f in R[X] is unit then f * g = 1 which means both deg f and deg g are 0, so the equation f * g = 1 is entirely inside R

[novel parrot]

why are all prime elements also irreducible? does this work? if x is prime, say $x = uv$ that means that $x | uv$ hence $x | u$ or $x|v$ say $x| u$ wlog. then $xy = u$ which makes z become a unit.

[cloud walrus]

ActiveChapter

[rustic crown]

nit pick, non-zero prime in an integral domain. and that makes y a unit, not z.

[novel parrot]

but y doesnt matter

[rustic crown]

oh oopsie, meant v

you get vy = 1

[novel parrot]

yes

how many elements does $\mathbb{F_2}[X]/<X^2 + X + 1>$ have

i counted just four

1,0,x,x+1

[rustic crown]

F = F_2?

[novel parrot]

oop yes

[rustic crown]

yee then its 4

[novel parrot]

about $\mathbb{F}_2[X]/<X^2 + X + 1>$

[cloud walrus]

ActiveChapter

[novel parrot]

i have to answer question about if its a integral domain and/or a field

i say that it is a domain but not a field

is it correct?

I just verify by the 4 elements that i computed before

not a field because X + 1 has no inverse

[rustic crown]

(btw \langle \rangle)

what's x*(x+1)?

[novel parrot]

oh its 1

oops

[rustic crown]

(also in general, finite integral domains are automatically fields)

[novel parrot]

oh

[rustic crown]

(also finite dimensional algebras over a field which are integral domains are also fields)

try proving these

:3

[novel parrot]

i know the first result

but not the second

[rustic crown]

everytime you ask me something i'm reminded to check how long i need to wait for next ep of one piece >.<

[rustic crown]

oh the proof idea is very much the same.

in one case you use pigeon hole to say injective => surjective

in other you would use rank-nullity

[woeful sand]

this one is not different too

[chilly ocean]

read the manga

[woeful sand]

just like mr det said

[chilly ocean]

then you'll have to wait for both

[rustic crown]

mr det

[woeful sand]

im ofc joking im not that formal

[woeful sand]

[rustic crown]

hehe

[woeful sand]

i have a question, you guys have only one name for absolute value in own native language?

it's weird, on my country the word for absolute value is the same for "module"

it just feels wrong for me after studying abstract algebra to name absolute value as "module"

[chilly ocean]

two

[woeful sand]

what are they

[chilly ocean]

in original?

[woeful sand]

yes

or roman-latin alphabet (in preference)

[chilly ocean]

wartość bezwzględna
and
moduł

[woeful sand]

that's for absolute value right?

[chilly ocean]

yes

[woeful sand]

my head goes on a rollercoaster everytime i read module so it feels weird for me rn

i don't know if you all get me

[woeful sand]

i mean i start immediately remembering propeties of module

then i get confused

because it was actually talking about abs value

[chilly ocean]

moduł means module, absolute value and result of taking something modulo n

[woeful sand]

there's no compelling semantics interpretation for me unfortunately

[chilly ocean]

(though not sure who uses the last one)

[void cosmos]

how did he go from 2nd to 3rd line

[rustic crown]

that's just the definittion of tau i think

tau(l+i) = i for i = 1, ..., k
tau(i) = k + i for i = 1, ..., l

[void cosmos]

yea

but

how did he

put out the sng

sgn

[void cosmos]

wow ur smart

[rustic crown]

at the cost of making that sgn (sigma * tau)

[void cosmos]

wait 1/sgn(t) = sgn(t)?

[rustic crown]

[rustic crown]

yea sgn takes values in +-1 lol

[void cosmos]

yea

what a cool ass func

[rustic crown]

lol

[void cosmos]

so he multiplied both

sides

no

he multiplied by 1 sgn(t)*1/sgn(t)

then used multiplicative property right

[rustic crown]

yee

[void cosmos]

cool proof

ty

[rustic crown]

[void cosmos]

fuck i cant follow this

in first line

or nvm i get first line

[tribal moss]

It's especially hard to follow when one cannot see what is being proved, or the assumptions it is being proved under.

[void cosmos]

mb

how did he get 3.7

i understand that if tao fixes the indices of g then u get the multiplication equality of tao

but then how does he just factor out these sgns

and also

[brittle pebble]

if f is a ring morphism, then a=b ==> f(a)=f(b) right?....

[void cosmos]

f just being well defined implies this

[rustic crown]

because summations

[void cosmos]

yea but the summation is taken over tao

so he cant treat it like a constant?

or am i stupid or what

[rustic crown]

the summation is over pairs (sigma, tau)

now it doesnt' matter if you first some over tau for a fixed sigma, and then sum these over sigma

like finite sums commute right >.<

[void cosmos]

they do

why am i blue

nvm

[rustic crown]

am i misunderstanding your question?

[void cosmos]

let me

reread

again

[rustic crown]

you apply sigma to that big summation

[void cosmos]

okay

[rustic crown]

by definition of the action, this is same as applying sigma to each summand

[void cosmos]

okay

so now

you have sigma(sum(...))

which means just sum(sigma the indices of whats being summed )

right?

[rustic crown]

ooops gomen gomen

i'm back

[rustic crown]

yee

[void cosmos]

what

yea okay

so

now i have

[rustic crown]

so you apply sigma to sgn(tau) tau(f)⊗g

sgn(tau) is just a scalar

[void cosmos]

yea

oh i got it

he applied the phi insied the sum

then he just double summed them

wow S_k is finite

damn

k!

okay

[rustic crown]

hehe

[void cosmos]

ok so now

[rustic crown]

what's phi

[void cosmos]

sigma

sorry

[rustic crown]

yee so you wanna see what's sigma(tau(f)⊗g)

so they ask you to think of tau as a permutation in S_{k+l} which fixes k+1 and beyond

so thats sigma(tau(f⊗g))

because you don't touch inputs of g

[void cosmos]

yea

i get that

now im trying

to understand

the main counting shit

[rustic crown]

okie so you swap the order of sum

for a fixed tau, sum over all sigma

as you vary sigma over all S_{k+l} then sigma*tau also varies over all of S_{k+l}

[void cosmos]

wait u can do that

[rustic crown]

you're summing exactly sgn(sigma*tau) sigma*tau

[void cosmos]

sum over x sum over y = sum over y sum over x

[rustic crown]

a+b = b+a

everything is finite

[void cosmos]

yea yea

[rustic crown]

yee it works uwu

[void cosmos]

it doess

thank god

[rustic crown]

just induct the hell out of it

[void cosmos]

no fubini tonelli fuckfest

[rustic crown]

lol

tonelli still nice tho

cause no hypothesis

just non-negative measurable

which are like all the functions

[void cosmos]

yea but u have to like calculate one of the integarsla and make sure one of them is atleast finite

thats fubini right? ( im new sowy )

[rustic crown]

oh it doesn't matter

for fubini you need to ensure convergence and finiteness and all that

but if you're integrating non-negative measurable functions then order doesn't matter at all

[void cosmos]

yeaa

[rustic crown]

if one side is infinite, then so would be the other

[void cosmos]

so anywayss

[rustic crown]

yea lol

finite sums

uwu

[void cosmos]

he is saying

[rustic crown]

so you're summing over m = tau*sigma

and you're summing sgn(mu) * mu(f⊗g)

[void cosmos]

lmao yea so he is saying

[rustic crown]

A(f⊗g)

and now you sum this over tau's

[void cosmos]

every tao has a unique like extension to S_k+l call it mu

[rustic crown]

to get that k!

[void cosmos]

and there are k! taos

[rustic crown]

not tau

well yea

but that's not what he calls mu

mu is the product

[void cosmos]

yea

its the element

in S_k+l

that arises from tao times the rest of the l elements

is that right?

[rustic crown]

mu = sigma * tau

[void cosmos]

yea so sigma has to be in S_l rihgt?

[rustic crown]

where you pretend tau is in S_{k+l}

na

you're composing two perms

both are in S_{k+l}

[void cosmos]

oh tau is in S_k+l fixing the l elements

as we said above

forgot about that ok

[rustic crown]

[void cosmos]

and then what

oh is he just saying

composing two permutations is just one

in the same group

[rustic crown]

he calls this composite mu

and then says that for a fixed tau, this mu would vary over all perms

[void cosmos]

ok but why not (k+l)!

[rustic crown]

as sigma varies over all perms

[rustic crown]

because tau has to fix the last l things

[void cosmos]

lmfao yea

xd xdsxd

got it

[rustic crown]

[void cosmos]

k+1 tho

nvm vnm

[rustic crown]

yea k + {1, ..., l}

[void cosmos]

got it

tysm tysm tysm

[rustic crown]

[toxic zephyr]

anyone know where i can find a list of groups that have an abelian normal subgroup? looking for as many examples as i can

[rustic crown]

[coral shale]

all of them

[timid yarrow]

rusty on algebra (took first course 2 years ago) and now am in second course...

want to clarify this -- Factor (polynomial) in Q[x] --> factors will have only rational coefficients

particular example has a zero at 3 --> (x-3) * (quadratic) but quadratic has two complex roots, so that cannot be factored further in Q[x]?

[toxic zephyr]

errr a nontrivial one i mean

[coral shale]

All even groups contain C2?

right?

iirc

Hmm

[rustic crown]

normality would be a prob

[coral shale]

o

[rustic crown]

yee

[delicate orchid]

the trivial group is abelian and characteristic, thus normal, ergo all groups contain a normal abelian subgroup Q.E.D.

[coral shale]

wew being slow as usual

[south patrol]

oops reacted to wrong thing

[sonic coral]

what does characteristic mean

[delicate orchid]

I still think it's all of them btw, take the centre

[rustic crown]

normal means phi(H) = H for phi any inner automorphism

[delicate orchid]

any automorphism, not just inner

[rustic crown]

char means phi(H) = H for every automorphism

[delicate orchid]

yurrrr

[sonic coral]

guess i havnt learned that terminology yet

[delicate orchid]

it's usually not mentioned

[sonic coral]

i just had my first test, we covered groups up until lagrange theorem and cosets

[south patrol]

but ye uh i think this is basically unreasonably big of a question

unless you mean examples

in which case like, just take any abelian group lol

[delicate orchid]

any group with non-trivial centre will do

[coral shale]

ooo

[delicate orchid]

as I previously said

[coral shale]

centre is normal yes yes

[south patrol]

any non-perfect group

[delicate orchid]

even stronger, it's characteristic

[south patrol]

jk

needn't be abelian

[chilly ocean]

yes by cauchys thm. 2 is prime, and divides all even numbers, so theres an elt of order 2 which in turn generates C2

[toxic zephyr]

yeah. worked extensively with abelian groups, dihedral-esque groups/semidirect products of abelian groups, and metabelian groups already, and i'm looking for more.

[delicate orchid]

semidihedral groups?

generalised quaternions?

[toxic zephyr]

yeah

[delicate orchid]

dang....

[rustic crown]

[toxic zephyr]

going to start by looking at groups with a nontrivial center

[delicate orchid]

more what btw, lol

[rustic crown]

sounds scary

[toxic zephyr]

honestly, they're basically the dihedral group with like one or two slight changes lol

[rustic crown]

yea but imagine working with them >.<

dihedral are already too much for me

i remember in our group theory course the prof asked us to write a presentation of S_n

and i was like why

[toxic zephyr]

more example of groups with a nontrivial abelian normal subgroup

[delicate orchid]

semidihedral are fine, what's scary is an arbitrary extension by C_2

ok so we have non-trivial centre, uhhh

if a subgroup is the only subgroup of that isomorphism type it will be normal

[chilly ocean]

do you remember what the presentation was?? i wouldnt rly have an idea of how to do this

[delicate orchid]

hint: S_n is a (real) reflection group

[chilly ocean]

like like c_k^k = 1 for cycles of length k?

[chilly ocean]

maybe i dont understand the hint, but wouldnt this be useful for a representation not a presentation?

i figured he meant with like generators and relations

[delicate orchid]

it's useful for both, real reflections groups are generated by elements of order 2

[rustic crown]

yea use a transposition (12) and the n-cycle (12....n)

show that you can get any (1k)

then show you can get any (ij)

then show you can get any perm

[delicate orchid]

you can get the presentation from the coexter diagram

[chilly ocean]

lmao my group theory isnt coxter diagram hardcore

[delicate orchid]

not surprising, they're pretty niche

[rustic crown]

+1

[toxic zephyr]

thank you to everyone who helped! i have some things to look at now 🙂

[formal ermine]

what's a coexter/coxter/whatever diagram

does it have something to do with quiver theory

[delicate orchid]

nope, although they are graphs

they're to do with reflection groups

[formal ermine]

what's that

[delicate orchid]

they denote relations between the generating reflections basically

hence why you can recover a presentation for the group from them

[formal ermine]

idk much about symmetric groups

what's a reflection

[delicate orchid]

an element of order 2

there's multiple ways of thinking about it

[formal ermine]

ah lol

I see

[delicate orchid]

you can actually think of them as linear transformations that fix a hyperplane

but that's far too geometric for my tastes

[formal ermine]

that hyperplane is their axis of reflection?

[delicate orchid]

the hyper plane is fixed, I don't know what an axis of reflection is

the normal vector to that hyperplane is mapped to it's negative

[formal ermine]

yes wait let me draw something

[delicate orchid]

so take uhh

S_3

one mo I need to check the rank

yeah it's rank 3 cool

[formal ermine]

D4

the red line is the axis of reflection

of that reflection

[delicate orchid]

so the 3 hyper planes would be the xy-plane, yz-plane, and xz-plane

[delicate orchid]

yeah a line is a "hyperplane" in 2D

really I should say "space of codimension 1" instead of hyperplane but who cares

[formal ermine]

yes but that's what I mean with axis of reflection

[delicate orchid]

in that case yes

[formal ermine]

ok nice

how do we think of groups as vector spaces here? by reps?

[delicate orchid]

it's more that you get a rep of a group via it being a reflection group

one min lemme get my book to make sure I get the definition right

a reflection "along" a vector $\alpha$ (here alpha would be the normal vector to the hyper plane) is a linear transformation of $\bR^n$ that maps a vector $\lambda$ to $\lambda - \frac{2\langle \lambda, \alpha \rangle}{\langle \alpha, \alpha \rangle}$

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

which is like gramm-schmidt but twice if you know what that is

[formal ermine]

ah so like, we look at some vector space and then form a group by taking some reflections (linear transformations that fix a hyperplane)?

[formal ermine]

yes I know what it is

[delicate orchid]

yes exactly

[formal ermine]

ahhh

I thought you meant the other way around lol

[delicate orchid]

these vectors are called "roots" I won't get into (mainly because I don't fully know about it myself) but they are STUPIDLY important for rep theory of lie groups

[formal ermine]

what's the difference between a lie group and a lie algebra

is it just like

both are differentiable manifolds

but one's a group

[delicate orchid]

well for one lie groups have inverses

[formal ermine]

and the other one's an algebra?

[formal ermine]

inverses for what?

[delicate orchid]

elements

[formal ermine]

wait with lie algebra

[young fiber]

Are lie groups groups

[formal ermine]

do we mean

vector space with vector multiplication

so like

ring over a field?

(and some other condition of course)

[delicate orchid]

yes but no

the multiplication isn't that nice, it doesn't even have to be associative

for example, R^3 with the cross product is a lie algebra

[young fiber]

Just a bilinear map

[delicate orchid]

I'll screenshot the properties from wikipedia

[formal ermine]

wut

isn't bilinear

like

A x B -> C

not A x A -> A

[delicate orchid]

B = A

C = A

[formal ermine]

yes

[young fiber]

Could be any

[delicate orchid]

major "lol" moment

[formal ermine]

that's what I meant with vector multiplication

[young fiber]

Just linear in each argument

[formal ermine]

why make it complicated

[delicate orchid]

because you then said a ring over a field

this is not a ring

[formal ermine]

ah

lie bracket

ic

[delicate orchid]

yeah, fun exercise is to show the cross product(s) in R^3 satisfies these

[formal ermine]

I was gonna ask something

but I forgor

[delicate orchid]

an even more fun thing to do is to show the cross products in R^7 do

[young fiber]

I think only alternativity and the jacobi identity make a lie algebra different from a normal algebra

[formal ermine]

isn't cross product an inner product

[delicate orchid]

it's an outer product iirc

[formal ermine]

the results of outer products are matrices though

and those of inner products are scalars

[delicate orchid]

well a cross product gives us a vector back so that throws a spanner in the inner product theory

such an awful operation

[formal ermine]

ah I forgot what cross product meant for a second lol

[young fiber]

Are inner products the same as bilinear maps

[formal ermine]

I confused cross product with dot product

dot products are inner products

[delicate orchid]

an inner product is an example of a bilinear map yus

[young fiber]

So no

[formal ermine]

inner products are symmetric alternating bilinear forms or whatever

[young fiber]

But okay

[formal ermine]

not every bilinear map is an inner product

[young fiber]

Wait vector multiplication is not a bilinear map

Bc the output is a vector

[delicate orchid]

the cross product is definitely bilinear

[formal ermine]

^

[delicate orchid]

R^3xR^3 -> R^3

[south patrol]

but isn't a bilinear form

[young fiber]

Don't think they're alternating, through they're positive definite

[south patrol]

In the sense of like lol

[formal ermine]

yes

there were some like

big words

for it

I read in some article

I forgor

[young fiber]

Ok i was looking at different sources

[south patrol]

Positive definite symmetric bilinear form

[young fiber]

Ig the codomain can be whatever

[formal ermine]

yes that's what I was looking for

[young fiber]

Just should be linear in each argument

[south patrol]

Linear in each argument

Being bilinear in an argument is incoherent

[young fiber]

Depends on book

[south patrol]

Uhh

[delicate orchid]

problem with bilinear maps is that (ax, ay) = a^2(x, y) which makes me angry... if only there was some way to fix this :trollshiro:

[glossy crag]

What's a concrete example of an infinite perfect field of char. p that is not algebraically closed (alg. closed fields are perfect)? I know that F_p(x) is imperfect, but I don't know any perfect examples. I was thinking maybe take an imperfect field K and adjoin all the p^k roots of all its elements, but idk if that's guaranteed perfect (i.e. L=L^p).

[rustic crown]

what about the colimit of F_{p^{2^n}}

this is not the alg closure because any element here will have degree of a power of 2

so in particular stuff like cbrt(some non-cube) won't be there

what i've done is basically adjoin square roots as much as possible until you can't do it anymore

[glossy crag]

Interesting

Is the colimit of fields always a field? I am not yet familiar enough with limits.

[rustic crown]

the important word is "filtered colimit"

but for now just think of it as a union

construct extentions of Fp by induction

and make sure that these are actual inclusions

F_p subset F_p^2 subset F_p^4 ...

and then actual union

from this it should be clear this is a field

uwu

[novel parrot]

does someone know how they prove that c(f_0g_0) = 1

if irreducible element divide c(fg) then why does showing that it divides c(f) or c(g) make c(f_0g_0) = 1

[rustic crown]

they'll do it in the next proposition or something

they reduced it to proving prod of primitive is primitive

[novel parrot]

ok

but i dont understand that last sentence

"that if an irreducible element of R divides c(fg) then it divides c(f) or c(g)"

how does that imply that c(f_0g_0) = 1

[rustic crown]

oh

it doesn't imply that

basically you might also have to keep track of powers

saying just for irreds not enough

like 12 = 2 * 3 * 2

every prime dividing 12 divides either 2 or 3

so the last 2 isn't forced to be 1

[novel parrot]

i see

thats the full proof

i dont understand why c(f_0g_0) = 1

[rustic crown]

oh wait i see what they wanted to say

[rustic crown]

they want to show c(fg) = c(f)*c(g)

so it suffices to show c(f0g0) =1 right

do you agree with that?

[novel parrot]

yes

[rustic crown]

so we want to show that product of the primitive polynomials f0 and g0 is also primitive

that is we have reduced to the case where c(f0) = c(g0) = 1

and we want to show that c(f0g0) = 1

they rename these f and g

so now you have c(f) = c(g) = 1

that's why they start by taking a prime divisor of c(fg)

[novel parrot]

i dont understand

[rustic crown]

which part

if c(fg) was not a unit, then a prime would divide it

and tehy show this means at least one of c(f) or c(g) is also divisible by that prime

so you can't have c(f) = c(g) = 1

[novel parrot]

yes

but why does that show that c(f_0g_0) = 1 where f_0 and g_o are priminitve

[rustic crown]

because it has no prime divisors

so it must be a unit right

[rustic crown]

this was the confusing part

[novel parrot]

i see

and in a UFD there is no distinction between irreducible and prime elements yeah?

[rustic crown]

non-zero prime and irred yea

[novel parrot]

understand

[rustic crown]

uwu

[novel parrot]

@rustic crown but do not we also have to prove that it doesnt divide c(f_0g_0)

since a prime can divide c(f) or c(g) but still divide c(f_0g_0)

[rustic crown]

you wanna show c(f0) = c(g0) = 1 implies c(f0g0) = 1

you prove the contrapositive

c(f0g0) not unit should imply c(f0) or c(g0) not unit

[novel parrot]

oohhh

[rustic crown]

which is same as proving if c(f0g0) is divisible by some prime, then at least one of the other two are too

det will go

good night uwu

[novel parrot]

good night

[green shell]

something about this proof is just not registering for me

it looks like the inductive hypothesis p(r) is

$p(r):$ id $=\tau_1\tau_2,...,\tau_r \rightarrow$ $r$ is even

[cloud walrus]

cryptic-spinach

[green shell]

but it sounds to me like we're just verifying that the inductive hypothesis holds for even number

[formal ermine]

just use the sign function lol

[rustic crown]

That's one way to prove sgn is well defined

[formal ermine]

[rustic crown]

det didn't sleep after all

[green shell]

but we're not necessarily disproving that id can't be written as a product of odd transpositions

[rustic crown]

.<

[formal ermine]

bad

[rustic crown]

yea det is bad

[green shell]

uh oh idk what sgn is

[rustic crown]

if you write a permutation as a product of transpositions, then sgn is 1 if there are even number of them and -1 if there are odd number of them

[green shell]

ok just looked it up and its not so scary

[rustic crown]

you're proving that this function is well defined

at least one way to prove it is that

[rustic crown]

what

you need to prove r is even

by induction you get r-2 is even

you didn't know r was already even, so doesn't make sense to say that you're only verifying it for even r

for even r, there is nothing to verify

cause that's what you want to prove

if you assume r is even, then what will you do lol >.<

:p

[rustic crown]

btw that's not exactly the induction hypothesis, as not all cases reduce to a smaller p(r)

only the first case reduces to that

[green shell]

yeah I'm looking through the rest of the proof and I'm not sure they actually prove that r is even

[rustic crown]

btw just a question, what is your group here?

S_n

(want to know what you call it, is it n?)

[green shell]

I'm good with calling it S_n

[rustic crown]

okie

this is what we're actually proving.

if id = product of transpositions t1 * t2 ... tr and a is in {1, 2, ..., n} then you can write id = s1 * s2 * ... * sk such that parity of r and k are same but this time these s_i doesn't move a at all. and the stuff these s_i move are a subset of what t_i moved

writing it out makes it eww

but that's the idea

so if you keep doing this, after a while you can write id as a product of transpositions which doesn't move any element of {1, 2, ..., n}

that means you had no transpositions

so r and 0 have same parity

let me know if that makes sense

or else i'll try to phrase it a bit better

.<

once you understand the proof, writing it won't be a big deal

[green shell]

when you say s_i doesn't move a at all do you mean that a is not included in any of those transpositions

[rustic crown]

it's just tedious

[rustic crown]

yee, that's what i meant

(there are better to write proof out there, i can tell you one if you're interested)

[green shell]

yus please I more interested in understanding the result than understanding this particular proof 🙂

[rustic crown]

it's not a bad proof, like if you're stuck and scratch your head for an hour... more likely this would be the proof you end up coming up with

i like the other proof because it defines the sgn homomorphism directly

but then your work changes to proving that transposition have sgn -1

so you would still have to work a little

basically "conservation of work"

try understanding this proof first maybe

[green shell]

cool yeah I'll spend a bit of time with the one you provided

thanks 🙂

[rustic crown]

i'll say the idea once more

if you write id as a product of r transpositions, then you can do one of two things... either write it as a product of r-2 transposition, or decrease the index where the letter "n" appears in the product

if the first thing happens then you're done

if the second thing happens, by induction you can assume that none of the transpositions had an "n" in them

so this is actually a permutation of S_{n-1}

and so r would be even, this time by induction on n

sort of like double induction

[glossy crag]

So I was toying with the $K[x]/(f)\otimes_K L\cong L[x]/(f)$ isomorphism and I was wondering whether it generalises. If $A,B$ are algebras (over a field or comm. ring) and $I\subset A$ is a 2-sided ideal, does $A/I\otimes B\cong (A\otimes B)/(I\otimes 1)$ hold, where $(I\otimes 1)$ is the ideal generated by $I\otimes1$ in $A\otimes B$?

[cloud walrus]

Ocean Man

[rustic crown]

yep it holds

[glossy crag]

Exactly as stated?

[rustic crown]

wait lemme think again

i had commutative stuff in my head

basically this should reduce to "right exactness of tensor"

you have exact sequence
0 --> I --> A --> A/I --> 0 of k-modules

[glossy crag]

If it holds for comm. algebras, good enough.

[glossy crag]

Ah.

[rustic crown]

then you tensor with B

to get
I ⊗_k B --> A ⊗_k B --> (A/I) ⊗_k B --> 0

the left most map may not be an inclusion though

you can't say much about the left most object in general

but you only care about it's image

which is the ideal generator by stuff like i⊗b

or simply i⊗1 (because you generate it as a A⊗_kB-module)

which is what you had written

only thing i would have to check in the non-commutative version is left-sided ideal vs two-sided ideal

[glossy crag]

That's much cleaner, thanks.

[glossy crag]

Well you want a 2-sided ideal so that A/I is an algebra, no?

[rustic crown]

all this works for modules, which is why i haven't thought about it much

so if I is an ideal in A

then the stuff generated by i⊗1 would also be an ideal in A⊗_k B right

[glossy crag]

Great, thanks.

[rustic crown]

the image would be linear combination of i⊗b

and this is an ideal

yea it works

k here is just a commutative ring, not necessarily a field

[rustic crown]

okie this was a bad thing to say, i didn't really say what i'm generating it as

[glossy crag]

as a 2-sided ideal, no?

[rustic crown]

yea but then that statement has no content in it

the ideal generated is an ideal

the actual thing is the left B-submodule generated by i⊗1 where B acts by multiplication on the second entry is actually a 2-sided ideal in A⊗_k B

but this has too many words

[rustic crown]

which is why i like this more

i need to show that image is an ideal, that's all.

after that check that this ideal is generated by those i⊗1

[glossy crag]

I just did it by hand.

I⊗B is the image of the first map, right?

[rustic crown]

no

not necessarily

the map may not be injective

[glossy crag]

Sure, but if you tensor the inclusion I->A and identity B->B, don't you get as a result I⊗B in A⊗B?

i⊗b is sent to i⊗b

[rustic crown]

yea but the point is this

unless like k is a field, where it would be injective

lemme give an example maybe

0 --> 2Z --> Z --> Z/2Z --> 0
and B = Z/2Z

then (2Z)⊗Z/2Z --> Z⊗Z/2Z is the 0 map

but the left object is non-zero

(tensor is over Z)

[glossy crag]

I've seen such examples before, but I don't see why this doesn't hold, just unwinding definitions

the map is inclusion tensored with identity

[rustic crown]

yep

[glossy crag]

each elementary tensored is unchanged

[rustic crown]

well kind of

[glossy crag]

sure, the resulting sum of elementary tensors might be 0, but that's beside the point, no?

[rustic crown]

notice that teh tensor 2⊗1 is sent to 2⊗1

but on the right this 2 lives in Z

not in 2Z

so you can write it like 2*1

and (2*1)⊗1 = 1⊗2*1

[glossy crag]

I see, but that's still an afterthought, no?

[rustic crown]

but multiplication by 2 in Z/2Z gives 0

[glossy crag]

The image is still I⊗B

it's just it happens to be 0 inside A⊗B

does that not make sense?

[rustic crown]

yea but I⊗B need not be 0

in the above example (2Z)⊗(Z/2Z) is isomorphic to Z/2Z

[glossy crag]

By itself maybe, but inside A⊗B it might? IDK, maybe I've got it wrong, I'm kind of sleepy

[rustic crown]

image can't be both Z/2Z and 0 right

[glossy crag]

I mean if inside Z⊗Z/2Z you take the set 2Z⊗1, then the ideal generated by will be 0, right?

[rustic crown]

but you won't call it that set right

[glossy crag]

When I say "inside A⊗B", I mean the set of sums of elementary tensors with first coordinate in I

[rustic crown]

yea this is true

[glossy crag]

So the elements might have some relations between them by virtue of sitting inside A⊗B that they might not ahve otherwise

[rustic crown]

the confusion arises because there is a difference between writing A⊗B vs a⊗b

first refers to some module

second is an element

[glossy crag]

OK, let me restate what I meant: the image of the module I⊗B in A⊗B is the set I⊗B, i.e. the sums of elementary tensors i⊗b.

Is this correct now?

[rustic crown]

so when you say I⊗B, this is not same as ⊗ of elements inside A⊗B

yep

[glossy crag]

Sure, I understand. I just explained it poorly

[rustic crown]

I⊗B contains sums of i⊗_1 b

and the image is sums of i⊗_2 b

[glossy crag]

Sure, it's different tensors

[rustic crown]

sowwy >.<

you were sleepy >.< and i made you say all that >.<

[glossy crag]

and the set I⊗B in A⊗B is 2-sided ideal that happens to coincide with the 2-sided ideal generated by the set I⊗1.

Correct?

[rustic crown]

yee

[glossy crag]

Nah I should've worded it better, you don't know what I might've meant.

[glossy crag]

and finally we get (A⊗B)/(I⊗1)\cong A/I⊗B as algebras

the projection map tensored with identity A⊗B-->A/I⊗B is easily seen to be multiplicative

[rustic crown]

yee

[glossy crag]

Wonderful, thanks for all the help.

Gonna go to sleep now, gn.

[rustic crown]

oyasumi

[rustic crown]

why can't det sleep

[novel plover]

What's the identity element and inverse of nonzero complex numbers under multiplication?

1+i is the identity?

[chilly ocean]

do you think it is?

what is the identity by definition?

det go to sleep

close discord and your eyes

[rustic crown]

okie i'll try the first thing

can't say for eyes

[chilly ocean]

i phrased this strangely. rather: "what is the definition of the identity?"

[vivid tiger]

let us see.
z*x = z where x is identity
z * z^-1 = x
Remember that identities are unique as complex numbers - {0} are a group under multiplication. So find an identity, and that's the only one.

[novel plover]

Well, I'm looking for a c+di such that when I multiply it by a+bi gives a+bi

[vivid tiger]

correct

but you are overthinking this

[novel plover]

Nope, 1+i wont work

[vivid tiger]

you are correct

[novel plover]

Oh 1+0i

[vivid tiger]

yep

what about inverses? also, don't overthink this.

(really you could quickly get to knowing 1 is the identity by realizing the reals - {0} is a subgroup and therefore has the same identity)

[novel plover]

I'm looking for c+di that gives 1+0i when multiplied by a+bi

Got it. Thank you everyone

[white oxide]

is it true that if G is a finite cyclic group with n elements, then if a is a generator a^n = e?

[primal tusk]

I need help proving part c

One way

I’ve proved if gcd = 1 that it’s an isomorphism

But I need help proving if it’s an isomorphism that the gcd = 1

[void cosmos]

just chinese remainder theorem it

u have the idea ig

[primal tusk]

I don’t know what that is 😂

I’ll look into it

[void cosmos]

its like

suppose u have some equivalence y = x mod n

then umm

let p_i be the prime factors of n

then there is a solution in each mod inside the p_i

its basically something similar to what ua rgued

not only that but is x

lmao

its actually what ur trying to prove

but the algebraic version of it

ie Z/nZ is isomorpmihc to Z/n1Z x Z/n2Z.....

where each ni and nj are coprime to each other

[primal tusk]

Okay I’ll try to understand that

im not really seeing the connection

[lapis latch]

If the gcd isn’t 1 try finding a non trivial element of the kernel

[primal tusk]

ahh thanks ill try that

[delicate bloom]

for this problem that'd be circular

[void cosmos]

Yea ig

[inner wave]

Reading Dummit and Foote as an accompaniment to my abstract algebra class (because the actual class textbook isn't vibing with me) and want to do some exercises but like...theres so many...

andy advice on picking what problems to do/which are important?

[chilly ocean]

Then it fits since Z/nZ is on the circle

go through each one and flip a coin

[inner wave]

lol

that would probably still leave me with too many lmao

There are 36 exercises for section 1.1

[chilly ocean]

i usually just pick the ones that look interesting

[inner wave]

alright fair

[chilly ocean]

36 is not that much

i don't know which one to pick

[inner wave]

36 for like a whole chapter is fair

I think around 50 is good for an entire chapter

but for like the 1st section of a chapter with 6 sections, 36 is a lot lol

[chilly ocean]

Try to avoid computational ones and pick anything that looks interesting or you don't know how to solve

[inner wave]

alright

[chilly ocean]

Typo

[primal tusk]

ive spent the last 14 hours doing 4 problems so honestly 36 problems sounds really bad

[chilly ocean]

Two problems don't have to be equal in difficulty

Since it's the very first section 36 exercises doesn't sound that bad

Sure you might have problems if they constantly throw 30 exercises at you

[inner wave]

Everytime I self study something it goes well until I reach the exercises and think "now what"

[inner wave]

Most of them are proofs

I think most books should have a "recommended exercises" list

or at least star by importance

[toxic zephyr]

say you have some product of cyclic groups $G=C_{n_1}\times\ldots\times C_{n_t}$. does there always exist a positive integer $m$ such that $G$ is isomorphic to $(\bZ/m\bZ)^\times$?

[cloud walrus]

nilpotent nix

[coral shale]

chinese remainder thm?

[toxic zephyr]

hmmm. what if the n's aren't coprime?

at least i think that's what you need to use the CRT

was messing around with Z/7Z^x and Z/9Z^x and how it seems they're both isomorphic to C6. but i was wondering if there was some m where the product group is isomorphic to C2xC3, and if there's a way to find it in general (if it does indeed exist). obviously, phi(m) would have to be equal to n1*...*nt. but i was hoping there was a better way than just testing every integer that has phi(m)=6.

[coral shale]

sorry this was just me guessing

i cant remember the statement, but looked v similar

hm yh it needs pairwise coprime

So whats got C2 x C2 structure? 6?

Oh wait

[toxic zephyr]

maybe it is sometimes impossible. i saw that U(p1^k1...pt^kt) (my professors notation) is isomorphic to U(p1^k1)x...xU(pt^kt), and since all of them have order (Euler phi function image) which is even (or 1), then only products of even integers can possibly work.

[toxic zephyr]

yeah

[coral shale]

so now u want to...

look at the image of the totient function

[coral shale]

oops

8 more likely