groups-rings-fields

[delicate orchid]

yeah, the index of the subgroup in the normaliser is (p-1)^2

subgroup is order p^{p+1}

so it should be (p^2)!/(p^{p+1}(p-1)^2)

there we go

[teal vessel]

thoughts: given a permutation σ=(1 2 3 4 ... n), the group generated by σ should be isomorphic to the nth cyclic group, yes? so this question is really asking me to first take all all these values mod 12 and then (potentially) write out what remains (if I feel like it afterwards).

[delicate orchid]

yeah that's correct

not sure what you mean by "what remains"

[teal vessel]

the actual permutations, but for example 13 is congruent to 1 which is congruent to -1211, so I'd only do it once and say "see earlier case"

[delicate orchid]

yeah that's fair

[teal vessel]

huh, I haven't dealt with permutations this big before, interesting little pattern to see that squaring it splits it into an even cycle and an odd cycle

oh, meta-pattern, performing a permutation n times splits it into x cycles where x is the gcd of n and the size of the permutation

I love how abstract algebra and number theory are just the same thing (joking, but the overlap is beautiful)

[delicate orchid]

yeah it's cool isn't it. Can you see a pattern in the lengths of those cycles as well?

[teal vessel]

haven't seen an example of an iteration where the gcd is both nontrivial and less than n yet, but I might do an example of one just to see

[delicate orchid]

try sigma^7

[teal vessel]

right now they're just generating arrays, which is pretty obvious given that every element is originally getting mapped somewhere, and no element gets home early with this permutation.

[delicate orchid]

this also generalises in a really swag way. If you have an n-cycle (1,2,...,n) and raise it to the power of k, then the sets of numbers in the each cycles it splits into will be exactly the cosets of (Z/nZ)/(k)

for example, n = 12, k = 2, this splits into (1,3,5,7,9,11)(2,4,6,8,10,12), and the cosets of (Z/12Z)/(2) are exactly 0+(2) = {2,4,6,8,10,0} and 1+(2) = {1,3,5,6,9,11} (we have to set 12 = 0 cause we're working mod 12 lol)

[teal vessel]

yeah

[delicate orchid]

I even think this works if you take an infinite cycle in S_{\infty}

[teal vessel]

but each sub-permutation is also cyclic, and since every element gets used up in these cyclic permutations, it seems to me that it must be the case that if there are n sub-perms, each one will be m long, where nxm is the length of the original generating permutation, so for example if I take sigma^8, 8 is congruent to -4 and so it should be 4 sub-perms of length 3, with the permutations cycling in reverse order (just doing it off the top of my head, so that last statement about the nature of each cycle is a guess), which should track with the fact that (8,12)=4, and 12=4x3

it's like the camera shutter effect that makes wheels look like they're spinning backwards when the framerate is just right, 8 skips so far forward that it's the same as just going 4 backwards

[delicate orchid]

yes, exactly

this is true for powers of any permutation it's just really nice for cycles
good thing the cycles generate S_n huh

like, (1234)(567)^2 = (13)(24)(576), same princple at play

[teal vessel]

I'm sure I'll get more used to it if I play with more permutations, I just haven't worked with large order permutation groups much

I suppose to get a really good idea for it I should mess around with S_12 or S_24 for a little bit, since those are highly composite numbers and can give better insight

these little insights are the reason I'm doing every example of D&F, not because every example is exciting, but any example could lead to little discoveries.

[delicate orchid]

this is how I'd picture sigma^3

[teal vessel]

yep, that's exactly what I did lol

which then makes it make perfect sense why (sigma6)^2 has to be identity. It simultaneously is acting on six unique sub-cycles of length two, and since they're skipping every other one, each one has no place to go but back to itself.

[delicate orchid]

you can even go CRAZY!!!

the number of sides of the polygons are the cycle lengths wooohooo

this is getting group actiony now

[teal vessel]

would it not be more... aesthetic... to place 1 at the top and have identity be just off to the left?

[delicate orchid]

this is how i think about subgroups of cyclic groups

[teal vessel]

since all the cyclic groups here have to start at 1 anyway?

[delicate orchid]

except there are other cyclic groups hidden away

[teal vessel]

well, at that point, showing "all possible trangies" on the clock would hit the 12 anyway

so yeh

when abstract algebra is just geometry

[delicate orchid]

so these squares are cosets of C_12/C_4 which is really cool

also a good way of showing that D_n is a subgroup of D_m for each divisor n of m

[teal vessel]

I've told my students so many times in the last week "oh, I was just learning about this yesterday" when they're doing 4th grade work. expertise in a skill really is just mastery and exploration of the basics, just like my horn teacher said.

literally just "hey, you remember how adding commutes? what if it didn't?" boom, abstract algebra.

[coral spindle]

Yo that's cool, the left cosets of <g> in G are the orbits of the map h |-> gh it all makes sense

Writing this down to put on an exam in 15 years

[delicate orchid]

Yeah whenever I think about group actions it’s either via the lens of a representation or through these funny shapes

Or a disjoint union of perfect graphs

Depends what mood I’m in

[teal vessel]

something about the number of star polygons possible seems significant to me.

given x generates a pentagon, I suppose x^2 would generate a pentagram, then x^3 a negative pentagram because x^3=x^-2, and x^4=x^-1

[delicate orchid]

interesting observation, however if we take a hexagon in this dodecagon, then the square of that hexagon is a triangle

but you're on the right track I think

[coral spindle]

Well it's quite simple if you know the sizes of subsets of Z/n

[teal vessel]

well, even n-gons don't have regular star variants

[coral spindle]

The subgroup generated by m is of size n/gcd(m,n) = lcm(n,m)/m

And of course, as noted, the orbits are cosets. They're all of that size.

What you're doing by squaring this, for example, is doubling m. So it will change depending on the divisibility of n by powers of 2

For example if we're working with n = 12, with m=3 we get 4-gons and with m=6 we get 2-gons.

I.e. lines :)

[delicate orchid]

wtf non-simple graph MODS

[coral spindle]

However with, say, n = 15, and m = 3 we get 5-gons, and with m = 6 we still get 5-gons since 15 is not even. Very simple, very epic.

It really boils down to that.

[delicate orchid]

I do actually think that 2-gons are different than lines but that's a different convo

[coral spindle]

Well they visually appear to be lines so I thought I would mention that

I agree in principle

[delicate orchid]

the simplical sets!!!! they're different!!!

[coral spindle]

"simplicial complexes" names dreamed up by the deranged

How can it be SIMPLE and COMPLEX?

[delicate orchid]

cosimplical mplexes

[coral spindle]

Anyway, whether or not sigma^m generates a star depends, therefore, on the difference between m and gcd(m,n), which is a minimal (in Z) generator of <m + nZ>. We can draw a k-gon by skipping some number of vertices every time, and as it happens the number of vertices we skip is going to be m/gcd(m,n). You get more starriness the more distant this is from a multiple of k, which I remind you is k = n/gcd(m,n).

I have digressed.

[lime depot]

I have the set ${i\in\mathbb{Z}_n;|;\operatorname{order}(i)=d}$ for some $d\in{1,\dots,n-1}$ dividing n. How do I show that the size of this set is the order of $\mathbb{Z}_d^\times$?

[cloud walrus]

Beous

[lime depot]

I really just want to know where to start

[coral spindle]

What does order mean here? In Z mod n or the group of units of Z mod n?

[lime depot]

in Z mod n (not the units)

order of units of Z mod d is cardinality of units of Z mod d

[coral spindle]

I knew what you meant for the second use.

Are you aware of a way to calculate order(i) given n and i?

Perhaps that would be a good way to start.

[lime depot]

I know it is n/gcd(i,n)

so then order(i)=d iff n/d=gcd(i,n)

i want to show this is equivalent to gcd(i,d)=1

[coral spindle]

Sounds like something to work on

[teal vessel]

Q is dense, and therefore not cyclic, and the cartesian product of non-cyclic groups is not cyclic. ■

[coral spindle]

Groups have no notion of topology, so this argument is invalid.

You cannot talk about density without adding a lot more structure than just a group structure.

[night onyx]

If QxQ was cyclic think about (1, 0) and (0, 1)

[coral shale]

bruh thats the solution

!nosols

[flat tree]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[lethal dune]

give an example of noncyclic groups s.t. taking product becomes cyclic

just curious

[coral shale]

whats the definition of cyclic first of all. generated by 1 element?

to be sure we re on the same page

[coral spindle]

Yes

[lethal dune]

ye

[teal vessel]

yes, and you can't generate Q in one element, let alone QxQ

[coral spindle]

Note that you haven't proved that Q isn't cyclic yet

[teal vessel]

"Q is dense" is me being cheeky because I already proved Q isn't cyclic in an earlier problem and don't feel like retreading it

[coral spindle]

OK

[south patrol]

Dense in what lol

In itself?

[lethal dune]

valid argument but it assumes you already know it's subgroup of R

which has a topology

[south patrol]

Tbf dense in itself has a technical meaning too

[coral shale]

no, thats dense-in-itself

[coral spindle]

You would have to show a lot more to finish this

[lethal dune]

everything is dense in itseld

[lethal dune]

you do

[south patrol]

Yes in one sense

[lethal dune]

also another argument, Q is divisible, no divisible module over PID is finitely generated

[south patrol]

If Q x Q were cyclic then its endomorphism group would be abelian, but M2(Q) isn't

[lethal dune]

QxQ is again divisible

[south patrol]

Or Q isn't even of finite type over Z by using the Jacobson property

[coral spindle]

Q is an injective Z-module yet no cyclic Z-module is injective

[night onyx]

Actually interesting question, if a topological group G has a dense cyclic subgroup is G also cyclic?

[lethal dune]

same thing I said

[coral spindle]

True true

[rotund aurora]

fake

[coral spindle]

No, take the indiscrete topology.

[night onyx]

wow fake. Now it makes sense

[coral spindle]

Maybe if it's Hausdorff it would work.

[teal vessel]

but more formally, Q cannot be cyclic with addition because for any size step n (which would be the generator) I can always find another element of Q between n and 2n

[south patrol]

I like how that still isn't formal lol like

Idk isn't that the intuition

[teal vessel]

hence "more." I suppose "concretely" would be the better descriptor

[rotund aurora]

uwu

mmh

[lethal dune]

show Q has no finite index subgroup

[topaz solar]

Q

[lethal dune]

congo you just broke math

what's the hardest way to prove Z is cyclic

[rotund aurora]

[teal vessel]

yes, but it's like 1 step off of the exact mathematical statements needed to genuinely write it formally, rather than like 7 steps

[south patrol]

Sure lol I didn't know what'd be less formal then this

But maybe there is smth I am missing

Anyway

[summer path]

Start counting...

[south patrol]

Another funny proof: the exterior square of Z vanishes, whilst the exterior square of Q x Q doesn't

[lethal dune]

what's exteruor square?

[south patrol]

As in like lambda^2

[teal vessel]

q is not cyclic because I can't roll a circle and hit every element of Q, QED

^less formal

[south patrol]

Fair

A tautology lol

[lethal dune]

why

you can have a QxQ sided dice

[south patrol]

If Z x Z were cyclic then since Q is a flat Z module, Qx Q would be a 1 dimensional Q vector space

Lol

[teal vessel]

tried, dumb thing doesn't stay upright when extrapolated to 3d

[coral spindle]

"The field Q(sqrt(2)) is of dimension 1 as a Q-vector space since it looks like a line and a line is of dimension 1"

[rotund aurora]

?? Fake reasoning

[teal vessel]

precisely

[rotund aurora]

Does Q look like a line?

[teal vessel]

"epicycloids on circles of positive radius cannot have a period of 0"

[coral spindle]

Sure it looks like a line when I draw it in C

[teal vessel]

an infinitely densely dotted line. Those pesky irrationals seem to poke holes in it tho

[summer path]

Wew can't spell?

[delicate orchid]

I can, it's a desync issue

[teal vessel]

it's spelled RTUE, can't you read?

[delicate orchid]

this never used to happen a few years ago, discord have actively made their reaction system worse as time has gone on

[teal vessel]

probably helps prevent reacts sending bad messages

at least that'd be a public facing reason to make a system intentionally worse

[rocky cloak]

Fun fact: the Galois group of any algebraic extension of a finite field is "topologically cyclic" generated by the Frobenius automorphism, in the sense that it has a dense cyclic subgroup.

So for example the p-addic integers or the profinite completion of Z would be such examples.

(these examples are also Hausdorff if you were asking about that)

[delicate orchid]

that's pretty crazy

[night onyx]

that's really cool!

is there a class of spaces where it's true? (like topological groups with a metric topology)

like it feels "intuitively" true if the group topology is nice enough

lol only intuitively though

[rocky cloak]

I mean the p-addic integers is a compact metric space. Not sure what other niceness conditions one would ask

[coral spindle]

Oh lmao yeah the p-adics are hausdorff I completely forgor

[night onyx]

lol ngl, I never studied p-adics in school

nothing but analysis and rep-theory related courses after undergrad

[delicate orchid]

I just learnt about completions instead of p-adics specifically

if you know what a categorical limit is you know what the p-adics are

[night onyx]

I did a bit of reading after school just to know roughly what the hell people are talking about

yeah at this point I know what they are and basic stuff, but thats it

we had a great prof who did p-adic analysis and representation theory, I always wanted to take a course but it never happened!

[rotund aurora]

p-adic analysis > real analysis > complex analysis

[south patrol]

The third is just studying boring functions, the second is first year undergraduate so this is clear

[rocky cloak]

I feel like p-adics falls under the umbrella analysis and rep theory

[night onyx]

lol that doesn't change that I never studied them

[coral spindle]

And not number theory?

[rocky cloak]

Number theory is it's own umbrella containing all the other branches of math

[coral spindle]

Alright the number theorists already have an inflated ego, don't encourage them

[teal vessel]

but set theory tho

[sly crescent]

Every branch of math is like that.

[rotund aurora]

no

[delicate orchid]

if a mf walked up to me and said I was doing set theory I would roundhouse kick that mf to Enceladus bro

[white oxide]

Lang says "it is important to emphasize here that there are some categories for which the set of morphisms is not an abelian group" intuitively, is this not the case for most categories? i'm thinking of mappings as the morphisms in my head rn, which are obviously not commutative right? idk

[delicate orchid]

yes it's true for basically every category

in the statistical sense of "basically every", the probability of a random category being preadditive is 0

don't ask me what distrubution I'm using I don't care

[white oxide]

lol okay

[delicate orchid]

but.....

a good proporition of the interesting ones are

[rotund aurora]

the operation when Hom(A,B) is an abelian group is not composition

[rocky cloak]

I thought we agreed all categories are additive, I will not stand for this

Don't take my lovely biproducts away

[delicate orchid]

no no they're additive. Just not preadditive

[delicate orchid]

good luck making that into a group zoo we mama...

[white oxide]

please help is this problem doable without prior knowledge of tensors

[delicate orchid]

the COPRODUCT??

:uponthewitnessing:

[delicate orchid]

and yeah if you do the second thing you show the first

alternatively, show that any bilinear map f : Z[t^\pm] x Z[t]/(t) -> R must be 0 for any ring R

where the x is the standard direct product, not tensor

[white oxide]

damn okay i'll just look up what a bilinear map is then

[delicate orchid]

...

did u skip ahead

also the use of the word "coproduct" is REALLY throwing me off

and these aren't rings - they're Z[t]-modules! Wtf is going onnnnnnnnnnnnnnn

[white oxide]

we didn't even go over bilinear maps and now we're doing some category theory

then again

it's our ta teaching this week since our prof is out for a conference

so he's taking this opportunity to try to murder us

[delicate orchid]

are they using \otimes for the coproduct? but then why is there a _{Z[t}}?

a pushout?? some kinda van kampen lookin ass

[night onyx]

lol tell your TA to chill, you can't do much with tensor products without first talking about bilinear maps

[white oxide]

lol yea... this homework looks fucking brutal

[night onyx]

thats a homework question?

[delicate orchid]

tell ur TA that tensor products aren't fucking coproducts as well

[night onyx]

have you guys learned about tensor products?

[white oxide]

nope

we're covering that later in the syllabus

[night onyx]

not even in an abstract way, just R^n \tensor R^m

ahhh gotcha

[delicate orchid]

then I would complain to the prof about this

[white oxide]

yeah maybe, i would have to do so anonymously since he's also our grader 💀

[night onyx]

ohh, unless that tensor symbol is a mistake, and it should be a sum symbol

[delicate orchid]

but then why the subscript

[night onyx]

like it seems like the homework is talking about coproducts

lol dunno!

[delicate orchid]

if it's an amalgamated coproduct (aka a pushout) then I have even more questions

[night onyx]

it seems like a "one of these questions is not like the other" sort of thing

[rocky cloak]

I mean it's the coproduct in the category of commutative rings, which seems to be what the exercise is getting at

Except it's amalgamated, so should probably be called a pushout

[delicate orchid]

therefore, I have futher questions

[rocky cloak]

It's the coproduct in the category of commutative Z[t] algebras

So you could use that I guess

[delicate orchid]

that's probably what we're after then

what a mess!

[chilly ocean]

What is an group-Grothendieck??

[rocky cloak]

Given a monoid, the grothendick group is what you get by formally introducing inverses

[delicate orchid]

^

or

[rocky cloak]

aka the left adjoint to the inclusion of the category of groups into the category of monoids

[delicate orchid]

there's some definition to do with exact sequences of modules, where we set M_3-M_2-M_1 = 0 in the free abelian group generated by isomorphism classes of R-modules iff there is a split SES M_1 -> M_3 -> M_2

I've also heard this be called the Grothendieck group

I understand that this is just formally adjoining inverses to the monoid given by isomorphism classes and direct summation but

some people specifically mean this so it's context dependant

and would instead call formally introducing inverses "groupification"

[chilly ocean]

think that the Gothendieck Group does not only discriminate the cohomology. For example, there are Chow-rings that in P^{1} produce Grothendieck-Groups with some new modular-extension

[rocky cloak]

I've also seen this without requiring the sequence to be split. In which case I guess there isn't really a natural monoid in sight.

[delicate orchid]

I guess that's why some authors care so much

[south patrol]

coproduct of algebras tbf tho

[delicate orchid]

grrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr

[south patrol]

Wait now I am confused lol

Eh it's okay nvm

[delicate orchid]

oh goodness. I was ever so worried

[south patrol]

LOL

[slate forge]

Given a prime p, the polynomial $P(X) = X^{p-1} + X^{p-2} + ... + X + 1$, how to prove that P(X+1) is monomial mod p pls

[cloud walrus]

RéaleSansChampi

[coral spindle]

Try using the binomial theorem. See what you get.

[slate forge]

The obvious thing is that the monomial is of degree p-1 but i developped for example the coefficient for x^2 and cant proove that its divisible by p

Something like $p-2 + \sum_{k=3}^{p-1} \frac{(k+1)(k-2)}{2}$

[cloud walrus]

RéaleSansChampi

[slate forge]

For p>=3 obviously else not working

[rocky cloak]

So the obvious candidate monomial should be X^p-1

P(X+1) = X^p-1 iff
P(X) = P(X-1+1) = (X-1)^p-1

So you could try expanding (X-1)^p-1 using the binomial theorem.

And worth noting that when a/b is an integer then a/b mod p is the same a times the multiplicative inverse of b mod p

[slate forge]

How would u reduce (p-1)(p-2)/2

[open sluice]

if p > 2 then 2 is invertible so 2^-1 exists
rewrite it as (p - 1)(p - 2)2^-1 and now reduce this mod p

[rocky cloak]

p-1 is -1 mod p, p-2 is - 2 mod p, so this would be

(-1)(-2)/2 = 1 mod p

[slate forge]

Thank u

[tiny jolt]

Why is the opposite ring of the quaternions isomorphic to itself?

[coral spindle]

If you're just looking for a proof, you can just note that the quaternions are the only 4-dimensional real division algebra.

[tiny jolt]

Hold on I think I got it, is it because it is the unique 4 dimensional divison algebra over r?

ok

[coral spindle]

Wow

[tiny jolt]

lmao

[rocky cloak]

If you're looking for an explicit isomorphism swapping j and k works.

[chilly ocean]

${SL(2,\mathbbm{Z}) = \left\langle x,y ;\vert; x^4=y^6=1 \right\rangle}$

[cloud walrus]

Gev

[chilly ocean]

is this correct?

shouldn't we have like x^2 = y^3 too?

I don't see how it follows from the given presentation

[spice whale]

i don't think SL(2,Z) is finite

so like

[chilly ocean]

yes its not finite

[tribal moss]

Definitely not, since it contains (1,n;0,1) for all n.

[spice whale]

yeah

that presentation gives a finite group

wait

no

it doesn't

[chilly ocean]

I dont think so since we can have xyxyxyxyxy... something

[spice whale]

hm

[chilly ocean]

I think this presentation is a simple free product but we want to amalgamate two Z2s

[spice whale]

the presentation is C_6 coprod C_4

can't see why that'd be SL(2,Z)

[chilly ocean]

yeah we can't derive x^2=y^3 so I mark this as incorrect...

[delicate orchid]

It’s C_6 \ast_{C_2} C_4 iirc

[chilly ocean]

SL(2, Z) yes

[spice whale]

what are the generators

[chilly ocean]

but this presentation doesnt seem like an amalgam

[delicate orchid]

No I agree you’re missing a relator

That’s just the free product

[spice whale]

like of SL(2,Z)

going for this kind of presentation

i can see
0 -1
1 0
potentially

[delicate orchid]

So we should have an amalgamation isomorphism, identifying the C_2 copies inside each subgroup, so I completely agree we should have x^2 = y^3

Or the other way around, I’m not scrolling up

[chilly ocean]

the generators

[spice whale]

makes sense

then YX is the funny shear

[chilly ocean]

what is a funny shear?

[spice whale]

1 1
0 1

[delicate orchid]

Or as I like to call it, Z

[spice whale]

or
1 0
1 1

[tribal moss]

What would a boring shear be, then?

[delicate orchid]

Anyway was the question just asking for a presentation of SL(2, Z)? Slightly confused

[rocky cloak]

The identity is a very boring shear

[delicate orchid]

1 0
0 1
Can’t get much more boring

[spice whale]

i wouldn't use a shear to bore
there are much more effective tools

such as drills

[chilly ocean]

no it was just included in a bigger homework question and I wasn't sure if it's a mistake

[summer path]

interesting way to spell point-set topology

[crystal turtle]

topology uwu

[summer path]

point-set

[spice whale]

a drill is just a closed ball

[hollow mica]

for most objects in algebra, they [their operation(s)] usually come with associativity, and then a certain class of them are also commutative

how come we rarely see the other way around

Like if commutativity is there then associativity is almost always there as well

bruh the top answer to this MSE question on examples of operations that are commutative but not associative is the outcome of rock-paper-scissors… bffr

[open sluice]

because life without associativity is not living

life with commutativity but without associativity is not significantly better

[rocky cloak]

The reason is simply that there are very few non-associative operations we study, but if you look at the ones we do you can come up with commutative examples.

The most common non-associative structure is a lie algebra. These are anti-commutative, which besides the name is basically the same as commutativity, and in fact in characteristic 2 lie algebras are commutative.

Besides lie algebras the non-associative structure become a lot more niche, but you have Jordan algebras, which are always commutative. You have quandles which can be commutative, and then I don't really know many more examples.

As for why associativity is so important it's that it's often convenient to think of
y |-> x*y
as a function. And then you'd want * to correspond to something like function composition. Since function composition is associative you'd need associativity to do this.

Lie algebras get away with something similar by having the bracket correspond to commutators of functions instead of composition.

[slate forge]

How would you say that a polynomial irreducible on Z/pZ[X] is irreducible on Q[X]
I got trouble to understand what would be the minimal set of conditions

And if somebody got some notes on details about reductibilities of polynomials that would be good

[coral shale]

Z/pZ and Q[X] are different wut

[slate forge]

Plz

[coral shale]

Z/pZ[X] ???

[slate forge]

Yeah but you use reduction mod p

Yes my bad

[coral shale]

Can you give an example of a polynomial irreducible in some Z/pZ[X] but not in Q[X]

I dont see how thats possible rn

https://math.stackexchange.com/questions/271327/irreducibility-in-finite-field-implies-irreducibility-in-mathbbz

ah ok, so u can put in coeffs which are 0

think the answers are in there

[slate forge]

Alright lets be clearer on my problem

For example i dont see how to go from Z[x] to q[x]

[coral shale]

euclids lemma

[slate forge]

?

[coral shale]

im high

gauss' lemma

the other useful ones are eisenstein, and rabins criteria

[rocky cloak]

So you have a monic polynomial with integer coefficients, then if the polynomial is reducible in Z[x] it must also be reducible mod p.

A monic polynomial is irreducible in Z[x] iff it is irreducible in Q[x], this is Gauss lemma.

[slate forge]

Yeah but how would you know that the coefficient you extract in the gauss' lemma is not null mod p

[coral shale]

thats necessary in the assumption, see what i linked

[rocky cloak]

If you don't require the polynomial to be monic you can do things like (2x + 1)(x + 1) which is clearly reducible in Q[x], but is irreducible mod 2

Really the leading coefficient being relatively prime to p should be enough.

[slate forge]

Thanks, i wish there were examples in my notes for these, any knows some notes/books that has exercices/examples with solutions on polynomials?

[teal vessel]

"find a presentation for Z_n with one generator." Uhh... ⟨x⟩? Am I missing something?

[lusty marlin]

A presentation consists of the generators as well as the relations they satisfy.

[teal vessel]

Ok, so {x | xⁿ=1}

[coral shale]

<x | x^n>

[topaz solar]

It’s < G | R > stylized, since it’s not just a set

[topaz solar]

(Putting aside how this set is ill formed to begin with)

[hollow mica]

ty

[topaz solar]

<G|R> should be the largest group with those generators which satisfies those relations

So <x |> is just Z

[south patrol]

Using G rather than S or smth seems a little misleading to me aha

[coral shale]

< S | R >

[topaz solar]

Gemerald smh

I was gonna say Generator but autocorrect trolled me

So I’m just gonna leave it because it’s way funnier

[coral shale]

Setofgenerator

[rocky cloak]

Is gemerald even a word, like what is your autocorrect up to

[south patrol]

Lol

[topaz solar]

Lil bro is cooking nothing

[south patrol]

Here are my daughters Gemerald and Gapphire

[coral shale]

< a | a^n, a^{2n}, a^{4n}>

[south patrol]

Pain!

[coral shale]

clearly the better answer

[south patrol]

< a | a^2n, a^3n> is preferable

[coral shale]

[topaz solar]

G = <G|R(G)>

[teal vessel]

⟨x | xⁿ=1⟩ then, since that ensures the cyclic nature, and provides a single generator.

[open sluice]

i think it parsed **G**enerator as **G** + enerator ~ **G** + emerald
somehow

[topaz solar]

Ye

[delicate orchid]

Gemerald is a word but only if you have a very specific cultural background

[alpine island]

should be Jenerator

< J | R >

[chilly ocean]

so true

[vagrant zinc]

guys I was doing this exercise but I'm not getting it.

[chilly radish]

Is there more context?

It isn't clear what the group here even is

Who are x,y

[rocky cloak]

x/(1 + ex) / (1 + s(x/(1+ex))) = x / ((1 + ex) + sx) = x / (1 + (e+s)x)

[vagrant zinc]

i fout my mistake thank you

[vagrant zinc]

Search for symmetry analysis differential equations, I think you can find what you are looking for.

[chilly radish]

You should generally try to provide context when asking for help

[alpine island]

$\frac{\frac x{1+\epsilon x}}{1+s\frac x{1+\epsilon x}}=\frac x{1+\epsilon x + sx}=\frac x{1+(\epsilon+s)x}$

[cloud walrus]

nHail

[vagrant zinc]

YES!

I was putting another group transformation into my argument.

[rocky cloak]

I don't really understand what this notation is supposed to mean though. Is the point that these expressions form a group under composition? If so why not just give them as functions or whatever?

Or do the expressions describe one parameter families that generate the group? If so what is the identity?

Either way I find this a very weird notation

[vagrant zinc]

It is a transformation under a composition that is to say I have to prove that it is a group, the book gives some examples of EDO already with its transformations that in turn are groups, it does not give them as compositions because you have to replace them in the EDO to perturb that system, perturb it under the transformation that is to say make the change but without changing anything.

[rocky cloak]

So these are transformations of differential equations and form a group under composition?

[vagrant zinc]

The answer is yes, it should be a function since it is a binary operation that sends to an operation under the sum or the multiplication, in this case it can be said that it is a sum, now if you already know group, it is not necessary to place it so trivial, since I need to make an invariance in an EDO.

[crystal turtle]

Who are x and y? ShiN is asking the real question here.

[vagrant zinc]

yeah

[rocky cloak]

So like number (i) in your example, is there a family of ODEs associated with it, and how would it act on them?

[vagrant zinc]

It's like taking the circumference, every transformation you do to it will always make the same circumference, so it's invariant.

The transformation I'm going to do is a composition group.

Let's see right now I'm doing an exercise

I have my transformation where it's a group, which is a composition of functions that defines a binary operation, I've already tested.

now i need to evaluate that composition of functions in my EDO, to know if it's invariant under that transformation, so i'm seeing if it is, since it's not spitting out my transformation.

now i can take albitrary functions as a reparametrization so doing that will give it an invariance in which i am left with an easier EDO :D where i would get to the same result if i do my EDO normally.

I didn't explain the last part well, but I take my EDO and turn it into an easier one because one my EDO is invariant I base its transformation (Composition of functions which is a group). and that's it.

Later on I will add Galois theory, it is for my final thesis.

I'm just starting with the subjects :((

[vagrant zinc]

look jagr2808

[rocky cloak]

I'm still not much wiser. Like what transformation is being described by "x = e^s"? What is being transformed in what way?

[delicate orchid]

I've never seen this notation before

odd

[south patrol]

Yeah I was very confused last time too lol

[vagrant zinc]

It doesn't use the group notation that one usually sees, it just says that it is a lie group, which can in turn be a transformation, which is why I said that the book already takes as trivial or less notation when writing.

[ocean raven]

How do I categorize all ideals of Z[X]

[chilly radish]

all ideals?

That's a lot

[delicate orchid]

A fair few indeed

[crystal turtle]

At least 2.

[rocky cloak]

A third one is conjectured to exist, but no one has been able to pin it down yet

[delicate orchid]

Can we bound the number of generators at least

[rocky cloak]

Don't think something like
(p^n x, p^n-1 x^2 , ..., x^n+1)
can be written with fewer generators

[delicate orchid]

I might cry but just a little bit

[rocky cloak]

This is just x(p, x)^n though, so maybe you can factor out ideals that are products of primes or something...

I dunno, it's probably super complicated

[delicate orchid]

What’s the height of Z[x]?

[rocky cloak]

Like Krull dimension? That's 2

[delicate orchid]

Hmmmmmm

[rocky cloak]

So describing the primes is not too hard

[delicate orchid]

Yeah

Can we do primary decomposition or something

Apologies if this is gibberish I’m drunk

[next obsidian]

This doesn’t follow because it’s dim 2, you really need that it’s a polynomial ring over a PID to have a shot at doing this

[rocky cloak]

I guess polynomial ring over any ring where you know the primes should be fine.

But yeah dim 2 rings in general are probably complicated.

[next obsidian]

I don’t think so. I think this requires a Jacobson hypothesis to make good sense of

[rocky cloak]

Hmm, like if I is a prime in A[x], then p = I\cap A is a prime in A, so I corresponds to a prime in A/p[x] which doesn't intersect A/p. Then localizing at A/p - {0} gives a polynomial ring over a field, so can be described by an irreducible polynomial over Frac(A/p).

I guess you don't necessarily get like a nice description of the generators of the ideal from this... idk

But that should at least be a classification.

[warm saffron]

This message is for Jagr and Ryx (not gonna ping them but oh boy you guys are going to love this). So my prof said that for problem 7, that I need to show it's a binary operation...lol it literally says "define a binary operation on S by..." so I can assume it is a binary operation. When I talked to him about it (because he marked me down some points for not showing the operation is closed), he said that "I've read many a paper where authors will say 'define a binary operation...' and it turns out not to be a binary operation. Since the operation is not trivially a binary operation, you should show it is."

[crystal turtle]

Well

A priori you don't know that it's a binary operation just because you say it is, because you don't know that it actually takes values in S

[chilly ocean]

prof writes sloppy and then takes points away from you for not being precise lmao

[warm saffron]

Yea it's not a textbook written by him, but he did agree that there's good reasons for believing it's a binary operation based on the wording of the problem

[chilly ocean]

i had an analysis class with the exact same problem and it was the worst fucking class i've ever taken

[warm saffron]

Oh I bet

[chilly ocean]

textbook full of loose and sloppy writing and the prof wanted a ridiculous level of precision

it suuuucks

[warm saffron]

Problems like these aren't helpful. Like literally just say "define an operation by..." and then I would've showed it's a binary operation lmao

and yea if there's good reasons for believing we can assume it's a binary operation and good reasons for not, just lemme have the point loll

Also, Ryx, actually we know it takes values in S because it was at the very least an operation defined on S! I think you meant to say, we don't know it sends values to S

[elder wave]

A friend of mine had that exact problem in an exam

[crystal turtle]

That's what I mean by "takes values in S"

[open sluice]

takes values in S = range is contained in S

[warm saffron]

takes seems like you meant domain

[elder wave]

For one a in R … defines a riemann distribution

So naturally one would assume that and show that a= whatever

But got points taken off at the end for not proving again that it is in fact the thing it was assumed to be

[warm saffron]

range is contained in S...that means the image of the operation star is a subset of S...I'm lost on the takes part...

LOL Timo

anyway, now I know you meant where it gets sent so we're on the same page now

[open sluice]

it's idiomatic in english math
maybe you've heard "the function f takes on only positive values"

[warm saffron]

ahhh takes on definitely triggers the range for me

[open sluice]

well I suppose the correct phrasing should be "takes on values"

although I thiiink "takes values" is also occasionally fine?

[crystal turtle]

[open sluice]

i hate english

[crystal turtle]

No other languages have problems

[warm saffron]

yea...okay either way...we're on the same page now. Thanks for the explanation Bladewood...I forgot about that takes on phrase kinda like the function f achieves values...

[open sluice]

ye
it's just funny

[warm saffron]

fr fr

luckily the prof ended up being prety chill about it and said "i'll make a deal with you and give you two options. Option 1, prove it's closed and I'll give you your point back. Option 2, if at the end of the semester you differ from a better grade by 1 point, I'll bump you up." I ended up taking option 2 and proving it's a binary operation just for my own practice (by showing it's well-defined and closed...I accidentally did closure first lol) but he only really cared about the closure part.

[coral shale]

Yh I think its rather silly to say "G is defined to be the group with the underlying set S and operation *"
if you don't know G is a group

or similarly, f is the function that... without knowing it is one

You surely wouldn't use the same phrasing for "let f be an isomorphism defined by..." without having shown it is one first, would you? idk

[crystal turtle]

"it's an isomorphism because I said so"

[rocky cloak]

I guess this is why some professors emphasize "closure" as an axiom even though that's already part of the definition of being an operation. Just a helpful reminder to check that what you defined actually is an operation.

But yeah, there's definitely more clear ways one could have worded this exercise.

[warm saffron]

100% I agree. The wording is strange if we can’t assume the operation is binary

That's one of those things about proof problem statements...if the problem statement is worded in a way where I would be led to believe I can assume something, I'm not even going to bother verifying it's true. I'm just going to assume it.

[white oxide]

i'm not familiar with presentations, but is $\langle a, b \mid a^2 = b^2 \rangle$ just saying the group generated by $a$ and $b$ with the property that $a^2 = b^2$?

[cloud walrus]

okeyokay

[wraith cargo]

yes

[white oxide]

huh okay, thanks

[white oxide]

can somebody give me some guidance as to if I'm on the right track of what to prove/how to approach this problem (all of a sudden our prof left for this class so the ta took over to force us to do some category theory)? so since G is a coproduct in the category of groups, I need to consider some infinite group Q and exhibit a surjection from G to Q in order to show that G is infinite. so my idea was to use the concept of a push-out; that is, I need to find maps j_i such that the following diagram commutes

(replace P w/ G and X and Y w/ Z lol)

and then somehow use the universal property or smt

so I would have to come up with a homomorphism from Z to G that makes the diagram commute and allows us to use the universal property of G right?

[crystal turtle]

Coproduct or pushout in category of groups?

pushout ok

[white oxide]

yeah pushout

[crystal turtle]

Sure yeah that works

[white oxide]

could I just take Q = Z?

[crystal turtle]

In fact ||you just need one of the j_1 or j_2 to be surjective, since then u is surjective as weill [this is g o f surjective ==> g surjective]||

Yeah that works

[white oxide]

oh shit rlly

okay thanks

[white oxide]

[\begin{tikzcd}
{\mathbb{Z}} \
& G & {\mathbb{Z}} \
& {\mathbb{Z}} & {\mathbb{Z}}
\arrow["\psi"', from=2-2, to=1-1]
\arrow["id", curve={height=-6pt}, from=3-2, to=1-1]
\arrow["id", curve={height=6pt}, from=2-3, to=1-1]
\arrow["2"{description}, from=3-3, to=2-3]
\arrow["{i_2}"{description}, from=2-3, to=2-2]
\arrow["{i_1}"{description}, from=3-2, to=2-2]
\arrow["2", from=3-3, to=3-2]
\end{tikzcd}]
i hope this works... here I let $\psi$ be such that $a \mapsto 1$, $b \mapsto -1$ and $i_1: n \mapsto a^n$ and $i_2: n \mapsto b^n$

[cloud walrus]

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

[white oxide]

ok tikz doesn't work whatever

probably doesn't work anyways

[crystal turtle]

but yes that gives surjective onto Z, which means it's infinite

[white oxide]

oh damn was it that easy lol

i thought it was gonna be hella hard

well i guess nonabelian is gonna be more annoying

honest question

do you really need all this category theory to see that G is infinite?

[crystal turtle]

No

[white oxide]

lmfao that's funny

alr then

uh

like why couldn't we just let f: a --> 1 and b --> -1 that would show that G's infinite right lolll

yeah i'm not sure what the purposes of the diagram are/considering G as a coproduct lol

maybe it'll come in handy to show that G is nonabelian

[hollow mica]

Well then

[crystal turtle]

I suppose we were just talking about that recently huh

[hollow mica]

Idk what a coproduct of a diagram is so I can’t appreciate it tho rip

[void cosmos]

what is the space of projective modules over a ring

like whats that

is it just a set of all such modules

how is it a monoid

[lusty marlin]

In question 3, what does the alternating subgroup of H₃ mean?

[void cosmos]

A_3

[rocky cloak]

I haven't heard it called a "space" before, but the projective modules form a monoid under direct sum.

[void cosmos]

oh yeah ofcourse

i think it was in the context of its completion

as a monoid

but idk

tysm tho

[lusty marlin]

How? That isn't isomorphic to the alternating subgroup of S₅.
Also I haven't seen the term 'alternating subgroup' being used for a subgroup of any group other than a symmetric group.

[rocky cloak]

I guess they just mean the intersection of H3 with the alternating group.

Sort of a weird problem though

[dim widget]

It would be easier to say if you said what H_3 is in your context!

[safe meteor]

Can someone provide a little bit of intuition behind

why all elements of F[x]/<p(x)> is written as that in the 2nd img?

[terse crystal]

What happens to x^m for m>=n in F[x]/<p(x)>?

[lusty marlin]

Yeah, I just realised that I didn't provide any context.

[safe meteor]

tbh im not sure (F[x] is a UFD?)

[terse crystal]

You are overthinking I think

[safe meteor]

uh

UFD stuff wouldnt even matter nvm xd

[lusty marlin]

H₃ is the Coxeter group with Coxeter graph as shown in the diagram.

[safe meteor]

is x^m in <p(x)> but the x^n-1 not in <p(x)> or smth

im so lost lmao

[terse crystal]

p(x)=a(0)+a(1)x+..+a(n)x^n
x^m=-a(n)^-1(a(n-1)x^(m-1)+..+a(0)x^(m-n))

Power decreased

[safe meteor]

ngl im even more lost xd

[rocky cloak]

The alternating subgroup of a Coxeter group is the subgroup of things that can be written as a product of an even number of generators

[safe meteor]

How do you know that x^m=.....

same for p(x)

im clearly missing smth basic idk what

[rose pasture]

division algorithm of polynomials

[terse crystal]

Because x^(m-n)p(x) is in <p(x)>

[lusty marlin]

Thanks!

[safe meteor]

ah ic now ty

[mossy lintel]

Hello guys, i need some help with this problem

[rocky cloak]

Try expanding (xy - yx)^3

[rocky cloak]

Wait, maybe that's not so helpful

[south patrol]

Interesting problem

[rotund aurora]

what are some motivations for projective modules in pure module/ring theory?

ik they satisfy neat properties, arise in questions about exactness and their characterization also gives information about the base ring (eg, asking when projective modules=free modules)

[mossy lintel]

It seems that 6x = 0 for all x \in R (by expanding (x+x)^3), anyway i'm going to have some rest for now

[elder wave]

There’s the projective modules <-> vector bundles analogy

But i guess that’s more towards the realm of geometry/k theory

[rotund aurora]

I think I did it

its not a one liner lol

oh nvm Im still missing something

[solar totem]

(\mathbb{F}_2[X]/(X^2+1) \cong \mathbb{F}[i])

is this untrue?

[dim widget]

it is true by definition if you define things the right way

[solar totem]

cool

[dim widget]

but also \cong F_2[x]/x^2

[solar totem]

ahh

https://math.stackexchange.com/questions/472235/describe-r-mathbbzx-x2-3-2x4
in the first answer, I was confused why he didn't just write F_2[i]

[solar totem]

rhs was F_2[i] sry

[dim widget]

because F_2[i] is not a field it looks weird i guess

[summer path]

wow tteg is not ded

[solar totem]

if x,y are non zero then x^2=y^2=1
(xy-yx)^2=(xy)^2-2xyyx+(yx)^2 =1-2+1
is this valid?

[rotund aurora]

no

x^3=x doesnt imply x^2=1

it is not true in general that if xy=0 then x or y is 0

[solar totem]

mb

[rotund aurora]

Example: In Z/4Z, 2*2=0 but 2 is not 0

I could prove 3xy=3yx

then I made a mistake and thought I proved x^2y^2=y^2x^2. From here I can prove xy=yx, but I think that strategy is doomed now

[frank cosmos]

Why is not sufficient to show that each $x^i y^j$ for $0 \leqslant i \leqslant 1$ and $0 \leqslant j < n$ is distinct?

[cloud walrus]

Hello1

[coral spindle]

Because there may be other elements

[frank cosmos]

D_2n only has 2n elements though

[coral spindle]

Yes, but you cannot assume that: that's part of what you need to prove

Anyway I was going to elaborate

[frank cosmos]

no D_2n was defined to have order 2n

earlier in the chapter

[coral spindle]

Yes, but you need to find generators and relations, which do not specify the order.

Are you sure you understand the exercise?

[south patrol]

You want to show that the relations determine D_{2n}, which means that like

If you take those generators and relations, you get D_{2n}

So in particular part of the problem will involve (or at least entail) that you get a group with 2n elements

[frank cosmos]

if i can get 2n distinct elements of the form x^i y^j then i have the entire group right

[coral spindle]

No

That's what you need to prove!

[frank cosmos]

wait so is it possible for like random x,y to get more than n distinct elemnets x^i y^j (where x,y in group with order n)

that would not make sense, since the group is closed

[frank cosmos]

so couldnt you just show that every element in the group is of the form x^i y^j

[south patrol]

Yup

That is usually a wise first step when working with generators & relations

Show they're all in that form, then show that you never wrote the same thing down twice

[frank cosmos]

oh, why would it be bad if u got the same thing twice

[south patrol]

Well you wanna show there are 2n things

And there are 2n things of the form x^i y^j

[frank cosmos]

there are trivially at most 2n things right

since the group is of order 2n

[south patrol]

You don't know that yet

You ware working with a different group

I'm not sure you fully understand the question

[frank cosmos]

every x^i y^i is in the group D_2n since the group is closed,

[coral spindle]

Indeed you do not understand the question.

[frank cosmos]

oof

[south patrol]

Basically the deal is uh

D_2n is a group and we can take generators for it and we know certain relations hold.

Then we create a new group G with certain generators and those relations

As in like, the group defined by those generators and relations.

Then we need to show that G is isomorphic to D_{2n}

Since there is a surjective homomorphism G -> D_{2n} (why?), it suffices to show that |G| = 2n.

So in fact the question boils down to something completely internal to G

[frank cosmos]

oh

uhh

i think im just restating what you said but we let G be group of all products of x,y,x^-1,y^-1 subject to those relations and then show this iso to D_2n?

[south patrol]

Yes

[frank cosmos]

and x^-1=x and y^-1=y^(n-1) so just products of x and y

im still not sure i understand why it would be conceivable for G to have order larger than 2n given that we are using the same binary product as D_2n

do you have a counter example where x,y in a group G and the order of the group generated by x and y subject to some (or none?) relations has order larger than |G|?

[south patrol]

Yes

[frank cosmos]

oh wait, we're just forgetting about the structure of G?

thinking of x and y as random shit

[south patrol]

Yeah sort of

So liek

[frank cosmos]

and multiplying them and usign the relation

to get something iso to G

[south patrol]

Take Z/2 with generator x

The free group on x

The elements x, x^2, x^3,... are all distinct for example

[frank cosmos]

got it

ez problem still

thanks for help

[south patrol]

np

group presentations are confusion

[rotund aurora]

if a ring is non-unital, do you have a notion of characteristic?

[rocky cloak]

Yes, just n such that nx=0 for all x

[rotund aurora]

like if 6x=0 for all x does it follow that either 3x=0 for all x or 2x=0 for all x or nah?

[coral spindle]

Z/6Z

[rocky cloak]

No that doesn't follow, just consider Z/6Z

[rotund aurora]

mmh tru

wait right the characteristic is only prime for fields

[coral spindle]

That's not true either

(Z/3Z)^2

[rotund aurora]

sry I meant that it is not true in general

it is true in general for fields, but not for rings

[rotund aurora]

but could it be that all elements are torsion over Z but there is no common anihilator?

[rocky cloak]

In this case we say the characteristic is 0

[rotund aurora]

yeah but it can happen right

mmh yeah

nvm

just take prod Z/nZ where n runs through integers

[coral spindle]

Try producing an example

[rocky cloak]

Like prod_p Z/p

[coral spindle]

Yup

[rotund aurora]

faster

ig I should have said coprod

[coral spindle]

But notice that not all elements are torision in that product

in particular (1, 1, ...) is not torision

[rotund aurora]

only finitely many entries are nonzero

[rocky cloak]

direct sum then

nonunital baby!

[coral spindle]

The direct sum is no longer a unital ring

The presence of 1 is so useful, huh

[rocky cloak]

Yeah croq asked specifically for nonunital rings

[rotund aurora]

yeah I was doing the x^3=x problem above

I think I did it

but I was trying to see what happened if I divided into different chars, but there is no need to

[rotund aurora]

I think its correct lmao

I may be heavily clowning, cuz there may be a way shorter solution

I will give a cookie to whoever finds a mistake 🍪

if there is one, I think I will be able to fix

[open sluice]

is equation 8 supposed to hold for a specific x or for all x

[rotund aurora]

for all

it is true in Z/6Z at least

[rotund aurora]

it is derived from (7) by setting y=x

I was trying to be careful at first with what Im doing, but since the argument ended up being not exactly linear, I stopped giving careful indications

[open sluice]

maybe it's the case that y^2 = y holds for all elements of the form y = 3x but you still can't commute them because 3R isn't a ring or something spooky like that
maybe it is a ring idk

[rotund aurora]

its non unital

so you do this (3x-3y)^2=3x-3y

btw

there is a mistake in writing (6)

because one of the yxy should be yx^2y

I almost panicked

[rotund aurora]

and use 9x=3x for all x

[mossy lintel]

[uncut girder]

Broz studying abstract rings 😭😭

[warm saffron]

I had a question: I'm supposed to prove or disprove that every subgroup of the integers has finite order. I'm thinking this is false because the subgroup (Z,+) of the group of (Z,+) is a subgroup with infinite order. Can anyone confirm?

[mossy lintel]

The order of proper subgroup can be infinite too, like (2Z, +)

[warm saffron]

yea I was just looking for one counterexample for disproving it

mine is good, though, right?

[crystal turtle]

That's an odd question, since all subgroups of Z are infinite except for {0}.

[warm saffron]

this is judson again

LOL

and then the previous question asks me

every subgroup of the integers has finite index. I think this is true? I intend to show that given a subgroup of H of the integers and a,b in Z, we have aH=bH.

would that work, I just learned about left cosets so still getting familiar with the topic

oh wait aH=bH doesn't work when a=0 and b is nonzero unless we can assume the operation is addition?

so I'll need to prove it someother way....

[mossy lintel]

The index of {0} is not finite?

[warm saffron]

Hmm

because a{0} for any integer a, gives the same thing?

[tribal moss]

Every non-trivial subgroup, perhaps?

[warm saffron]

I'm kind of lost on what the operation is

is this addition?

"subgroup of the integers"...they including multiplication too?

[tribal moss]

Yes, when Z is considered a group it is always under addition.

[warm saffron]

oh so they're not taking Z_p when p is prime

[tribal moss]

It can't be multiplication, because Z doesn't have inverses.

[south patrol]

Hm, if $V$ is an $n$-dimensional vector space over $\mathbb F_2$, what does the space $TV/I$ look like ,where $TV$ is the tensor algebra and $I$ is the ideal generated by all the $x^2, x \in V$?

[cloud walrus]

potato

[south patrol]

It seems one ought to get some big infinite dimensional thing, if i read smth correctly

[tribal moss]

Hmm, that's the alternating algebra, isn't it?

[warm saffron]

I see. So then I would just state that H={0} is a subgroup of (Z,+) and given z in Z, z{0}={z+0}=Z. right?

[south patrol]

Yes though the characteristic 2 sort of messes it up

[tribal moss]

Oh foo, didn't notice char 2.

[south patrol]

For example I don't think this should be commutative (or anticommutative) but I may be being silly

[tribal moss]

Yes, almost, you wouldn't say {z+0}=Z at the end; the coset is just {z}, not the entire set of integers.

[warm saffron]

oh right because z was fixed

there are infintely many integers z and each one gives the set {z} which only contains that integer

gotcha

[tribal moss]

(And in the name of avoiding horrible confusion, write both the group and the quotient with additive notation, so z+{0} instead of z{0}).

[crystal turtle]

dammit tropo beat me to it

[warm saffron]

ah yea will do. Thanks for the help guys

knowing my prof he prolly wants me to show also that given a,b in Z if a+0=b+0, then a=b.

[tribal moss]

But if I remember correctly, the x²=0 is the correct way to express alternativity for char 2, rather than xy = -yx which becomes trivial there.
We do get 0 = (a+b)(a+b) = a²+b²+ab+ba = ab+ba so ab=ba, however.
So if you select a basis for V, then a basis for TV/I is just the set of all subsets of the original basis, and TV/I is 2^n-dimensional.

[south patrol]

Oh very good point, thanks

with the 0 = (a+b)(a+b) thing

The dimension was mostly what I was interested in

[warm saffron]

<8> in Z_{24} is precisely the set <8>={[8],[16],[0]}, right? having a major brain fart for some reason

[tribal moss]

Yes.

[warm saffron]

thanks

If an element g of a group G has order n, then <g>={e,g,g^2,...,g^{n-1}} is a subgroup of G, right? regardless of G being cyclic or not?

[crystal turtle]

Correct

[warm saffron]

okay perfect thanks

[delicate bloom]

prove it

[warm saffron]

I was asked to show that ig g has order 5 and h has order 7, then |G| is at least 35. I showed that <g>={e,g,g^2,...,g^4} is a subgroup since it is closed and certainly nonempty. I showed closure using the division algorithm

here's what I did. I proved a lemma before hand that states e^z=e for any integer z. And another lemma that states that if n,a,b>0 are integers with a|n and b|n, then ab|n if gcd(a,b)=1

[crystal turtle]

Did your class not already show that <g> is always a group? If not, I don't imagine there's much point to writing the same proof twice; either show one and state the other is analogous, or show it in generality once

also

$\langle g \rangle$

[cloud walrus]

grothendieckfan1 (Ryx)

[summer path]

This last thing as stated is not true

[warm saffron]

edited it

I forgot to write that but I said it in my proof that gcd(7,5)=1

[warm saffron]

I agree. I'll just prove another lemma LOL when I assume an element has order n

[summer path]

You could also just have done it once and said the other one was similarly

[warm saffron]

I remember we talked as a class about <g>=G means g is a generator, but I don't know if we showed that if g has finite order, then <g> is a subgroup of G

[warm saffron]

Agreed. I was thinking about saying "by a similar argument <h> is a subgroup". However, I think I'll prove a lemma for this in it's generality and cite the lemma

[delicate bloom]

the way I'd write it is basically like not much more than this
|G| = [G:<g>]|<g>| = [G:<g>] 5 and similarly |G|=[G:<h>]7 so 5, 7 divide |G|, lcm(5,7)=35 -> done

[summer path]

I'm not really sure why you'd use <g> notation without saying it's a group

[delicate bloom]

if you ever wrote a program before, think of your lemmas as like that, you're trying to clean up the code by putting it into "functions" defined earlier on in the code

you don't want to be writing your program as one long mess of everything from first principles every time, it obfuscates the thing you're actually proving

[warm saffron]

<g> was just defined as all integer powers of g...and we read in class that if <g>=G, then g is a generator of G. But yea, again, we haven't shown <g> is a group.

[crystal turtle]

Also if you try to do everything from the basics in an exam... (unless you are meant to)

[warm saffron]

yea I'm typing up my lemma now. i know lol that's why I wrote those other lemmas first prior to this proof

Remember guys this is the same prof that likes to be picky on and off so I never know LMAO

[summer path]

Is it not some ta that grades homework

[warm saffron]

not for this class. There's only 9 of us in this class

and he know this is the only class I have this semester so he's prolly thinking "this guy has nothing but time, why not just prove a lemma and use it rather than assuming without justification?"

[delicate bloom]

try presenting both to him during office hours to ask which he would prefer on hw/test, curious to see what his rationale is

[crystal turtle]

Me when my supervisor grades my homework in one of my courses

i worry if I make a mistake he will think I am dumb

[warm saffron]

To give you an idea of his mindset...there was one test I remember he graded for my vector calc class where I wrote 1/a and he marked me down for not saying this only works when a is nonzero. But the problem statement implied a was nonzero in the first place....

[crystal turtle]

You gotta find the right balance tho

[delicate bloom]

did you follow up with him after pointing that out

[warm saffron]

still working on it

[delicate bloom]

I've had my profs give pts back after pointing out stuff like this, they're human too

[warm saffron]

this was 2 years ago for that vec calc thing. I think I did point that out and he said, you should still cite why you can divide by a"

hard to absorb things as assumptions when randomly we're required to cite minor things we would've cited the first week of class

[summer path]

The same prof teaching vector calc and groups huh

[warm saffron]

yup and my discrete math class inbetween. He's a nice guy just can't get a fix on when he wants us to cite minor details and when not. The whole class can't figure this guy out

[delicate bloom]

the whole class is probably wrong and you should stop deferring to them tbh

[warm saffron]

I'm not deferring lol. If I don't know if I can assume something, I either ask him or just prove a lemma.

[summer path]

Or do both

[warm saffron]

anway, thanks for the help guys, just proved that lemma... that way I can also cite it in other hw problems for this assignment as just lemma 1.2 or whatever I named it.

[crystal turtle]

I assert it as true and leave the exercise for the grader

[summer path]

Ah yes the super reaction misclick

[warm saffron]

happened to you too??

Yea that's the first super reaction I've ever sent lol

I don't even know how it happened

[open sluice]

it's how you lose all of your super reacts

[summer path]

blame discord for adding them and then moving it to be immediately under the normal reaction button

[warm saffron]

ripp

[elfin prairie]

how does one prove that smth which we claim as group is associative?

[kindred oracle]

show the group multiplication is associative, that is for all a,b,c from the group: f( f(a,b), c) = f(a, f(b, c))

[elfin prairie]

what is f?

binary opr?

[kindred oracle]

yes

[elfin prairie]

but would you verify this for every single possible case?

[kindred oracle]

if it is false for any ordered triple then the opr isn't associative

https://en.wikipedia.org/wiki/Light's_associativity_test

[elfin prairie]

but would you hunt for that single counter-example?

[elfin prairie]

hmm ill give this a read...

[kindred oracle]

most of the time groups are given by functions or presentations, both which automatically associate for cayley table it is tiresome to verify that the op is associative

[elfin prairie]

I see

are there any structures which are not associative

[kindred oracle]

consider the cross product on 3d vectors

[elfin prairie]

hmm

[drifting plover]

If I am a group of order |Z/qZ|*|GL(2,p)|, and I surject to Z/qZ and to GL(2,p), then I am a direct product of Z/qZ and GL(2,p), where p and q are odd primes, possibly equal

^ is this even true? Seems to be true for the first few primes q I tried, with p = 3

[wraith cargo]

isn't this equivalent to having a split SES

[drifting plover]

Sure, but why does it split?

[wraith cargo]

take it to be 0 -> Z/qZ -> G -> GL(p,2) -> 0
And you also have a surjection G->Z/qZ that's the splitting homomorphism

[drifting plover]

Do I need to need to compute an Ext group lol

Ohhh I get you

I mean this is certainly not true for arbitrary (finite) groups

[wraith cargo]

And I think since Z/qZ has prime order you're guaranteed to have a subgroup of that order in G

[rocky cloak]

The surjection G -> Z/qZ doesn't necessarily have to be a splitting of the inclusion. But maybe it will be in this case...

[wraith cargo]

that is a good point hm

[wraith cargo]

Wouldn't the subgroup isomorphic to Z/qZ map isomorphically onlto Z/qZ via any surjection?

this might be my brain playing tricks but that sounds realistic-ish

[rocky cloak]

It can be contained in the kernel

[wraith cargo]

okay good point lol

[south patrol]

Dont split exact sequences also correspond to semidirect products in general

[rocky cloak]

Splitting on the right is equivalent to semidirect, splitting on the left (and right) is equivalent to direct

[drifting plover]

That what I thought originally but ^

[south patrol]

Rather than just direct ones

[south patrol]

Ah

[rocky cloak]

How about this: consider p=7. Then GL(2, p) surjects onto (Z/7)^* = Z/6 with the determinant map. Thus surjects onto Z/2 = (Z/3)^* so has a semidirect product with Z/3. And since Z/6 surjects onto Z/3 this group surjects onto both GL(2, 7) and Z/3.

Now I don't think this group is isomorphic to the direct product, but I would have to think about it more carefully...

[drifting plover]

Right... so it might just be a semidirect product in general, and the examples I tried all have trivial action because p = 3 is too small?

[rocky cloak]

I don't think it has to be a semidirect product in general. At least that's not obvious to me, but those would be the easiest examples to construct.

[tidal plover]

Could someone explain c, and d? I dont really understand. A_n is the set of all even permutations

[coral shale]

show {t_ij} is a generating set?

= {s_is_j}

= {(i i+1)(j j+1) : i, j between 1 and n-1}

[tidal plover]

im sorry could you further explain?

[coral shale]

idk what ur not understanding

the definition of a generating set?

how to make sense of the composition of 2 permutations?

[coral shale]

or like what this set is?

[tidal plover]

yeah my bad, what is the set that ur describing

[coral shale]

(i i+1) is a transposition

is that ok

[tidal plover]

yes

[coral shale]

(i, i+1) in your text

(i i+1)(j j+1) is the product of 2 of them

thats "swap j with j+1, then swap i with i+1"

Then we consider the set of all of these for i, j ranging between 1 and n-1

Thats what the set builder representa

[coral shale]

thats what they mean by "the set of all t_ij"

[tidal plover]

ohhh i see thank you!

[rocky cloak]

I think maybe I have an example that is not semidirect.

Consider GL(2, 7) again, and adjoin a central element s such that s^3 is a primitive root modulo 7 (like 3 for example). Call this group G

Then (s^6) -> G -> G/s^6 is a central extension, so not split. (s^6) is isomorphic to Z/3, and I believe G/s^6 is isomorphic to GL(2, 7), but maybe I'm wrong about that.

Of course it could be that G is isomorphic to some semidirect product of GL(2, 7) and Z/3 in some other way...

[drifting plover]

What's your argument that G/(s^6) is isomorphic to GL(2,7)?

[teal vessel]

Usually I try to show that following the operations in both orders yield the same expression at the end for some arbitrary members a,b,c. Since those members are arbitrary, the structures they form don't depend on any of their properties specifically.

[rocky cloak]

It's more of a hunch than an argument really, since I haven't thought about it very carefully.

[rocky cloak]

Thinking about it more, I guess it probably isn't true. Oh well.

[chilly ocean]

why can't we say that if $G=\langle X\vert R\rangle$ and $G'\le G$ is generated by some $X' \subseteq G$ then $G'=\langle X' \vert R \rangle$

[cloud walrus]

Gev

[chilly ocean]

wait

it doesn't work

[coral shale]

why doesnt it work

i think it surely does by definition of " G' is generated by X' "

that surely means G' = <X' | R>

[chilly ocean]

I mean the relations should do their work but what if there are relations on elements that are outside of X'

and we don't have them in G' somehow

but we have relations on them

can we have elements in the presentation of G' that aren't in G' ?

[coral shale]

Then im not sure I would say " G' is generated by X' "

it all depends on what is meant by that phrase

[coral shale]

hmm.

im not sure actually because you always have

X' = <X' | R>

and now taking away elements from the generating set gives you more relations? Dont think so...

But then again like, if G = <a, b | R>

<a, b | R, b = e>

is a subgroup

[coral shale]

but im not sure if this actually "counts"

i dont think it does.

[chilly ocean]

but if we want to talk about this subgroup we can just take X'={a}

[coral shale]

oh i dont think thats a subgroup anyways unless a independent of b

[coral shale]

yes exactly

i think its actually not a subgroup most technically

and is isomorphic to some quotient

[chilly ocean]

what about the case when X' is not a subset of X

or can we always pick 2 presentations such that X' is a subset of X

[coral shale]

nah u cant always do that

take <a | a^4>

<a^2 | a^4> is a subgroup

but <a | a^2> is only isomorphic to the above, its different.

===
So yeah to conclude, from what I can see, your group and subgroups should be considered with the same underlying relations

[chilly ocean]

but we can take <a, a^2 | a^4> instead of <a | a^4>

[coral shale]

if you change the relations, thats some quotient stuff going on

yes but a^2 is not in the generating set

X' is not a subset of X

like technically <S | R> = <S>/<<R>> right

<<R>> smallest normal subgroup

[chilly ocean]

yes

[coral shale]

and so your subgroup must be of the form <S'>/<<R>>

[chilly ocean]

I don't think there's even a way to describe a subgroup if we are given the presentation of the group

what about the fact that any countable group is a subgroup of some group generated by 2 elements

[coral shale]

so to answer, theres only one way to interpret " X' generates G' "

[coral shale]

nah

[chilly ocean]

then the subgroup of this group generated by 2 elements might not even be finitely generated

[coral shale]

at worst you stick all the elements of the subgroup in your generating set

< G' | R >

[coral shale]

the "is" in this sentence is an isomorphism

yes the subgroup might not be finitely generated

but your generating set of it

is going to be a set of strings consisting of a and b

the 2 generators

[chilly ocean]

wait

this makes sense

[coral shale]

{ab, a'bbb'bbabaa'a, ...}

[chilly ocean]

I agree

thx

I'm still worried about the excess elements that appear in relations

[frank cosmos]

Is this proof that x^3=9 has no solutions in Z/31Z correct?

First, [9] has order 15 in (Z/31Z)^*. So if x is a solution then x^45=9^15=1 mod 31 so then the order of x in (Z/31Z) (under mult) divides 45. But, since (Z/31Z) has order 30 the order of x must also divide 30. So x^15=1 mod 31 but this contradicts 9 having order 15

[coral shale]

you may get symbols in your relations that dont appear in your generating set

but i dont see the issue with that

<a, b | c> = <a, b> right?

i doubt it becomes a definitional issue

[coral shale]

looks legit at a glance

[chilly ocean]

I believe some element might be a part of a relation that affects our subgroup but we can't use this relation because of this symbol that we don't have

[coral shale]

can u construct an explicit example of what ur worried about

[lusty marlin]

Yeah seems right

[coral shale]

like to me, i have no problems with the left object

[lusty marlin]

As long as the order of 9 is actually 15

[frank cosmos]

part 1 of the problem was to determine order of [9] in (Z/31Z)^*. Was there an easier way then trying all divisors of 30?

[coral shale]

[frank cosmos]

like idk some theorem i forgot

[coral shale]

i wouldve gone 9, 81 = 19, ...

which is slow but does the job

fermats little might help

[frank cosmos]

not really

[chilly ocean]

G=<a,b | b^6=1, a^2=b^3>, G'=<a | b^6=1, a^2=b^3>

[frank cosmos]

that just tells you 9^30=1 mod 30 which is trivial

dosent tell you about 9^(divisor of 30)

[coral shale]

i think aab'b'b' dies

like it has no effect

[chilly ocean]

in G we get a^4=b^6=1 but in G' we cant do it without using b

[coral shale]

hmm..

not sure

<a, b | aabb, bb> ->
<a | aabb bb>

u come into some issues here

u cant just ignore the relations just because they have b's

theyre supposed to give u aa

[chilly ocean]

so we actually use b without having him in G'?

[coral shale]

it seems like an issue.

i guess the overall issue is that while G is some quotient of the free group

the subgroup isnt also necessarily one

so u get slightly less than well-defined objects like that. Which is a problem idk how to resolve

[chilly ocean]

I see

[coral shale]

Id then go back and say <X' | R> may not be well defined as things stand...

[chilly ocean]

I'll just use it like that for the time being

[coral shale]

im not 100% sure whether this is legit, feels like it is in these cases highlighted

maybe there is always a way to "simplify" your relation set so the redundant clashes dont occur, and u can just ignore relations with unused symbols, but have doubts.

https://math.stackexchange.com/questions/1607005/every-subgroup-of-a-quotient-group-is-a-quotient-group-itself

oh ok, maybe we're good and that set of relations can always be simplified

this feels like the 3rd iso now i think about it

so we're invoking 2

[south patrol]

Bsst theorem

[chilly ocean]

this is some good stuff

[coral shale]

so i think we re good, just gotta simplify the relations

[south patrol]

Well like the snappiest way to think is that there is a correspondence between subgroups of G/N and subgroups of G containing N, given by K/N <-> K

[coral shale]

mmm.

[south patrol]

Like 2 says one of the maps is surjective and 4 says the correspondence preserves normal subgroups

[coral shale]

So going back to free groups and quotients

Youre thinking of subgroups of <S>/<<R>>

and these must be in bijection with subgroups of <S> containing <<R>>

Oh ok, so technically i think ur doing this