groups-rings-fields

[old lava]

(or you can start with b and induct on a)

[crystal vault]

is it obvious?

[old lava]

you don't have to use induction on a to prove x^(a+0) = x^a x^0

that is very obvious yes

[crystal vault]

i mean i think it's obvious but i've gotten burned before when i tried to assume stuff like that

alright

[old lava]

I mean x^(a+0) = x^a = x^a 1 = x^a x^0 if you really want to "prove" it

but that is a very obvious statement that you can take for granted

I feel

[crystal vault]

general associativity has already been shown in a preceding exercise

i just thought it was a bit strange to require induction for something that seems so straightforward, especially given that we're allowed to just count the number of terms in each product

[chilly ocean]

I agree yeah

[crystal vault]

alright sweet, thanks so much (:

[mild laurel]

@crystal vault this is a common idea

For some statements, you can prove them directly without induction

Like the statement that 1+2+...+n = n(n+1)/2 you can prove directly with no induction

[crystal vault]

ooooh yeah i was gonna ask about that at some point

just because something has to do with natural numbers doesn't mean you need induction right?

[mild laurel]

So for these, you take an arbitrary n and prove it for that arbitrary n

[crystal vault]

like you can just say "let $n \in N$"

[mild laurel]

Yeah exactly

[crystal vault]

ok perfect

[mild laurel]

So they're taking an arbitrary a

[cloud walrus]

kirafa

[mild laurel]

But using induction on b

[crystal vault]

oh ok

perfect, that clears up the rest of the confusion

but all the same, i don't necessarily need to use induction for this, right?

[mild laurel]

Yeah agreed

[crystal vault]

ok sweet, thanks so much

this has been bothering me for a while

so i really appreciate all your guys' help

[celest brook]

bro I thought that was a shitpost that's an actual book 💀

haha

[chilly ocean]

@mild laurel why did you give up honorable?

[carmine fossil]

Let's say the integral domain has a minimal ideal

Now,This ideal is principal (otherwise it's clearly not minimal)

Call the generator a

Let's say there is a nonzero element b,now consider the ideal generated by ab

ab is not 0, since this is an integral domain

and (ab) is contained in (a)

[bronze snow]

oh, and that is not possible since this minimal ideal cannot (ab) in it?

[carmine fossil]

Yes

Although you need to check if (ab) is a proper ideal of (a)

Just pick b to be a nonunit(This requires the ring to be an integral domain)

If all elements are units,(AKA a field) your statement is true, since there are no proper ideals

[carmine fossil]

||Let's say b is a non unit and (ab)=(a) this implies abk=a for some k,which implies a(bk-1)=0 implying b is a unit,a contradiction||

[next obsidian]

What???

Every ring has a minimal ideal

Every ring has a minimum ideal even

[carmine fossil]

minimal ideal means ideal which doesn't contain a proper nonzero ideal

[next obsidian]

Your proof has a slight flaw. You’d have to show there’s no ring in which you have exactly one nonzero non-unit

Err...

No you’re fine

I retract my statement because you can look at a^2

[hot lake]

you still need to show that (ab) is not (a) I think

[next obsidian]

Yeah actually

I’m thinking now and

[hot lake]

ah right b noninvertible

[next obsidian]

Nvm

Chmonkey

Also I really dislike the idea that a minimal ideal is necessarily non-zero

Since for AG the fact that an integral domain has generic point the zero ideal is because that’s a minimum prime

Feels kinda weird

[rich ravine]

In AG we only look at prime ideals though

minimal prime ideals are minimal among the prime ideals

[dire bramble]

not the best at abstract algebra here

let $\textrm{res} : \hat{A} \to \hat{B}$ be defined by $\chi \mapsto \chi \mid_B$ where $A,B$ are finite abelian groups and $\hat{A}$ is the dual group of $A$ and $B$ is a subgroup

[cloud walrus]

bacono

[dire bramble]

prove that $\ker \textrm{res} \cong \widehat{A / B}$

[cloud walrus]

bacono

[dire bramble]

I'm thinking some finagling with the first isomorphism theorem of groups but honestly I'm not sure how to approach this

[mild laurel]

I don't think it's really first isomorphism

I think using the idea that anything in the kernel of this map takes the value 1 on all of B, you can get a map from the kernel to \widehat{A/B}

[cloud walrus]

slimvesus

[dire bramble]

I think I see, I'll try that

[latent anvil]

What?

[next obsidian]

0 ideal

[latent anvil]

oh right, was thinking of only primes

[cloud walrus]

squirtlespoof

[hazy leaf]

$\frac{\mathbb{Z}_{M}}{{Nj,j= 0,\dots,M/N}}=\frac{{0,1,\dots,M-1}}{{Nj,j= 0,\dots,M/N}}$

[cloud walrus]

亜城木 夢叶

[hazy leaf]

it is under addition, dont see how do they form a group

[cloud walrus]

Neil_P

[real panther]

I guess ( { 1, e^{2 i \pi /3} } ) makes a spans the extension field?

[cloud walrus]

Neil_P

[delicate bloom]

not quite

well, it factors considerably

you should think about what the irreducible polynomials it factors into are to determine the degree

[real panther]

Any tips? I'm trying to prove that, in general, the degree of the splitting field of ( x^n - 1 ) over ( \mathbb Q ) is ( \phi( x ) ), where ( \phi ) is the euler toitent function

[cloud walrus]

Neil_P

[real panther]

would this be ( (x^3 -1)(x^3 +1) = x^6 -1 ) ?

[cloud walrus]

Neil_P

[delicate bloom]

it factors more than thatj

(x-1)(x+1)(x^2+x+1)(x^2-x+1)

[real panther]

Gotcha - how does that generalize?

[delicate bloom]

well from the geometric series you can see x^n - 1 is divisible by x^d - 1 when d|n

[real panther]

Right - because they're roots of unity

[delicate bloom]

eh?

no, I'm saying because the geometric series simply factors that way

[real panther]

My bad - there's more to the question that I forgot to mention

[cloud walrus]

Neil_P

[mild laurel]

Uh

I think you want

pi i/(2n)

[real panther]

hmm - I think I misread the problem then. it does actually say ( e^{ \frac{n \pi i }{2}} )

[cloud walrus]

Neil_P

[mild laurel]

That's uh

Probably a typo

[real panther]

Probably on the instructor's part then. You're right

wait - isn't it (2 pi i / n)? Because a full turn around the circle is ( 2 pi i), and you want (1/n) th of that

[mild laurel]

Yeah that

[cyan marten]

This is equivalent to showing that the cyclotomic polynomials are irreducible.

[real panther]

Thank you

[delicate hawk]

could some1 plz explain to me what R / P actually is?

it cant be Z[-2]

[carmine fossil]

R/P will be iso to Z_18

as rings

[delicate hawk]

why is that

dont we set 2 + \sqrt{-14} to be = to 0

[carmine fossil]

Let $w=\sqrt{-14}$ then
a+bw=(a-2b) mod P=
rem((a-2b)/18)) mod P

[cloud walrus]

DrunkenDrake

[carmine fossil]

18 is the smallest natural number in P

So consider the map
phi:Z[√-14]->Z/18
phi(a+bw)=(a-2b)+18Z

[delicate hawk]

the set of ideals of Z_18 is just {0, 3, 6, 9, 12, 15} and {0, 2, 4, 6, ..., 16}

meaning the # of ideals of R containing P is just 2

[carmine fossil]

So,You can find the ideals of R/P

By which you can find ideals of R containing P

[delicate hawk]

was what i wrote correct

im still wondering how an ideal containing P would look like

[carmine fossil]

You left {0},{0,6,12} and {0,9}

[viscid pewter]

dumb q, in ring theory do we not consider 0 and R as subrings

or however the terminology works

[carmine fossil]

They are subrings

Tho,we don't usually call R a ideal of R

The corresponding ideal to say {0,9} would be P U 9+P

Where P=<2+√-14>

An ideal of R containing P is the preimage of some ideal in R/P(the map being R->R/P phi(a)=a+P)

[stiff jacinth]

hey guys

any ideas on wat to do next? this is wat i got so far

[carmine fossil]

If a is not 1,y is uniquely defined by x

[stiff jacinth]

im just not quite sure what to do about a, b, x, and y being in Z_p and a, x being in the units of Z_p

[carmine fossil]

Z_p here is p-adics?

[stiff jacinth]

its the group of integers mod p

[carmine fossil]

Otherwise, that statement just says a is nonzero

[stiff jacinth]

dont think its p-adics cus we didnt get that far

[carmine fossil]

you end up with y=b(x-1)(a-1)^-1

If a!=1

[stiff jacinth]

but a can be 1 because gcd(p, 1) = 1

[carmine fossil]

Take that as a separate case

When a=1 ,it depends on b

If b is zero,x can be anything

If b is nonzero,x is 1

And y can be anything

[stiff jacinth]

ok ill try to do it case by case

but theres only two possible cases for a and x anyway: 1 and p (mod p)

[obsidian loom]

Hi, how would i go about listing all the cyclic subgroups of < Z_10, + > ?

my first thought was that every set under Z_10 is a cyclic subgroup, since they're all groups under addition

i guess the bit that's confusing me is how a set like Z_6 is a cyclic group under addition, when it has more than 1 generator

[sharp sonnet]

every set under Z_10?

a cyclic group can be generated by multiple elements, that's fine

i mean: multiple elements in a cyclic group can generate that group

[obsidian loom]

so what is the definition of a cyclic group then?

[sharp sonnet]

generated by a single element (doesnt have to be unique)

[obsidian loom]

oh

so kind of looking at each element separately

so there's groups which can be generated by more than 1 element?

[sharp sonnet]

sure

Z/6Z is generated by either 1 or 5

but for your original question first think about which order those subgroups can even have

(or actually you can just look at the subgroup generated by each element)

[harsh pagoda]

Hey !

I'm a bit stuck here, let me explain

I'm working on clifford algebra here,

and i need to find a condition for a,b,c, and d such that A is an automorphism

And after that, i need to describe all the matrix-A that satisfies thoses conditions

any idea on where/how to start here ? :)

[cloud walrus]

K_

[harsh pagoda]

Where,

[cloud walrus]

K_

[harsh pagoda]

Finally found it )

thanks :)

[cyan marten]

In R, any "uncountable infinite sum" is essentially countable, in the sense that almost all terms vanish. Is this true in any topological abelian group?

The proof I know uses the ordering of R.

[latent anvil]

how do you define an uncountable sum without an ordering?

the definition I know in R involves taking a supremum

[cyan marten]

I only need to order the index set, not my group.

[cyan marten]

I don't know what the standard way is, but I saw it defined as the sup over all finite sums (which also doesn't need ordering).
I am not sure how the approach by nets work, but it seems more satisfactory.

Oh, my bad

What if I consider all accumulation points of the set of finite sums?

Also, does the approach via nets use ordering? From what I understood, we define a net whose elements are finite sums (I am thinking of a well-ordered uncountable index set), and then one can speak of its convergence.

Yeah, as you said, for this we need an ordered group.

[cloud walrus]

slimvesus

[cyan marten]

I was thinking of the sequence 1/n, which has a single accumulation point and thus converges to it. But this is misleading, because if we consider the net (?) 1/x for x in (0, 1), say, then every point is an accumulation point, and we can't read off the "limit".

I see

I am still not sure how to define f_a though

Aha

[cloud walrus]

slimvesus

slimvesus

[cyan marten]

And we use the natural ordering, right?

I mean an element is farther off in the net if it corresponds to a larger finite set.

I see

To make sure I follow, this is equivalent to "sup of all finite sums" formulation, right?

When we're working in an ordered group.

But it seems strange that we could drop the ordering and get the same thing..

Maybe it works but it's trivial in the sense that it rarely ever converges

Me neither

What's the sum of the set X = (-1, 1)?

I think it just doesn't converge

It's just sup{s1 + s2 +... + sn : s_i in X}

Here the elements are distinct

X is a set - we don't have indexing, so there are no "repeated elements" (which can happen if we have a sequence, say).

In the definition I saw, yes, but I don't see why it's necessary.

Ohh, good point.

Makes sense.

Let I be an ideal of a commutative ring A. Define I^\infty = the intersection of all powers of I.

I was thinking of products whose terms are elements of I

Let A have the I-adic topology. A convergent (countably) infinite product whose elements are in I is in I^infty..

Is it true that elements of I^infinity are such countable products?

More accurately, linear combinations of them, just like the situation of I^n.

For context, this is (trivially?) true when A is Noetherian, by Krull's intersection theorem.

[steel saffron]

http://math.hws.edu/eck/js/symmetry/wallpaper.html

This is a cool visual for the 2 dimensional crystallographic groups

[next obsidian]

I^infty is a really curious object

[next obsidian]

Pretty sure you're gonna need to force localness of the ring as well, or at the very least I < J(R)

At least if you want to use Krull's intersection theorem

[cyan marten]

Ohh, I had just learnt it, so I forgot.

Let's go with I <= J(R)

[frank fiber]

hi

[frank fiber]

I have a question with exercise 1.8

[next obsidian]

In order for this problem to make sense I think you need to assume that G is abelian

I don’t think that product is wel-defined otherwise

[frank fiber]

If G is trivial group, then f=e, but order of e is 1

[next obsidian]

That doesn’t satisfy the hypothesis of the problem

The trivial group doesn’t have any elements of order 2

You don’t need to say nontrivial

The trivial group does not have exactly one element of order 2

[cyan marten]

Maybe it is meant in the sense that no matter which ordering we used, the result would be the same. That would be pretty interesting.

[next obsidian]

I don’t think this is true tho

I don’t have a counterexample, but I would be mad surprised

Maybe look at Q_8

[cyan marten]

No, such a group is necessarily of order 2n with n odd, I think.

By Sylow's theorem

[next obsidian]

No

[cyan marten]

Ohh, my bad

[next obsidian]

Q_8 fits the bill

[cyan marten]

Yup

[next obsidian]

I see where your brain was headed

But not every Sylow-2 is made of only order 2 stuff haha

[cyan marten]

What 2-groups have exactly one element of order 2?

I think it's possible that Q8 is the only one

Maybe the generalized quaternions are a generalized counterexample

[next obsidian]

Highly doubt this

[cyan marten]

But then, in such a group G, we will have a^(|a|/2) = z, where z is of order 2.

For every element a in G

I think this is a relatively strong restriction, at any rate.

[next obsidian]

Yeah

[oblique river]

https://groupprops.subwiki.org/wiki/Subgroup_structure_of_generalized_quaternion_group:Q16

that's a group of order 16 with only one element of order 2

[cyan marten]

Yeah

Well.. That doesn't really help

To classify these groups, it suffices to classify 2-groups with this property. But the hypothesis never holds in a 2-group.

Let z be of order 2. Then <z> is normal, and thus we can consider G/<z>, and try to prove that it has the same property..

This could ultimately help in proving that only generalized quaternions have this property (if it's true)

[cyan marten]

If that's true, and G is of order 2^(n + 1) and has a unique involution, then by induction G/<z> is isomorphic to Q_(2^n)

So G is an extension of Q_(2^n) by C2. I choose to believe that this forces G to be the next generalized quaternion group.

[next obsidian]

Why is the r module not free

[latent anvil]

Which R module?

[next obsidian]

Yeah

[latent anvil]

[south storm]

twitter reference

[obsidian sleet]

ah jeez

i have small question

suppose i have the following permutation expressed in the um

( ) notation

forgot the name

anyway

ab = (13)(245)(6)(124356)

i want to reduce this to product of disjoint cycles

oh its called cycle notation 💀 lmao

ok

yeah so i got to this so far

= (14)(25....?

idk how to deal with that 6

i know we usually omit the 6 but i just left it in there cus why not

5 will go to 6 but 6 just.. stays there

i want to put 6 after 5 and just close the cycle but 6 doesn't go to 2??

maybe i am over tinking this
😔

[scarlet estuary]

see where each number is mapped

and make your cycles from that

rather than trying to do an ad-hoc method

so:
1 -> 4
2 -> 5
3 -> 2

and so on

once youve completed the list you can just follow each element to make a cycle

and these are guaranteed to be disjoint.

[wild sapphire]

Very quick question. With a cyclic group (Z_n, +) (modular arithmetic). Are all subgroups covered by the set of subgroups formed by all valid generators?

[scarlet estuary]

covered?

[wild sapphire]

equality really

[scarlet estuary]

oh i thought you meant like, topological covering

uh

every group has generators

and certainly if its a subgroup of Z_n those generaators will be from Z_n

im not sure what you mean by "valid generators" admittedly

[wild sapphire]

I guess it is better re-phrased. Can a cyclic group have a subgroup that is not a cyclic subgroup, just a group.

Like, if I find all generators of these finite groups. Can I simply use them to find all subgroups?

[obsidian sleet]

ohh nami that worked ty

[scarlet estuary]

oh no, subgroups of cyclic groups must be cyclic

this is actually a good exercise

[wild sapphire]

ah, implying they have a generator

?

[scarlet estuary]

a single generator yes

but yeah this is a good exercise, although iirc the most elementary proof requires the division algorithm

[wild sapphire]

nice, I was hoping that was the case. I'm still pretty shakey with groups and subgroups. the common rotational symmetry examples make perfect sense but when it gets all numerical it starts to get a little less confortable

[obsidian sleet]

i feel that

im in my first course in group theory

and it was fun and games with the dihedral group

[wild sapphire]

me too 🥴

[obsidian sleet]

but then they did cyclic groups

[scarlet estuary]

in fact

[obsidian sleet]

and i shat myself

[scarlet estuary]

a stronger theorem is true

this is a Very Important Result ™️

[obsidian sleet]

yeah

the proofs are very cool

[wild sapphire]

yeah, I'm in both my first course on groups, and first course on graph theory. All of these new proof styles are hurting my brain 🥴 I'm coming from just a bunch of analysis courses

[scarlet estuary]

(an analogous result holds for the infinite cyclic group btw)

(the only subgroups of the infinite cyclic group are [isomorphic to] the infinite cyclic group itself)

(and the trivial group)

[obsidian sleet]

ah

right

[scarlet estuary]

(in fact, if you really wanna nerd out, you can prove the theorem in the screenshot i posted using that fact! since the infinite cyclic group is Z and its subgroups are nZ, just consider quotient groups Z/nZ)

[wild sapphire]

@scarlet estuary when you said every group has generators, how does that work? I've been learning about cyclic groups and their generators. But if the generators aren't a property of this cyclic setting, it starts to confuse me.

Like if a group isn't cyclic, does that just mean that it contains a cyclic group generated by certain elements, but then there's also some other stuff that is in the group but can't be in a cyclic subgroup?

[scarlet estuary]

to clarify

every group has some generating set

its just that cyclic groups have exactly one generator

other groups have more than one generator

[wild sapphire]

ohh, so it's just a lil nuance with the definition of a cyclic group. gotcha

[scarlet estuary]

(except the trivial group which has an empty generating set)

this fact follows from groups being closed BTW

[wild sapphire]

yeah, I get that

[scarlet estuary]

suppose there is an element g not generated by the generating set S of a group. then we could construct a new generating set S' = S U {g} and it would now generate g (and it still only generates group elements since they're closed).

then S U {g} would be a generating set that actually generates g

contradiction

this argument can be applied if S is empty.

(again, unless your group is the trivial group! in fact the assumption that g isn't the identity is necessary)

[wild sapphire]

So with a group in general. Would it be valid to say that every element in a group G is in at least one cyclic subgroup?

[scarlet estuary]

yes!

every g in G is in the cyclic subgroup generated by g

[wild sapphire]

And this is just the group of integer powers of g with respect to whatever group operation?

so just to be clear, if we generate a group from an arbitrary g in G, it doesn't ensure in any way that this cyclic subgroup is finite or contains the identity right?

[scarlet estuary]

in fact, more generally, for any finite set of elements of a group G, there exists a smallest subgroup H of G given by intersecting all subgroups containing each of those elements

And this is just the group of integer powers of g with respect to whatever group operation?
yes
so just to be clear, if we generate a group from an arbitrary g in G, it doesn't ensure in any way that this cyclic subgroup is finite or contains the identity right?
it must contain the identity but it may not be finite

it must contain the identity because g^0 = e

but it won't be finite if g^n is not equal to g for any n > 1

this can, of course, only happen if the parent group is infinite as well

[wild sapphire]

ah, so g^0 is like a trivial case?

[scarlet estuary]

yes

the identity can be generated by any set

by just taking empty products

and this should make sense, since after all, these are subgroups

so they must have an identity

the fact that they coincide with the group identity is perhaps less immediately obvious

but it can be proved from uniqueness of inverses

[wild sapphire]

yeah that makes sense

[scarlet estuary]

sorry if i kinda threw a ton of facts at you there

its just that cyclic groups are VERY well-behaved

so its easy to rapid-fire rattle off a lot of info about them

[wild sapphire]

no no it all makes sense. That H and G smallest subgroup part was a little confusing but everything else makes sense

in fact, more generally, for any finite set of elements of a group G, there exists a smallest subgroup H of G given by intersecting all subgroups containing each of those elements

[scarlet estuary]

basically suppose you have a set of elements {g_1, g_2, ... g_n} from G

i want to find a "smallest" subgroup H of G that contains g_1, g_2, ... g_n

by "smallest" i mean

is a subset (and thus a subgroup) of any other subgroup containing g_1, g_2, ... g_n

since intersections of subgroups are subgroups (a good exercise!), what we can do is just

take ALL the subgroups of G that contain {g_1, g_2, ... g_n}

and take the intersection of all of 'em

and that MUST be the smallest subgroup containing them all

if your set {g_1, g_2, ... g_n} only has one element [so it actually looks like {g_1}], we get the fact we were talking about earlier

[wild sapphire]

Ah, I haven't seen the fact about intersections before (or maybe I missed it 🤔 )

that makes sense

[scarlet estuary]

actually i dont think finiteness is required here either

so your set of elements could be infinite if you wanted

since arbitrary intersections of subgroups are subgroups

thats harder to prove though

[wild sapphire]

yeah, interesting

[scarlet estuary]

thats a side tangent to the discussion about cyclic groups in any case

but it does show that in many ways, facts about cyclic groups are just "simpler" versions of facts about general groups

this is a theme that shows up frequently in algebra: facts about principal ideals are "simpler" versions of facts about finitely generated ideals

facts about groups are "simpler" versions of facts about groupoids

[wild sapphire]

oh god groupoids

sounds like the final boss

🥴

[scarlet estuary]

theyre just a weird categorical generalization of groups

you dont have to worry about them, theyre not that useful if youre not doing pure category theory

basically its like, the defining/interesting property of groups is the fact that we can take inverses, right?

closure, associativity, identity are "common" features of operations we study

since these show up in function composition

being able to take an inverse of every element is "special"

groupoids are the category theory version of groups essentially

[wild sapphire]

ah, interesting

[scarlet estuary]

a category where all the morphisms are invertible

by the way, when people say "group theory is the study of symmetry", this inverse thing is usually what they mean

that's what makes groups special: theyre the "simplest" structure in which we can take inverses, hence the natural setting to discuss symmetry

(this isnt technically true, there are actually simpler structures, but they're far less nice than groups)

[wild sapphire]

ah, like "morphisms" branching out to other terms like isomorphism and homeomorphism that are used all over the place?

[scarlet estuary]

yeah, a "morphism" is a function that preserves some sort of structure

theres a precise categorical definition but thats the intuition

homomorphisms preserve the structure of the group operation (or whatever youre taking a homomorphism of)

a homeomorphism preserves the topological structure of the space

(in the sense of being continuous)

etc

[wild sapphire]

yeah interesting. I have all the linear algebra, real/complex analysis, and calculus I could ever need so I know those contexts. but its a bit weird to go to pure algebra and work on the "lower level" structures

[scarlet estuary]

sets have no structure, so "morphisms" in the context of pure sets

are just functions

since there's no structure to preserve

only pretentious people call them "morphisms" in this context

sane people call them "functions"

sadly category theorists arent sane

[latent anvil]

Really pretentious people call them set-functions 😌

[wild sapphire]

ah, so if I say like "bijection", that's in the context of a function on a pure set and the lack of structure. but using the word "isomorphism" is implying that we have the same kind of mapping but some sort of structure is being preserved?

I always wondered why the two words were used almost interchangeably at times.

[obsidian sleet]

well in linear algebra we have isomorphism as a consequence of invertible (bijective) linear transformation and so the dimension of the vector spaces are the same right

so is this a similar thing

but instead of dimension it's just

size of the group or something

[scarlet estuary]

you can have bijective maps between groups that arent isomorphisms

[obsidian sleet]

o

[scarlet estuary]

theyre not particularly interesting though

since they dont preserve structure

[obsidian sleet]

i didn t know that :3

[scarlet estuary]

but like, if i have two groups of 8 elements

[wild sapphire]

yeah, so you would say that "the bijective mapping sometimes carries the isomorphic property"

[scarlet estuary]

and just pair up their elements randomly

this is a bijection

but (probably) not an isomorphism

[wild sapphire]

ah okay yeah that makes sense.

[scarlet estuary]

thats a bit of a weird way to phrase it but sure

[obsidian sleet]

wait oh do i have it backwards

[scarlet estuary]

to be precise, an isomorphism is just:

• a homomorphism
• that is a bijection

[obsidian sleet]

isomorphism implies bijective

[scarlet estuary]

thats all

in a linear algebra context when we talk about linear maps

"linear map" is a synonym for "vector space homomorphism"

so we're already talking about homomorphisms

hence bijective linear maps are already isomorphisms

the terminology can get a bit criss-crossy.

[wild sapphire]

ah yeah, I think the linear algebra context was what confused me but "linear map" is a synonym for "vector space homomorphism" clears it up a lot

[latent anvil]

For groups or linear transformations, Isomorphism = bijective + homomorphism

In namington's examples of groups the bijection won't preserve multiplication

[obsidian sleet]

ohh

okey

[lunar hedge]

Could someone help me with this linear algebra problem? im lost

not really sure where to start with this, i know that it's vectors and i need to put them into a 2x2 matrix since it's R^2 but not sure where to go from there. any help is appreciated

[north widget]

what doesnt satisfy ascending chain condition?

or should I just use google

[mild laurel]

Z[x_1, x_2, \dots, ] for example

[north widget]

oh

um

damn

ill try and prove it ig?

polynomials of infinity variables?

[mild laurel]

(x_1) \subsetneq (x_1, x_2) \subsetneq (x_1, x_2, x_3) \subsetneq \cdots

[north widget]

o

what is that ring called?

[mild laurel]

the ring of integer polynomials in infinitely many variables

[north widget]

anything else interesting about that beast?

oh

ig i can read more about polynomial rings on wikipage

[next obsidian]

Countably many

Smh my head zoph

[mild laurel]

has the ring of integer polynomials in continuum many variables ever come up in your life chmonkey

[next obsidian]

I’ve thought about it yes

Z[continuum many] isnt countavle

While Z[countably many] is

Which makes it smaller than Z[[x]]

Which I hate

Also not for integers but for alg closed fields of char 0 this is kinda important I think

Cuz if you have only countably many stuff you can do an embedding into C or something

[golden pasture]

do people even care about such esoteric rings

i mean id assume people do

but like

any decent reason to

[sturdy marsh]

poly ring in countably many variables (with a different grading) represents formal group laws

[golden pasture]

:o

wait i always thought formal groups is from finite poly rings/formal power series or am i thinking of a diff thing

[sturdy marsh]

yes, they are related to formal power series

a 1-dim commutative formal group law over a ring R is an element of R[[x,y]] satisfying a bunch of things

There's a ring L which represents the functor which sends a ring to formal group laws over the ring

and it's a theorem that this ring is a polynomial ring

https://ncatlab.org/nlab/show/Lazard's+theorem

[golden pasture]

ahhhh cool

[sturdy marsh]

weirder polynomial-ish rings also turn up

like the novikov ring

which turns up while defining Floer cohomology

[vestal snow]

How do I find a matrix for 4?

I mean I know how

But is there any way other than guessing the linear combination?

[chilly ocean]

What do you mean guessing the linear combination?

[vestal snow]

For example, I want to know the basis representation of $\phi(\xi)=\sqrt{2-\sqrt{2}}$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

How am I supposed to get that

[chilly ocean]

I would probably write it in terms of xi (allowing division by xi) then reduce it to a polynomial in xi

[vestal snow]

I'm an idiot

I computed the powers of \xi and forgot that they were... powers of \xi

I got that it is equal to $\frac{\sqrt{2}}{\xi}$

[cloud walrus]

Have a Banana, Bitch

[chilly ocean]

Then there are two things to address, sqrt(2) is not in Q, and 1/xi is not a polynomial in xi

[wild sapphire]

Can someone verify this for me?

What is the order of z = (1 + i) in the abelian group: The non-zero complex numbers with respect to multiplication?

I think the order is infinite 🤔 because regardless of proper complex rotations, this vector is not a unit vector and so it will increase in magnitude with each power, hence it never reaches the identity.

[vestal snow]

How do we address these?

[chilly ocean]

How do you write sqrt(2) in terms of powers of xi and elements of Q?

[vestal snow]

\xi^2-2

Ah got it

[mild laurel]

yes thats right

[wild sapphire]

thanks !

[golden pasture]

basis representation?

like

how 1,ξ,ξ^2,ξ^3 are linearly related?

they are linearly independent tho

ξ^4 can be seen directly from minimal polynomial

Ohhhhhhh

okok so basically what you do is

You kinda "reverse engineer" it

so an example is like

$$x=\sqrt{1+\sqrt2}$$
$$x^2=1+\sqrt2$$
$$x^2-1=\sqrt2$$
$$x^4-2x^2+1=2$$
$$x^4-2x^2-1=0$$

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[golden pasture]

ah oops misread

[harsh pagoda]

Hey guys, anyone here familliar with Clifford or Weyl algebra ?

Let $Q_{ij}$ be a $n \times n$ symetrical matrix , Clifford algebra :
\begin{align}
Cl(Q) = \braket{\xi_i}{i=1,...,n \quad ; \quad \xi_i \xi_j + \xi_j \xi_i =Q_{ij}}
\end{align}
Find the condition on the matrix $A_{ij}$ so that
\begin{align}
\xi i \to \sum_i A{ij} \xi_j
\end{align}
Is an automorphism

[cloud walrus]

K_

[celest brook]

[prisma ibex]

there are 5 isomorphism classes of groups of order 8

there are a little more than 1.8 billion isomorphism classes of semigroups of order 8

[rigid cave]

Hello! I am currently reading about the Banach-Tarski paradox here http://math.stmarys-ca.edu/wp-content/uploads/2017/07/Alex-Lowen.pdf. I am confused with the proof of theorem 6. The red areas are the parts that I do not understand. When defining $A_i^*$, why do we need to have braces around the set? It seems unnecessary. I am also confused with the last sentence. How does that follow directly?

[cloud walrus]

おねえちゃん

[rigid cave]

What does even $g_i(A_i(R))$ mean?

[cloud walrus]

おねえちゃん

[wispy scaffold]

||I don't like how it's written. || wrt the long red line, you can think it as making G act on R ||like multiplying the previous eqs by R||, so if you write the action you could see how does it works with the union. Why braces? Idk, as they write it as a union (and then they use A_i(R) ).

g_i( A_i(R)) is g_i acting over A_i(R), and A_i(R) is the set of g.r with g in A_i and r in R (i.e. A_i*)

[rigid cave]

Oh okay! Thanks for your answer. I am still a little confused why the set X equals the union of g_i(A_i^*). Would you care to explain this? I have also posted this question on math stackexchange (here: https://math.stackexchange.com/questions/4038452/confused-with-a-proof-about-paradoxical-groups) in case you want to answer there!

[golden pasture]

im already losted at the

2 multisets

since when did multisets appear

[wispy scaffold]

That's why it's ugly notation

[golden pasture]

but i shall ignore that

[rigid cave]

Man, I've been sitting here for hours trying to understand this...

really frustrating

[golden pasture]

A_i looks like some subset of the group

[wispy scaffold]

This M is your R

[golden pasture]

so naturally $A_i R = \bigcup_{g\in A_i}gR$

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[wispy scaffold]

(idk why they use a choice function)

[golden pasture]

the big brackets are probably typo

[rigid cave]

Yeah I thought so as well

[golden pasture]

you need a choice function to construct R i believe?

[wispy scaffold]

Yes

I mean

[rigid cave]

Homeomorfismo, that proposition in your image looks like the theorem I'm looking for!

[wispy scaffold]

Why choice function, just state the set of representatives

[golden pasture]

wdym?

[wispy scaffold]

Yes, it is

[rigid cave]

Can you send a picture of the proof?

[golden pasture]

you still need a choice function for this dont you

[wispy scaffold]

It's the same, you need to do what Ariana did

[golden pasture]

to construct the set of representatives

[wispy scaffold]

Yeees, but

Aesthetic, ig

[golden pasture]

im guessing the objective of that thesis is jus show banach tariski from zfc lol

[wispy scaffold]

Idk, doesn't matter

[rigid cave]

What book is that?

From your picture?

[wispy scaffold]

The Banach Tarski Paradox

Wagon, Tomkowicz

[rigid cave]

Well your book uses the same method as my picture.

[wispy scaffold]

I know

But the picture helps

[rigid cave]

Oh well, It's getting late here so I will try to figure it out tomorrow. Thanks for your help!

[wispy scaffold]

X=G.R

[rigid cave]

Yes, I understand that. But I am not sure how to go from there

[wispy scaffold]

Replace G by the union of gi Ai, then do what Ariana did

[mystic jungle]

alright so from lin alg i remember the ker being equal to like null space or something, its analogous to the abstract algebra def right? like how would I find the kernel for this perhaps

[rigid cave]

What did Ariana do?

[mystic jungle]

where ker(f)={g in G | f(g)=e'}

[wispy scaffold]

This

[unique juniper]

oh wait

[wispy scaffold]

Why not?

Oof

[unique juniper]

hm

[mystic jungle]

they literally say it lol

[wispy scaffold]

K is non empty

[unique juniper]

ye i realise now lol

mb

[mystic jungle]

ok so ker(f)={A in G | f(A)=det(A)=ad-bc} is that right or

[next obsidian]

you haven't said anything about what f(A) equals

you've merely given the definition of what det(A) is

the kernel is gonna be the stuff for which that value is the identity

which in this case is 1

I think...

I must use my chmonkey brain

to remember how multiplicativity of determinat works

Yes, it will be {A in G| f(A) = 1}

[mystic jungle]

oh so det of id matrix is what its asking

thank you

[pine patio]

is this true?

S^1 is abelian but S0(2) isnt

so how are they isomorphic

S^1 is the group of complex numbers of size 1, with multiplication

S0(2) is the group of orthogonal 2x2 matrices

0(2) is same as S0(2)

[mild laurel]

how is SO(2) nonabelian?

[pine patio]

matrix multiplication isnt commutative

[mild laurel]

No two matrices commute?

[pine patio]

some dont

some do

shouldnt they all commute for SO(2) to be abelian?

[mild laurel]

then how do you know that SO(2) isn't comprised of matrices that commute?

Yes

[pine patio]

hmm

oh they do commute

idk why i assumed they don't

how do i do the next one?

[mild laurel]

also O(2) is not the same as SO(2)

[pine patio]

what is O(n) then?

[viscid pewter]

jesus christ

SO is subgroup of O with determinant 1

[pine patio]

wth i need to tell my lecturer to fix the error

[mild laurel]

https://en.wikipedia.org/wiki/Orthogonal_group#SO(n)

[viscid pewter]

anyway if two groups are isomorphic, their orders must be equal, if one is abelian then the other is and most relevantly the mapping of an element under the isomorphism has the same order as the original element

is that too opaque? it's clearer with the actual notation

[pine patio]

we havent done much with orders yet

they both have infinite order so idk how do distinguish between them

[viscid pewter]

yeah those first two things aren't really relevant

try the third

look at orders of elements

[pine patio]

so in 0(2) the orders are all finite

the matrices represent either rotations or reflections

[viscid pewter]

welllll

what about a rotation of sqrt2 radians

that has infinite order

[pine patio]

oh so order has to be an integer then

[viscid pewter]

does it?

[pine patio]

rational?

[viscid pewter]

i thought i just gave you one without integer order

what does rational order mean

[pine patio]

im just thinking of rotations

so like e^(ix) times something

[golden pasture]

(btw it's O(n) not 0(n))

[pine patio]

• raised to the power

[viscid pewter]

oh. well that's really unfortunate, i gtg

someone else should be able to help you out tho

[golden pasture]

O(2) isnt abelian

[viscid pewter]

oh wait duh

[latent anvil]

also, you won't be able to distinguish the groups just by looking at rotations

[pine patio]

yes i know now. i thought SO(2) is same as O(2)

[viscid pewter]

i think i was still thinking about SO(2) also

[latent anvil]

Since SO(2) ≈ S^1

[pine patio]

what's that symbol? isomorphic?

[latent anvil]

Ari is saying this shows O(2) isn't isomorphic to S^1 × {±1}

[latent anvil]

yeah I'm just be lazy lol

[golden pasture]

SO(2) is immediately isomorphic to S^1 cuz like both are just rotations of the 2D plane

[pine patio]

yes we did that in the first part

[pine patio]

here

[viscid pewter]

i think i was just going to go for infinite elements of order 2

[golden pasture]

then the second part is done as well

[latent anvil]

Also works

[viscid pewter]

ok time to be productive

[golden pasture]

[pine patio]

what's the operation for S^1 x {pm 1} ?

[golden pasture]

\pm 1 is a group

s^1 is a group

take product

[pine patio]

termwise multiplication?

[golden pasture]

you get a group

yesh

[latent anvil]

> what's the operation on this product?
> take product

[pine patio]

oh so S^1 x {pm 1 } is abelian but O(2) isnt

now it makes sense.

my lecture notes said O(2) is same as SO(2)

u can imagine how good they are at explaining then

also what was the argument about infinite order?

or infinite elements with order 2

so in O(2) we have infinite reflections

[latent anvil]

Right

How many elements of order 2 are there in SO(2) × {1,-1}?

[pine patio]

infinite?

(I, 1) is the identity right?

[latent anvil]

Yup

[pine patio]

so (A, 1) * (A^-1, 1) = e

[latent anvil]

Yes?

But this doesn't mean (A, 1) has order 2

[pine patio]

oh so its just 1

[latent anvil]

also no

[pine patio]

oh theres -1 too

(A, -1) and (-A, 1) and (-A, -1)

[latent anvil]

What's A?

[pine patio]

A represents rotation with pi

[latent anvil]

right, which is just -I

[pine patio]

yes, so is the answer 4?

[pine patio]

3 here too

[latent anvil]

There's only 3

(-A, 1) = (I, 1) has order 1

[pine patio]

(A, 1)

oh true

i get it now, thanks

what book do u recommend for groups and group actions?

today's my first day starting it

[sturdy marsh]

dummit and foote

[pine patio]

and how is groups and symmetry by armstrong

[golden pasture]

jacobson

[chilly ocean]

DUMB FOOT

[uncut girder]

DUMB FOOT

I READ DUMB FOOT CHAPTER 13-14

[sturdy marsh]

dumb foot chapter 13 is dope

there's a super cool proof of wedderburn's little theorem in the exercises

[latent anvil]

Wedderburn's little theorem doesn't get enough hype

It's incredibly based

[rustic crown]

Aluffi had a really cute proof in the exercises

[south temple]

Let $G$ be a group such that $(ab)^n = a^bb^n$ for all $a,b\in G$ and for some $n\in\mathbb{N}$. Define
$$
H = {a^n : a\in G}.
$$
Does anyone have a hint as to how I can show that $H$ is a normal subgroup?

[cloud walrus]

vov&sons

[south temple]

I'm having trouble showing that $ga^ng^{-1}\in H$ for any $a,g\in G$.

[proud bear]

do you mean a^nb^n?

[cloud walrus]

vov&sons

[carmine fossil]

g a^n g^-1=(gag^-1)^n

[south temple]

ah yeah sorry, my bad

[golden pasture]

doesnt this imply G is commutative already tho

cuz abab=aabb

ohh for some

oops

[south temple]

oh my god, I can't believe I didn't see this

thank you

so it's kind of like a telescoping sum

[golden pasture]

haha yea once you learn this you cant unsee

i rmb that problem in herstein was stuck for so long

then searched online and saw taking conjugate

and

[south temple]

ya I've been trying to do some weird stuff for like an hour and it just was not working 😔

[latent anvil]

ari there is a cool thing related to this

let k be some integer

Call a group k-abelian if (ab)^k = a^k b^k for all a, b

If a group is k-abelian for three consecutive k then it is abelian

I think there's other stuff like k-abelian iff (1-k)-abelian

[chilly ocean]

Sounds kind of creepy

[latent anvil]

It is weird

[vestal snow]

Let $F'/F$ be an Artin-Schreier extension of function fields over $k(x)$ given by the equation $y^p-y=f(x)$ where $f(x)\in k[x]$. and $k$ is algebraically closed. Consider the differential $x^iy^jdx$ where $i,j$ are non-negative. For what values of $i,j$ is this differential holomorphic?

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Is the answer to this none?

$x^iy^jdx$ has non-negative order at all finite places

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

But i think it has negative order at the ramified place of infinity

Here's my reasoning

Let $\bar{\infty}$ be the ramified place over $\infty$. Then notice that $x^iy^jdx=x^{i+2}y^jd(\frac{1}{x})$. Thus, $ord_{\bar{\infty}}(x^iy^jdx)=ord_{\bar{\infty}}(x^{i+2}y^j)$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Can someone verify this?

[unique juniper]

Prove that the subgroup generated by any two distinct elements of order 2 in $S_3$ is all of
S$_3$.

[cloud walrus]

Yes

[unique juniper]

Am not sure if i understand

The elements of order 2 in S3 are (1,2) (1,3) (2,3)

[carmine fossil]

Yes

[unique juniper]

$\langle (1,2), (1,3) \rangle = \bigcap_ {(1,2) \cup (1,3) \subset H, H \leq S_3}H$

[cloud walrus]

Yes

[unique juniper]

is this the correct definition? how could this ever be the whole group S3 ?

[carmine fossil]

S_3 is just D_6 tho

Use a presentation to show the said generated subgroup will be iso to D_6

and hence S_3

[unique juniper]

is using the definition not helpful ?

[pine patio]

can someone answer a quick question? what happens when k is 1?

[carmine fossil]

I mean,just writing down the terms works too

[unique juniper]

i guess so

[carmine fossil]

k can never be 1

In that notation,fixed elements are ignored

[pine patio]

oh ok ty

[chilly ocean]

I must say banana bitch has an interesting background, the other day you came here about a simple field extension question, today it sounds like some crazy algebraic number theory shit

[unique juniper]

so if a and b are of order 2, and d8 is of order 8, why that does mean its not possible to write every element in the form of a's and b's

[chilly ocean]

Because there are four such elements

1, a, b, ab

[unique juniper]

oh right

ty

[summer geyser]

Is it enough to say that, if alpha was algebraic then K_0(alpha) would be an algebraic extension of K that is larger than K_0 contradicting the fact that K_0 is the largest? I feel like this argument is too simple and I'm missing something. New argument below, but still seems too simple (especially because this question was labelled as a question that's supposed to provide bonus points).

[cloud walrus]

Athurus

[pine patio]

how do i show that sign of a permutation with odd order is even?

ive managed to show that the sign is (-1)^m where m is the number of cycles in its cycle decomposition

[carmine fossil]

Decompose it into transpositions

[carmine fossil]

Use this

[pine patio]

that's how i proved this

[pine patio]

so i decomposed it into cycles and then i decomposed each cycle like this

[carmine fossil]

In an odd order cycle,There will be even number of such cycles

And each transposition is odd

[hollow imp]

that's not exactly true. m should be the number of cycles of even length in its cycle decomposition.

[pine patio]

ah i made an error

just realised

i thought order meant the number of terms and got confused

we proved that sign(ab)=sign(a)sign(b) previously

so if a has odd order then (sign(a))^(2k+1)=1

doesnt make sense

sign of the identity is 1 right?

and a has odd order so a^(2k+1)=e

oh yh sign(a) must be 1

but if a has even order then its sign can be 1 or -1 right?

[obsidian sleet]

hello

i am once again returning to ask for help 😔

for two things this time

i need to prove that the inverse of an odd permutation is odd

i think i know how to do this but im so worried about missing something

[hollow imp]

if you have a cycle with an even # of terms, then the sign of that cycle is -1.

[pine patio]

do u know sign(a)sign(b)=sign(ab)?

[obsidian sleet]

in my proof i said basically this: Take some odd permutation K = k_1k_2...k_j where each k is a 2 cycle

sign?

wait what is sign

order?

[pine patio]

if the permutation is even then sign is 1

similarly for odd

[obsidian sleet]

o .o

we never learned about sign in class

idt i can use that

[pine patio]

then ur not meant to use it

[obsidian sleet]

i then took the inverse using shoes and socks

[hollow imp]

shoes and socks?

[obsidian sleet]

its a silly thing to mean u reverse the order

(ab)^-1 is b^-1a^-1

shoes and socks

anyways i said the inverse was k_j^-1...k_2^-1k_1^-1

but each 2 cycle is its own inverse so

its jst

k_j...k_2k_1

so K^-1 has j number of 2-cycles

[hollow imp]

ye

[obsidian sleet]

oop i forgot to mention

j is odd

from assumption

so ... is that it

that's the whole proof right

im not missing some strange case

[carmine fossil]

Have you seen A_n?

[obsidian sleet]

ya that was my next thing to talk about

yes

[hollow imp]

sign is just (-1)^(num of transpositions)

[obsidian sleet]

i was gonna ask to show that A_n (set of even permutations) is a proper subgroup of group of permutations

S_n

and uh i was just gonna use the finite subgroup test

here n is finite in the problem

[carmine fossil]

sgn(a)sgn(a^-1)=sgn(1)

[obsidian sleet]

nuuu but i can't use sign

[carmine fossil]

How are you defining odd permutations?

[obsidian sleet]

odd permutations are those that can be written as a product of an odd number of 2-cycles

and we showed in class that it was well defined, just not with sign or anything

we somehow just

didn't talk about that

[mild laurel]

The inverse of a product of 2-cycles are just the same 2-cycles but in reverse order

[obsidian sleet]

right zoph

so i wrote in my hw that the inverse is just the same number of 2-cycles just arranged in reverse

so the inverse is also odd

[carmine fossil]

Btw, A_n is also the subgroup generated by 3 cycles

[obsidian sleet]

what

3 cycles...

o.0

oh

also

so to show A_n is a subgroup of S_n i just said it was nonempty subset cus identity is in A_n, then i took two even permutations A and B and now i want to show that AB is also in A_n (finite subgroup test)

so the thing is

i feel like

there are many ways for this to go

since on one hand maybe all of these cycles are distinct which is fine

automatically we have even permutation

but then maybe there are some non-distinct 2 cycles , one in A and one in B

and i can't just commute them obv cus not necessarily disjoint

and maybe if these two identical 2-cycles appear in a manner where they can cancel each other out

like the last 2-cycle of A and the first 2-cycle of B

if those are the same cycle then identity and so we just took away 2 cycles

which also seems fine

but

this seems very hairy to me

[mild laurel]

uhhhhh, if your definition of even permutation is just can be written as product of even number of 2-cycles, then you're just done

[obsidian sleet]

idk how to organize my thoughts

oh

oh right

i dont

i don't need to worry about

any of this

omg

hgggggggg

okay ty

[pine patio]

what's the operation is i) ?

[carmine fossil]

Take a wild guess

[pine patio]

how do i compose (12)(34) with (13)(24)

composition?

[carmine fossil]

Yes

[pine patio]

ok so is this (12)(34)(13)(24) ?

1 goes to 2 then 2 goes to 4

2 goes to 1 then 1 goes to 3

have i got the notation correct?

[pine patio]

this looks a bit dodgy

[obsidian sleet]

in cycle notation

taking the "product" is the same as composing the permutations

(i hope im right)

u just have to always have those rules for reading the product of cycles in mind ofc

[carmine fossil]

You read that from right to left

1 goes to 3,and 3 goes to 4

[pine patio]

i thought we're supposed to feed in from left

[next obsidian]

Which direction you read it from is convention

[pine patio]

so 1(sigma) instead of sigma(1)

idk my notes mentioned this convention

[hollow imp]

huh?

[obsidian sleet]

Pain

[pine patio]

convention is to feed in from the left

is there a quick way to show associativity from the cayley table?

[carmine fossil]

That's a subset of S_4

[pine patio]

what is

[carmine fossil]

V_4

[pine patio]

so u mean i should use that to show associativity?

[carmine fossil]

Yes

Associativity holds for any subset of a group

[pine patio]

oh that makes sense

thanks

[pine patio]

do we have 6 isomorphisms for part iii)?

[obsidian path]

can anyone give me the thought process or layout for this?

[carmine fossil]

#linear-algebra

[obsidian path]

oh shit sorry

[pine patio]

how do i do ii) ?

we know alpha must have a cycle of order atleast 4

[carmine fossil]

Order of alpha is 12

Cycles can be of lengths 2,3,4 or 6

[pine patio]

how? is that lagrange?

oh no i get it

[carmine fossil]

Have you shown order of a permutation is LCM of orders of disjoint cycles present in it?

[pine patio]

yes

[pine patio]

is this true?

[carmine fossil]

Why do you think that's true?

[pine patio]

if we have all of them of order 2 then apha^3 will not be a 4 cycle

[carmine fossil]

*can be 4 or 12

alpha should be (1432) suppose order is 4

1,2,3,4,6,12 are possibilities according to order dividing 12 condition

[pine patio]

what about 5 6 and 7

[carmine fossil]

5,6,7 will be fixed

[pine patio]

we havent done that yet

[pine patio]

how are they fixed under alpha?

[carmine fossil]

alpha^4=1

[pine patio]

theyre fixed under apha^3

[carmine fossil]

(1234)alpha=1

alpha=(4321)

[pine patio]

are we assuming that for now?

[carmine fossil]

This

Wait, It's not even lagrange

[pine patio]

ok so if the order is 4 then alpha is this

[carmine fossil]

It's just order of a element g divides k where g^k=1

[pine patio]

this?

[carmine fossil]

That's for this

Now,6 isn't a possibility here, because to get a LCM of 12 you need a disjoint cycle of length 4

[pine patio]

yes

[carmine fossil]

You can have 1 4-cycle,1 3-cycle

Now,There are no elements of order 12 without a 4-cycle

(because you need some cycle with 2^2 as a factor for 12 to be LCM)

[pine patio]

yes because we know 6 isnt an option

oh wait even if it was we still need 4

[carmine fossil]

So you end with
{(4321),(4321)(567),(4321)(576)} as your possible elements

[pine patio]

yes that makes sense, thanks

permutations of prime order must be cyclic right?

[viscid pewter]

what if it's just two disjoint prime cycles

[carmine fossil]

Possible

Like (234)(567)

[viscid pewter]

so not necessarily

[cyan marten]

But they're necessarily products of disjoint cycles, each of which is of order p (p fixed), right?

[carmine fossil]

Yes

Because otherwise LCM won't be p

[cyan marten]

Yeah

[pine patio]

how do i do this part?

i was trying to work backwards

decomposing aphabeta

and assuming it to be cyclic

i should have cases right?

for when n is prime and when its not

[carmine fossil]

Ever heard of D_2n?

[pine patio]

i think so

[carmine fossil]

That's your answer

[pine patio]

its the dihedral group right?

[carmine fossil]

Yes

[pine patio]

so two reflections that have a product which represents rotation by 2pi/n

[carmine fossil]

Take the permutations corresponding to those reflections

[pine patio]

how do i show existence

[carmine fossil]

Take a polygon of n sides and mark some points

Apply a rotation and note where the points land,that would be your r(similarly for reflection)

For example (123) is your "rotation"(r) for a triangle

(23) will be your reflection(s)

Now just consider elements rs and s

(123...n) will be your rotation (r) and (2 n) (3 n-1) (4 n-2) .... will be your reflection (s) in n case

[woven obsidian]

I'm not sure if this is a question that's solved through algebra or number theory, but I'll try asking it here

[pine patio]

thanks! i seriously need to practice more questions. i wouldntve guessed to use D_2n

[woven obsidian]

I have 2 irreducible, distinct, polynomials in Z[x] with degree larger than 1. I want to show that they cannot have a root in common in $F_p$ for infinitely many $p$.

[cloud walrus]

AoiKunie

[oblique river]

do you mean that you want to show that the set {p | f and g have a common root in F_p} is finite? or that {p | f and g don't have a common root in F_p} is infinite?