groups-rings-fields

[sinful mirage]

no?

[hidden haven]

The induced injection is

G/~ = G/G_x isn't

[hidden haven]

Do you see why this does the job?

[sinful mirage]

sadly no. G~ means I partition G into cosets, right?

how are cosets at all related to G_x?

[hidden haven]

Cosets of what?

~ doesn't necessarily mean identifying cosets btw, it could be an arbitrary equiv relation

Here it happens to identify cosets, the cosets of G_x

[sinful mirage]

if it identifies that,hten the LHS is a partition of G into cosets

[hidden haven]

Of G

[sinful mirage]

so would G look like g G_{x}?

[hidden haven]

I don't get it

G looks nothing like that

Maybe try working with an explicit example?

[sinful mirage]

i'm super confused of what this equaliy is supposed to mean

[hidden haven]

It means that ~ defines the same partition as the cosets of G_x

So they give the same quotient

[sinful mirage]

how is ~ defined?

[hidden haven]

g ~ h iff along the map G → Gx, g and h map to the same thing

ie gx = hx

[sinful mirage]

ahh

[sinful mirage]

well this is exactly he condition to partition in cosets

but now, why of G_x?

[sinful mirage]

used in Lagrange's theorem oto

[hidden haven]

Partition into cosets condition is
g ~ h iff gG_x = hG_x

[sinful mirage]

because for that G_x would need to be a subgroup of G

[hidden haven]

G_x is a subgroup of G

[sinful mirage]

why is g G_x=h G_x the same as gx=hx?

[hidden haven]

That's what you have to prove

That is the only non trivial part in the problem, the rest is just deciphering language

We went from surjective map
G → Gx
To a bijective map
G/~ → Gx
And we wanted a bijective map
G/G_x → Gx
And we said G/~ and G/G_x are the same set
Which is true iff (gx = hx iff gG_x = hG_x)

This is the entire argument

I have to go to sleep now, but I'd suggest that you just look at this outline and maybe work with explicit examples until this is utterly obvious, and then try to prove the last iff

[sinful mirage]

yep,the last iff is the only part missing

thanks a lot for patience

[wise igloo]

why are elementary algebra and abstract algebra both considered algebra when they are very distinct? I am aware of the relationships but I don't get how they can be under the same roof

[rapid bramble]

From my understanding (ie someone who hasn't formally studied abstract algebra, bar linear algebra) it's cause they're both algebra.. algebra is just the branch of math for dealing with manipulating symbols and includes number theory, geometry, etc.

[wise igloo]

it includes number theory and geometry too? damn

alright thanks

[fossil shuttle]

algebra is the branch of mathematics which deals with structures whose elements can be added, multiplied, or otherwise combined to yield new elements. an algebraic structure also often has distinguished constants like 1 or 0 which have special roles with regards to addition, multiplication and the other objects. Both elementary algebra and abstract algebra study polynomials, for example, which can be added, multiplied and have distinguished elements 0 and 1. The difference is that abstract algebra applies to a more general class of structures.

[wise igloo]

oh alright

[fossil shuttle]

linear algebra is part of algebra because it deals with systems of linear equations, which can also be added together, scaled by constants, etc.

[wise igloo]

oh ok, thanks guys

helps a lot

[fossil shuttle]

it also deals with matrices, which can be multiplied, added... there is an identity matrix and so on

[wise igloo]

true

[fossil shuttle]

geometry is not part of algebra

[wise igloo]

ok so abstract algebra is just an abstraction of elementary algebra

[wise igloo]

ok yeah that seemed odd to me

[fossil shuttle]

concepts like "continuity", "differentiability" don't belong to algebra really

the notion of limit is also not really an algebraic concept

[wise igloo]

they're all more of calculus concepts

[fossil shuttle]

Number theory is arguably a branch of algebra.

[wise igloo]

hmm

couldn't number theory be viewed as kind of an extension of arithmetic

?

[fossil shuttle]

Yes.

[wise igloo]

(of integers)

nice

[fossil shuttle]

I would say that the integers, as equipped with 0, 1, + , x, and these operations form an algebraic structure.

[wise igloo]

because they're a group right

[fossil shuttle]

Yes. The integers are a group under the binary operation of addition. the 0 is the additive identity.

This is an Abelian (commutative) group.

[wise igloo]

I've been learning about abstract algebra for the past few days and im learning all about groups, is group theory a subset of abstract algebra?

[fossil shuttle]

Yes, definitely. It's one of the main parts of the subject taught in a first class in algebra.

[wise igloo]

why does the word abelian exist if we can just say commutative

sounds redundant

[wise igloo]

oh okay cool, now im starting to learn about rings and fields

lol

[fossil shuttle]

They are named after Abel. I don't remember what exactly he used them for. Abel contributed a lot to group theory and field theory, but I'm not sure why commutative groups in particular would be named after him.

The name "Abelian" is so well established in the mathematical community that it would be going against the grain and idiosyncratic to speak of "commutative groups" in most contexts.

[wise igloo]

oh alright

so saying "commutative groups" is essentially mathematical blasphemy

[fossil shuttle]

Haha. It will be annoying.

[wise igloo]

time to piss everyone off then ;)

btw once I learn some more algebra, would I be well-equipped to begin algebraic topology?

I tried studying it but after watching two lectures I am bored out of my mind

ugh

[lavish nexus]

you'll learn point set topology in analysis

[wise igloo]

what type of analysis

I've learned some real analysis

[lavish nexus]

so you know basic topology

[wise igloo]

I do?

[lavish nexus]

what's an open set

[wise igloo]

uh

I forgot lol

[lavish nexus]

lul

[wise igloo]

oh wait is it just a shape in space that doesn't contain the boundaries

[lavish nexus]

aight just go through rudin

pretty much

[wise igloo]

oh ok

alright I looked up the pdf and there's some that I know but a lot that I don't

this should help me

tyvm

[lavish nexus]

suppose G is not abelian

is the free group generated by all finitely generated subgroups of G G?

[terse crystal]

Of course not for example let G=S_3 then the group you are asking about isn’t even finite

[lavish nexus]

ah right I forgot the textbook example

ty

[fossil shuttle]

study analysis, it's more interesting

[wise igloo]

alright thank you

[fallow plume]

I ended up trying this but ended up proving something stronger than the original statement, which makes me think I did something wrong in my proof:

Let $n\in\mathbb{Z}$, and $\phi_n ~ : G\to G$ be defined by $g\mapsto g^n$. Prove that if $G$ is abelian then $\phi_n$ is a homomorphism of groups.

[cloud walrus]

Thomas

[fallow plume]

I noticed that since $\operatorname{Im}(\phi) \leq \mid G \mid$ that it suffices to show that $\operatorname{Im}(\phi)$ is a subgroup as that implies it's a homomorphism with that alone.

[cloud walrus]

Thomas

[fallow plume]

I did assume that the operators are the same for each group. Lemme know if this is a bad assumption.

[fallow plume]

So now we use the one step subgroup test and of course trivially $\phi_n(1) = 1$

[cloud walrus]

Thomas

[fallow plume]

So now we hope to show ab^-1 is in the set for every a and b in the new group

So we show that $\phi_n(x) = a$ and $\phi_n(y)=b$ just by having a preimage.
Thus $x^n = a$ and $y^n=b$ by definition of phi.

Now we try $\phi_n(xy^{-1})$. Of course this maps to $x^n y^{-n}$ and that's equal to $ab^{-1}$ by simplification.
So it is in the group.

[cloud walrus]

Thomas

[fallow plume]

QED?

I'm confused because I didn't use the fact that G is abelian.

[marsh fossil]

so i have a T/F question: if n = a + b + c does there always exist a permutation in Sn where the order of that permutation is abc? and i think it's false because it's only order abc if the gcd of abc is 1 bc of product of disjoint cycles but i'm having a hard time finding a counter example bc if i find one permutation that i can write as abc that then turns into abc/gcd(a,b,c) does it prove that i can't find one with order abc?

[rustic crown]

This isn't true... also use the word "homomorphism"... "homeomorphisms" are a completely different thing

[fallow plume]

Thanks, and lemme check my reasoning there

[rustic crown]

if that were true, then any surjective function would be a group homomorphism

another problem is that (xy^-1)^n = xy'xy'xy'....xy' does this equal x^n y^-n? you see some sort of Abelian-ness is used here?

[fallow plume]

Oh wow yea that was literally me assuming it's abelian bahaha thanks for that catch.

[rustic crown]

to verify it's a group homomorphism, all you have to do is check phi(xy) = phi(x) * phi(y)... seems like you're already doing more work which isn't required

[fallow plume]

Okay yea I see it. THat's embarrassing

And now the whole thing about it being a subgroup not working makes sense now as that doesn't imply it preserves group structure.

[rustic crown]

didn't fully get you... if phi is a homomorphism, then it is true that image of phi is a subgroup... but not the other way round

[fallow plume]

Yea the last statement was me saying "Ah yes, I see where I went wrong now and I agree"

[rustic crown]

oh okie

[rustic crown]

the order would be lcm(a, b, c) if you take disjoint cycles. so if you want to write that in terms of gcd, it will be a little longer expression. But if you're just looking for an example, take a look at S6, 6 = 2+2+2, but can you find an element of order 8?

[final pasture]

Is it obvious you can't ?

Oh hum ig it is

factor into disjoint cycles

[rustic crown]

yep

[final pasture]

and then cases

yeah ok

[rustic crown]

cycle length would have to be powers of 2, so that reduces that cases by a lot

[marsh fossil]

why do cycle lengths have to be powers of 2?

can't i have a cycle of length 3?

[rustic crown]

but then raising to the power of 8, won't kill that 3-cycle

[final pasture]

Remember the cycles are disjoint

so you can think cycles per cycle

they can't interact with each other

said another way, the order of the product of the disjoint cycles will be the lcm of the order of each cycles

[rustic crown]

largest is 4
4+2
4+1+1
largest is 2
2+2+2
2+2+1+1
2+1+1+1+1
largest is 1
1+1+1+1+1+1

well you don't really have to write it down, but ig that might make things a little more concrete

because if the largest cycle has length 4, then raising to the power of 4 will already kill everything

[marsh fossil]

so i get that the lcm of the orders of the disjoint cycles is the order of the permutation, and see why if i have a product of disjoint cycles of length a, b, c and they're not coprime, it's just lcm(a, b, c)

[final pasture]

and it can't have anything bigger than 4, because well, 4 is the biggest power of 2 smallest than 6

even if they're coprime, it's still lcm(a, b, c)

[marsh fossil]

but what i don't get is if i have that, how does that show i can't possibly have one

[final pasture]

it's just that if they're coprime, lcm(a, b, c) = abc

[marsh fossil]

which is the condition

[rustic crown]

pairwise

[final pasture]

do they need to be pairwise coprime for this to be true ?

Oh wait yeah

thanks

[rustic crown]

yep, 2, 2, 3 are coprime

[final pasture]

yeye

Also I don't get your question @marsh fossil

could you rephrase your last sentences ?

[marsh fossil]

like this question is false because its only true if they're all coprime, because then i can write them as disjoint cycles of lengths a, b, c where lcm(a, b, c) =abc but if i take an example, and show it does not have order abc for some a, b, c i don't get how that shows that there can't be any permutation of that order, i guess my question is there some m, n, k such that (a-m)(b-n)(c-k) = abc? and i guess not just because how algebra would work so is that a sufficient proof? showing that say i have a product of cycles of these lengths, then the permutation has order lower than abc and i can't have another because changing the cycles orders to still sum to a + b + c changes the value of the product so that it'll never equal abc

[rustic crown]

okie say we're working with S6 and a = b = c = 2

we need to show that there is no permutation in S6 with order 8

so lets assume there was such a permutation

and say if you write it as a product of disjoint cycles, then the lengths of the cycles are c1 c2 ... ck

so we know that order of this permutation is lcm(c1, c2, ..., ck) and this should equal 8.

from this what can we say about all these ci?

[marsh fossil]

they're all multiples of 2

[rustic crown]

first that ci divides lcm(c1, ..., ck) = 8

[rustic crown]

well some ci could be 1

[rustic crown]

and next as these are disjoint cycles we must have ci <= c1 + ... ck = 6 as this was a permutation of S6

so each ci is <=6 and divides 8

from this we can conclude ci divides 4

but if all these ci divide 4, this means that 4 is a common multiple of c1, ..., ck

so 8 = lcm(c1, ..., ck) <= 4

this can't true

[marsh fossil]

okay that actually makes a lot of sense, and thanks that clarifies a lot, but my teacher's problems specifically say find a counter example, idk if contradiction is valid

[rustic crown]

umm... we found a counter example right?

6=2+2+2

abc = 8, but we proved that there are no elements of order 8 by means of contradiction

[marsh fossil]

okay, so is it okay to use contradiction for a counter example? like i guess the "example" is s6 is that the idea?

[rustic crown]

yea so counterexample is S6 with n = 6 and a = b= c= 2

but how do you know this is a counterexample?

that requires some sort of reasoning

we just happened to use contradiction to prove that it is indeed an counterexample

[marsh fossil]

okay, so the key is understanding that it says there exists an element, typically with counter examples i just produce one that follows the hypothesis but fails the conclusion but bc this one says show that for any Sn there exists an element in it of order abc thats the cue to actually go forth and prove it. alright so we're basically reversing quantifiers, there exists an Sn where every permutation does not have order abc

[rustic crown]

looks right

[marsh fossil]

okay great, thanks a lot for all your help!

[rustic crown]

[fickle brook]

alright so i suppose this could go here but maybe it's better in #combinatorial-structures or whatever

there was a problem in yesterday's class that i was almost able to solve but not quite

the problem: let A and B be nonempty subsets of Z_p. prove that |A+B| ≥ min(p, |A|+|B|-1)

the instructions were to use this result

i wrote $f(x_1, x_2) := \prod_{s \in A+B} (x_1+x_2 - s)$, noticed it vanished on $A \times B$ by construction, and thus the coefficient on $x_1^{t_1} x_2^{t_2}$ had to be zero, where $t_1 = |A|-1$ and $t_2 = |B|-1$

then i wrote $f(x_1, x_2) = \sum_{j=0}^{|A+B|} c_j (x_1+x_2)^j$

so this coefficient is in fact $\frac{(t_1+t_2)!}{t_1!t2!} c{t_1+t_2}$

and something went wrong when i tried to examine what happens if it's equal to zero
i've computed the highest and second highest coeffs of f (the c_j) as, respectively, 1 and sum[s in A+B] (-s)

this is where i got stuck

[cloud walrus]

Ann

[fickle brook]

ping me if replying

[wooden ember]

What does ct_1 + t_2 mean?

Oh nvm I see

[fickle brook]

underscore nonsense

$c_{t_1+t_2}$

[cloud walrus]

Ann

[fickle brook]

this is what it was meant to be

[rustic crown]

I think I got it

[cloud walrus]

det

[rustic crown]

@fickle brook

[fickle brook]

ah

of course

ty

[rustic crown]

[sinful mirage]

I figured out! I just had to use the homomorphism property, i.e. that G_x is a subgroup of G.

(which I should prove )

[hidden haven]

nice

In general gS = hS iff there's some s in S such that gs = h, where S is a subgroup of G

This is an important property to remember

[sinful mirage]

just to be more explicit,when I write it in my notes: here x~y if f(x)=f(y)

right?

[hidden haven]

iff yeah

[sinful mirage]

yep

[sinful mirage]

a short sketch in one direction. Suppose $g_1,g_2$ are in the same equiv class. That is $g_1x=g_2x \implies g_2^{-1} (g_1 x)=x \implies (g_2^{-1} g_1)x=x \implies g_2^{-1}g_1 \in G_x$

is this ok?

[cloud walrus]

ProphetX

ProphetX

[hidden haven]

Nice yes

You also have to show the converse, and that is just going back along the same implications

Everything is iff

[sinful mirage]

and the other direcion I can just reverse the arrows by assuming the last setp

yes

exactly

[hidden haven]

[sinful mirage]

also,a short note: g_1x=g_2x here the operation is group action, while g_1 G_x=g_2 G_x operation is group operation, correct?

different operations I mean

[sinful mirage]

since G_x are elements of G

[hidden haven]

Yes

[sinful mirage]

[hidden haven]

This in fact proves that the G/G_x and Gx are not just in bijection, but isomorphic as G-sets

Where G/G_x is viewed as a G set by (g, hG_x) ↦ ghG_x

So as G sets, all orbits look like quotients of G (notice that these are quotients by subgroups, not by normal subgroups, so these quotients are merely left G-sets, not groups)

So the quotients of G are called the canonical orbits of G

Or something similar I might have the name wrong

[sinful mirage]

this was exactly the prof's next remark

[hidden haven]

oh nice lol

[sinful mirage]

now the only part which I don't think I 'need to' understand is why is this identification natural

my prof said naturalness plays an important role later on

[hidden haven]

😵‍💫

Natural identification between what functors

Is he saying that there's an equivalence of categories between category of transitive G sets and canonical G orbits

[sinful mirage]

also:the identification of left-cosest with orbits is this bit, correct?

LHS=orbits,RHS=cosets

[sinful mirage]

not sure,it was just a remark that this identificaiton is natural

[hidden haven]

Not with orbits, but with elements of the orbit

[sinful mirage]

but he never defined natural

[sinful mirage]

yes

[hidden haven]

Interesting lol

Naturality is cat theory stuff

[sinful mirage]

waitt there's some neat stuff

using this identificaion and that G_x is a subgroup + Lagrange theorem,I immediately get the orbit-stabilizer theorem? :PogChamp:

[hidden haven]

Yes 😌

[sinful mirage]

a free action is defined by the fact that all point stabilizers are trivial

how does this imply the fact that the orbit maps are injective?cause their kernel is just the identity?

i.e. gx=x iff g=e => orbi tmaps are injective

[hidden haven]

What's an orbit map

[sinful mirage]

Le $x \in X$. he orbi tmap is $G \times {x}\to X, (g,x) \mapsto gx$. The image of the orbit map is the orbit of $x$.

[cloud walrus]

ProphetX

[sinful mirage]

it's basically the action,where x is fixed

[cloud walrus]

ProphetX

[hidden haven]

ah I see

[sinful mirage]

free is defined as all sabilizers are trivial. this somehow should imply that orbit maps are injective

[hidden haven]

It's not a group homomorphism so you can talk about kernels

[sinful mirage]

ahh yikes

[hidden haven]

But notice that that map is exactly the map G → Gx in the previous thing (× singleton doesn't matter)

So it induces a bijection G/G_x to Gx (really from (G × {x})/(G_x × {x})

Here you have to use the commutativity of the diagram

Since the diagram commutes, the horizontal map = quotient map then bijection

But because G_x is trivial the quotient map is a bijection too

So the orbit map because a composition of bijections hence a bijection (onto Gx)

[sinful mirage]

why?

how does G_x triviality imply quotient is bijection?

ah cause G/G_x={g G_x|g in G}={g e|g in G}=G?

[hidden haven]

Yep

The last one is a bijection not an equality

Because it should be g{e} not just ge

[sinful mirage]

can one say that transitive action means:each point in X can be reached via an action of g on it?

in words

[hidden haven]

from any other point

[bronze jay]

shouldn't an element in Zm x Zn be (x^a, y^a)

how do we know that x^ay^a is well defined?

[hot lake]

they identify Zm with Zm x {1} and Zn with {1} x Zn

[bronze jay]

interesting, so x^a = (x'^a, 1^a)= (x'^a, 1) for some element x' in Zm?

[hot lake]

well the x' is still a generator of Zm

but now if you identify Zm with Zm x {1}

x becomes a generator of Zm x {1}

it's uuh weird to look too hard into it lol

I guess they could have said "let Zm = <x'>, Zn, = <y'> ; let x = (x',1) and y = (1,y')"

[bronze jay]

if A is isomorphic with Zm x Zn and B is isomorphic with Zk x Zs, can we necessary conclude that A x B is isomorphic to Zm x Zn x Zk x Zs?

[hot lake]

yes

[bronze jay]

hmm, why is that?

[hot lake]

???

well uh if A isomorphic to C and B is isomorphic to D

then A x B is isomorphic to C x D

[bronze jay]

ah

i see

thanks

[hot lake]

you take isomorphisms f : A -> C and g : B -> D

[bronze jay]

does this stem from CRT?

[hot lake]

you define (x,y) -> (f(x),g(y)) and check that it is an isomoprhism

[bronze jay]

the reason we can say that Z6 is isomorphic to Z2 x Z3 is because we know that (2,3)=1

[hot lake]

Z6 = Z2 x Z3 and Z15 = Z3 x Z5 come from the CRT

[bronze jay]

ah thanks

When it is asked "Determine all of the qoutients of a group G", what does that mean?

Say G is cyclical with order 10

[hot lake]

determine what groups are isomorphic to a quotient of G ?

there you would get Z10,Z5,Z2,Z1

also it's not clear if they want you to count how many different ways you can get a particular quotient

[bronze jay]

wouldn't that be to find what the qoutients of G are up to isomorphism?

[hot lake]

when doing group theory you only care about groups up to isomorphism

you do not care at all if the elements of your group are numbers or stuff you ate for breakfast

[bronze jay]

But how did you determine this?

[hot lake]

I went through all the subgroups H of Z10 then determined what Z10/H was isomorphic to

[bronze jay]

oh, and we know from lagrange that a subgroup must have an order that goes up in the order of Z10

so the possible subgroups are of order 1, 2 5 and 10?

[hot lake]

yes

and there are in fact exactly one subgroup of each order

in this case

[bronze jay]

Ah, and if we mod G out with a subgroup of order 1, we get a qoutient group of order 10, if we mod it out with order 2, we get a qoutient group of order 5, etc. etc.

[sinful mirage]

I am trying to prove direcly that lef mult. is simply transitive => right mult. is simply transitive,but I don't get anywhere

I can prove separately that both left and right mult. are simply transitive. thus if left is simply transitive=>right is simply transitive

i.e. let A,B be two propositions. if both A,B are true,then A=>B is true.

however,I want to do a proof without assuming anything about righ tmult.

so a direct proof. Let left mult. be simply transitive=> right mult. is simply transitive

does anyone have hints?

[bronze jay]

Is it true that Z(G) <= C_G(S) <= N_G(S) <= G ?

[small marsh]

Hi, this question's official answer is "false" but I don't see why, could anyone explain pls? Thanks

Isn't N an inverse element to every element in P(N)?

Regarding intersection

[rustic crown]

umm... the inverse of an element is unique...

also the easiest way to go about this would be to use that in any group you have a cancellation law. i.e. if ab = ac in a group G, then b = c

but notice that for any two subsets B and C of N, taking A = empty set gives,
A intersect B = A intersect C = A

but clearly you can start with different subsets B and C

[small marsh]

Thanks!

[rustic crown]

the advantage of this is that, you don't need to wonder about what inverses and identities are

[small marsh]

But why N doesn't work in this example?

[rustic crown]

"work" as in?

are you saying that N is the identity for the operation?

[small marsh]

Works as an inverse to every element
You said that the inverse of an element is unique , I think N could be an inverse to every element though

Because in P(N), every set intersection N is N, isn't it?

[rustic crown]

no

A intersect N = A

[small marsh]

Ah ok

[small marsh]

Isn't N included in every set in P(N)?

[rustic crown]

P(N) is (one of) the standard notation for the power set... i.e. set of all subsets

an element of P(N) is a subset of N

[small marsh]

Alright just realized I got a big confusion lul

[small marsh]

I knew this before but idk why was so confused 🤷‍♂️

Thanks for the reminder

[rustic crown]

also if inverse of two elements are equal, then the elements themselves are equal

so all elements can't have the same inverse

[small marsh]

Yep got it thanks

I also had another question

Why is the product of 2 groups cyclic iff the order of these groups are relatively prime?

[rustic crown]

you mean product of two cyclic groups?

[small marsh]

Yes

[rustic crown]

if G = <a> and H = <b> are cyclic with orders relatively prime, then show that G x H is cyclic by finding the order of the element (a, b)

[small marsh]

Got it thanks!

[rustic crown]

conversely, if they are not relatively prime, say have orders m and n, then show that any element of G x H satisfies g^(lcm(m, n)) = identity in GxH

so order of no element can be m*n as lcm(m, n) < m * n if they are not relatively prime

[small marsh]

I understand now

Thanks

[rustic crown]

[simple valley]

oh I have a big brane proof

\begin{tikzcd}
\bZ \arrow[rd] \arrow[rdd] \arrow[rrd] & & \
& R \arrow[d] \arrow[r] & {\Hom_{Ab}(N, N)} \arrow[d, "\circ f"] \
& {\Hom_{Ab}(M, M)} \arrow[r, "f \circ"'] & {\Hom_{Ab}(M, N)} \
& M \arrow[r, "f"] & N
\end{tikzcd}

[cloud walrus]

mniip

[simple valley]

Z -> R epic in Ring implies U : R-mod -> Ab full

wait... not all of those are rings tho

...back to the drawing board

[sinful mirage]

.

is it even provable directly? Haven't seen proofs like that in books

[hidden haven]

Aren't left and right multiplication always simply transitive

If g ≠ 1 then gh ≠ h so it is free

If x and y are in G then (yx^-1)x = y so it's transitive

I'm assuming that by left multiplication you mean the group acting on itself by multiplying on the left

[dull root]

I'm having trouble understanding this: If $C = \langle c \rangle$ is a cyclic group of infinite order, then $\lvert Aut(C) \rvert = 2$. I don't see how this is the case. If I have an automorphism $\varphi: C \to C$ then I understand $\varphi(c)$ completely determines the automorphism, but $\varphi(c)$ can be anything in $C$, except for identity so that gives infinitely many automorphisms. What is the issue with this?

[cloud walrus]

Mr.Hahn-Banach

[hidden haven]

C is isomorphic to Z. Do you see the issue now?

[hot lake]

isomorphisms have to be surjective

[hidden haven]

Unless you're following herstein 🤕

[viscid pewter]

yeah, just consider Z and find the only two automorphisms

[dull root]

So $C \cong Z$ and the only two automorphism should be the indentity and the inverse map $a \to -a$

[cloud walrus]

Mr.Hahn-Banach

[viscid pewter]

yes

[hidden haven]

Yeah, though if you're writing it down then involving Z just makes the argument more complicated, and it's just for intuition for what's happening

[dull root]

Yea i see now, thanks

[hidden haven]

@sinful mirage

[sinful mirage]

tthey are

[hot lake]

in groups ?

[sinful mirage]

i have other wishes tho

I want to ignore right being simply transitive

I proved left is

[hidden haven]

Group actions

[sinful mirage]

and I want to show,that left is => right is

[hidden haven]

Ah

Use duality

[sinful mirage]

without proving right is true

[hot lake]

so you have a set with two group actions on it ?

[sinful mirage]

yes

[hidden haven]

If a statement is true about G, then the dual statement is true about G^op

[hot lake]

and they commute ?

I'm not sure it's even true

[hidden haven]

A group is acting on itself by left and right mults

[hot lake]

ah on itself

[hidden haven]

Want to prove that both actions are simply transitive

[sinful mirage]

@hidden haven the issue is accepting left is transitive

[hot lake]

that uuh

[sinful mirage]

I can't prove right is transitive

[hot lake]

is obvious from the definition ?

[hidden haven]

Depends on the definition

But they've checked left mult is, somehow

And want to get same for right mult without extra work

Duality works here

[hot lake]

extra work

I guess duality works yeah

[hidden haven]

The dual statement of "action is simply transitive" is itself, while the dual group G^op makes left mult into right mult action

In simpler words I guess we could say

You proved this statement "left mult is simply transitive" for all groups, so in particular for G^op

G^op acting on itself by left mult is same as G acting on itself by right mult

[sinful mirage]

G^op=?

[hidden haven]

G but with reversed multiplication, opposite group

Or wait

You already proved that correspondence between left and right G actions

You can show that a left action is ST iff it's corresponding right action is ST

That's essentially what I'm saying

Wait it's corresponding right action is not right mult is it

It's just isomorphic as G sets

[rigid cave]

this really is a dumb question but I don't see how A tensor B is isomorphic to B tensor A. I know that the isomorphism should obviously be f(a tensor b) = b tensor a but how is f(a tensor b + c tensor d) = f(a tensor b) + f(c tensor d) when a isn't c or when b isn't d?

[sinful mirage]

ahh

makes sense

I use the proof from yday

[stark sigil]

You’re free to extend f linearly so that holds

[hidden haven]

How is tensor product defined for you? Universal property, or explicit construction?

[stark sigil]

Or rather, if you didn’t specify f is linear then you’ve incompletely defined f

[rigid cave]

just explicit construction I guess, it's free abelian on basis AxB and blablabla

[hidden haven]

Well definedness is an issue though

[rigid cave]

ye I get that you extend stuff linearly but I don't know how that homomorphism thing holds

the definition in Hatcher

[hidden haven]

Oh Hatcher does this? oof

[rigid cave]

yee lmao he mentions this quickly

on page 215

[stark sigil]

Once you’ve extended f linearly doesn’t your equation become a tautology?

[hidden haven]

You have to show that there is a well defined linear extension

[stark sigil]

A tensor B is a vector space

Give f in a basis

[hidden haven]

Oh are taking vector spaces

[stark sigil]

I think so ya

[rigid cave]

wait what?

what's happening

[hidden haven]

{a ⊗ b} is still not necessarily a basis though

[stark sigil]

Ok maybe they’re not vector spaces, they’re modules?

Or abelian groups?

[rigid cave]

ye abelian groups in my case

[hidden haven]

I see

[rigid cave]

free abelian on this

[stark sigil]

Ok so yeah there’s a massive quotient

[rigid cave]

sorry I didn't mention that

[stark sigil]

By bilinear relations

However the simple tensors DO generate the tensor product

[rigid cave]

so the relations are (a+a') tensor b = a tensor b + a' tensor b and the same thing on when you tensor on the other side

[hidden haven]

So tensor product is free group on product modulo stuff
So first define f on the free group on product set rather than on the tensor directly

[stark sigil]

So saying f is linear and specifying it on simple tensors

[hidden haven]

Then show that f maps the things you are quotienting by to 0

[stark sigil]

Automatically specifies f on the entire space, provided f is well defined

[hidden haven]

Thus f factors through the quotient uniquely

And linearity will be there from this being a factoring lol

[rigid cave]

oh lmao

[rigid cave]

I thought that you could just define f directly from A tensor B to B tensor A

[hidden haven]

Yeah that's annoying in general

Because the well definedness of maps out of a tensor is hard to check explicitly

Because the quotient is absolute non sense, we don't know what the subgroup we are quotienting by looks like

So tensor products without universal properties are

But at least you can use univ prop of quotients here

[stark sigil]

A map from A/B to C is a map from A to C such that B maps to 0

[hidden haven]

Just send the sticker bro

Imagine typing it out, that's so 2020

[rigid cave]

but yeah okay I think I get it now. So in general when you want to show some isomorphism of tensor products, you just do this?

like define the map on the underlying set etc.?

[hidden haven]

In general you use univ prop of tensors instead of univ prop of free then of quotients

But in the absence of that you use these 2 yes

Univ prop of tensors bundles the useful stuff together

[rigid cave]

bundles 😵

okay thank you so much!

[hidden haven]

🥴

[dull root]

So I know that if $C$ is an infinite cyclic group, then there are two possible homomorphisms $\psi: C \to Aut(C)$, namely $\psi(c) = id \in Aut(C)$ or $\psi(c) = \varphi \in Aut(C)$ where $\varphi(c) = c^{-1}$. How can I tell if the semidirect product $C \rtimes C$ w.r.t those two homomorphisms yield two non-isomorphic groups? I was trying to contradict the existence of a isomorphism between those two semidirect products, but without success. There should be an easier way to see the two are non-isom.

[cloud walrus]

Mr.Hahn-Banach

[sinful mirage]

is this map smh famous?

we used it yday too

for set iso

[dull root]

I doubt the two semidirect products are actually isomorphic, since $C \rtimes C$ w.r.t the hom $\psi = id \in Aut(C)$ gives us a group operation that works coordinatewise, namely $(a_1,b_1) \cdot (a_2,b_2) = (a_1a_2 , b_1b_2)$ while the operation w.r.t other other automorphism behaves differently

[cloud walrus]

Mr.Hahn-Banach

[hidden haven]

It is the first isomorphism theorem/universal property of quotients. It pops up whenever you talk about quotients, which is very very often

[hidden haven]

One is commutative, the other isn't

[dull root]

.... yep, that does it....

[hidden haven]

And I mean quotients of anything, not just of groups, even though the diagram above is the specific instance of the univ prop of quotient groups @sinful mirage

I guess that diagram works for any object that has kernels really, not just groups

Kernels and all fibers of a map are translates of the kernel in some way

[stark sigil]

The quotient is the universal “C along with a map A -> C such that everything in B goes to 0”

[hidden haven]

Yeah when you have kernels and cosets, but you can do it for an equivalence relation on a set too

The quotient is the map which is initial among maps that respect the equivalence classes of a given equivalence relation, idk if there's a more general way of saying this outside of concrete cats

Your definition is the one for abelian cats though with some change 😌

[cloud walrus]

Mr.Hahn-Banach

[marsh fossil]

I’m having trouble showing the only central element of Sn is the identity.. I know disjoint cycles commute but I’m having trouble using that

Oh wait do I show that if an element has order higher than 1 it cannot be central?

[next obsidian]

Just write it as a product of cycles

And come up with something that doesn’t commute with it

This shouldn’t be too hard

Like if I gave you (1534)(26) in S_6 what’s something that won’t commute with it?

Maybe try a few examples and try to note a pattern

[thorn delta]

Take a nonidentity permutation and try to construct another that does not commute with it

Oops discord being strange

[next obsidian]

Too late mr roblox man

Also worth noting that this is false for S_2, lol

But I think only for S_1 and S_2

[viscid pewter]

no, it's true for S_1

:P

[next obsidian]

Oh

True

Chmowned

[viscid pewter]

f's in chat

[small bison]

Try looking at the conjugation action

[next obsidian]

This guy fux^^

That’s a good solution

[marsh fossil]

okay, thanks all for the suggestions

[dull root]

I'm trying to find out how many non-isomorphic groups are of the form $Z_7 \rtimes Z_3$. There are 6 possible groups in total of that form noting $\lvert Aut(Z_7) \rvert = 6$, given by the following homomorphism $\varphi_n: Z_3 \to U_7$ (the multiplicative group) that maps $1 \to n \in U_7$. I was able to show the two semidirect products are isomorphic w.r.t to the homomorphism $\varphi_2$ and $\varphi_4$ and $\varphi_3$ and $\varphi_5$ since they are multiplicative inverses in $U_7$. That narrowed it down to at most four possible non-isomorphic groups, but the idea does not allow me to reduce it anymore than that.

[cloud walrus]

Mr.Hahn-Banach

[chilly ocean]

Hi, I am working our this proof and I am running into trouble. I have a lot of chicken scratch but can’t seem to budge.

Here’s is what I have but I know there must be another direction. Is there a way to relate o(gh) and o(g)o(h)?

<@&286206848099549185>

[lethal cipher]

Wait, it's abelian?

[chilly ocean]

I don’t think the group G is necessarily Abelian

[lethal cipher]

"order in G that commute"

[chilly ocean]

I think that these particular elements g and h commute

It doesn’t mean every element in G commute

[lethal cipher]

Yea, that's all we care about in this case

Note, the claim is true regardless of commutativity

[chilly ocean]

What claim?

[lethal cipher]

The order of ab divides the least common multiple of ord(a) and ord(b)

[chilly ocean]

Ah

[lethal cipher]

Commutativity is probably just to make the proof slightly easier to deal with.
Important thing to note here (gh)^m=g^mh^m

[proud bear]

just use the fact that if for some element x in a group, if x^n=1, then o(x) divides n

[chilly ocean]

Will we utilize that $lcm(a,b)=\frac{ab}{gcd(a,b)}$?

[cloud walrus]

Mega Euler

[proud bear]

if you prove that $(gh)^{\lcm(o(g),o(h))}=1$, then you will have proven that $o(gh)\divides\lcm(o(g),o(h))$

[chilly ocean]

Ohhhh wow

That’s great

[lethal cipher]

What Stain said. And this is a way to do it (for these commutative elements anyways)

[chilly ocean]

So what if the elements g and h doesn’t commute. What is the different in the proof?

Difference*

[lethal cipher]

I am not entirely sure off the bat. But I do know this equality is generally invalid and cannot be used in that proof.

[proud bear]

i don't think it's true if they don't commute

[chilly ocean]

Right. $(gh)^m = \underset{\text{$m$ times}}{(gh)\cdots(gh)}$

[cloud walrus]

Mega Euler

[lethal cipher]

Perhaps the group needing to be finite would be a necessary condition

[chilly ocean]

I see

Thank you

[hidden haven]

All the phi_n that you describe are not homomorphisms. The order of an element cannot become larger in the image

[simple valley]

so wait

going back to the problem of whether U : R-mod -> Ab is full

we showed that if it's full then the ring inclusion Z->R is epi

is the opposite actually true?

oh

if we M, N in R-mod and we have f : M -> N an Ab-morphism but not an R-mod morphism, then we have some m, r such that f(rm) =/= rf(m)

consider the left ideal {s | f(sm) = sf(m)} in R

it clearly includes all of Z but not all of R which is impossible

because we have a map R -> R/I that's zero on Z but not everywhere

wait is it an ideal

it is a subring

not even that. Only if we quantify "forall m"

[rustic crown]

does this work? Say you had such a ring R such that R-Mod --> Z-Mod is full.
Then consider two maps f, g from R --> S. We need to show these are equal in order to say Z --> R is epi.
using f, g we can give R-Mod structure on S. (forget about the R-Alg thing... for that we require other stuff like R maps to center of S and all that which we don't need.) To be precise I'll call them S_f and S_g. Now consider the "identity" map of abelian groups.
S_f --> S_g
By full-ness this should also be an R-Mod map. so r * 1 in S_f equals f(r) * 1 = f(r) so this should be sent to r * 1 in S_g which equals g(r).
but since this was just the identity map on the abelian group S, we can conclude f(r) = g(r) showing f = g. And so Z --> R is epi

oh oops you wanted the other way round

[weak oriole]

I don't think so

[stark sigil]

I love the way it says "The next theorem shows that we can say nothing at all"

[final pasture]

That's a funny theorem

[marsh fossil]

i've got a very simple question, is any element of the group of integers Z infinite order?

i think it would be so because it would imply the group is finite if it wasnt

[final pasture]

Except 0, yes

[final pasture]

no

There are infinite groups with only finite order elements

[marsh fossil]

okay, for cyclic groups then

i guess is the case im missing

[final pasture]

Yes, cyclic groups with finite order elements are the finite cyclic groups

[marsh fossil]

so if a cyclic group has a finite order element then it has finite order

[gritty sparrow]

Take the tensor product $End_{Z}(M) \otimes End_{Z}(N)$ over Z, notice that the maps $Z->R->End_{Z}(M)-> End_{Z}(M) \otimes End_{Z}(N)$ and $Z->R->End_{Z}(N)-> End_{Z}(M) \otimes End_{Z}(N)$ commute, so as the map from Z to R is an epi, we can omit the Z->R map in both cases and it will still commute. now define an abelian group map $End_{Z}(M) \otimes End_{Z}(N)->Hom_{Z}(M,N)$ by taking (a,b) to bfa (and using bilinearity to get a map of tensors). now just chase an element r in R both ways $to Hom_{Z}(M,N)$ and that will exactly be bilinearity.

[cloud walrus]

saketh

[final pasture]

Yeah this is correct

The simplest way of seeing it is:
pick a generator, denote it by a, and say g is your finite order element.
There is some n such that a^n = g.
If g has finite order k, then g^k = (a^n)^k = (a^{nk}) = 1, meaning that a has finite order, so the group must be finite since the order of a generator is finite

[marsh fossil]

okay gotcha, thanks for the clarification

[rustic crown]

maybe say finite order element other than the identity

[final pasture]

oh and yes this ^

[fallow plume]

Am I reading this correctly?

"assume phi hom[omorhism] show that for all x in G, |x| = |phi(x)|"
Totally makes sense and logical ^^^

And then it says
"Show that |phi(x)| divides |x|"???

My prof says "in general this a little harder to prove" but that makes no sense. Literally they're equal so of course it's divisible?

My best guess is she's talking about a different proof method but Idfk

[vast quiver]

the picture is very blurry at least on my end, so I can’t tell what she wrote.

It’s probably not |x| = |phi(x)| because it’s not always true
example: || Z/4Z -> Z/2Z by “moduloing again” (ie quotienting by 2Z/4Z). In our starting group the order of 1 is 4, but in the image group, the order of 1 is 2 ||

My guess is that the first part is just |x| >= |phi(x)|

or maybe the first part is supposed to assume that phi is injective

[wraith obsidian]

Is there a systematic way to see what the (non-proncipal) prime ideals in a ring extension of Z (say Z adjoined w/ finitely many algebraic elements) look like?

I'm trying to understand the whole dedekind and ramification business better but it's hard to come up with examples if I don't know what the spectrum looks like in its entirety

[rustic crown]

weren't ideals in a dedekind domain always generated by 2 elements?

[wraith obsidian]

I never heard about that, so can neither confirm nor deny that :P

[wraith obsidian]

Nice, then I should probably limit myself to the integrally closed case 🙃

[wraith obsidian]

What's this from if I may ask?

[rustic crown]

Oh, this is from Chapter 3 of Marcus Number Fields

[wraith obsidian]

tyvm

[bronze jay]

what are the 3 subgroups of index 2?

[viscid pewter]

generated by r, r^2 and s, and rs?

seems plausible

[bronze jay]

you've named 4?

[viscid pewter]

no

'r^2 and s' is one

[rustic crown]

i think the last one is also <r^2, rs>

[viscid pewter]

i think (rs)^2 = r^2

so it's the same as <rs>

wait

[rustic crown]

srsr = 1?

[viscid pewter]

yes

ok so yeah

1. <r>
2. <r^2, s>
3. <r^2, rs>

[bronze jay]

how can we use the fact that the 3 subgroups are abelian to see that x \nin Z(D8) ....

[bronze jay]

from the example here..

[viscid pewter]

ok so any element is in one of those three subgroups

it definitely commutes with everything in its own subgroup, since they're all abelian

so if it doesn't commute with everything

then it can't commute with anything more than its own subgroup

since the centraliser is a subgroup?

and the largest non-trivial subgroup is of order 4

which is its own subgroup

if it commuted with anything more, the centraliser would have to be the whole group, by lagrange's, and it would be central

[bronze jay]

oh, so if the index are 2 for each of these subgroups where it is abelian, that means that the subgroup must have order 4

[viscid pewter]

trivially

[bronze jay]

the conjugacy classes, how are those calculated?

do we calculate for example, for an element s, gsg^{-1} for all g in D8?

[viscid pewter]

i think so

it's not as bad as it looks bc equivalence classes

so would only need to do around half the elements, not all

still not incredibly fast, but possibly slightly faster than brute-forcing it

[bronze jay]

But i get for s,
sss^{-1}=s
r^3sr^{-3}=r^6s=r^2s
(sr)s(sr)^-1=srsr^{-1}s=sr^2s=r^2

there are 3 elements

and the conjugacy classes all only contain 2 elements

[viscid pewter]

gut says last one is wrong

srsr^{-1}s=sr^2s
how

[bronze jay]

srsr^{-1}s=s(rsr^{-1})s=s(rrs)s=sr^2

okay i guess i did it wrong

[chilly ocean]

Any book recommendation that focuses on cringe? (Commutative ring endomorphisms).

[sinful mirage]

how would a mathematician compute the orbits of a group using the definition?

for example Lorentz group

why are the orbits classified by the bilinear form?

the orbit is defined as: O_{p}:={lambda p|lambda is a lorentz trafo},p is in R^{1,3}

[limpid edge]

someone ik brought up this cool question

is it okay to consider the vector spaces of all vector spaces over a field?

where this vector space would be over Z/2Z

where multiplying by 1 corresponds to including a vector space in a sum and multiplying by 0 means not including (or just getting the 0 vector space)

does this vector space lead to paradoxes?

[wooden ember]

Sounds like it should

[limpid edge]

similar to how "set of all sets" does

[wooden ember]

And anyways what does a sum of vector spaces mean

What is V + W as a vector space?

[limpid edge]

the vector space Z whose vector addition operation is just V + W

[wooden ember]

Yeah but V and W might be over different fields

[limpid edge]

no they're over the same field

I should've specified

[wooden ember]

Ah over a particular field

[limpid edge]

"vector space of all vector spaces over a particular field"

[wooden ember]

My bad I misunderstood

[limpid edge]

you won't run in to a russels paradox type thing if the specific field is something other than Z/2Z

[wooden ember]

Might be possible but I’m not sure

Would need the opinion of someone more qualified

[hidden haven]

The class of vector spaces over a specified field is not a set

But you can consider the vector spaces only upto some cardinality

Also that doesn't seem like a vector space operation
If you are defining the addition as V + W = V ⊕ W then adding W twice doesn't equal adding nothing so this can't be a Z/2Z vector space

Or I didn't get your addition operation

[lethal cipher]

I am having a bit of trouble figuring out how to deal with part 2

[chilly ocean]

a normal subgroup is one such that gKg^-1 = K for all g. what automorphism does this suggest you use?

[lethal cipher]

Oh, the conjugacy map.

so here's the question, how do I know that is indeed surjective?

[chilly ocean]

try proving it

[lethal cipher]

le sigh

[chilly ocean]

no, really

prove that conjugation is an automorphism if you haven't

[lethal cipher]

oh...yea, that is easy :p

okay, I have two more questions.

So I know part 1 is not true, but I suck at finding counter examples. If possible, I could use some help figuring out a good counter example here

I can do part 2 and 3 on my own

[sturdy marsh]

try matrix groups

[lethal cipher]

Actually, I think subgroups of integers works just fine.

2Z+3Z is not 5Z

[sturdy marsh]

it is true for abelian groups

if AB = BA, then AB is a subgroup

[lethal cipher]

Hmmm, well that put a damper on things

[rustic crown]

you can try finding the size of AB when A and B are finite subgroups

some theorem by some guy starting with L might help

[lethal cipher]

I know that guy!

[lethal cipher]

oh duh. I already know how to prove (iii) and normal just means the subgroup is in the center. Silly me.

[sturdy marsh]

I was being silly, there's an easier example

think free groups

[rustic crown]

||using this i think S3 works fine... <(12)> and <(13)> both have size 2. their product has size 4 which doesn't divide 6.||

[sturdy marsh]

||<s,t | s^2 = t^2 =1>, <s><t> is not a subgroup||

[lethal cipher]

doesnt the product have size 5?

[rustic crown]

the product is (1), (12), (13) and (12)(13)

[lethal cipher]

Oh yes. AB, not the combination of AB and BA. My b

[rustic crown]

it's easy enough to check by hand that this isn't a subgroup... but Lagrange

[lethal cipher]

For some reason, I was looking at all of the possibilities for AB and BA. Live and learn I guess

Also, yea. Lagrange is doing wonders here

Ty det

[rustic crown]

[sturdy marsh]

evee butt

[rustic crown]

[wooden ember]

The last sentence requires each omega_i be of equal size right

[lethal cipher]

Okay, last question. I am not gonna lie, I do not even know where to begin with this

My only thought is all elements in (Z_2)^n have order 2

So if we can find the U(k)'s that are isomorphic to (Z_2)^n (for any n), we solved the problem.

Hmmm, I got excited

[rustic crown]

U(16) = Z_2 x Z_4

[lethal cipher]

Yea, I got excited for a second.

[rustic crown]

you know CRT right?

[lethal cipher]

Depends, what does CRT mean?

[rustic crown]

for coprime m, n we have U(mn) = U(m) x U(n)

[lethal cipher]

Oh yes, I know that

[rustic crown]

with this... you can analyze one prime at a time

i love saying this lol

[lethal cipher]

I also know what U(p^n) is isomorphic to when p is odd or even

[rustic crown]

one prime at a time

[rustic crown]

Oh awesome

[lethal cipher]

Okay, so one thing I know is U(4) is such a group.

It is isomorphic to Z_2

[rustic crown]

yep

[lethal cipher]

Moreover, U(12) is such a group, since U(3) is also isomorphic to Z(2), you end up with the whole thing being isomorphic to (Z_2)^2

[rustic crown]

yee

what's left?

[lethal cipher]

U(8), as we know, is also that group.

However, there is an issue. If we looked at (Z_2)^3, then it is built by some combination of external products of U(3), U(4) or U(8), but U(4) is not relatively prime to U(8)

So... I think that's it

[rustic crown]

ah... we're still missing a few

[lethal cipher]

Well shoot.

[rustic crown]

maybe think it like this... what are the nice powers of 2? they are 1, 2, 4, 8... beyond that we don't get good stuff

[lethal cipher]

I may be missing something then

Yea, I agree with that.

[rustic crown]

for 3, we only like 1, 3. after that stuff is bad

for other primes only 1 is good

what are all the product of these things?

basically divisors of 24

[lethal cipher]

Wait, 9 works?

[rustic crown]

nah

[lethal cipher]

Ah, U(6) is U(3)xU(2), which is indeed isomorphic to (Z_2)^2

[wooden ember]

Did you start by first looking at which values of m give phi(m) = 2^n btw

Nvm me it’s basically what you’re doing now

[lethal cipher]

No. I do not see how that immediately relates. However, we did find them all.

[rustic crown]

this will also include those fermat primes... so thought of not doing that

[rustic crown]

this requires some work though. What i'm using here is that U(p) is cyclic

[lethal cipher]

Okay, so we know all primes greater than 3 aren't gonna cut it. So we have to look at factors that are a combination of 3's and 2's. But anything larger than 2^3 is gonna fail, and anything larger than 3^1 will fail. So that exhuasts our options.

So the largest thing we could have is U(8*3) because of this.

[rustic crown]

yep

so the answer to the problem is all the divisors of 24 except for 1, 2

[lethal cipher]

Yep, I see what's happening now. I was on the right path, I just was having a hard time seeing how to exhaust the info. Thank you again det.

[rustic crown]

[wooden ember]

I was just saying this since |U(m)|=phi(m)

Of course what you did is better since my condition isn’t sufficient, only necessary

[lethal cipher]

real quick question. How do you write the isomorphic symbol in latex (a bar with a ~ over it).

[rustic crown]

\cong

wait not = with ~ over it?

[lethal cipher]

no, just one bar

[rustic crown]

$\simeq$

[cloud walrus]

det

[lethal cipher]

Thank you det 😄

[rustic crown]

[lethal cipher]

Z has infinite order, does it not?

So then no. The question does not apply if both orders are infinite

[final pasture]

if any of the order is infinite*

[lethal cipher]

Yes, that too

Keep in mind, Z and 3Z have the same exact cardinality, so there is no difference in 'size'

Do you mean Z/3Z

So {0,1,2}?

Ah, well the issue is still the same. Z is infinite

[lethal cipher]

My bad, I realize this claim is not true.

[rustic crown]

it's true right? for abelian every subgroup is in the center

[lethal cipher]

Well, I was generalizing it to any group. But that just isn't the case.

xH=Hx, does not mean that there is an h in H so that xh=hx

[viscid pewter]

yes it does

1

[lethal cipher]

True, 1 is trivial.

[wooden ember]

doesnt make it be in the center though

in general normal subgroups are not in the center

but the center is normal

[lethal cipher]

Yep, that is the assumption I made foolishly.
xH=Hx means for all h in H, there is an h', and h'' in H so that xh=h'x and hx=xh''

I was not paying attention to that subtle distinction when I made that claim.

[wooden ember]

you noticed it after though that's good 👌

[eternal furnace]

I'm a bit confused with how to relate these two descriptions of Schur's lemma

for a given complex and irreducible matrix representation D, if a matrix Q commutes with D(g) for every group element g, then the matrix Q is proportional to the identity

vs

[stark sigil]

Such a Q is an element of Hom_G(V(D), V(D))

[eternal furnace]

Yeah I get that but where does the commuting statement come from?

[stark sigil]

That's what the _G under Hom indicates

[gritty sparrow]

yes because a homomorphic image is just G/N for some normal subgroup N and has order |G|/|N| so it is a factor of |G|

oh sorry I misread your question to be about the homomorphic image of a group not a subgroup

but it is still fine, the image of subgroup H of G is a subgroup of G/N so its order divides |G|/|N| which divides |G|

[marsh fossil]

so i have to find the left cosets of S4 = H, S5 = G, am i right in saying that all the sets would be found by taking all the powers of cycle of length 5 and finding the coset formed by those? since for every power i choose, that cycle times another power of that cycle is not in s4 so the cosets of them would not be equal right?

[next obsidian]

S4 isn’t well defined in S5

There’s… a chance that the cosets ang copy of S4 generates is the same

But there’s ambiguity as to what S4 means

[marsh fossil]

s4 is the symmetric group of 4 elements

[next obsidian]

Yes but there’s many of those

You could permit 1234 inside 5

Or 1245

Or 2345

Etc

[marsh fossil]

but no matter what power of the cycle i choose, barring the identity, it's never in s4 right?

since it's always a cycle of length 5

[next obsidian]

I don’t know, I’m not convinced you can’t embed S4 into S5 in a really weird fucked up way

Which does permutr all 5 numbers but in a weird way

[marsh fossil]

im so confused, arent all cycles of length 4 or less?

in s4?

[next obsidian]

Why?

But you aren’t specifying what copy of S4 it is

You’d need to show any subgroup of S5 which is isomorphic to S4 doesn’t involve all 5 numbers

This isn’t immediate

[hidden haven]

will never be a 5 cycle by lagrange, would be some 2 cycle and 3 cycle kinda stuff

[next obsidian]

Okay well there’s a proof so you’re good there

But it isn’t as simple as just

“S4 acts on 4 things”

[hidden haven]

is saying that there are many ways to put S_4 inside S_5, and we don't know (a priori) if they give the same quotient

[next obsidian]

Because S4 as a subgroup of S5 isn’t just “the elements that permute 1 through 4” because that’s a choice of a specific subgroup

But there’s multiple subgroups of S5 that are isomorphic to S4

[hidden haven]

another way to say it is that S_4 is not a subgroup of S_5 (only isomorphic to a subgroup of S_5) so S_5/S_4 doesn't make sense because you can only take cosets of a subgroup

[marsh fossil]

my class hasnt even talked about ismorphisms lol

[next obsidian]

Well the term S4 inside S5 is ambiguous is all I’m saying

[hidden haven]

Instead you should be saying cosets of i(S_4) in S_5 where i is an embedding

[next obsidian]

If you defined that to be the set of things which only involve 1 through 4

[marsh fossil]

but is it fair to say any 5 cycle is not in s4

[next obsidian]

Then your proof is immediate

[hidden haven]

so you have to say what i is

[next obsidian]

You have to argue why though

Moldi gave an argument using order

[hidden haven]

or prove that different choices of i give the same answer

[next obsidian]

5 doesn’t divide 4!

But it isn’t just 5 > 4 so it isn’t in there

If S4 means “the set of elements of S5 which only permutr 1 through 4” or like “which fix 5”

Then yes it’s fair to say that without any more argument

[marsh fossil]

well im trying to show the cosets are all different for the all powers of a 5 cycle

since every 5 cycle taken to a power is a 5 cycle

[next obsidian]

But you say cosets

[hidden haven]

cosets of what though

[next obsidian]

You haven’t fixed what you’re taking cosets of

[marsh fossil]

cosets on S4

[next obsidian]

You aren’t addressing the issue

What does S4 mean??

What do the elements look like

You have to be precise

[hidden haven]

so S4 could be the subgroup of things that fix 1, or the subgroup of things that fix 2, and so one

all of these are S_4

they permute 4 things

but since you haven't done isomorphisms I am guessing you are looking at S_4 as the subgroup of permutations that fix 5?

[marsh fossil]

okay. the proof for if cosets are different, is if for 2 elements of G, if g1xg2-1 is not in the subgroup im taking cosets of, S4, then their cosets are not equal

so regardless of the type of s4 im taking, the product of the powers of different powers of a certain 5 cycle, is never in any permutation of 4 elements, since it's a 5 cycle

[next obsidian]

But you have to prove any copy of S4 has to fix one element of {1,2,3,4,5}

[hidden haven]

are products of 5 cycles always 5 cycles? Their powers are

[next obsidian]

To make this argument tick

Unless you fix a specific copy of S4 you’re dealing with

[hidden haven]

You can multiply 2 different 5 cycles and get a 3 cycle

[marsh fossil]

yes but powers of a specific 5 cycle will always be a 5 cycle

i dont understand why i need to fix a copy of s4.. no matter what copy i pick it never contains a 5 cycle

[hidden haven]

You have to prove that these are the only 5 copies of S_4

there may be more, ones that contain the 5-cycle too

(though you can prove there aren't)

I am assuming that you are supposed to look at the copy which fixes 5

Are you working on a problem? Could you send that? That might clarify things

[marsh fossil]

it says find all the left cosets of H = S4 G = S5

the index is clearly 5

[hidden haven]

bruh

So either say the problem is wrong or assume they are talking about the copy that fixes 5

but you should probably bring this up with the instructor and ask for clarification, if you understood what problem we have with this

[marsh fossil]

i kind of do, but with that aside my approach is fine? take some 5 cycle and take powers of it for the different cosets?

[hidden haven]

if you change the embedding, the cosets change too

Yeah assuming that embedding it does

[marsh fossil]

okay cool, thanks

[versed ruin]

quick question

I have a group G of order 105 and I am assuming that there is a unique subgroup for every divisor of 105

I want to show that G must be cyclic

I know by Lagrange's theorem that the subgroups of order 2, 3, and 5 must be cyclic since they're of prime order

so I have 3 subgroups <a>,<b>, and <c> where |<a>|=2, |<b>|=3, and |<c>|=5

I also know that |<a>|=|a|=2, and by the same logic |b|=3 and |c|=5

I believe my next steps would be to show the following

|a| |b| |c| = 105 = |abc| so <abc> = G

However, I believe |abc| only divides |a| |b| |c| and that's only if commutativity holds

How would I go about showing that |a| |b| |c| must equal |abc|

[hot lake]

are you saying that 105 is even

how did you get |abc| = 105

did you also assume that G was commutative ?

ah nevermind

in general |abc| can be almost anything it wants

[versed ruin]

I don't know that |abc|=105 but I think it's what I need to show

[hot lake]

I would try to show that the subgroups are ordered by inclusion as they are ordered by divisibility

then it would make things easier

but using sylow's theorem just for that seem super heavy handed

[versed ruin]

I haven't learned sylow's theorem yet

[hot lake]

yeah

[versed ruin]

Also no, I haven't assumed commutativity unfortunately

[hot lake]

well you can show all the subgroups are normal

not sure how that helps though

ah it might show |ab| = |a| |b|

[versed ruin]

oh

[hot lake]

since |bcb^-1| = |c| = 5, bcb^-1 is in <c>, so it's c^k for some k
do the same with cbc^-1 to show it's equal to b^l for some l
then bc = c^kb and cb = b^lc so eventually you get c^(k-1)b^(l-1)=1, and that implies c^(k-1) = b^(l-1) = 1

so k=l=1

and so b and c must commute

[versed ruin]

alright, i'll give it a try

thank you!

[long obsidian]

Is there a general thing I might try if I wanna show that no groups of some order n are simple?

[willow mason]

lagrange + the sylow theorems

[tame grove]

Is this true? I dont know if I am allowed to alter the value of y, but by definition I dont think x^2=x^2(mod6) can be an equivalence relation

it isnt reflexive is it?

[hidden haven]

Checking reflexivity means putting x = y and checking if that satisfies the condition

[tame grove]

Yea, that is what I did and I dont think it is reflexive for all integers (where the problem is defined). Is this correct? I am doubting myself due to how the problem is stated.

[hidden haven]

x² ≡ x² (mod 6)

This is true for any x by reflexivity of ≡

I think you might be reading it as
x ≡ (x (mod 6))

[tame grove]

hmm alright. is that because a mapping can be made from Z6 to all of Z?

[hidden haven]

It is supposed to be read like
(x ≡ x) (mod 6)

[tame grove]

ooooooooooohhhhhhhhhhhhhhhhhhhhhhhhhh. that makes much more sense thank you

[viscid pewter]

(1, 1) should do it

oh duh

ok

<(1, 0), (0, 1)>

[chilly ocean]

I was trying to solve an exercise from a book but i got stuck in my calculation.
I'm trying to show that foundamental groups of a class of orientable 3-manifold is not left orderable where by this i mean that they admits a strict ordere invariant for left multiplication.
The groups are Seifert manifold built over a surface where the surface is $\mathbb{R}P^{2}.$

They admit the following presentation: $$\pi_{1}(M)=\langle\gamma_{1},...,\gamma_{n},y,h;|$$
$$yhy^{-1}=h^{-1},\gamma_{j}^{\alpha_{j}}=h^{-\beta_{j}},\gamma_{j}h\gamma_{j}^{-1}=h,y^{2}\gamma_{1}\cdots\gamma_{n}=1\rangle.$$

With some calculation I got to show that if $k>0,h>1$ the following holds: $h^{-k}<y^2<h^k$ and $h^{-k}<y^{-2}<h^{k}.$ In the book I was given the hint to show that $yhy^{-1}>1()$ and this in facts concludes as it is a contradiction with the assumption that $h>1$ given the first relation but I could not find a way to prove the key fact $()$. Any suggestion in what kind of relations I should look for?
Thanks in advance.

[cloud walrus]

Stephen

[oak grove]

question

[toxic vessel]

Is there some kind of duality between groupoid actions and intransitive group actions? I can't exactly put my finger on why, but they seem very analogous in some sense.

[bronze jay]

How do I quickly see if a cyclic finite group has an element of a given order?

[toxic vessel]

check if the given order divides the order of the generator

[barren sierra]

So for the second part

say I have phi: R -> S

do I need to show phi([x, y]) maps to [phi(x), phi(y)] is an isomorphism?

[past temple]

Frac(Q) = Frac(Z) = Q

but Q is not Z

is that a sufficient example? not sure

[thorn delta]

spam is talking about the second part

[past temple]

ah ok

yeah

u do need to show that phi

extended to the fraction fields is a well defined isomophirsm

[barren sierra]

so basically what i said above?

[barren sierra]

that's what I used for the first part lol

[past temple]

yeah

[barren sierra]

oh I shouldn't have called the map from frac(R) to frac(S) also phi

that's confusing lmfao

my bad

ok cool that's easy

[thorn delta]

anyway, you have the map f : S -> Frac(S) and a map f phi : R -> Frac(S) which you have to show induces a ring hom f phi : Frac(R) -> Frac(S)
and then show its iso

[past temple]

are u using this bracket notation to denote the equivalence classes?

[barren sierra]

yea

[past temple]

u can just denote the elements as x/y

[barren sierra]

oh true

[past temple]

well-definedness and injectivity and such

is gonna follow from like

the usual fraction cross multiplication

and such

putting the elements in fraction notation will make this all more cler

clear

[thorn delta]

you can do this without ever writing a fraction

[deep nova]

hey so if you're given two groups as such:

[rustic crown]

[deep nova]

how would you figure out if they are isomorphic exactly? What I did was just list out the elements and sort of just do the math and I've encountered a few cases where it is not isomorphic

and from the examples of my book, it doesn'

doesnt give a way to figure this stuff out

[rustic crown]

oh so there are many ways to do it

one of the first ones that you see are checking whether both are cyclic or not

like Z4 is generated by a single element which would have order 4

but not the same with U_8

U8 is {1, 3, 5, 7} but square of everything is 1

[viscid pewter]

check order of group, check order of elements, check abelianness and things...

[deep nova]

ah ok i see

i sort of did the slow and brute way of just doing the whole math and it turned out it wasnt isomorphic, but then obviously such a way is just too inefficient

so would the cyclic method always work for instance?

or does it depend on the problem?

[rustic crown]

the checking order of elements works for any two finite abelian groups as far as i remember

[deep nova]

👍 ok i see

[rustic crown]

have to confirm this though

maybe Kaisheng has something nice in mind?

[viscid pewter]

?

nah

[rustic crown]

but yea the cyclic thing won't work because what if both groups are not cyclic... that doesn't give a whole lot of information

[viscid pewter]

if anything at all is different between the two of them, they're not isomorphic

[chilly ocean]

I was trying to solve an exercise from a book but i got stuck in my calculation.
I'm trying to show that foundamental groups of a class of orientable 3-manifold is not left orderable where by this i mean that they admits a strict ordere invariant for left multiplication.
The groups are Seifert manifold built over a surface where the surface is $\mathbb{R}P^{2}.$

They admit the following presentation: $$\pi{1}(M)=\langle\gamma{1},...,\gamma{n},y,h;|$$
$$yhy^{-1}=h^{-1},\gamma{j}^{\alpha{j}}=h^{-\beta{j}},\gamma{j}h\gamma{j}^{-1}=h,y^{2}\gamma{1}\cdots\gamma{n}=1\rangle.$$

With some calculation I got to show that if $k>0,h>1$ the following holds: $h^{-k}<y^2<h^k$ and $h^{-k}<y^{-2}<h^{k}.$ In the book I was given the hint to show that $yhy^{-1}>1()$ and this in facts concludes as it is a contradiction with the assumption that $h>1$ given the first relation but I could not find a way to prove the key fact $()$. Any suggestion in what kind of relations I should look for?
Thanks in advance.