groups-rings-fields

[solemn rain]

no not all groups are cyclic

[shy bluff]

Oh ok

[solemn rain]

every finite group is finitely generated tho but not by 1 element

[shy bluff]

Ah, ok yea thats' what I was thinking

[solemn rain]

with the obvious G=<G>

[shy bluff]

but uh yea no idea how to do this

[solemn rain]

okay

[shy bluff]

I thougght that it had something to do with Laggrange's TheoreM?

[solemn rain]

yea it does

think about the order of the elements

in a group of order 4

[shy bluff]

My reasoning so far is that if |G| = 4, then there are only 3 possible subgroup sizes

either its subgroups are of size 4, 2, or 1

Right?

[solemn rain]

yea what about the elements

same right?

[shy bluff]

What do you mean?

[solemn rain]

so we want to show this is an abelian group

what about the orders of the elements

they have to divide the order of the group no?

[shy bluff]

Z/4Z is abelian right

Uh

Order of elements have to divide the order of the group, yes this is true

[solemn rain]

okay

so the possibilities for the orders

of the elements are 4 2 and 1 right?

[shy bluff]

Yes

[solemn rain]

okay so suppose if we have an element of order 4

then the group is cyclic hence abelian

right?

[shy bluff]

Sure

[solemn rain]

so

the group has x^2=1 for all x

right?

[shy bluff]

Yea

[solemn rain]

so with this

can you show that xy=yx

for all x and y in G

i think you maybe have done this b4

as an exercise

showwing that H = {x in G |x^2=1} is an abelian subgroup of a group G

[shy bluff]

Yea

[solemn rain]

okay

so now youve proved G is abelian

right?

[shy bluff]

Yes

[solemn rain]

first of all

if G had an element of order 4 it would have been isomporphic to Z/4Z right?

[shy bluff]

Yes this is true

[solemn rain]

so that solves the first part of the problem

okay

so assume G is not cyclic

define the function f:Z/2 x Z/2 --> G as f(a,b) = x^ay^b

x and y in G

is this an isomorphismm

[shy bluff]

Hrm

[woven delta]

lol do you know the theorem that index 2 subgroups are normal?

[shy bluff]

a = 0, b = 0, then you get ee = e, a = 1, b = 0 then you get xe, a = 1, b = 1 then xy and yea that looks like a bijection which is an isomorphism

@woven delta yes

[solemn rain]

okay cool

[shy bluff]

we're supposed to use that somehow

I dont' know how but we are

[solemn rain]

wait really

[woven delta]

okay here's some facts

1. if your group is not cyclic then every element has order either 1 or 2

2. if an element has order 2 in a group of order 4, then the subgroup generated by that element has index 2

[shy bluff]

2 is just by definition oforder right?

[woven delta]

it's because the index is the order of the group over the order of the subgroup

so 4/2=2

[shy bluff]

Yes

[woven delta]

okay so we know that all the subgroups are normal

and we know that given 2 distinct nonidentity elements a, b, the subgroups <a>, <b> have <a>\cap <b>=e

[shy bluff]

why is that?

[woven delta]

and we also know because of cardinality reasons that the order of the product of the subgroups is 4

because <a>={e,a} and <b>={e,b}

and a\neq b

[shy bluff]

Yes

[woven delta]

okay so we know that <a><b>=G and <a> and <b> are normal

and they have trivial intersection

boom direct product

[shy bluff]

Ok

[woven delta]

do you know the definition of the internal direct product?

[shy bluff]

Yes

[woven delta]

and how it agrees with the external direct product

[shy bluff]

I dont' nkow what an external direct product is?

Is it an internal direct product but just with any group rather than restricting it to subgorups?

[woven delta]

it's the one that given groups G and H we let GxH be the group with elements {(g,h)|g\in G,h\in H}

[shy bluff]

Ok

[woven delta]

and the group law is (g_1, h_1)(g_2, h_2)=(g_1g_2,h_1h_2)

and it turns out that if there are 2 normal subgroups G, H of K so that G intersects H trivially and GH=K, then K is isomorphic to the external direct product of G and H

[shy bluff]

Ok

[brisk granite]

Im having trouble showing prime and irreducible are the same in this ring

how would I go about showing this?

[steady axle]

Hint The given ring is an ufd

[brisk granite]

yea, I can see how that implies the thing I want to prove but idk how to prove that

also, this was in chapter one of the book im using and the term "ufd" hasnt

been introduced yet

[bleak abyss]

Which book is this? Just in case like, it assumes background in algebra such that you'd know k[x_1,...,x_n] is a UFD

[brisk granite]

uh, my teacher wrote it. So, Im not sure if that's something I should know

I'll just ask him if I should know this

[steady axle]

Btw if A is ufd then A[x] is ufd

[brisk granite]

Could it be possible the question is design such that I should try to prove this?

Hint The given ring is an ufd

Btw if A is ufd then A[x] is ufd
oh, I see

I'll try to prove this

[steady axle]

Do you know gauss' lemma

[brisk granite]

which one?

ik one but I kinda doubt that's the one youre referring to

[steady axle]

Given the context if you ask which one then probably you don't know 😂

[brisk granite]

yea, probably lmao

so, um, what is gauss's lemma?

[knotty mason]

A prime ideal is an ideal such that ab in P => a or b in P

To show that (f) is a prime ideal in k[x1,x2,...] pick ab in (f) so you know ab = fg for some g and show that a or b is in (f)

this will use irreducibility

[green pine]

hey

I need some help in this question

Prove or give a counterexample: if U is a subspace of V that is
invariant under every operator on V , then U = {0} or U = V .

I know one counter example and wish to know any simpler ones

Counterexample for one such problem can be given by the fact that ”Every operator on a finite-dimensional,Non-zero,Complex Vector space has a eigenvalue

[olive mirage]

operator here just means a linear transformation from V to V?

[green pine]

yes

[olive mirage]

In that case, I do not believe there is a counterexample

[green pine]

are you sure?

[olive mirage]

I agree that every particular operator on a finite dimensional complex vector space leaves a subspace invariant

but that's just one, where as your question asks about being invariant under EVERY operator.

[green pine]

we need not define every operator on V

[olive mirage]

(by under every operator, it means under every operator simultaneously)

[green pine]

yes

all operators defined on V

if we restrict operators we can find satisfying examples, but will this hold for every operator??

[olive mirage]

no, it won't.

Here's a nudge:

if v1 and v2 are non-zero vectors in V, there is always an operator A s.t. A(v1)=v2.

[green pine]

ok

[olive mirage]

(The intuition here is that no non-zero vector in a vectorspace is special, they're all the same. The same is true of subspaces of a particular dimension. )

[green pine]

Consider Vector Space V(F^2)

ok, trying

consider u belongs to U
and v belongs to V but not U
Consider operator T(w) = v
then
T(u) = v which implies U is not invarient on T

😅

this is pretty dumb but does this qualify as a proof

yes

T(w) = v + a(w-u)

ok

pls have a look at this

can this help in proof

we can show that two operators have disjoint null spaces(other than 0)

ok, lemme try

https://cdn.discordapp.com/emojis/669700075751997442.gif?v=1

actually I got confused due to something

see this

so we have subspace for each operator

we may have hope that their intersection is not {0}

that may be be invarient on all

😅

nope

I am just thinking

I was reading this book and wanted to see If I can prove

I want to prove this?

or have an idea what this means

9.8 says we have a subspace for each operator, so to have a subspace that is invarient on all operators we need this subspace to be in all subspaces

idk what I am thinking most the time it is wrong

can you pls help

yes

we need to show that there exist an opertor

that maps from u to not u

yes

we can take u, e1,e2,...en

and find an orthnormal basis of this

v is non zero cause v does not belong to U

so we have a basis of V containing u

in a way i think we must show Tu = 0

or belongs to U

ok we have two now

yes may be

or what about we take basis of U ={u1,u2....um}

extend this

what can be the linear operator, we do not need to define opertor just telling what it does mus be enough right?

T({a,b,c,d}) = {d,a,b,c} how does this look?

T({u1,u2,0,0}) = {0,u1,u2,0} which does not belong to U

does this sound like a proof ?

https://cdn.discordapp.com/emojis/715656211709690058.png?v=1

you can think of F^n

for now

cordinate system

[buoyant relic]

you know we have a sad cat emoji

[solemn rain]

whats the problem

[green pine]

whats the problem
@solemn rain https://discordapp.com/channels/268882317391429632/496784958430380033/718843240668987493

yes

what i was thinkig is any element in v ca be written as (given basis {u_1,u_2,..,v_n-1,v_n}) a1(u_1),a2(u_2)....

element can be written by {a1,a2,...a_n-1,an} and operator T({a1,a2,...a_n-1,an} ) = {an,a1,a2,...a_n-1}

@chilly ocean

is operator clear now

{1,1,..,0,0} belongs to U

yes sort of

{1,1,..,0,0} belongs to U
T({1,1,..,0,0}) = {0,1,1..,0} does not belong to U

atleast do I have a right idea??
can this lead to a proof

what is the portion needed for this?

:thankscat:

[shy bluff]

For this,

When they say that the order of [g] in G/N is min {n >= 1: g^n in N} do they mean any n in N?

[olive mirage]

I do not think n in N makes sense

n is in Z^+

[shy bluff]

Hrm

Then do they mean any element of Z^+ such that g^n is in N?

[solemn rain]

the smallest n such that g^n in N

@olive mirage maybe book has N={1,2..}

[olive mirage]

@solemn rain any reference to the natural numbers in a math text should be groups for banishment

but it is perfectly reasonable, upon seeing a small n and a big N to assume the small n is an element of big N. That's how math notation usually works. It just so happens that here you're meant to think of n as an integer.

[shy bluff]

Oh it might be natural numbers

That would make sense

so this is asking me to prove that for any g in G that its order in G/N is going to either be that of the smallest integer that sends it to N or infinity if it isn't sent to N at all?

[olive mirage]

that's exactly right

[shy bluff]

Thank you!

And I'm just going gto hazard a guess here, the infinity one is just whenever G cyclic

Err isn't cyclic

[shy bluff]

Wait another question

Is the order of g just the result of taking q(g), where q is defined as q: G -> G/N, q(g) = gN?

[stone fulcrum]

The order of an element g? It's the amount of times you have to multiply g×g×g×g... = e

We write that concisely as gⁿ = e

[shy bluff]

Yes

Well

Here we want g^n in N

that'd just be .... Anything in N?

Or rather

All of N?

~_N such that for all a, b in N, a ~_N b?

Ok I don't know

[olive mirage]

A fun follow-up exercise once you get this to make sure you have it all down: A) Prove that if G has a normal subgroup N of index k, then x^k is in N for all x in G. B) Use this to prove that A_4 has no subgroup of order 6.

[shy bluff]

Yes

Oh

Ok

Yes

So are we looking for a way to write [N]?

e

Identity

What I dont' really understand is, what is the order of[g]?

Is the order of [g] n such that [g]^n = e?

Yes, the minimal n such that that happens

Oh then by definitino the order of [g] in G/N is min{n >= 1: g^n in N}

Ok but what's with the infinity?

Is that for when G/N isn't cyclic?

Or rather G isnt' cyclic?

Because if G isn't cyclic then wer''e never goging to get around to e right?

[cloud walrus]

slimvesus:

[shy bluff]

Uh

You lost me on that last one

I'm somewhat sketch on the bit wrt infinite order

But the bit that's finite order makes sense

It's like

Let g be some element in G such that g^n in N

Then by definition of G/N, [g^n] = [e]

Right?

that isa good question

Uh

Not necessarily?

An integer?

Yea

Then g^{n - k} is not in N

Thus it's not equal to [e]?

[shy bluff]

Oh then it's infinite

[shy bluff]

I see

[shy bluff]

Is there a way to do this without a "tedious consideration of all cases"

[olive mirage]

it is much cleaner with generators and relations, but those are a more advanced topic.

[shy bluff]

Oh ok

[ruby pine]

I'm looking for a good book on Lie Groups and Lie Algebra. Any recommendations?

Just trying to build up my understanding to eventually get to a general notion of what an algebra is

[scarlet estuary]

do you know any differential geometry?

if not, Lee's Introduction to Smooth Manifolds is probably a good place to look

the focus is, of course, on the diff geo content, but it also covers Lie groups and Lie algebras (since Lie groups are smooth manifolds)

if you dont want to do any differential geometry for some reason, you could try Hall's Lie groups, Lie algebras and Representations but

i think cutting out the diff geo content from the subject does it a fundamental disservice

Hall's text certainly has substantially less prerequisites and is more "focused", though

well, it's "focused" at least on matrix lie groups - i dont think it covers general lie groups

[olive mirage]

Big thumbs up for INtroduction to Smooth Manifolds

[stone fulcrum]

Loving Tu's smooth manifolds atm, finding it a bit easier than Lee. Don't know what the differences are yet

[latent anvil]

I really liked ISM

Just finished my course using it and it was great

[ruby pine]

@scarlet estuary I know some differencial geometry, but I wouldn't say that I mastered it yet

From a quick search I'm also not exactly seeing the difference between a Lie Group and a Lie Algebra

As far as background. I've taken: Modern Algebra, Linear, Functional, and Differencial Geometry.

[scarlet estuary]

Well, a lie group is a group (and a smooth manifold) while a lie algebra is a vector space with added structure

that said, we built lie algebras from lie groups

(indeed lie algebras are tangent spaces at the identity of lie groups)

[ruby pine]

Okay, so for a Lie Group there aren't any other added axioms from regular groups

[scarlet estuary]

a lie group is just a "smooth continuous" group

[ruby pine]

Got it

[scarlet estuary]

the more proper definition is

a lie group is a group that is a smooth manifold

if you consider the tangent space of a lie group at the identity, you get a lie algebra

and the correspondence, by lie's third theorem, goes both ways

anyway uh, you could try reading some of the relevant chapters of the Lee text i mentioned

it sounds like you have the prerequisites

[ruby pine]

Yeah, I could try.

For that definition of (x,y) would the multiplication just be any regular group multiplication?

[scarlet estuary]

yeah, just the group operation

[ruby pine]

I think seeing it and DiffGeo along side eachother would be helpful to me too. Since I'm a bit more on the applied side

[scarlet estuary]

ah, lie is AFAIK a bit more on the pure end but it still might work

im afraid im not familiar with applied sources

[ruby pine]

No worries

It's just easier to follow the pure stuff if there is an application fo fall back on for intuition

[mild laurel]

I've had humprey's recommended to me for an introduction to lie algebras

but I think that it might be a problem because lie groups/algebras are usually studied using rep theory

[ruby pine]

Rep theory is something I'm also kind of interested in.

But it may be due to a misconception.

I tend to do a lot of fourier and wavelet analysis. Which feels like a "representation" of what I'm studying, but I'm guessing that it's not the same thing

Looks like Humprey might be a bit easier for me to stomach than Hall

[mild laurel]

I don't know a ton about this, but I've heard it said that fourier analysis is the representation theory of S^1 in some sense, so I don't think its too far off

[ruby pine]

Cool :)

I guess my last question is if there is a good definition of an "Algebra"?

My education of Abstract Algebra stops at Fields and vector spaces. Subtraction, division, and the cross product not being included in those is kind of what lead me to wanting to learn about lie stuff

[scarlet estuary]

an algebra is a vector space with a notion of "multiplication of vectors"

specifically a bilinear product

[ruby pine]

Ahh, okay. So, I'm almost there. Thanks @scarlet estuary

[scarlet estuary]

yeah its not as "fancy" as it sounds

but the theory is quite deep

[ruby pine]

Yeah, I'm guessing the inner product would count?

[mild laurel]

Inner product doesn't work because you don't get another vector when you take the inner product (unless you're dimension 1)

[ruby pine]

Understood

[mild laurel]

Think like k[x] is an algebra over a field k

[ruby pine]

Okay, so it sounds like I need a book to be introduced to what an algebra is first. Then Lie algebra's would be a good second

Alright, thanks for answering my questions and pointing me in an interesting direction

[shy bluff]

Question

If we have a group G

With subgroup H

And we go and make G/H

Everything in h is removed from G/H right?

[scarlet estuary]

if by "removed" you mean "belongs to the same equivalence class as the identity", yeah

[shy bluff]

Ye athat

Ok then if I want to count the number of subgroups of G that contain the subgroup H

For one thing we have that G/H has to contain it H right?

Like my question is this

And we know that Z(D_16) = {e, s^8} right?

So I want all subgroups that contain s^8

Would that just be every subgroup that contains s^2, s^4 or s^8?

And yes, Finitely Degenerated, that's waht I meant

[shy bluff]

OH and subgrgoup that contains s too

[woven delta]

liria there is a theorem

there is a correspondence between subgroups that contain a normal subgroup and subgroups of the quotient group

[shy bluff]

Correspondence Theorem!

[elder valley]

The subgroups of G/H are in correspondence with the subgroups of G which contain H. So counting your subgroups is the same as counting the subgroups of G/H

[shy bluff]

Ah, that's what I was thinking I think

Or at least what I was thinking is that all subgroups of G/H will contain H

[upbeat juniper]

is there some result stating that every order p^n group has a normal order p subgroup

it sounds vaguely familiar but I can't quite pin the theorem down

[wind steeple]

the center of a p-group isn't trivial

[upbeat juniper]

could you elaborate, I'm not sure how to construct a normal order p subgroup using the center

[wind steeple]

bcp it is non trivial, you have a non trivial element of ordre p in the center, the subgroupe generated by this element is normal

[upbeat juniper]

oh yeah didn't think of that, thanks

[shy bluff]

Oh wait so the correspondence theorem says that the set of all subgroups of G containing N are in bijection with G/N?

[wind steeple]

if N is normal yes

[shy bluff]

ooh ok

[shy bluff]

How do I do part c here? I think that this has to do with the correspondence theorem?

That or the first isomorphism theoreM?

[knotty mason]

If you find a homomorphism from B whose kernel is N then the iso theorem tells you the image of that is iso to B/N

[shy bluff]

From B to where?

B to Q^x x Q^x?

[knotty mason]

it could be B -> B or B -> GL_2(Q)

[shy bluff]

Oh ok

[knotty mason]

Q^x x Q^x could work too

[shy bluff]

Kernel being N would mean that any element of N maps to identity right

[knotty mason]

yeah

[shy bluff]

Yea I'm jsut thinking gthat mapping ot Q^x x Q^x would be easier because having something to gog from B to GL_2(Q) wouldn't tell me anything about Q^x x Q^x

I think?

Like if I have this, its kernel is N right?

[knotty mason]

yeah that seems good

[shy bluff]

Wait but this is'nt surjective?

[knotty mason]

yeah it's not surjective

the image of this is a subgroup of GL_2(Q) but you can also see this subgroup as isomorphic to Q^x * Q^x

[shy bluff]

I can see that it's isomorphic visually and intuitively but don't know how to show it mathematically

[knotty mason]

to demonstrate the isomorphism formally you can say (G, +) is isomorphic to (H, *) since we have a map f : G -> H that respects identity the operation and inverses

so in this case you'll be mapping [a, 0; 0, c] to pairs (a, c)

and verifying that the multiplication of matrices matches up with the multiplication you specify for the group of pairs

[shy bluff]

Wait question

So given the function that I defined above, we actually have that $B/N \cong Im(\phi)$ right?

[cloud walrus]

Liria ^(;,;)^:

[shy bluff]

And now I need to show that $Im(\phi) = Q^x \times Q^x$?

[cloud walrus]

Liria ^(;,;)^:

[knotty mason]

yeah

[shy bluff]

And how can I show that? Do I just have t oshow that multiplication of matrices matches up with the multiplication for pairs of integers?

[knotty mason]

showing that it matches up proves that phi is a group homomorphism, this was needed to apply the isomorphism theorem

then Im(\phi) = Q^x \times Q^x proves that B/N is isomorphic to Q^x * Q^x

[shy bluff]

but how do you get that Im(\phi) = Q^x \times Q^x?

Wait can I just go like "Let Phi: B -> Q^x \times Q^x" and then show that it's a homomorphism with kernel of N

[knotty mason]

yes

[shy bluff]

Oh wait is that what you were tellingg me to do

Oh

Okie

Thank you!

[knotty mason]

sorry I realize now you were talking about the phi earlier that mapped to GL_2(Q)

[shy bluff]

Yaaaa thats' why I was confused lol

Wait for this question, can you just go and use the correspondence theorem with the quotient map and map from K to K/N and G to G/N?

[knotty mason]

yes that will follow as an application of the correspondence theorem

[shy bluff]

Ah ok

For once I know what to do 😅

[shy bluff]

Ok wait no I'm still lost

So we have that tho quotient map q: G-> G/N , with kernel N

Then we satisfy a here

And then we can say that phi(N) <= phi(K)

And we can also say that phi(K) <= phi(G) is really just K/N <= G/N

But that doesnt' establish the normality?

[half nebula]

Hi! Could I have some help with this question. I was told earlier by someone on here to solve the first one with proof by induction and I kinda get it but I'm also still confused. Like how would I do a base case (plugging 1 in somewhere).

Here's the proof I think I need to do but I'm not 100% sure.

Also I emailed my professor about this, and he sent me back this if it helps any. "Just remind you that the induction might be only working for the positive integers k. That is, you prove this statement in the positive case first, and then use the positive case to show the negative case (just one more line, think about how). k=0 is trivial. Then you are done with the whole proof. "

Or well actually the base case is already proven to be true so never mind. ><

[scenic sage]

@half nebula for a base case literally plug in k=1 in a

and reread the task

[half nebula]

Well ywah so that's already done in the top part.

So I just need to do the induction part.

[scenic sage]

do u know that

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

[cloud walrus]

Commander Vimes:

[half nebula]

Sorry I keep being dragged away by family @scenic sage

And kinda. Like I know that you distribute the -1. Why does it switch?

[scarlet estuary]

intuitively:

the definition of an inverse is that $xx^{-1} = 1$, right? where $1$ is the identity

[cloud walrus]

Namington:

[scarlet estuary]

so $(ab)(ab)^{-1} = 1$

[cloud walrus]

Namington:

[scarlet estuary]

but note that if we rewrite thi sas

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

[cloud walrus]

Namington:

[scarlet estuary]

we can apply associativity to get $a(bb^{-1})a^{-1} = a1a^{-1} = aa^{-1} = 1$

[cloud walrus]

Namington:

[half nebula]

is ab not equal to ba though?

[scarlet estuary]

this isnt exactly a proof but it's the idea behind the proof

uh

this is a dihedral group

so $ab = ba$ is not necessarily true

[cloud walrus]

Namington:

[scarlet estuary]

not all groups are commutative (aka abelian).

[half nebula]

Ah okay.

Is the (ab)^-1=b^-1a^-1 something I have to prove first?

[scarlet estuary]

i'd honestly expect that to have been proved previously

but yeah, if not, it might be a good idea

it follows from uniqueness of inverses

[half nebula]

He'll probably let it slide. Worst case scenario, I lose a point or two.

[scarlet estuary]

we can see that $(ab)(ab)^{-1} = 1$, and also that $(ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = aa^{-1} = 1$

[cloud walrus]

Namington:

[scarlet estuary]

so both $(ab)^{-1}$ and $b^{-1}a^{-1}$ are inverses of $ab$

[cloud walrus]

Namington:

[scarlet estuary]

you can prove the other direction too, ie $(ab)^{-1}(ab) = (b^{-1}a^{-1})(ab) = 1$

since inverses are unique (not hard to prove, maybe the first proof you did), it follows that $(ab)^{-1} = b^{-1}a^{-1}$

[cloud walrus]

Namington:

[half nebula]

That makes sense. But how would that help with this specific problem with k?

Oh wait

I can just say that $(ba^k)^{-1}=a^{-k}b^{-k}$

Or wait

$(ba^k)^{-1}=a^{-k}b^{-1}$

[cloud walrus]

Soulgiver831:

[half nebula]

Right?

[scarlet estuary]

sure, what does that tell you

[half nebula]

Well that would have to mean that $(a^{-k}b)^{-1}=a^{-k}b^{-1}$

[cloud walrus]

Soulgiver831:

[half nebula]

Or $b^{-1}a^k=a^{-k}b^{-1}$

[cloud walrus]

Soulgiver831:

[half nebula]

so a^k is equal to a^-k. I think.

[half nebula]

Okay so I almost have this down. I'm on the last problem here, C.

since I already know the statement from part b, I can say that $ba^mb=(ba^m)b=(a^{-m}b)b.$

[upper pivot]

yeah thats the solution

[cloud walrus]

Soulgiver831:

[half nebula]

But then what?

Do the two bs cancel each other somehow?

[upper pivot]

whats b^2

[half nebula]

It's b^2 I don't know ><

An element or order 2?

of*

[upper pivot]

b is an element of order 2 yes

what does that mean

[half nebula]

It has 2 subgroups.

Or it generates 2 subgroups.

I think.

[upper pivot]

no

[half nebula]

Or no, it generates a subgroup with 2 elements

[upper pivot]

yeah

[half nebula]

Which would be itself and the inverse.

Or itself and the identity?

[upper pivot]

well how did you solve A

cause you kinda have to know this idea to do the former

[half nebula]

We only had to do b and c. He said a was optional.

[upper pivot]

well dont skip it

[half nebula]

Oh. Well okay.

[upper pivot]

alright so i will define order for you

[cloud walrus]

JohnDoeSmith:

[half nebula]

Oh wait so a^n =1.

[cloud walrus]

JohnDoeSmith:

[upper pivot]

yeah

[half nebula]

So you can multiply it by 1 which is the same as multiplying it by a^n

[upper pivot]

mhm

[half nebula]

But wouldn't that be a^-nm?

like x^a multiplied by x^b is x^ab.

Or wait no it isnt

[upper pivot]

umm no

[half nebula]

its x^a+b

Sorry I'm tired as heck rn.

[upper pivot]

its fine

[half nebula]

So that would be a^n-m

I got it.

what does that have to do with b^2 though?

Oh wait

b^2 is the identity element right?

Because it has an order of 2.

[upper pivot]

yep

[half nebula]

But even though wouldn't it just be a^m

what makes it -m?

like you have $b^2a^m$

[upper pivot]

umm remember (b)

[half nebula]

Thank you ><

Also sorry for being so nudgy tonight I really just wanna get this homework out of the way. I think I got this one though.

Would the element just be a with order of 6 and b with order of 2 still?

[upper pivot]

well theres 6 elements

or umm 12

[sharp sonnet]

ye, give your definition of D_n

i was about to ask if D_6 only contains 2 elements a and b?

[half nebula]

Well the previous problem said this.

So that's why I assumed it only had two elements, a and b.

[sharp sonnet]

no

it's generated by 2 elements and the given relation

[half nebula]

Ooooooh okay

Does that ba=a^-1b thing still stand?

[sharp sonnet]

yes?

[half nebula]

Okay cool.

I just figured I have to use that I'm guessing.

Right?

[sharp sonnet]

yeah

i mean, if you don't even know how D_6 looks like

and this is the definition you have

you first have to figure out what all the elements in D_6 even are

and how to represent them

[half nebula]

Different combinations of a and b? Or are a and b even elements in d6?

[sharp sonnet]

a and b are elements of D_6

and all other finite combinations of a and b

but you can use the relation to simplify

and thus get a finite amount of elements

[maiden ocean]

Loch do you think it would be a good idea to explain the free group word thingy

Since that's basically what this is

It might make it make little more sense

[sharp sonnet]

yeah, but like

go ahead

[maiden ocean]

Yeah sure

[sharp sonnet]

this question is simple

very simple

[half nebula]

Okay

[sharp sonnet]

and the hardest part here is figuring out how D_6 looks from the definition

[maiden ocean]

When they say that D_n is the group given by elements a of order n and b of order 2

[sharp sonnet]

so i'm kinda thinking that they covered that somewhere?

[maiden ocean]

Basically what they are saying is that elements of D_n are like "words" made out of the "letters" a and b

[half nebula]

Maybe. I don’t have classes so we don’t really cover stuff. It’s more so just our professor giving us homework, the section to read and from there it’s all on us.

[maiden ocean]

A word being some sequence of a and b

like, abaabaaa

is a word

[half nebula]

That makes sense.

[maiden ocean]

so what "elements a of order n" and "elements b of order 2" means is that a^n and b^2 are the same as the identity right?

[half nebula]

Yeah, the identity element 1.

[maiden ocean]

Yeah

So what they're saying is that any time a sequence of 2 b or n a comes up

which we can shorten to b^2 and a^n

we can basically "reduce" our word by collapsing that string to the identity

so in the case of D_6 if we have like, a^3b^2a^6b

we can reduce this to a^3b

Does this make sense?

[half nebula]

Yeah I remember that. Like how with permutations, to get the order of each element of like S_3 for example, we had to take the permutations of that group and multiply them by each other until we reached the identity element.

[maiden ocean]

Yes

[half nebula]

So if you had to square a permutation to get to the identity element, the order would be 2.

Okay cool.

Would it just be different combinations of powers of a and b?

[maiden ocean]

Yes

Have you seen geometric representations of the dihedral group?

[half nebula]

I can’t say I have. (Mainly because I don’t even know what a geometric representation is ><)

[maiden ocean]

That might help make it a bit more intuitive

Oh okay

[half nebula]

Wait hold on. Maybe I do.

Is it the rigid thing?

Like hold on

[maiden ocean]

It's as symmetries of the n sided polygon

[half nebula]

I had this for a problem earlier.

[maiden ocean]

Yeah it's sort of like this

[half nebula]

Okay cool.

[maiden ocean]

For reference when they talk about elements a and b

a corresponds to rotating our shape once

in this case it would be sending 1 to 2, 2 to 3, 3 to 4, 4 to 1

so it makes sense that we say that a^n = 1, because if we have an n sided regular polygon and we rotate it around the origin n times we end up exactly where we started, right?

[half nebula]

So the possible combinations would be
a^1
a^2
a^3
a^4
a^5
1 (a^6 or b^2)
b^1
ba^1
ba^2
ba^3
ba^4
ba^5

So all 12 right?

[maiden ocean]

Yep, should be!

[half nebula]

Okay great!

So now the order of each one. Is the order just the power of a? (or the power of b in b^1)

[maiden ocean]

the order of each element?

[half nebula]

Yeah. The question saya to find the order of each element in d6

says*

[maiden ocean]

not quite

the order of each element g is the smallest positive integer n such that g^n = e right?

[half nebula]

Ah okay so I just have to keep multiplying until its b^2a^6

Like a^2 would have an order of 3.

Right?

[maiden ocean]

yeah

[half nebula]

Cool

[maiden ocean]

But you should remember that the dihedral group isn't abelian (for n > 3 anyway)

[half nebula]

I know, from when I was doing b in the last one. (where I had to prove $a^mb=ba^{-m}$

[cloud walrus]

Soulgiver831:

[maiden ocean]

Right

so for each ba^r for 1 <= r <= 5

what is the order

[half nebula]

Well it has to be an even number.

like if you have ba^3 the order would be 2 since (ba^3)^2=b^2a^6. Right?

[maiden ocean]

hmm see

that only holds for abelian groups

[half nebula]

Ah okay.

[maiden ocean]

so lets say we have (ba^3)(ba^3)

can we flip one of these ba^3?

[half nebula]

Well I mean I did prove from before that ba^m=a^-mb

[maiden ocean]

yep

thats the key

so how can we use it to prove that (ba^3)^2 is the identity?

[half nebula]

Oh wait.

All of them would have an order of 2 since ba^m=a^-mb so you'll always have $ba^{m}a^{-m}b=b^2=1$

[cloud walrus]

Soulgiver831:
Compile Error! Click the reaction for details. (You may edit your message)

[half nebula]

Hold on ><

[maiden ocean]

right!

[half nebula]

But yeah ^^

[maiden ocean]

(heads up texit bugs anytime you put in a ^ outside of math mode lol)

[half nebula]

Ah I gotca.

[golden pasture]

^ in latex is something weird lol

[half nebula]

Oh and sorry, I almost forgot. Thank you very very much @maiden ocean

Lol true.

I'm new to the whole texit thing.

[golden pasture]

https://tex.stackexchange.com/questions/77646/how-to-typeset-the-symbol-caret-circumflex-hat

[maiden ocean]

<3

[golden pasture]

it isnt texit

[maiden ocean]

Oh is it?

[golden pasture]

it's cuz it's rendered with latex

[maiden ocean]

Interesting

[golden pasture]

latex ^ is special character for idk what

[maiden ocean]

oh huh

interesting

[golden pasture]

if texit parses latex wrongly that is quite a interesting problem cuz it literally runs vanilla pdflatex

[half nebula]

I also thank John and Loch but I don't wanna @ them since they're offline ><

[upper pivot]

oh np

[half nebula]

^^

[indigo ravine]

Hey in an exam I just had I had to prove that for a normal operator f on a finite-dimensional space F, F is the direct sum of Ker(f) and the orthogonal complement of Im(f). It seems impossible since if you take f=0 it leads to a contradiction (both set are F in that case). Am I wrong ?

also sorry if that's the wrong channel I didn't know whether I should post it in #linear-algebra or here but there's an ongoing conversation there

[upper pivot]

umm f=0 has ker=F

[sour plume]

(but the image of f = 0 is zero)

[upper pivot]

oh right

orthogonal complement

yeah hmm this doesnt make much sense, because orthogonal complement of im(f) contains the kernal(f)

in this case

[sour plume]

why does it not? the orthogonal complement of the zero space is the whole space

I'd have to think about this a bit to prove it, but I can assure you that f = 0 is not a counterexample unfortunately

[indigo ravine]

but that means the direct sum of the whole space and the whole space is the whole space which doesn't make sense right ?

[upper pivot]

actually i was thinking about the general case not the counterexample specifically

but i realize what i said is wrong

[sour plume]

That's not what you have, though; you have the direct sum of the zero space and the whole space

[upper pivot]

because i made an error

[sour plume]

Because the kernel of the zero map is the whole space, but the image of the zero map is just zero

[indigo ravine]

the formula is $Ker(f) \oplus (Im(f))^\perp = F$

[cloud walrus]

plougue:

[sour plume]

Oooh sorry I misread

[indigo ravine]

if Im(f) is 0 isn't its complement is F ?

oh np

[sour plume]

Yeah that doesn't sound right at all

[upper pivot]

i think they might mean just sum, not direct sum?

i am not sure if that works in general though

[sour plume]

That's weird though, i remember a similar statement from a functional analysis course I've once seen. Lemme quickly go through my documents

[indigo ravine]

I think the identity function is a counter example for the sum ? Both terms are {0}

[upper pivot]

yeah thats true

doesnt sound true at all

[sour plume]

this was the statement i remembered

(ignore the overline, that's infinite-dimensional stuff)

But yeah, kernel of the adjoint plus image, that's a thing you can prove

[indigo ravine]

i think that's it it would be coherent with the previous questions

i will ask my teacher then ! thanks

[sour plume]

Yeah it could be that the teacher miswrote when he put the orthogonal complement over the image, who knows

At least showing that ker T^* is a subset of the orthogonal complement of T is super straightforward, yeah

you're prolly gonna need a basis argument for showing that that's everything

[oblique river]

it's also true that im(T*) is orthogonal complement of ker(T). I wonder if that's the statement that was meant to be on the exam

[sour plume]

oh yeah right it's the definiteness of the scalar product

[indigo ravine]

Isnt it really easy to prove that Ker T* = orthogonal complement of Im T?
@chilly ocean that was the previous question of the exam

yes ! that seems like a pretty straightforward consequence if that's what he meant

[half nebula]

So I have two permutations here. permutation x which can be any permutation and r which is equal to the permutation (a,b,c). I need to prove that xrx^-1=(x(a)x(b)x(c))

Oh I forgot I can just pose the question too.

[olive mirage]

where does xrx^-1 send \sigma(a)?

[half nebula]

sigma?

Oh is that the symbol that looks like an o?

[scarlet estuary]

yes

$\sigma$

[cloud walrus]

Namington:

[half nebula]

Okay.

[olive mirage]

baby sigma

[half nebula]

So I do it from right to left though right? Like when multiply permutation AB, you go from B to A. So like if you have (123)(135) you see 1 maps to 3 and then 3 maps to 1 so the first part of that permutation would be
(1
(1

So like here first I multiply sigma inverse by r.

[eager willow]

Given symmetric matrices $A$ and $B$, is it always possible to find some orthonormal matrix $Q$ such that $QAQ^TB$ is symmetric? i.e. if $C = QAQ^T$ then $C$ commutes with $B$? I ask this looking for representations of $\mathbb{Q}(\sqrt p, \sqrt q)$ for non-squares $p$ and $q$

[cloud walrus]

datorangeguy:

[eager willow]

I guess that I should also add that A,B, Q, and thus C would have to be matrices with rational entries...

[icy yarrow]

@chilly ocean

[echo olive]

Please let me know if I should be asking this in a different channel, but I have a question about notation

Only mildly familiar with Group Theory, so I looked up what Z_2 is and found it's the additive group of integers modulo n, e.g. {1,-1} under addition.

However, in this context I'd need to promote the 1 and -1 to I_4 and -I_4 where I_4 is the 4 x 4 identity matrix. Does that mean Z_2 can mean different things under different contexts? Moreover, I'd need to consider the group operation to be matrix multiplication now

E.g. it's just a cyclic group of order 2, but the elements themselves are not necessarily integers?

[woven delta]

So Z_2 here is identified with the subgroup {I,-I}

[echo olive]

👍 {I, -I} under matrix multiplication

[woven delta]

yeah

[echo olive]

Thank you! So Z_N can have different realisations depending on context?

[woven delta]

so this is a bit of an abuse of notation here

[echo olive]

since wikipedia seems to call Z the additive group for integers and Z_N additive groups for integers modulo n

ah ok, I just got a little confused since it's so hard to find anything talking about this explicitly

[woven delta]

you can kind of formalize this notation with group actions

the group Z_2 acts on SL(2,C) by sending a matrix M to -M

then you can mod out by this action, written SL(2,C)/Z_2

[echo olive]

Does Z_2 only send M to - M or also back to M? i.e. M --> M and -M?

[woven delta]

oh sorry

so there are 2 elements of Z_2, lets call them 1 and -1

[echo olive]

ok

[woven delta]

and they just act by scalar multiplication

[echo olive]

ooh wikipedia refers to Z_N as additive though

[woven delta]

lol we view it as an abstract group

so we can view it in whatever context we want

[echo olive]

ok I see

(sorry I have a physics background :P)

[woven delta]

you should just thing of Z_2 as the unique group with 2 elements

instead of in any specific context

[echo olive]

that makes me more comfortable with then using G/Z_2 in different contexts

I guess strictly the groups SO(1,3) and SL(2,C) themselves are more abstract than the 4 x 4 and 2 x 2 matrix reps but in some contexts the matrices are just said to form those groups.

In a similar way, Z_2 is more abstract than just {1, -1} and I was incorrectly seeing Z_2 as defined by {1,-1} (mostly because of the quote "Every finite cyclic group of order n is isomorphic to the additive group of Z/nZ, the integers modulo n" from wikipedia)

that helps a lot, thank you!

[kindred mist]

Is this correct?: Let R be a ring. Then any R-module is the epimorphic image of a free R-module. proof. Let P be an R-module, then let <P> denote the free R-module generated by the set P. Here I am using X to denote some basis of <P>. Here we may have X = P (and f = 1), and f^ being the natural projection epimorphism (<P> is isomorphic to R^(rank<P>) --->> P). Is it a problem that we don't know if <P> is finitely generated, even though it must be freely generated (no, right?) ?

(Yes I proved earlier that any R-module is the homomorphic image of a free module, that was trivial, this ^^^ idk if I'm using all the concepts correctly)

Should I just use the fact instead that <P> is an R-module containing the R-module P, so that P must be <P>/K for some submodule K of <P>, then just use projection <P> --->> <P>/K?

<@&286206848099549185> ^^^

[elder valley]

it seems that f^ is all you need

if P is infinite then <P> is not finitely generated

[kindred mist]

wouldn't <Z> = Z be a counterexample to that though?

[elder valley]

my understanding is that the sums in <P> are purely formal. so e.g. in that example, 2+3 and 5 are different

[upper pivot]

yeah^

[elder valley]

your map f^ collapses things down so that they become equal

f^ would map an element r_1 p_1 + ... + r_n p_n over to r_1 p_1 + ... + r_n p_n, which looks like the identity map, except that the sum in the domain is just a symbol whereas the sum in the codomain is the operation of the R module P

[kindred mist]

Ah ok that makes sense now

thank you

Are the sums formal for free objects in any category?

[elder valley]

i don't know enough about category theory to answer that

[half nebula]

On the last lap here (Until tomorrow when I get another thing of homework because summer classes suck.)

I know what permutations are in S_4 but I’m a bit unsure of how to do this. Lemme grab theorem 10 though.

I’ll be honest I just suck at notation reading.

[elder valley]

do you understand how that group action works?

[half nebula]

Group action? Like the binary operator?

[elder valley]

S_4 acting on polynomials in 4 variables

[half nebula]

I’m afraid not really no.

Sorry.

[elder valley]

for example (13) would map (x_1 - x_2)(x_3 - x_4) over to (x_3 - x_2)(x_1 - x_4). it just acts on the subscripts

[half nebula]

Oh okay so I’m looking for a permutation that keep it the same then?

[elder valley]

yeah, it wants you to find all of them

[half nebula]

Or permutations that keep it the same.

I know one is the identity element.

Could another one be just (4321)?

[elder valley]

what polynomial do you end up with when applying that

[half nebula]

Wait sorry that would just make it backwards.

Oh could another one be (13)(24)?

Because it would be the same since it just swaps both of them right?

[elder valley]

yup

[half nebula]

Okay cool!

How many are there? Or is there a way I can figure that out?

[elder valley]

there won't be that many

but here's a hint to make it slightly easier. what you're actually computing is called the stabilizer of that polynomial. and stabilizers are always subgroups

so at the very least, however many you find, it should divide |S_4| = 24

[half nebula]

So a subgroup of (12)(34)?

[elder valley]

a subgroup of S_4

[half nebula]

Ah okay.

[elder valley]

that also means that if you find 2 answers, you can compose them in S_4 to get another answer

take powers of them, or invert them

[half nebula]

Okay cool. So right now I can’t do that though since the only ones I know are the identity element and (13)(24)

[elder valley]

right. and there may not be enough of them to really need to do that. it's just for your info

the real point was that the number of them should divide 24

[half nebula]

So if I somehow find 5 I know there’s at least 1 more.

[elder valley]

exactly

[half nebula]

Just being sure I get this, the subgroups of s4 are e,(12),(23),(34),(13),(14),(24),(123),(134),(234),(321),(431),(432),(1234),(4321),(4312),and (2134) I think that’s all of them.

[elder valley]

those are the elements you are listing. are there are 24 of them

[half nebula]

So I’m missing 7,

(1324),(4231) as well.

Is (13)(24) different from (1324)?

[elder valley]

yes they are different

in the first case 3 is sent to 1, but in the second case 3 is sent to 2

[half nebula]

Okay so then there’s also (13)(24), (12)(34), (23)(14). I can’t think of what the last 2 could be.

[elder valley]

you missed 124

[half nebula]

Ah okay (124) and (421)

Okay so I should just start going through them all?

[elder valley]

it would probably be good practice. you'll probably notice some shortcuts after you get going through them

[half nebula]

Okay so how would (123) work? Like would it just map them as such so like (x2-x3)(x1-x4)?

[elder valley]

yes exactly

[half nebula]

Okay cool.

Idk if I’m doing this entirely right but so far I got that none of these outside of e are correct.

[elder valley]

you found a non identity one earlier though

[half nebula]

I know I just meant out of those.

[elder valley]

oh. yeah should be correct

[half nebula]

Okay I just went through them all. Those 2 are the only ones I got.

(13)(24) and e.

[elder valley]

there are 2 others. they are a little subtle

[half nebula]

2 that I listed?

[elder valley]

the 3 cycles and the 4 cycles don't work

a 2 cycle also doesn't work

[half nebula]

So it has to be two more of the dual 2 cycles like (13)(24) was?

[elder valley]

yeah check them again

[half nebula]

(12)(34) would make it (x2-x1)(x4-x3)

And (23)(14) would make it (x4-x3)(x2-x1)

So it can’t be either of those right?

[elder valley]

you did it right. but these are polynomials, so you are checking if the answer you get is equal to what you started with as polynomials

[half nebula]

Oooooooh.

So foil?

[elder valley]

that would be one way. but manipulate through algebra, yeah

[half nebula]

Ah okay.

Okay great all done with a.

How would the second one work? B.

I can post it again so we don’t have to scroll up.

[elder valley]

it's just a different polynomial

that polynomial is (x_1+x_2)(x_1 + x_3)(x_1+x_4)(x_2+x_3)(x_2+x_4)(x_3+x_4)

[half nebula]

Sorry but could you help me with this one? I know how to do it but I have to turn this in in 15 minutes.

[elder valley]

there's probably some shortcut you can use, like applying a theorem or making some observation. but i think every permutation works there

it's a symmetric polynomial i believe

[golden pasture]

hm i wonder if there is a neat way with determinants

[half nebula]

So for these first few questions, all I have to do is just tell if they're true or false but I'm a bit weary since he's marked it as "extra credit" for our upcoming exam next week. That is to say that we don't actually have to do these but if we do, extra points. But I also feel like because of that some of these are going to be really stupid trick questions, like the answer is super obvious and I'm going to end up over thinking it.

Like for that first one. It really really feels like that isn't right because I remember the one on the right was like the LCM of the two on the left but again, I might just be stupid rn.

[upper pivot]

Well try disprovingit

[half nebula]

Well like the first one that popped into my head was p=3

But like the lcm(3,3) is still 3. So it is also isomorphic to Z_3 right?

[upper pivot]

No? Z3xZ3 has order 9 while Z3 has order3

[shy bluff]

Is your notation of Z_p the same as Z/pZ?

[golden pasture]

yea people use Z_n=Z/nZ ;-;

[shy bluff]

Yea then it's true

[golden pasture]

uhh

[half nebula]

Okay that's what I was thinking. Idk where I was getting the lcm thing from.

[upper pivot]

Lol i dont like it either

It isnt

[shy bluff]

I get it if you're talking on discord but on homework

[upper pivot]

Look at orders of element

In z3xz3

And z9

[shy bluff]

In any case, yes that is true, in fact I had that question on my assiggnment this week lol

[golden pasture]

it isnt lol maybe yours is a slightly different statement

[shy bluff]

Oh ok

[half nebula]

So it isn't true then right? <->

XD sorry I'm a bit tired still

[upper pivot]

Try what i suggested

[half nebula]

Well z3xz3 is all combinations of (1-3,1-3) while z9 is just 1-9 night?

[upper pivot]

Umm this isnt a good way to think of this tbh

[half nebula]

Oh sorry ><

[upper pivot]

The numbers dont really mean anything. Its the operation thats important

[half nebula]

Oh is it that one creates a set of pairs while the other creates a set of just numbers?

[upper pivot]

Umm im not sure what you mean

Gimme a second ill get on my laptop

[half nebula]

Okay, if it makes it any easier btw, I can get into voice. I know it's easier to talk than type (at least for me it is <->)

[upper pivot]

alright back

[half nebula]

Coolio

[upper pivot]

hmm so tell me what Z3xZ3 is, and its operation

[half nebula]

The operation pairs the two numbers.

So for example if you had z2xz2 the set would be (1,1),(1,0),(0,1),(0,0)

Or wait no.

[upper pivot]

well we usually like calling 2=0

but that doesnt matter

[half nebula]

Yeah

[upper pivot]

(because we care about the classes and class of 2 is the same as that of 0)

now what is the binary operation

[half nebula]

Is it not pairing them? ><

Like I know what a binary operation is.

[upper pivot]

well every group must have a binary operation right

[half nebula]

Yeah

[upper pivot]

well ok so ig let me make it clear

lets say we have $G, G'$ two groups with operations $\cdot_G, \cdot_{G'}$ then $G\times G'$ has the underlying set be the usual product of sets and operation $(g,g') \cdot_{G\times G'} (h,h') = (g \cdot_G h, g' \cdot_{G'} h')$

[cloud walrus]

JohnDoeSmith:

[upper pivot]

i.e you do the operation component wise

if this makes sense

[half nebula]

Kinda. I think I'd understand it better with an example. Like Z2xZ2. How is it doing that there?

[upper pivot]

so operation for Z2 is addition module 2 right

[half nebula]

Yes

[upper pivot]

so like 1+1=0, 0+1=1,0+0=0

so then you would have in Z2xZ2

(1,0)+(0,1)= (1,1)

and like (1,1)+(1,0)=(0,1)

[half nebula]

Okay so you're just doing that with all the combinations?

[upper pivot]

yeah that is the group operation

[half nebula]

Okay cool.

[upper pivot]

so do you know what the order of a group and element is

[half nebula]

Yeah. Its n where a^n=e

at least for an element.

[upper pivot]

yeah thats order of an element

[half nebula]

For a group I think it's the number of elements within it.

[upper pivot]

order of a group is simply its carinality

yep

[half nebula]

Yeah

[upper pivot]

so if two groups are the same, they must have elements of the same order right

[half nebula]

Yeah I remember that was some sort of theorem.

[upper pivot]

(also its smallest n such that a^n=e just incase you didnt mean that)

well actually try proving it

precisely that the image of an element under an isomorphism has to have the same order

[half nebula]

well (0,0) is the identity in Z2xZ2

then (1,0) has an order of 2, (same with (0,1) and (1,1)

[upper pivot]

yep

[half nebula]

So 1 element of order 1, 3 elements of order 2

[upper pivot]

what about Z4

[half nebula]

While z4 though has 3 which has an order of 4. 3,6,9,12 right?

[upper pivot]

yeah 3 has order 4

[half nebula]

or 3,2,1,0

[upper pivot]

(so does 1)

[half nebula]

Well there you go, they can't be isomorphic then right?

[upper pivot]

mhm

[half nebula]

Okay perfect!

[upper pivot]

do you see how this exact logic would extend to all p

[half nebula]

Well yeah since zpxzp will only have an order up to p but p^2 can have an order up to, well, p^2

[upper pivot]

yup

[half nebula]

Okay cool!

Want me to post it again for the other 3?

[upper pivot]

sure

[half nebula]

I'm a bit confused as to what 13Z means.

[upper pivot]

its just ${13 n: n\in \bZ}$

[cloud walrus]

JohnDoeSmith:

[half nebula]

Ah okay.

[upper pivot]

and operation is addition ofcourse

[half nebula]

Well then it's already wrong because the first set has 13 and the other set has 17 right?

They do both have infinite order I believe though.

[upper pivot]

well we said isomorphic

they dont need to have the same elements

[half nebula]

Just the same order?

Or that is to say, elements of the same order.

[upper pivot]

thats not sufficient for an isomorphism usually i think

but umm do you know what an isomorphism is

[half nebula]

It's a homomorphism that's also onto and one-to-one.

[upper pivot]

yeah

can you think of a homo like that between the two groups

[half nebula]

Just being sure, a homo is where phi(m)+phi(n) is equal to phi(m+n) or something like that?

Sorry it's been a bit,

[upper pivot]

yeah

[half nebula]

Okay well that would be like phi(17)+phi(13) has to equal phi(30)

But how do I know what phi is doing?

[upper pivot]

17 is not in both sets

[half nebula]

Oh it has to be an element in both sets.

[upper pivot]

well let me guide you along a bit

take $f:13\bZ\longrightarrow 17\bZ$

[cloud walrus]

JohnDoeSmith:

[upper pivot]

can you first think of just a bijection

that does this

[half nebula]

Isn't bijection like the onto and one to one part?

[upper pivot]

yeah

[half nebula]

Well onto is like phi(phi(x)) is equal to x right?

and one to one is if phi(x)=phi(y) then x=y

[upper pivot]

umm one-to-one (aka injection) is that f(x)=f(y) => x=y and onto (aka surjection) is that everything in the codomain is in the image of f

[half nebula]

image?

[upper pivot]

well the image of f, say f(G) is everything of the form f(g) for g in G

that means everything in the codomain is of the form f(g) for some g

[half nebula]

Oh okay I think I remember that.

Well there's 0, the identity element since n can equal 0.

Right?

[upper pivot]

yeah

[half nebula]

Okay so do I need to somehow prove it for all n?

Or if I can show 1 of them is that good enough?

[upper pivot]

umm what do you mean

i am asking you to find such a map

[half nebula]

Like f(0)=0

Sorry if that not what a map is?

[upper pivot]

yeah that tells you what f(0) is

but you need to know it for all values of 13Z

[half nebula]

So like a formula or something? Like f is taking 17n-4n to get 13n?

[upper pivot]

sure, just tell me how f maps each 13n to an element of 17Z

[half nebula]

Is it that like you add 4n in that case? so if n =2 then 13n=26, +4n is +8 which would be 34 which is 17n.

[upper pivot]

i mean yeah sure that is the map, just more concisely say f(13n)=17n which is what i assume you mean?

[half nebula]

Yeah

[golden pasture]

(another way to show instead of checking injective and surjective is there exists some f:G->H and g:G->H such that fg is the identity on H and gf is identity on G)

[upper pivot]

show this is a homomorphism now

and then show bijection by either what i said or what ari said

[half nebula]

Well for homomorphism I need elements n and m that are in both groups right?

[golden pasture]

(try both ways it's good practice)

[half nebula]

Okay. First for homomorphism though, like is that right?

[upper pivot]

oh umm homo means f(a+b)=f(a)+f(b) for a,b in the domain group

[half nebula]

Okay.

Can they both be the same element?

Also does a+b have to be in the domain?

[upper pivot]

a+b by definition has to be in the domain since G is closed under the operation (+ in this case)

and yes they can be the same

[half nebula]

That makes sense.

Well yeah it's the same because that would just be f(13n+13n)=f(13n)+f(13n)

So that would be f(26n)=34n which is on the right side as well.

So it's an isomorphism.

[upper pivot]

umm show f(13n+13m)=f(13n)+f(13m) to show its a homo

[half nebula]

Well that would be f(26n) on the left which is 34n.

On the right f(13n)=17n and 17n+17n is also 34n.

[upper pivot]

n and m being disctinct

you need to show that f(a+b)=f(a)+f(b) for all pairs a,b \in 13Z

[half nebula]

well yeah that covers them all because 13n is the whole set right? So no matter what you put in for n. It works.

[upper pivot]

13n+13n isnt all possible combos

a and b can be distinct

[half nebula]

Oh okay.

Don't worry I blocked him.

Well wait isn't 13Z 13 n for all n in Z?

[upper pivot]

yeah

[half nebula]

and f is 13n to 17n right?

[upper pivot]

mhm

[half nebula]

I'm sorry I just can't get this. What could I put in for all 13n for both a and b?

Other than 13n

[upper pivot]

so all elements are of the form 13n