groups-rings-fields

[cloud walrus]

yeshua

[inner steppe]

Can someone explain with words what the last point is saying. I thought I understood it but I keep using it in wrong places so probably don't.

What I think it means is

for every d that divides the group order |G| = n, then the number of elements in G whose order is d is the the Euler function phi(d)

If I am asked "How many generators does the multiplicative group in Z2[x]/(x3 + x + 1) has? Is it correct to solve it using this theorem considering the fact that all multiplicative groups of finite fields are cyclic thus |G| = 2^3-1=7, the answer is phi(7) = 6. Is this correct reasoning?

[inner steppe]

Also aren't ii) and iii) saying basically the same thing, if d is a divisor of n then the number of elements in G whose order is d is

but how come phi(d) for iii) and d for ii)

[chilly ocean]

ii), in that set element has not necessarily order d

[chilly ocean]

Multiplicative group in Z_2[x]/ (x^3 + x + 1 ), how it can be multiplicative group?

[inner steppe]

oh shit, I keep forgetting that d has to be the minimum number, thanks

[inner steppe]

but I have to ask then what exactly is ii) saying in words

what is the point?

[chilly ocean]

You can interpret in the way as the number of zeroes x^d - 1 is exactly d when d | n.

It is an interesting question, when G is a finite group and for every m | |G|, there are at most m solutions of x^m-1 = 0 then G is cyclic.

[chilly ocean]

iii) part helps you to prove ii.

[golden turtle]

I am wondering if I did this correctly

I thought that the system would have to arise because the different e_i tensor e_j would be like basis elements in the tensor product space?

But I'm not sure how correct my intuition is

[serene dune]

$\varphi(f(x)) = f(x)

\ \

\varphi(f(x)) = f(-x)$

[cloud walrus]

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

[serene dune]

are these the only possible elements of Aut(Z[x])

[chilly ocean]

You mean ring automorphism or group?

[serene dune]

ring

[inner steppe]

I have been meaning to ask, shouldn't this be x ∈ {1, 2, . . . , (n-1)}?

[chilly ocean]

Including n doesn't change anything

[rocky cloak]

I guess it would cause you problems if n=1

[chilly ocean]

Yes that's what I meant

[rocky cloak]

Well, I'm saying it does change something when n=1. In that in that one case you would get the wrong answer

Tbf, I'm not sure whether phi(1) = 1 or phi(1)=0 is more appropriate

[chilly ocean]

Let A and B be finite groups and let p be a prime. Prove that any Sylow p-subgroup of A × B is of the form P×Q, where P is Sylow p-subgroup of A and Q is Sylow p-subgroup of B.

I don't see how I can prove that ?

Guess: do some homomorphism construction

Any hints would be appreciated

[grave sedge]

There is zero reason for it to be 0

Any single thing involving phi i can think of needs phi(1)=1

[grave sedge]

P x Q gives some p-subgroups, now use the fact that the others are all conjugate to this one

[chilly ocean]

Yes P ×{1} will be p-subgroups

[grave sedge]

No, i mean P x Q are sylow p-subgroups

(if P is sylow in A and Q in B)

[grave sedge]

Because it has the right order

[chilly ocean]

So first we are proving that when P is Sylow p-subgroup and Q is Sylow p-subgroup then P × Q is Sylow p-subgroup

[grave sedge]

Yes

And then use the fact that any other sylow subgroup of AxB is conjugate to a "fixed" PxQ

[chilly ocean]

We can argue on cardinality

[chilly ocean]

Yes, Sylow theorem

Thus others are gPg^-1 × hQh^-1.
And gPg^-1 is a Sylow p-subgroup of A and hQh^-1 is a Sylow p-subgroup of B

[grave sedge]

Indeed

[chilly ocean]

Thank you @grave sedge

[wild jasper]

What is the "inital term" in (a)? It is not defined...

[thorn jay]

Consider the endomorphism generated by
f(1) = 1
f(x) = x + 1

sends p(x) to p(x + 1)

This is an automorphism

[serene dune]

hmm so translation by a constant too

alr

[thorn jay]

Ye but i think those are the only ones lol

[dull ginkgo]

Do you know how to prove this?

Or anything of the like

[serene dune]

yeah i have

thanks

i was just checking if there are any other shi

[thorn jay]

Z[x] / (p) ~ Z[x] / (f(p))
For an automorphism f, as f fixes Z

Z[x] / (f(x)) must then be isomorphic to Z

That means Z[x] / (f(x)) is generated by 1, which means that x = n for some integer n => x - n in (f(x)) => f(x) = x - n or f(x) = -x + n

Aut(Z[x]) as a group under composition is therefore isomorphic to Z x C2

[topaz solar]

Isn’t x -> -x one?

[thorn jay]

Yes

Wait

Does this look good?

[mighty kiln]

No

[thorn jay]

Aw :(

[inner steppe]

but it does, it increases the size of phi(n) by one

[mighty kiln]

That's not the group operation

[inner steppe]

True but still not but does phi(1) defined?

[thorn jay]

Oh let me change the wording

[topaz solar]

I don’t think that composition on those are commutative?

[mighty kiln]

It should be the ||Infinite dihedral group D_2∞||

[thorn jay]

Aw

Oh no ofc not

Z semdirprod C2 then

[topaz solar]

You’d wanna make sure you didn’t miss any automorphisms

But that group you gave is indeed Z \rtimes C_2 with C_2 acting by negation

[thorn jay]

All automorphisms fix 1 and send x to either x + n or -x + n for some n

[topaz solar]

I’d want more details on the Z[x]/(f(x)) \cong Z part too, but that’s just me

[topaz solar]

Well I meant to say show those are the only choices, but yes

[thorn jay]

Z[x] / (p) ~ Z[x] / (f(x))
For any automorphism f, any polynomial p.
This is because
Z[x]/(p) ~ f(Z[x])/(f(p)) = Z[x] / (f(p))
I suppose you'd have to prove that, shouldn't be too hard
Then
Z ~ Z[x] / (x) ~ Z[x]/(f(x))
for any automorphism f

[topaz solar]

Ye

[thorn jay]

But that means Z[x]/(f(x)) is generated by 1 + (f(x))
Which means that
x + (f(x)) = n + (f(x))
For some n, and therefore
x - n = f(x) * q
Now as multiplying polynomials adds degrees, this must mean deg f(x) = 1 or deg f(x) = 0
deg f(x) = 0 makes it impossible for f to be an automorphism, so f(x) = ax + b
But the leading coefficient of x - n is 1, so a has to be either 1 or -1, giving the desired result

[topaz solar]

Ye

[dull ginkgo]

Any linear substitution x |-> ax + b where a is a unit is an automorphism

[thorn jay]

Yee

Just happens only 1 and -1 are units

In Z

[dull ginkgo]

With inverse x |-> a^-1(x - b)

[topaz solar]

it was a question to be like “don’t forget this one”

[thorn jay]

:P i work too fast too often

[dull ginkgo]

Maybe every automorphism of R[X] is of a substitution x -> \alpha(x) + b or smth

Plus one over the coeffs

[thorn jay]

Whats alpha here?

[dull ginkgo]

Nevermind

(It’s incorrect)

[chilly ocean]

No, you can verify it. Since gcd(n,n) ≠ 1 when n ≠ 1

And Jagr said you can also find phi(1) = 1 by this definition but if you exclude n then for n =1, what would be the value of phi(1)?

The author defines the rank of free abelian group.
Question: is it related to somehow the rank of matrices ?
So when we said G is a free abelian group of rank r, does it imply G = Z × Z ×..×Z ?

[vapid vale]

yes, isomorphic to Z^r

[bitter rover]

If G is finitely generated, yes.

[chilly ocean]

Any intuitive thoughts on the free abelian group?

[bitter rover]

Every abelian group is a Z-module.

[chilly ocean]

Yes

[vapid vale]

what do you mean by intuitive thoughts

[chilly ocean]

But at this moment, I don't want to use module theory. I am interested to know about the invariant factors of group G

[rocky cloak]

Does that not qualify as module theory?

[chilly ocean]

I don't get it, free abelian means? Is it related to free group stuff?

[vapid vale]

yes

wait

no

[thorn jay]

It is related, but not the same

[rocky cloak]

It satisfies the same property as free group of you just replace "group" with "abelian group" throughout

[chilly ocean]

I am reading Dummit and foote's chapter 5 but module theory is introduced in chapter 10

[vapid vale]

are you familiar with how bases work in linear algebra

[chilly ocean]

Yes

[thorn jay]

Abelian groups should really be thought of as modules, in the end

But you can see the free group on n generators as the product of n copies of Z
(This only works for finite n)

[chilly ocean]

No idea about free groups

[vapid vale]

its the same thing then. but your scalars are integers. and for an integer n and a basis element b in your free ab gp, nb is b*b...b n times

[bitter rover]

I feel like people are taking turns typing at Baire's keyboard or something.

[rocky cloak]

An abelian group (or module) is free iff it has a basis

[chilly ocean]

Is it just the definition or why do we consider the copies of Z?

[chilly ocean]

Yes

[vapid vale]

do you agree that if you have an R-vector space with n basis vectors, we can write it as isomorphic to RxRx...xR

[chilly ocean]

Yes

[bitter rover]

There's a bijective correspondence between Z-modules and abelian groups, and between free Z-modules and free abelian groups.

"Invariant factors" are a statement most straightforwardly made in the language of modules.

[chilly ocean]

Should I do module theory now ?

[bitter rover]

Are you working through a textbook?

[chilly ocean]

Yes Dummit and foote

[vapid vale]

so your free abelian group is given by some basis set. if an integer k acts on it, you multiply that basis by itself k times, as i explained before. then, like in linear algebra, you can express each element of the group as a sum of k_ib_i

[bitter rover]

They'll get to it.

[chilly ocean]

Yes

[rocky cloak]

I would rather say that if you're studying invariant factors you are doing module theory now.

But you don't need to do anything differently than what you're doing

[bitter rover]

Dummit and Foote probably introduces the concept of "invariant factors" in the context of abelian groups first

[chilly ocean]

Yes

[bitter rover]

Then later when you get to rings and modules, they'll go, "BTW, this connects back to the concept for abelian groups because every abelian group is a Z module."

[chilly ocean]

I know basic stuff about module

[chilly ocean]

Okay

[vapid vale]

so think of ..., b^-2, b^-1, e, b^1, b^2, etc as a long strip of powers of b indexed by Z. if you have n of these strips for n basis elements, that is some intuition for why it is isomorphic to Z^n

you already know sufficient "module theory" to understand this stuff

[chilly ocean]

Sorry, I don't get it this one

[vapid vale]

this should be very similar to how you construct R^n in linear algebra. you have a bunch of 1-dimensional subspaces generated by the basis elements

[chilly ocean]

Yes

[thorn jay]

What are invariant factors?

Things that stay the same under isomorphism?

[vapid vale]

so think of the corresponding 1-dimensional subspaces here (even though this is not a vector space, as Z is not a field) as some basis element b and all of its powers

[chilly ocean]

So we are treating a set of power of b as subspace?

Intuitively

Subspace generated by b

[vapid vale]

well again, think back to vector spaces. the span of a basis elemnet is every coefficient in the base field times that basis element

[chilly ocean]

Yes

[vapid vale]

here the "coefficients" are just Z

[chilly ocean]

So that's why we are taking powers

n•a -> a^n

[vapid vale]

well if you think of the group operation as addition, it really is just multiplication by Z

[bitter rover]

If you have a PID R then you can decompose any finitely generated module over R into direct sums of R and quotients of R.

Because R is a PID, those quotients look like R/(a_i) for some a. The list of a_i are unique for a given module.

[chilly ocean]

Z acts on G ?

[vapid vale]

yes

[thorn jay]

Ohhh okay thanks :3

[vapid vale]

and an abelian group with a ring acting on it is precisely what a moduleis

[chilly ocean]

Yes

Thank you hk

[inner steppe]

x+1 wasn't hard to find but how in the hell is x^3+2x+1 a monic polynomial?

[void cosmos]

a monic polynomial is a polynomial with leading coefficint 1

a_n = 1

[tardy hedge]

Struggling with finding example of a torsion R-module with zero annihilator

[thorn jay]

The annihilator is an ideal of R right?

[tardy hedge]

Yea

[thorn jay]

Just take a torsion module T over R, then take the induced module T' over R/Ann_R(M)

So you have the scalar multiplication defined as:
(r + Ann_R(M)) * m = r * m

You can check for yourself this is well-defined

[tardy hedge]

T’ over R/Ann(M) is a torsion module?

[thorn jay]

Oh, i didnt think of that

Shdhsh

Maayybee??

a module M over a ring R is equivalent to a ring homomorphism
h : R -> End_Z(M)
The endomorphism ring of the underlying abelian group
Ann_R(M) is just ker h, so a module with annihilator 0 is just a module M over any subring of End_Z(M)

[tardy hedge]

Hm. Yea

[thorn jay]

Not super helpful..

[tardy hedge]

Cool tho

Lol

[thorn jay]

Yeah!

Ring representations

I stumbled upon them while doing some stuff with pairs of MOLS

[thorn jay]

Not generally >.< ughh

[rocky cloak]

A hint is that the annihilator of a direct sum is the intersection of their annihilators

[thorn jay]

Ohhh and the direct sum of torsion modules over a commutative ring is a torsion module

[rocky cloak]

At least over an integral domain. Might depend on how you define torsion otherwise

[thorn jay]

||dirsum_p prime Z/pZ over Z||

[thorn jay]

:3

Yayyy

[tardy hedge]

Yea my friend went to office hours and prof said think about infinite direct sums

[thorn jay]

I've worked with an analogous module over Z[x] a while ago, interesting object

[bitter rover]

A finitely generated torsion module over an integral domain has non-zero annihilator, so you'll have to look at something that isn't finitely generated.

You can find such a module over ℤ.

Make sure you write down what the definition of an infinite direct sum is, though.

Or rather, what elements of such an infinite direct sum look like.

[chilly radish]

Sorry for the ghost pings I keep sending before fully verifying

[chilly ocean]

Question: Let A be a finite Abelian group.
A^p = { a^p | a in G } and A_p = { a in G | a^p = 1 }. Show that G/A^p isomorphic to A_p.

Observation: G/A_p is isomorphic to A^p

Any hint would be appreciated

[rustic crown]

is G = A?

[rocky cloak]

Finite abelian groups are products of cyclic groups, so it's enough to show for the them. Or even for cyclic groups of order q^m for prime q

[chilly ocean]

You mean first I have to show for when |A | = q^m, for some prime number q

[chilly ocean]

Yes

[rocky cloak]

Yes, and A cyclic.

The cyclic part is more important really

[chilly ocean]

Yes then we can find the order of A_p, because the given hint is to show that order of G/A^p is same as the order of A_p, and both are Elementary abelian group

A_p gives the order p

But since G/A_p is isomorphic to A^p, shows that G/A^p has the order p

Hence both are isomorphic

I assumed |G | = p^m, for some m

Correct? Jagr

[rustic crown]

also need to handle the case where |G| = q^m where q is a prime different from p

[chilly ocean]

But then A_p = {1}

[rustic crown]

after that you just need to check if the statements holds for two finite abelian groups, then it also holds for their product

[chilly ocean]

But, let G be an abelian group and H and K are subgroups of G.

If G/H is isomorphic to K, does it implies G/K is isomorphic to H?

If not, then let me think.

[rustic crown]

nup

[stone vault]

This is not true in general

[chilly ocean]

Now let me think

[inner steppe]

I am not really sure what they mean but the facotrization in Z_3 is unique?

[stone vault]

Are you familiar with the notion of a UFD

[inner steppe]

depends on what UFD is?

[stone vault]

Unique Factorization Domain

[inner steppe]

I am gonna say no

[stone vault]

UFDs are integral domains where non-zero elements have factorizations into irreducibles that are unique up to units and ordering

[rustic crown]

Z_3[x] :3

(they're unique in fields too, but that's not what they wrote ><)

[stone vault]

Basically, this means that up to unit factors, if we have an element of our UFD, there is some unique list (up to shuffling) p_1,…p_n of irreducibles such that the product of the p_i’s is the element

F[x] is a UFD for a field F

[inner steppe]

okay so like

numbers can be factorized into primes

like for example 30 = 2x2x3x5 no matter how we factorize it, it will always boil down to these

[inner steppe]

If I am not mistaken then what does Z being a field has anything to do with it?

[stone vault]

Yes it’s a very similar idea

[stone vault]

Well R[x] is not always a UFD

[rustic crown]

(Z isn't a field, we were talking about Z_3 there)

[inner steppe]

I meant Z_3

[rustic crown]

field[x] is a ufd

[inner steppe]

I guess you mean the polynomial ring by R[x]?

[stone vault]

Yes, where R there is any ring

[inner steppe]

But isn't every field is a ring?

[rustic crown]

and fields are important because any non-zero polynomial after multiplying with a unit has leading coefficient 1, so you can perform division with remainder. then essentially the same way one proves Z is a UFD, we get that field[x] is a ufd

[stone vault]

But not every ring is a field

[inner steppe]

just out of curiosity why wouldn't it work if Z was just a ring?

Z_3*

[stone vault]

Well if it were itself a UFD then it would still work too

But R[x] is not a UFD if R is not

[inner steppe]

okay so in the solution what exactly do they mean by "Therefore, the common factors are exactly the polynomials that are factors of the greatest common divisor of gcd of f and g"

[stone vault]

GCDs must exist in UFDs

The jargony version is “UFDs are GCD domains”

[golden turtle]

I got a hint on how to solve this problem but I guess I'm not understanding the idea

I was suggested to use the projection map pi from Zn to Zn-1 just by killing off the last coordinate

and then for some submodule M of Zn to show that

M is isomorphic to ker(pi restricted to M) direct sum pi(M)

and I guess I'm just confused about like, say I did show this, what does it tell me about the rank of M?

Would its rank be equal to rank of ker(pi restricted to M) + rank of pi(M)

[inner steppe]

no but like if we ignore group theory for a second and work with numbers

What are they saying exactly lets say f is 20 and g is 30

the common factors are exactly the "NUMBERS" that are factors of the greatest common divisor of f and g

[rustic crown]

yepsie :3

[stone vault]

Yes this is by definition of the gcd

[rustic crown]

so you reduced the problem from n, to n-1 and 1.

[golden turtle]

I'm not sure why this is necessarily true, is it supposed to be obvious

I mean in like a linear algebra setting it is obvious to me

[rustic crown]

depends on your definition of rank, but it follows reasonably easily from that. i mean the behavior of rank w.r.t. direct sum

[inner steppe]

Oh wait I think I get it now, I have been reading the question wrong lol I never realised what they meant by "factors of the greatest common divisor of f and g" until now, feel stupid

thanks

btw

[topaz solar]

Why tf is n a subscript

[topaz solar]

It’s a direct sum of those so it sure should be

[golden turtle]

typoooo

I'm a bit lost too on how to start showing that isomorphism

like because I don't know what M is so I don't know how to work with ker(pi restricted to M)

[topaz solar]

Start with the base case

Is it true for n=1

[golden turtle]

any submodule of Z is free of rank at most 1

well

Z is generated by like one thing a, so any submodule is either generated by a or doesnt exist? and if its generated by a then its just Z and has rank 1

does that work

[topaz solar]

2Z is not generated by 1

But it sure exists

[golden turtle]

right okay

[topaz solar]

Anyhow, any subgroup of Z is either 1 or iso to Z yes?

[golden turtle]

yes

[topaz solar]

What about we look at Z^2

How does your suggested method handle, say, Z•(1,1)

[golden turtle]

you can have subgroups isomorphic to 1, Z, and Z^2 now

[rustic crown]

either 0 ><

[golden turtle]

I'm not sure what this notation is

Z*(1,1)

[topaz solar]

That is, if I pick some M < Z^2, how do you know this

[rustic crown]

it's the subgroup of Z⊕Z generated by (1,1)

[golden turtle]

Oh I think we use <1,1> for that

[golden turtle]

this might be stupid but does Lagranges theorem apply

[topaz solar]

how

[golden turtle]

well modules are abelian groups with like some extra stuff

so like |M| | 2

if it is applicable

[topaz solar]

It’s infinite and no

[golden turtle]

🤦‍♂️

[topaz solar]

0 is for empty

[rustic crown]

nu ><

[topaz solar]

:3

[golden turtle]

if you pick one, I'm not sure. But I know like you can just take the subgroup with a 0 in the first (or second) coordinate and get one isomorphic to Z, and then you can take just the identity subgroup to get one isomorphic to <e> and taking Z^2 itself as a subgroup will give you one isomorphic to Z^2

but

maybe it is from the hint

we can show M isomorphic to ker(pi|M) direct sum pi(M)

possibly

[topaz solar]

Try it on Z^2, what’s pi applied to (1,1)Z

[golden turtle]

Z? Because Z/<1,1> will be everything besides elements like (a,a) and so we can have any (a,b) with a!=b and so once we apply pi we just kill of the b term and we get elements (a,0) for any a

[topaz solar]

Why are you quotienting

[golden turtle]

I thought that's what the notation meant

[golden turtle]

nvm I see

its not that

pi applied to (1,1)Z then would just be isomorphic to Z

it'd be the subgroup of Z oplus Z generated by (1,0)

[topaz solar]

So the image has rank one, what about the kernel

[golden turtle]

it'd be the subgroup of Z oplus Z generated by (0,1), also rank one

[topaz solar]

Your direct sum clearly don’t work then

This is generated by 1 point so it’s rank 1

[topaz solar]

This is wrong also

It’s a subgroup of Z for one so

[golden turtle]

oh right pi maps into Z

pi( (1,1)Z)= Z

[golden turtle]

I'm not sure why, didn't we just show that Z^2 is isomorphic to the direct sum of 2 things with rank 1

isn't that what we wanted

well, for the image of pi((1,1)Z ) we have to append an extra coordinate

[topaz solar]

<(1,1)> is a rank 1 submodule mf

[golden turtle]

I'm such a bot

[thorn jay]

beep boop
being a bot is boring, try being a protogen instead

[golden turtle]

Ok. We're looking at Z^2. The submodule is (1,1)Z the submodule of Z oplus Z generated by (1,1)
If we apply pi to this, we get any element of Z
So pi((1,1)Z) =Z
If we look at the kernel, it is elements of (1,1)Z with 0 first coordinate, so the subgroup of Z oplus Z generated by (0,1) is the kernel

My claim was that M iso Ker(pi restricted to M) direct sum pi(M)
In this case that'd be saying
(1,1)Z iso Z direct sum Z
but then (1,1)Z has rank 2 which isn't true

I see now I think

I'm not sure how to approach it then

[topaz solar]

Right, so clearly something is wrong

So, I don’t think adding that kernel works

It’s not even a submodule of M after all

(Hint)

[golden turtle]

Are you suggesting that I make it into a sub module somehow, or that I try something else?

[topaz solar]

Yes

Rank nullity needs a submodule so idk

Morally anyway since you’re kinda proving some aspects

[golden turtle]

Wait

Isn’t the kernel restricted to <1,1>Z trivial

[topaz solar]

Yeah

[golden turtle]

So what’s the issue then

Maybe I didn’t just say it clear enough

But

It should be M iso to ker(Pi restricted to M) direct sum pi(M)

And append a zero onto pi(M)

[topaz solar]

You didn’t restrict that kernel

[golden turtle]

I think if we do then it works

[golden turtle]

I meant to from this

[topaz solar]

Try and verify that M is iso to that sum

[golden turtle]

In the case that we just did it is

But in the general case I’m not sure

[topaz solar]

well, consider if it is

[toxic zephyr]

if a nontrivial subset of a group is closed under conjugation by the whole group then is it necessarily a subgroup?

[tardy hedge]

to show f nilpotent in R[x] u can show a_n is nilpotent quickly and then argue f-a_n x^n is nilpotent over and over to show all other coefficients must be nilpotent right

[toxic zephyr]

trying to verify if the subset of all products of exactly two disjoint transpositions is normal in A4 and not Ai for i>4.

[rocky cloak]

Have you tried to look at some small groups?

Say Z/2 to take the smallest possible example

[thorn jay]

Conjugacy classes

Such a set must be a union of conjugacy classes

But all conjugacy classes except { 1_G } do not contain the identity and therefore can never be a subgroup

[south patrol]

(union of conjugacy classes + subgroup) = (normal subgroup)

[proven mango]

I had a question regarding left and right cosets for this problem. How would I approach this

[thorn jay]

Now suppose two elements g, g' in N_G(H)
Then look at gg'H
Can you show that
gg'H = Hgg'

You need to use the fact that gH = Hg and g'H = Hg'

[proven mango]

Maybe I'm not seeing it but why are we trying to get gg' = Hgg'. Wouldn't gHg'=gHg' work too?

[thorn jay]

No

Because gHg' = gHg' is true for all g, g' in H

[thorn jay]

We're trying to get
gg'H = Hgg'
to show gg' in N_G(H)

[proven mango]

So since gH=Hg and g'H=Hg', we have gg'H = g(g'H) = g(Hg') = (gH)g' = Hgg'

[thorn jay]

Yes

So gg' in N_G(H)

Clearly the identity is in N_G(H), as
1_G * H = H = H * 1_G

I'll leave showing inverses up to you

[serene dune]

for subgroup test i have always used...

for every x,y \in H, xy^- is \in H

[thorn jay]

I guess

[lethal cipher]

Okay, so some how I need to show that every set in SL_2(F) can be generated with matrices of that form.

Intuitively, I see why: these are the only row elementary matrices that preserve the determinant...but I am having a hard time actually showing this is true...

[thorn jay]

g'g^-H = g'g^-Hgg^-
= g'g^-gHg^-
= g'Hg^-
= Hg'g^-

[lethal cipher]

I'm also kinda just having a hard time figuring out how to start this...because just toying with these matrices seems to be pretty difficult

[thorn jay]

Toy around with powers of these matrices

See if you can find some structure in them

[lethal cipher]

These form individual subgroups, and powers just add the off diagonal entries by a that many times

I am familiar with that.

That at least let's us simplify to any matrix being a product of upper*lower*upper*lower*... (I do believe you only need 4 though)

[serene dune]

$\begin{pmatrix}
n & * \
0 & 1/n
\end{pmatrix}$

[cloud walrus]

yeshua

[serene dune]

in F every nonzero n has inverse(multi) soo what about these matrices ?

[lethal cipher]

those aren't of the desired form though

[serene dune]

oh i thought they asked to show those form only can generate

mb

[lethal cipher]

Yea, for part (1) we want to show that any element in SL_2(F) (det=1) can be written as a product of strictly upper/triangular matrices

[serene dune]

i think u need upper/diagonal/lower

[proven mango]

I just wanted to ask a question regarding Abstract Algebra courses. Is the undergraduate course for Abstarct Algebra I supposed to just cover groups and not rings and fields? If so, would the sequential course Abstract Algebra II explore rings and fields?

[serene dune]

depends ig

[vapid vale]

there is no standard for this sort of thing. ask the professor teaching the course

[proven mango]

Yeah was just curious about this but really depends on the textbook/resources/professor in how the topics are arranged.

[vapid vale]

not even the tetxbook

there are many standard introductory textbooks on abstract algebra and they will cover far more material than your usual semester-long introductory course

you could use lang or artin or dummit and foote but could end up with extremely similar or extremely different courses depending on what you pick out of it

[serene dune]

there exists no such map like $S_5 \twoheadrightarrow \mathbb{Z}_5$

any hint

[chilly ocean]

Which kind of map?

Always exists trivial mapping

[cloud walrus]

yeshua

[serene dune]

surjective homom

notice the double arrow

[chilly ocean]

I didn't know about it

[serene dune]

now u know it, XD

[chilly ocean]

If there exists then there is a normal subgroup of S_5 which has order 24, correct

[serene dune]

sly theorem ?

should i show that there doesnt exist unique order 24 subgroup

[chilly ocean]

I think there is a result that there is no normal subgroup of order 24 in S_5.
A_5 is simple, correct?

https://math.stackexchange.com/questions/2016157/understanding-a-proof-about-the-only-normal-subgroups-of-s-5

The only normal subgroup of S_5 is, {1}, A_5 and S_5

[serene dune]

that's true ig

also

there exists no such map like $S_4 \twoheadrightarrow \mathbb{Z}_4$

[cloud walrus]

yeshua

[serene dune]

order 6

[chilly ocean]

And there is the result that the non-trivial normal subgroup of S_n is A_n for all n ≥ 5.

[serene dune]

oh yeah i now recall doing that somewhere in galois theory

am i tweaking?

[chilly ocean]

You don't need Galois Theory for this result

[serene dune]

u need this ?
to prove the quintic statement or smth

[chilly ocean]

I see

[serene dune]

https://math.stackexchange.com/questions/157876/normal-subgroups-of-s-4

hmm

there exists no such map like $S_5 \twoheadrightarrow S_4$

[cloud walrus]

yeshua

[serene dune]

there exists a map $S_4 \twoheadrightarrow S_3$

[cloud walrus]

yeshua

[chilly ocean]

Let this be true for the cyclic group A and B, such that |A| = q_1^m and | B | =q_2^n, where q_i are prime numbers.

Now assume G = A × B, what exactly is G^p ? It is same as A^p × B^p.
Similarly, G_p is same as A_p × B_p.

Now we assume our result is true for A and B, thus there exists surjective homomorphism mapping, f_1: A -> A_p with kernel A^p and f_2: B -> B_p with kernel B^p.

Now define homomorphism mapping f: G-> G_p such that f(a,b) = f( f_1(a), f_2(b) ).

It is a surjective homomorphism mapping with kernel G^p.

det, correct?

[chilly ocean]

What is the orthogonal complement of subgroup H of finite abelian group G?

If it is { g in G | gh = 1 for all h in H } then it would not be a group when H ≠{1}.

[restive birch]

sanity check:
let S_3 act on the set of ordered pairs of elements of {1, 2, 3} by σ(a, b) = (σ(a), σ(b)). find the orbits of this action
i got 2 orbits, {(1, 1), (2, 2), (3, 3)} and everything else. is this correct?

[chilly ocean]

Yes

[rustic crown]

two thingies

• not enough to say this for cyclic groups A and B, because in the next step we want to apply this to (A x B) and C and A x B is no longer cyclic.
• f(a,b) = (f1(a), f2(b)) is the def, you had typo on the right.

[chilly ocean]

Yes it is typo

[chilly ocean]

So we need to do more things to counter 1

[rustic crown]

nup

we don't need to change anything the proof, just the claim changes

if the statement you're proving holds for A and B which are finite abelian, then it holds for A x B.

[chilly ocean]

Yes there is no need for cyclic

[rustic crown]

now to finish the proof, we just need to do a very simple induction on the number of cyclic summands

[chilly ocean]

Yes

[rustic crown]

if you want, you could also show the staement for cyclic groups A1,A2,...,Ar at once. the proof changes just a lil.

we'll just define f(a1,...,ar) = (f1(a1),...,fr(ar))

[chilly ocean]

So we proved the base case n = 1. Now use induction hypothesis for n-1 cyclic results also true.

Now for n, let A = direct product of n-1 cyclic and B = A_n.

Now used above result to show the result is true for A × B.

Just an outline

[rustic crown]

[chilly ocean]

Yes this is working

Thank you det

[rustic crown]

np

[chilly ocean]

But I am stuck at other Dummit's problems

I have to show the number of subgroups of A of order p equals the number of subgroups of A of index p, where A is a finite Abelian group.

Now I proved this for when A is an elementary abelian p-group.

Now I have no idea how I can proceed further

[hasty skiff]

I have been reading Topics in Algebra.... This is an example from chapter group theory.... I didn't quite get what did the author mean by x*sigma ≠x

What is x.... Is it the no. Of elements that are being moved in S?

[chilly ocean]

x is an element of S

Herstein used x(sigma) as sigma(x), where sigma is a function, in this case it is a bijective function

[rustic crown]

show that an order p subgroup is a subgroup of A_p.
And an index p subgroup is a subgroup containing A^p.

by correspondence theorem, index p subgroups correspond to index p subgroups of A/A^p which is iso to A_p.

[chilly ocean]

Now sigma is a bijective function from S to S. Let sigma = f, so f: S -> S, x*sigma≠x is f(x) ≠ x

[chilly ocean]

I proved the first one, wait let me prove the second one

[hasty skiff]

Ahh okay... Makes sense... These notational conventions🥲

[chilly ocean]

A^p contains the elements of the form a^p. Now assume H be a subgroup of index p, by quotient group A/H, a^p in H.
Hence H containing A^p

[rustic crown]

yee

A/H has order p, so a^p is the identity in A/H

[chilly ocean]

Why index p subgroups correspond to index p subgroups of A/A^p?

[rustic crown]

that's correspondence theorem,

if A is any abelian group and B a subgroup, then there is a correspondence
{subgroups of A conaining B} with {subgroups of A/B}

[chilly ocean]

Yes

[rustic crown]

C corresponds with C/B

by third iso thm, C has index p in A iff C/B has index p in A/B

(A/B)/(C/B) = A/C

[chilly ocean]

Yes I got it the point

[rustic crown]

:3

so yee, so we need to show there is a bijection between subgroups of A_p of order p and subgroups of A/A^p = A_p of index p

so it reduces the question from A to A_p

and A_p looks like a finite direct sum of Z/pZ = F_p

[chilly ocean]

Yes that was my confusing point index of subgroups of A_p of order p or subgroups of A_p has index p

[rustic crown]

mb ><

[rustic crown]

anyway, this is a much nicer place to ask that question.
A_p is just a finite dimensional F_p-vector space

and subgroup of an F_p vector space is precisely a subspace.

i'll let you finish the proof

(btw if you know about pontraygin duals already, you can do this problem in another way)

[chilly ocean]

And A_p is an Elementary p-subgroup

And I already proved for an elementary p-subgroup

Correct, det?

[rustic crown]

what's an elementary p-subgroup?

i thought it meant Z/p^n

[serene dune]

elementary abelian ?

[chilly ocean]

Z/pZ × Z/pZ ×..×Z/pZ

Yes elementary abelian p-group

[serene dune]

i knew it as elementary abelian

[serene dune]

right!

baire, may i know the mother problem here

[chilly ocean]

This one

[rustic crown]

(i don't know if you did that ><)

it reduces to the linear algebra statement that if V is a finite dimensional k-vector space, then there is a bijection between subspaces which have dimension d and subspaces which have codimension d.

we only need d = 1

[serene dune]

do anyone use it here

[chilly ocean]

Actually I showed that the number of subgroup of order p is p^n-1/p-1 is same as the number of subgroups of index p

[rustic crown]

oh okie

[chilly ocean]

Sorry, co-dimension

[chilly ocean]

Why

[rustic crown]

A_p = (F_p)^n so it's an n-dimensional F_p vector space

[chilly ocean]

Yes

[rustic crown]

we need to show subgroups of order p and subgroups of order p^(n-1) are in bijection

[chilly ocean]

Yes

[rustic crown]

that's same as subspaces of dimension 1 and subspaces of dimension n-1

(codim = n - dim)

[chilly ocean]

I see

So the complement works here in case of linear algebra?

But it has an unique complement?

[rustic crown]

not quite, since we don't have a nice inner product

but a very similar idea works

you take duals

[chilly ocean]

But some hyperplane stuff

[chilly ocean]

Oh

[rustic crown]

if W is a subspace of V, then W^perp is the subspace of V* defined by W^perp = {f in V* : f(W) = 0}

that's a more canonical thing

[chilly ocean]

Yes

[rustic crown]

so you get that dimension d subspaces of V are in bijeciton with dimension n-d subspaces of V*

[chilly ocean]

Yes

[chilly ocean]

Annihilator of W?

[rustic crown]

but since V and V* are non-canonically isomorphic, we get a a non-canonical bijection between dimesnion d and dimension n-d subspaces

yee

[chilly ocean]

Sorry non-canonically isomorphic?

[rustic crown]

the isomorphism between V and V* requires a choice of a basis

there isn't a way to define an iso without making a choice

[chilly ocean]

I see

[chilly ocean]

That's interesting

[rustic crown]

so yee, overall we reduced to some linear algebra fact and used nice properties of the dual operator

you can do that directly by staying inside finite abelian groups, it also has a nice dualizing operator

given a finite abelian group G, you can define G* := Hom(G, ℂ\{0})

[chilly ocean]

Yes after some exercise Dummit introduced a dual group of G.

[rustic crown]

:3

[chilly ocean]

And thank you for helping, I don't know why Dummit directly asked the problems, you helped me a lot to break my problems.

[serene dune]

would u give me the definition

[serene dune]

oh got it

[inner steppe]

Is there like any theorem that I can use to simplify modules like 71^17 mod (143)

[mighty kiln]

CRT and FLT

[inner steppe]

how can I use CRT here?

[mighty kiln]

143 = 11*13

[inner steppe]

but 5^17 is not easy to do either

[rocky cloak]

FLT simplifies it.

Then with binary exponentiation it's pretty easy.

Like
5^17 = 5^7 (mod 11)
5^2 = 25 = 3 (mod 11)
5^4 = 3^2 = 9 (mod 11)
5^7 = 5*3*9 = 4*9 = 3 (mod 11)

[chilly ocean]

Let G be a group of order p^3, where p is an odd prime number. I tried but I don't see why x^p in Z(P) for all x in G.

Yes when G is abelian it is trivial and when G is not abelian then Z(G) = [G,G] and |Z(G) | = p

[serene dune]

$G$ is a finite group and $H_1$,$H_2$ are two distinct subgroups of order $2$. $H$ is the subgroup of smallest order that contains both $H_1$ and $H_2.$ What is the cardinality of $H$ $?$

$A.$ always $2$.

$B.$ always $4.$

$C.$ always $8.$

$D.$ none of the above.

[cloud walrus]

yeshua

[chilly ocean]

If the group is abelian then answer 4 but if the group is not an abelian then D_4 is a counterexample

[serene dune]

well u mean V4 =Z2 x Z2 or D8

[sharp ice]

think about the group G/Z(G), which is not cyclic and is of order p^2, and think about the element x^p Z(G)

[chilly ocean]

D_8

[serene dune]

right

so as per hypothesis the answer should be D. ig

[chilly ocean]

G/Z(G) is isomorphic to Z/pZ times Z/pZ

and x^pZ(G) has order 1 or p, we have to show it has order 1

[sharp ice]

i was thinking more about how x^p Z(G) = (x Z(G))^p

[chilly ocean]

got it thanks @sharp ice

[inner steppe]

It is my understanding that a field is by default a group given both multiplicative and additive operations. On a lecture slides it says the requirements of a group "(F, +) is commutative group with 0 as identity element, and (F, x) should be commutative group"

Why did they mention that additive should have identity element if it is a group then shouldn't it by default have identity element considering we are saying additive forms "commutative GROUP"

My questions boils down to must multiplicative (F, x) have identity element?

[serene dune]

\raisebox{.2em}{\small{$G$}}\big/\raisebox{-.2em}{\small{$Z(G)$}} can't be non-trivially cyclic.

[cloud walrus]

yeshua

[sharp ice]

should be F - {0} that forms a group under the multiplication of the field. 1, the identity under the multiplication of the field, is the group identity

[inner steppe]

Correct I forgot

[inner steppe]

so we require that the multiplicative part of a field to have identity element

then what is the difference between ring and field?

[mighty kiln]

The full definition is probably

A field (F, +, ×, 0, 1) is such that (F, +, 0) is a group and (F-0, ×, 1) is a commutative group and × distributes over +

[sharp ice]

a field is ring under more conditions. a field is a commutative division ring

[glad osprey]

you're right that they don't need to specify that a group has an identity element, when they write "(F, +) is a commutative group with 0 as identity" it's just to clarify which element is the identity

[inner steppe]

Okay so both require identity elements for both + and *?

[mighty kiln]

Meanwhile

A ring (R, +, ×, 0, 1) is such that (R, +, 0) is a group and (R-0, ×, 1) is a monoid and × distributes over +

[sharp ice]

yes. in a ring u need multiplicative and additive identities

[inner steppe]

same goes for fields no?

The difference is that multiplication does not need to be commutative in rings

[sharp ice]

yes. as fields are rings

[inner steppe]

but does it have to be non zero elements in mutliplicative group for rings for it to have identity?

I would guess so since it is impossible otherwise since

0*a !=1 for any a

[sharp ice]

the identity is the same for all elements in the ring/field. even 0. 0 * 1 = 0

[sharp ice]

r u confusing identity with inverse ?

[inner steppe]

so it is not an equivalence relation? If something has an inverse it needs to have identity tho right

[sharp ice]

im not sure i understand what u mean. which part is the equivalence relation

[inner steppe]

I said in order for an inverse to exist there must exist an identity element

existent of inverse -> existence of element

I am saying Ig this relation is not equivalence relation

But anyways I look at my book and they say that a ring is where the operation + is a commutative group and then for R2. The operation x has closure, associativity and identity properties.

So the multiplicative part is not a group in ring?

They never said it was a group or has inverse property

[sharp ice]

yeah, it doesnt necessarily form a group. but in fields it does since u have inverses. this is the division part in commutative division ring

[inner steppe]

But that is because we say F \{0}

if we make the same statement couldn't we say the F \{0} of a ring is a group?

The only difference I'm seeing so far is that multiplicative part of ring doesn't need to be abelian whilst it needs to be for fields.

[sharp ice]

no. they dont necessarily have multiplicative inverses. take the integers Z. Z - {0} under multiplication do not form a group

[inner steppe]

Yeah you are definitely right so in a ring we don't require inverses to exist. I was solving a problem where I was asked to prove a specific field was a field this whole time I totally forgot everything else. Thanks.

But to summerize the main difference is that multiplicative needs to be abelian in fields but not in rings. Inverses must exist for nonzero elements in fields but inverses don't have to exist at all in rings.

But if we ignore numbers for a second

what does 0 in F\{0} mean?
Or are only numbers of some sort fields?

[sharp ice]

yep

[sharp ice]

0 denotes the additive identity of F here

[inner steppe]

That is what I thought so if I am ever given a set of "things/numbers" given and was told to prove it is a field. e as additvie identity and a as multiplicative identity. I would then define it as F\{e} should be a commutative group

[sharp ice]

its just convention to use 1 as the multiplicative identity and 0 as the additive identity of a ring. its easier to make a distinction between them, and its just easier to use than just e as the additive identity and a as a multiplicative identity

[chilly ocean]

if G = HK where H and K are characteristic subgroups of G with H cap K = {1}, then Aut(G) is isomorphic to Aut(H) times Aut(K).

Proof: Since H and K are characteristic subgroup of G therefore they are normal in G, their intersection is trivial implies they commute each other. Hence G is abelian.

phi : Aut(G) --> Aut(H) times Aut(K)
phi(f) = (f_H, f_K), f_H denotes the function restrict to H, it is well-defined because H and K are characteristic subgroup
Now it is injective because if f_H = g_H and f_K = g_K then f = g because G = Hk.

I proved this is surjective by using G abelian and G = HK.

now i have to verify that phi is homomorphism.
am i in right direction?

[thorn jay]

can you assume G is abelian?

[bitter rover]

You don't need to

[thorn jay]

I proved this is surjective by using G abelian and G = HK.

[bitter rover]

Oh hah

uhhh

that's what I get for not reading closely

[thorn jay]

loll

[bitter rover]

Actually, if the intersection is trivial then G is abelian, I think. So never mind.

Wait no

Yeah the direct product is a counterexample.

[thorn jay]

characteristic subgroups need not be abelian

for example, if Aut(G) = Inn(G), then any normal subgroup is characteristic

[thorn jay]

hell, any symmetric group

[bitter rover]

You get what commutativity you need, though.

[rocky cloak]

You don't need G abelian to show it's surjective.

Not really sure how it would help either

[chilly ocean]

but G is abelian, right/

[bitter rover]

No

[rocky cloak]

Not necessarily

[bitter rover]

Not necessarily

yeah

[chilly ocean]

H and K are normal and their intesection is trivial

H and K commutes each other

[thorn jay]

what if H and K aren't normal

abelian*

[chilly ocean]

but in given question they are

[thorn jay]

then state that

[rocky cloak]

Not the one given to us 😛

[chilly ocean]

no i mean given is they are characteristic subgroup

so they are normal,right?

[thorn jay]

normal doesn't mean they're abelian

[rocky cloak]

They're normal, but not necessarily abelian

[chilly ocean]

and their intersection is trivial

[rocky cloak]

Like S3 x C5 for example

Then H=S3, K=C5 works

[bitter rover]

Just because HK=KH doesn't mean HK is abelian

It means elements from one commute with elements from the other, which is enough for this proof.

[chilly ocean]

hkh^-1k^-1 is in both H and K so h and k commutes, i think my mistake is i assumed H and K are subgroups of G

[bitter rover]

That's weaker than HK being abelian.

[thorn jay]

i think my mistake is i assumed H and K are subgroups of G
? what would they else be?

also, how do you prove hkh^-k^- is in both H and K

[rocky cloak]

Elements of H commute with elements of K - correct
Elements of H commute with other elements of H - no not necessarily

[chilly ocean]

i got it my mistake

so let me finish my proof for surjectivity

[bitter rover]

(I don't think you need to use this for surjectivity.)

(You definitely do for proving it's a homomorphism.)

[chilly ocean]

so let f and g are automorphism of H and K, respectively.
Now t defines the automorphism of G as t( hk) = f(h)g(k), it is well defined because intersection is trivial.
also injective and surjective, now t(h_1k_1*h_2k_2) = t(h_1h_2k_1k_2) = f(h_1h_2)g(k_1k_2) = f(h_1)f(h_2)g(k_1)g(k_2) = f(h_1)g(k_1)f(h_2)g(k_2) = t(h_1k_1)t(h_2k_2)

correct?

we only need H and K commutes each other

[bitter rover]

Hard to parse, but yes, that's the idea.

[chilly ocean]

thanks Jagr, cufflink

[bitter rover]

(As in, there might be a small mistake in there, but that's the place where you'd use the fact that elements of H and K commute w/ each other.)

[chilly ocean]

yes i used this fact

[lethal cipher]

Okay, so I have shown that the subgroup of commutators for SL_2(R) is itself.
Now...I need to somehow prove that -I is not a commutator, which will tell me the set of commutators is not a subgroup.
That being said....I can't see a nice way to do this without an extensive amount of busy work...

Though, not even sure I understand how to do it with the busy work either.

[rocky cloak]

I mean, 2x2 matrices are not that big. You could just write out what a commutator looks like

[bitter rover]

yeee

[coral steeple]

Thanks.

[inner steppe]

If I have like a polynomial x^3+2x^2+2x+1

If I am asked to check if it is irreducible or not. Is it enough just proving that it doesn't have roots in like Z_3?

Because it cannot be a product of a polynomial of higher than 1 degrees

But what if the degree of the polynomial was like x^4x^3+2x^2+2x+1, I have no clue if this has roots but assume it didn't have roots. I am guessing I cannot just conclude it is irreducible right

But how can otherwise check?

[grave sedge]

Maybe mod something it factors in polynomials with degree 2+2 and mod something else 1+3

[inner steppe]

I am not sure I understand compeletely

[grave sedge]

Take something like x^4+3x^3-x^2+1

Mod 2 It becomes x^4+x^3+x^2+1=(x^3+x+1)(x+1), and both factors are irreducible

Mod 3 it becomes (x^2+1)^2, again both factors are irreducible

Now assume it can be factored in Z and get a contradiction

[lethal cipher]

I get these relations

,w rot

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by dackid

[dawn quest]

What is everyone's favorite group homophobe $f: G \to H$ ?

[cloud walrus]

plexcty

[inner steppe]

can you do that tho

can you change modes?

[lethal cipher]

Point is...it doesn't get to anything very useful

[grave sedge]

I'm not going from mod 2 to mod 3, i'm looking at the same polynomial first mod 2 and then mod 3

[restive birch]

okay im a bit stuck and also want to check what i have to so far:
problem: show that a transitive group G is primitive iff the only subgroups of G containing G_a for any a are G_a and G
one way (primitive -> proper subgroup) isnt bad but the other one is tripping me up. so far what i have is:
let f be in H with G_a < H < G, with f not in G_a, and let f(a) = b. then G_b = fG_af^-1 <= H. If f(b) = a, then {a, b} is a block. otherwise, say f(b) = c and then G_c <= H. continue like this to show that all G_x <= H for all x ∈ G, and f is not in any stabilizer
kinda stuck on where to go from here. ideally i want to show that this implies that H = G

and also unsure if what ive done so far holds

[bitter rover]

If a polynomial f is reducible in ℤ then it is reducible mod p for any prime p.

For your polynomial, if it were reducible in ℤ then the degree places constraints on what the irreducible factors can look like.

For example, say your polynomial x^4 + 3x^3 - x^2 + 1 were divisible by some degree-2 polynomial over ℤ[x]. Like Matteddy pointed out, over (ℤ/3ℤ)[x] the polynomial factors as (x^3 + x + 1)(x + 1) with both factors irreducible.

Is there a problem with that?

[bitter rover]

G_a is the stabilizer of a?

[restive birch]

yes

is this not standard notation?

[bitter rover]

It's common enough, just wanted to make sure. You might also see Stab_G(a).

Or you can say: A transitive group G acting on S is primitive if and only if the stabilizer of every element in S is a maximal subgroup. Just wanted to clarify that that was what you were asking for sure.

[chilly ocean]

Am I correct? If S and T are subset of a group G, then the commutator subgroup of < T, S > contains the commutator subgroup of < S >

Hence if H and K are perfect subgroups of G, then < H, K > is perfect subgroup.

[< H, K >, <H, K> ] contained in < H, K>. Now we observe that [H, H] contained in [< H, K >, <H, K> ], and [K, K ] contained in [< H, K >, <H, K> ].

Since H and K are perfect, implies that < H, K > contained in [< H, K >, <H, K> ].

Hence, <H, K> = [< H, K >, <H, K> ].

Is my reasoning correct?

[inner steppe]

Can anyone recommend a simple exercise on finding an isomorphism between groups with solution. I have solved multiple questions on book and recommended questions but there is only one exercise that asked me to find a bijection function so I feel very unsure on that part. I would really appreciate it.

[bitter rover]

If G acts on A and B ⊆ A, denote by Stab_G(B) the set-wise stabilizer of B, i.e., the the set of all g ∈ G such that gB = B.

If a ∈ B then Stab_G(a) is a subgroup of Stab_G(B) (and a proper subgroup if B has more than one element).

If G acts transitively on A then Stab_G(B) = G implies B = A.

Show any block containing more than one element must be all of A.

[chilly ocean]

If isomorphic groups are regarded as being the same, prove, for each positive
integer n, that there are only finitely many distinct groups with exactly n
elements.

My reason: basically the binary operation is a function so we have to find distinct binary operations which follow the group axioms. Now for the worst case there are n^(n^2) binary functions and I know we can eliminate many binary operations but these are the finite number of distinct groups of order n.

Can anyone check this one ?

[rustic crown]

yee that worksie

[thorn jay]

Yes

[inner steppe]

Guys when we talk about the multiplicate group of a field, do we always ignore the 0 element? As in 0 doesn't exist right in the muliplicative group of an element

but if that is the case we cannot claim that 0 is distributive tho

I guess what I am asking is whenever we include 0 and use both + and *, how do we think?

[bitter rover]

Let R be a ring with identity (0 and 1, respectively). Assume 0 has a multiplicative inverse. What can you say about R?

[inner steppe]

I am not sure I understand the question. But according to my understanding, I'd say R's multiplicative part is a group.

[bitter rover]

In any ring, a·0 = 0·a = 0. You agree with that?

If yes, write out what it would mean for a to be the multiplicative inverse of 0.

(And I mean actually write it out.)

[inner steppe]

I don't know what you want me to write out. 0 cannot have multiplicative inverse but prolly a*0=1 where a= 0^-1*1 which is the same as saying 1/0 which is impossible/ undefined

[bitter rover]

It's not a trick question. I wanted you to do just what I asked!

If a is a multiplicative inverse of 0 then you have 0·a = 1. That's what it means.

So together, you have the following:

a·0 = 0   (from the basic ring axioms)
a·0 = 1   (because a is an inverse of 0)

Do you see any issues with that?

[inner steppe]

Yes I agree but my question was does 0 exist in multiplicative group of a field or do we ignore it totally. So far whenever I did exercises where I made tables, I have been ignoring it in cayle's table and so has the solutions.

Which is fair given the definition

But what I don't understand is a field is a mix of both operations + and *, so how does it work whenever we have to multiply

[bitter rover]

You don't see an issue with both of those equations being simultaneously true?

[inner steppe]

No I meant that yes I agree there are issues

But I am just not seeing where we are going with this

[bitter rover]

What's the issue with both a·0 = 0 and a·0 = 1 being true?

[inner steppe]

inconsistency?

[bitter rover]

There's no inconsistency.

Remember, 0 and 1 are avatars for elements of R. They're not our literal numbers 0 and 1.

[restive birch]

okay that makes sense, was i on the right track?

[inner steppe]

It is ambigous tho a·0 can be two things at once

[bitter rover]

No. You're assuming 1 != 0.

There's no ring axiom which says that's the case.

[inner steppe]

so are you saying that this is true a·0 = 0 and a·0 = 1 for rings

[bitter rover]

I'm saying that if 0 has a multiplicative inverse then

0 = a·0
  = 1

i.e., then 0 = 1.

There's exactly one ring where 0=1 and that's the trivial ring {0}.

Because if 0 = 1 then for any element r:

r = 1·r   (def'n of 1)
  = 0·r   (because 1=0)
  = 0     (from basic ring axioms)

@inner steppe Now, look up your field axioms and somewhere it will say that F contains an additive identity 0 and a multiplicative identity 1 and that 0≠1

If 0≠1 then 0 can't have a multiplicative identity.

For a variety of reasons, we don't want to call the trivial ring {0} a field.

[bitter rover]

The idea of looking at conjugates can be made to work, but the idea of building up to G one element at a time won't quite work. There's nothing that says G is finite.

@inner steppe In other words, you can't just toss in a multiplicative identity for 0. Something else has to give.

And if you permit {0} to be a field then virtually all field theorems that began with "Let F be a field..." would now begin with "Let F be a non-zero field...".

Just like we don't want 1 to be prime. We could call it prime. Many famous mathematicians did call it prime. But then most theorems involving prime numbers become "Let p be a prime other than 1..."

(There are deeper reasons at work than not wanting to make theorems awkward, but hopefully that sufficies.)

[restive birch]

oh, i see

thanks!

[south rain]

I have a pretty simple question. I don't understand why, if a function has no right inverse, and the source space has at least two elements, it must have more than one left inverse.

According to Aluffi's book I should see it in the proof of injectivity is equivalent to having a left inverse, but I just don't see it.

I understand the proof though I just don't see the argument.

[rocky cloak]

So I think what you're saying is you have an injective function that isn't surjective.

Then a left inverse just maps the elements in the image where they came from, and the remaining elements wherever.

If the domain has more than one element, then you have several choices for where "wherever" could be

[south rain]

Sorry, yea injective but not surjective. That makes sense. If it would be surjective though I would not need to define an inverse with that "wherever" element right, and could just use the map back to the element in the domain?

[rocky cloak]

If every element is in the image, then there is no "the remaining elements".

If something is injective and surjective, it has a unique inverse (both left and right)

[fast crescent]

Yea of course. Thanks, that’s all!

[rocky cloak]

Looks like there was a parity error with the account

[inner steppe]

Thanks for this!!

Don't we like need to know what operations are being used here? Z_4 under what? ZxZ under what

[prisma ibex]

lol no

[inner steppe]

y

[prisma ibex]

I mean there are literally only two groups of order 4 up to isomorphism

and it's clear which one each of these refers to

[stone vault]

Without additional context “Z_4” or “Z/4Z” is going to be the cyclic group of order 4

In the same way that someone will refer to say the ring of integers without specifying the algebraic structure

Or open sets in R without specifying a topology

[inner steppe]

well it is not clear for me

[inner steppe]

for both multiplication and addition?

[stone vault]

Well it’s not exactly a group under the standard multiplication is it

[inner steppe]

Sure but still don't get it

[prisma ibex]

you should review the definition of groups and what Z_n refers to then

[south patrol]

For some mathematicians heh

[prime arch]

unless you want to define your own wacky operations, there are only two operations: multiplication and addition. they're not groups under multiplication, so they must be groups under addition. there aren't any other reasonable choices. so you know what operations are being used just based on this

[stone vault]

I mean yeah I also have problems with this notation but it is certainly common and it’s basically always obvious what it is in context

[ember field]

If f in K[x_1,...,x_m] and f is in (g_1,...,g_n) is f integral over K[g_1,...,g_n] ?

[golden turtle]

say M is a submodule of Z^(n)

and I want to show this isomorphism

specifically say i have a map f, and I want to show the f(rx)=rf(x) property

where does r come from?

r in Z^(n) or r in Z or r in M? I'm confused by that

[inner steppe]

is this true

If (F,+,×) is a finite field with characteristic p, then the additive group (F,+) is isomorphic to a cyclic group of order pr, where r≥1

[unkempt stream]

what's the context

[golden turtle]

M is a submodule of Z^(n)

[unkempt stream]

what is pi?

and the asterisk

[golden turtle]

its not important for the question but pi is the canonical projection from Zn to Z(n-1) and the asterisk is indicating to append a 0 onto the last coordinate of the elemetns in pi(M)

[unkempt stream]

you could use iota for the latter because that's the canonical injection

[golden turtle]

ok

[unkempt stream]

for finite sums/products it's a "biproduct", any module in the product you can both inject into and project from the product

[golden turtle]

I think the r comes from Z and we call it a Z module homomorphism if I had to guess but I'm not sure

[prime arch]

yeah, a field is, in particular, an additive group

[golden turtle]

M and the direct sum are both Z^n submodules

[unkempt stream]

this is true pretty much for any finite product/sum of modules lol

namely look at the intersections

[golden turtle]

Do you know what my question is?

[unkempt stream]

with the embedding of A into A (+) B

[unkempt stream]

Is that not from the module def

[golden turtle]

I sent the definition above

and I am confused

[unkempt stream]

pi_A(r(a,b)) = pi_A((ra,rb)) = ra = r(a) = r * pi_A(a,b)

[golden turtle]

My question is where is r coming from

Z?

[unkempt stream]

Yes, Z^N viewed as a Z-module

[golden turtle]

so we say that this is a Z-module isomorphism

and r is from Z

[unkempt stream]

yep

[golden turtle]

thank you

[unkempt stream]

it's "r" standing for the ring

[golden turtle]

I was hoping someone could check my work, not the middle bit about f being an isomorphism because I think that's fine, but mostly just the last paragraph and the last is included for context

I am wondering if I showed enough

[unkempt stream]

This can be generalized to every submodule of a free module is free when the ring is a PID, since the annihilators are an ideal (principal) :3

[golden turtle]

What step does the ring being a PID come into play for that

I’m interested

Oh I guess Z^n gets replaced with just something more general and the base case is different?

[tardy hedge]

why does that statement seem straightforward but the proof is so long

Its probably not straightforward then

[golden turtle]

I feel like many “obvious” things I’ve had to prove lately have very non obvious proofs

[tardy hedge]

I should probably look at that cause im also in module theory rn

[golden turtle]

If you spot a mistake please let me know

[glad osprey]

(F, +) is cyclic only if |F| is prime (in which case p = |F|)

[cobalt vessel]

(problems are from fraleigh), im so cooked guys i shouldve started this sooner 🤦‍♂️

wish me luck i will see how many i can finish by 3am

[unkempt stream]

Yes, basically you use Axiom of choice in combination with Well ordering principle to get a well-ordered basis of the module. Then you can use the projection onto the first basis element to get a map into R, with the image being an ideal.

This ideal is principal by definition of a PID, so has a generator. You can consider elements in the fiber of that generator, and use what you proved. Transfinite induction completes the proof, but the rest requires a bit of algebra to show that the "remaining part of the module" outside the kernel is generated by the remaining basis elements

[ember field]

If S is a finitely generated R-module, then is S integral over R?

[dull ginkgo]

What is your defn of integral

TTerra what does that mean bro because I thought that was for rings

Isn’t that an equivalence

Oh wait no

Thinking of a different statement

[unkempt stream]

any wacky algebra shit you learned

anything polynomial identity related and you explode violently

[golden turtle]

Awesome

[unkempt stream]

wait that's the one that you told me about lol

[golden turtle]

[unkempt stream]

seems fine

basically what the general PID form uses, since Z is a pid lol

[golden turtle]

thank you

[chilly ocean]

an infinite group not necessary has a cyclic group which is infinite.
example : Z/2Z[x] additive group, every element has finite order so any cyclic subgroup is finite.

is this one works?

[south patrol]

Yes

The additive group is just (Z/2)^N

[wild jasper]

I'm trying to do (b), but I'm having trouble seeing that the difference with f has the same degree as f

[marsh scaffold]

What does it mean by lexicographic initial term

[wild jasper]

here is some more context...

[barren sierra]

Well (a) shows you have the same initial term

So when you subtract suppose the result is nonzero

The leading terms are the same (same coeff, same degree) so intuitively they should cancel no?

If you want more details look at Theorem 1.1.1 in Algorithms in Invariant Theory by Sturmfels

I like the proof there (due to van der Waerden)

But really just look at how the initial monomial (the largest monomial) cancels

[wild jasper]

@barren sierra thanks 🙂

[barren sierra]

What book is this from btw

[wild jasper]

@barren sierra Dummit and Foote

[barren sierra]

nice

[dawn quest]

What is everyone's favorite group homophobe $f: G \to H$ ?

[cloud walrus]

plexcty

[reef trench]

???

[marble mural]

homophobic?

[serene dune]

🥶

[marble mural]

im black if that counts

[thorn jay]

Assuming you mean homomorphism probably the exotic inclusion of S_3 in S_4

Or S_4 in S_5

I forgot

[grave sedge]

(S5 in S6 probably)

[thorn jay]

Yeah thanks

But yeah that

I also like the surjective homomorphism
S_4 -> S_3
Which has V as kernel

[hot wadi]

S4 to S3 is an All-time classic

I really like the outer automorphisms of S6

[next obsidian]

S5 inside of S6 be like 🫨

[thorn jay]

They scare me

Haha

Outer automorphisms in general

I really want to know how they proved that for n > 6 Aut(S_n) = Inn(S_n)

What makes n > 6 special?

Is it a too small/too large thing?

[thorn jay]

I like it because V is my favourite group

[grave sedge]

Basically, yes

The usual argument is by looking at elements of order 2

Since automorphisms preserve conjugacy classes and order of the elements, transpositions are sent to some conjugacy class with cycle type of order 2

But for n>6 the transpositions are fewer than any other cycle type of order 2

While for n=6 the numbers align and you have the (12)(34)(56) type permutations which have the same cardinality as the transpositions

Ah yeah i forgot to mention, it's quite easy to see that an automorphism sending transpositions to transpositions is an inner automorphism

[pseudo mesa]

Hia peeps I have a question

let rings R,S be isomorphic to each other

If R is a field (has identity, and is commutative)

Fk lemme just write it

Can I assume r is also non zero? I’m not sure if I can or not

Sorry the photo is sideways :’)

[surreal dagger]

Yeah, you can assume r to be non zero

[pseudo mesa]

It’s because all isomorphisms are bijective homomorphisms

Right?

[weary frost]

also because a field has a unique 0 element

[surreal dagger]

Yes

[pseudo mesa]

Okay, thank you guys!

[thorn jay]

Is it? Haha

Hmm

Maybe i got it

[grave sedge]

It's basically a looking at which transpositions commute with which others thing

[thorn jay]

(n m) = (1 m)(1 n)(1 m)
Right?

Yeah

For n ≠ m

[grave sedge]

That too, i guess

[thorn jay]

Right I must admit i havent properly worked with permutations much so im just kinda playing around haha

[grave sedge]

I guess you can use the fact that if n≥5 a set of n-1 transpositions none of which commute with any other is necessarily the n-1 transpositions containing some fixed element

Which means that such a set is necessarily sent to another set with the same property

And this basically defines the whole automorphisms from what you said above

[thorn jay]

Such a set is a set where every transposition shares one element right?

[grave sedge]

Yes

[thorn jay]

Sorry im just trying to wrap my head around this lol

[thorn jay]

Wait seriously is that true?

I see the proof now that's awesome!!

[grave sedge]

I think so

Assume you have a transposition (ab)

Now every other must have either a or b

If there's one with a and another with b, since they don't commute they are necessarily (ac) and (bc) for some c

Now n-1≥4 so there's at least another one which must commute with at least one of the three we already have

This is a contradiction, so all the others contain either a or b

[thorn jay]

Mhm!

Thanks

Wouldnt it just need n-1 >= 3 though?

[golden turtle]

Does this work so far?

for (a) and (b)

[rocky cloak]

Looks good to me

[golden turtle]

thank you

[rocky cloak]

Is part c to show that M/N1 has a divisible subgroup or something like that?

[golden turtle]

This is the remaining parts, I don't know how to do (c)

[rocky cloak]

Well, if you have a basis and you remove some basis vectors, then what are you left with?

[golden turtle]

in LA terms a linearly independent set that does not span

like a sub basis

[rocky cloak]

It doesn't span M, but it spans something

[golden turtle]

a sub-module of M

?

[rocky cloak]

Yeah, and how is it related to N1?

[golden turtle]

it's like the complement

sort of

[rocky cloak]

Indeed

So that's M/N1

[golden turtle]

the complement submodule of N1 in M

i proved this on a later question, maybe it helps

[rocky cloak]

Don't think it helps really

[golden turtle]

Well atleast it gives us that Mbar is a free Z module, given that its a submodule of M

[rocky cloak]

Yeah, I mean you already knew it had a basis, but sure

[golden turtle]

intuitively I understand this, but when it asks me to show M bar is a free Z module is that enough to say?

like

M/N1 is the submodule generated by the basis vectors in B set minus the basis vectors of N1

and so it is a free Z module

[rocky cloak]

I'd say, you have B is the disjoint union of B1 and B2.

Then you can either argue directly with the definition of basis, that the image of B2 is a basis for M/N1.

Or you can say that M = N1 (+) spanB2, so M/N1 = spanB2

[golden turtle]

that makes sense

the thing about xbar seems random to me

[rocky cloak]

If it seems random you may want to look ahead at d and e

[hot wadi]

n=6 has a weird numerical coincidence where the number of transpositions is the same as the number of triple transpositions

[thorn jay]

ohhh

[hot wadi]

This provides an opportunity for an outer automorphism which swaps these, and indeed one exists

[thorn jay]

symmetric groups are reallyy nicely structured, but it's just a little wonky at small values

haha

[hot wadi]

I’ve never worked through the details myself but there’s some combinatorics you can do to show that there aren’t analogous coincidences in any higher Sn’s

[golden turtle]

Any idea how to show S is uncountable in (d)?

[thorn jay]

mhm

[rocky cloak]

I mean, the cardinality is 2^N

[hot wadi]

Contains power set of naturals (sorta)

As in you can inject the power set of the natural numbers into this in a nice way, showing it’s uncountable

[golden turtle]

how?

[hot wadi]

Maybe start with a smaller example

Consider the set of binary strings of length 10

Do you see a bijection between this set and the power set of {1,…,10}?

[golden turtle]

No

[hot wadi]

Alright

Say you were going to have to design some computer software which would work with subsets of {1,…10} and you wanted to encode these subsets as binary strings

[thorn jay]

||ah, because the subobject classifier in Set is the two-element set right? or explicitly the direct sum of two copies of {*}||

[rocky cloak]

But you see how there are 2^10 choices for a sequence of 10 +/- signs?

[hot wadi]

One thing you could do is store 1 bit for each element of your big set, and record what is contained in the subset with 0’s and 1’s

So like

The set {2,3,5} goes to
0110100000

[dreamy edge]

Can there be a ring with infinite elements but a finite number of units and irreducibles?

[next obsidian]

It necessarily can’t be a UFD at the very least

Does an infinite product of F_2 work?

there’s only a single unit

Actually maybe like, everything is irreducible lol

[vagrant zinc]

I was afraid of dummit when I saw it in my pdf files, I've been giving it to dummit for 1 month and it's great, although the exercises, without giving me for ***, hehe but it explains wonderfully and the construction roars :3 , I just wanted to comment this

[tardy hedge]

You’ve been… giving it to dummit? 😳

[vagrant zinc]

hahaha, it's a metaphorical expression for “studying”.

[golden turtle]

it's a metaphorical expression for something alright

[vagrant zinc]

Guys a question, is the first time I read about fibers, basically what is the function, here I can understand what it says, I do not know if this is correct, is a natural isomorphism of the function defined and already ?, or ...?

or what the fibration does is to take the ring and turn it into a unitary element, as well as the dual space ?

[tardy hedge]

ehat do u mean by natural isomorphism of the function?

When dummit and foote talks about "fiber above a" they mean all elements in domain that map to a

[vagrant zinc]

That is to say that an equivalence class with respect to each element of the domain ?

[tardy hedge]

yeah since the relation a ~ b if phi(a) = phi(b) is an equivalence relation, you can consider the "fiber above a" to be the equivalence class that maps to a

[vagrant zinc]

That is to say that a single element is representative of all the elements of the domain, therefore, we have a fibrated

[tardy hedge]

hm i mean the way you are saying it isnt really precise

"we have fibrated" also doesnt really have meaning

[tardy hedge]

Yeah so "the fiber above a" means the set {f in Q[x] | phi(f) = a}

[bitter rover]

Metamucil bundle

[vagrant zinc]

Something like this

$\varphi:\mathbb{Q}[x]\to \mathbb{Q}\
\varphi(p(x))=a\
1+x^2\sim 1+x+x^2+x^3+x^4+x^6 \\Rightarrow \varphi(1+x^2)=\varphi(1+x^2)=\varphi (1+x+x^2+x^3+x^4+x^6)\
\Rightarrow a=a$

[cloud walrus]

Homology

[tardy hedge]

yea

[vagrant zinc]

Then that's a fibrado

[tardy hedge]

ok i mean a = 1 in that example yea?

your example showed that 1+x^2 and 1+x+x^2+x^3+x^4+x^6 is in the fiber above 1

[vagrant zinc]

Now why above ?