groups-rings-fields

[coral spindle]

The presentation is correct though.

[rocky cloak]

Shouldn't the quotient be S3? The quotient of a group by its center is never cyclic.

[coral spindle]

Ah yes ofc

I was just checking that the quotient worked, but this makes it simple

Right, so the only group of order 12 I haven't checked yet is A_4. The dihedral group Dih(12) does have a normal subgroup isomorphic to Z/2Z, but its quotient is S_3.

And according to GroupProps at least, A_4 has no cyclic normal subgroup, so we can kick that out easily.

OK! So an argument.

The generator of $\bZ/6\bZ$ lifts to an element $g \in G$ of order either $6$ or $12$. If it's order $12$, then $G$ is cyclic and we're done, so assume it is order $6$. Let $h \in G$ be the generator of the normal subgroup isomorphic to $\bZ/2\bZ$. Since this is indeed normal, we must have $ghg^{-1} = h$, since $h$ is the unique nontrivial element of the subgroup. So $g, h$ commute and since $G = \langle g,\ h\rangle$ the group is Abelian.

[cloud walrus]

Boytjie

[final dragon]

how can i proove that if a non comutative ring if element in R has 2 different right inverse then a has infinilty many right inverses

[coral spindle]

Hint: what property does the difference of these inverses have? If there are infinitely many elements with this secret property, you can find infinitely many right inverses.

[delicate orchid]

gross...

[coral spindle]

No u

[thin bough]

I am trying to prove that the set of integers modulo 13 does not form a group with respect to multiplication

Did I do this right ?

[coral spindle]

No.

[thin bough]

oh

[coral spindle]

Because the operation with respect to is multiplication,
It is not multiplication; it is multiplication modulo 13 which is significantly different.

There is no reason that a^-1 would be equal to the inverse as a rational number.

Hint for actually showing this: think about the element 0.

[thin bough]

can you give an example ? I don't understand

[coral spindle]

Modulo 13 we have 4x7 = 1, but in Z we have 4x7 = 14. These are evidently completely different operations.

[thin bough]

hmm ok

[coral spindle]

This is one of those times that pedagogy annoys me. It is so much easier to see this as the quotient group Z/13Z, yet we are chained to remainders...

[thin bough]

is my inverse element correct ?

a^-1

[coral spindle]

That is just the notation for an inverse element

Saying a^-1 = 1/a is incorrect, as I stated.

[thin bough]

why did you cross out with respect to ?

[coral spindle]

Because it was ungrammatical

[coral spindle]

I suggest you think about my hint here.

[thin bough]

oh but that is the question in my textbook

[thin bough]

ok

i don't get why I would need 0

i know I would need 0 during the identity axiom

but don't get why i would need it during the inverse axiom

oh is it because there is no inverse of 0 ? @coral spindle

[coral spindle]

Prove it

[thin bough]

that is tough

isn't the inverse of 0 jjust 0

[rocky cloak]

Does 0*0 equal 1?

[thin bough]

oh why does it have to be 1

[barren sierra]

Recall the definition of inverse

The inverse of g is an element g^-1 such that g * g^-1 = g^-1 * g = e, the identity element

[thin bough]

oh yes. But if I was doing addition, then it would be 0 right ?

[barren sierra]

Addition yes

So the additive inverse of 0 is 0

[thin bough]

i see that's where i got mixed up

[barren sierra]

0 has no multiplicative inverse

Since the multiplicative identity is 1

And there no number x such that 0x = x0 = 1

[thin bough]

I see

thanks a lot

what does this mean?

I know the Z is the integer set

but what does the 23 denote

[coral spindle]

Z modulo 23.

[celest cairn]

Integers mod 23.

[thin bough]

oh ok thanks

[celest cairn]

If you don’t know how basic arithmetic in Z_n works, if you have a number (say k) greater than or equal to n, it’s the remainder of k/n. Furthermore, Z_n can be denoted (n/nZ, +), as it contains all integers from 0 to n-1 (has order n), and is also an additive group.

[coral spindle]

n/Zn
You mean Z/nZ, I think.

[celest cairn]

Sorry typo

[formal fractal]

How can i find all composition series for S3 x Z3

How does this help with the direct product

[rocky cloak]

Because GxH/H = G

[formal fractal]

You mean Gx(H/H)=G

Or

[thin bough]

thank you

[coral spindle]

No. If you are being especially stringent with notation, you may prefer to write (G x H)/({1} x H} \cong G.

[formal fractal]

Ok so N = {e} x H

Is N solvable?

[coral spindle]

What do you think

Perhaps it is isomorphic to a group you already know

[formal fractal]

Im an idiot

[formal fractal]

Sorry i can’t see this

It seems its just G x H

Not G

Er wait

[coral spindle]

You should be able to prove this yourself and I suggest you take the time to do so. I assure you there is no mistake.

[thin bough]

Find the multiplicative inverse of each nonzero element in ℤ23.

I used Fermat's little thereom for this. Is this the right approach? Seems too easy

all the way to 23^(23-1)

[coral spindle]

That's technically correct, but the question probably wants you to calculate the remainder of the inverse modulo 23, so simply saying that the inverse of 5 is 5^21 is not sufficient.

[thin bough]

so it would be better if i did this ?

[coral spindle]

I feel like you haven't responded to my comment at all

[thin bough]

but i am calculating the inverse modulo 23

because it equals to 1

[coral spindle]

You're saying that the inverse of 3 is 1?

What's 1 x 3 modulo 23?

[thin bough]

I'm saying the inverse is 3^(23)-1

[thin bough]

3

[coral spindle]

Firstly, this is simply wrong. Secondly, from my comment again:

[...] the question probably wants you to calculate the remainder of the inverse modulo 23, so simply saying that the inverse of 5 is 5^21 is not sufficient.

[thin bough]

but there can exist more than one inverse for a number right ?

[coral spindle]

No.

Exercise: prove that in any group, there is only one inverse for any particular element.

[thin bough]

but how come 3^(23-1) mod 23 = 1 ?

[coral spindle]

Because they are the same mod 23! They are the same element of Z_23.

[severe linden]

This means that $3^{22} \equiv 1 \mod 23$, i.e. the inverse of $\overline{3}$ is $\overline{3^{21}}$ in $\mathbb{Z}_{23}$. But what he's saying is that you probably want a more ``satisfactory" expression for $\overline{3^{21}}$.

[cloud walrus]

Yuese

[severe linden]

So you would want it to be represented by some number between 0 and 23, i.e. you want to calculate $3^{21} \mod 23$.

[cloud walrus]

Yuese

[thin bough]

oh i see

[coral spindle]

Note that we're still stuck at the stumbling hurdle of believing that 3^22 is the inverse of 3 in Z_23, but thank you for elaborating.

[thin bough]

yup still don't get that part

[rocky cloak]

Maybe you should carefully think about what the definition of inverse is

[severe linden]

So, in the above example, you can calculate that $\overline{3^{21}} = \overline{8}$ in $\mathbb{Z}_{23}$, so $\overline{8}$ is a good representation for the inverse of $3$. However, I think you know that $\overline{8} = \overline{31} = \overline{54} = \dots$. My guess is that you are confused in thinking that these are \textit{different} inverses of $\overline{3}$, which isn't true, as they are all the same element.

[cloud walrus]

Yuese

[severe linden]

And as the others pointed out, the inverse in a group is always unique.

[thin bough]

I see. So, I just want the smallest possible value x that gives me x mod 23 = 1?

[severe linden]

More or less. There exist infinitely many $x$ such that $3x \equiv 1 \mod 23$, and they all differ by a multiple of $23$. So you want to choose the (unique!) one which lies in $[0,22]$.

[cloud walrus]

Yuese

[thin bough]

I see

[severe linden]

But if you think about $\mathbb{Z}_{23}, \cdot$ as a \textit{group}, then $\overline{x + 23k} = \overline{x}$ for all $k \in \mathbb{Z}$. In that context these are all one and the same element.

[cloud walrus]

Yuese

[thin bough]

i see yeah

i need to learn how to reduce this large number now

[severe linden]

Yep. As a hint, you might want to see that $3^{21} = (3^3)^7$. What's $3^3 \mod 23$?

[cloud walrus]

Yuese

[thin bough]

4

why did you take the 7 out ?

[severe linden]

Because in this way, we see that $3^{21} \equiv 4^7 \mod 23$ (we can replace the $3^3$ by $4$ as they are the same $\mod 23$). This reduces the number already by a considerable amount.

[cloud walrus]

Yuese

[thin bough]

ok

did you do that in your head or what

[thin bough]

I can't understand that intuitively

[severe linden]

What I did is
[ 3^{21} = (3^3)^7 \equiv 4^7 \mod 23].
Because $3^3 \equiv 4 \mod 23$ (which you found), we can ``replace'' them mod $23$.

[cloud walrus]

Yuese

[severe linden]

It takes some getting used to. The mechanism at work is that when you multiply two numbers, their remainders upon division also get multiplied, in a sense.

[celest furnace]

What is the automorphism group of $\mbb Z/p^n$ for primes $p$ and any integer $n > 0$?

[cloud walrus]

rbmuk

[coral spindle]

The automorphism group of Z/n is (Z/n)^\times

I.e., the group of units.

(I suppose if we're being pedantic it's isomorphic to that group of units!)

[rocky cloak]

It's the cyclic group of order p^n-1(p-1)

[celest furnace]

Is it $Z/\cph(n)$ in general?

[cloud walrus]

rbmuk

[coral spindle]

No

The smallest counterexample is n = 8 iirc

[hot lake]

it's not for p=2

[rocky cloak]

Only for n equal to p^k or 2p^k for odd prime p

(and 2 and 4)

[celest furnace]

Oh dang

What is the automorphism group of Z/8? Z/2 x Z/2?

[hot lake]

yeah

[rocky cloak]

Yeah

[celest furnace]

Wack

Just for curiousity what goes wrong with p = 2 in the odd prime proof

[rocky cloak]

If g is a primitive root mod p^2 (exists for any prime), then write

x = g^(p-1) = 1 + pa

Where a is relatively prime to p. Then

x^p = 1 + (pC1)pa + (pC2)(pa)^2 + o(p^3)

If p > 2, then (pC2)(pa)^2 is a multiple of p^3, so x^p is not 1 mod p^3, thus g is a primitive root mod p^3 and you can inductively show that g is a root mod p^k. If p=2, then pC2 = 1, and the argument doesn't work.

[jaunty void]

Hi everyone I'm going to start studying Abstract Algebra by David Summit anyone care to study it with me

[kindred tree]

How is a fraction field Q(x_1, ... , x_n) in indeterminates x_1, ... , x_n defined by? Is this like a fraction field of the polynomial ring Q[x_1,...,x_n]?

[formal ermine]

Yes, precisely

[kindred tree]

And something like Q(a_1, ... ,a_n) would be a field extension of this if the a_i's are roots of all possible polynomials here?

[formal ermine]

no

[stark helm]

for p(x)=a(x)*b(x), why we can say both a(x) and b(x) are non constant polynomial?

I know that if R[x] is Z[x], then it is definitely correct

but if we can factor that into a constant like 1/3?

I just consder must some constant like 1/3 have inverse in R?

[rocky cloak]

If say a is constant, then it would also be a common factor for the coefficients of p, but you assumed the gcd was 1

[stark helm]

yes, the gcd is 1, but the divisor might also be 1/2, 1/3 since it is the common divisor less than 1?

[rocky cloak]

'Greatest' is with respect to divisibility, so 1 is not a 'bigger' divisors than 1/2.

Saying that the gcd is 1, means that the only common divisors are units.

[rocky cloak]

Surely 1/3 has 3 as an inverse, or what does 1/3 even mean if not

[stark helm]

because I don't know gcd is 1 means the only common divisors are units

[rocky cloak]

We say that d is a gcd of a and b iff both a and b are multiples of d and for any common divisors d' of a and b we have that d is a multiple of d'.

So in particular if the gcd is 1, then any common divisors d' must have 1 as a multiple, i.e. there exists a k such that d'k = 1.

'unit' just means something that has an inverse

[formal fractal]

Confused on how to find all composition series of S3 x Z3

[rocky cloak]

What part makes you confused?

[formal fractal]

I mean what’s the process

Do i just start by finding normal subgroups of S3 x Z3

[rocky cloak]

You can start by finding all the normal subgroups sure

Then just look for chains where the composition factors are simple

[formal fractal]

Well how would you start

Is there a more efficient way

I mean what even are the normal subgroups of a direct product

Wrt the normal subgroups of S3 and Z3

[rocky cloak]

Certainly the product of a normal subgroup from each factor will work, but there can be others.

Another thing you can do (at least if you know the Jordan-Hölder theorem) is to first find a single composition series and identify what the composition factors are. Then a maximal normal subgroup will be the kernel of a surjective map into one of the composition factors, so you can find that instead.

But for an example like this, it's so small I guess just finding all the subgroups and checking which are normal is reasonable

[formal fractal]

Will all composition series have the same length?

[rocky cloak]

They will have the same length and the same composition factors (possibly in different order)

That's the Jordan-Hölder theorem

[formal fractal]

But isn’t {e} < G a composition series

[rocky cloak]

A composition series is usually defined as a chain of submodules

{e} = G0 < G1 < ... < Gm = G

Such that Gi+1 / Gi is a simple group. So this would only be a composition series if G is simple.

If you're working with another definition, then I guess you might have even more series to worry about.

[formal fractal]

Are there any theorems on how to get composition series of a direct product if you know composition series of the components

[rocky cloak]

So if N is a maximal normal subgroup of G, then NxH is a maximal normal subgroup of GxH. So given a composition series of G and H you can interleave them to get a composition series for GxH.

But there can be other composition series. For example in Z/2 x Z/2 the subgroup generated by (1, 1) is a maximal normal subgroup.

[formal fractal]

Ok so hold on {e} < A3 < S3 is a composition series?

[cosmic niche]

Can you guys help me figure out what number 7 is even asking? What is the operation of the group?

I know what a power set is, is taking the power set the operation of the group? It doesn’t make a whole lot of sense to me

[rocky cloak]

The operation could be symmetric difference. I'm guessing whatever you're reading has defined a group operation on the power set earlier.

[lethal depot]

Any chance P refers to permutation group or something else? I've not personally come across power set notation where the set is written as a subscript.

[formal fractal]

@rocky cloak does upvote mean yes

[rocky cloak]

Upvote means yes yes 🙂

[formal fractal]

Ok

So

Lengths of c series of S3 is 3

Z3 only has trivial subgroups so {0} < Z3

Can we say what the length of series will be in direct product ?

[rocky cloak]

Yes, see if you can compose these two series together into a series for S3xZ3

[formal fractal]

Uh

A3 x {0}?

Then {e} x {0}?

Is the length also 3

[rocky cloak]

You're close to something, but what is
S3xZ3 / A3x(0)?

[formal fractal]

Uh

Z3?

[rocky cloak]

Not quite

So in general GxH/NxM = (G/N)x(H/M)

This should feel reasonable

So then the quotient here becomes Z2xZ3, so we should be looking for a bigger subgroup.

[formal fractal]

What do you mean bigger

[rocky cloak]

Like containing A3x(0)

[formal fractal]

Man im lost

[rocky cloak]

Alright, I have to go to bed soon, so I'll just tell you.

(e) < A3 < S3 and (0) < Z3 compose to give the composition series

(e)x(0) < A3x(0) < S3x(0) < S3xZ3

The first has 2 inclusions, the second has 1 and their composition has 2+1 = 3

[formal fractal]

So this can be generalized

[rocky cloak]

Yeah, it even works for any extension, not just direct products

[formal fractal]

Ok

And so

That single composition series is all the composition series?

Theres no others?

[rocky cloak]

No, there's a bunch of them

Even with this composing trick you can do

(e)x(0) < A3x(0) < A3xZ3 < S3xZ3, or

(e)x(0) < (e)xZ3 < A3xZ3 < S3xZ3

Then you also have subgroups that are not of the form GxH for G and H subgroups of S3 and Z3 respectively

[formal fractal]

Ye

As u said before

But they all have 3 inclusions

And is there a way to tell how many there will be

[rocky cloak]

Probably not, idk

Finding all, then counting them does the trick

[formal fractal]

I mean

I cant find any more than 4

Nvm

3

Cant find any more than 3

[rocky cloak]

I count 5

[formal fractal]

Damn

So you found a “nontrivial” subgroup of S3 x Z3 then

[rocky cloak]

Sure

[formal fractal]

Is Z(S3 x Z3) normal?

[rocky cloak]

The center is always normal if that's what you're asking

[formal fractal]

So thats what you found then

Er

[rocky cloak]

The center is just (e)xZ3

[formal fractal]

Ye

Dam

Well what is it lol

[rocky cloak]

So you have A3xZ3 which is just isomorphic to Z3xZ3.

If you look at the subgroups in there you have the two trivial subgroups and then you have 4 more.

Those generated by (1,0), (1, 1), (1, 2) and (0, 1) respectively.

The first and last are Z3x(0) and (0)xZ3, but the other two are non-trivial.

In general when you're looking for subgroups it can be an idea to just look at the subgroups generated by a single element to start with.

[formal fractal]

So i think (123)x(1) is one and

uh

Ths shit is time consuming

[long obsidian]

Do the polynomials in the variable x squared inject into polynomials in a single variable? $k[x^2]\xhookrightarrow k[x]$

[cloud walrus]

HausdorffT1

[slim kayak]

Uh, wdym by inject here

send f(x^2) to g(x) with each exponent doubled?

[long obsidian]

Yeah that's what I mean $(x^2)^3+x^2\mapsto x^6+x^2$ for example. Basically what you said

[cloud walrus]

HausdorffT1

[slim kayak]

Like, y=x^2 gets send to x^2

[south patrol]

I mean yeah

[long obsidian]

Yeah

[slim kayak]

argue by degree ig, only element send to 0 is 0

[south patrol]

lol

Like

[slim kayak]

Yeah

[long obsidian]

Oooooh okay actually I guess that works thank you!

[south patrol]

This is trivial basically

i mean

it is just an inclusion

of a subset

(if you define it as such)

[slim kayak]

this is definitely "freedom of identification" territory

[south patrol]

Well it depends

on how hausdorff is defining k[x^2]

to me that is the subset of k[x] generated as a k-algebra by x^2, so this is trivial

[slim kayak]

My thought was k[y] and iota sends y to x^2

[south patrol]

Oh sure

[slim kayak]

I mean, i stressed about more "trivial" stuff before

[south patrol]

Yeah sure, I just thtought it'd important to ask cause like

Depednign on how it's defined there may be a different answer

[slim kayak]

what is the answer in the subset case?

[south patrol]

Well then it is pure set theory

[slim kayak]

Inclusions of subsets are always injective?

[south patrol]

Ye

Lol

[slim kayak]

now that would be something that would confuse me "wdym there is something to prove here"

[south patrol]

Hm i'veb een thinking recently like a lot of problems are best solved just by thinking about what you are actually talking about more

like with harder problems too i mean

making everthing very explicit and hands on

[slim kayak]

The singular value decomposition of math problems

[south patrol]

lol

Well the example I was thinking about recently was cellular homology

Like if you unpack exactly what we mean when we say it is free on the cells then the cellular boundary formula becomes almost a formality

but otherwise the proof is unenlightening imo

[slim kayak]

the mitochondira is the representing object of the cellular homology functor

[south patrol]

He's so real for that...

[slim kayak]

I feel this might be even clearer with singular homology

There you are hard stuck until you find the right perspective

[south patrol]

Wdym

[slim kayak]

When you are dealing with all continuous functions it becomes essentially to proceed besides with some canonical choices which makes things work

[thin bough]

Find the multiplicative inverse of each nonzero element in ℤ_23

so, I am working on 3 right now. I figured out that (18*9) mod 23 = 1. So, does that mean 9 and 18 are multiplicative inverses of 3?

[south patrol]

No, it means they are inverses of each other

[thin bough]

so, I am doing this wrong ?

[south patrol]

I'm not sure what you are doing to be honest

what are you trying?

[thin bough]

I put the question here

[coral spindle]

I figured out that (18*9) mod 23 = 1.
But you're trying to find the multiplicative inverse of 3. What does this have to do with 3?

[south patrol]

Sure but what are you doing

[coral spindle]

I want you to write down the definition of the multiplicative inverse y of an element x. What property does y satisfy?

[thin bough]

according to me, the multiplicative inverse states that for each value a in G, there exists an element a' such that a*a^'=1.

[coral spindle]

That is the inverse axiom

And indeed, this element a' such that a * a' = 1 is called the inverse of a.

So you are looking for some element — let's call it y instead of a' — such that 3y = 1 mod 23.

Clear?

Search for such an element y.

[thin bough]

I see

is fermets little theroem the wrong way of doing this

is there any other easy way

[south patrol]

Well you only need to do up to like 12 as e.g. 13 = -12

[coral spindle]

Yes, you can just check the options.

[south patrol]

And then I would just stare at it honestly

Like e.g. you can see that 2*12 = 24

Fermat's little theorem would be a pain as there are elements of very high order

[thin bough]

maybe a table would be better?

[south patrol]

I do think just inspection is fastest

[thin bough]

okay.

[south patrol]

Even if you restrict to like 1 up to 12 you'd be drawing a 12 x 12 table

and doing lots of unnecessary computations

Though you could try a table in our head I guess lol like go through options as boytjie said

[formal fractal]

What about <(2,2)>?

[south patrol]

same as the one generated by (1,1)

since (2,2) + (2,2) = (1,1) and (1,1) + (1,1) = (2,2)

[thin bough]

is this a better definition of multiplicative inverse: The multiplicative inverse states that for each value a in G, there exists an element a' such that a*a^(-1)=1.

[south patrol]

I guess a funny way to think about it is that Z/3 is a vector space and a subgroup of (Z/3)^2 is just a subspace

[formal fractal]

Yeah but

[south patrol]

This is a worse version of what you previously had

[formal fractal]

He said this

[south patrol]

I know

Lol typo sorry

[formal fractal]

Why didn’t he include (2,2) in that list

[coral spindle]

No, that's a malformed version of the inverse axiom, which states that an inverse always exists.

[south patrol]

I am well aware what an inverse is

[thin bough]

is this the wrong wiki page ?

[south patrol]

No the wiki page is fine (if slightly informal)

[thin bough]

is it different for number theory

oh

[south patrol]

But you quoted it imprecisely

I mean for example you said a' and then a^-1

[thin bough]

because wiki said a^-1

[south patrol]

Tbf I would also add that a' a = 1 as well

[thin bough]

i guess i messed up on the for each part?

[coral spindle]

Are there any spare hammers around with which to bash my head in

[south patrol]

Well that is just notation

[cobalt heath]

I have nothing to add but I wanna say, that nickname of "The-Elite" is quite unsettling to me

[thin bough]

sorry guys

[south patrol]

Writing a' is better from some points of view, since you are just saying an inverse exists (and then you can prove it exists)

rather than explicitly having to give an inverse

but i wouldn't worry about it much at this point

[thin bough]

ok

thanks

[stark helm]

I am wondering that why d can be decompose uniquely into product of finite irreducibles can imply that p' can be factor uniquely into irreducibles in R[x]? Also why we can also assume that the gcd of coefficient of p(x) is 1?

[barren sierra]

The decompositon of d doesn't imply that p' also decomposes uniquely

They're just saying that since we know d decomposes uniquely, then it suffices to show that p' decomposes uniquely

And then if the gcd of the coefficients isn't 1, we can pull out the gcd as an element of R

@stark helm

[viral mountain]

Consider the map $\phi : \mathbb{Z}_2 \to S_5$ defined by $\phi(0) = \sigma_0$, where $\sigma_0$ is any arbitrary permutation in $S_5$ and $\phi(1) = i$, where $i$ is the identity permutation (i.e. identity element in $S_5$)

When we consider $\phi(0 +_2 1) = \phi(0) = \sigma_0 = \sigma_0 i = \phi(0) \phi(1)$, I can consider this a valid homomorphism right?

[cloud walrus]

BlazingSaber

[viral mountain]

If so, since there are $5! = 120$ possible permutations in $S_5$, and since I can choose $\phi(0) = \sigma_0$ to be any permutation out of these $120$ while fixing $\phi(1) = i$, am I right in thinking that there are at least 120 possible homomorphisms from $\mathbb{Z}_2$ to $S_5$?

[cloud walrus]

BlazingSaber

[dull marsh]

Wait so you are mapping 1 to the identity permutation?

[viral mountain]

To show a visual description of what I mean, here's what I mean by choosing to map $0$ to all 120 of the possible permutations 1-by-1, while fixing the mapping of $1$ to $i$ (in the image, I choose to represent $\sigma_1$ as the identity

[cloud walrus]

BlazingSaber

[viral mountain]

yes

[dull marsh]

That makes phi(0) = phi(1 + 1) = phi(1)phi(1) = identity permutation

[viral mountain]

hmm, fair
I didn't realise I'd need to satisfy the homomorphism property $\phi(a *_1 b) = \phi(a) *_2 \phi(b)$ for all possible $a, b \in $\mathbb{Z}_2$

[cloud walrus]

BlazingSaber

hmm, fair
I didn't realise I'd need to satisfy the homomorphism property $\phi(a *_1 b) = \phi(a) *_2 \phi(b)$ for all possible $a, b \in $\mathbb{Z}_2$
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[dull marsh]

So, when you are constructing homomorphisms

Keep in mind that a homomorphism is uniquely defined by the image of the generators

Here you have a homomorphism from Z2 to some group

1 is the generator of Z2, so you only need to decide where 1 goes

And the rest will be built up on that

And I think homomorphisms should preserve identities as well hold on

Yeah

So 0 gets mapped to the identity permutation anyway

[viral mountain]

Yeah you're right

[dull marsh]

And 1 to some element of order 1 or 2

[viral mountain]

Why order 1 or 2 specifically?

[dull marsh]

In general when you have a homomorphism, order of the image of an element is a divisor of the order of that element in the domain

It's not necessarily the same order in the codomain as in the domain

[viral mountain]

I see

I think these properties would be handy to keep in mind while trying to construct the homomorphisms

[dull marsh]

Yeah

I'd say phi(x^n) = phi(x)^n is the most important property to keep in mind

Besides the definition itself

[viral mountain]

right fiar

[viral mountain]

So if I want to find the number of homomorphisms from $\mathbb{Z}_2$ to $S_5$, I'd need to consider only those maps which map 0 to the identity permutation

And since we must have $\phi(x^{-1}) = \phi(x) ^{-1}$, It is the case that the inverse of 1 must map to the inverse of the permutation which 1 maps to

And since the inverse of 1 is also 1 in $\mathbb{Z}_2$, 1 must map to only those permutations which are their own inverses, correct?

[cloud walrus]

BlazingSaber

[dull marsh]

Yeah

Which are again the elements of order 1 or 2

Basically you can identify the homomorphisms from Z2 to S5 with the set of elements of S5 with order 1 or 2

There is a one-to-one correspondence

[celest cairn]

How do I find the quotient group
$\frac{\mathbb{Z}{6}^*}{\mathbb{Z}{3}^*}$?

[cloud walrus]

Sapphire

[dull marsh]

$\bZ_3^\times$ is not a subgroup of $\bZ_6^\times$ though

[cloud walrus]

孤独な豆

[dull marsh]

What do you mean by that exactly?

[south patrol]

I assume you mean include Z/3 -> Z/6 via x |-> 2x

Then you get an inclusion (Z/3)^x -> (Z/6)^x ig

But I mean like just use the chinese remainder theorem on Z/6 ig

or count

[lime lagoon]

Hello! How can I prove that a group $G$ has exactly one element of order 2?

[cloud walrus]

juan

[south patrol]

Well that is very dependent on what G is, and how G is given to you

(it certainly isn't true for all G)

[lime lagoon]

I did show here the existence and uniqueness

but i am lost at the uniqueness part

fyi, $G$ is abelian group of order $2n$, where $n$ is odd

[cloud walrus]

juan

[south patrol]

Sure, that's an important hypothesis

[sonic coral]

cauchy

[south patrol]

Cauchy proves existence, not uniqueness

and juan has already got existence

[lime lagoon]

i did not use Cauchy but Lagrange

[south patrol]

Uh how did you use Lagrange?

[lime lagoon]

idk if these arguments are valid but...

Let $ a\in G$. Since $G$ is a finite group, then the order of $a$ must divide the number of elements in $G$. Thus, the order of all elements of $G$ is either 1, 2, $n$, and $2n$, which are the divisors of $2n$. And thus, $G$ has an element of order 2.

[cloud walrus]

juan

[lime lagoon]

isn't it wrong?

[dull marsh]

Pretty sure 2n could have other divisors

Even if n was prime, in which case the divisors of 2n would indeed be only 1, 2, n and 2n, your conclusion still doesn't follow immediately

Anyway, you may want to refer to Cauchy theorem for showing existence

Hmm hold on

S_3 is 2 * 3 elements, but there isn't exactly 1 element of order 2

Are you sure what you are trying to prove is true?

[south patrol]

Yeah

[south patrol]

It isn't abelian

What juan is claiming is true e.g. by classification of finite abelian groups

[dull marsh]

Ah I missed that part

[south patrol]

Though of course there must be an easier way

[south patrol]

It's teh wrong way round.

(at least, without further work)

However you could just do a count of elements, namely: only one element has order 1. If there's an element of x order 2n, then x + ... + x (n times) has order 2

and then maybe you can show that you can't have all other elements of order n

Wait yeah there are other divisors of n anyway

[lime lagoon]

this is what the problem says

[south patrol]

It is true dw

[rocky cloak]

Let ||x and y be two elements of order 2, then (x, y) is a subgroup of order 4, so the order of G is a multiple of 4||

[south patrol]

Yeah

nice

Okay sure that is what I had in mind except I realised I'm pretentious lol

like ||pick x of order 2, then G/<x> has order n, hence no order 2 elements lol||

[lime lagoon]

do you mean those parenthesis in $(x, y)$ as generator, like$ \langle x, y \rangle $?

[cloud walrus]

juan

[south patrol]

Ye

[lime lagoon]

okayy, okay and, how do you come up that it is a subgroup of order 4?

[dim widget]

well maybe think about the possible multiplication tables of size 4

[rocky cloak]

You just compute explicitly what the subgroup is, i.e.

{0, x, y, x+y}

And then count the elements, and see that it's 4

[lime lagoon]

i think i got what it means

so, does it mean i will use the contradiction method, where there's no other element of order 2?

[slim kayak]

in an abelian group if you had two element of order 2, then they produce a subgroup of order 4. You know that G has order 2n with n odd. Can you see the problem?

[delicate orchid]

I can't get over this it's too funny

[slim kayak]

I am not sure even if I was paid for this id recover

[coral spindle]

Nah this doesn't work. 1 gets sent to 2, which isn't a unit in Z/6. The question just doesn't make sense to me, oh well.

[south patrol]

True

[delicate orchid]

They’re both just C_2 OmegaLUL just isomorphism them

[south patrol]

I was like eh group of units is a functor but the map I said is not a map of rings

Lol

Rings with identity at least

[coral spindle]

Yeah this is true, but this is an xy problem. There's some horrifying misunderstanding that has occurred lmao

[delicate orchid]

I don’t know what that means. Just do it and stop complaining

[coral spindle]

https://xyproblem.info/

[delicate orchid]

Nice virus

[coral spindle]

Bish

[delicate orchid]

Just compute the quotient lil nerd…

[coral spindle]

I'm a big nerd actually

[rocky cloak]

Z/6 = Z/2 x Z/3, so (Z/6)^* = (Z/2)^* x (Z/3)^*

So there is a pretty natural inclusion of (Z/3)^* into (Z/6)^*

[coral spindle]

Ey there we go

[fading field]

lol my friends and I were talking about how much we hated this website

[delicate orchid]

Adjoints would never commute with biproducts :upon: they would never do this

[fading field]

that and nohello

[barren sierra]

what's wrong with it

[rocky cloak]

This joke is too advanced for me

[chilly radish]

Also lmgtfy

At least that one is sometimes appropriate

It's usually just people trying to make conversation though

[cloud solar]

Let A be a ring and b an idempotent element. Prove that for every x in A there exists 2 uique elements y in Ab and z in A(1-b) s.t. x=y+z

I proved that Ab intersected with A(1-b) is only 0

Idk if that helps

[agile burrow]

In proving that the intersection is 0, you might've noticed that b(1 - b) = 0. Using this, there's really only one natural choice for y and z

[rocky cloak]

You may also notice that b + (1-b) = 1

[elder wave]

:uponthewitnessing:

[cloud solar]

xb+x(1-b)=x so y=xb and z=x(1-b)

And what argument should I use to say that is unique

[rocky cloak]

That's existence, then you can use Walters hint for uniqueness

[cloud solar]

Okay thanks guys

[delicate orchid]

hey guys. Wew Lads Tbh here to explain the joke. The joke is that -^* is an adjoint functor and so commutes small limits/colimits, and so what you wrote holds.

[rocky cloak]

Nah, mate. They would never

They promised me not, and adjoints wouldn't lie to me

[delicate orchid]

EXACTLYYYYY

[slim kayak]

Still don't get it, I thought you could view biproducts as either products or coproducts and kick them through the adjoint

[delicate orchid]

adjoints would NEVER do this

they are better than stooping to such levels

[celest furnace]

Is there a theorem like the folllowing: Let F = <a_1,…,a_n | R> be a presentation for a group F. If G is another group generated by n elements b_1,…,b_n satisfying the same relations, is there an embedding G -> F?

Seems super true by sending like a_I to b_I but not sure how to show injectivity

[slim kayak]

take a word
assume its reduced
send it through the map
reduced or not?

[delicate orchid]

^

there was something in my course called "the substitution test" which is annoyingly nonstandard but basically answers this question directly

[slim kayak]

on my end you reacted in 100 ms wtf

[delicate orchid]

I'm good at what I do

[slim kayak]

you are right, adjoints would never do this

[delicate orchid]

proof by bashing two diagrams together

[slim kayak]

the solution set condition was made by category theorists to sell more complete cats

x \theta?

[final dragon]

how would R/I look like if R is a ring that consist of continuuos functions that go from the interval [0,1] into real numbers and the ideal contains functions that at 1/2 they are 0.

evaluated at 1/2 are 0

i dont understand how would i create R/I

like what is the proces

[slim kayak]

Think about it geometrically

[final dragon]

but i dont even understand what R/I is supposed to be

[slim kayak]

what does R/I means? It means the ring consisting of "cosets" r+I. So you should ask yourself what would be sensible representatives of these cosets?

[slim kayak]

w h a t

were you that guy who had abstract algebra before linear algebra?

[final dragon]

correct

[slim kayak]

of course

given a normal subgroup, do you remember what G/H was?

[final dragon]

so how would i get the cosets

[final dragon]

we do rings

before groups

i dont know groups

[slim kayak]

Right

You remember what ideals were

[final dragon]

yes

[slim kayak]

Given a ring R and an ideal I you define the quotiuent ring R/I as the set of r+I with r an element of R. You can check that this forms a ring

[final dragon]

so its where i add all the elemnts r from R with the Ideal of R

[slim kayak]

{r+I : r in R}

[final dragon]

so that is the quotiuent ring

[slim kayak]

yes, with the obvious structure

[final dragon]

but like i have a task determine R/I

what is there to determine

[slim kayak]

(a+I)+(b+I)=(a+b)+I
(a+I)(b+I)=ab+I
you should check that this is well-defined

the ring, up to isomorphism

[rocky cloak]

Do you know the first isomorphism theorem?

[final dragon]

yes

[slim kayak]

if I asked what is "2+5+3+6" you would give me the number this simplifies to, in this case you are meant to find what "simple" ring R/I is isomorphic to

[rocky cloak]

Can you imagine a ring homomorphism that maps functions that are 0 at 1/2 to 0?

[celest furnace]

Awesome thanks

[final dragon]

i dont know

[rocky cloak]

Here's a hint

[final dragon]

its the ideal

[coral spindle]

Can you think of a function that maps functions like that to 0? You can check if it's a ring homomorphism later.

[slim kayak]

Maybe start with a simpler example. Take functions from {1,2,3} to Z. Can you see what this ring mod the ideal of functions that are 0 on "2" should correspond to?

[final dragon]

no sorry i dont understand

[coral spindle]

I'll just assume that you need to think about it for a bit

[final dragon]

is it like ker

[coral spindle]

Is what like ker

[slim kayak]

cant even spell my name right smh so rude

[coral spindle]

The kernel of a ring homomorphism is an ideal, if that's what you're asking

[slim kayak]

Any ring homomorphism induces a quotient ring by the isomorphism theorem and vice versa, they give you different way of thinking about problems

take the ideal (3), what is Z/(3)?

[final dragon]

is it like 0+ideal,1+ideal,2+ideal

[coral spindle]

Those are the elements of Z/(3)

[final dragon]

yea

[coral spindle]

What I think Kerr is really asking is

what familiar ring is Z/(3) isomorphic to

[slim kayak]

Yes, but how did you arrive at this?

[final dragon]

because at 3 module 3 its like starting at 0 again

[slim kayak]

You probably noticed that given n+(3) you can just remove multiples of 3 until it is one of those three, right?

[final dragon]

correct

but how would i do this with functions

[slim kayak]

well, try with this first

the ideal has functions that are 0 on 2 but whatever on 1 and 3 right

[final dragon]

yea

so what are they on 1 and 3

[slim kayak]

any element in Z

R = functions on {a,b,c} sending each element to an integer
I = functions on {a,b,c} sending a and c element to some integer, and b to 0.

[final dragon]

okay and that would represent like my problem here

[slim kayak]

somewhat? the point is that you get some warm-up practice

[final dragon]

so is R/I the set of all functions that are not 0 in the value 1/2

[slim kayak]

No, how did you conclude that

[final dragon]

idk

[slim kayak]

idk what else to say atm, try working this out

remember that functions correspond to pairs of points, or equivalently:
A function f is determined by what f(a),f(b) and f(c) are

[final dragon]

can i deffine some mirroring s.t fi(f(x)) where f(x) is in R if x is 1/2 its 0 and if x is different from 1/2 its 1

and use the first homomorfism theorem

like fi:R->{0,1}

idk i go to sleep i cant do this anymore

[delicate orchid]

how is this anything other than... just what it is?

[slim kayak]

re-doing something you know the answer of by hand to get some practice

[delicate orchid]

fair enough

[coral spindle]

||Often when people see Z_3 for the first time it is defined (annoyingly) as the set of literal integers {0, 1, 2} with addition being composed with taking remainders. People often do not know it as a quotient ring immediately, and I think it is obvious that dzonzas does not know it as a quotient ring.||

[delicate orchid]

elementary number theory ruins yet another thing

[coral spindle]

BAD PEDAGOGY

[rotund aurora]

Doesnt this dude play chess

[coral spindle]

No he plays with my heart

reply to my dms, levy....

[delicate orchid]

boytjie... your reaction to uponthewitnessing being added to mathcord is most underwhelming...

[coral spindle]

Holy shit

[delicate orchid]

all pedagogy is bad. Make students want to learn themselves ykwim

[coral spindle]

I assumed u just got nitro

I have unlimited power now

[rocky cloak]

[delicate orchid]

I don't like supporting the millitary industrial complex via a bunch of freakazoid furries

[slim kayak]

whats the lore of

is there a 5h wendigoon video on it

[coral spindle]

It's when u witness something

And it's a funny face that some chess player makes idk

[summer path]

why is the name so short

i thought it was max length name

[coral spindle]

"the name" is 7 characters

so quite short yeah

This is a certified algechill moment

[summer path]

this is basically algechill

[slim kayak]

rip 😔

[rotund aurora]

free algechill

[delicate orchid]

:heyguyspleasenamethisemojisomethin:

my personal favourite atm is ":emoji:"

[summer path]

that sounds like something i would name an emoji (because i suck at naming things)

[delicate orchid]

:thecosmoshumswithatunemostsweet:

anyway lets not shit up the channel too much

[summer path]

where should we shit up instead

[delicate orchid]

ermmm.. please keep memes in #chill or I will be forced to hand out mutes

[rocky cloak]

Damn, wews power really went to their head

I guess we'll have to rename the channel to #nochill

[slim kayak]

#algechill^op

[south patrol]

Real imo

This has caused many tedious conversations for me

[coral spindle]

Btw someone just yesterday thought that the multiplicative inverse of 2 in Z_3 would have to be 0.5, hence it doesn't exist (or something to that effect)

Wonderful pedagogy, thank u so much

Make 👏 Z_3 👏 a 👏 quotient 👏 again

[slim kayak]

yikes

[coral spindle]

You can totally see why they'd think that

Because they've been taught that the elements of Z_3 are just integers, when it's deeply unhelpful to think of them that way

[slim kayak]

The argument for introducing i'd make for introducing Z_n or at least some examples before quotient rings is to get something you can actually interact with instead of having to go through a kinda tedious sequence of working through ring theory up until quotients

the ideal is thinking of some rings that arent quotient rings for the time being, but some people cant be bothered

[coral spindle]

Yes butttttttt I think this should really be fixed at the level of modular arithmetic, not rings

So many people think that mod just means remainder

I see this mistake constantly

If the equivalence class approach was taught first this could be avoided

[slim kayak]

modular arithmetic became more interesting after learning ring theory lol

probs cuz the remainder pedagogy kinda sucks

[south patrol]

Oh and also don't call ti Z_3

:)

[coral spindle]

Also facts but more forgivable

[summer path]

I think if you're seeing mod for the first time in a discrete class or something in high school, it's much easier to end up with thinking it is remainder

[slim kayak]

idc number theorist smh

[south patrol]

The annoying thing is in topology where it somehow seems standard to use Z_n a lot, but then p-adics are also common enough

Lol

[coral spindle]

Aside: programming languages calling the % operator 'mod' and then having things like -6 % 5 = -1 is truly criminal and frankly should be punished by 10 years in the tickle chair

[slim kayak]

what language did that

[summer path]

many of them?

[slim kayak]

this is terrible

friendship with oop's ended

[coral spindle]

#include <stdio.h>

int main() {
    int result = -6 % 5;
    printf("-6 'mod' 5 = %d", result);
    return 0;
}

This C code outputs -6 'mod' 5 = -1

Just tested it now

Javascript also does this, at least in my browser.

[rocky cloak]

I remember writing some python code once, where I had to use

(( x % p) + p) % p

To not get weird bugs

[coral spindle]

I just checked python now actually and it seems to work correctly, probably due to a newer version than the one you were using then. I was pleasantly surprised to see this, and ruby also does this correctly which is nice.

(By correctly, I mean that -6 % 5 = 4 ofc)

[rocky cloak]

Maybe it wasn't even python, I don't remember. But I've certainly had to write that horrible expression

[summer path]

what would happen if you did not use that expression

[rocky cloak]

Then it would say things like -6 and 4 not being equal modulo 5

[celest cairn]

Hi

Can I say that $\frac{\mathbb{Z}^{}_6}{\mathbb{Z}^{}_3} \cong C_2$?

[cloud walrus]

Sapphire

[coral spindle]

It took us a while to see how these groups embed

Sapphire, do you know a nice formula for |G/N| in general?

[celest cairn]

Nope

[coral spindle]

Well I'll tell you.

|G/N| = |G| / |N|

Nice huh?

[celest cairn]

Oh yeah ik that

[coral spindle]

So what's |Z_6^\times| and what's |Z_3^\times|

Then tell me what the order of the quotient is

[celest cairn]

3 and 2?

[coral spindle]

How did you calculate that

If you think that Z_3^\times is a subgroup of Z_6^\times, then you should be able to immediately see the problem with what you've just written.

Do you know a nice name for |Z_n^\times|? This is a function of n which people talk about a lot.

[celest cairn]

A function? Not sure.

[coral spindle]

It's called phi(n), aka Euler's totient function.

[celest cairn]

Oh ok, yes I’ve heard of it

[coral spindle]

If you are at all familiar with this, you should be able to calculate |Z_6^\times| immediately.

I'd like you to try working out |Z_6^\times| and what's |Z_3^\times| again, this time showing your working.

Last time you threw out numbers and you were wrong, so I'd like to see how you get your numbers.

[coral spindle]

By the way, did you see the problem that I was alluding to? I hope you did.

[celest cairn]

Oh wait. Does Z_6^\times have order 2? Z_6^\times = {1, 5}?

[coral spindle]

It does indeed have order 2.

Now you can tell me the order of the quotient.

[celest cairn]

Is it just 2/2 = 1?

[coral spindle]

So what is the quotient, up to isomorphism?

[celest cairn]

Z_6^* / Z_3^* is iso to {e}?

[coral spindle]

It is the trivial group, yes.

[celest cairn]

Dang ok. Wrote something interesting but now ik it’s wrong :/

[coral spindle]

You haven't answered this, so I assume you didn't see the problem

The problem is the Theorem of Lagrange.

2 does not divide 3, so a group of order 3 cannot contain a group of order 2 as a subgroup.

[celest cairn]

Oh that’s simple, alright

[coral spindle]

Yes, it's all very simple

[celest cairn]

Thanks for the help

Also, how does the Galois correspondence work for $Q(\zeta_3) \subseteq Q(\zeta_6)$?

[cloud walrus]

Sapphire

[uneven bobcat]

It doesn't tell you much, because the extension is trivial; they're the same field.

$2=[\mathbb{Q}(\zeta_6):\mathbb{Q}]=[\mathbb{Q}(\zeta_6):\mathbb{Q}(\zeta_3)][\mathbb{Q}(\zeta_3):\mathbb{Q}]=[\mathbb{Q}(\zeta_6):\mathbb{Q}(\zeta_3)]2$,

hence $[\mathbb{Q}(\zeta_6):\mathbb{Q}(\zeta_3)]=1$, and thus $\mathbb{Q}(\zeta_6)=\mathbb{Q}(\zeta_3)$.

[cloud walrus]

kr1staps

[uneven bobcat]

Therefore $\text{Gal}(\mathbb{Q}(\zeta_6)/\mathbb{Q}(\zeta_3))={\text{Id}}$.

[cloud walrus]

kr1staps

[celest cairn]

Ah, alright thank you

[south patrol]

Also generally $[\mathbb Q(\zeta_n):\mathbb Q] = \deg \Phi_n$ (nth cyclotomic) $ = \varphi(n)$ (Euler phi function) and in particular $\varphi(6) = \varphi(3)$ (in fact more generally $\varphi(2n) = \varphi(n)$ for odd $n$)

[cloud walrus]

potato

[south patrol]

Or like, if c is an nth root of unity (n odd) then -c is a 2nth root of unity (since (-c)^n = (-1)^n c^n = -1)

[crystal vale]

If G has order p^n for some p -prime number then how can I proof that G has normal subgroup of order p^(n-1)

[next obsidian]

Inductionnnnnnnnnnnnn

[limber sequoia]

So groups have these property that a homomorphism totally embeds a substructure

As in G/kerN is like G if there was a "bigger trivial member"

And the isomorphic theorems I,III show that it works to all th things you'd think it does

Is there an abstract name for this property? That every homomorph also defines a full substructure?

[coral spindle]

This first isomorphism theorem, arguably

[cloud solar]

I have a question

If G is a finite group of order p^n with n>=2 it contains a subgroup of order p^2 by sylow 1?

[south patrol]

Well @crystal vale and @cloud solar - both your questions can be answered if I mention the following technique for finite p-groups (groups of order p^n). Any such nontrivial group G has non trivial centre (use the class equation) Z(G). If Z(G) is a proper subgroup, then since it is normal we can consider the non trivial, smaller p-group G/Z(G). Subgroups of G containing Z(G) correspond to subgroups of G/Z(G). If Z(G) = G then you can just pick the subgroup generated by a single element to get a normal subgroup anyway.

You can use these things just to induct on stuff - maybe have a go

But basically you use the centre to cut down the size of ur group

[cloud solar]

Oh so we cant use sylow. Sylow tells us that we can find a subgroup of order p^k with k the muliplicity of p in the order of the group

[south patrol]

Well if your group is order p^n Sylow won't tell you much

(unless you use better refinements of it which include this as part of the statement)

But the point is that you can use induction to find those subgroups of any order p^k (k <= n)

By considering either Z(G) or G/Z(G)

(except the abelian case, which is easier)

tteg is this the first example of devissage in the ug curriculum

[mighty kiln]

They are isomorphic as Lie groups

[dim widget]

Here is a nice alternative solution. Let $H$ be a subgroup of a finite group $G$ of index $p$, with $p$ the smallest prime dividing the order of $G$, then $H$ is normal (in particular any $H$ of index $2$ in any finite group is normal). Proof: the action of G on G/H induces a map $G \to \text{Aut}(G/H)$ by left translation. Since $G/H$ has order $p$ this induces a map $G \to S_p$, but because $G$ has no elements of order less than $p$ and this map cannot be trivial this map has to map all elements of $G$ to either the identity or $p$-cycles, in particular the image is a p-sylow subgroup of $S_p$ and thus of the form $\mathbb{Z}/p$. But the kernel of this map is $\cap_{g \in G/H} gHg^{-1}$, whence $H$ is normal.

Now let $G$ be a group of order $n$, let $p^a|n$. Then there exists a subgroup of $G$ of order $p^b$ for all $b \leq a$. For $b = 1$ this is Cauchy's theorem. So suppose $G$ has a subgroup $H$ of order $p^{b-1}$. Then $G$ acts on $G/H$, a set of size divisible by $p$, and the kernel is contained in $H$. Thus the image is a subgroup of $S_{|G/H|}$ of order divisible by $p$, so it contains an order $p$ cycle by Cauchy's theorem. Thus $G$ contains a subgroup of order $p^b$.

[cloud walrus]

kålrot

[dim widget]

putting the two paragraphs together we have your result, and indeed something much stronger.

They are the same in every conceivable way except considered as specific sets of matrices in a very specific way.

[mighty kiln]

A Lie group is a manifold with smooth group structure

Isomorphic as Lie groups means there is a diffeomorphism that is also an isomorphism of groups

[dim widget]

There is even a holomorphic isomorphism between the two groups, which is what underlies all of the richness of complex analysis

[mighty kiln]

What do you mean the "same" group anyway?

[coral spindle]

there is a more "abstract" definition of a group, which is as a manifold.
You're misunderstanding a bit maybe. A group alone is not typically a manifold. As Arki says, the 'abstract' definition of a Lie group is a group which is also a manifold in a compatible way.

We talk about Lie groups being 'the same' when there is an map which simultaneously preserves the group structure and also the manifold structure between the two. In this sense, U(1) and SO(2) are 'the same'.

[dim widget]

U(1) \cong S^1 \cong SO(2) where the isomorphisms have every nice structure you could ask for

[coral spindle]

Idk what you mean by that

[mighty kiln]

^

There's no canonical "sameness" built into math

Other than maybe equality of sets in ZFC

[coral spindle]

They are the same in any reasonable sense of the word

Have you seen finite groups? Are you aware of what we mean by an isomorphism?

[dim widget]

although for objects like these with only two automorphisms, they are pretty canonically the same

[coral spindle]

I don't really know what you mean by that. Any representation of SU(2) also gets you a representation of U(1), and vice versa.

[dim widget]

it is similar, although in one situation you have faithful representations of different dimensions, and in the other situation you have one Complex representation and the underlying real representation

So there is a natural representation of $S^1$ as the set of $(a, b) \in \mathbb{R}^2$ with $a^2 + b^2 = 1$

[cloud walrus]

kålrot

[coral spindle]

Oooh you didn't mean like, linear representations. Lol.

[mighty kiln]

(Representation means a homomorphism G → GL(V) for some vector space V)

[dim widget]

considering this as describing complex number $a + bi$ and letting this act on $\mathbb{C}$ as a one-dimensional representation $\rho_{\mathbb{C}}$ by multiplication we get an isomorphism of $S^1 \to U(1)$

[cloud walrus]

kålrot

[south patrol]

I haven't seen this, nice

So like an inductive proof of Sylow

Well the 2nd part I mean

[dim widget]

Now since this representation is $\mathbb{C}$-linear it is also $\mathbb{R}$-linear, so there is also a map $S^1 \to Gl_2(\mathbb{R})$ from taking the basis $1, i$ for $\mathbb{C}$ as a real vector space

[cloud walrus]

kålrot

[dim widget]

This induces the usual isomorphism $S^1 \to SO(2)$.

[cloud walrus]

kålrot

[dim widget]

Coming from the underlying real representation $\rho_{\mathbb{R}}$

[cloud walrus]

kålrot

[dim widget]

yep, it's nice

[cloud solar]

But this technique proves that for a p-group of order p^n it contains a subgroup of order p^k 1<=k<=n or just says that the number of p-subgroups of G is compared to something?

[south patrol]

No it gives ou subgroups of all n

Like, if Z(G) is of order k and G of order n where 1 < k < n, then Z(G) gives you subgroups of order p^l for all l <= k by induction, and G/Z(G) gives you subgroups of each order between Z(G) and G

[cloud solar]

I understand. Thank you

[hidden wind]

mwahaha i attended the lectures for a course in abstract algebra two years ago, but i was not ready at the time, lol

i feel so powerful looking over the material anew, clearing up old confusions and actually understanding most of it this time

mathematical maturity go brrr

[glossy crag]

What's the motivation for nilpotent groups? Solvable groups and their connection to radicals and composition series are clear to me, but the defining condition for nilpotency feels strange to me.

[dim widget]

is this finite groups or lie groups?

or just abstract groups?

[glossy crag]

Does it make a difference if the group is finite or not?

[dim widget]

not really i guess

[glossy crag]

Regular groups, although I was reading about the definition for Lie algebras, so if you could explain the connection for that too that'd be nice. I'm guessing a Lie algebra is nilpotent if and only if the associated group is or something like that?

[dim widget]

i added abstract groups as an afterthought

[dim widget]

I think of nilpotent groups as being like some generalization of p-groups in general, in that they are the maximal class of subgroups where the technique of "considering the center (which must be nontrivial) and use induction" is useful for resolving properties of the group

For finite groups nilpotent groups are exactly just direct products of p-sylow subgroups so this really holds up

In lie theory solvable lie groups morally correspond to subgroups of the upper-triangular matrices of Gl_n for some n, and nilpotent ones correspond to subgroups of the strictly upper-triangular matrices (i.e. trivial diagonal). However if you take this analogy too literally it may lead you astray

[glossy crag]

That's a neat explanation, thanks. Is there some kind of "length" analogue for nilpotent groups that we can induct over for when the group is infinite, like for groups with composition series?

[dim widget]

yeah you can resolve nilpotent abstract groups by cyclic groups, and you can ask about the length of this filtration

just like with all solvable groups

or you can ask about the length of the ascending central series or the descending central series or about...

I think the most natural filtration on nilpotent groups is the classic one $N_i = [N_{i-1}, N_{i-1}]$

[cloud walrus]

kålrot

[dim widget]

and $N_0 = N$

[cloud walrus]

kålrot

[glossy crag]

Muchas gracias

[crystal vale]

If Z(G)=G then how I am sure that there is an element of order p^(n-1)

[south patrol]

Well that is smth I think I mentioned there

You pick an element of order p by Cauchy, call it g

Then <g> is a normal subgroup of order p

[crystal vale]

Yes

[south patrol]

Then do the same thing with <g> instead of Z(G)

Really all that matters is that every p group of order > p has a proper, non-trivial normal subgroup

[crystal vale]

And I don't understand what's mean of Subgroup of G containing Z(G) correspond to subgroup of G/Z(G) ?

[coral spindle]

You may know this result as the correspondence theorem. If you don't already know it, commit it to memory now as it is very important!

Every normal subgroup of G/N is of the form H/N where H is a normal subgroup of G containing N.

The map sending H to H/N is a bijection {normal subgroups of G containing N} <-> {normal subgroups of G/N}.

[native sequoia]

Does anyone know where I can find a good introduction to skew group rings/algebras?

[crystal vale]

And how I use this in solution?

[coral spindle]

I think you should think about that in your own time. Potato has elaborated enough, I think.

[crystal vale]

Okay I will figure out

Thank you

[rocky cloak]

When I was writing my bachelor thesis I read Cohen Macaulay representations by Leuschke and Wiegand (chapter 5).

I think it was a good enough introduction, but maybe depends why you're interested idk

[crystal vale]

Thank you, I need to study more theory to understand this........

[native sequoia]

I'll try that

[hidden wind]

i love making these

[dull marsh]

Find all pixel arts that represent a cayley table of some group

[hidden wind]

lol

these two are isomorphic (group of units mod 36 and the direct product Z2 x Z6)

[dim widget]

Z5 I hope

[hidden wind]

wdym?

[dim widget]

there are 30 units mod 36

so it should be Z5 \times Z6 not Z2 \times Z6

[hidden wind]

no there are are 12

[dim widget]

oh lol

true

i thought 6 was prime for some reason

sad

[hidden wind]

we've all been there

[rocky cloak]

The smallest Grothendick prime has been found!

[crystal vale]

In last paragraph I have a doubt, if I am correct then G acts on G/H by g->gH then kernel=H and if order of image of this mapping is divisible by p , how we conclude that it contains an order p cycle....(what p cycle) ..... cyclic group of order p?

[dim widget]

the kernel is not necessarily H

it would be H if and only if H is normal

[crystal vale]

Is mapping correct?

[dim widget]

in general the kernel is the intersection of gHg^{-1} for all g

yes the mapping is correct

but your "kernel" is just the set of elements which fix the identity coset

[crystal vale]

Yes

[dim widget]

that's not the kernel of the homomorphism G \to S_{|G/H|}

but anyway the point is that the kernel is contained in H

whence the image has to have order divisible by p

since p||G/H|

[crystal vale]

Yes

[dim widget]

But the image is also a group

and any group of order divisible by p has an element of order p

[crystal vale]

So it has an element of order p

Yes

[dim widget]

okay but we know more

because S_p has only elements of orders divisible by primes q where q <= p

so in fact the only elements of S_p which can be in the image of G are the identity, and the p-cycles

[crystal vale]

But for mapping G to G/H if g belong to Kernal then g maps to what ...... H?

[dim widget]

So then the image of G in S_p consists only of elements of order p or 1, i.e. it is contained within a p-Sylow subgroup

since p|p! and p^2 does not divide p! this means that the image of G is just Z/pZ

ah wait this is the second part that youre asking about

[crystal vale]

Yes

[crystal vale]

What p cycles...... Permutation of p elements?

[dim widget]

yes the transitive permutations which take any element to any other

[dim widget]

okay I think this is a good point, I need to do a little more work here:

so the fixed points of H acting on G/H are exactly the cosets of N_G(H)/H

under the natural inclusion

but if a p-group acts on a set of size x where p|x then the number of fixed points for the action is divisible by p

(you can see this from many things, for instance the class equation)

So this implies that p|[N_G(H):H]

so now without loss of generality we can let G = N_G(H) and assume that H is normal

[crystal vale]

What N_G(H)

[dim widget]

the normalizer of H in G

[crystal vale]

Oh

[dim widget]

the set ${g \in G| gHg^{-1} = H}$

[cloud walrus]

kålrot

[crystal vale]

So from first paragraph you mean kernel is of order p^(n-1) ?

[dim widget]

i don't understand what you mean by "from first paragraph"

and the kernel of what

[crystal vale]

Kernel of mapping G to G/H

[dim widget]

it's not the mapping from G \to G/H

it's the mapping from G \to Aut(G/H)

[crystal vale]

First paragraph of this

[crystal vale]

Yes

[dim widget]

the kernel of that map has order p^{b-1} = |H| if and only if H is normal

[crystal vale]

I am not sure about this action and mapping

[dim widget]

Okay

the mapping is this

if $xH$ is a coset of $x \in G$

then $g \in G$ takes $xH$ to $gxH$

this gives an automorphism of $G/H$ as a set

i.e. just a bijective map $T_g: G/H \to G/H$

if$ xHx^{-1} = H$, then for all $h \in H$, we have $hxH = xh'x^{-1}xH = xh'H = xH$

[crystal vale]

Sorry but I am not familiar with this notations if xH is a Coset of x \in G what is meaning of ?

[dim widget]

so the fixed points of the action of $H$ on $G/H$ are exactly the cosets $xH$ where $x \in N_G(H)$

[dim widget]

$xH$ is the set of all elements $xh$ where $h \in H$

that is the definition of a quotient of a group by a subgroup

it's the set of those subsets of G

[crystal vale]

Yes

I am asked for \ this

[dim widget]

what?

[cloud walrus]

kålrot

kålrot

kålrot

kålrot

kålrot

kålrot

kålrot

[dim widget]

okay I typeset everything in latex

now you can read it like this

[crystal vale]

Thank you

[summer path]

many lines

[dim widget]

we are proving the sylow theorems

[crystal vale]

So kernel of mapping G to Aut(G/H) is N_H(G)?

[dim widget]

no

the kernel is the set of $g \in G$ such that $gxH = xH$ for all $x \in G$

[cloud walrus]

kålrot

[dim widget]

this only happens when $x^{-1}gx \in H$ for all $x \in G$

[cloud walrus]

kålrot

[dim widget]

so the kernel is $\cap_{g \in G} gHg^{-1}$

[cloud walrus]

kålrot

[dim widget]

this is the largest subgroup of $H$ which is normal

[crystal vale]

Thank you, I will try to understand

[final dragon]

is the order of group of permutations S3 3

[wraith cargo]

no

[celest furnace]

Why is it 6?

[final dragon]

yea its 6

[celest furnace]

Do you know why?

[final dragon]

i didnt want to ask for order but degree

my bad

[slim kayak]

degree?

[final dragon]

is it cause there are 6 ways to rearenge the permutations

[final dragon]

yea

[slim kayak]

whats that

[final dragon]

like when you multiply elements how many times to get to neutral element

what degree

[slim kayak]

thats also called order

[final dragon]

idk how to say it

well then how its not order 3

the higest order of the element in the group is 3

[slim kayak]

why would it be lol

[final dragon]

arent degree and order different

[celest furnace]

What is the general order of S_n, the permutation group on n elements?

[final dragon]

S3 degree 3 and order 6?

[celest furnace]

Not sure what degree is

[final dragon]

idk how to say it in english

[celest furnace]

Like the maximal order of any element?

[final dragon]

is it factorial

[slim kayak]

how do you know that

[celest furnace]

Yes

[final dragon]

idk

when you go throu all elements

idk how to say it

is it like the imprit of group S3 is 1-2-2-2-3-3

and the higest degree of an element is 3

there is no element degree 6

i have a task to find all the subgroups of S3

there are 4?

[slim kayak]

Fair enough. Can you see why the maximal order of an element in S5 isnt 5?

[final dragon]

idk

[rocky cloak]

I think the word your looking for might be 'exponent of the group', but the answer is still 6

[final dragon]

how is it 6 tho

cause it has 6 elements?

[rocky cloak]

The exponent is the smallest number such that x^n is the identity for every x in the group.

So if you just check you can see that no number less than 6 works

[final dragon]

ohhhh

for every x

but im talking about if you take 1 element from S3

then 6 is not the lowest

okay but asside that what is the degree of S5

cause 5 is prime

i dont want to type 120 elements and see

[celest furnace]

You can do it without computations with some tricks

For example the order of a product of disjoint cycles is the lcm of each ones order

So for example (12)(345) has order 6

And you probably can’t get much higher than this since there are so few elements

[final dragon]

what elements consists group Q8

idk i just found it in a task

it doesnt say the deffenition

[outer valve]

Q8 is the quaternion group. It's elements are
{1, -1, i, -i, j, -j, k, -k}

The rules it follows are i^2 = j^2 = k^2 = ijk = -1

It's like an extension to the complex number system with 2 added imaginary components j, k.

It's non-abelian btw

[final dragon]

i need to find subgroups that are normal to Q8

[barren sierra]

you mean subgroups of Q8 that are normal in Q8?

idk what you mean by "to Q8" and suspect you mean "in Q8"

anyways, can you think about subgroups of Q8?

there aren't many

and there's some good symmetry

so my hint to you is don't overthink it, just think about possible generators of the subgroups

[opal osprey]

I'm trying to understand how to use this highlighted fact to prove the above fact that every symmetric polynomial expression on the roots of a monic polynomial can be obtained by a polynomial expression on its coeficients.

well, if the coefficients belong to a field K

and if the roots belong to the splitting field of the polynomial

then the fact that K is precisely the field that is fixed by all elements of the galois group of this extension would just imply a priori that S(x_1,...,x_n) belongs to K, right? since S being symmetric implies in particular that it is fixed by all permutations of the galois group (thus belonging to the fixed field of the galois group which is precisely K).

but how can we use this to show that S(x_1,...,x_n) must equal some polynomial expression P(a_1,...,a_n)?

any polynomial expression P(a_1,...,a_n) belongs to K of course

but I'm not sure if every element of K is a polynomial expression of the a_1,...,a_n

this is stronget than what is actually needed, but I was thinking about this as well

[crystal vale]

If H is proper subgroup of finite index in a Group G then there is some elements x belong to G which is not in any subgroup conjugate to H in G

I use a result here , if H is a subgroup of finite index in a group G, then H contains a subgroup N which is of finite index and normal in G....thus now I use this to make finite group G/N