groups-rings-fields

[tacit hemlock]

I did

[tropic moth]

any book recommendations for abstract algebra?

[tender wharf]

#book-recommendations

[tacit hemlock]

Artin's Algebra

[tender wharf]

doesn't artin have the notation

[tacit hemlock]

$\mathbb{Z^+}$

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

that clearly denotes the additive group

[tender wharf]

no

that's the positive integers

anyway arguing about notation is not really abstract algebra

[tacit hemlock]

that was supposed to be sarcastic

it's what's used in Artins

also

is there a lemma/theorem that implies this explicitly

[tender wharf]

this follows directly by the definition of gcd

the gcd is the largest thing that divides both of them

if not suppose for contradiction that d divides both m, n such that d does not divide the gcd

[tacit hemlock]

how do you actually prove that lol

do you need to use bezout's

[tender wharf]

this is more #elementary-number-theory at this point

[tacit hemlock]

is the trivial subgroup (order 1) still valid?

I can't think of a reason why it wouldn't bve

[tender wharf]

yep

why wouldn't it be

[tacit hemlock]

yeah exactly

lol

i got that question wrong

it won't tell me why though

I typed 2,3,1,6

[tender wharf]

maybe it wanted it sorted

[tacit hemlock]

as in ordered smallest to largest?

[tender wharf]

yea

[tacit hemlock]

https://math.stackexchange.com/questions/4654801/coset-of-18-in-bbb-z-55-11-bbb-z-55?noredirect=1#comment9829938_4654801

apparently u were wrong

[tender wharf]

how was it defined

I was going to ask whether 11Z_55 meant 11 + Z_55 or literally multiplying the elements

[tacit hemlock]

I thought addition modulo n was always the default operator for $$\mathbb{Z}_n$$

[cloud walrus]

normalAtmosphericPa=101,325

[tender wharf]

honestly I have no idea what 11Z_n means here given they keep using multiplication to denote the group operation

[obsidian loom]

how much subgroups does (Z/10Z, +) have?

[lethal dune]

Hint: ||It’s a cyclic group||

[obsidian loom]

so 3? Z/2Z, Z/5Z and Z/10Z?

[summer path]

also the trivial subgroup

[tacit hemlock]

for this question, can (436)(5124) be composed together to form a single cycle?

I am struggling to remember how to compose them

4->3->3, 3->6->6, 6->4->5??

or are these two cycles disjoint

but there is a 4 in both cycles right?

[near hawk]

Hey! Do I understand correctly that:
if we take the category used to define a free group through universal property (where objects are functions from a fixed set into groups), then the terminal object would be the function that projects all elements of the set into the sole element of a trivial group?

[tender wharf]

if there's a 4 in both cycles, can it be disjoint?

[tender wharf]

maybe not a single one, but you can certainly rewrite them as disjoint cycles, yes.

[tender bough]

I'm learning quatum information theory, what could the caligraphic 𝑆 mean here? I can't find it in the lecture notes.

symmetric matrices? self-adjoint matrices?

[topaz solar]

The terminal object [1] is whatever you have 1 and exactly 1 arrow from X to [1] for any X

I have no idea what category you’re using, but if there’s only one arrow towards that trivial one, then that’s terminal

[near hawk]

The category is $$\mathfrak{F}^A$$ defined as follows: objects are pairs $(i: A \rightarrow G, G)$ where G are groups and $i$ is a set-function from $A$ to $G$, and morphisms are commutative diagrams that morph one $i$ into another through group homomorphisms

[cloud walrus]

klay

[topaz solar]

I don’t know why you use morph there, do you mean A->G->H = A->H kind of commutative diagram?

The terminal object, if it exists, has exactly one arrow to it from every object (and if there’s more than one then they have unique iso morphisms between them)

[rustic crown]

(yee that's the final object, and the initial object would be A --> F(A) the free group on A)

[topaz solar]

Yeah, since any group has the homomorphism x |-> e

[topaz solar]

This is a very strange way to define free group imo

[rustic crown]

is it? :p

[near hawk]

yep I know what a terminal object is, I just want to check if in this specific category the terminal object is what I think it is: a pair ( A -> {e}, {e} )

(self studying with Aluffi)

[topaz solar]

Yeah that would be terminal, since any function to {e} is equal

Function extensionality saves the day fr

[rustic crown]

(if you're cat minded, then that's the general recipe to define a free object or get a free-forgetful adjunction)

[topaz solar]

It’s a very categorical way, it’s odd to take it as a definition rather than an equivalent property

[near hawk]

Aluffi is a bit categorical in his approach, but I wonder whether I could digest it if I did not know a bit of categories already (and did not have an intuition about free groups either)

[topaz solar]

Relating it to more set-based math would be hard without knowing it

I’d have to read the book in particular to give anything more tailored, but I’d expect it to be feasible to follow on its own?

At least formally

[rustic crown]

well giving the definition via a universal property doesn't tell you that object exists

you still need to carry out the usual work with words (or a bit more categorically with the adjoint functor theorem)

[topaz solar]

🤔 I wonder how self-contained the relevant constructions and such are (I’d have to comb through the book though which is )

[near hawk]

if you are curious, in Aluffi this is on the next page after the basic intuition part and the universal property part

[topaz solar]

Yeah that’s what I was expecting about as far as a definition goes

Since definition via universal property still needs something to assert it exists

So might as well just make it anyway

[near hawk]

so I would say he s sprinkling everything with categories (and he even starts the book with some basic category theory) but it is still accessible-ish: the intuition and explicit constructions are there. Still, I am not a clean slate when it comes to university math and thus can't tell whether you can actually use it as your first algebra book.

[topaz solar]

Can and should are very different questions too

[rustic crown]

aluffi is a good book :3

[near hawk]

easier than Lang that much I am sure about

[topaz solar]

It was nicely readable for the first little bit I saw, but as a first book idk

That’s a lot stronger of a statement and baits kids to becoming category brain

Better than category’d real analysis

[solar glacier]

question

in finding all the elements of

$K=\Bbb{Z}_2 / (x^3+x+1):={a+b \theta+c \theta^2: a,b,c, \in \Bbb{Z}_2}$

[cloud walrus]

MyMathYourMath

[solar glacier]

are these some of the 8 elements

[formal ermine]

how did you get 8

[solar glacier]

${0,1,\theta,\theta^2,1+\theta, 1+ \theta+\theta^2,...}$

[cloud walrus]

MyMathYourMath

[formal ermine]

what are the ...?

[solar glacier]

are those 6 of them

im trying to find the rest

[formal ermine]

good luck because there are only 6

[solar glacier]

is it those 6 i listed

[formal ermine]

wait

am I being dumb dumb

[agile burrow]

bruh

there are 8 elements

[formal ermine]

oops

yes I am being dumb dumb

[solar glacier]

i thought there were 8

[south patrol]

8

[agile burrow]

You're good, you have to list two more

[solar glacier]

do i have the first 6 down correct

ok

[agile burrow]

Yes, those 6 elements are distinct

[formal ermine]

I thought 2^3 = 6 lmao

[solar glacier]

then 1+\theta^2

and \theta+\theta^2?

[agile burrow]

Right

[solar glacier]

sweet thanks

[agile burrow]

Note that these correspond to the elements in Z_2[x] of degree less than 3

And this should make sense, because if you have some polynomial p(x) of degree greater than or equal to 3, the division algorithm in Z_2[x] tells you that you can rewrite p(x) = q(x) * (x^3 + x + 1) + r(x) for some polynomial r(x) of degree less than 3

Then p(x) corresponds to r(x) in the quotient

[summer path]

what would a basis for K[x]/(x^5-1) look like

[solar glacier]

would theta times theta^2+theta+1 just be theta^2+theta+1 again

or would it be 1+\theta^2

[rotund aurora]

Let R be a ring, S a multiplicative system, I a prime ideal. Are there nice conditions under which Frac(R/I)=Frac(RS^{-1}/I) holds?

Assume I and S are disjoint

Frac(R) is the quotient field of R

[rustic crown]

isn't that always true?

[rotund aurora]

idk

but it would be nice

[rustic crown]

also in RS^{-1}/I i think you wanna quotient by the extension of I

[rotund aurora]

sure

by I there I mean the ideal generated by the set I in the ring RS^{-1}

I haven't really played with routine facts with localization tbh

[south patrol]

RS^-1

[rustic crown]

if you let S' be the image of S under R --> R/I then (S^-1R)/I^e is same as (S')^-1(R/I) right

[south patrol]

Or are these like non comm rings lol

[rotund aurora]

nah nah, all commutative

[south patrol]

Idk I either see S^-1 R or R[S^-1] lol

But i am baby

[rustic crown]

so here you're first inverting fewer elements of R/I and then later inverting all, and asking is this same as if you inverted everything to begin with

[rotund aurora]

ah I see

thanks

[teal saddle]

is it a fact that the order of elements in Zp* is symmetric if you look at them in a list?

like for Z_7^* you have 1 4 2 2 4 1 for 1 2 3 4 5 6

[rotund aurora]

the order of 6 is 2

what's going on is simply that 1,2,3,4,5,6=1,2,3,-3,-2,-1

[teal saddle]

mistyped, i meant when squaring elements

[rotund aurora]

ah

[rotund aurora]

this

(-1)^2=1

[teal saddle]

makes sense ty

[rotund aurora]

🐧

[solar glacier]

can someone verify if i did the multiplications correct thus far

is (1+theta)(1+theta)=1+theta+theta^2?

[coral spindle]

No.

[solar glacier]

i think i caught my mistake

is it 1+\theta^2 @coral spindle

since 2 is cong to 0 mod 2

[coral spindle]

Yes.

[solar glacier]

got it

thanks

so \theta^3 = \theta+1

so then would $(1+\theta+\theta^2)(\theta)=\theta + \theta^2 - \theta - 1 = \theta^2 - 1$ which is $\theta^2+1$ mod 2

[cloud walrus]

MyMathYourMath

[solar glacier]

and is $(\theta)(1+\theta^2)=\theta+\theta^3=\theta-\theta-1=1$ mod 2

[cloud walrus]

MyMathYourMath

[void cosmos]

just do long division

and get our the remainder

thats what mod means

[white oxide]

does this problem reduce down to into showing that $|G| = |G'|$? Since $\phi[G]$ is finite and is a subgroup, we have to prove that it divides the order of $G'$. Therefore all we need to show that $|G| = |G'|$?

[cloud walrus]

okeyokay

[coral spindle]

No, typically |G| and |G'| will not be equal

Hint: use the first isomorphism theorem

[white oxide]

what's the first isomorphism theorem

[coral spindle]

Good luck

[white oxide]

uh thanks

idk man maybe my book called it smt else??

[celest furnace]

Do you know langranges theorem?

[toxic zephyr]

so would Q(sqrt2,sqrt3) be the set of a+bsqrt2+csqrt3+dsqrt6 where a,b,c,d are in Q?
and so would iv be 4? and iii be 2?

[white oxide]

oh you mean the fundamental homomorphism theorem

well that comes after this chapter

[white oxide]

yes

[celest furnace]

Then you are done right if you showed it is a subgroup

Or well isomorphic to a subgroup of G

[white oxide]

well clearly $\phi[G]$ is a divisor of $|G'|$ since $\phi[G]$ is a subgroup

[celest furnace]

G’ need not be finite

[cloud walrus]

okeyokay

[celest furnace]

So divisor here doesn’t make sense always

[graceful dome]

I tried solving this using the Euclidean algorithm but 25 and 5 are not relatively prime so I don't know how else I can form the Diophantine equation to find the general solution.

[celest furnace]

Ah i see where his hint comes in

Try using the 1st iso theorem to build an isomorphism from a quotient of G (also finite) to the image

[white oxide]

i'm not familiar with quotient groups yet

they're introduced in the next chapter

[celest furnace]

Ah i see

[white oxide]

so i have to prove this without the first isomorphism theorem or whatever or any knowledge of quotient groups

[coral spindle]

Think about the kernel and how that relates to the size of the image.

In doing so you will get close to proving the first isomorphism theorem. Swings and roundabouts.

[celest furnace]

For finite try arguing that the image has size <= |G|

[white oxide]

would this line of argument work?

Since $|G|$ is finite, we may number its elements $a_1, \dots a_n$. Then for each $a_j$, $\phi(a_j) = \phi(a_je) = \phi(a_j)\phi(e)$, so $\phi[G]$ is finite.

or nah

[cloud walrus]

okeyokay

[celest furnace]

Yes that works

[white oxide]

ok sweet

[coral spindle]

The argument is flawed

why does phi(a_j) = phi(a_j)phi(e) imply that phi[G] is finite?

Simply say that the elements are listed (possibly non uniquely) by phi(a_1), phi(a_2), ..., phi(a_n).

That's all there is to showing that phi[G] is finite.

[white oxide]

i mean that's basically the same thing

i perhaps did it a bit overkill

[coral spindle]

No I disagree, your argument made no such mention and also talked about something non-relevant

[celest furnace]

Should say one more sentence like and as 1 <= j <= n then it is finite

[white oxide]

right

[white oxide]

okay you're right

[celest furnace]

This is better fitted in #elementary-number-theory ? Maybe #proofs-and-logic

Anyways for divisors if it’s cyclic then it should be clear enough but not quite sure how to do it if it’s not cyclic

[graceful dome]

Yea I realized, I posted in #elementary-number-theory

[coral spindle]

Elaboration on this hint: when is it true that phi(a_i) = phi(a_j)?

[white oxide]

wait bruh isn't this kinda trivial or am i tripping

Since $|G|$ is finite, we may list its elements $a_1, a_2, \dots a_n$. Then $\phi(a_1), \phi(a_2), \dots, \phi(a_n)$ is finite, so $\phi[G]$ is finite. But then $\phi[G]$ = n, and clearly $n \mid n$.

[cloud walrus]

okeyokay

[celest furnace]

No

[coral spindle]

No

[celest furnace]

It can map to not distinct elements

[white oxide]

oh right

my b

[coral spindle]

You should consider the hints I have given

[white oxide]

i will look right now

good man for that

[coral spindle]

Or you could just move on and come back when you've learned the first isomorphism theorem

[white oxide]

nah i'm tryna prove this without that

cuz it's in the chapter before that

so i should be able to prove it with the tools i've learned from basic properties of homomorphisms

if it even requires that

[celest furnace]

Maybe try constructing an isomorphism from the image back into a subgroup of G

We know right away there’s a right inverse…

[coral spindle]

This will rarely exist.

[white oxide]

ah that's smart

[coral spindle]

There is not a right inverse, no.

[celest furnace]

Ah

But it’s an onto map?

[coral spindle]

Yes

[celest furnace]

Onto iff right inverse?

[coral spindle]

Not for group homomorphisms, no.

[white oxide]

perhaps i can relate it to the coset of the kernel of G

and then it would have the same cardinality of a subgroup

which would divide the order of G

[celest furnace]

This is just 1st iso theorem

[white oxide]

is that what you were going for @coral spindle or something like that

[coral spindle]

What you say about phi[G] having the same cardinality of a subgroup is not correct

this typically does not occur

[white oxide]

well isn't by definition phi[G] a subgroup

[coral spindle]

No.

Not of G.

[celest furnace]

Of G’

[white oxide]

oh okay

[celest furnace]

My right inverse misconception messed up a lot of things

[white oxide]

i thought you were talking about G' my b

hmm..... this might be a little bit stupid or silly but we know phi[G] is a subgroup of G', so phi^-1[phi[G]] is a subgroup of G

so maybe i could play with that

[coral spindle]

Yes, phi^-1[phi[G]] = G.

Maybe you should think about this on your own for a while.

At the moment it seems like you're checking ideas with us instead of trying to think independently

One last hint for you to take away: phi(a) = phi(b) if and only if phi(ab^-1) is the identity in G'.

[white oxide]

perhaps i can construct a homomorphism $\tau: G \mapsto G\H$ given by $\tau(g) = gH$ where $H$ = Ker($\tau$)

[cloud walrus]

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

[white oxide]

oops compling error

anyways

compiling

oh well we know that $\mu: G\textbackslashH \mapsto \phi[G]$ given by $\mu(gH) = \phi(g)$ is an isomorphism

[cloud walrus]

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

[coral spindle]

There is no group G\H.

It does not contain the identity.

[white oxide]

wdym

i thought the quotient group was always a group

by def

[coral spindle]

Don't confuse G/H for G\H

[white oxide]

oh

my bad

wrong backlsash

backslash

I lied

Gonna use the first isomorphism theorem

[celest furnace]

Why?

[white oxide]

because it provides an isomorphism from the image of G to the factor group of G

$|\phi[G]| = n$, correct?

[cloud walrus]

okeyokay

[white oxide]

by my way of construction

wait no

that's not true

two elements could map to the same element

so i would have to define or use from the first isomorphism theorem

$\mu: G/H \mapsto \phi[G]$

[cloud walrus]

okeyokay

[white oxide]

which shows that $G/H \simeq \phi[G]$

[cloud walrus]

okeyokay

[celest furnace]

try to be careful on notation, this picture is nonsense

you should use \to instead of \mapsto

[white oxide]

oh what's the difference

didn't even know there was a \to

[empty rose]

$\to \mapsto$

[cloud walrus]

bee

[celest furnace]

say $f(x) = 3x$

[cloud walrus]

Geopchad

[celest furnace]

$2 \mapsto 6$

$f: \bR \to \bR$

[cloud walrus]

Geopchad

Geopchad

[white oxide]

ah i see

okay

fuck i've been doing it wrong all along then

my prof prolly thinks i'm an idiot lmfao

[celest furnace]

anyway the map myu would be myu(gH) = f(g)

[white oxide]

right

[celest furnace]

and then you have to show this is well-defined, a homomorphism, a bijection, and then you have 1st isomorphism theorem

[white oxide]

right

and also cosets are disjoint

[celest furnace]

this is for injective

[white oxide]

so could we perhaps plug all the elements of g into the isomorphism

which would show that \phi[G] is of order n?

since a bijection shows the cardinalities are the same

[white oxide]

wdym

[celest furnace]

it would show \phi[G} iso to G/ker(phi)

[white oxide]

isn't the isomorphism injective

by def

[celest furnace]

which has order |G| / |ker(phi)|

and then clearly |G| / (|G| / |ker(phi)|) = |ker(phi)| which is an integer as desired

[celest furnace]

yes it is but you would need to prove that about myu

[white oxide]

ah

but i'm relying on it already being proven in the textbook LOL

[white oxide]

ah okay

[celest furnace]

wait have you learned this in class?

[white oxide]

nah i've read ahead

section 13 is about homomorphisms

section 14 is about factor groups

i'm doing a problem from section 13

from fraleigh

[celest furnace]

ah

[white oxide]

i would suppose this problem is to motivate the defintiion of a factor group

or the first isomorphism theorem

[celest furnace]

is the book in portugese?

[white oxide]

wha-

oh nah

[celest furnace]

what the hell did i find lol

[white oxide]

bruh what did you find

[celest furnace]

what exercise in section 1 3?

[white oxide]

44

i'm doing it before 47

[celest furnace]

why is 45 so easy

but this one so hard

[white oxide]

math moment

[torn warren]

no, this problem is before the chapter introducing 1st homo theorem. So he should try to prove it without using that powerful weapon

[tender wharf]

I agree, We don't need the

[topaz solar]

That or just prove it

[pastel cliff]

linkint this convo

[tender wharf]

tbf if the exercise was given before the theorem you probably should not be invoking it

[topaz solar]

I think I’ve seen books that blatantly say some exercises shouldn’t be done on like a first pass through

But that’s definitely not this book

[tacit hemlock]

(436)(5124) = (124365)

did I compose these two cycles correctly?

[torn warren]

wrong

[tacit hemlock]

where did I go wrong wat

[torn warren]

(123645)

[tacit hemlock]

how did you get that

i did it again and got the same answer lol

[torn warren]

just by eyes

[toxic zephyr]

start with 1
1->2->2
2->4->3
3->3->6
6->6->4
4->5->5
5->1->1 done
so its (123645)

[tacit hemlock]

oh wait

why are you going from right to left

[toxic zephyr]

you always go right to left

[tacit hemlock]

I thought cycle composition is multiplcative notation

from Mr Artin's Algebra

[toxic zephyr]

it is. but its right to left. pretty much everything in algebra is right to left

think of it like functions

[tacit hemlock]

Yeah I know

[toxic zephyr]

$f\circ g$ means $g$ then $f$

[cloud walrus]

nilpotent nix

[tacit hemlock]

but like for permutations

[toxic zephyr]

a permutation is a function

[hollow mica]

um why does this guarantee the field can't be trivial

nvm Lang defines division ring with the condition 1 \neq 0

[tacit hemlock]

people often write $f \circ g$ as $g\cdot f=gf$

[cloud walrus]

normalAtmosphericPa=101,325

[upper pivot]

I think in artin they made permutations go the other way

[tacit hemlock]

oh that might b why

[toxic zephyr]

who tf does that

[upper pivot]

so its like x (permutation) rather than (permuation) x

[tacit hemlock]

Artin's algebra

and my lecturers

[toxic zephyr]

that's rancid

why

when everything else is right to left

[torn warren]

yes, just use the fact that G is partitioned by its cosets

[upper pivot]

some reason related to reading out cycles or something

[tacit hemlock]

yeah

[upper pivot]

it doesnt really matter

[torn warren]

then just switch to another book

[upper pivot]

well its a really small thing id hardly say enough to switch books over lol

[white oxide]

too bad

[tacit hemlock]

the answer was left to right btw

so I was right

i think

[hollow mica]

what happens when a function f is 0 at some value? What would its multiplicative inverse be in Map(S, A)

if f is 0 everywhere then it is the additive unit/identity so we don't need it to have an inverse, but for functions that are 0in some places, we need there to be a multiplicative inverse, but what do you multiply by 0 to get 1

later in the text they say this:

But I don't see why this issue doesn't affect the general definition too

[tacit hemlock]

can someone please explain why this is a negative parity?

a 4 cycle composed with a 3 cycle is an even permutation right?

so +1 parity?

[topaz solar]

I usually see f \circ g as fg or like f(g) or such, I rarely see it as gf except on like nLab sometimes

[topaz solar]

3 cycle is even, 4 is odd

Even number of transposition in an odd number length cycle since (123) = (12)\circ(23)

[tacit hemlock]

oops

ok but like

[hollow mica]

if you want to give the set of some class of functions the structure of a ring, it is common to make addition function addition and multiplication function composition. In this way, f * g = g ○ f makes more sense

[tacit hemlock]

do I just memorise these rules

like I understand the logic behind it

but do I really need to think of the entire permutation sign proof every time I do a sign computation

[hollow mica]

also, in this case, what would be the additive unit if the ring does not permit functions that are 0 somewhere?

[charred crescent]

can someone help me with the following:
dihedral group Dn with a of order n, and b of order 2, and ba = a^(-1)b.
show that ba^m = a^(-m)b

i just can't seem to show it no matter what i try

[celest furnace]

induction would work well

and BIG hint: remember that b = b^{-1}

[charred crescent]

i though about that, but wouldn't i need to show it in the negative integer direction as well or nah?

[hollow mica]

ok problem solved. I forgot that rings need not have multiplicative identities lol

[south patrol]

Depends on context lol

I would say to most people they do

[tacit hemlock]

Q: Let $f$ be a permutation on the set ${1,2,3,4,5,6,7,8,9}$, defined as follows $$f=(576)(642835).$$
Write the permutation $f^4$ as a product of disjoint cycles.

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

A key question that I have: is it typically better to first rewrite $f$ as a product of disjoint cycles so that taking powers will then be easy:
$$f=(576)(642835)=(283574)\implies f²=(283574)^2$$ Or, should simply keep them separate: $$f^2=(576)(642835)\cdot(576)(642835).$$ In the latter case, would $$f^2=(576)(642835)(576)(642835)=(576)^2(642835)^2$$ be true? Because in this instance, these 2 cycles are no longer disjoint, meaning that shouldn't be commutative right?

Meanwhile, I can write $f²=(283574)(283574)=(237)(485)$, whereby because they are disjoint cycles, I can do $f^4=(f²)²=(237)^2(485)²$.

[cloud walrus]

normalAtmosphericPa=101,325

[summer path]

you should be able to just compute both sides and see for yourself whether such a statement is true or not

[white oxide]

suppose $G$ is a finite abelian group. then it's isomorphic to $\mathbb{Z}{p_1}^{r_1} \text{ x } \mathbb{Z}{p_2}^{r_2} \text{ x } \dots \text{ x } \mathbb{Z}_{p_n}^{r_n}$. it would be incorrect to say $a \in G$ is of the form $(a_1, a_2, \dots a_n)$, right? because we could send $a \in G$ to some element in this direct product, but $a \in G$ is not necessarily of the form $(a_1, a_2, \dots a_n)$

[cloud walrus]

okeyokay

[white oxide]

oops compiling error

should be Zp_1^r_1 you guys know what i mean

[tender wharf]

sure

but you can use the isomorphism

algebraically they are the same

[white oxide]

because i'm trying to describe the kernel of phi if G is a finite abelian group

[white oxide]

yeah true

[tender wharf]

You can talk about it in terms of the homomorphic image if you want I guess?

[white oxide]

hm ok

i'll have to think about that

[tender wharf]

Oh, maybe what you want is

The kernel is isomorphic to

a subgroup of those products

But if that theorem (classification of finite abelian groups) is later on in the book I wouldn't invoke it yet

[tacit hemlock]

for this Q

[white oxide]

well i think it's pretty much the same

the fundamental theorem of finitely generated abelian groups

is what my book says

[tacit hemlock]

$f = (6537412), f^{-1}=(2147356)$

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

I can decompose this any further right?

[chilly ocean]

Z is euclidean domain with norm N(n)=|n|.. and every nZ is ideal of Z... and every ideal of euclidean domain is (d) where d is nonzero element with smallest norm...
So here's contradiction cause every nZ is ideal...
what did I wrong

[white oxide]

ok i'm too lazy but you know what i mean

G is finite and abelian then it's isomorphic to a direct product of set of integers mod blah blah

is it true that if G is a finite nonabelian group, then the order of any element a in G is n where n is the order of G?

i know it's true for cyclic groups

but i'm not sure for nonabelian groups

and i'm too lazy to prove it

or disprove it

lol

[topaz solar]

e is order 1 right?

[white oxide]

ye

well

okay

[topaz solar]

3 in integers mod 6 is order 2 for a less haha identity one

[white oxide]

what

[topaz solar]

Also, consider Z_2 x Z_2, which is order 4, but I'm pretty sure each nonzero element is order 2

[topaz solar]

3+3 = 6 = 0 mod 6

[white oxide]

wait

i'm talking about nonabelian groups

[topaz solar]

oh nonabelian

[white oxide]

ye

[topaz solar]

dihedral group of order 6

flip * flip = e

order 2

[white oxide]

ah

hm

[topaz solar]

[white oxide]

well

[topaz solar]

also it has no order 6 element

[tender wharf]

err if we had such an a

it would generate G

since <a> is a subgroup of G but has the same order as G

[topaz solar]

since powers of a gives all n objects

[white oxide]

oh wait i'm stupid

[tender wharf]

contradicting your assumption of it not being abelian

[white oxide]

wait

[tender wharf]

take a nap

or a break

[chilly ocean]

?

[tender wharf]

its already a disjoint cycle

unless you want to turn it into 2 cycles

[topaz solar]

isn't n of smallest norm

[white oxide]

lil bro snapped

[topaz solar]

for things that make nZ

norm of k for k in nZ is |k|=|nm|=n|m|

so m=1 and get n idk

[lethal dune]

where's contradiction

[chilly ocean]

2Z=(2), 3Z=(3) and N(2)=2=/=3=N(3)

[lethal dune]

So??

[chilly ocean]

2=3 but 2=/=3

[lethal dune]

Since when 2=3

[chilly ocean]

Every ideal in eculidean domain is (d) where d is nonzero element with minimum @lethal dune norm

[lethal dune]

d is the non zero element of the ideal with minimum norm

[celest furnace]

The fact that Z is a PID is not a contradiction

[solemn shoal]

For every ideal in a Euclidean domain there exists some element d s.t. the ideal is (d)

[chilly ocean]

I got it

[celest furnace]

You got what???

[chilly ocean]

Got what I did wrong

[celest furnace]

What did you do wrong? What are you even trying to prove?

[solemn shoal]

I think they interpreted as
For every Euclidean domain there exists some element d s.t. every ideal is (d)

[celest furnace]

Aah

Algebra is fun 😁 I too cleared up a misconception today

[chilly ocean]

1.1.1 easily implies 1.1.2 but how to prove the other direction

[chilly ocean]

sucks when you're stuck on the first problem of the book

the rest of the problems are a lot easier tho

[torn warren]

the definition1.1.1, it didn't say the binary operation is closed

[chilly ocean]

aight let's just assume it is

[torn warren]

it is not assume, it is a must...

[chilly ocean]

why didn't they specify it tho

[flint cave]

Binary operation is closed by default

It's a map G×G to G

[torn warren]

if it is default, why doing hw exercise prob need to prove it is closed

[flint cave]

You can prove that 1.12 means it has an identity element right? After this showing that each element has an inverse is also quite straightforward

[flint cave]

That might be when you need to show something is a subgroup

[chilly ocean]

I was only able to show left and right identities but not a unique identity

[flint cave]

Did you use associativity?

[chilly ocean]

lemme see what I did

[topaz solar]

What if you multiply identities

[flint cave]

There's a related fun question, given a finite set with an associative binary operation having cancelation property (ax = ay implies x = y and xb = yb implies x = y) is a group.

[tender wharf]

is this a nosols moment

[chilly ocean]

thanks

[hollow mica]

is this just saying that the direct sum decomposition has length 2

if so it seems extra that they "index over the ring Z/2Z" instead of just saying "direct sum of two subspaces"

[tacit hemlock]

Let $H,K$ be proper subgroups of a group $G$. If $|K|=39$ and $|H|=65$, what are the possible orders of $K\cap H$? Enter your answer as a comma separated list.

My answer: 13,1

**Motivation:** 

Let $Q=K\cap H$; by Lagrange's Theorem, $Q\over 39$ and $Q\over 65$, so this effectively becomes a basic number theory question; I need to find all common divisors of 39 and 65.

Thus, I first find the $gcd(39,65)=13$, since all common divisors are also divisors of the $gcd$.
(how do I prove this? I tried.) Since 13 is prime, the only possible orders of $Q$ are 13 and 1.

Have I done this question correctly / is there a better way? I have not taken a formal number theory course yet, so I am quite weak at it.

where is the TeX bot

[hollow mica]

probably sleeping at this hour

[hollow shore]

Anyone confident with abstract algebra can help me over dm?

[glass vine]

Abstract algebra is a broad term but I might be able to help depending on whats the problem

[hollow shore]

I failed my abstract algebra course, 47% had to redo it this semester;(

Depressed

[tender wharf]

do you want to like

provide specific questions or something

[hollow shore]

I need to revise first

But im so depressed rn

[tender wharf]

as in, could we help you with specific questions that you did wrongly

if you want

[hollow shore]

yes please lemme try them first tho

[tender wharf]

that's the idea of this channel

[formal ermine]

yea

[tender wharf]

time to practice a bit of number theory then

[hollow shore]

Hm im a bit stuck

[tender wharf]

,rccw

[tender wharf]

oh you're trying to make them into disjoint cycles

[hollow shore]

what happens next when 1 is the 1 cycle from b

[tender wharf]

what is a and b

[hollow shore]

a = (132)(45) b = (2543)(1)

[tender wharf]

(1) can be ignored

basically

[hollow shore]

yea so where does it go

[tender wharf]

because its the identity permutation

in b?

b(1) = 1

[hollow shore]

nah like when i multiplied it

[tender wharf]

compute ab(1)

[hollow shore]

from a) 2 -> 1 then what after that?

[tender wharf]

that's it

[hollow shore]

so what do i write when it done?

[tender wharf]

write it as disjoint cycles I guess

[hollow shore]

can u show me how to do a.b?

[tender wharf]

you're probably asked to find ab in terms of disjoint cycles

a trick to find disjoint cycles is as such

ab(1) is 3 so we'll see what ab(3) is

ab(3) = 1 so (13) is a cycle

now we check the other numbers

[hollow shore]

what do u mean by ab(3)?

[tender wharf]

ab is a function

so we evaluate it on 3

[hollow shore]

okay

[tender wharf]

can you evaluate ab(2) ab(4) and ab(5)

[hollow shore]

isnt that just (13)(45)

a * b = (13)(45)

[tender wharf]

^

[hollow shore]

i havent been taught to calculate ab(1) or ab(2) like how u did it

i been taught to go from a to b

[tender wharf]

now you know

ok do you want me to demonstrate how i found ab(1)

ab = (132)(45)(2543)

(2543) fixes 1 and so does (45)

(132) sends 1 to 3 and thats the last cycle so ab(1) = 3

note that I am composing permutations right to left

can you demonstrate how to compute ab(3)

[hollow shore]

ab(3) sends 3 to 2 and 2 to 5?

[tender wharf]

well it does send 3 to 2

(2543) sends 3 to 2 yes

but how does 2 go to 5

[hollow shore]

yes

wait my bad

it goes back to 1

[tender wharf]

cool so you've already discovered a cycle in the final thing

(13)

we're still missing information about 2, 4,5

[hollow shore]

so i have to start from b now right

cuz 2 send back to 1 in a

[tender wharf]

no no

you evaluate ab(2)

etc

you must treat the whole thing as a function

because it is

you cannot treat them as separate here because they are not disjoint and so don't commute

[hollow shore]

so 2 -> 1 and 1 is identity

so we got (13)(2) so far

[tender wharf]

one sec

try again

[hollow shore]

cuz i said we start from b, if thats the case then 2 -> 5 -> 4

[tender wharf]

so 2 goes to 4

[hollow shore]

so ab(2) = 4

[tender wharf]

now check ab(4)

yep thats correct

[hollow shore]

is that what u meant

[tender wharf]

yep

[hollow shore]

so start from b yea okie

[tender wharf]

now find ab(4)

no no

[hollow shore]

now ab 4

[tender wharf]

oh sure if ur thinking about it like that

[hollow shore]

4 -> 5 then 5 ->4

[tender wharf]

so

[hollow shore]

yea i need to think of them as systematic thing

cuz i cant wrap my head around them otherwise

[tender wharf]

this is the system you use

you see each cycle

you do them one by one

right to left

[hollow shore]

wait

not left to right?

[tender wharf]

what does your textbook use

[hollow shore]

Write 𝑎 and 𝑏 in cycle notation, and compute their product. (Remember that for us,
𝑎 ∗ 𝑏 means “first do 𝑎, then 𝑏”)

[tender wharf]

ah okay sorry

yeah i misled you

[hollow shore]

"first do a, then b" confuses the hell outta me

[tender wharf]

we need to evaluate b(a(1))

b(a(2)) etc

[hollow shore]

so back to start or

[tender wharf]

yeah unfortunately

but now you know

[hollow shore]

so i start from a right

[tender wharf]

one sec please

how would you evaluate this permutation

(132)(354) on 3

like

based on your class

[hollow shore]

so 1 -> 3 -> 5

[tender wharf]

not based on what I said

errr

[hollow shore]

wait

[tender wharf]

are you sure

theres no one

ok let f = (132) g=(354)

i want to know if fg (3) means

f(g(3)) or

g(f(3))

[hollow shore]

hm im practicing looking at notes rn

this is the example from lecture

[tender wharf]

here, do you evaluate permutations left to right

anyway the idea here is just

start with some element

say 2

see where that permutation sends it

say 2-> n

then evaluate the permutation on n

eventually youll get to some point which goes back to 2 and that completes the cycle

you repeat until you got everything

[untold cloud]

Hi, guys, suppose A is a finitely generated k-algebra, and A is an integral domain, then is it true that Frac(A) is also a finitely generated field extension of k?

where k is algebraically closed

[tacit hemlock]

Did I do this question correctly

[atomic mesa]

quick question: in proving "every integral domain is a field", do i need to prove that the integral domain is commutative?

[tender wharf]

yes

and not every integral domain is a field

[atomic mesa]

sorry, every finite integral domain

[tender wharf]

ah

no

an integral domain is already a commutative ring

[atomic mesa]

oh i see

[hollow shore]

wow i worked it out

i think

Is this correct?

[tender wharf]

,rccw

[tender wharf]

(45)(2543) is that (45) * (2543) or is there some other multiplication

that I'm not aware of

[hollow shore]

(132)(45) * (2543)

[tender wharf]

no

that is not what I am asking

I am asking if (45)(2543) is implicitly doing (45) * (2543)

[hollow shore]

no

it a * b where a = (132)(45) and b = (2543)(1)

[tender wharf]

that is not my question.

[hollow shore]

what do u mean by implicitly doing (45)*(2543)?

[tender wharf]

(45)(2543) has an operation in the middle

what is it

is it

[coral shale]

whats goin on D:

[tender wharf]

I'd assume it's *

the OP's class uses

left to right composition

[hollow shore]

YES

[coral shale]

they should leave it

[tender wharf]

ah okay

[hollow shore]

i should change my name to i am confusion

[tender wharf]

ok now i get it

one moment

[coral shale]

no one does this

change

[hollow shore]

i been told to do this so

[coral shale]

[tender wharf]

does your class insist on

left to right composition

[coral shale]

tell em theyre -&:-&'--&_':&£##"''*

politely

[tender wharf]

nice pound symbol

[hollow shore]

i cannot change

[tender wharf]

let me just evaluate it with the sane composition first

[hollow shore]

my applied math major only got 2 more units left

i suffered through 11, 2 more left and im DONE

[tender wharf]

(2543)(45)(132)

[coral shale]

(12)(13) = (321) and nothing else

[tender wharf]

wow

[coral shale]

so stoopid

[tender wharf]

$a * b = b \circ a$

[tender wharf]

[hollow shore]

a = (132)(45) b = (2543)

[coral shale]

taking these as functions

[hollow shore]

i been told to do left to right tho

[coral shale]

a(b(x)) = (b * a)(x)

[tender wharf]

so cursed

i need a break

[coral shale]

in which case the notation is nonsensical and u should use

[tender wharf]

left to right composition makes me want to cry

[rustic crown]

hewwo #abstract-chill

[tender wharf]

[coral shale]

xab = x(a * b)

[hollow shore]

um so did i do it right?

[coral shale]

^ i cant remember what this notation is called

xf

to mean f evaluated at x

[tender wharf]

yup

you are correct

[tender wharf]

yes

[coral shale]

well nowhere do i see functional notation being used in their notes

so also: don't

It is worth asking whoevers teaching how they would notate
"a * b evaluated at x"

And see what they do

[hollow shore]

nty

[coral shale]

hopefully itll be xab

[hollow shore]

i want him to not pick on me

[coral shale]

youre there to learn

[hollow shore]

he was so strict that im 1 mark away from passing, yet he didn't give it to me

[coral shale]

well no... thats how things work usually

[hollow shore]

well no im sure i did decently okay in my final

[tender wharf]

did you get your grade back for your final?

[hollow shore]

but they were being too picky when marking it

that was last year im redoing it again

[coral shale]

this could just be your opinion

[hollow shore]

😱

[tender wharf]

this is technically no longer abstract algebra

[coral shale]

you should get the opinion of someone else before thinking this

[tender wharf]

if you really must, see your prof in person and ask

[hollow shore]

well, i dont get my test back, only result

[tender wharf]

[hollow shore]

and my total assignment marks were quite high, like 38/50

[coral shale]

sounds absurd schooling

[uneven sundial]

are they using left to right only for permutations? or just in general

[hollow shore]

hm good question, so far thats what i been taught this week

[coral shale]

functions

[hollow shore]

we'll be back next week

[coral shale]

???

what about functions in general

[hollow shore]

um no? not yet

thats why i was so confused when pigeon used function

[coral shale]

well it is worth realising permutations are just functions

In particular they are bijections from sets to themselves

[uneven sundial]

it is weird but it does make it 'easier' for permutations so you can read left to right

[tender wharf]

the definition of a permutation is a bijection

lmao

im not a pigeon btw

[coral shale]

yes u r, shuddup.

[uneven sundial]

but its not like reading right to left is really that much harder

that you would introduce special notation or w.e

[hollow shore]

i need stressball

[hollow shore]

hm can someone explain this to me?

How many elements are there in 𝑆10 of cycle-type (×××)(×××)? And how many elements
of cycle-type (××)(××)(××)? Explain your answers.

[rustic crown]

make me explain my answer

[chilly ocean]

Are you DerpZ

[tender wharf]

im not ryc

[rustic crown]

are you rycool?

[tender wharf]

maybe

[chilly ocean]

I didn't ask who you weren't

[tender wharf]

you could click on my profile

[chilly ocean]

I'm on phone

[hollow shore]

someone pls halp 😱

[chilly ocean]

Oh. You can click on avatar. TIL

[chilly ocean]

You definitely need to pick 6 elements first

Right

[hollow shore]

yes

ngl im quite confused by what he mean (xxx)(xxx)

[chilly ocean]

And then figure out how to make a product of two disjoint 3-cycles from those

[chilly ocean]

They mean two disjoint 3-cycles

I'm pretty sure

[hollow shore]

should i figure out the lcm first then

lcm(6,4)

[chilly ocean]

Why are you doing this

I'm confused

[hollow shore]

finding the order?

how would u do it?

[chilly ocean]

Why are we finding order of something

[hollow shore]

okay, explain to me how u would do it

[chilly ocean]

We need 6 numbers from 10

So it's going to be 10 choose 6 times amount of how many products of two disjoint 3-cycles we can make from 6 numbers

Right

[hollow shore]

how many disjoint 3-cycles can we make with 6 numbers?

[chilly ocean]

Well then we need to take 3 numbers from those 6

To make into a 3-cycle

But product commutes so we need to divide by 2

So it's 6 choose 3 divided by 2

[hollow shore]

10

[chilly ocean]

And now we need to make those 3-cycles from the choosen 3 numbers

So it's going to be times 2^2

2 for each 3-cycle in the product

[hollow shore]

40

but dont we need to account for the rest?

[chilly ocean]

Wdym

[hollow shore]

because some cycles are similar

[chilly ocean]

Oh!

Yeah right

[hollow shore]

so should be 40 x (3-1)!

i think 80?

[chilly ocean]

Wdym actually

[hollow shore]

like the order of the cycles repeat

[chilly ocean]

Like ab and ba?

[hollow shore]

wait...

so if i got

(123)

that could also be (231)

[chilly ocean]

Yeah

[hollow shore]

they are distinct right

if we get the same ones, do we count them as well?

[chilly ocean]

They're the same

This was when I counted 2^2, with 2 for each cycle

We can have (123) or (132)

Right

[hollow shore]

eh

oh right

yes

so in total should be 10 choose 6 times 40

= 210 x 40

[chilly ocean]

Sounds about right

[hollow shore]

nice thanks

what about (xx)(xx)(xx)?

still need 6 numbers from 10

[chilly ocean]

Yeah. Because there is 6 x

[hollow shore]

but multiply by the product of 3 2 - cycles

[chilly ocean]

Hm?

[hollow shore]

(xx)(xx)(xx)

[chilly ocean]

We need to choose 2 for each batch

[cloud walrus]

Blitz

[chilly ocean]

But count the commutativity of the product

[rustic crown]

you also have to divide by 3! as the order of cycles don't matter

[hollow shore]

righto!

[chilly ocean]

I am getting to this

[rustic crown]

oh okie >.< mb

[chilly ocean]

Here. This means we need to divide by 3!

[hollow shore]

yes

so 210 x 15

for (xx)(xx)(xx)

[chilly ocean]

Now, if you have 2 numbers, there's only one way to make a 2-cycle from them

So that's it

[chilly ocean]

Now I want to point out that I interpreted this to mean product of disjoint cycles

But I think that part is correct

[coral shale]

cycle type would refer to disjoint.

[tender bough]

Consider the Hilbert space ℋ = ℂ² ⊗ ℂ², what is the rank of the operator: 𝜌 = |00⟩⟨00| + |10⟩⟨10| + |11⟩⟨11|?

I just want a quick way to check that the rank is greater than 2.

[untold cloud]

Hi, guys, is it true that if L'/L is algebraic field extension, L'/k and L/k will have the same transcendental degree?

[formal ermine]

what is k

arbitrary (proper?) subfield?

[untold cloud]

yes

[formal ermine]

proper or not?

[rustic crown]

that doesn't matter right

[untold cloud]

it does not say, so both cases i think

[rustic crown]

yea they have the same transcendence degree

[prisma thunder]

I might’ve asked this but I can’t remember

[untold cloud]

how do i show this, suppose the transcendental basis of L/k is x_1,...,x_d, i hope to say that if x_{d+1}\in L' is transcendental over k, then x_{d+1} is transcendental over L

[prisma thunder]

Is every Dedekind domain that’s finitely generated as a Z-algebra isomorphic to some ring of integers of a number field (as Z-algebras)?

[rustic crown]

okie i'll assume you have verified the usual stuff like the existence-uniqueness statement.
if B is a transcendence basis of L/k then by definition B is a maximal alg-indep set, so any bigger set can't be alg indep, so in particular this means that L/k(B) is an alg-extension. now show that B is still a maximal alg-indep set for L'/k using the fact that L'/k(B) is algebraic.

[untold cloud]

Thank you!

[rustic crown]

idk much theory of dedekind domains, but a bad counter example is F_p[x].

i changed the characteristic so can't be something obtained from number fields

[prisma thunder]

I mean

You’re not wrong lol

[white oxide]

bruh can someone give me a hint, i'm bad at number theory

clearly it follows from lagrange's theorem

and you can use induction of some sort

i just don't know how to a) show that (n - 2) does not divide n

i know you can do (n - 2)k = n and then expand that

[celest furnace]

can n-2 divide n for n > 4?

[white oxide]

no, of course not

i'm just having trouble proving it

number theory wise

[celest furnace]

divisors can't be larger than sqrt(n) i believe <- try proving this

[coral spindle]

Try arguing via the size of n. Remember that a | b implies |a| <= |b|.

[rustic crown]

if n-2 divides n, then it divides n - (n-2)

[white oxide]

god i hate number theory

okay i'll try those

[rustic crown]

[white oxide]

thanks

[celest furnace]

this is nice

[coral spindle]

det's suggestion is the best

[rustic crown]

(oh they can be... the smallest non-trivial divisor can't be larger than sqrt(n))

4 is a divisor of 8 >.<

[celest furnace]

ah

[rustic crown]

but yea ig you wanted to say that a proper divisor would always be <= n/2

[pastel cliff]

[white oxide]

this worked perfectly, thanks!!

ig i should have thought about it a little more though

but my approach is to always do (n - 2)k = n and then try to find a contradiction there, but it's quite annoying to do so

[rustic crown]

yea but that's not a fun example :p

okie what about this... consider Z[1/2] i believe this is a pid, hence dedekind.. but since it's not a free Z-module it shouldn't be ring of integers of some number field.

[white oxide]

or use some form of induction ig but that's annoying too

[rustic crown]

it's not going to be that bad... you know k = 1 doesn't work, so k >= 2. this inequality gives you what you want.

[torn warren]

start from here, $k=\frac{n}{n-2}$, now take $k$ as a function of $n$, $k(n)=\frac{n}{n-2}$, find the range for this function, then you will find the contradiction if $n>4$

[cloud walrus]

Witness

[white oxide]

ah ok thanks

i suck at number theory shitttt

[rustic crown]

.<

that just definition of "|"

[white oxide]

would you have to prove this with means of induction then

[rustic crown]

nah, k >= 2 means n = (n-2)k >= 2n-4 which is same as n <= 4

[white oxide]

something tells me real analysis is gonna fuck me in the ass if i can't see these inequalities

i see

thank you

[celest furnace]

you get used to it

[fallen trout]

can i get a hint for this?

[torn warren]

haven't learned ring part, is ring more difficult than group or less difficult?

[rustic crown]

sqrt(R)? i think they mean the nilradical of R, which shoudl be denoted sqrt(0)?

[fallen trout]

we don't necessarily have that R is an integral domain so i'm not sure how to show that all the non-constant coefficients couldn't theoretically multiply out to 0

[fallen trout]

our class has defined sqrt(R) to be the nilradical ¯_(ツ)_/¯

[rustic crown]

normally sqrt(I) for an ideal means the collection of elements such that a certain power of them lies in I.

[fallen trout]

yeah ik it's weird

we use sqrt(I) as you do, but sqrt(R) for some reason is the nilradical

[rustic crown]

okie det bacc

[rustic crown]

to show that right thing consists of units, use that unit+nilpotent is a unit and sum of nilpotents is nilpotent.

[fallen trout]

and it remains to show the hell direction

[rustic crown]

to show that these are all the units, ig one way is to take a prime p of your ring R, and then consider that unit in (R/p)[T]. since this new ring is an integral domain, so the units are constants and hence all the non-constant coefficients of a unit lie in every prime.

and you should have seen that intersection of all primes is exactly the nilradical

[fallen trout]

ohh that makes sense!!

tysm!!

[rustic crown]

[rustic crown]

i wonder if we can avoid using AoC

[rotund aurora]

God exists axiom --> AoC

[rustic crown]

i wonder if we can avoid using God

[torn warren]

NoGod exists axiom --> AoC

[rustic crown]

[torn warren]

is this correct?

[agile burrow]

Yes, that's right

[torn warren]

but why they didn't claim this

[agile burrow]

They did - they say that it's a map G x X -> X

in particular, it takes (g, x) to some element in X which we denote gx

[torn warren]

but that is the map for $*(g, x)=y, y\in X$

[cloud walrus]

Witness

[torn warren]

but we don't know how $*( , )$ is defined in details

[cloud walrus]

Witness

[agile burrow]

Right, and we typically denote the image of *(g, x) by gx, which should invoke the idea that g is acting on x

[torn warren]

ok, got it!