groups-rings-fields

[barren sierra]

it's just *

or in what you sent they're omitting the *

[warm shoal]

ah yes

implied to just be some operation *

[barren sierra]

exactly

what's the operation? Who knows, but it's the operation from G and that's all you need

[warm shoal]

like okay this must be so trivial

for all x, y in Ca

WTS x*y in Ca

[barren sierra]

hmmmm

I should have picked a better variable name

wait no this is fine

[warm shoal]

yeah

i just shouldnt pick a ahahah

both x and y commute with a

[barren sierra]

ok yea yea we want to show x *y in C_a yes

so what does that mean, for x*y to be in C_a?

[warm shoal]

THAT x*y must commute with a

[barren sierra]

yea so axy = xya

or written in the second one, xya(xy)^-1 = a

what do we know from x and y being in C_a? What tools do we have

[warm shoal]

thinking thinking

we know that x and y both commute with a individually

[barren sierra]

yea exactly

so xa = ax and ya = ay

and we want to show axy = xya

[barren sierra]

can you think how to use this

[barren sierra]

to show this

[warm shoal]

im not sure

only thing i can think of is doing something with

xa*ya or something!

[barren sierra]

I don't think that'll get you anywhere

cause that's introducing two a's and we only want one a

ok start with axy and think about any substitutions you can make or whatever

[warm shoal]

axy = xay = xya

[barren sierra]

perfect

using the fact that ax = xa and ay = ya

thus, x, y being in C_a implies that x*y is in C_a

[warm shoal]

yay 🙂

[barren sierra]

try something similar for inverses now

if x is in C_a, show x^-1 is in C_a

do the same thing we did above.

[barren sierra]

write out the end goal

[barren sierra]

and write out what tools we have given the assumption

[warm shoal]

well

since a = xax-1

then x-1a = ax-1

since x-1 commutes with a

then x-1 must be in Ca

😄

YAY

[barren sierra]

bingo

[warm shoal]

is this a typo?

how to do a proof with 2 iff statements?

[chilly ocean]

ugghh how?

last square part

[barren sierra]

prove (1) iff (2) and (2) iff (3) or (1) implies (2) implies (3) implies (1)

[south patrol]

Or many such things lol

[barren sierra]

I mean (2) iff (3) should be immediate from what you said here

(I'm using the numbers to represent each statement here cause I'm too lazy to type it all out)

[barren sierra]

you have the map described in the problem right? Can you show it's injective for example?

[chilly ocean]

i have the map G/K -> G/H

[barren sierra]

Oh wait no I'm dumb don't show it's injective

Ok do you know the first isomorphism theorem?

[chilly ocean]

no

[barren sierra]

Hm

What about the homomorphism theorem?

[chilly ocean]

i may know it

[barren sierra]

https://en.m.wikipedia.org/wiki/Fundamental_theorem_on_homomorphisms

This

[chilly ocean]

yeah thats what i have in the number 2

[warm shoal]

bet thank you

[chilly ocean]

i have this

but how is it isomorphism?

[warm shoal]

what would Cax and Cay look like here?

[delicate orchid]

{cx : c in C_a} and similar for C_ay

yeah...?

what point are you making here

[warm shoal]

that im still confused af brotha

[delicate orchid]

you just multiply everything in that from x or y on the right

[warm shoal]

so like

Cax = {(xa)x = (ax)x} ?

@delicate orchid

[delicate orchid]

let c be in C_a
then cx is in (C_a)x

[pliant forge]

can someone reference me to this or if its trivial explain

[hollow mica]

bro is that a lewis structure

[pliant forge]

i wish

[hollow mica]

uhh, aren't modules by definition abelian groups (with respect to their addition operation)

we call M an R-module if it is an abelian group and there is a map R x M -> M satisfying certain properties (1*m = m, (r+s)*m = r*m + s*m, r*(m+n) = rm + rn, (rs)m = r(sm))

[upper pivot]

yeah thats odd, i am sure they wanted to tell you the converse

that an abelian group is a Z-module

what they did here is very wierd

[hollow mica]

the proof is also wrong too then right

[upper pivot]

i have no idea what the properties they mention are

[upper pivot]

hmm i think they got some wires crossed here for sure

ill let you think about how to recover that an abelian group is a Z module from what they wrote

[hollow mica]

given an abelian group G, if M is a Z-module of G, then necessarily, for every n > 0 \in Z, m \in M we have n * m = (1 + ... + 1) * m = m + ... + m.
Similarly, for negative n we have n * m = (-m) + ... + (-m). Finally, 0m = (0 + 0)m = 0m + 0m so 0m = 0.

ok I see now

[next obsidian]

b -> (g(b), b - f(g(b)) is a group homomorphism B -> Im g (+) ker f and is bijective

[pliant forge]

oh theyre modules

[hollow mica]

just to clarify, in Example 39, for V <-> ⊕ kv_i to be an isomorphism, the decomposition must be unique, unlike the definition right after, right?

[next obsidian]

@pliant forge okay wait what I said isn’t quite true, I’m not 100% sure this is true lol

[hollow mica]

I imagine the isomorphism ⊕ kv_i -> V sends (k_1 v_1, ..., k_n, v_n) -> their sum, and if this is an isomorphism it must be injective, i.e., each v in V is written uniquely as a linear combination of the v_i

oh I was comparing this definition and the definition above, not example 39, but the point still holds

[next obsidian]

If it is true I’m pretty sure it should be like

im f (+) ker g

[pliant forge]

i know thats true

but not the other one

however my prof used the other for a proof

[next obsidian]

Yeah okay so look at this

R -> R^3 -> R

Inclusion into the first coordinate

Then projecting out the first coordinate

ker f = 0, im g = R

[pliant forge]

hmm

[next obsidian]

Or even R^2

[pliant forge]

ill check the notes again

tomorrow

ill tag u

weird

[next obsidian]

Yeah this is absolutely false

[pliant forge]

no wonder i was trying to prove crankery

[torn warren]

@warm shoal

another way is to use equivalent class, similar proof as in the proof for cosets @warm shoal

[white oxide]

here, does multiplication in F refer to function composition?

[static needle]

guys, is there a neat visualization of Abelianization?

[torn warren]

what is this prob asking to do?

[white oxide]

asking if it's a homomorphism

[torn warren]

it is just *

just multiplication

simple multiply

actually either case, the answer is false

[tender wharf]

no

it's literally multiplication

f(x)g(x)

[white oxide]

good man that derp

[white oxide]

okie

[tender wharf]

Check it for yourself though!

witness nosols in the future please

[white oxide]

just did

[white oxide]

nosols?

love my sweet derpz

[tender wharf]

!nosols

[flat tree]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[tender wharf]

for advanced channels imo saying whether something is correct is kind of almost giving out sols

[white oxide]

oh ok yea

[tribal moss]

It's not like that single bit of information can be handed i as homework ...

[tender wharf]

is phi a homomorphism? true or false

but imagine not having a "prove your claim" after that

[warm shoal]

Hey!

Isn't "the set of all the conjugates of a" just Ca itself?

[tribal moss]

If a=1, they're clearly different ...

[warm shoal]

but TROPO

isnn't that second set

(which yes is the same)

literally the set of all conjugates of a?

[white oxide]

nah i prove it regardless

[barren sierra]

no

the set of all conjugates is of $a$ is $\left{ xax^{-1} \colon x \in G \right}$. Note this includes $xax^{-1} \neq a$

[cloud walrus]

Spamakin🎷

[white oxide]

are there infinitely many homomorphisms of $\mathbb{Z}$ into $\mathbb{Z}_2$?

[cloud walrus]

okeyokay

[tribal moss]

As rings or groups or?

[white oxide]

as groups

my thought would be yes

[barren sierra]

why?

[tribal moss]

I can only think of two ...

[barren sierra]

yea but why do you think there would be infinitely many? Worth exploring where the logic fails

[warm shoal]

ah gotcha i see.

[white oxide]

because one homomorphism is $\phi(g) = g \text{ mod } 2$ is one, and same with $\phi(g) = 2g \text{ mod } 2$ is another, and you can keep on going along with $\phi(g) = 4g \text { mod } 2$

but would those all just be the trivial homomorphism

[cloud walrus]

okeyokay

[tribal moss]

(Z is cyclic, so the only choice you have when making a homomorphism from it is where to send the generator).

[barren sierra]

^

[white oxide]

why

[barren sierra]

ok so the generator is 1 right?

[white oxide]

yea

and -1 right

[barren sierra]

or well lets say g for ease of notation

[white oxide]

ok

[barren sierra]

yea that too

[white oxide]

which would explain your guys logic for 2

[barren sierra]

ok so like 2 = 2g right?

[white oxide]

ye

[barren sierra]

phi(2g) = phi(g + g) = phi(g) + phi(g) = 2phi(g)

so phi(2g) is completely determined by phi(g)

[karmic moat]

Fix a prime $p \in \bZ$. Let $R$ denote the set of rational numbers whose denominator is not divisible by $p$. Show that $R$ is a discrete valuation ring with parameter $p$.

[cloud walrus]

anamono for anamono

[karmic moat]

for some reason this is way harder than it feels like it should be

i assume the goal is to show that any prime in R is p

[white oxide]

i'm having trouble understanding what you mean

would it help to prove that if G is a homomorphism with G' and G is cyclic, then G' is cyclic?

because by then maybe i could say that a homomorphism always has to send a generator of G into a generator of G'

of which there are two generators of Z, namely 1 and -1

and they both have to get sent to 1 in Z2

or am i lost here

but then there's also the trivial homomorphism

don't crucify me

[next obsidian]

Once you know what phi(g) is you know what phi(2g) is

It’s forced to be 2phi(g)

So phi(g) determines phi(2g)

This holds for any k, so you can see that any map out of Z/nZ is determined by what it does to 1

[next obsidian]

What’s your definition of a DVR

and/or do you know multiple

[karmic moat]

Let R be a PID with the property that, given any two prime elements p_1 and p_2, <p_1> = <p_2>, i.e., up to unit multiple, there is just one prime element.

[next obsidian]

This is such a Hurb definition lmfao

But sure

Do you see why it’s a PID

Also, do you know what localization is

[karmic moat]

is it a PID because, given a non-prime element q in R, it has a unique factorization as a product of primes; but there's only one prime, and so that ideal is principal?

i don't think that makes any sense

man that is a good question

also nah i dont know what a localization is

[next obsidian]

This doesn’t say anything about an arbitrary ideal

So I don’t see how you can possibly turn this into a proof

[karmic moat]

yea u right

[next obsidian]

Figure out why every ideal is principal, then do the following

Prove that the other condition is equivalent to saying R has one maximal ideal.

Let R be a ring with a single maximal ideal m. We say R [or (R,m) if we want to emphasize that m is the maximal ideal] is a local ring

Prove that (R,m) is a local ring is equivalent to showing that R\m is exclusively units

[not needed but bonus: same as saying the set of non invertible elements form an ideal. Also equivalent to saying this set is closed under units]

Once you’ve done this, you just have to show that the complement of (p) is formed of units, but this is obvious

[torn warren]

homomorphism map subgroup to subgroup, so for Z2, there are two subgroups, Z2 and {e}, for Z->Z2, only one generator for Z2, so you have one homomorphism, for Z->{e}, you have one homomorphism, so totally 2 homomorphism

[karmic moat]

tbh i can't figure out why a DVR is PID; i'm not sure how to assign a value to an arbitrary element of an integral domain R (e.g., for Z you can compare size, for F[x] you can compare degree, etc)

[next obsidian]

Wtf

You said your definition of a DVR involves being a PID

[karmic moat]

oh i thought you were asking me to show that an ID that has that unique prime property was a PID

oops lmao

my bad

[next obsidian]

I meant to prove that this ring is a PID

[karmic moat]

ah okay

[next obsidian]

Then you can show it’s local separately

Although technically if you show it’s local and the maximal ideal is principal you can actually show it’s a PID from that so

Maybe do that first

[next obsidian]

This is very false btw

[karmic moat]

yea that's why i was getting really stuck lol

[next obsidian]

The easiest way to do it is to use localization which I guess you can describe concretely

[karmic moat]

we haven't defined localization nor local rings :(

[next obsidian]

So take rational functions over a field where the denominator is not in (x,y)

you can show that it’s local with maximal ideal (x,y)

[karmic moat]

Let $\mathfrak I$ be an ideal of $R$. Let $\alpha \in \mathfrak I$ be non-zero. Then $\alpha$ and $-\alpha \in \mathfrak I$, and one of them is positive. Thus $\mathfrak I$ contains strictly positive elements. Take $\beta \in \mathfrak I$. Write $\beta = \alpha q + r$, where $r = 0$ or $r < \alpha$. By definition, $\beta - \alpha q \in \mathfrak I$. Thus $r \in \mathfrak I$. If $r$ is non-zero, then contradiction to assumption that $\alpha \in \mathfrak I$ is minimal. Thus $\mathfrak I = \langle \alpha \rangle$.

[cloud walrus]

anamono for anamono

[karmic moat]

I'm trying to mimic the proof for Z, but i'm pretty sure it doesn't hold since positive rationals aren't well ordered (?)

[karmic moat]

oh wait. take a/b in R. we have that gcd(b, p) = 1. if gcd(a,p) = 1, then (a/b)^{-1} = b/a is in R. furthermore, ba is not divisible by p, so (a/b)(b/a) = ab/ba is in R and is equal to 1. Thus a/b is a unit in R. Thus is not in any proper ideal of R.

Thus, given a proper ideal J of R, every element in J is a fraction a/b, where a is divisible by p. write a = (p^i)c, where i is a natural number. Thus each value of i generates a principal ideal in R?

[white oxide]

can i think of it this way? to satisfy the homomorphism property, we have to have $\phi(ab) = \phi(a)\phi(b)$ where $\phi(a)\phi(b) \in \mathbb{Z}_2$. But the only elements $\phi(a)\phi(b)$ can be are 0 and 1, so one is the trivial homomorphism, but i'm not so sure how you would define the one which sends things to one

[cloud walrus]

okeyokay

[white oxide]

idk if i'm on the right track

[torn warren]

no

[white oxide]

i'm still struggling to see how this shows that there are exactly two homomorphisms

[next obsidian]

Let G be a group, f:Z -> G and g:Z -> G homomorphisms. Show that if f(1) = g(1) then f = g

this shows you have at most two possible maps

show both of them constitute actually homomorphisms

[white oxide]

oh wait

that problem probably requires this right

then

[torn warren]

to determine the number of homomorphism, it is like to specify the initial value, after the initial value is settled, all other values are settled, the f(ab)=f(a)f(b) is like the algorithm to compute other values, like a domino, but you need to specify where you trigger it

@white oxide

[white oxide]

LMFAO thanks for the domino image

[white oxide]

ohh okay

that makes sense

i should probably prove 46 to gain a rigorous intuition tho

[tender wharf]

there are also overkill ways to do it like with first iso

[white oxide]

first iso?

do you mean first isomorphism?

[tender wharf]

yeah i was just lazy to type it out

[white oxide]

oh is that the fundamental homomorphism theorem or smt

[tender wharf]

eh sure

[white oxide]

good man

well i would have no clue

[tender wharf]

the kernel of the homomorphism is a normal subgroup so

how many normal subgroups of Z do you have

ok theres a lot whoops

[torn warren]

$f(n)=f(n1)=nf(1)=ng(1)=g(n1)=g(n)$

[cloud walrus]

Witness

[white oxide]

noooo please don't spoil :(

[tender wharf]

Z->Z_2
| /
v /
Z/ker phi

well this assumes phi is surjective

you can always use the trivial homomorphism

the one that just sends everything to the identity

[torn warren]

is this Pi-homomorphism unique?

[tender wharf]

its the natural homomorphism

or natural mapping whatever

[torn warren]

yes, I know it is natural, but I am curious if this is unique one

[tender wharf]

where pi(g) is the left coset of ker phi

yeah this should be the unique one

[torn warren]

I am thinking maybe not unique, because you can do a permutation

[tender wharf]

i mean sure but

the cosets are an equivalence relation

so

it probably doesn't matter too much

so long your pi still remains a homomorphism

[torn warren]

what I propose is , for example, H is the kernal, $xH={x, y}$, can we define phi as $\Phi(x)=yH$ ?

[white oxide]

here, would i be allowed to prove the contrapositive? Namely, assuming $\phi \neq \mu$ and then trying to prove that $\phi(a_i) \neq \mu(a_i)$ for all $a_i \in G$

[cloud walrus]

okeyokay

[tender wharf]

you can always prove the contraposition so long you negated it correctly

[tender wharf]

but will your permutation preserve the homomorphism

[cloud walrus]

Witness

[torn warren]

right, so the imaged object is the same coset

[tender wharf]

I think it's fine because of how cosets work

but didnt check whether your permutation is a homomorphism

but yeah it should

[torn warren]

what about this: $\phi(x)=x^{-1}H$

[cloud walrus]

Witness

[tender wharf]

fine too but these are really the same map

because you're still mapping it to the same coset

[white oxide]

and can homomorphisms be extended to more than two elements, i.e. if $a_1, a_2, \dots a_n \in G$ and we have a homomorphism $\phi: G \mapsto G'$, is it true that $\phi(a_1a_2\dotsa_n) = \phi(a_1)\phi(a_2)\dots\phi(a_n)$

[cloud walrus]

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

[tender wharf]

actually this is by definition

to see this consider 3 elements

f(xyz) = f(x(yz)) = f(x)f(yz)

[white oxide]

oh right

then f(yz) = f(y)f(z)

[tender wharf]

yeah yeah

[white oxide]

ok awesome

that makes perfect sense

[tender wharf]

the understanding here is that the "yz" is a single element in G anyway

because that multiplication is closed

alternatively we can also use proof by contradicting obviousness: man it would suck if that wasn't true

[torn warren]

wait, $\phi(x)=xH$ and $\phi(x)=x^{-1}H$ they are mapping to different coset

[cloud walrus]

Witness

[white oxide]

right

[white oxide]

LMFAO

[formal ermine]

[tender wharf]

no they are the same coset

recall that xH = yH iff x'y is in H

[tender wharf]

$xx\inv = e \in H$

[tender wharf]

I would recommend a quick refresher on the properties of cosets

[tender wharf]

recall that homomorphisms preserve the group operation

[torn warren]

no, $x~_Ry$ iff $xy^{-1}\in H$

[tender wharf]

so certainly it must preserve the group operation

sigh

[white oxide]

to prove that $\phi = \mu$, would it suffice to take some arbitrary $a \in G$ (which is of the form of a finite product of integral powers of the $a_i$) and show that $\phi = \mu$? or would this be a priori reasoning

[tender wharf]

let y = x

[cloud walrus]

okeyokay

[white oxide]

oh ok

[cloud walrus]

Witness

[tender wharf]

oh whoops i remembered that one wrong

you're right my mistake

but anyway

its easier to see its the same if we do something like this

pretend G/N is just some group

with elements a b c d etc..

I'm popping off to eat

[torn warren]

no, if y=x, then it is just xRx

[tender wharf]

yeah I said I made a mistake already

[torn warren]

what I am saying is $\phi(x)=xH$ and $\phi(x)=x^{-1}H$, they are different cosets

[cloud walrus]

Witness

[torn warren]

in general

[tender wharf]

I am saying I will go and eat lunch

[torn warren]

sorry I didn't see it, too many messages

[tender wharf]

no worries

let me have lunch and I may or may not return to this discussion

[white oxide]

thank you derpz

[tender wharf]

np

[white oxide]

ur a good man

respect

[tender wharf]

But the DEFINITION of natural homomorphism is

f(x) = x Ker phi

[feral agate]

I am a bit confused to how to get the unit element based on this description. Can anyone nudge me i nthe right direction?

[tender wharf]

the homomorphism onto G/Ker phi may not be

[torn warren]

but the problem is , if define $\pi(x)=x^{-1} H$, then $\pi(xy)=\pi(y)\pi(x)$ this is not homomorphism anymore, unless it is abellian @tender wharf

[tender wharf]

well yeah then that won't work

[cloud walrus]

Witness

[torn warren]

but is that true the selection for $\pi$ is unique?

[cloud walrus]

Witness

[prime sundial]

if f is the transformation x \mapsto Ax + a, then we want a transformation e(x) = Bx + b such that
e(f(x)) = f(e(x)) = f(x)

[tender wharf]

we take pi to be the natural homomorphism

like I said

[prime sundial]

though the + b is throwing me off. is b fixed?

[tender wharf]

but I'm quite sure any homomorphism onto G/N will work

like if you replace it with gamma, gamma!=pi and gamma is a homomorphism yeah I think itll work

think

[next obsidian]

Yes

[prime sundial]

fixed

[next obsidian]

The matrix describes rotation, scaling, and reflection

But affine transformations also have a component which is just a shift

b describes that

[next obsidian]

Wait no!

b is fixed for each separate transformation

[feral agate]

Ah I see thanks

[prime sundial]

[prime sundial]

oh

[next obsidian]

f = Ax + a

[feral agate]

Probably not by that reasoning

[next obsidian]

e = Bx + b

[prime sundial]

i see

[feral agate]

Let $x \rightarrow Ax+b \in Aff_2$ for some invertible two by two matrix $A$ and $b \in \R^2$.
We know that $x \rightarrow A'x+b'$ is the unit element, where $A = \begin{bmatrix}1 & 0 \ 0 & 1\end{bmatrix}$ and $b = \begin{bmatrix}0 \ 0 \end{bmatrix}$.
The inverse is $A''x+b''$, where $A''$ is the inverse of $A$ and $b''=-A''b$.

[cloud walrus]

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

[feral agate]

This is what I have currently

[next obsidian]

Do you have a guess at what it should be?

[torn warren]

how is gamma defined?

[prime sundial]

your A and b look good for an identity

[tender wharf]

any homomorphism that isnt the natural one

[prime sundial]

chmonkey you can take it from here

[next obsidian]

Do you know what the group operation is in this group?

[next obsidian]

No I have to sleep

[prime sundial]

me too

[next obsidian]

Oog

Anyway, the group law is just composition

[torn warren]

got it, and is there a formal definition for "natural" ?

[tender wharf]

well yes I just

gave it to you

[next obsidian]

So try showing your guy is the identity

[tender wharf]

the natural homomorphism is f(x) = xN

N is your normal subgroup

[next obsidian]

Hint: what you’ve written down becomes the map x -> x. Try showing this is the identity in this group

[tender wharf]

now I really should eat

[torn warren]

great, thank you!

[torn warren]

yes

[toxic zephyr]

probably a stupid question, but how do we know that C is not properly contained in any other field? like there isn't another field extension or something. or is it actually?

[white oxide]

does anyone know any good visualizations for homomorphisms that i can possibly find online?

[white oxide]

go eat

[tender wharf]

let me take a screenshot one sec

[next obsidian]

It’s contained in C(t)

There’s no algebraic (which also means finite) extension because C is algebraically closed

[toxic zephyr]

what is t?

[next obsidian]

An indeterminate

This means f(t)/g(t) where g(t)≠ 0

It’s rational functions

[white oxide]

oh shit that's beautiful

thanks!!

[toxic zephyr]

oh rational functions with complex coefficients i see. yeah that makes sense

[tender wharf]

source is contemporary abstract algebra (gallian)

[white oxide]

i'm working through fraleigh rn, do you think it would be a good idea to go through gallian once i finished fraleigh

like is it more rigorous

[toxic zephyr]

I'll look into this because it seems to be what I'm looking for. thank you once again for schooling me in algebra, @next obsidian 🙂

[tender wharf]

It has more examples

Idk whether it would be more "rigorous"

[torn warren]

pic is beautiful, shit cannot be beautiful

[white oxide]

but shit is beautiful

[white oxide]

good man for that

[torn warren]

nice pic, imo, the key to understand homomorphism is to understand not-injective

[tender wharf]

I mean you can have injective homomorphisms

you can inject an entire group into a group bigger than it

[torn warren]

yes, but that's not interesting, right?

[white oxide]

what does it exactly mean for a homomorphism to be completely determined? does it mean that mapping one element in a group G by a homomorphism phi implicitly assigns a value to all other values in G?

and if so, how?

[torn warren]

non-injective means we regroup them into blocks which as the elements in the imaged group

[tender wharf]

tbh the only uninteresting homomorphism is f(x) = e

[tender wharf]

completely determined means if we know the value of f(x) for some x in G, then we know the value of f(y) for all y in G

in particular, this is a property that cyclic groups have

[white oxide]

huh ok

i guess that makes sense

because if you take the generator = x in G then all other y in g are in the form x^n for some integer n

[torn warren]

you can think that domino example

[white oxide]

if this is rhgt

[tender wharf]

recall the defining property of a homomorphism

[torn warren]

how do you determine a solution to a diff eq? initial values or boundary conditions

[tender wharf]

if G = <x> is cyclic and f: G -> H is a homomorphism, and you know f(x)

do you know f(g) for any g in G?

[white oxide]

well it would just be f(x)f(x)....f(x) however many times that g is a power of x right

ahhh i see

thank you!!

ur goated

[tender wharf]

that's what it means to be "completely determined"

you know the ENTIRE homomorphic image of G because you know the image of the generator

[tender wharf]

not really lol

[white oxide]

yes ur my savior

[next obsidian]

DerpZ is goat

[tender wharf]

i am bird

[white oxide]

ur birded

[white oxide]

ohhh ok that makes so much sense

[tender wharf]

a tip is to get another abstract algebra book

[white oxide]

i have dunnitt and foote

dk if i spelled that right

[tender wharf]

yeah you can cross reference

[white oxide]

but i dont look through that

but i should

[tender wharf]

god dummit I stubbed my foote

dummit & foote yeah

[white oxide]

oops dummit

good way to remember that LOL

i spend my free time working through steven roman now that i know some algebra lol

very very slowly

[sharp imp]

Trying to get an idea of the projective special linear group, for example PSL(2,9). Is there an intuitive way to represent an element of this group, and to see how they act on each other? Are they just special "representative" matrices in SL(2,9)?

[next obsidian]

You want to think about projective space

The action of SL factors through PSL

On projective space

[sharp imp]

I see. I don't know about projective space and was looking for a workaround 😅 I guess I go learn about projective space.

I'm trying to do some work with the automorphisms of each finite simple group, and PSL is next on the chopping block

[tacit hemlock]

the inverse in cycle notation is just an inverse permutation right?

[tender wharf]

yes

[tacit hemlock]

Any good beginner's abstract algebra recommendations guys?

or resources in general

our class uses Artin's Algebra, but I find the explanations a bit poor at times

[coral shale]

I would not recommend Artin

idk though

https://media.discordapp.net/attachments/917213985365983253/1079508885846638663/image.png

horrific notation.

[tacit hemlock]

a bit unusual to use a superscript yeah

but not too big of a deal imo

[coral shale]

R^+ usually refers to the positive reals, same for Z

[tacit hemlock]

yeah

[coral shale]

The convention is just to denote the group using the same symbol as the set

[tacit hemlock]

what book do you use for abs. alg?

[coral shale]

he doesnt write G^* to mean G under * elsewhere, so this is inconsistent.

I didn't use any, don't have a recc

u can try #book-recommendations

and also #books-old #books is a list i think

[blissful crystal]

from what i hear artin is good at fields/rings if thats what you're doing atm

although i cannot attest to that

our class used jacobson for group theory, and continues to use it through fields/rings/modules

i think jacobson is nice, but takes a while to get used to and its probably better if you have someone going through the book with you in a course (ie not the best for self-study)

Herstein was good for group theory, but he doesn't do group actions iirc, which were a major part of our course

you'll also need to read it with a mirror since it does function composition backwards

[tender wharf]

our notation is more compact, and bad.

[delicate orchid]

Not covering group actions = bad

[tacit hemlock]

what is $\sigma_x(i)$ supposed to be?

[cloud walrus]

normalAtmosphericPa=101,325

[tacit hemlock]

this is for proving cayley's theorem

that all finite groups G are isomorphic to a subgroup of S_n

Why is $\sigma_x(i)=i \forall i$ needed? I thought that an injective homomorphism meant mapping the identity to the identity

[cloud walrus]

normalAtmosphericPa=101,325

[delicate orchid]

all homomorphisms have to map the identity to the identity

[cloud walrus]

WKABC

[pliant forge]

I was wondering if we have a quiver $Q\colon 1 \to_a 2 \to_b 3$ and a representation of $Q$, given as $k \to_0 k \to_0 k$ where $k$ field, would the projectives be given as $kQe_1 \colon k \to_1 k \to_1 k$, $kQe_2 \colon 0 \to k \to_1 k$, $kQe_3 \colon 0 \to 0 \to k$

[cloud walrus]

Eso

[pliant forge]

anyways the main point is for me to figure out the projective cover of M: k \to_0 k \to_0 k

I'd assume in this case $M/rad(M) \cong M$ as $rad(M) \colon 0 \to_{f_a|k} Im(f_a) \to{f_b|_k} Im(f_b) = 0 \to 0 \to 0$

[cloud walrus]

Eso

[pliant forge]

the arrows are just confusing me tbh

we have a result that if $f:P \to M$ is an epi and $\bar{f}: P/rad(P) \cong M/rad(M) \implies f$ essential epi, and since P here should be projective it's also a projective cover

[cloud walrus]

Eso

[pliant forge]

i can only imagine that there are relations making this true

[stuck cosmos]

Given a quadratic space $(V,q)$, what exactly does the ideal generated by elements of the form $v \otimes v +q(v)$ look like?

The algebra considered is the tensor algebra.

[cloud walrus]

SK2099

[warm shoal]

does it matter if I define a set of conjugates like xax^-1 OR x^-1ax?

[agile burrow]

No, because you'll have both once you conjugate by an element and it's inverse

[warm shoal]

right, thank u mr walter.

[tribal moss]

It matters in the sense that xax^-1 makes conjugation a left action on the group by itself, whereas x^-1ax gives a right action, which is notationally less convenient.

[warm shoal]

ahhhhh interesting, okay.

Now for #5 here, I proved injectivitiy, but now trying to prove surjectivity.

[tribal moss]

Wonderful. Apparently your book switches between conventions paragraph-for-paragraph.

[warm shoal]

ik for f:X-->Y is surjective if for all y/Y there exists x/X s.t. g(x) = y

[warm shoal]

yes so annoying!

lots of typos

and my friends have diff ones too.

[warm shoal]

though how would u go about proving surjectivity here, like, can i just say that for every conjugate there exists a coset?

[tribal moss]

Without thinking it completely through: won't that just be the coset that contains that conjugate?

Every group element is in some coset.

[kind jacinth]

can someone help me out with understanding this?

i get stuck when they do ab^(-1)

how can they expand it to theta(ab^-1) to theta(a)[theta(b)]^-1 ?

if they applied the homomorphism definition for that arent they assuming both groups have the same binary operation in tht case?

[elder wave]

no, homomorphisms don't require the binary relations to be somehow equal to each other

that's just using the definition plus that inverses map to inverses

[kind jacinth]

yeah the binary operations arent required to be equal but in the example above they are assuming its equal right?

since they arent separating the operations

[chilly ocean]

it's just that we denote different operations by the same thing

it's abuse of notation

[kind jacinth]

are they using the same operation symbol in both cases even if its supposed to be different?

[chilly ocean]

they're using different operation, but they denote in the same way

[kind jacinth]

ahhh ok, and since b is in H1 this implies b is in G1so we can do theta(ab^-1)

but how do they get that ab^-1 is an element of H1 in the end?

the way they are trying to prove ab^-1 is an element of H1 is confusing for me

[lament dawn]

phi(ab^-1) is in H2. Then read the definition of H1

[warm shoal]

so i could lowkey just say this ahahah right?

[kind jacinth]

oh wait.. is it phi(ab^-1) = phi(a)phi(b^-1) therefore b^-1 is an element of H1? (since phi(b^-1) is an element of H2?)

why did they rewrite phy(b^-1) as [phi(b)]^-1?

and why would phi(b^-1) be an element of H2? it could be an element of G2 without being in H2

[kind jacinth]

im assuming by the definition of H1?

[lament dawn]

You don't know phi(b^-1) is in H2 immediately. You do know that phi(b) is in H2, since b is in H1. Then you note that H2 is a group, and thus closed under inversion. So [phi(b)]^-1 is in H2.

Then, since phi is a group homomorphism, you can take the inverse inside to say phi(b^-1) = [phi(b)]^-1. Thus phi(b^-1) is in H2

Then you can probably complete/follow their argument.

[kind jacinth]

ahh ok and from phi(b^-1) we can deduce that b^-1 is in H1 right?

so ab^-1 is in H1

[lament dawn]

Yes

Well

Yes on the first not really on the second

You haven't shown H1 is closed under multiplication yet really. What you want to say is that phi(ab^-1) is in H2, and thus ab^-1 is in H1

That shows closure under multiplication

And inversion

Just a and b^-1 being in H1 is not enough to conclude ab^-1 is in H1, until you know H1 is closed under multiplication. Which is what you are trying to show

This is the one step proof of subgroup, if you are familiar. Instead of showing closure under multiplication and inversion separately, you can do both at once by showing ab \in H1 \implies ab^-1 \in H1

[kind jacinth]

yes i am familiar with it, it was to show the set isnt empty and if a, b is in H1 then ab^-1 is in H1

[lament dawn]

Yeah

Anyway, does that clear it up for you?

[kind jacinth]

i think im am getting it, need a minute to process it😂

[lament dawn]

Ping me if something is unclear

[kind jacinth]

ok but to show this do they individually show that phi(a) and phi(b)^-1 are in H2 and then conclude that φ(ab−1) is in H2?

because of φ(ab−1) = φ(a)[φ(b)]−1

[lament dawn]

Yes

You can do this since H2 is a group

[kind jacinth]

so because its a group we can multiply them both and we still know its in the group since a group is implied to be closed

[warm shoal]

For #6, do I have to define a bijection?

If so - what is the intuition behind doing so?

[kind jacinth]

did i get this right? if so I understood it thank you very much!

[lament dawn]

yes

a group by definition is closed under multiplication

[kind jacinth]

alright thanks for ur time 😄

[warm shoal]

Can someone help the boy 🙏

[lament dawn]

Which part?

First or second sentence?

First follows from 5

[warm shoal]

Gotcha @lament dawn

So the index of Ca in G is the number of distinct cosets of Ca ?

[warm shoal]

second sentence

[lament dawn]

second sentence is lagrange's

[warm shoal]

right

from 1) we know Ca is a subgroup

so does that tell us that the set of all conjugates of a is a subgroup as well?

like we're using lagrange

but we want to show that the set of all conjugates of a is a subgroup as well, no?

[lament dawn]

the logic you want is as follows:
C_a is subgp of G (1). So [G:C_a] divides |G|, by lagranges.
In (4) you showed that there is a one-to-one correspondence between set of conjugates of a, and set of cosets of C_a. But [G:C_a] counts the number of cosets of C_a in G (definition). Thus, |{conjugates of a}| = [G:C_a]
And [G:C_a] divides |G|, so |{conjugates of a}| divides |G|

[warm shoal]

Ah, I understand.

Thank you ❤️

The insight is very helpful

[delicate orchid]

lol

[warm shoal]

the kernel of f is {0,0} and {x, -x} right?

or just 0, 0

because it says f(x,y)

[delicate orchid]

well what does f(x, -x) map to

[warm shoal]

maps to 0

x + -x = 0

[delicate orchid]

sure seems like it's in the kernel to me

[warm shoal]

but my question is

it says f(x, y)

is that implying that

x and y must be distinct values?

well i guess x and -x are distinct

nevermind :)_

[delicate orchid]

no?

why would it be implying that

it's mapping from RxR not RxR \{(x,x) : x in R)}

alternatively, f(x, y) = 0 <=> x+y = 0 <=> y = -x so ker(f) = {(x, -x) : x in R}

[prisma ibex]

\ not /

[delicate orchid]

yeah I couldn't get it to work

[south patrol]

Eh same thing

[delicate orchid]

discord kept making it go invisible

[south patrol]

commutative

[delicate orchid]

no not the same thing lmfao

[prisma ibex]

write double \ or just write -

[warm shoal]

than u ❤️

[south patrol]

Oh wait I didn't even read the thing

LOL

Yeah ew

[delicate orchid]

ah, ty

[south patrol]

The function $f:_\mathbb R \times _\mathbb R \to _\mathbb R$

[cloud walrus]

potato

[delicate orchid]

lollll

[lament dawn]

but also the \bR look blurry. included graphic??

[prisma ibex]

yeah it looks like bad typesetting

[warm shoal]

alright - this is what i always get confused by - for f:R* --> R+ defnd by f(x)=|x|, what am I to infer the operation is?

multiplication?

[prisma ibex]

what?

[formal ermine]

you forgot to put a \bR subscript under f

[prisma ibex]

R* is a group under multiplication, R is a group under addition

[south patrol]

Lol

[warm shoal]

okay

but then when proving that f is homomorphic

i wanna say like

f(a) = |a| and f(b) = |b|

then f(ab) = |ab| = |a||b| = f(a)f(b)

or do I do addition?

[south patrol]

That is correct

[topaz solar]

Well is it true for addition

[south patrol]

The group operation on R^x and R>0 are both given by multiplication

[delicate orchid]

^

wow potato ur so smart and cool

[topaz solar]

Addition really doesn’t seem right there too, since it’s hard to have an additive inverse without a negative

[delicate orchid]

it's also just not a homomorphism

[topaz solar]

And |ab| usually isn’t |a|+|b| yeah

[warm shoal]

word okay

[lament dawn]

whyre yall adding

[warm shoal]

so they both multiplication

[delicate orchid]

maffs innit

[chilly ocean]

define |ab| to be |a|+|b|

[lament dawn]

did yall get baited into trying to endow R^+ with an additive group structure

[delicate orchid]

POP QUIZ: find a homomorphism from $(\bR^+, \times)$ to $(\bR, +)$

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

you have 12 minutes

[warm shoal]

ahahah ❤️

[lament dawn]

cartesian product

[chilly ocean]

easy. ||Trivial homomorphism||

[warm shoal]

and since these groups are both multiplicative (R* and R+), I reckon I am to assume that the identity element is 1 (the multiplicative identity)

or has my poor intuition failed me again.

so impressive how u guys know all this stuff

[delicate orchid]

I'm not being rude but knowing 1 times anything is that thing isn't that impressive

[warm shoal]

haha

[delicate orchid]

I did, however, just get 0 and 1 backwards

so maybe it is

[lament dawn]

wew moment

[coral shale]

what do ppl call the elements of generating sets?

[topaz solar]

generators?

[coral shale]

I thought so too, but then what about what they taught at the start of group theory

"A generator is an element that has the same order as the group. If such an element exists, the group is 'cyclic'"

or something like that

idk. Just seen it a few times recently and was confused when I see questions like 'find all generators of ???'

[topaz solar]

that's specifically something that can generate the cyclic group, which is a group with one generator

[topaz solar]

There's usually a few different possible choices of generators, but in the case of cyclic groups thats finding the ones with order the size of the group?

[coral shale]

It just throws me

I think the terminology is surely overloaded

[chilly ocean]

this is only for cyclic groups

elements of a generating set are called generators

[topaz solar]

Generators being elements of a generating set is the usual meaning

think basis of a vector space

[chilly ocean]

"find all generators of ..." only really makes sense when you're in a cyclic group tbh

[topaz solar]

or a spanning set in general, though that's not minimal

[chilly ocean]

because otherwise your generating sets aren't one-element sets

[coral shale]

I mean I get all this, and it's clear when you think about it for a moment

but idk - it just feels very odd.

[topaz solar]

that's because the name they used for cyclics is weird

[coral shale]

Such questions would be better phrased as 'find all generating sets of G which have 1 element', but then the ppl answering them havent seen the definition of a generating set yet...

[chilly ocean]

I think "find all generators of a cyclic group" is fine

[topaz solar]

"Find all elements of order |G|"

[coral shale]

Are there any texts which refer to them as 'cyclic generators'

that would be better

[coral shale]

but anyways, I was asking because seeing such misled me into thinking I might be calling them wrong here

[chilly ocean]

I see your point but this is kinda standard way of referring to this

[coral spindle]

A fun problem in elementary group theory, for those interested:

Suppose G is a FINITE group, and that Aut(G) acts transitively on G\{1} via the natural action. Classify all such groups G. Hints for doing so: (1) ||Use Cauchy's theorem to show that all elements have the same order.|| (2) ||Recall that the centre of a group is a characteristic subgroup.||

Another problem is to classify all groups such that Aut(G) acts simply transitively (aka sharply transitively). Hint for doing so: ||Count the size of Aut(G) by seeing G as a vector space||

[delicate orchid]

this is fun

[coral spindle]

You can't make me!!!

Give me a moment and I'll come up with a hint to help

Well I haven't spotted why yet

But notice that it suffices to show that (x+y)^2 = x^2 + y^2.

[chilly ocean]

this seems to be a variation on "Boolean rings are commutative"

[chilly ocean]

the challenge for me would be unravelling the definitions alone

[coral spindle]

Haha yes well that is part of what's interesting about it

[delicate orchid]

the "||characteristic subgroup||" part is a BIG hint

wait, I'm not sure your first hint is true boytjie

oh nevermind, same order under the action is what I think you mean

[coral spindle]

true tbh but you still have to USE that hint

[coral spindle]

||automorphisms certainly preserve the order of an element in the group, i.e. the size of <g>||

[delicate orchid]

||ok then I have absolutely no idea what you meant, there are characteristically free groups with elements of different order - take any finite non-abelian simple group||

[coral spindle]

Oh whoops clearly I should have specified finite!

I have not thought about the infinite case

Oh wait no

Sorry Wew, you are mistaken

I can tell you for sure that the example you give does not have Aut(G) act transitively on G\{1}

[coarse stag]

are all abelian groups also free abelian?

[south patrol]

No

Free abelian groups are isomorphic to a direct sum of Zs

[coarse stag]

so only countable infinite order?

[rustic crown]

~~yes since you don't have to pay ~~

[chilly ocean]

Z/2Z

[south patrol]

Well at least infinite

Could be as big as you want

[rustic crown]

0 is free abelian

[chilly ocean]

can be any infinite cardinal

[south patrol]

Is 0 actually free abelian lol p sure it isn't

[coarse stag]

I thought they can be infinite or finite cardinal

[chilly ocean]

with 0 generators, sure

[south patrol]

Maybe im dumb

Oh okay sure

It corresponds to free group on empty set

Nice

[chilly ocean]

I think you mean orders of abelian groups

[coral spindle]

An example of an infinite Abelian group which is not free is Q. It's actually quite easy to show that it is not free. Try it!

N.b. it is also torsion-free :)

[coarse stag]

oh wait nvm, im dum

[rustic crown]

(whuts n.b.?)

(i feel i have asked thsi before, but can't remember what it was)

[delicate orchid]

groupprops agrees with me

[coral spindle]

I think you have maybe misunderstood the question. Can you link the groupprops page which you think contradicts what I'm saying?

[south patrol]

Nota bene

[delicate orchid]

uhh how do I spoiler a link ||https://groupprops.subwiki.org/wiki/Characteristically_simple_group|| in the examples section

[coral spindle]

Nota bene literally means "note well"

[delicate orchid]

||"and the characteristically simple non-abelian groups are precisely the direct powers of the simple non-abelian groups."||

[rustic crown]

(yee det was right)

[coral spindle]

...

I'm not sure why you think this contradicts what I was saying lol

[delicate orchid]

what am I missing here

A_5

is characteristically simple

and there are elements with different orders in it

[coral spindle]

The automorphism group of A_5 does not act transitively on A_5\{1}

Idk what else to say

[delicate orchid]

ok now I'm very very confused

[coral spindle]

You seem to think that an equivalent property is that G is characteristically simple, and this is just not true

[rustic crown]

this was cute

[delicate orchid]

||https://groupprops.subwiki.org/wiki/Automorphism_group_is_transitive_on_non-identity_elements_implies_characteristically_simple||?

[coral spindle]

Yes, this implies characteristically simple, but it is not equivalent.

[delicate orchid]

ah I see now

[coral spindle]

It should be immediately obvious that Aut(A_5) does not act transitively on A_5\{1}, because ||there are elements of distinct order in A_5 !||

[rustic crown]

i kinda forgot that ||if all elements have order p, then it's a p-group :p||

[coral spindle]

Hey if I put too many hints in, then it'd be too easy

There's another theorem on that kind of group that you have to use for the 2nd hint to apply

[coral shale]

|| ||

[warm shoal]

when u search injective function

that image comes up

but that's cap lol

[south patrol]

Lol

The summary js also wrong

[warm shoal]

math is so lit

like i struggle and dread doing this

but when i do it

lowkey fun sometimes

like this a wild little vibe

[prisma ibex]

it is fun!

[warm shoal]

So i've shown that f^-1(J) subgroup of G

but for the kerf part, im stuck

so i know that f(eg)=eh

...

eg in f^-1(J)

and eg obviously in kerf

but do i need to do a whole new subgroup test?

[topaz solar]

What is the kernel

Define it for me rq

[warm shoal]

x in kerf if f(x)=e where e in codomain

[topaz solar]

So it’s everything such that f(x)=e

[warm shoal]

yes

[topaz solar]

J is a subgroup

So e is there

[warm shoal]

oh i see

and f(e)^-1 = x

and f(e)^-1 is obviously in f(J)^-1

thus ker(f) subgroup

@topaz solar

[topaz solar]

e is in J

So f^-1({e}) = ker f < f^-1(J)

{e} is also a subgroup

[white yoke]

Does anyone have a copy of the book 'A First Course in Module Theory' by Keating?

[torn warren]

is $G'\cap\phi[G]=\phi[G]$ ?

[cloud walrus]

Witness

[torn warren]

$G'$ is the codomain, $\phi[G]$ is the range, which should be inside codomain, right?

[cloud walrus]

Witness

[topaz solar]

That’s odd to intersect

[torn warren]

right, this is from the textbook

[white oxide]

would Ker($\phi$) just be $4\mathbb{Z}$ x $4\mathbb{Z}$?

[cloud walrus]

okeyokay

[tender wharf]

[white oxide]

please mr. derp

save me

oh wait

it would be 2Z

[torn warren]

so for #4, what is the correct way to write ? $K'$ is a subgroup of $\phi[G]$? or $K'$ is a subgroup of $G'$ ?

[cloud walrus]

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

[versed ingot]

It would be $K'$ is a subgroup of $G'$ since K' is in G' so you want the inverse image of K' and that would be a subset of G

[white oxide]

brah wtf is that a typo or something

[versed ingot]

Wait i dont know how to read

[stable orbit]

weird, what book is that?

[white oxide]

fraleigh

mine doesn't have that for some reason

[stable orbit]

the presentation my abstract algebra teacher made is just like this

"si" is if and "entonces" is then

[torn warren]

Fraleigh, 7ed

So this is the correct picture?

[cloud walrus]

MasakaBakana

[barren sierra]

I feel this is a loaded question

but

how does one compute this

[cloud walrus]

Spamakin🎷

[barren sierra]

how do I get the isomorphism

and the table

is the only way to get the isomorphism via the proof of Wedderburn Artin? That seems quite complicated

[white oxide]

how do we define the binary operation on a group $\mathbb{Z}_m$ x $\mathbb{Z}_n$? if $m, n$ are not coprime would it just be addition modulo $min(m, n)$?

and what if they are coprime?

[cloud walrus]

okeyokay

[white oxide]

like for instance, in $\mathbb{Z}_2$ x $\mathbb{Z}_4$ would we add components modulo 2?

[cloud walrus]

okeyokay

[tribal moss]

Same as for any other product of groups.
$$ (a,b) *{G_1\times G_2} (c,d) = (a *{G_1} c, b *_{G_2} d) $$

[cloud walrus]

Troposphere

[white oxide]

ohh okay, thank you!

[tribal moss]

So add the left components modulo 2 and the right compoments modulo 4.

[white oxide]

yup got it

thanks!

[south patrol]

Tbh in practice I would find the character and then that lets you deduce the decomposition

[barren sierra]

ok how do I find the character then lol

I should preface that I've never done group representations before

they didn't come up in my undergraduate level algebra class nor my first graduate algebra class

they're coming up rn as an application of Wedderburn Artin

[torn warren]

Is this pic correct?

[barren sierra]

and my prof just gave me a handout with these character tables for S4 and A4 but I want to actually understand how to find them

oh I think I found a youtube video

[white oxide]

for 43 would no such homomorphism exist, since it would essentially require you to "cut" out 4 in the domain of any permutation in S3, and that would violate the homomorphism property? (i'm about to prove this rigorously if that's the case, which i think it is, but it definitely has something to do with permutation multiplication?)

since we know the operation in S4 is permutation multiplication

and same with S3

and S3 is defined to act on the set {1, 2, 3}

and S4 {1, 2, 3, 4}

[white oxide]

sorry shouldn't be act

but have permutations with domain of 1, 2, 3

[barren sierra]

nvm video doesn't make sense but that's because I think I'm missing some baseline knowledge of representations so I'll try to figure all that out

[white oxide]

would exercise 47 rely on the results of 44 and 45?

[formal ermine]

yes

[torn warren]

no, it doesn't need 44 or 45

just definition of kernel and Lagrange theorem

[formal ermine]

44 and 45 follow directly from those lol

[torn warren]

doesn't rely on the result of 44 or 45, so you can solve 47 independently without using the conclusion of 44 and 45

[white oxide]

Oh no

I’ve started a huge feud

I’ll try both approaches gentlemen

[kind jacinth]

Example 4.4.2 The free group on one element a denoted by F({a}) is iso-
morphic with Z
is it because i can map empty word to 0, a to 1, aa to 2, ... (and vice versa/inverse would work as well)?

but this would just show its bijective, i would still need to prove homomorphism

[coral spindle]

There are many ways to prove it; this is acceptable.

[kind jacinth]

but is it complete?

i only showed its bijective

[chrome ravine]

does Z_17 have any non trivial sub group ?

[formal ermine]

Z_17 is a non trivial subgroup of Z_17

[chrome ravine]

lol

how is (1 2 3 4 5)=(1 2)(13)(14)(15) ? permutation groups

[timber oxide]

It does not have any non-trivial propper subgroup, which is trivially true for any $\mathbb{Z}_p$ as any non-zero element has a multiplicative inverse mod $p$ hence, any non-trivial subgroup has to contain 1 and thus it has to be the entire group, as $\mathbb{Z}_p$ is generated by 1.

[cloud walrus]

Timwestlund

[tender wharf]

your message is phrased poorly. the permutation (12345) can be broken down into 2 cycles like you showed

[tender wharf]

I think they were asking about it under addition

but yes still true

[timber oxide]

I answered for the additive group, but multiplication is just repeated addition, hence the existence of a multiplicative inverse is equivalent to that the group generated by that element contains 1. In mod 5 we have that 2*3=1, hence the group generated by 2 contains 1=2+2+2.

[tender wharf]

I mean Z_p is a finite field anyway so

[formal ermine]

also like, lagrange lol

don't overcomplicate things

[tender wharf]

to make it even simpler we note that cyclic groups have subgroups for each divisor of the order

[timber oxide]

Fair point

[tacit hemlock]

did I do this question correctly?

[tender wharf]

your H is wrong

recall aH = H iff a in H

[tacit hemlock]

ah true

[tender wharf]

I hope you weren't planning on manually typing out the entire gH lol

[tacit hemlock]

Hold on

[tender wharf]

just write a quick python script or smth

[tacit hemlock]

why is H wrong

[tender wharf]

recall aH = H iff a in H

[tacit hemlock]

yeah but how can 18 be part of H

[tender wharf]

H is arbitrary

here your H = 11G

is 11 in G?

[tacit hemlock]

yeah

[tender wharf]

so what is H

[tacit hemlock]

just Z_55?

[tender wharf]

yep

so what's gH

[tacit hemlock]

also just gH?

[tender wharf]

gH = gH sure

but what is gH

[tacit hemlock]

just Z_55 I meant

[tender wharf]

cool

[tacit hemlock]

gH = H = G?

[tender wharf]

I hope you weren't planning to manually enter everything in

yeah

[tacit hemlock]

wtf

[tender wharf]

just write a python script

[tacit hemlock]

i used notion ai

or chatgpt could work ig

[tender wharf]

to do what

just use python to write the list

python3 -c 'print(",".join(list(map(str, range(55)))))'

[tender wharf]

ugh

[tacit hemlock]

lol

[tender wharf]

it's literally wrong

[tacit hemlock]

i typed it afterwards lol

[tender wharf]

it included 55 lmao

[tacit hemlock]

the 55 is a typo

i typed zero to 55 initially lol

[tender wharf]

my python solution is superior

[tacit hemlock]

where do you go next with this question?

$|K\cap H| \vert 18$ and $|K\cap H| \vert 24$ from Lagrange's Theorem

[cloud walrus]

normalAtmosphericPa=101,325

[tender wharf]

consider the theorem

|HK| = |H| * |K| / |H cap K|

ok that's overkill

[tacit hemlock]

wat

I don't see it

[tender wharf]

this is all you will need actually

[tacit hemlock]

I have not done a number theory class

[tender wharf]

just find all common divisors of 18 and 24

[tacit hemlock]

yeah

[tender wharf]

you don't need number theory for that

[tacit hemlock]

is there a fast way to do that

asides from listing

[tender wharf]

find their gcd first

[tacit hemlock]

uh huh

[tender wharf]

then every common divisor of 18 and 24 certainly must be a divisor of the gcd

[tacit hemlock]

so prime factorise that

also

[rustic crown]

gcd is the product in the poset category :3

[tacit hemlock]

is there a way of finding the gcd without intuition either

[tender wharf]

prime factorize 18

and 24

it's small enough we can just observe it

[rustic crown]

oh lol

[tacit hemlock]

"no number theory", "use euclid algo"

[tender wharf]

well there's a way

,w gcd(18, 24)

[cloud walrus]

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Requested by DerpZ#9141

[tender wharf]

if you don't want to do a single computation

but surely you can perform a prime factorization of 18 and 24 by hand