groups-rings-fields

[torn warren]

Do I intepret correct? I don't see that writing before

I mean your writting (12) x C_3 union e x C_3

is it equivalent to ${e, (12)}\times {e, a, a^2}$?

[cloud walrus]

Witness

[quiet pelican]

Yes

[torn warren]

ok

[white oxide]

here, what do they even mean about setting (2, 3) equal to zero?

i know what they're trying to go for, but what would that even look like lmfao

[next obsidian]

Quotienting by <(2,3)>

[white oxide]

what does that mean

[next obsidian]

It’s like

Written down

8 words before the highlighted thing

[white oxide]

do we necessarily have to prove that (1 + 1 + ... + 1) for n summands is equal to n?

[next obsidian]

That’s the definition of n

[white oxide]

gotcha, thanks

[solar shore]

can someone help me with the finite case here?

so far i have that assuming $\phi(a^x) =\phi(a^y) \implies x \bmod n = y \bmod n$

[cloud walrus]

blanket

[solar shore]

but i dont really see how it follows that x = y

oh wait

order n

never mind x mod n is x

woops

.

[chilly ocean]

Why (pi) intersection Z=pZ

[rotund aurora]

they explain it

the only thing they don't explain is that actually (pi) intersection Z is non-empty

but you can think about it

let (pi) intersection Z=P. Then for a,b in Z, the author showed ab in P implies a in P or b in P, which is exactly the definition of a prime ideal

and like "contraction" of ideals are again ideals

[wooden ember]

im probably being an idiot but if real embeddings exist won't s=0?

[coral shale]

that depends on n?

take a quartic with 2 non-real roots, 2 real roots

[south patrol]

Lol 0

[gilded wigeon]

Not sure how to approach 2

Just not sure how to work with cosets… if $a \in bH$ implies $b \in aH$, why is it that $H = H^{-1}$?

[cloud walrus]

Iago

[tender wharf]

what is H inv

the set of inverse elements of H?

this is obvious from what H is

H is a subgroup

you dont need the hypothesis at all

[gilded wigeon]

H is an arbitrary subset

Not necessarily a subgroup

[tender wharf]

ah

[gilded wigeon]

I think that I’ve got an argument

Lemme type it up on my phone

[tender wharf]

my bad

[gilded wigeon]

My current argument goes that there exists $h_1,$ $h_2$ such that $bh_1 = a$ and $ah_2 = b$. It follows that

$bh_1 = a$

$ = bh_1 h_2 = a h_2 = b$.

Thus $h_2 = h_1^{-1}$, which gives us that $H = H^{-1}.$

[cloud walrus]

Iago

[gilded wigeon]

The fifth proposition is actually the trickier one

[tender wharf]

I am going to eat some food maybe ill read it later sorry

[gilded wigeon]

“Leibniz Theorem” 🤔

[tender wharf]

time to email the textbook author

"lagrange is rolling in his grave"

[tribal moss]

"I'm using the German spelling, duh!"

[chilly ocean]

Who proved ex68

[rotund aurora]

I was supposing pi was not zero, and that the intersection was pZ with p prime (therefore p not zero). So should of said "it contains integers besides 0".

[south patrol]

Yeah that's what I meant

[lethal dune]

π ∈ ℤ

[livid willow]

How would I approach to find minimum polynomial over Q(root 2)? I know how to approach to this over Q (The video has set x = root 2 + root 3, and then found the irreducible polynomial). But I am fairly confused on how to approach this over Q(root 2).

[tribal moss]

Probably you can follow the same procedure as you've already seen, just consider coefficients in Q(root2) instead of Q. Hopefully you can already do arithmetic in Q(root2).

[livid willow]

Would this work?

I have been asked if a the degree of irreducible polynomial is 2 and I think I have gotten a degree 2 polynomial.

[formal ermine]

what minimal polynomial are you trying to find

[livid willow]

Minimal polynomial of sqrt(2) + sqrt(3) over Q(sqrt(2))

[formal ermine]

your terminology is a bit off

[livid willow]

Hmm thought so, can you please say why?

[formal ermine]

you'd say "minimal polynomial of ... over ..."

[livid willow]

Would that be correct now?

[formal ermine]

do you know that Q(sqrt(2) + sqrt(3)) = Q(sqrt(2), sqrt(3))?

[livid willow]

No

[formal ermine]

try showing it

[livid willow]

Okay

[livid willow]

Sorry for taking quite long, I went for dinner.
I think this works? I showed inclusion of elements from one field to the other and said it exists both ways.

[north sand]

yes

[white yoke]

Is every one-dimensional real vector space isomorphic to the real numbers?

[tribal furnace]

yes

[formal ermine]

every n dimensional F-vector space is isomorphic to F^n

[formal ermine]

ye that looks good

now you should be able to see why [Q(sqrt2 + sqrt3) : Q] is 4 and why [Q(sqrt2 + sqrt3) : Q(sqrt2)] is 2

[livid willow]

Yep, I get it, thank you v much king

[formal ermine]

now to your original question, yes that should be the correct polynomial

[white yoke]

Thanks

[sonic coral]

if i’m trying to write a f, an element of Sym(7) as a product of transpositions, do i need to write it as a product of disjoint cycles first or can i do it directly from f

[coral shale]

How exactly is this f being described

or defined

That kindof matters?

[formal ermine]

do you mean Sym(Z_7)?

[coral shale]

permutations are usually given in disjoint cycle notation

[coral shale]

7 is a set containing 7 elements anyways if u want to be pedantic

[sonic coral]

i’m good so far

[coral shale]

I would go for disjoint cycle notation myself

[sonic coral]

that’s what i’ve done in parts c and d

[void cosmos]

is (3) a minimal prime ideal in Z[i]?

cuz Z[i] is a PID so assume (p) is a subset of (3) then p = 3k for some k but p is prime --> p is irred so either 3 or k is a unit but 3 isnt a unit tho lmfao so it must be k so 3k is in (p) but there is a v that can remove this k so 3k*v is in (p) as its an ideal so 3 is in (p) which says that (3) is a subset of (p) but (p) is a subset of (3) tho so (p)=(3)

and i think this doesnt have to do with the prime 3

is that correct boys?

[next obsidian]

The only minimal prime in Z[i] is 0

[void cosmos]

why is this wrong

[oblique river]

(p) being prime only implies that p is prime if p is nonzero

[void cosmos]

assume p is nonzero then

[oblique river]

...

(0) is a prime ideal which is contained in (3)

you can't just assume that away

[void cosmos]

does this work if we disregard 0

[coral shale]

maybe restate this properly

[void cosmos]

this sucks ass then

cuz in any integral domain trhis is jusut going to be true

tf

like (0) is minimal everywhere

[formal ermine]

ye

[oblique river]

yes... in an integral domain, (0) is a prime ideal

[void cosmos]

ik

i meant minimal

[oblique river]

and hence a minimal prime

[void cosmos]

yea

[oblique river]

what's the problem

[void cosmos]

i was to prove that the set of all prime ideals have a minmal element ( given that they contain each other downwards)

i did that using the intersection

and proved that the intersection indeed is prime ( and then used zorn's lemma to say that it exists over the whole ring )

and then the problem askjed for such an example for Z[i]

and it just doesnt amek sense that the naswer is just (0) lmfao

but its right so im dumb

[south patrol]

This is slightly sloppily worded right lol

I assume you mean you used intersection whilst applying Zorn's lemma

I'm not sure what you mean by "used Zorn's lemma to say that it exists over the whole ring"

[void cosmos]

i showed a chain has a minimal element

so this whole ring has a minimal element

i went from a chain to the whole ring

by zorn

is this correct

[formal ermine]

yes

[void cosmos]

cool

[chilly ocean]

?

[coral shale]

i was replying to this

[coral shale]

and this.

===
What?

[chilly ocean]

I honestly don't know what to say. They used "lmfao" in their proof.

[void cosmos]

proof

[white oxide]

im having a little trouble understanding why exactly the inverse image of any nontrivial proper normal subgroup of G/M must contain M. I know it's because all subgroups in G/M must contain the identity, and that the preimage of any subgroup in G/M is a subgroup of G, but how does the preimage contain M? doesn't it consist of selecting single elements in G which act as representatives for the cosets in G/M, and so for M in a subgroup of G/M, wouldn't the inverse image of that get mapped to a singular element in G which is in M?

idk if what i said makes sense

[next obsidian]

Inverse image is monotone

f^-1(N) contains f^-1(e) and the latter is M

Here f:G -> G/M

N is a subgroup of G/M

It also just follows directly from definitions. M is mapped into any subgroup of the quotient

[white oxide]

ohh okay got it thanks

[white oxide]

oh as in e = M you mean

[next obsidian]

No

The preimage of the identity in G/M is exactly M

[white oxide]

oh that makes sense

[white oxide]

or let me ask this question instead my bad

does (xHx^-1)^s has elements of the form (x^s)h(x^-1)^s where s is the order of H

[celest furnace]

No

because h^s would just be 1 always

So like suppose |H| = 2, then (xhx^-1)^2 = xhx^(-1)xhx^(-1) = xh^2x^(-1)=1 because the order of h divides the order of |H|

[coral shale]

[stuck fiber]

I figured I'd ask here but is there anything interesting about the quotient group whose elements are level sets? I'd imagine there are multiple of these groups too or no? If so are these groups named?

[south patrol]

What would that mean exactly

Quotient of what, for example

[stuck fiber]

Hmmm actually good point I guess I was just thinking of the level set as a fiber and quotient groups being made of fibers so a group made of level sets but I guess that doesn't entirely work yeah

[south patrol]

This works in a more topological way though

[stuck fiber]

How so?

[south patrol]

Well you can take a space A and some map A -> X and define ~ by a ~ b iff f(a) = f(b)

And do constructions like that

But you'd need smth special for it to give you a group structure

[grand cliff]

I'm trying to prove the left multiplication action and right multiplication action on a group G are equivalent

have I done something dumb here, for whatever reason I get a g^-1

[empty rose]

what's this?

[astral stream]

What is b)? I can't make any figure that matches this. for context the square has 8 symmetries: 4 rotations and 4 reflections

[summer path]

#linear-algebra

[tender wharf]

ask your question openly

[frigid lark]

Try to think about something that only has rotational symmetry

[analog zephyr]

No closed figure maybe?

[astral stream]

OK thanks Ill try

[white oxide]

Why are they showing $g^{-1}xg \in C$ for all $g \in G$ and $x \in C$ instead of $gxg^{-1} \in C$?

[cloud walrus]

okeyokay

[white oxide]

I thought the requirement was gxg^{-1} in C

For all g in G and x in C

It probably has to do with an expansion of c huh

[agile burrow]

It doesn't matter, since you'll have both g and g^-1

Meaning if you write your conjugation as gxg^-1, you recover the conjugation they describe when you plug in g^-1

[white oxide]

ohh right

[celest furnace]

? (Ah, H is a coset! It felt wrong when writing out...)

[agile burrow]

And since every element has an inverse, you have the same thing

[white oxide]

ok got it thanks!

[grand cliff]

Rho is a bijection between the sets G is acting on

When you want to show two group actions are equivalent you construct a bijection between the sets that the group is acting on and you prove the property which I have got in the image

Basically,
If G acts on Omega and Gamma. Then rho will be a bijection which just "relabels" the elements of Omega in the right way so that the group actions look the same

[white oxide]

what exactly is a nonzero ring? if R is a ring, then it is an abelian group under addition so it must have an additive identity, which my book defines to be 0. therefore, if a ring is nonzero, or does not have 0, then it can't be an abelian group under addition

what am i misunderstanding here?

[prisma bluff]

do they mean a nontrivial ring?

[white oxide]

i believe so, but here's the example

[prisma bluff]

the elements r and s are nonzero, not the ring

hmm

oh wait

[white oxide]

ye i just thought nonzero ring meant it didn't have an additive identity

or that the entire ring is not equal to the zero ring

is what they mean?

[prisma bluff]

yeah they meant nontrivial i think

[white oxide]

oh the zero ring is referred to as the trivial ring

okay

thanks!

[prisma bluff]

both are used but i like trivial better since the zero ring also contains 1, just turns out 0=1 though lol

[white oxide]

wat

im so confused

was that a joke that flew over my head

[prisma bluff]

in the trivial ring there is a multiplicative identity but it happens to be 0

1 is just the name of the multiplicative identity

[white oxide]

ohhh right

the additive identity is the multiplicative identity

and that's the only case

[white oxide]

wait so the original conjecture was right?

[agile burrow]

How/why is commensurability a useful notion of equivalence in group theory?

Responses don't need to be specific to group theory, really any place where commensurability shows up would be fine

[chilly radish]

'nonzero' doesn't mean "doesn't have zero" it means "is not just 0"

[white oxide]

ah i see

gracias

[chilly radish]

De nada

[flint cave]

Commensurability implies the groups are quasi-isometric, this is apparently a useful invariant in geometric group theory and hyperbolic geometry and tells us about the "large-scale geometry". Quasi-isometry is related to so many nice things in GGT : Growth rate of a group, Hyperbolicity, Ends of a group...

And things like growth rate are further important because they tell you information about random walks on the group!

Man GGT is so based

Unrelated PSA: the cayley graphs of a group are unique up to quasi-isometry!

[agile burrow]

Very neat, thanks

[hot veldt]

do homomorphisms have to be commutative

[paper flint]

Can you be more precise?

Like given a homomorphism f: G->H, do you mean to ask if it would be the case that f(ab)=f(b)f(a) for all a, b in G?

[hot veldt]

yeah

[paper flint]

No

[hot veldt]

thank you

[paper flint]

👍 the definition requires f(ab)=f(a)f(b), no guarantee that f(a)f(b) would be equal to f(b)f(a) in general.

[hot veldt]

i think my confusion is when the rings are comutative instead or something

im pretty dumb

[paper flint]

A ring is said to be (non)commutative if its multiplication operation is (non)commutative. The addition operation in a ring is taken to be commutative by definition.

[chilly ocean]

A function such that h(xy) = h(y)h(x) is called an anti-homomorphism

[formal ermine]

clopen moment

[rustic crown]

h : R^op --> S saves the day

[opal osprey]

If $\mathbb{F}$ is any field, is the symmetric bilinear form on $\text{Mat}_{n}(\mathbb{F})$ given by:
$$
(A,B) \mapsto \text{tr}(AB)
$$
non-degenerate?

[cloud walrus]

MisterSystem

[opal osprey]

When K = R this in fact defines an inner product on Mat_n(R)

[opal osprey]

But I am not sure how it goes for other fields.

[oblique river]

Yes it is

Basically you just need to show that for any nonzero matrix A there is a matrix B for which tr(AB) is nonzero

Which can be done explicitly

[opal osprey]

I wonder if B = A^t should do the trick

probably not since for certain fields we can have a sum of non-zero squares being 0 ig

[oblique river]

I think you can go even more primitive here

If tr(A) is nonzero take B = id

[warm urchin]

I have a group theory exam in a couple days , if anyone has some classical exercises or a problem pset or anything feel free to link me some (content includes usual basics , cyclic/symmetric/solvable/derivable groups , sylow theorems , group action , abelian group decomposition)

[oblique river]

Otherwise, if it has a nonzero element on the diagonal, take B to be the matrix which multiplies that row by 0

Otherwise… (fill in the details)

[formal ermine]

check pins

[warm urchin]

[void cosmos]

is Z[sqrt(d)] where d is square free integer being noetherian follows from hilbert basis theorem?

[white yoke]

What does this mean: 'nontrivial units all have support of cardinal 3' ?

Idk what 'support of cardinal 3' means

[oblique river]

what is the context

[white yoke]

The non trivial units of F_2[C_2 x C_2] have support of cardinal 3

[elder wave]

hint: quotients of noetherian rings are noetherian

[void cosmos]

yea i just said that this is jusut Z[x]/(x^2-d)

[white yoke]

They are 1+i+j, 1+i+k, 1+j+k, i+j+k

[void cosmos]

which is irred in Z so this is noetherian as Z is notherian ( hilbert basis )

is this enough of a proof?

[oblique river]

everything in that ring is of the form a[1] + b[x] + c[y] + d[xy]

I guess you're using i, j, and k instead

maybe it should say "cardinality 3"

andi t's just saying that the units only have three nonzero coefficients

if you're viewing F[G] as the set of functions from G to F, then the support of a function is the set of values where it's nonzero

so it would make sense in that context to say that the support has size 3

[quick moss]

hi guys, can you help solve this task: find all ideals in the ring R of real numbers

i think that it's nR where n \in R, but my lecture said that is something wrong in my solution

[solar shore]

quick question

for some $r \in U(n)$, is it true that $r \bmod n \cdot r^{-1} \bmod n = 1$?

[cloud walrus]

blanket

[solar shore]

oh wait

[south patrol]

What are your solution

Your answer is technically correct tbf but I'm not sure how you got to that (the solution is much simpler)

[oblique river]

i mean, "find all X" as an instruction typically means "write down a list such that each X shows up exactly once"

in which case this answer would not be correct because the proposed list has repeats

[quick moss]

I think that nR is all ideals (subrings in ring of real numbers)

[quick moss]

u r right, i know this fact, but idk how to fix this %((

[south patrol]

Oh I mean I don't think it should be accepted as correct for a question

[south patrol]

How did you get to that

[quick moss]

I think it's obvious because nR is contains some of numbers in R, it is subgroup by addition

[formal ermine]

do you remember what an ideal is?

[south patrol]

Yeah that isn't the full definition of an ideal, plus they are asking you to find all ideals, so just listing some obvious ones isn't enough (in general - here you've actually done them all but with repeats)

[quick moss]

how can i listing it without repeating

[oblique river]

can you give some examples of elements which are in 2R, and some examples of elements which are not in 2R?

[quick moss]

oh.. i agree that is the same

[oblique river]

that what is the same?

[quick moss]

i didn't understand the question(

[oblique river]

i dont understand you right now

what are you concluding

[quick moss]

I know that if answer is nR the items is repeating, but I don't know how to fix it( i cannot come up with an ideal

[oblique river]

you did not answer my question

[oblique river]

^

[quick moss]

no, because R is the real numbers, so for any item X i get X/2

or am i wrong?

[oblique river]

you're not making a claim right now

what are you trying to say about the ideal 2R

that it's equal to some other ideal maybe?

[quick moss]

2R = R?

[oblique river]

yes

[quick moss]

so...?

2R?

[foggy warren]

If R denotes the real numbers, is the quotient ring R[x] / (2x - 1) isomorphic to R ?

[oblique river]

what about 2R

[lethal dune]

Send x to 1/2

[median valve]

What’s the difference between Nullspace and Columnspace in linear algebra

Is it just that in the latter we have AX = 0?

[lethal dune]

#linear-algebra

[delicate orchid]

wtf is a column space

oh ok it's the span of it's columns

aka the image

[south patrol]

The image of the corresponding linear map x -> Ax 🤓

[void cosmos]

yo (3) is maximal in {m/2^n for m and n in Z and N respectively) right?

cuz 3 is prime and the localization of Z wrt any S is also a PID

[south patrol]

Well (3) being prime isn't enough to say it's maximal

[void cosmos]

it is

tho

cuz this is a pid?

[south patrol]

Okay sorry I thought you meant the ideal (3) lol

Cause (0) is prime but not maximal

But yeah sure

I would just use the correspondence for prime ideals of S^-1 R though

[void cosmos]

yes

thats what i did

but u would also use the fact that

S^-1 R is a pid if R is a pid

so u go from prime to maximal

right?

[south patrol]

Well you don't need that for this

You can just show that (3) is the only prime ideal containing (3)

i guess

Uhhh haven't thought about it lol

[void cosmos]

yea i man but i wrote this for an exam

soo i was just aksing if its true XD

[formal ermine]

what is sigma supposed to be here

at the end

like it's a typo

what was it supposed to be instead

there's this diagram

was it supposed to be tau_i = tau circ phi_i?

[elder wave]

????

thief

[formal ermine]

the sticker is just too good

[primal tusk]

anyone know what the notation on the right means?

specifically the x part of it

[south patrol]

Group of units

[summer path]

Invertible 2x2 integer matrices

[south patrol]

If $R$ is a ring then then $R^\times$ is the set of elements $a$ which are units i.e. such that there exist $b$ with $ab=ba=1$. This then becomes a group if you restrict the multiplication from $R$

[cloud walrus]

potato

[south patrol]

So, for example, $M_n(\mathbb R)^\times$ is just the invertible matrices, i.e. $\mathrm{GL}_n(\mathbb R)$

[primal tusk]

ahh okay thanks

[cloud walrus]

potato

[summer path]

Well here it's Z not R

[south patrol]

You may have seen this for (Z/nZ)^x before

Yeah, I was giving a separate thing

[prisma ibex]

this is probably referring to the group of 2x2 matrices with integer coefficients and determinant 1, most people write SL_2(Z) or SL(2,Z) for this group

[rustic crown]

det

[untold cloud]

Hi, guys

how do i show "<--" direction? f,g_1,g_2 are all field homomorphisms

[rustic crown]

field homomorphisms are automatically injective

[untold cloud]

Yes, but I guess this statement wants to give an alternative definition of injectivity which does not depend on elements in the field. "categorically injective"

[rustic crown]

like you're asking me to show monomorphisms in Fld are injective right

[untold cloud]

yeah, i do not how to show this

[rustic crown]

lol that's such a weird thing to do

[rustic crown]

yea you can give the definition "injective means it's a field homomorphism"

[untold cloud]

why

how can we show that f(x_1) = f(x_2) implies x_1 = x_2?

i.e. how can we find g_1, g_2 such that g_1(x) = x_1, g_2(x) = x_2

[rustic crown]

you can't always

so you have to specifically use something about fields

in rings you can do this because you have the polynomial ring Z[x]

there is no "free object on one generator in Fld"

[untold cloud]

Sorry, i don't understand

if i cannot always do this, then doesn't this statement is false

[rustic crown]

what i'm saying is that knowing that it's a monomorphism doesn't help much

because there are so few maps in the category of fields

it's a true statement because all field maps are injective

idk if i'm understanding your problem correctly

[untold cloud]

i am not sure, but thanks! i will think about it

[untold cloud]

oh, I see what you mean

yes, this makes sense

Thank you!

[primal tusk]

this is kind of a linear algebra question, but if detA*detB =1 can you conclude that AB=I by the use of some inverse determinant thing

[feral agate]

I have a feeling that you should use the first isomorphism theorem here. So we know that C7 is a simple group, so the kernel can only be of size 1 or 7. If there is a non-trivial homomorphism the kernel must be of size 1. Moreover, im(f) must be either of size 2 or 3. However, I am stuck at this step can someone help me out pls

[rotund aurora]

no

the determinant is not injective or anything, simply notice that det(A)=1 doesn't imply that A is the identity matrix (for example, let A be upper triangular, with all 1s on and above the diagonal, and the rest zeros).

[primal tusk]

ahh true

[feral agate]

nvm I think it's that there cannot exists such a homomorphism, because you have that 7 = 2,3 which is contradiction

[celest furnace]

You can show that the image of f will divide both 7 and 3!, of which only 1 works (showing that it divides 7 uses the first isomorphism theorem)

[feral agate]

So it cannot exist right?

[celest furnace]

yep the only divisor of both 7 and 3! = 6 is just 1 and the only homomorphism that has an image of size 1 is the trivial homomorphism

[brazen folio]

Hello, I've to prove that $G$ a group, $H \lhd G$, $K \It G$ and $H\subseteq K$ then
\begin{itemize}
\item $H\lhd K$ and $K/H \It G/H$ (done)
\item $K/H \lhd G/H \Leftrightarrow K \lhd G$
\end{itemize}

the first one is done and for the second one I only proved the $\Leftarrow$ implication and i'm stuck to the $\Rightarrow$ implication, I don't know how to mass from the quotient group to the actual group since $\pi_H : G \to G/H : x \mapsto xH$ is only injective if $H = {0_G}$.

Can someone help me ?

[cloud walrus]

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

[brazen folio]

don't tell me the resolution, just the intuition pls

[teal void]

i have done part (a), stuck on part (b). i just don't know what i'm eventually trying to get to - my progress so far is just messing around with the equation $aba^{-1} = b^k$ to $ab = b^ka$ but i don't understand how i'm even remotely close to showing $k$ must be certain values or something between $1 \leq k \leq 6$. from part (a), we know $o(a) = 3$ and $o(b)=7$ which sounds potentially helpful

[cloud walrus]

fluff

[frigid lark]

Do you have access to Sylow?

[teal void]

no sylow theorems

based on the next 4 parts of this problem, i think my goal is to show that k = 1,2, or 4

(next 4 parts)

well.... i guess i could say for part (b) that
Case 1: suppose G is abelian (then prove part c, that is, k = 1, which is pretty trivial)
Case 2: suppose G is nonabelian (then prove part f, haven't tried yet. but ultimately 1 <= k <= 6)

...but this feels like cheating?

[chilly ocean]

pi^-1(K/H) = K

Preimage of normal subgroup is normal

[tender wharf]

You don't really need this homomorphism here

[white oxide]

Is it true that if $G = G_1 \text{ x } G_2 \text{ x } \dots \text{ x } G_n$ for groups $G_1, \dots, G_n$ and G is not cyclic, then for a normal subgroup $N$, $G/N$ is always not cyclic?

[cloud walrus]

okeyokay

[frigid lark]

No, S_n/A_n

[sonic coral]

any hints on how to show that the intersection of any two sylow-p subgroups is never a sylow-p subgroup?

i know that the intersection can never be trivial, ie be the whole sylow p-subgroup (since they’re distinct) but that’s about all i’ve got

the intersection of the two subgroups is also a subgroup though, so maybe it has to do with lagrange… but the sylow-p subgroups don’t have to have an order of a prime, they can also have the order of a prime to some power

[quiet pelican]

What’s the definition of a sylow p subgroup?

[sonic coral]

if p^n divides the order of G then we have a Sylow-p subgroup of order p^n

if p^n+1 does not divide G

where p is prime obv

[quiet pelican]

So can a subgroup of size < p^n be a sylow p subgroup?

[sonic coral]

not be definition i don’t think. since if p^n-1 divided G then it wouldn’t be the biggest prime power that divides G

[quiet pelican]

Exactly

And you’ve proven that any intersection of sylow p subgroups has order strictly less than p^n

[sonic coral]

well that just comes from the sylow p subgroups being unique though

but then i guess we know that the intersection of two sylow p subgroups divides p^n-1

but any subgroup of order p^n-1 is never a sylow p subgroup since - sylow p subgroup is defined to be the maximum power of p?

thank you, you asked the right questions for it to make sense

[white oxide]

ok this may seem like a silly question, but is associativity of multiplication in the set of integers taken as an axiom?

is it some sort or related to the peano axioms?

[smoky cypress]

No you can prove it from the peano axioms

Let $P,Q$ be different maximal ideals of $R$, why is $P_Q=R_Q$? The book claims that ``since $P$ is maximal, $P\not\subset Q$, so $P_Q=R_Q$'' but I don't see why

[cloud walrus]

Whoever

[next obsidian]

P contains an element not in Q, so that becomes a unit in R_Q

[smoky cypress]

Ah

And also if $I\subseteq R$ is an ideal and ${\frak p}\subseteq R$ prime then $I_{\frak p}$ is an ideal of $R_{\frak p}$ and $I_{\frak p}=IR_{\frak p}$?

[cloud walrus]

Whoever

[smoky cypress]

Yeah that's clear

Ok

[next obsidian]

@smoky cypress in fact all ideals of R_p are of that form

But the correspondence is not bijective for non-prime / non-primary ideals

Specifically, if f is the map R -> S^-1R, and I an ideal then I = f^-1(I)S^-1R

But it’s possible two different ideals I and J have IS^-1R = JS^-1R

[smoky cypress]

Right

I like how borcherds just never mentioned anything but rather went straight to the geometric intuition for localization

[white oxide]

What is an example of a commutative ring without unity?

[prisma ibex]

compactly supported functions on a non-compact space. Functions still add and multiply point-wise, but the function with constant value 1 isn't compactly supported, so you don't have a unit

[white oxide]

what are compactly supported functions and a non-compact space

[prisma ibex]

I mean we could take for example compactly supported functions on the real line R, which isn't compact

[white oxide]

what does compact mean

[prisma ibex]

compactly supported means like

[white oxide]

is that an analysis thing

[prisma ibex]

the function is zero outside a compact set

it's analytic or topological yeah

[white oxide]

what is a compact set 🐒

[prisma ibex]

being a compact subset of R is the same as being closed and bounded

think closed intervals [a,b]

[white oxide]

ohh ok

[empty rose]

wouldn't almost any ideal of a commutative ring be a commutative ring without unity

[next obsidian]

Oog

So true

But actually it’s possible for the ideal to have its own identity

Like in R x S

R x 0 has an identity in the form of (1,0)

But if your ring doesn’t break apart into a product what you said is true

[empty rose]

yeah that's why i said "almost"

[next obsidian]

Ah

I thought that was to get rid of (1) and 0

[chilly ocean]

Are there 8 representations of 5?

[sharp sonnet]

depends what you count as a representation, but why dont you just check?

[chilly ocean]

1 2
2 1
-1 -2
-1 2
-2 1
-2 -1
1 -2
2 -1

[atomic mesa]

In Z[x], why is the ideal generated by (2, x) different from (x) ?

As I understand, (2,x) = 2*(any polynomial in Z[x]) + x*(any polynomial in Z[x])

[rustic crown]

try looking at their quotients :3

[topaz heart]

the ideal generated by (x) doesn't have 2 in

[elder wave]

that seems a bit overkill

[rustic crown]

yep. things in (2,x) have the constant term which is even. but things in (x) have 0 as their constant term >.<

[atomic mesa]

OH MY GOD

Thank you

By the way, is it possible to consider ideals generated by a_i’s where the “scalar” multiplier is different?

[atomic mesa]

So in this example, could I have instead interpreted (x) = x*(any element of the ring Z)

[rustic crown]

that's not true.. (x) contains x^2 for example

[atomic mesa]

Ah ok I see now

[south patrol]

That'd be the Z-submodule (i.e. subgroup) generated by x, rather than the ideal (or Z[x]-submodule)

[oblique river]

i dont think that using module language is going to be useful for someone who is asking questions about ideals

[rustic crown]

https://youtu.be/IYiULwc9Jz8

[south patrol]

P-potato

[south patrol]

Oh i learnt about modules and ideals at basically the same time lol mb

sozzles

[barren sierra]

Really?

Interesting

I didn't see modules in my undergrad course in algebra at all

But ofc ideals appeared alot

[quasi hatch]

heyy..for this equation x^5 + x + 1 = 0 is there way to find a root ..by the method of how we solve quadratic or cubic equation ??
i mean not just by trial or error method or any iterations
can we solve this symbolically without using numbers ???

[south patrol]

The course was called rings and modules

Well x^2 + x +1 divides x^5 + x +1 apparently?

[formal ermine]

we only did modules for the classification of finitely generated modules over pids in ug alg

[chilly ocean]

x^5+x+1 = (x^2+x+1)(x^3-x^2+1) then

Now... let my wipe out my Cardano formula...

[rotund aurora]

Let p, q be primes. If p is a quadratic residue modulo q, what does that say about the ideal (q) in (the ring of integers of the extension) Q(sqrt p) ?

[oblique river]

let O be the ring of integers of Q(sqrt p); compute O/(q) directly

it will be one of three options: F_q x F_q (in which case q splits) or F_(q^2) (in which case q is inert) or F_q[x]/x^2 (in which case q is ramified)

[rotund aurora]

okay got it

[oblique river]

(to be clear, by "directly", I mean if O = Z[x]/f(x), then O/q = F_q[x]/f(x), so you just look at how f factor mod q)

[rotund aurora]

yes, yes, that's what I did

well but O is not exactly Z[x]/f(x) ?

[oblique river]

👍

[rotund aurora]

I did this

[oblique river]

i mean, it is for an appropriate choice of f haha

[rotund aurora]

which is valid when p is relatively prime to the conductor of Z[theta]

I guess it applies in most cases here but maybe there's some minor exception lul

[oblique river]

I mean, choose theta so that O = Z[theta]

then the conductor is 1

[rotund aurora]

ah well

right right

[oblique river]

i mean at that point you dont even have to think abuot conductors

cuz if O = Z[theta] = Z[x]/(minpoly of theta)

[rotund aurora]

yeah definitely

[oblique river]

yeah

[rotund aurora]

this is monogenic

[oblique river]

indeed

if you want, you can either split into cases for p mod 4 and then either use f(x) = x^2 - p or f(x) = minpoly of (1+sqrt(p))/2

or you can just use f(x) = x^2 - p for all p, and this will work for all q \neq 2 because in the cases where O isn't equal to Z[sqrt(p)], the index of that order is 2

so for all q neq 2 you're still good

[rotund aurora]

here I think it works to simply consider O=Z[1/2*(d+sqrt d)] where d is the discriminant of Q(sqrt p)

wait maybe you still get little problem if q is 2 because it may divide the discriminant without dividing p

Thank you again

[primal tusk]

I need to prove that the symmetric group of size n>=3 is not abelian

should I go about this using induction?

[south patrol]

No, in fact if you show it for n=3 then the rest follow immediately

Think why

[primal tusk]

im thinking about this, its making me think about a couple of things

possibly unrelated

okay now i see it

but also how many bijections are there from a set to itself of size n

and also how many isomorphisms?

[south patrol]

So like how big is Sn?

[south patrol]

This is vague

[primal tusk]

yea i think im a little confused on this stuff

so for my initial question i see why its true now

but on an unrelated note how is Sn related to the number of isomorphisms i guess you could say

idek if that question makes sense

the way im thinking about it, a bijection from a set onto itself is almost just like a permutation of its elements

but like how does that differ from an isomorphism

nvm tbh idek what im saying

[south patrol]

Okay lemme try to clear stuff up

Basically when we have a set $X$, bijections $X \to X$ are equivalently permutations of its elements and form a group under composition (called for example $\mathrm{Sym}(X)$, standing for symmetries of $X$). This is cause the composition of bijections is a bijection (existence of inverses and identity etc is clear) Note that $X$ here is just a set, whilst $\mathrm{Sym}(X)$ is a group. One then in particular takes $X = {1,\dots,n}$ to form $S_n = \mathrm{Sym}(X)$.

[cloud walrus]

potato

[south patrol]

But the important thing is that X is just a set, so there is no notion of group isomorphisms X -> X

(Bijections are isomorphisms of sets)

So like it makes sense to ask how many bijections X -> X there are, i.e. how big is Sym(X)

and the answer is the factorial of the size of the set (can prove by induction - maybe I can help if you don't know why that's the case)

[primal tusk]

that makes sense

im realizing a bijection doesnt imply commutativity, which is obvious in hindsight

all these ideas are just fuzzy im not exactly sure how to think about them but im starting to get a grip

like i realizing an isomorphism is a map between a group or ring but i want to understand it on a deeper level but its just not clicking that well

[south patrol]

Not sure what you mean by this

Or like

Composition of biijections isn't commutative?

[primal tusk]

yea thats what i meant

[warm urchin]

Prove that if |G|=105 then G has a normal sylow 5 subgroup and a normal sylow 7-subgroup.

I managed to show that G need to have a normal sylow 5 "or" a normal sylow 7 using a sylow theorems argument
( without too much details If not we would get number of sylow 5 subgroups is s5=21 and s7=15 so we 21 x 4 = 84 elements of order 5 , similarely we have 90 elements of order 7 and so we have at least 84+90+1 elements in G a contradiction so one of them must be unique )

but how do i conclude that it needs to have both a normal sylow 5 and a normal sylow 7? i have seen some answers online that use semi-direct products or characteristic subgroups which im not too familiar with , is there other ways around solving this or nah?

[void cosmos]

yes there are

assume one is normal

and keep going

[warm urchin]

Let $H$ be the unique sylow 5-subgroup and $K$ be a sylow 7-subgroup , then we consider the subgroup $H.K$ of $G$ , since $H \cap K={e}$ then $|H.K|=35$. notice that $$K \trianglelefteq H.K$$ (its a sylow 7-subgroup of $H.K$ and a simple sylow argument shows its unique) moreover we have $H.K$ is a subgroup of the normalizer of K (since the normalizer is the largest subgroup of G where K is normal in it) finally if the normalizer is not equal to G , and thus has order 35 , we get $[G : N_{G}(K)]=3$ which is a contradiction since the number of sylow 7-subgroups cannot be 3 and hence it must be that $s7=1$

[cloud walrus]

Susilian

[warm urchin]

i think this should work

let me know if i made any mistakes in my argument

[atomic mesa]

In Z[x], is the ideal (2,x) the same same as the ideal (2,x^2) ?

[rustic crown]

nooope, the latter doesn't contain x

[atomic mesa]

ahhh, but (2,x) = (2,x,x^2) ?

[south patrol]

Yes, as x^2 is in (2,x)

[atomic mesa]

im just trying to figure ideals out but not so great at spotting non elements

OHH

ok got it

[glossy crag]

I don't get why the inverse of such an expression is again such an expression

[rotund aurora]

all of that is algebraic over Q

[glossy crag]

I mean, I get that the inverse is a polynomial

But why should it look like this

[oblique river]

if a is algebraic over Q then its inverse can be written using sums of powers of a. consider the minpoly of a, say c_na^n + ... + c_1a + c_0 = 0. we know that c_0 is nonzero as this polynomial is irreducible (and a is nonzero). subtract c_0 and factor out the a from what's left:
a (c_n a^(n-1) + ... + c_1) = -c_0
now divide both sides by -c_0 and we've found a^(-1) as a polynomial in a

now apply this to a = (alpha_1)(beta_1) + ... + (alpha_r)(beta_r). that's algebraic (as it's a sum/product of algebraic elements) and so its inverse is a polynomial in a

now just expand out all the powers in the polynomial

[rotund aurora]

tru

[glossy crag]

Ah, that's good. Thanks!

@oblique river quick question on composites: does |KL:Q|=|K:Q||L:Q| iff K\cap L=Q always hold (if ever) or does one of the fields have to be Galois?

[oblique river]

no, not necessarily

[glossy crag]

Meaning the equivalence is always true, w/o any restrictions on K&L?

[oblique river]

sorry

no it's not true even with that restriction

like, in some cases it is true

if |K:Q| and |L:Q| are relatively prime, then |KL:Q|=|K:Q||L:Q| (no other assumptiosn necessary; this assumption implies that K \cap L = Q)

if both are galois over Q then "K \cap L = Q implies |KL:Q|=|K:Q||L:Q|" is true

[glossy crag]

Thanks

[oblique river]

if only one is galois?

uhhhh let's see

i think it should still be true?

[glossy crag]

This you mean.

[oblique river]

yeah

[oblique river]

like here you asked "or does one of the fields have to be galois"

i think that if one of them is galois then it's true; i know it's true if both are galois

(because if they're both galois then one can prove Gal(KL/Q) = Gal(K/Q) x Gal(L/Q) which implies this)

[glossy crag]

I know about this isomorphism, but does this mean in practice that every \sigma in Gal(KL/Q) can be uniquely written as \tau_i\pi_j (meaning pointwise multiplication, not composition)?

[oblique river]

yes, assuming you have elements of Gal(K/Q) act trivially on L, and vice versa

fwiw that's also the same as considering it as a composition of functions

you can also think of it as a pair (tau, sigma) acting on a product alpha*beta, and you get tau(alpha)*sigma(beta)

[glossy crag]

This is reminiscent of tensor products

Is KL the tensor product of K and L?

[oblique river]

(brb in like 15m or so, you can keep typing and i'll respond when i get back)

[oblique river]

yes if K and L satisfy the condition we're talking abuot

people say K and L are "linearly disjoint" if |KL:Q|=|K:Q||L:Q|. I believe it's also true that K and L are linearly disjoint iff KL = K \otimes L.

but KL = K \otimes L can't always be true

e.g. if L = K then K \otimes K won't be a field (unless K = Q)

[white oxide]

Why is the quotient group in number 10 not isomorphic to $\mathbb{Z} \times \mathbb{Z}_8$? My logic was the only requirement was that we need $\mathbb{Z}$ in the direct product since the quotient group is of infinite order, and we also need $\mathbb{Z}_8$ since the quotient group has an element of order 8

[tribal moss]

(Use $\times$ to get a product sign).

[cloud walrus]

Troposphere

[white oxide]

oh ok thanks

[cloud walrus]

okeyokay

[tribal moss]

Would you agree that number 10 is definitely isomorphic to Z × Z4 × Z8?

[white oxide]

wdym

how can a number be isomorphic to a group

[tribal moss]

The thing listed as item number 10 in your picture.

[white oxide]

bruh i'm stupid

uh

does it have to do with all multiples of 4 in the second component getting collapsed

so we're left with Z4

[tribal moss]

Yeah -- I was trying you figure out if your question was something like "I know it's Z × Z4 × Z8, but why is that not also Z × Z8?".

[white oxide]

oh i suppose that too

because if you take out all the multiples of 4 in Z

it's still infinite right

[tribal moss]

No, there are only 4 elements left, because you're removing the distinction between elements that used to be separated by a multiple of 4.

You're aware that $\bZ_4$ itself is $\bZ/\langle 4\rangle$, right?

[cloud walrus]

Troposphere

[white oxide]

no i wasn't aware

wait

wait but Z4 is just {0, 1, 2, 3}

oh wait

nvm i'm wrong

ok now i'm aware

[tribal moss]

A bit harsh to ask you about quotients of products of Z's if you haven't seen the construction of Z_n as a quotient yet. That's the main intuitive example of how quotients work.

[white oxide]

oops i lied they did

brain not working today

I remember learning that $\mathbb{Z}/n\mathbb{Z} \simeq \mathbb{Z}_n$

[cloud walrus]

okeyokay

[white oxide]

oh right, and $\langle(0, 4, 0)\rangle$ is exactly $4\mathbb{Z}$ in the second component

[cloud walrus]

okeyokay

[white oxide]

so we only regard the second component?

[tribal moss]

In this case yes, because all elements of <(0,4,0)> are trivial in the two other components.

[white oxide]

got it, thank you!

[solar glacier]

random question

is $\Bbb{Q}(\sqrt[m]{2}) \subset \Bbb{Q}(\sqrt[n]{2})$ iff $m \mid n$

[cloud walrus]

MyMathYourMath

[solar glacier]

cause i know $\Bbb{Q}(\sqrt{2}) \subset \Bbb{Q}(\sqrt[6]{2})$

[cloud walrus]

MyMathYourMath

[solar glacier]

i had a prof who hated using Z_n for integers modulo n as z_p is the p adic integers lol

[south patrol]

Localisation of Z at p jk

[solar glacier]

LOL

[white oxide]

Does anybody know what the notation $R_n$ refers to in Gallian

[cloud walrus]

okeyokay

[white oxide]

I tried looking at the list of symbols

but couldn't find it

[frigid lark]

Well they both have the same maximal ideal

well kinda

[rotund dragon]

rotation by n degrees

Or at least thats what assuming based on ch.1
Idk without more context

[torn warren]

Is this correct?

[chilly radish]

Well one is the completion of the other

[untold cloud]

Hi, guys, how do i see that this definition does not depend on the choice of morphisms k\to E and k\to E'

[rustic crown]

it depends, because objects are the maps

for example consider k = F_p(x) then id : k --> k and Fr : k --> k are two different extensions

[untold cloud]

Oh I see, then is it true that suppose i have f_1, g_1: k\to E be two extensions, f_2, g_2: k\to E' be two extensions, then the size of Mor_{k,f}(E,E') is equal to the size of Mor_{k,g}(E,E'), where Mor_{k,f}(E,E') is the morphism of f_1 and f_2

[untold cloud]

or is there an isomorphism between f_1 and g_1

[rustic crown]

in general they need not be equal

one example is this, take k = C the complex numbers and then you can think of C as an extension over itself, by id : C --> C or also by doing some transcendence basis stuff to find a non-surjective map f : C --> C which isn't an algebraic extension. now automrophism of the first extension is a trivial group, but for the second it is non-trivial

[tribal moss]

(We can avoid the choicey transcendence-basis stuff by doing the same thing with R(x1,x2,x3,...) as an extension over itself instead.)

[untold cloud]

I see thank you, they in general are not isomorphic

[rustic crown]

yea, they're like different objects

so shouldn't expect a lot of things to hold

[spark crow]

I am trying to understand wreath products and wanted to write down the isomorphism from $D_8$ the Dihedral group of order 8, to the wreath product $C_2 \wr C_2 = (C_2 \times C_2) \rtimes C_2$. However since i don't yet understand how the semidirect product works with an example i wanted to look this up.
As far as I understand my wreath product here should be the same as $V_4 \rtimes C_2$ since $V_4 \cong (C_2 \times C_2$. My problem now is that Wikipedia states as an example of a semidirect product "The dihedral group $D_{2n}$ with 2n elemntes is isomorphic to a semidirect product of the cyclic groups $C_n$ and $C_2$".
Does this mean $D_8 \cong V_4 \rtimes C_2 \cong C_4 \rtimes C_2$
This seems to me like $V_4 \cong C_4$, but that is obviously wrong. SO what am i missing?

[cloud walrus]

achilles199703

[empty rose]

why would $V_4 \rtimes C_2 \cong C_4 \rtimes C_2$ imply $V_4 \cong C_4$?

[cloud walrus]

bee

[spark crow]

That was my question. Can different Semidirect products, create the same group?

[empty rose]

well apparently they can, given that in this case they do

i don't really see why that would be a problem

notably it's not necessarily true that the subgroups in which the part from the second group is the identity element, are mapped onto each other by the isomorphism

i'm pretty sure there's an isomorphism that maps the element (1, 0) of C4 \rtimes C2 onto ((0, 1), 1) of V4 \rtimes C2

so this group has a subgroup isomorphic to C4, and a subgroup isomorphic to V4, and they're not the same subgroup, so we don't conclude that C4 \cong V4

...now that i think about it there are also far more obvious cases where even distinct products produce the same group

let A, B, C be arbitrary groups, and consider (A x B) x C, and A x (B x C)

(or for that matter, A x B and B x A)

[spark crow]

ok thanks I see, I just have to look at semidirect products a bit more, I even have the subgroup diagram of D_8 drawn and now looking at it after your explanation I clearly see that there are {e,r^2,s,r^2s} = V_4 and {e,r,r^2,e^3} = C_4, which are subgroups of D_8. Really helped me, thanks

[empty rose]

...oh huh yeah i had completely forgot that this group is also D_8 and that the C4 and V4 were therefore going to be subgroups like that
but of course now that you mention it it's extremely obvious

[spark crow]

when you understand someting it always looks trivial from then on, but before it sometimes is just weird

[vestal lynx]

Write each of the following arguments in symbolic form. Then establish the validity of the argument or give a counterexample to show that it is invalid.

If there is any chance of rain or she loses her red hair ribbon, then Laura will not mow the lawn. Whenever there is a temperature above 20 degrees Celsius there is no chance of rain. Today the temperature is 25 degrees and Laura is wearing her red headband, so at some point in the day Laura will be mowing the lawn.

[formal ermine]

this is not abstract algebra

[vestal lynx]

Sorry, what topic is it?

[tribal moss]

Probably symbolic logic, part of the topic of #proofs-and-logic.

[rustic crown]

luffy

[lethal dune]

f-luffy

[rustic crown]

[lethal dune]

[primal tusk]

for this question would only 0 be in the kernel

because no others could have 19 in their prime decomposition

and if not is there a way to do this without testing all numbers

[rustic crown]

do you know the definition of kernel?

[primal tusk]

something that maps to 0

[rustic crown]

not exactly 0

and not something

[primal tusk]

an equivalence class that maps to the equivalence class of 0?

[rustic crown]

it's everything that maps to the identity element of the target

[primal tusk]

ohh

my mind was stuck in ring theory mode for some reason lol

okay but now is there a way to do this besides testing all numbers

[rustic crown]

yee

[primal tusk]

i think its all numbers that are their own inverses

[rustic crown]

you wanna find all x (mod 19) such that x^2 = 1

[primal tusk]

yea

is there a fast way to find those other than just cranking them out

[rustic crown]

how would you do that if it were reals?

(instead of Z/19Z)

[primal tusk]

take the square root

[rustic crown]

but that gives you only x = 1

how do you get the other root

and in general how do you solve quadratics

[primal tusk]

quadratic formula?

[rustic crown]

you can factor it >.<

[primal tusk]

ohh lmao

[rustic crown]

(x+1)(x-1)

so when does 19 divide a product :3

[primal tusk]

oh its just 1 and 18

[rustic crown]

[primal tusk]

https://tenor.com/view/ohhh-duh-why-didnt-i-think-of-that-gif-21849807

[formal ermine]

what do you mean with this hint? I would have solved it by factoring then either x - 1 = 0 so x = 1 or x + 1 = 0 = 19 so x = 18

[rustic crown]

oh so here you're using that 19 is a prime right

[empty rose]

either x - 1 = 0 [...] or x + 1 = 0
...well that's what the hint is

[formal ermine]

ah

[empty rose]

19 divides (x+1)(x-1), therefore either it divides x+1 or it divides x-1

[formal ermine]

wait

is 19 prime

[empty rose]

yes

[rustic crown]

yes, 57 isn't

[formal ermine]

oh

I was thinking like

3 divides it lol

[empty rose]

3 divides 18

[rustic crown]

and 3 divides 9

[empty rose]

also if 19 wasn't prime that logic wouldn't be valid

what if like, x-1 is 3, and x+1 is 6, and 19 is 3*6

then you can have (x-1)(x+1) be divisible by 19 even though x-1 and x+1 both aren't

[white oxide]

What would be the most efficient way to find a subgroup of order 8 in Z4 x Z4? No element in Z4 x Z4 can generate a subgroup of order 8, so would the best method be of trial and error?

If so that's a fucking pain

[empty rose]

well what if two elements generate a subgroup?

[hot lake]

you could look at the kernel of group morphisms into Z_2

there aren't that many group morphisms into Z_2

[white oxide]

do you mean homomorphism

if not i don't know what a morphism is

[hot lake]

yes

[white oxide]

oh right

but i'm trying to find a faster way to do it then trial and error, i'll have to think about it

[white oxide]

ok

[rustic crown]

in general you have a simple bijection between subgroup of order p and subgroups of index p for any finite abelian group A.
say f : A --> A is multiplication by p, then subgroups of order p are contained in the kernel of f and subgroup of index p contain image of f. so subgroups of order p are exactly subgroups of ker f of order p, and subgroup of index p are in bijection with subgroups of coker f of index p.

but both ker f and coker f are iso and look like a direct sum of Z/pZs

(rest is linear algebra, i.e. giving a bijection between dim 1 and codim 1 subspaces :3)

ig this is much more helpful if you only want to find the number of subgroups of order 8 (= index 2) than actually finding the subgruops

[torn warren]

Is this pic correct for the #4?

[rustic crown]

yee looks good

[torn warren]

ye, I want to confirm that those elements inside $K'\backslash \phi[H]$ is not defined under $\phi^{-1}$

[rustic crown]

that's a weird phrasing

[cloud walrus]

Witness

[rustic crown]

if f is a function from X --> Y, then f^-1 is by definition a function from subsets of Y to subsets of X.

so weird to use "not defined under phi^-1"

it's just phi^-1(K' \ phi(H)) = empty

[torn warren]

another way to say: #4 says $\phi^{-1}[K']$, it means $\phi^{-1}[K']=\phi^{-1}[K'\cap\phi[H]]$, right?

[cloud walrus]

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

[rustic crown]

nah, it means phi^-1(K')

and both of these things are equal

f^-1(B) = {x in X | f(x) in B}

if f : X --> Y

i.e. all the things in X that map into B

[torn warren]

$\phi^{-1}$ can only map where it is defined, here "where it is defined" is $K'\cap\phi[H]$

[cloud walrus]

Witness

[rustic crown]

yea and phi^-1 is defined on all of P(G') = power set of G'

[rustic crown]

notice here you're applying f on to stuff from X, so there isn't any "defined-ness" issue

[torn warren]

but if you pick an elment $x\in K'\backslash \phi[H]$, what is $\phi^{-1}(x)$? it is not defined.

[cloud walrus]

Witness

[rustic crown]

right

but f^-1 of an element and f^-1 of a subset are very different things

f^-1(y) is asking "what is the element that maps to y"
and
f^-1(B) is asking "what are the elements that map to something in B"

[torn warren]

if $f: A\rightarrow C$, and $B\subset C$ then why not directly use $f^{-1}(B\cap f(A))$ to replace $f^{-1}(B)$

[rustic crown]

well, why complicate the definition when it doesn't need to? and they're equal anyway, so it doesn't hurt to define something on a larger class of objects

i would say f^-1(B) is more direct, than the first thing you wrote

[cloud walrus]

Witness

[torn warren]

so I think this is just convension

[rustic crown]

[torn warren]

but if visualize in pic I prefer $f^{-1}(B\cap f(A))$

[cloud walrus]

Witness

[summer path]

if there are things in B that aren't even in the image of A, why does it matter that we specify that we want the preimage of things that are actually in the image of A

[torn warren]

so for $f^{-1}$ here it maybe not even a function, right?

[cloud walrus]

Witness

[torn warren]

because f might not be injective

[rustic crown]

f^-1 is not a function from Y to X unless f was bijective

but it is a function from P(Y) to P(X)

[torn warren]

do you mean $P(X)$ and $P(Y)$ are power sets?

[rustic crown]

nah

if f : X --> Y is a function then you can naturally define f^-1 : P(Y) --> P(X)

this is nice to do because you don't need any conditions of f to define this

and this has really nice properties

like it preserves unions and intersections

[cloud walrus]

Witness

[rustic crown]

yep

[torn warren]

wait, if $f:A\rightarrow B$, where $A={-1,1,2}$, $B={0,1,2,3}$ and $f(-1)=f(1)=0, f(2)=2$, what is $f^{-1} ({0})$

f^-1{0} can be {-1, 1}, {-1}, {1}

it is not a function

[rustic crown]

no

it is all the things that map in to {0}

so it has to be {1, -1}

[torn warren]

ah, i see, so it is by default, "all the things"

[summer path]

this feels like level sets

[cloud walrus]

Witness

[white oxide]

oh ok thanks i'll use this!

Can I get a hint for this question? My idea was to use contradiction, and assume that |Z(G)| > 1. Since Z(G) is a subgroup, it has to divide the order of G which is pq, so let k = |Z(G)| > 1. Then k | pq, and I'm trying to show that this means that p or q is not prime, but I can't come up with a case, because there is a case that k = 2 and either p or q = 2

[versed ingot]

Hey guys i need some help. How can I proof that $Z_{p^2}$ is not isomorphic to $Z_p$ $\times$ $Z_p$

[cloud walrus]

Stardustfalco

[versed ingot]

I tried assuming they are isomorphic but i dont know how to get to a contradiction using that

[remote dust]

you can try looking at the order of an element in Zp^2 and the order of one in Zp x Zp in particular the highest order element of Zp^2 is p^2, what about Zp x Zp?

[white oxide]

For example, G can have order 2q for some prime q and if k = 2 is the order of the center of G then k | 2q

[versed ingot]

The order should be p in the case of Zp x Zp isnt it?

[warm urchin]

what is the order of G/Z(G) ?

[remote dust]

exactly, and isomorphisms map elements of the same order to ones of the same order

[warm urchin]

If you assume its not trivial

[versed ingot]

Lets goooo, thanks a lot!

[white oxide]

Would it be of prime order?

[warm urchin]

Correct , what does that mean?

[white oxide]

i suppose there's a contradiction in there somewhere, showing that it's cyclic but how does it having prime order imply that it's cyclic

does that only follow for Zp where p is prime

maybe i could show that it's isomorphic to it?

[warm urchin]

If you have a group of prime order , and you pick an element thats not the identity, what is the order of that element? ( use || lagrange||)

So what im trying to say is that you can prove that : || every group of prime order is cyclic ||

[white oxide]

ohhh wait that was a corollary in the book

i should review taht

[white oxide]

rightt

okay thank you so much!

[primal tusk]

if f is a permutation on the set {1,2,3,4,5,6,7} and f^6 can be written as

(1,3), (4,5)

as disjoint cycles

what is the order of f^6 ?

[chilly radish]

Do you have a guess?

[primal tusk]

well im guessing its not just 7

[chilly radish]

Those are not disjoint cycles tho

[primal tusk]

i wrote it wrong

[chilly radish]

Ok

[primal tusk]

so maybe 5

[chilly radish]

Are you asking about the order of f^6 or f

[primal tusk]

since 3 things map to themselves and 2 cyclic groups

f^6

because the order of f would just be 7 right

[chilly radish]

No it's not 7, note that multiplying f^6 by itself is not the same thing as multiplying it by f

What you're saying is that f*f^6=id

[coral shale]

You should review some of these definitions

[chilly radish]

You can verify this is not true

[primal tusk]

i thought the order of the group was just the number of elements

[coral shale]

Why would the order of f be 7

[primal tusk]

i swear i just saw that in my book

[coral shale]

f is an element

[chilly radish]

This is a permutation

[coral shale]

of the symmetric group

[chilly radish]

Not a group

[primal tusk]

oh so theres only 2 here?

[coral shale]

???

2 what.

[primal tusk]

in f^6 i mean

since two cycles?!

[coral shale]

forget that

Whats the order of f

lets get this straight first

f is an element of S_7, how do you check its order?

[primal tusk]

by breaking it down into its cyclic groups?

[coral shale]

ok, but how does that help?

[primal tusk]

im not sure actually lol

[coral shale]

What does 'order' mean for an element of a group

[primal tusk]

number of elements

[coral shale]

No...

That's the order of the group

That is different from the order of a specific element in the group

Yes, 'order' is used in 2 contexts.

===
The order of an element is the smallest positive number of times you need to combine it with itself to get the identity element

[primal tusk]

ahh

[tribal furnace]

It's the order of the group generated by that element

[primal tusk]

generated by its powers right?

because i can see that

so im getting (1 5 3 4)(2 7 6) as the cycles for f

[coral shale]

Ok

So you have f = ab

where a, b are disjoint cycles

You should hopefully know that disjoint cycles commute

so f^n = (ab)^n = a^n b^n

[primal tusk]

so there are two cycles in f why do i know that its order is the same as its number of cycles

[coral shale]

It isn't, why do you think it is

[coral shale]

The only thing you know is the defn i wrote above

[coral shale]

or equivalently

[primal tusk]

true

[coral shale]

this is a hint to compute it

Do you know what the order of a cycle is?

[primal tusk]

the number of times you have to do it to get the identity?

[coral shale]

yes, but specifically for a cycle?

[primal tusk]

i dont know other than that

[coral shale]

Consider the cycle (1 2 3)

compute by hand its order (it wont take you long)

[primal tusk]

like in the numbers 1 through 7?

[coral shale]

it doesnt really matter

[primal tusk]

is it just 1?

[coral shale]

no?

Can I just check - how much has your course done so far

Group, Subgroup, Lagrange, Permutation?

[primal tusk]

a bunch of ring stuff ideals and modular stuff and we just started groups so we have done isomorphisms and subgroups

[coral shale]

I'm trying to help as directly as I can, but it isn't happening until we clear up a few definitions

[primal tusk]

ur good i just want to know what is happening

[coral shale]

How comfortable are you with the definition of a group - have you seen several examples

[primal tusk]

i know the definition and i know they have to do with symmetries

[coral shale]

Its best to check some small examples, and check the orders of elements in those

[primal tusk]

the orders of rotations of shapes makes sense

like of a square the order of 90 degree rotation is 4

i guess idrk what (1 2 3) means

[coral shale]

ok

So talking in general for a group

only one element has order 1

and that is the identity

because you have g^1 = e

and only g = e will solve this equation

===
The other 'special' order is when you have a group of n elements

and an element has order n

When this happens, your group is cyclic

And that element generates the group - all elements of the group can be written as a power of that element

[primal tusk]

okay that makes sense

[coral shale]

Ok, so back to permutations

You have a set of elements

(finite for introductory examples)

And a permutation is a bijection from that set to itself

In your questions, this set is {1, ..., 7}

===
So an example of a bijection is we fix all numbers except 6 and 7 which we will swap

Now this corresponds to (6 7) in cycle notation

Cycle notation is meant to be a shorthand to express these permutations

f = (a_1 a_2 ... a_n) is meant to mean
f(a_1) = a_2
f(a_2) = a_3
...
f(a_{n-1}) = a_n
f(a_n) = a_1

lmk if that notation as I defined it makes sense first

[primal tusk]

yea that makes a lot of sense

[coral shale]

ok, and then when we talk about products of cycles

we really mean function conposition

f = (1 2 3)
g = (3 4 5)

f o g = (1 2 3)(3 4 5)

"do g first, then do f"

[summer path]

Some textbooks do this the opposite way, so might be a good idea to consult your textbook about that

[coral shale]

So in this particular example
g maps 5 to 3.
f maps 3 to 1
So f o g = fg maps 5 to 1

[coral shale]

we burn those

[coral shale]

So in this way, that is how we compute products of cycles

[primal tusk]

that makes sense my book actually does it this way luckily lol

[coral shale]

Ok, so going back to my question before
(1 2 3)(1 2 3)

You should be able to compute this

You start by writing
(1 ...)

And think where 1 goes

[primal tusk]

so if im getting this right

1 goes to 2 which goes to 3

so 1 maps to 3 ?

[coral shale]

yep

so (1 3 ...)

then you continue by asking where 3 goes to

[primal tusk]

(1 3 2)

[coral shale]

yh so 3 goes to 2

and the important thing is to check where 2 goes to

which is 1

and this completes the cycle

(1 2 3)(1 2 3) = (1 3 2)

Then using this, you can compute
(1 2 3)(1 2 3)(1 2 3) = (1 3 2)(1 2 3)

(remember functions are associative, so order doesnt matter)

[coral shale]

You will find that this is the identity permutation

So (1 2 3)^3 = e

sometimes written ()

And thats the order of this cycle

If you think about it carefully (consider where 1 maps to), the general result is
(1 2 ... n)^n = e

You should be able to prove this

And you should also be able to show
(1 2 ... n)^k is not e for 1 =< k < n

===
These 2 results show the order of any cycle is equal to its length

And that finally lets you compute the order of f in the original question (or f^6)

[primal tusk]

so just to make sure, the order of the elements doesnt matter so (1 3 2) has order 3?

[coral shale]

yes

the numbers are just labels

[primal tusk]

and what about if there are multiple cycles?

[coral shale]

from above - you can have permutations on any set

[primal tusk]

disjoint cycles*