or alternatively it is the set $G/K \times X / \sim$, where $(xk,y) \sim (x,ky)$ for all $k \in K$.
#groups-rings-fields
sturdy marsh
cloud walrusBOT
Brofibration:
sturdy marsh
and define the G action by left multplication
brittle prism
can someone explain this notation $\mu_1, \mu_2 : {1,2,3,4} \rightarrow {1,2,3,4}$
cloud walrusBOT
Boom2o:
chilly ocean
probably mu1 and mu2 are each functions from {1,2,3,4} to {1,2,3,4}
brittle prism
Oh thanks
upbeat juniper
or alternatively it is the set $G/K \times X / \sim$, where $(xk,y) \sim (x,ky)$ for all $k \in K$.
@sturdy marsh I'm not really familiar with this haha
cloud walrusBOT
wy:
upbeat juniper
just to be sure it's supposed to be $(G/K \times X)/\sim$ right
cloud walrusBOT
wy:
sturdy marsh
yup
chilly ocean
Thanks for the suggestions, but I think I've solved it
chilly ocean
Can anyone recommend a text where I can read about Dedekind's domains that are UFDs?
wind steeple
coarse stag
What’s the name for the group like structure that’s uncountable again?
next obsidian
wat
a group can be uncountable
the word "group like structure" thinks maybe you mean a... groupoid...?
paper flint
I need to prove that every group is a Latin square. From the few examples of Cayley tables I've worked with, this seems to be true but I don't know why. Thus, I'd like to know why it is true, and what a reasonable proof should demonstrate.
chilly ocean
@paper flint im not that good but is this what youre talking about?
oh wait the last one is c not a
cbc
paper flint
Uh something similar but the proposition says each element appears only once in each row and each column
Kinda like a sudoku
chilly ocean
Oh wait
I think this is the right table
since a is the identity
Idk looks funky. if b * a = b and b * b = b doesnt that mean that and be are the same?
next obsidian
what's a latin square?
is that like
each row and each column has every element
uniquely
sudoku?
next obsidian
each column / row is just the image of the multiplication by a fixed element map
take the square above
the first row shows the image of multiplication by a
the first column is also the image of multiplication by a
(on different sides maybe, but both are bijections)
the second row is multiplication by b
third row is multiplication by c
they're bijective since the inverse is given by mult. by the inverse
next obsidian
Np
I assume the converse is not true
that any latin square gives a multiplication table for a group
chilly ocean
Sorry just real quick. Is this the correct formula for inverse element?
a*a^(-1)=e
that's the definition of the inverse element of a
slow osprey
paper flint
There are right and left inverses too?!?
chilly ocean
I only know of left and right cancellation using inverses
next obsidian
There are right and left inverses too?!?
@paper flint yeah, but for groups they coincide
consider functions
paper flint
Ah, I see.
next obsidian
I think that umm
associativity and presence of a left and right inverse is enough to say they coincide?
like if you have those two and consider a, a_l a left inverse and a_r a right inverse
you can calculate
a_laa_r = (a_la)a_r = ea_r = a_r
or a_laa_r = a_l(aa_r) = a_le = a_l
paper flint
Can you give an example of some structure in which the left and right inverses don't coincide?
next obsidian
No fucking clue
I just proved associativity guarantees they do
so you'd have to come up with a pretty shitty structure to not have them line up since it can't be associative
Like
Okay
I can arbitrarily come up with one lmfao by just writing a stupid ass multiplication chart on 3 elements
err
4 elements I guess
but it won't really mean anything since without associativity I can basically just define the product of two elements to be whatever the hell Iw ant
I guess the really weird thing is
you need the presence of an identity without being associative
and the non-associative stuff I can think of don't have an identity
chilly ocean
non-associative operations 
next obsidian
i.e. stuff satisfying jacobi identity like the cross product
paper flint
Oh, I see.
I'm stuck with another problem: Show that in a finite group G, the number of elements x such that x^3=e is odd, and the number of elements x such that x^2!=e is even.
next obsidian
so like
e^3 = 3 right?
so that gives one
so consider x not equal to e such that x^3 = e
paper flint
Makes sense.
next obsidian
then x^2 is not equal to x^3
but also (x^2)^3 = e
So they come in pairs
so that the number of non-unit elements satisfying that is even
then even + odd = odd
next obsidian
For the other one, idk maybe try the same idea
next obsidian
I would also try this problem
Show that if the order of G is odd, then the number of elements of order 2 is even
and if the order of G is even, the number of elements of order 2 is odd
this tells you that in any even-order group that there is at least one element with order 2
The proof is kinda similar
paper flint
I see, thanks! I'll try this problem too.
steep hull
||Every element of order not dividing 2 pairs up with its inverse (of the same order)||
maiden ocean
I think i'm being dumb but I'm kind of unclear what exactly this part of the question is asking me to do
cloud walrusBOT
Moth:
oblique river
what are your a_i?
maiden ocean
elements in F_2
so 0 or 1
I'm pretty sure the other polynomial this is supposed to be is g(x) = x^3 + x^2 + 1 right
idk i feel like im just confused because i dont understand why this linear combination would be zero in F_2[x]/(x^3 + x + 1)
oblique river
which is true -- beta^3 = alpha^2, beta^2 = alpha^2 + 1, and 1 = 1
maiden ocean
beta^3 = alpha^3 + 3alpha^2 + 3alpha + 1 right
before reducing
so just alpha^3 + 1
oblique river
:^)
oblique river
haha np
maiden ocean
next obsidian
F_2 is so fucked dude
woven delta
In contrast, 2^X is logic
maiden ocean
part 2 of the earlier question
the first part is easy
for the 2nd part i think you can map p(alpha) + (f) -> p(beta) + (g) but it sounds like a massive pain to verify that its a homomorphism
and stuff
is there a nicer way to do this lol
oblique river
@maiden ocean does that map work? if p(x) = x^3 + x + 1 then you'd have 0 --> p(beta) + (g), but p(beta) isn't 0
I think you basically want to identify what alpha is in F_2[beta]/g(beta) and send alpha to that
since beta = alpha + 1, alpha = beta - 1
so I would send alpha to beta - 1, or if you write it in terms of x, x --> x-1
maiden ocean
yeah but i was too lazy to specify all that when i was writng that quick msg
its like
ok i did this problem basically bc it was optional on the last pset
but verifying that its a homomorphism is af ucking nightmare
in that its rly ugly
so i want an elegant way
oblique river
Is it? The map F[x] --> F[x] --> F[x]/g(x) where the first map is x --> x-1 and the second map is quotient by g(x) is a homomorphism (as it's a composition of two homomorphisms). Note that: f(x) is in the kernel of this composition, and no poly of smaller degree is in the kernel (since the kernel is the set of all p s.t. p(x-1) is a multiple of g) and therefore the ideal is generated by f(x) since F[x] is a PID.
now apply first isomorphism theorem
maiden ocean
i dont think he ever showed in class that that first map is a homomorphism
is the thing
what if i just assert that it is because its obvious
oblique river
the thing about F[x] is that for any F-algebra R and any r in R, the map F[x] --> R given by x --> r is a homomorphism
and the proof is "just define the map so that p(x) goes to p(r)"
(just a handy fact to know about F[x])
maiden ocean
i dont remember if he ever stated that in class but honestly its probably fine
i feel like i can just say that the first map is a homomorphism and the TA will be like "ok"
oblique river
yeah
if you want to be precise about it, the proof just comes down to the definition of polynomial addition
(f+g)(x) = f(x) + g(x), so (f+g)(x-1) = f(x-1) + g(x-1)
i.e. if phi is our map F[x] --> F[x], phi(f+g) = (f+g)(x-1) = f(x-1) + g(x-1) = phi(f) + phi(g)
same for products -- (fg)(x) = f(x)g(x)
mint gulch
the first part is easy
@maiden ocean
Both have the same basis as a F_2 vector spaces, maybe map the two basis?
maiden ocean
i got it but ty
rain vigil
I've tried looking at the eigenvectors of A and could find a sort of general form to it. I also found that scaling each row i of the vector with x_i would give an eigenvector of the eigenvalue k+1, where k is the current eigenvalue. But i don't know how to proceed
sturdy marsh
Hmm, on each row the sum of the non-diagonal entries is equal to the diagonal entry. There should be a neat way to do row/column operations on the matrix to make it upper/lower triangular.
rain vigil
I'm not sure there is (at least from what I've tried), as when it gets to larger values of n, the matrix A becomes more complicated
cloud walrusBOT
Lartomato:
sour plume
i retract the question 'cus it's prolly too much to ask
hot lake
that sounds very unlikely
unless I'm mistaken the dimension of the space of invariant 3 forms is much larger in general than the dimension of the space of invariant 2 forms
golden pasture
@rain vigil Try the eigenvectors $v_i$ whose $k$th component is the $i$th symmetric polynomial in $x_1,\dots,x_n$ ommiting $x_k$
cloud walrusBOT
ariana:
golden pasture
*left-eigenvectors
as for how to actually find this uh
stare at sage output for a few minutes
rain vigil
i've been staring at it for hours ahaha
golden pasture
must train eye power
rain vigil
but yea i'll try left eigenvectors since my friend is also close to a solution using left eigenvectors
sour plume
@hot lake probably! i know for sl2(C) they are actually the same size, but it's probably too much to hope that that generalizes
jade crag
@golden pasture do you have any idea or intuition on creating a formula for the right eigenvectors?
jade crag
i managed to get that for the jth element of the eigenvector with eigenvalue = lambda and size = n,
$$\frac{x_j^\lambda}{x_1^\lambda}\prod_{k\neq 1,k\neq j}^n \frac{x_1-x_k}{x_j-x_k}
$$
cloud walrusBOT
yimong:
jade crag
<@&286206848099549185>
carmine fossil
Both parts?
snow cliff
I think I have the solution for the first part
So I proved that G_b = g G_a g^{-1} from b = ga
By showing that they are both within each other
I'm just very confused on the deduction
And I'm not certain I have a good grasp of what this stuff means, it feels like symbols
carmine fossil
Transitive here means every element can be written as g.a for some g
And kernel is intersection of stabilizers of all elements
snow cliff
does the first mean that a_1 = ga, a_2 =ga etc?
snow cliff
I see
so then is my argument something along these lines:
By definition we know that the kernel is the intersection of all stabilizers of all elements
We showed in the first part that for a, b in A that G_b = gG_ag^{-1}
Therefore, we for each a, b in A the stabilizer of a in G can be represented as G_b = gG_ag^{-1}
So since the intersection of all stabilizers, it must contain the intersection of all gG_ag^{-1}?
carmine fossil
Ye,But the kernel being intersection is not by definition
You have to prove it separately
carmine fossil
so then is my argument something along these lines:
By definition we know that the kernel is the intersection of all stabilizers of all elements
We showed in the first part that for a, b in A that G_b = gG_ag^{-1}
Therefore, we for each a, b in A the stabilizer of a in G can be represented as G_b = gG_ag^{-1}
So since the intersection of all stabilizers, it must contain the intersection of all gG_ag^{-1}?
*Equal to intersection
snow cliff
I have a question about this. For all stabilizers, we are only referring to the stab of each a in A?
carmine fossil
Yes
You take a element a and write down its stabilizers
You take a second element B and write down its stabilizers and repeat this process
And you take intersection of all stabilizers obtained
snow cliff
is the fact that G acts transitively on A useful here? Or is this proof completely separate from the previous part of the problem?
carmine fossil
Separate
You could also prove the other direction too
Kernel of action will always be in intersection of stabilizers
snow cliff
Would that be kinda the reverse of what I just did?
Except I do not have to mention the subgroups/identity part?
carmine fossil
Kernel fixes all elements so, any element of kernel also belongs to any stabilizer. So any element of kernel belongs to intersection
snow cliff
Yea that would've been simplier
That makes sense
I have to thank you a lot
This course has gotten way more complicated after my midterm
It's been a battle to even understand the questions at this point
So I greatly appreciate it
snow cliff
A few questions:
I do not quite understand what a permutation group is, or at least how it is different from a symmetric group. Is a permutation group just a singular bijection from the set to itself, while a symmetric group is the set of all bijections from the set to itself?
If a permutation group is acting on a set A, does that mean the map sigma (A -> A defined by a -> ga) is an orbit?
so sigma(a) = ga?
This works because sigma(a) is in A since sigma maps A back to A.
golden pasture
permutation group = symmetric group
snow cliff
Is wikipedia wrong?
mint gulch
permutation group = symmetric group
@golden pasture
I thought rhis
snow cliff
I'm reading it says that a permutation group is a subgroup of a symmetric group
glossy yoke
as per wikipedia, a permutation group is a group consisting of permutations of a set. the group of all permutations of the set is a group called the symmetric group. emphasis on the word "all" there.
words... you know? WORDS.
WWWWOOOOOOOOOOORRRRRRRRRRRRRDDDDDDDDDS
snow cliff
so a permutation group would only have some of the permutations of a symmetric group then?
glossy yoke
sure. maybe all.
snow cliff
Edit: so some, but not necessarily all bijections of a symmetric group
glossy yoke
if the permutation group acts on A, then the orbit of a \in A is the subset of A consisting of elements of the form ga for g in the group.
mint gulch
Yes, then, it's a subgroup
glossy yoke
so you basically push a around using your group.
snow cliff
since sigma(a) is in A then that would be correct?
if the mapping of sigma did not map back to A then it would not be true?
glossy yoke
what's sigma? it's an element of the group?
glossy yoke
so since the permutation group acts on the set A, you can view each element of the group as a permutation of A.
snow cliff
i don't understand how sigma could be part of a group, that seems confusing to me, since isn't it a mapping?
or should I in that context think of sigma as like g?
glossy yoke
sure. each $g \in G$ gets associated with $\sigma_g$
cloud walrusBOT
ball:
glossy yoke
sometimes people will suppress the $\sigma$ altogether.
cloud walrusBOT
ball:
snow cliff
what do you mean by suppress?
glossy yoke
they don't write it.
so, for instance, you might write $\sigma_g(a)$. But you might write $g(a)$ or maybe $g \cdot a$
cloud walrusBOT
ball:
snow cliff
would you ever write $\sigma \cdot a$?
cloud walrusBOT
camolot457656:
snow cliff
because of the provided definition of g in G?
Also, why is my latex so huge compared to yours?
glossy yoke
think of the group action as $\sigma: G \times A \to A$ satisfying certain conditions. Denote $\sigma(g,a) = \sigma_g(a)$
cloud walrusBOT
ball:
glossy yoke
if you fix a $g \in G$, then you get yourself a permutation of A.
cloud walrusBOT
ball:
glossy yoke
and that permutation is called $\sigma_g$
cloud walrusBOT
ball:
snow cliff
I think that makes sense. Visually speaking, can I think of this as a being moved to the left by g? I understand that may not necessarily be the case since it depends on g, but is this idea of motion correct? Kinda like a clock?
glossy yoke
well, if we're thinking of group $G$ acting on a set $A$ in general, then not every $g$ does something non-trivial to every $a \in A$.
cloud walrusBOT
ball:
glossy yoke
sometimes, a $g$ can map $a$ to itself.
cloud walrusBOT
ball:
snow cliff
would that only be in the case of the identity?
glossy yoke
no.
snow cliff
powers of a too?
glossy yoke
a doesn't have powers. it lives in a boring set.
snow cliff
I should clarify
By powers of a I meant hitting a with itself by the amount of some multiple of its order?
Or is that the same as the identity?
glossy yoke
hitting a with itself doesn't make sense. a lives in a boring set. not a group.
glossy yoke
yeah. it's fun and useful and important to say the wrong things though so someone can yell at you til it makes sense. for reals. then you won't do it in front of your beloved professors and they'll wonder if you're some kind of savant for not spewing utter nonsense to them like every other fu.. nevermind.
obsidian path
hey if anyone's awake
snow cliff
?
obsidian path
5f
I'm struggling to express D4 as a subgroup of S4
can anyone assist me in conceptualizing how to transform its elements?
snow cliff
D4 only has 2 vertices correct?
obsidian path
wait.. it does?
snow cliff
let me double check
I think so
D_4 is a dihedral group
D_2n = D_4 => n = 2 vertices?
which seems kinda wrong...
thorn delta
Not everyone subscribes to the D_2n notation
glossy yoke
sometimes $D_n$ is used to denote the group with 2n elements which arises as the symmetries of the regular $n$-gon.
cloud walrusBOT
ball:
obsidian path
checking my textbook
snow cliff
but doesn't that only apply when n>= 3?
that's my sudden confusion
Unless this is not D_2n notation
thorn delta
its not. D_4 is referring to the symmetries of a square here
obsidian path
right
okay
so you should take D4, and express rotations/reflections as permutations instead right?
cloud walrusBOT
ball:
obsidian path
so how would you express a counterclosewise rotation.. for example?
glossy yoke
well... S_4 is the group of all permutations of 4 things, yes?
obsidian path
true
glossy yoke
i wonder what 4 things D_4 permutes.
glossy yoke
oh, that sounds right.
glossy yoke
the problem you are doing is assuming $D_4$ sits as a subgroup inside of $S_4$.
cloud walrusBOT
ball:
glossy yoke
so i guess you want to write down the 8 permutations of 4 elements that are in this subgroup?
obsidian path
right.. let me think about that a little more i guess
obsidian path
yeah I.. don't get it please enlighten me @glossy yoke
glossy yoke
label the vertices of the square 1,2,3,4
obsidian path
so would it be such that (say just for imagination's sake) 1 would be top left, 2 is top right, 3 bottom left, 4 bottom right?
glossy yoke
sure.
obsidian path
I understand that it doesn't actually matter, but I've been trying to image the rotations and then try and express them in permutation form
glossy yoke
i guess, i'd prefer a cyclic ordering.
obsidian path
how exactly?
glossy yoke
go around clockwise labelling vertices 1,2,3,4 in succession starting from top left.
obsidian path
gotcha
glossy yoke
then how does a 90 degree clockwise rotation permute the vertices?
obsidian path
(1432)?
glossy yoke
that feels counterclockwise to me.
but yeah, something like that.
i was thinking (1234)
glossy yoke
doesn't really matter. the idea matters.
obsidian path
I guess because earlier I was working with a rectangle it was confusing me for some reason
but a square makes things obvious
glossy yoke
anyway, do this for all the symmetries of the square.
glossy yoke
well, im tired. but maybe.
glossy yoke
gogogogogogo
obsidian path
e,(1234),(13)(24),(1432),(12)(34),(14)(32),(13)(24).
3 minutes.. hahah
got confused by the end
glossy yoke
you wrote (13)(24) twice.
obsidian path
yikes
glossy yoke
oh no you didnt
obsidian path
right ok good
glossy yoke
oh yeah, you did
carmine fossil
2nd and last
obsidian path
so then what it's only 6 elements?
carmine fossil
Anyway,do 4 rotations first and 4 reflected rotations next
glossy yoke
i think you just forgot a comma.
e, (1234), (13)(24), (1432), (12)(34), (14)(32), (13), (24)
civic linden
guys, how comes GL(n,C) has a subset of R^2n^2 , where does the 2 come from?
oh
multiply the same matrix and we get the real part
sharp sonnet
you can define C as a set of certain 2x2 real matrices
then a nxn complex matrix is really a 2nx2n real matrix
chilly ocean
what is a nice algebraic proof of the fact that y² + x - x³ is irreducible in R[x,y]
i have already done the brute force one
queen vine
Eisenstein's criterion over R[x][y] with prime element x (or 1-x,1+x if the characteristic is not 2)
(At least when R is an integral domain)
snow cliff
Hmmm, could I get a check on my logic for this problem?
My solution so far is based on the conclusion of the previous exercise, where the intersection of all stabilizers (so the kernel) was shown to be the identity:
Assume $\sigma(a) = a$. This implies that $\sigma \cdot a = a \cdot \sigma = a$ becuase G is abelian.
From this we can deduce that $\sigma$ is either $a^{-1}$ or the identity.
However, since G is transitive, there is only one such orbit that given any two elements $a, b \in A$ there is some $g \in G$ s.t. a = gb
Furthermore, since the kernel was shown to be the identity, the only g that solves our orbit is the identity. So, the only way for $\sigma(a) = a$ is if $\sigma$ is the identity.
Therefore, $\sigma(a) \neq a$ for all $\sigma \in G - {1}$ and all $a \in A$.
I haven't done the deduction part yet, but I wanted to make sure that this is correct, or at least I am on the right track.
cloud walrusBOT
camolot457656:
glossy yoke
A is a boring set. it does not have inverses.
here sigma is a group element. the group action associates to sigma a permutation of A.
snow cliff
So I have to show that $\sigma(a) \neq \sigma \cdot a$ when $\sigma \neq 1$?
cloud walrusBOT
camolot457656:
snow cliff
or rather that $\sigma(a) \neq a$ if $\sigma$ can't be 1?
cloud walrusBOT
camolot457656:
snow cliff
why is it illegal to discuss $\sigma \cdot a$ and $a \cdot \sigma$?
cloud walrusBOT
camolot457656:
carmine fossil
Well $\sigma$(a) means $\sigma$.a
cloud walrusBOT
DrunkenDrake:
carmine fossil
Right action?
snow cliff
Is that not possible to discuss since a may not be in G? So I misunderstood the commutative part here?
or rather a is in a separate set?
carmine fossil
Let $g_1$ be a stabilizer of a and $g_2$ be an arbitary element,then $ (g_1g_2)a=(g_2g_1)a \implies
g_1(g_2(a))=g_2(a)\implies g_1(x)=x \forall x \in A$(Because Transitive action)
cloud walrusBOT
DrunkenDrake:
snow cliff
Sorry
carmine fossil
Np
snow cliff
I'm trying to make sense of the spoiler
snow cliff
I think I understand what you are doing
I'm just confused as to how it relates. I'm confused as to how showing $\sigma(x) = x$ for all x in A is helpful? Is this because it implies that $\sigma(x)$ would also be in the kernel? Therefore a contradiction with the previous problem stating that the kernel only has the identity?
cloud walrusBOT
camolot457656:
carmine fossil
I am showing IF sigma(x)=x, sigma has to be Identity
cloud walrusBOT
camolot457656:
snow cliff
That makes sense
snow cliff
that makes perfect sense
bit blown away by how simple that really was
I have another question
When you say $g_1$ is the stab of a, does that mean $g_1$ is the only element such that $g_1 \cdot a = a$ or that $g_1 \in Stab(a)$
cloud walrusBOT
camolot457656:
snow cliff
I see, wish my textbook would clarify these things a bit more
carmine fossil
"The stabilizer" usually refers to Stab(a)
(forget the "a stabilizer part",i don't think anyone uses that. I don't know why I used that)
snow cliff
is it correct to say that $\sigma(a) \in G$?
cloud walrusBOT
camolot457656:
chilly ocean
question - this is the intro-level Abstract Algebra one-semester course for my uni
How much is this for one semester, aka should I be taking this alongside other work-intensive courses
further reference - homework set 4 of 5
next obsidian
Seems pretty reasonable
for an intro undergrad course it's maybe just a bit quick, but nothing unreasonable
As for if you should take it with other courses that are hard, it's kind of up to you since none of us know how much math you've done before, your work ethic, background, etc.
It could be reasonable to say you could take another heavy course if you're prepared to spend a lot of time doing hw and studying and stuff, but equally reasonable to say you could afford to sit back a bit ¯_(ツ)_/¯
olive mirage
that's a lot for a semester, but not an heinously unreasonable amount. The homework in particular looks very reasonable.
scarlet estuary
the homework is long but not particularly intense i'd say
especially if that's 4/5
olive mirage
well, it's 20% of the homework for the semester
scarlet estuary
fair haha
quaint ivy
bit tight but still reasonable for one course, just don't halfass it and you should be fine. You're actually skipping many "basic" things from group and ring theory (I'm thinking semidirect products, nilpotent groups, maybe some more profound facts about PIDs and UFDs) that presumably are reserved for later courses
oblique river
Chiming in late, but that syllabus is basically equivalent to what it was at my last institution
we were on the quarter system, so 10 weeks each
and this is basically "1 quarter of group theory plus half a quarter of ring theory"
although the ring theory does seem to ramp up pretty fast in your schedule
snow cliff
For the deduction of this problem. Is this valid?
For each $a \in A$ we have $|G : G_a| = 1$.
Since the orbits partition A, and each of them have order 1, we get $|A| = |O_a||G| = 1 |G| = |G|$. Thus, $|G| = |A|$.
cloud walrusBOT
camolot457656:
snow cliff
I feel nervous about this solution because I fear I might be doing the same mistake I was above, tangling sets and groups together
sturdy marsh
What is G_a?
if it is the stabilizer of a then the index of G_a in G is not 1 (unless A is a 1 element set)
as the action is transitive, [G: G_a] = |Orb(a)| = |A|
snow cliff
It is the stab(a)
snow cliff
Could you break this down a bit more for me? I am really struggling with the concept of orbits and index
sturdy marsh
do you know the orbit-stabilizer theorem?
you dont need to deal with any of this anyway, the solution is a lot more straightforward
You have a group G acting transitively on a set A. What is a lower bound on the order of the group?
snow cliff
I think 1? Just the identity?
sturdy marsh
that is a lower bound sure, as any group must have an element
but can you do better?
note that the trivial group cannot act transitively on a set with 2 elements
snow cliff
Is it |A| because I need a sigma for each element in a s.t. $\sigma(a) = \sigma \cdot a$?
cloud walrusBOT
camolot457656:
sturdy marsh
yup
so G has at least |A| elements
now if you have two elements of A, then their stabilizers are conjugate
as they are in the same orbit
but the group is abelian
so what can you say?
snow cliff
the stablizers are equal?
sturdy marsh
why is it 1?
We know that the stab at every point is the same
so the stab acts trivially on A
So G/Stab acts transitively on A
sturdy marsh
the G/Stab part?
snow cliff
yes
sturdy marsh
For any g in Stab, we have ga= a
for all a in A
So we now have an action of G/stab on A
where you define [g]a = ga
this is well defined as stab acts trivially
the action is still transitive
so we have |G/stab| = |A|
snow cliff
hmm, let me think a second
sturdy marsh
oh wait im confused now
snow cliff
I'm confused by what this is: G/stab
sturdy marsh
the quotient group
but nvm
it's simpler
G is a subgroup of S_A
we have a subgroup stab of S_A that fixes A
so stab = 1
any non-trivial element moves an elt of A lol
lmao
snow cliff
I'm not sure I follow
sturdy marsh
Okay let's go from the start
fix a in A
Consider stab(a)
for any b in A, stab(a) and stab (b) are conjugate, so equal (as G is abliean)
therefore stab(a) fixes all of A
so stab(a) = 1
done
snow cliff
that part makes sense
Don't follow the last bit
why is the stab(a) = 1 significant?
sturdy marsh
what is the problem asking you to prove?
snow cliff
|G| = |A|
sturdy marsh
well that's the second part
snow cliff
I see the disconnect
I got the first part earlier from drunkendrake
He helped me out on that
sturdy marsh
oh im sorry lol
sturdy marsh
the second part is just orbit-stabilizer
|A| = |Orb(a)| = [G: Stab(a)] = |G|/|Stab(a)| = |G|/1 = |G|
as stab(a) = 1
snow cliff
So is the orbit stabilizer thm what allow us to say that |Orb(a)| = [G: Stab(a)]
sturdy marsh
yup
snow cliff
Let me think about the index for a sec, I still struggle with that conncept
sturdy marsh
the idea is, Orb(a) = {ga| g in G}
snow cliff
okay yes that makes sense
sturdy marsh
aight cool
snow cliff
I do have one concern
Does it matter if G is infinite or finite?
when we do the division by 1?
sturdy marsh
I was assuming everything was finite all this while lol
snow cliff
ah
sturdy marsh
is S_A defined to be all permutations or ones with finite support?
eh it's probably all
snow cliff
I actually am not sure, this is what my book says
sturdy marsh
yeah I think it works always, G acts transitively and with no stabilizers on A. Fix an element a in A. Define a bijection g --> ga
injective as the stabilizer is trivial
surjective as the action is transitive
yeah there's no problem even if they are infinite
snow cliff
Ty for the help on that one
chilly ocean
Probably addition for R, and multiplication for gl
cinder bone
hmmm ok
yeah that works
is it enough to simply observe that for any x=y f(x) must equal f(y) for (b)
thorn delta
in GL(2, R) the operation is matrix multiplication. In R, its addition
cinder bone
👍
thorn delta
is it enough to simply observe that for any x=y f(x) must equal f(y) for (b)
no, you would have to show that f(x) = f(y) implies that x = y
thorn delta
ah yea thats fine too
cinder bone
x!=y -> f(x) != f(y)
but just an observe matrix (1 0 a 1) != (1 0 b 1) for any a!=b
seems too simple
thorn delta
It is simple. That line of reasoning is correct though
cinder bone
ok cool
for (c) it doesnt look surjective
there are plenty elements of GL(2,R) that don't have (1 0) in their top row
thorn delta
yep
cinder bone
so i guess i can just find an example of one that doesnt
and show there does not exist blah blah
thorn delta
yea, thats the idea
thorn delta
late reply, but yes, that is correct
snow cliff
what distinguishes orbits from each other?
I guess, perhaps a better question, is how can an acting group have multiple orbits?
Is one orbit defined by the action on one element?
mint gulch
Yes, $G \cdot x := {gx : g \in G}$
cloud walrusBOT
Enigsis:
mint gulch
The relation in the set $A :$ $x \sim y ,\iff \exists g \in G : x = g\cdot y$
You can make a partition of the set A with that relation, the elements of the partition are the orbits
cloud walrusBOT
Enigsis:
snow cliff
Okay, then as an example. If $S_3$ acts on the ordered pair (1, 1) such that $\sigma((1, 1) = (\sigma(1), \sigma(1))$ then would I simply be apply the following mappings to 1 for each of the ordered pairs:
()
(1 2)
(2 3)
...
(1 3 2)
So the orbit of (1, 1) in this case would contain 6 elements?
cloud walrusBOT
camolot457656:
snow cliff
Which I suppose will only generate (1, 1), (2, 2), and (3, 3) now that I think about it...
snow cliff
$\sigma((i,j)) = (\sigma(i),\sigma(j))$ where I believe $\sigma(i)$ is defined by a permutation of $S_3$
cloud walrusBOT
camolot457656:
snow cliff
The full set that $S_3$ acts on is defined as ${(i, j) | 1 \leq i,j \leq 3}$
cloud walrusBOT
camolot457656:
thorn delta
the full set that S3 acts on is defined as {(i,j) : 1 <= i, j <= 3}
I think you mean {1, 2, 3} x {1, 2, 3}, right? Not {(i,j) : 1 <= i, j <= 3}
The orbit of (1,1) would be {(1,1), (2,2), (3,3)}, yes.
snow cliff
I suppose {1, 2, 3} x {1, 2, 3} would be the more formal way of writing it. But my book actually gave it to me in that for loop like notation.
thorn delta
well, that just means you have elements like (4, 1), (2012, 3) and stuff
cloud walrusBOT
camolot457656:
thorn delta
ohh i didn't read it like that at first lol
anyway, if ur just looking for a simple example of orbits, let H be a subgroup of G and let H act on G by left translation. Then the orbits are just the left cosets of H in G.
snow cliff
Would that be invalid for right translation and the right cosets of H in G?
thorn delta
right translation would work the same way. right translation gives you right cosets of H in G
snow cliff
but the orbit part would not work?
thorn delta
why not?
snow cliff
because when right translating you no longer have the form gx?
it's now xg?
Or is that irrelevant because x is in G?
thorn delta
oh wait, eh it might have to be a right action
it would have to be a right action for it to work.
but same concept
this is because you need (ab).x = a(b.x) for the axioms of a left group action, and if you define a.x = xa, then this no longer holds: (ab).x = a(b.x)
other examples like this are g.x = gxg^{-1} which defines a left action of G on itself and x.g = g^{-1}xg which defines a right action of G on itself
slate forum
Anyone have any pointers for the first question here? I honestly have no idea how these problems work, I know that this corresponds to two planes in C^4 that don't intersect along with a line that doesn't pass through either of them, but I don't see what to do
scarlet estuary
trivial cycle
fickle brook
any string is a cyclic perm of itself
scarlet estuary
"apparently" according to whom?
scarlet estuary
i mean its possible the text defines it differently
and i'd go with the text's definition over wikipedia's, even if the latter is more common
carmine fossil
I don't think d and f defined cyclic permutations
chilly ocean
ok, what is the identity?
pure flume
Umm idk
For part i, I can say that closure does exist
But how do I show associativity
thorn delta
Well, you have (A * B) * C = (A u B) u C. What can you do from here?
thorn delta
empty set != phi
but i think you get the idea
pure flume
Yeah
thorn delta
part iii of that question is a doozy. I had to do it for homework once
namely, proving associativity of the symmetric difference
thorn delta
the notation for the empty set is not phi. $\emptyset = \varnothing \neq \phi$
cloud walrusBOT
kxrider:
thorn delta
that's all
chilly ocean
looks okay, although there is a much faster way to show that.
paper flint
I see. What is it?
chilly ocean
saying its obvious, just draw a triangle and see for yourself.
paper flint
Yeah I mean I drew a square to convince myself first lmao, but for a formal argument I needed to be more general.
chilly ocean
lmao what are you going to prove formally next? 1+1=2? lmao homeboy thats just easy induction by verifying the base case
paper flint
Induction...I didn't think about that.
chilly ocean
no I mean induction by just checking the base case, then saying its trivial.
sharp sonnet
lol
paper flint
I have to prove that the set {1,2,3,...,(n-1)} is a group under multiplication modulo n iff n is a prime.
I'm currently working on the backward implication, and have proven closure, associativity and existence of identity. However, I cannot find a systematic way to assert the existence of inverses and would like a hint for that.
carmine fossil
Do you know bezout's identity?
If a,b are coprime integers x,y exist such that ax+by=1
paper flint
Argh that Bezout's Lemma again...I should've learnt that.
Thanks, I'll take a look at it and figure it out!
carmine fossil
(You can show it with euclid's algorithm)
paper flint
Gotcha
carmine fossil
Now take p,i such that,i is coprime to p
pa+bi=1 implies bi=1(mod p)
paper flint
So for some a in {1,2,...(n-1)}, by Bezout's Lemma I have ax+ny=1, and taking mod n on both sides I get ax mod n=1, which is the identity(However, how do I prove that x is in {1,2,...,(n-1)}?)
carmine fossil
If x is not in that,just subtract an appropriate multiple of n from it
paper flint
Owww makes sense, thanks!
I figured out an alternate way to 'prove' the above proposition: For n not prime, {1,2,...,(n-1)} contains all the prime factors of n. The product of these primes factors mod n is 0 and hence not in {1,2,...,(n-1)}, which means it is not a group. (I'm not sure if this obeys the 'iff' structure of proof)
Is this correct?
carmine fossil
Sorry,not that
It's because the product is not closed
Therefore it cannot be a group
paper flint
Oh, okay, didn't pay attention haha.
civic linden
this question looks like it links to something in quadratic reciprocity
sharp sonnet
you also have an element without inverse though
paper flint
I'm not sure, I've just started with group theory, although the author mentions this result will be used in a later chapter.
carmine fossil
Yea,but that is not obvious from this
paper flint
you also have an element without inverse though
@sharp sonnet I can't think of one?
sharp sonnet
quadratic reciprocity needs a lot more machinery
carmine fossil
@sharp sonnet I can't think of one?
@paper flint take an element not coprime to n
paper flint
Hmm, suppose n=4, then I get 2 as such an element...ah gotcha
civic linden
hint: an inverse exists mod n iff there exists b,c: ab+cn=1 iff coprime of a and n is 1
paper flint
Makes sense, and it isn't possible for any choice of b,c if a is not coprime to n.
Thanks everyone!
carmine fossil
I'm not sure, I've just started with group theory, although the author mentions this result will be used in a later chapter.
What were you doing for the last 2 weeks or so?
paper flint
Shitposting here 😓
slate forum
@prime gale can you help with my AG problem?
cinder bone
R^* is a group under division or multiplication right
im guessing i should use multiplication?
chilly ocean
division is literally multiplication
cinder bone
okie
chilly ocean
division is not associative
chilly ocean
do you know what you're supposed to show?
cinder bone
that the orders are the same
i guess since its injective, f(a)=f(b) -> a=b
so maybe there's something there with the orders but idk
chilly ocean
write out f(a)^k for k >= ord(a)
i have a feeling you just need to unravel definitions here
more explicitly
if n is the smallest positive integer with a^n = e (so n is the order of a), you want to show that n is the smallest positive integer with f(a)^n = e (so n is the order of f(a))
that's just writing down the definitions of the things involved (which should be what you do if you don't know where to start)
cinder bone
oh shit, i can just show f(a^n) = (f(a))^n right
so then really I just need to show f^-1(eH) = eG??
is that on the right track
since f(a^n) = eH
chilly ocean
right, but f(a^n) = e_H alone doesn't show that n is the smallest positive integer with f(a^n) = e_H
you now know it's a positive integer satisfying that, so now you want to show it's the smallest one
chilly ocean
that's just a general property of homomorphisms
cinder bone
oh ok. cool
chilly ocean
f(e_G) * e_H = f(e_G * e_G) = f(e_G) * f(e_G), now cancel f(e_G)
cinder bone
yeah i have that in my notes as a property, i forgot.
chilly ocean
hint: so far you haven't used injectivity once in your proof, so you should get a feeling that the next step (showing n is the smallest positive integer with f(a)^n = e_H) will use it in some way
cinder bone
i guess with injectivity i can show that some a_1 is the only a for which f(a^n) = e_H
does that show that a_1 is the only a such that (a^n) = e_G?
chilly ocean
what is a_1? and how does that show n is the smallest positive integer with f(a)^n = e_H?
sorry for the late response, rereading a particularly messy problem set before i hand it in
one way to show n is that integer is ||to show that no positive integer m with f(a)^m = e_H can be smaller than n||
cinder bone
no worries finishing up another problem before i go back to it
oh, i let the order of f(a) be n
should i do it the other way around?
in my proof i was trying to prove n is the smallest positive integer with a^n = e_G
chilly ocean
you need to show the order of f(a) is n
ok well i guess that can work too
use injectivity somehow
cinder bone
😭
okay I’m gonna study for my LSAT some and come back to it
thanks for the advice it’s been helpful
cinder bone
I got it!!!!!
chilly ocean
cinder bone
I didn’t use injectivity though!!
😭
Might’ve done something wrong
Oh well will check it tomorrow
chilly ocean
this definitely fails if you omit injectivity (e.g. the trivial homomorphism)
b careful
solemn rain
u didnt show the order of f(a) is order of a
you showed order of f(a) is atleast or equal order of a
@cinder bone now show that this is the least possible
green locust
@cinder bone are you a pre-law math major?
green locust
I think math & philosophy is probably the best prelaw combo, good choices 👍
cinder bone
yes in another channel another time
solemn rain
haha okay
mint gulch
I think math & philosophy is probably the best prelaw combo, good choices 👍
@green locust
Yes, Oxford offer that
green locust
That’s pretty neat, I didn’t know that
thorn delta
anyone have a hint on how i would show that any finite group is isomorphic to a subgroup of the alternating group A_n for some n?
we have by cayley's theorem that any finite group of order n is isomorphic to a subgroup of S_n, so ig you would need to make "n" a little bigger than the order of the group to get an isomorphism to a subgroup of A_n, but idk im not even sure how to start
glossy yoke
perhaps you can shove an S_n into a S_BIG in such a way that it's entirely in A_BIG
i wonder if there is a way of "doubling" a permutation.
thorn delta
h m m
glossy yoke
S_n permutes {1,2,3,4...., n , 1', 2', 3', .... n'} maybe?
thorn delta
huh, yea i think that would work. The parity of every permutation would be doubled, giving u an embedding into A_{2n}
thanks!
vestal snow
Can someone explain why, after (3.8), $v_{P'}(a_n)=0$ and $v_{P'}(a_i) > 0$ implies $v_{P'}(y) > 0$?
cloud walrusBOT
The formerly edible banana:
vestal snow
I think it might be using the strict triangle inequality, but I'm not sure
chilly ocean
what book is this out of curiousity?
thorn delta
Doesn't the top statement follow immediately from proposition 4.8?
or am i being dumb?
oh wait the kernel of the homomorphism might not have finite index in G hmm
glossy yoke
it is though.
thorn delta
it is though.
wdym? [G : ker] is finite?
glossy yoke
i think so.
thorn delta
ig it should, but it doesn't follow immediately. If I can prove [H : ker] is finite, then I'm done because [G : ker] = [G : H][H : ker]
glossy yoke
A(S)... what can we say about it?
thorn delta
If [G : H] = n, then A(S) is isomorphic to Sn
glossy yoke
nice.
thorn delta
oh yea, duh, [G : ker] = |Im| < n!
glossy yoke
nice.
thorn delta
thank
paper flint
For elements $a,b$ in a group $G$, find $x$ such that $$xabx^{-1}=ba$$
cloud walrusBOT
TedNowKaczynski:
paper flint
Had G been Abelian it would've been true for all x, but as it stands I'm getting a bit confused.
glossy yoke
what have you tried for x?
paper flint
Nothing much besides multiplying both sides once by x, once by x^(-1) so that some cancellation shows up. That way I got xab=bax but I don't know how to proceed.
(or alternatively abx^(-1)=x^(-1)ba)
carmine fossil
Are you aware of centralizers?
paper flint
Yes...aah okay, but for which element would x be a centraliser? I have ab on one side and ba on another?
carmine fossil
Ok,I mean elements x such that xax^-1=a
glossy yoke
no need to be fancy. heh.
mint gulch
Take x=kb
@carmine fossil
Can you always do that? that b divides x
mint gulch
Ah, yes hahaha
So, you get that, if xa = xa, then, you find the k, for every x, but, are they the only ones?
carmine fossil
Should be
Because if there are elements k not of that form(x(ba) not (ba)x) but kb satisfies,you can just substitute k here to show they satisfy,giving you a contradiction
glossy yoke
gotta say, i don't understand why centralizers are coming into the picture.
in particular, i don't see that k(ba)k^-1 = ba.
carmine fossil
ba=(kb)(ab)(b^-1 k^-1)=(kba)(b b^-1)(k^-1)= k(ba) k^-1
Implies ba=k(ba) k^-1
So,k=xb^-1 should satisfy this given condition
carmine fossil
Associativity?
glossy yoke
ba=(kb)(ab)(b^-1 k^-1)
carmine fossil
Because inverse of (kb)(b^-1 k^-1)=e implying b^-1 k^-1 is the inverse of kb
So (kb)(ab)(kb)^-1 reduces to (kb)(ab)(b^-1a^-1)
glossy yoke
and why is that ba?
carmine fossil
By associativity (kb)(ab)(b^-1)(k^-1)=k(ba)(b b^-1) (k^-1)
glossy yoke
and why is that ba?
carmine fossil
Because x ab x^-1 =ba for some x
glossy yoke
but that's what we're trying to prove.
carmine fossil
(Do you want me to write down a complete proof?)
glossy yoke
no. im concerned you're assuming the conclusion.
carmine fossil
You are concerned that I assume such a x exists?
glossy yoke
the question was to find such an x.
carmine fossil
1)Let x exist such that it satisfies our condition,by algebraic manipulations we conclude (xb^-1)(ba)(bx^-1)=ba,i.e,xb^-1 has to satisfy the ka=ak condition
Implying if x is a solution, (xb^-1)(ba)=(ba)(xb^-1)
2)Let x be such that (xb^-1)(ba)=(ba)(xb^-1)
xab=bax which implies
xabx^-1=ba
Implying x satisfies our condition
x satifies the condition iff (xb^-1)(ba)=(ba)(xb^-1)
We know atleast one such x exists (x=b)
knotty mason
it's not too hard in a finite group
you want a'b'xab = x, so start with identity and keep wrapping your element in a'b' ab
you should be able to find a cycle eventually
i don't know how to prove it in general though
I tried looking at commutators
didn't really help
oh wait it's actually really easy
||just put x = a'||
paper flint
Damn, didn't see that coming. Thanks!!!
Here's another "proof" of mine I'd like to get verified. I was wondering if I should consider another element c besides a which is in H_1\cap H_2, and then show that ac, ca and their multiples/inverses are also in H_1\cap H_2. Is that necessary?
carmine fossil
Ok, Here's a shorter formulation,let a and b be 2 arbitary elements of H intersection K. Then a and b are elements of H,so ab^-1 is in H, similarly a and b are elements of K ,so ab^-1 is in K. Therefore ab^-1 is in H intersection K
paper flint
Oh, okay, makes sense. It seems you did it more concisely but our ideas concur, right?
carmine fossil
Yes
paper flint
Great, I'll use your argument while writing down the proof though 
carmine fossil
Have you proved the subgroup criteria?
paper flint
No, not yet, this just popped up in Rudenko's lecture I'm watching, I paused the video and proved this.
I'll be doing the chapter on subgroups from Gallian once I'm done watching the lecture.
paper flint
Is this proof okay?
carmine fossil
Sure
paper flint
paper flint
Is this proof okay?
Guess I could've been more precise about which elements were being mapped and that the domain of phi' is equal to the range of phi and so on.
hot lake
your usage of => is very dubious
south storm
You have to show that it‘s also a bijection G_1 to G_3 I think, but forgetting about that seems fine.
paper flint
I assumed composition of bijective functions is a bijection(but now that you tell me about it I should've mentioned it explicitly).
your usage of => is very dubious
@hot lake Which ones in particular seem dubious?
I'm still new to proof writing so I just use implications in succession without giving much thought to it.
south storm
Yes also someone mentioned when you use => it doesn’t really imply it because only from knowing for example that φ’(x)=φ’(yz) it doesn’t imply that φ‘(x)=φ’(y)φ‘(z). You also need to know that φ’(xy)=φ’(x)φ‘(y).
paper flint
Just show it's a homomorphism
@carmine fossil How does that help?
south storm
Also I think it would be better if you‘d use different signs for all the operations instead of just writing xy
paper flint
Uh I mentioned previously that phi' is an isomorphism, isn't that sufficient?
south storm
Not for the use of =>
south storm
Yes
paper flint
Alright, noted. Thanks everyone!
south storm
There is one tho, that‘s not how one’d apply =>
south storm
Because it doesn’t imply it, yes but => doesn’t take previous information into account
No it doesn’t
Look example
Let phi‘(x)=3
Then phi‘(yz)=phi‘(x)
But that doesn’t imply that phi‘(y)phi‘(z)=phi‘(x) which is easy to see since it‘s not true
Yes
I know that with the information of Phi‘ being a homomorphism
It does imply it
But that information isn’t supplied in the statement
No
phi‘(yz)=phi‘(x) doesn‘t say that phi‘ is a homomorphism
But the => doesn’t take that into account
No it‘s just wrong
There is a definition of =>
And you have to apply it onto
That‘s the issue
Yes that‘s basically what a proof is, applying a bunch of logic on statements.
And if you want to write => you have to use it correctly
No problem, understandable
slate forum
Can someone please help me with my AG problem, I have no idea how any of this works
south storm
AG? @slate forum
slate forum
Algebraic geometry is really an algebraic kind of class, uses a lot of commutative algebra
south storm
It’s still not pure abstract algebra imo, as for example you need a lot of algebra in analysis, but you still don’t ask an complex analysis question in #precalculus
slate forum
People have asked about ag here before, but ok I guess, let's see what happens
sturdy marsh
@slate forum it's probably not too bad after you start looking at the affine cones
L1 is a 2d- subspace in C4
slate forum
So, I should be turning it into a linear algebra problem?
That was my first instinct
sturdy marsh
yeah and the planes intersect at the origin
slate forum
oh... alright
slate forum
ooooh....
sturdy marsh
same for span(L2, P)
slate forum
That's a neat way to think about it
sturdy marsh
now intersect them and count dimension
slate forum
The P's are in common....
slate forum
They give us the whole space?
sturdy marsh
yup
slate forum
If that's the case, then the intersection just straight up is one dimensional
err wait
sturdy marsh
so we have 4 = dim span L1,P + dim span L2, P - dim intersection
slate forum
2 dimensional....
slate forum
but that's really a line
ohhh
Dude, this problem has been making me hate myself for like days
I hate projective geometry
This is only the first problem too
sturdy marsh
ahahaha projective geometry is a lot nicer than affine
slate forum
So you say...
sturdy marsh
yeah pretty much every AG theorem that we were taught has a projective/proper assumption
so if the thing isnt projective i wouldnt know what to do
slate forum
oh shit
Ok, so, main deal, these are really two planes and a line through the origin in C4
sturdy marsh
yup
sturdy marsh
yup