For the mini course
#groups-rings-fields
mint gulch
Hmm, that makes sense
Probably that's why people think I'm a woman
Hahaha
But not
Enigma and paraphrasis
I like those words
@chilly ocean xD
solemn rain
a problem i saw and looks mad fun but mad hard
and i just need a hint
Prove that for every finite group G the number of group homomorphisms f:Z^2 -->G is n|G| where n is the number of conjugacy classes in G
carmine fossil
Why is it not just |G|^2? Because to define f ,you need f(1,0) and f(0,1) . You can choose f(1,0) in |G| ways and f(0,1) in |G| ways
solemn rain
idk
wind steeple
Bc f(1,0)f(0,1) = f(1,1) = f(0,1)f(1,0)
sharp sonnet
yeah, so you choose the image of wlog (1, 0), then the image of (0, 1) is in the centralizer of that, so you count up those and thats it?
solemn rain
odlk
so each function can be mapped
to an element not in the centralizer?
@sharp sonnet
sharp sonnet
the map is uniquely determined by the images of (1,0) and (0,1)
once you map either, you have to map the other to an element in the centralizer of the first due to what zak said
so you just have to compute $\sum_{g \in G}\lvert C_G(g)\rvert$
cloud walrusBOT
Lochverstärker:
sharp sonnet
use orbit stabilizer to compute the sum
queen vine
Hint: $(\alpha\alpha'\beta)^2 = (\alpha(\alpha'\beta))^2$.
cloud walrusBOT
leoli1:
zealous mortar
The product of all monic irreducible polynomials in a finite field $\mathbb{F}_q$ is $(x^q-1)$ or finite field $\mathbb{F}_p^m$ is like $(x^{p^m} -1)$.. yes? or is it something else?
cloud walrusBOT
jm:
sturdy marsh
I am a bit confused on what you're trying to say, there are monic irreducible polynomials over F_q of arbirtrarily high (in fact every) degree
Every element of F_{q^n} satisfies x^{q^n} - x, and a definition of F_{q^n} is the splitting field of x^{q^n} - x over F_q
So the minimal polynomial of every element in F_{q^n} over F_q divides x^{q^n} - x if that's what you're trying to say
zealous mortar
Maybe that's what I'm trying to say. I antrying to get the product $\prod{i\in \mathbb{F}_{2^4}, i\neq 0, 15}(x-\alpha^i)$
cloud walrusBOT
jm:
zealous mortar
$$\prod{i\in \mathbb{F}_{2^4}, i\neq 0, 15}(x-\alpha^i)$$
cloud walrusBOT
jm:
cloud walrusBOT
jm:
zealous mortar
I think it's x^{2^4} - x.
sturdy marsh
Well the product of all the $x - \alpha$ has degree $2^4$, and they are all coprime, and they all divide $x^{2^4} -x$
zealous mortar
not 24, 2^4.. or 16.
cloud walrusBOT
Brofibration:
zealous mortar
Right. Ok.
Then, if get rid of the x-0 term we have the $2^4-1$ and then, can we get rid of $(x-\alpha^{15})$. I'm trying to figure out where that term went. is $(x-\alpha^{15}) = (x-1)$ if $\alpha$ is a root of $x^4 + x^3 +1$. Then i can factor $(x-1)$ to get the product of all these min.pols.
cloud walrusBOT
jm:
zealous mortar
Thanks.
sturdy marsh
I am super confused on what you're trying to do, we always have $(x- \alpha^{15}) = (x-1)$
cloud walrusBOT
Brofibration:
sturdy marsh
Im assuming $\alpha \in F_{2^4}$
cloud walrusBOT
Brofibration:
zealous mortar
ya. I'm trying to figure out BCH codes.
Which means I need the product of all the polynomials $(x-\alpha^i)$ for $1 \leq i \leq 14$, which I am concluding is $x^{14} + x^{13} + \cdots + x + 1$.
cloud walrusBOT
jm:
zealous mortar
Where $\alpha$ is a root of $x^4 + x^3 + 1$, the generator.
cloud walrusBOT
jm:
cloud walrusBOT
jm:
sturdy marsh
well if x^4 + x^3 + 1 is irreducible then it is
yeah it is irreduducible
it has no quadratic factors
and no roots
so $F_2[x]/(x^4 + x^3 + 1) \simeq F_{2^4}$
cloud walrusBOT
Brofibration:
sturdy marsh
for dimension reasons
which means that the root will generate the multiplicative group
or does it hmm
yeah it should, as x^4 + x^3 + 1 probably does not divide x^7 - 1
nvm im being stupid the root cannot live in a smaller field extension
yeah it should be a primitive
zealous mortar
.Thank you so much for all your help.
stoic dirge
Hey would someone be willing to help me out with some homework for my abstract algebra course? Not looking for full solutions, just want to be pointed in the right direction
Let $\phi : G \rightarrow H$ be a homomorphism of groups with $K = ker(\phi)$. Suppose $m \leq G$. Prove $\phi^{-1}(\phi(M)) = MK$
stoic dirge
What's tripping me up is the $\phi^{-1}(\phi(M))$ because intuitively I would just think that's M but there must be some reason the question is written in this way
stoic dirge
Yeah I just intuitively assumed that would be M
cloud walrusBOT
slimvesus:
stoic dirge
Ah ok so it also includes the elements of G that map to an element in $\phi(M)$ but aren't also in $M$?
cloud walrusBOT
Dunceular:
chilly ocean
Hi, I was wondering if anybody could help me on this homework problem. I'm guessing we have to use First Isomorphism Theorem, but not sure how to construct a homomorphism going from a space of hom's to a cartesian product.
mint gulch
I think you have to send the function with f(1) = r1, f(2) = r2 and f(3) = r3 to (r1, r2, r3)
Prove that is a homomorphism
chilly ocean
I see
vestal snow
Let K be a field and a be an algebraic element such that min_K (a) is not a separable polynomial. Is it true that the field extension K(a) will be purely inseparable?
sturdy marsh
I think it's not true
There's some example with the rational function in 2 variables over F_2 iirc
but im blanking on it
vestal snow
Hmm okay
Can someone explain how this definition of perfect is equivalent to the definition that K is perfect iff every algebraic extension of K is separable?
proving every alg. extension is seperable => there are no non-trivial purely inseparable alg. extensions is easy
I can't get the reverse implication though
Suppose that there are no non-trivial purely inseparable extensions. Let F be an algebraic extension with a in F being inseparable over K.
If I could somehow show that K(a) is purely inseparable over K then we would be done
But as Brofibration said, that does not seem to be true
Hmm
I don't see how exactly I would prove that
Let F be an inseparable extension w/ a an inseparable element
The only way I can think of is to show that K(a) must be purely separable
which is what I was trying earlier
I should say that Brofibration might be wrong about it so it is worth thinking of a proof
because I don't really see any other way of proving this
sturdy marsh
Try proving it this way: No inseperable algebraic eextensions iff K = K^p iff no purely inseperable extension
By K = K^p I mean you have all pth roots
char K = p
vestal snow
Viburnum are you assuming that there is an inseparable algebraic extension?
and then trying to show that that gives us a non-trivial purely separable algebraic extension?
sturdy marsh
im pretty sure you can have an inseperable extension with no purely inseperable sub extension
inseparable*
Something Artin-schreier-ish
vestal snow
Let F be an inseparable extension over K and let L be the set of all separable elements over K
What do you think I should do next Viburnum?
Hmm okay. Maybe there's a different way to prove the equivalence
sturdy marsh
Im pretty sure the equivalence K = K^p iff no purely insep shouldnt be bad
vestal snow
Is K^p the set of everything raised to the power of p?
sturdy marsh
yup
If there is nor purely insep, then x^p - a must have a root
for any a in K
If K = K^p, then assume we have an extension L/K. Fix a in L. Assume we have a^p^n in K for some n. Assume n is minimal. Then (a^p^n-1)^p in K
so a^p^n-1 in K
contradiction
So that does K = K^p iff no purely insep extension
Now we need all finite extensions are seperable iff K = K^p
Assume all finite extensions are separable, then we must have a root for X^p -a for all a in K
I think the other direction might be a wee bit hard
Yeah we would just need to show that all irreducibles are separable.
An irred poly is separable iff deerivative is not zero
Assume we have an irred poly p
If the polynomial has a term of the form X^l where l is not a power of p, then the derivative is not zero
as l \neq 0 in K
Therefore assume the poly is of the form $X^{p^n} + a_{n-1}X^{p^{n-1}} + ... + a_0$
cloud walrusBOT
Brofibration:
sturdy marsh
But now as we have $K = K^p$, we can write thee polynomial as $X^{p^n} + a_{n-1}X^{p^{n-1}} + ... + a_0 = (X^n + b_{n-1}X^{n-1}+ ... + b_0)^p$
where the b_i are pth roots of a_i
cloud walrusBOT
Brofibration:
sturdy marsh
which means the poly was not irred
and we are done
ye i think that does it lol
phew
sturdy marsh
I guess
K = K^p is a pretty neat definition of perfect
i don't think the proof used anything fancy
sturdy marsh
That's what I just did
vestal snow
You showed all finite ext. are separable iff K^p = K
sturdy marsh
oh you mean infinite ones
vestal snow
Yeah
sturdy marsh
it doesnt matter, if you have an inseperable algebraic extension, then you have an inseperable finite extension
but just picking the field generated by some inseperable element
vestal snow
Wait so
why does all finite extensions being separable imply that x^p - a has a root for every a?
sturdy marsh
hmm
yeah
if it had no roots
adjoin root alpha
then alpha is all the roots
as x^p - a = (x-alpa)^p
so not seperable
contradiction
sturdy marsh
?
sturdy marsh
we dont care
all factors of x^p -a will only have alpha as roots
no linear factor as no root in K
so we have some irred poly with repeated roots
thee min poly of alpha divides x^p - a
the*
vestal snow
but it cannot be x-a as otherwise alpha is in K
sturdy marsh
no problem
sturdy marsh
We showed K = K^p iff no purely insep at the start
vestal snow
@sturdy marsh I think there might be a problem with no non-trivial purely inseparable extension => K^p = K
You say that given a in K, x^p - a must have root
I'm assuming that you used contradiction for this
Suppose there are no roots in K
Let alpha be a root
then the minimum polynomial of alpha must be of the form (x-alpha)^n where 1 < n =< p. Therefore alpha is not separable
I'm guessing that the contradiction you went for is that K(alpha) is then a purely inseparable extension
However, it is not obvious why if alpha is inseparable then K(alpha) is purely inseparable.
coarse stag
Not formally taking a class in this subject but I wanted to ask if X=G/N does that mean N x X=G?
hot lake
no
queen vine
We indeed get a bijection $N\times X\to G$, $(n,Nx)\mapsto nx$ (after selecting representatives for the cosets) of sets, but in general we don't get an isomorphism as groups. Take for example $G=\mathbb{Z},N=2\mathbb{Z}$, then $2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\not\cong\mathbb{Z}$.
cloud walrusBOT
leoli1:
coarse stag
Nvm
obsidian path
It says to compute the permutation compositions, I don't understand how you would compute the 100th exponent though, and also what change does absolute value make?
stone fulcrum
Obviously there's a better way to do that haha. Do you know what the order of an element is? Do you know what the order of (1254) would be?
prime gale
When you get a problem like this and you don't know what to do, just start multiplying. You might notice a pattern.
mint gulch
It says to compute the permutation compositions, I don't understand how you would compute the 100th exponent though, and also what change does absolute value make?
@obsidian path
I think you have to find the order of (1254) then applied 100 mod |(1254)|
obsidian path
Ah I get it now sorry I should've said something
but yeah you're right
since you're here may I ask for confirmation
is the answer (3254)?
When you get a problem like this and you don't know what to do, just start multiplying. You might notice a pattern.
the issue is that I was confused with the multiplication earlier
still kind of am
25 is cancelled out because the order brings it back to the identity
then you get 3 goes to 1, 1 goes to 2, 2 to 5 and 5 to 4
is that correct?
obsidian path
Oh ok.. true
what about this? 1 goes to 2, then 2 is stopped, then you get 2 to 5, 5 to 3 right? so (253)
mint gulch
Yes, ans 3 to 2
sturdy marsh
@vestal snow alpha is purely inseparable over K as alpha^p is in K. Every element in K(alpha)/K will also have this property. Any element is a K linear combo of 1, alpha, alpha^2, ... raise it to the pth power and you land back in K. So the extension is purely inseparable.
obsidian path
2
scarlet estuary
right; next, where does 2 get mapped to?
scarlet estuary
but (23567) disagrees since 1 gets mapped to 2
obsidian path
right.. so do you do (123567)?
scarlet estuary
no...
i mean okay listen
im confused what you mean by
"compute"
whats the end goal
obsidian path
that's fair
scarlet estuary
since this is already a product of disjoint cycles
which is what we usually look for
scarlet estuary
there's no way to write it as a single cycle, if that's what you're looking for
obsidian path
ok well
what if you were trying to find the composition of the two cycles?
is that any different?
or sorry, compute the composition rather
scarlet estuary
"product" and "composition" of cycles are the same thing
obsidian path
right
scarlet estuary
the resulting permutation cant be expressed with a single cycle
scarlet estuary
what do you think? do you have any guesses?
obsidian path
well.. I assume inverse just means you're flipping the pointers
so instead of 1 maps to 2, it should be 2 maps to 1
scarlet estuary
yeah, essentially
obsidian path
well.. that's why I was trying to compute the two first and then seeing if I can just do that instead
scarlet estuary
the inverse of a permutation is just that permutation backwards
now if these weren't disjoint permutations
you'd have to be careful
since $(ab)^{-1} = b^{-1}a^{-1}$
cloud walrusBOT
Namington:
scarlet estuary
so you'd have to make sure to place the two cycles in the correct order
but compositions of disjoint cycles commute so
thats not necessary here
obsidian path
right
scarlet estuary
i'll write it that way anyway though
obsidian path
so that made me think that it's actually harder
so clearly I don't understand lmao
scarlet estuary
$[(1\ 2\ 3\ 5)(4\ 6\ 7)]^{-1} = (4\ 6\ 7)^{-1}(1\ 2\ 3\ 5)^{-1}= (7\ 6\ 4)(5\ 3\ 2\ 1)$
cloud walrusBOT
Namington:
scarlet estuary
but we traditionally rewrite cycles to start with the lowest number so
$(4\ 7\ 6)(1\ 5\ 3\ 2)$
cloud walrusBOT
Namington:
scarlet estuary
and its easy enough to check that the composition of this with the original yields us the identity permutation ()
obsidian path
so then you're saying
if they weren't disjoint
we wouldn't be able to use that law?
$(ab)^{-1} = b^{-1}a^{-1}$
cloud walrusBOT
Deku:
obsidian path
ay
scarlet estuary
we'd be able to use that law
thats fine
its just that
becuase they're disjoint, that isnt NECESSARY
if we wrote $a^{-1}b^{-1}$ instead
cloud walrusBOT
Namington:
scarlet estuary
it'd still work
obsidian path
oooh I see
scarlet estuary
since disjoint cycle composition commutes
but if they werent disjoint, that isnt necessarily true
consider eg
(1 2)(1 2 3) vs (1 2 3)(1 2)
[it shouldnt be too hard to see why disjoint cycle composition commutes, though]
scarlet estuary
yeah thats the idea
if we have two cycles ab
if a affects some element x
then b doesnt affect x
by disjointness
(more properly, b maps x to itself)
and vice versa; if b affects x, a doesnt
obsidian path
oooh that makes more sense
scarlet estuary
so it doesnt matter what "order" in which we do these since
no matter what, they'll never "get in each other's way"
obsidian path
yes no worries hahah, thanks for checking on me :'))
obsidian path
oh hi
mint gulch
If I don't know that H is a group but I found that
f : G -> H is a homomorphism, H is a group?
slow osprey
Well wouldn't that depend on what kind of structure G is?
static temple
a function is a morphism in its own category
a function is automatically a morphism of sets
vestal snow
Let N_p denote the number of p-Sylow subgroups of a group G. Prove that if G is simple then |G| divides N_p! for all primes p in the factorization of |G|.
Isn't G=Z/pZ a counterexample to this?
leaden finch
can someone help a sistah out
stone fulcrum
Go for it!
chilly ocean
I guess that is true, but subfield implies subring
mint gulch
It says there "Subfield"
You have to prove that is close under sustraction and division
$a,b \in \mathbb{Q[\sqrt{2}]}\implies a - b, a/b \in \mathbb{Q[\sqrt{2}]}$
stone fulcrum
Basically go through the axioms that define what a field is, and ensure that your set has them
There do exist quick tests. Be sure you know what they are if you are going to use them
cloud walrusBOT
Enigsis:
stone fulcrum
It is inherited from G
cinder bone
I can show closure by manipulating what it means to be in K but
associativity is the one thats confusing me
stone fulcrum
G has an associative binary operation. K uses it as well
cinder bone
I guess since (alphabeta)^2 = (betalpha)^2?
cinder bone
ah gotcha
stone fulcrum
Just because it operates on less elements doesn't mean it stops being associative
cinder bone
so in a proof that K is a subgroup i could just observe it is associative
yeah ive shown its closed and not empty
mint gulch
That's it, associativity property follows from be a subset of G
slow osprey
"division" that's inacurate
You have to prove closure and inverse elements (which can be done in a single thing)
mint gulch
Identity elemets follows from take a/a
Division isn't inaccurate
ab^{-1}
You can think of a divisin
a/b
cinder bone
this is what my teacher gave me for the existence of an inverse
stone fulcrum
Agree that using / as a symbol and calling it division is an odd choice in group theory
But it does work
cinder bone
but it seems to skirt any mention of an identity
stone fulcrum
Prove identity yourself! Take G's identity. Is it in K?
slow osprey
Well if a is in the set and a^-1 is in the set and its closed then a*a^-1 = Id is in the set
cinder bone
actually im having trouble with the inverse part too
I get that inverse of any Beta is also a member of G
but that seems kind of redundant because to prove the existence of an inverse i have to be working with the same element of G the entire time
any suggestions?
thorn delta
you don't have to work with the same element of G the entire time, but you need to end back up with beta in (beta a inverse)^2 at the end @cinder bone
cinder bone
yeah
thorn delta
are you stuck somewhere?
cinder bone
i guess the problem is getting from [(beta-inverse * a)^2]-inverse to [(a * beta-inverse )^2]-inverse
it looks like i need a commutative property
like i need to switch the beta inverse and a on this
thorn delta
yep. check your assumption. For alpha, we have (alpha beta)^2 = (beta alpha)^2 for any beta in G
thorn delta
np
slow osprey
$$f: R \to S$$
cloud walrusBOT
Tobii:
abstract rock
damn beat me to it
leaden finch
ty ty
chilly ocean
Okay, so an isomorphism would be map R to that set of matrices, what do u think this map should be
leaden finch
oo i got it now loll
someone know how i can move this == sign below the matrix
Consider $f ( a + b) = f(a) + f(b)$\\
Let
\begin{equation*}
f(a + b ) =
\begin{pmatrix}
0 & 0\\
0 & a +b
\end{pmatrix}```ampersand
don't know if the equation environment supports this or if you explicitly need align
leaden finch
oh
slow osprey
,, \begin{equation*} a &= abc\ test &= lol\ \end{equation*}
leaden finch
lol
slow osprey
Okay either I'm way too tired or the bot is trolling me
cloud walrusBOT
Tobii:
\begin{align*} a &= abc\\ test &= lol \end{align*}
if you just want a bad looking line break \\ will do the thing on it's own
leaden finch
oo
brittle prism
and i need help with this. i've already gotten the identity element which seems to be 0, but then i get $a^{ln(a^{-1})}=0$
cloud walrusBOT
Williamg:
slow osprey
It can't be zero as 0 isn't part of G
brittle prism
frick, i dumn. that means i must have made a mistake
slow osprey
well you want that $\forall a\in G: ea = ae = a$
cloud walrusBOT
Tobii:
brittle prism
the identity element a * e = a^ln(e)=a
slow osprey
so if you write that with the definition it should be apparent what the identity is :'D
brittle prism
which means ln(e)=1
slow osprey
And yes I chose e because it was our convention for the identity and not because it fits here
slow osprey
it's eulers number as:
$ae= a^{ln(e)} = a^1 = a$ and $ ea = e^{ln(a)} =a$
cloud walrusBOT
Tobii:
brittle prism
yea
thanks
is G an abelian group?
oh, ich sehe erst jetzt, dass du deutsch bist
slow osprey
It's easier to show when you remember the identity $a^b = e^{b*ln(a)}$
cloud walrusBOT
Tobii:
slow osprey
I have no formula for the inverse, but you can use the properties of the log function that such a value must exist in G
glossy yoke
solve $a^{\ln(b)} = e$ for, say, $b$
cloud walrusBOT
ball:
brittle prism
I've solved the main question. But I wonder what to say to the question "What can you say if S is infinite?"?
carmine fossil
Idts
Because aren't you using finiteness of S to say there is an element e such that e*a=a for an a?
slow osprey
Nah that has nothing todo with finiteness
I can tell you that 1*x in the real numbers is also x
and R is infinite
carmine fossil
How do you know there is such an element 1,which does that?
Without explicitly defining R to have such a 1.(Or constructing it to have 1)
I was using the fact that the function f:S->S f(g)=g.a is an injection and also a surjection because of same cardinality
slow osprey
That leads down a whole rabbit hole about maths was "invented" and such I guess
am I reading that correctly as g times a?
carmine fossil
Yes,(assume we are fixing a for the rest)
Since that's a bijection,there should be some e such that e.a=a and left inverse b such that b.a=e
Similarly e' and b' such that a.e'=a and a.b'=e'
With algebraic manipulations,we can show e=e' and b=b'
And ultimately show a.e=e.a=a for all a in S
dull shard
@slow osprey @carmine fossil is this a ring???
How do I formally prove that this is a ring????
dull shard
Example lol
carmine fossil
Say the axiom of additive inverse
dull shard
I don't know how this works for matrices
carmine fossil
It works the same as it does for anything else
dull shard
How do I formally show closure though?
carmine fossil
Also, what are the elements in your ring?
dull shard
$a,b \in Z$
carmine fossil
set of Upper triangular matrices?
cloud walrusBOT
nzrpi:
carmine fossil
Is entry in 1st row 2nd common always even ?
carmine fossil
No, The question is vague
carmine fossil
Nvm,Jt's a typo
Anyway for example take the axiom of additive identity
Take a matrix A=
[a 2b]
[0 a]
[0 0]
[0 0]
Is the additive identity
Check it by applying + on A and 0 matrix
carmine fossil
Manually compute the multiplicative inverse and say that doesn't lie in the set
dull shard
So counter example?
carmine fossil
Yes
dull shard
$\begin{pmatrix} 1 & 2\0& 1\end{pmatrix}$*x=
$\begin{pmatrix} 1 & 0\0& 1\end{pmatrix}$ where x =
$\begin{pmatrix} 1 & -2\0& 1\end{pmatrix}$
cloud walrusBOT
nzrpi:
dull shard
x isn't in the set, is it??? Of course not right?
coarse stag
Is the span of all nontrivial normal subgroups of G = G?
glossy yoke
what does span mean?
the answer is probably no whatever its definition -- consider a simple group.
coarse stag
Except for simple groups
carmine fossil
Try D6
coarse stag
Yeah lol
glossy yoke
so the span of H_1 and H_2 is like the subgroup generated by them?
glossy yoke
it's a good question i think. good for developing intuition around groups.
tight otter
I'm trying to show that, if $H_1$ and $H_2$ are subgroups of a group $G$, and $H_1\cup H_2=G$, then either $H_1=G$ or $H_2=G$. The furthest I've gotten is showing that if $g_1\in H_1$ and $g_2\in H_2$ then $g_1,g_2\in G$ and therefore $g_1g_2\in G$, which means that either $g_1g_2\in H_1$ (which infers $g_2\in H_1$) or $g_1g_2\in H_2$ (which infers $g_1\in H_2$). I'm not sure how to exactly develop this into showing that this means either $H_1$ or $H_2$ is equivalent to $G$. Any help?
cloud walrusBOT
bela:
mint gulch
I think you can use the same argument as Vector Spaces
tight otter
i asked my prof and he said to basically extend the logic and propose that $g_1$ is not in $H_2$ and $g_2$ is not in $H_1$, but then by showing that that cannot be true, then it follows that either $H_1$ and/or $H_2$ must contain all elements of $G$
cloud walrusBOT
bela:
chilly ocean
I think there was almost this problem in chill channel a few days ago
Er, NVM, it was a little more different than I remembered
mint gulch
And get the contradiction
If $g_1$ is not in $H_1$, then it have to belong to $H_2$ and if $g_2$ doesn't belong to $H_2$ it have ti belong to $H_1$, but, just do $g_1g_2$
cloud walrusBOT
Enigsis:
mint gulch
Just think of subspaces of a vector space, is the same
tight otter
i haven't learned anything about vector spaces yet
mint gulch
Oh, I see
chilly ocean
You can argue this via cardinalities pretty easily
mint gulch
Well, just do g_1g2
tight otter
yeah i just did g_1g_2
mint gulch
And find the contradicction
I'm trying to show that, if $H_1$ and $H_2$ are subgroups of a group $G$, and $H_1\cup H_2=G$, then either $H_1=G$ or $H_2=G$. The furthest I've gotten is showing that if $g_1\in H_1$ and $g_2\in H_2$ then $g_1,g_2\in G$ and therefore $g_1g_2\in G$, which means that either $g_1g_2\in H_1$ (which infers $g_2\in H_1$) or $g_1g_2\in H_2$ (which infers $g_1\in H_2$). I'm not sure how to exactly develop this into showing that this means either $H_1$ or $H_2$ is equivalent to $G$. Any help?
@tight otter
Use this
cloud walrusBOT
Enigsis:
mint gulch
You can argue this via cardinalities pretty easily
@chilly ocean
How?
chilly ocean
||if H1 is more than n/2 elems, then it must be G, so assume H1 and H2 both have cardinality n/2. But H1 and H2 intersect at the identity, so union is less than G, contradiction||
mint gulch
Actually, I think you can do the problem more general, if you suppose that $H_1 \cup H_2$ is a subgroup, then one is contanied in the other
cloud walrusBOT
Enigsis:
chilly ocean
Ah, I totally missed that, good point
mint gulch
I understand you, hahaha, I take a class in Group Theory and the focus is in Finite Groups
chilly ocean
Maybe instead have to argue by index
tight otter
ooh that's a good solution 88ddda for finite groups
not sure if i'd be allowed to use that technique in this specific homework though
really clever though
mint gulch
Maybe instead have to argue by index
@chilly ocean
Finite is more simple because every proper subset have less cardinality, but, at infinity proper subsets can have the same
@tight otter did you figure out the argument?
static temple
and its easier to prove isomorphisms
tight otter
yep enigsis, thanks 
mint gulch
👍
mint gulch
Which is your argument to show that those are the elements of Z(M)?
chilly ocean
Z(GL(n, C)) is indeed the set of non-zero scalar multiples of the n x n identity
what was your argument for the 2x2 and 3x3 cases?
and then i'm guessing you made clever choices of B to show that A had the desired form?
that same strategy should work in the general n x n case
the 3x3 case is on my algebra problem set right now lmao
i am pretty sure the answers will be very similar, the former the non-zero real multiples of the identity and the latter the non-zero complex multiples
chilly ocean
$\Omega \subset \bm{R}^{n+1}$
cloud walrusBOT
88ddda:
Compile Error! Click the 
reaction for details. (You may edit your message)
chilly ocean
maybe bm requires some package?
outer dust
Hey would this be a reasonable place to ask about representations and actions of SU(2)?
mint gulch
Yes
leaden finch
ty
leaden finch
can someone help me in ;atex
latex
my code came out weird
If $\Omega$ \subset $\mathbb{R}^{n + 1}$ is a smooth domain (i.e.,\partial $\omega$ is locally representable as the graph of a smooth function), ist boundary is well approximated by n-dimensional affine spaces. For $Q$ $\in$ ${R}^{n + 1}$ and $ r \textgreater 0,$ $B(r , Q)$, denotes the $(n +1)$ - dimensional ball of radius $r$ and center $Q$. The fact that \partial $\Omega$. The fact that \partial $\Omega$ can be well approximated by n-dimensional planes can be expressed as follows: for a point $Q$ $\in$ \partial $\Omega$ and a radius $ r$ $\textgreater 0$ sufficiently small there exists a number $\theta$( r, Q) $\textgreater 0$ such that the intersection of the boundary with the ball of the radius $r$ and center $Q$, \partial $\Omega$ $\cap$ $B (r, Q)$ is contained in a tubular neighborhood of width $ r\theta$ $( r, Q)$ about the tangent plane of \partial \Omega at $Q$, $T_{Q}$```Whoever:
about the tangent plane of $\partial \Omega$ at $Q$, $T_{Q}$
smoky cypress
Put $ around the math part please
leaden finch
how can i fix this
\par
The Hausdorff distance $D$ between two sets $A$ ,$B$ $\subset$ ${R}^{n + 1}$ is defined by:\
\begin{align*}
D[ A,B] &= max {sup {d(a,B): a \in A},
\end{align*}
cloud walrusBOT
SUNSHINE:
Compile Error! Click the reaction for details. (You may edit your message)
cloud walrusBOT
Whoever:
leaden finch
hmm i tried it this way but idk whats wrong
it comes out weird
The Hausdorff distance $D$ between two sets $A$ ,$B$ $\subset$ ${R}^{n + 1}$ is defined by:\\
\begin{align*}
D[A,B] &= \max{\sup {d(a,B): a \in A}}, sub \max{d(b, A): b \in B}.\\
\end{align*}
D[A,B]=\max{\sup {d(a,B): a \in A}}```SUNSHINE:
Compile Error! Click the reaction for details. (You may edit your message)
mint gulch
I think you should ask those questions in #discussion
leaden finch
how do you prove that something is a nonempty for a subring
leaden finch
for this one
we have to show its a subring first right ?
one of the coditions says 1. show thats nonempty
chilly ocean
that set looks super nonempty to me
nova plank
Yeah, literally choose any values for r and s and say that that element belongs to it, so it's nonempty.
leaden finch
oh like what values?
chilly ocean
your favorite
leaden finch
2 and 6?
chilly ocean
if those are your favorites
leaden finch
yes but we pick any number since its rational numbers
chilly ocean
well any pair of them would show that the set there is nonempty
there you have it, a proof that the set is nonempty
leaden finch
okay, how can i start my proof
chilly ocean
are you sure you posted the right problem?
leaden finch
yes, i only showed that its a subring
mint gulch
The problem says that you have to prove is a subfield
leaden finch
i just need to prove if its a subfield too
so we dont need to prove if its a subring first?
i think im confused
mint gulch
Yes, you can
You need to prove then, that inverses exists for every element and that's it
leaden finch
oh okay, can you check if my steps for subring is right?
mint gulch
It seems good to me
leaden finch
okie
leaden finch
to show that something is a subfield
it has satisfy addition, multiplication, multiplicative identity and additive inverse?
mint gulch
And multiplicative inverse
slow osprey
🤦♂️ I was confused what you meant by field for like 2 days, I just now realized it translates to the German "Körper" in a math context :-)
mint gulch
Yes
I think only in English is called field
By a bad translation (That was I read)
@leaden finch You could check too that if a,b in S then
a - b, ab^(-1) in S, that's a criterion
chilly ocean
@solemn rain yo homeboy where did you learn group stuff from?
smoky cypress
Books
leaden finch
can someone help me with this
creating a table
but my teacher said that theres a fast way to do it ?
hmm no
isnt Z2_2 : { 0, 1}
mm okay
so then
leaden finch
hmmm not sure
leaden finch
isnt like a * x = a?
thorn delta
you just need a triple (a,b,c) such that for any element (x,y,z) of your ring, (a,b,c)(x,y,z) = (x,y,z)(a,b,c) = (x,y,z)
where (a,b,c)(x,y,z) = (ax,by,cz)
@leaden finch
leaden finch
are we suppose to find the units ?
thorn delta
nope. but you know the multiplicative identities in Z2, Z3, and Z5
err, maybe you don't know. If you have Z3 = {0,1,2} for example, then 1 is the identity since 1x = x1 = x mod 3 for any x in Z3.
Does that make sense? Because by definition of equivalence mod 3, we have x = y mod 3 iff x - y is a multiple of 3. well, since 1x - x = 0, and 0 is a multiple of 3, we must have 1x = x mod 3
thorn delta
feel free to ask if you still have any questions, sunshine, but I'm just going to leave a quick question here too.
In the theorem, when they say "G is the direct product of a finite number of indecomposable subgroups," they are including the case when G is indecomposable, right?
thorn delta
alright, that makes sense. thanks
oblique river
np
supple marsh
Hello ! I couldn't find anything on the internet, so perhaps it's no known result... Is there some kind of formula for either 1) the characteristic polynomial, or 2) the spectral radius of a Vandermonde matrix ?
@chilly canyon Did you find an answer? I thought it was an interesting question but idk.
chilly canyon
Nope, nothing at all... Maybe I should open a MSE question, but I managed to solve my problem differently soooo :)
But yeah I'd also like some kind of answer, even though I doubt there's some "nice" formula
For cases where the coefficients are all positive, Perron Froebenius gives a tiny tiny bit of answer, but it's far from being efficient in practice
chilly ocean
Is there more context? I'm curious about what you were looking at
past temple
if a group G acts transitively on A
is it fair to assume that the stabilizer of any element a in A
is just the identity in G
<@&286206848099549185>
bleak abyss
So first off you're not supposed to be pinging helpers until 15 minutes go by
past temple
my bad i didnt know the rules
bleak abyss
That said, the answer is no, imagine you have an even bigger group G' and a quotient map G'->G
past temple
okay that makes sense
im stupid lmao
if A = {a}
then obviously G is transitive
but then the stabilizer of a is the entirety of G
static temple
1/3
past temple
wait
so does this mean that the identity of a transitive group action is not always unique?
im confused again
ah okay
woven delta
@chilly ocean s^2 = e
Expand (srs)^j and show that it is sr^js
What does e^X mean here?
Lol so you're talking about a particular type of group
If you're talking about the exponential then it seems like you are
glossy yoke
the idea is that a small neighborhood of identity is a generating set for G. the exponential map has a nice derivative at 0, and maps 0 to the identity in G, so the exponential map on lie algebra elements generates G.
but you might be interested in the G-product of two exponentials: exp(X)exp(Y).
So map this group element down to the lie algebra.
turns out this lie algebra element can be computed by the Campbell-Baker-Hausdorff formula.
and it also turns out the CBH formula involves only commutators in X and Y.
so the brackets somehow encode the group product.
@frigid umbra
carmine fossil
Can a finite abelian group be p divisible for some prime p?
carmine fossil
(for every element y,there is an element x such that y=px)
chilly ocean
i think this was an exercise in dummit and foote, my guess is no
carmine fossil
That is for divisible groups(well, because the crux step involved a multiple of order of the group n):nG=1, for our p divisible case,we can simply choose p such that p is not n,if n is prime)
Not p divisible ones
chilly ocean
i recall something about the map (in fact, homomorphism) x mapsto x^p (frobenius map)
carmine fossil
I think that works
mint gulch
(for every element y,there is an element x such that y=px)
@carmine fossil
pZ, works, isn't?
carmine fossil
No? Take y=p,x=1 is not in pZ
mint gulch
Ah, right
carmine fossil
Z/pZ works
mint gulch
I'm thinking then in a finite field
Yes
A finite integral domain
Z/pZ works with p and abelian
leaden finch
can someone help me with this one
what would be an example of this one
determine if a given subset is a subring
mint gulch
Which?
leaden finch
hmm not sure thats a question what my teacher told us
im not sure what example it would be
chilly ocean
thorn delta
{1} is a subset of Z but {1} is a not a subring. An example liek that?
leaden finch
hmm maybe
thorn delta
yea so in problem 1, to show S is a subring, you just have to show that S is nonempty, and closed under multiplication and subtraction
glossy yoke
there's only 6 elements.
pretty hard to go faster than just checking each one 🙂
you could use the fact that that thinger is a ringer for Z_6.
leaden finch
i wanted to do the table but my teacher told me that it will take longer
glossy yoke
like, a multiplication table?
that does seem like too much work. i was thinking you just go element by element and classify it.
leaden finch
oh can you show me loll
glossy yoke
so i think (1,0) is a zero divisor. what do you think?
leaden finch
hmm how do you know thats a zero divsior?
glossy yoke
can you find a non-zero element $(a,b)$ so that $(1,0) \cdot (a,b) = (0,0)$?
cloud walrusBOT
ball:
leaden finch
do we always pick ( 1,0)?
glossy yoke
nope.
we're thinking about 1/6 of your exercise.
however, i think if you understand (1,0), you'll understand a whole lot of others.
soft elm
it isnt? read it properly
scenic sage
can someone explain too me why 3 is an idempotent?
@leaden finch 3^2=9 = -3 (mod 12)
(or just 9)
thus it is not an idempotent
leaden finch
i meant *not idempotent lol
scenic sage
what
leaden finch
i got confused why 3 isnt idempotent
scenic sage
because 3*3 = 9
and 9 is not 3 mod 12
idempotent squared should be the same
prolly if you find out that idem potent roughly is translated as the same power it would make it intuitively more clear
scenic sage
wdym
nova plank
Go look up the definition of idempotent in your notes/book, and give the precise definition.
nova plank
So are you guessing the definition?
x is idempotent with respect to an operation * if x*x=x
chilly ocean
Does anyone have the time to check the output for my Python code? I have written a function which is supposed to generate the multiplication tables for the groups (Z/11Z)* and (Z/12Z)* but I'm not sure if the output is correct and I am not sure how I can check this:
slow osprey
Why is it posting 11 columns?
chilly ocean
That may be a small bug.
Ignore that column
If it's wrong, that is
I can easily cut it out if it's wrong
slow osprey
Z/11Z has 11 elements, but you seem to omit the 0 element, so that should be a 10*10 table?
otherwise the values look good while glancing over them
chilly ocean
Oh, I forgot that Z/11Z includes 0. I guess just remove that last column and add a column of zeroes to the beginning?
carmine fossil
(Z/11Z)* has 10 elements
slow osprey
11 is prime so the unit-group(is it called that in english) well the thing that only has the invertible elements has 10 elements
carmine fossil
If you include 0,it wouldn't have an inverse
chilly ocean
That's what I thought. I was thinking I just need to remove that last column where all the values are the same?
slow osprey
I'm not sure about the 2nd table though, but yes
chilly ocean
Isn't (Z/12Z)* = {1, 5, 7, 11}?
carmine fossil
Yes
chilly ocean
So all the elements should be elements in that set
Because it's a multiplicative Abelian group
slow osprey
(Yeah i just failed calculating eulers phi function, my result was 5 every time lol)
So it seems to be correct but for that last column
chilly ocean
Oh, that reminds me. I've seen Euler's phi function defined as phi(N) = |(Z/NZ)*| for abstract algebra, but I've also seen it in RSA crypto defined as (p-1) * (q-1). Are they both the same?
carmine fossil
Yes(assuming n=pq and p,q are primes)
chilly ocean
Thanks for clearing that up! 👍
slow osprey
Well that's a special case as if p is a prime then phi(p) = p - 1
and phi(p*q) = phi(p)*phi(q)
golden pasture
specifically in rsa we dont use p^2 cuz trivial to factor and pqrst... cuz security is size of largest prime factor so pq is best in size and hardness to factor
usually we also do LCM(p-1,q-1) cuz thats the order of elements in group of coprime integers mod pq
glossy yoke
@frigid umbra sup
vocal depot
So I've got this problem I'm working on. I'll latex both the problem and what I have so far just to make sure we're clear that I am not just fishing for an answer.
So first, the problem:
Let $G$ be a finite group of order $n$. Also, let $m\in\mathbb{Z}^{+}$ and $gcd(m,n) = 1$. Suppose that $a$ is sn element of $G$ satisfying $a^{m} = e$. Prove that $G$ is cyclic.
cloud walrusBOT
HisMajestytheSquid:
vocal depot
What I have:
Now since $a^{m} = e$ we have that $|a| = |\langle a\rangle| = m$. By corollary $2$ of Lagrange's theorem we have that $|\langle a \rangle|\vert |G|$. That is, $|G| = |\langle a\rangle|k$ for some $k\in\mathbb{Z}$.
cloud walrusBOT
HisMajestytheSquid:
vocal depot
@open torrent, Something like $mx + ny = 1$ for some $x,y\in\mathbb{Z}$?
cloud walrusBOT
HisMajestytheSquid:
glossy yoke
be careful: $a^m = e$ doesn't imply that the order of $a$ is $m$.
cloud walrusBOT
ball:
vocal depot
I guess it could be some multiple of the order then, correct?
glossy yoke
yeah
vocal depot
There is a corollary I have here that goes something along the lines that if $G$ is a finite group for an element $a\in G$, $a^{|G|} = e$. Is this something I can use to simplify my problem a little?
cloud walrusBOT
HisMajestytheSquid:
glossy yoke
seems doubtful to me.
glossy yoke
G is supposed to be cyclic. I wonder what the generator is.
oh, hm. actually, the fact that $a^{|G|} = e$ might come in handy!
cloud walrusBOT
ball:
vocal depot
Oh!
Okay then
The generator thing is something I considered but I wasn't sure how to go about using that. The corollary that I mentioned seems like if I can show that |G| is some multiple of m then I cam perhaps find a generator? Or show that a is a generator of G
Jfc
I wrote the problem down wrong
thorn delta
i was about to say...
vocal depot
Sorry for bothering you guys
thorn delta
this only happens when a = 1
glossy yoke
har.
thorn delta
proof. a^m = e implies that |a| divides m. But |a| divides n. Since gcd(m,n) = 1, |a| divides mn. So mn is at least |a|, which implies m = 1 <- doesn't follow
correct proof: ||since gcd(m,n) = 1, cm + dn = 1 for some integers c,d. Then a = a^1 = a^{cm + dn} = a^{cm}a^{dn} = e using ur corollary and assumptions||
vocal depot
Yeah, that last bit at the end is what I'm actually supposed to be proving.
vocal depot
You're all good. Thanks for the help anyway guys.
slate forum
How does the map being R-bi-additive give me an R hom
I thought it only gives me a Z hom
glossy yoke
doesn't it follow from the universal property? or are you saying that the universal property only gives you a Z-linear map?
slate forum
Yeah
This map isn't R bilinear
If I define $\psi: R \times M \rightarrow M$ by $\psi(r,m)=rm$, I understand that the map satisfies $\psi(rs,m)=(rs)m=r(sm)=\psi(r,sm)$, but I don't see why $s\psi(r,m)=\psi(rs,m)$
cloud walrusBOT
HelixKirby:
slate forum
Like, it is true if R is commutative, or even if s is in the center of R
glossy yoke
i got made fun of once for thinking about non-commutative rings. i haven't gotten over it yet.
slate forum
Honestly, I don't really care much for them, they just make everything less nice
compact needle
In some references (wikipedia), bilinearity is right-linear for the first argument and left-linear in the second.
slate forum
Apparently it just works..
$\phi(s(r\otimes m))=\phi(sr\otimes n)=\phi(h(sr,m))=\psi(sr,m)=(sr)m=s(rm)=s\psi(r,m)=s\phi(r\otimes m)$
cloud walrusBOT
HelixKirby:
slate forum
Even though the map $\psi: R \times M \rightarrow M$ is not $R$ bilinear...
cloud walrusBOT
HelixKirby:
compact needle
Yeah, are you sure the sided-ness of the action in each argument?
glossy yoke
perhaps they are emphasizing that $R$ is an $(R,R)$-bimodule to show that the resulting $\mathbb{Z}$-homomorphism from $R \otimes_R M \to M$ can be promoted to a morphism of $R$-modules?
cloud walrusBOT
ball:
glossy yoke
the right $R$-module structure on $R$ gets 'eaten' by the left $R$-module structure on $M.$
cloud walrusBOT
ball:
slate forum
Yeah, but the book said that we could only make R-bi-linear maps into R homs
I guess this is an exception...
Even though $\psi: R\times M \rightarrow M$ is not R bilinear, the tensor product one is
cloud walrusBOT
HelixKirby:
glossy yoke
yeah, im not sure this represents an exception.
slate forum
It seems like it, normally there's no guarantee that taking this R bi additive map and turning it into the map on the tensor product turns it into an R map
The only thing for sure you're guaranteed is an Abelian group hom
How bizarre
glossy yoke
i guess i mean it doesn't strike me as an exception to the universal property.
the universal property gives you a Z-hom. this can be promoted to a left R-hom.
by using the left-R structure on R which didn't get eaten.
slate forum
Yes, but you also supposedly had to use that the R bi additive map was also R bilinear
For the promotion
compact needle
I think for exactly the reason you outline above, the tensor product of a right R-module M and a left R-module N is itself an R-module only if R is commutative
compact needle
But the action doesn't associate correctly to be a left action (unless right actions and left actions are the same, ie R is commutative)
slate forum
Ugh, I wish I could just deal with commutative rings
compact needle
In order to be eaten "correctly" $r(m\otimes n) = (mr \otimes n)$
cloud walrusBOT
Turgul:
slate forum
Btw, what does a fortiori mean?
glossy yoke
i don't understand that. i would define r(m \otimes n) to be (rm \otimes n). is there something incorrect about this?
slate forum
I don't think it would necessarily be an R module
Or maybe it is
I don't know
I mean, what is rm?
That's not even defined
compact needle
Well, at the very least, you would lose the isomorphism $R \otimes N \cong N$
glossy yoke
M is an (R,R)-bimodule.
cloud walrusBOT
Turgul:
glossy yoke
sorry, we might be talking past eachother. fuckin non-commutative rings. har.
compact needle
For sure. It's telling that most people that take tensor products assume R is commutative (ie in Lang's Algebra)
slate forum
Like, it got upgraded, normally I'd get a Z hom, but in this case, I actually got an R hom
It's sufficient but not necessary I guess
An R bi additive function can sometimes lift to an R map
But for sure it will be a Z map
glossy yoke
i think the magic is in R.
because R is an (R,R)-bimodule.
in general, if M is an (S,R)-bimodule, and N is a left R-module, then M \otimes_R N is a left S-module.
slate forum
I see, why don't we just allow the tensor product to be of two left modules and just say $r(m\otimes n)=rm\otimes n = m \otimes rn$
cloud walrusBOT
HelixKirby:
slate forum
Does something go wrong?
compact needle
If you look at the link I posted, it forces $(rs)(m \otimes n) = (sr)(m \otimes n)$
cloud walrusBOT
Turgul:
slate forum
Like, why do you have to shift from left to right
No, I made it so that doesn't happen
It's not the same as a tensor product
glossy yoke
you want some kind of associativity in your ring action though.
compact needle
$(rs)(m \otimes n) = r(m \otimes sn) = (rm \otimes sn) = s(rm \otimes n) = (sr)(m \otimes n)$
cloud walrusBOT
Turgul:
slate forum
Oops, didn't mean to delete
$rs(m\otimes n)= rsm\otimes
n= sm \otimes rn = m\otimes rsn$
cloud walrusBOT
HelixKirby:
compact needle
But from $sm \otimes rn$ you are allowed to pull the r or the s out of the tensor product in either order
cloud walrusBOT
Turgul:
compact needle
So, you can, but this definition kills off a lot of things you might not want dead
But we're stuck with something that's not an R-module
Moral of the story, be careful around tensor products :p
Anyway, cheers
slate forum
Yeah, I know you should never try to define a function out of a tensor product
vestal snow
How can I show that S_n acting on Z[x_1,...,x_n] is a group action?
say $ \sigma , \tau \in S_n$ and $f(x_1, \cdots , x_n) \in \mathbb{Z}[x_1, \cdots , x_n]$
cloud walrusBOT
The formerly edible banana:
vestal snow
applying $\tau$ first gives us $f(x_{\tau(1)}, \cdots , x_{\tau(n)})$
cloud walrusBOT
The formerly edible banana:
vestal snow
Say $p(x_1,\cdots ,x_n) = f(x_{\tau(1)}, \cdots , x_{\tau(n)})$
cloud walrusBOT
The formerly edible banana:
vestal snow
How do I show that applying $\sigma$ on $p(x_1,\cdots ,x_n)$ gives me $f(x_{\sigma \tau(1)}, \cdots , x_{\sigma \tau(n)})$
cloud walrusBOT
The formerly edible banana:
nova plank
Lol, when you type a whole essay (i know I'm exaggerating, leave me alone) and answer your own question
vestal snow
Haha yeah
I realized the action literally was "apply sigma to every indeterminate's index"
brittle prism
Is the answer c?
brittle prism
Thanks
chilly ocean
Hey, I had this question about the number of orbits of a group when acting on a set and the size of the group itself. In particular, whenever you have G < Sn and an n element set, it seems like the more orbits you have, the less permutations there are. This certianly seems true whenever the number of orbits is n (because they are disjoint etc etc so G must only have the identity), but I cannot seem to find any good upper bound on the order of G dependant on the number of orbits.
I did calculate examples for all subgroups of S3 acting on {1,2,3} and got (# of orbits)/3 which just can't be a general thing.
Moreover, I figured a necessary lower bound must be (# of orbits), but I won't write that argument out.
glossy yoke
I don't really understand what you're saying. the result i know of for counting orbits is the Burnside Lemma. you might give that a look.
snow flint
i alr saw that @vital quail
sturdy marsh
@chilly ocean this probably doesnt answer your question but in general for a group G, subgroups H,K of G and a K-set X we have $(X \uparrow^G_K)\downarrow^G_H = \bigcup_{x \in [H \backslash G / K]}(^x(X\downarrow^K_{H^x\cap K}))\uparrow^H_{H \cap ^xK}$.
cloud walrusBOT
Brofibration:
sturdy marsh
The usual action of S_n on the usual n element set is the induced G-set from the action of S_n-1 on a singleton
oh wait does this do it
the restriction $(X\downarrow^K{H^x\cap K}) = *$ as X is a singleton
cloud walrusBOT
Brofibration:
sturdy marsh
then after you conjugate and induce then the H-set you get is $H/H \cap ^xK$
cloud walrusBOT
Brofibration:
sturdy marsh
which is transitive
so the number of orbits is in bijection with the number of double cosets H\S_n/S_n-1
Im considering Sn-1 to be the subgroup that fixes n btw
oh wait this did not really do anything lol
counting double cosets H\S_n/S_n-1 is the same as counting orbits ahahahahahaha
upbeat juniper
@chilly ocean this probably doesnt answer your question but in general for a group G, subgroups H,K of G and a K-set X we have $(X \uparrow^G_K)\downarrow^G_H = \bigcup_{x \in [H \backslash G / K]}(^x(X\downarrow^K_{H^x\cap K}))\uparrow^H_{H \cap ^xK}$.
@sturdy marsh what does the arrow notation mean
cloud walrusBOT
wy: