Yepp!!!
#groups-rings-fields
paper flint
next obsidian
You can phrase that proof directly as well
Going for contradiction is fine (if you assume LEM lol), but this proof is equally easy directly IMO
paper flint
Yeah, the contradiction seems superficial.
next obsidian
Yeah, direct proofs are also aesthetic
There's also something to be noted that in a proof by contradiction you don't prove any lemmas really. Since you prove stuff about something that can't exist.
paper flint
I'll phrase this as a direct proof. It's just that being able to write down "we prove this by contradiction..." gives me something to think about and proceed with.
next obsidian
Whereas in a direct proof you might prove stuff in little steps which are true in general
Yeah that's a good way to go.
If I don't know what to do I usually try contradiction first. It gives me more stuff to assume, and then you can play around to figure out how to get what you want.
If I feel like I can see the direct proof though I usually switch over and do it directly.
paper flint
That's nice, I'll keep this mind. Thank you!
fading wadi
given a group G of order 35, and define f:G->G as f(g) = g^29, how do you show that f is an isomorphism?
carmine fossil
Order of a element in that group could be 5,7 or 35
If an element has order 35, it's cyclic and hence this is trivially true
If an element g has order 7,it gets mapped to itself
If g has order 5, it gets mapped to g^{-1}
chilly ocean
Maybe all groups of order 3t are abelian?
35
I mean, it has to be
If the statement you posted is true
I think(maybe not? Hmm)
fading wadi
Ok, @carmine fossil so do I just deal with showing that in all 3 cases it's an isomorphism?
f is an isomorphism
carmine fossil
Yes
fading wadi
thanks
Wait I'm sort of confused still since i'm not good at this
"If an element has order 35, it's cyclic and hence this is trivially true" what is the theorem that says mapping will be isomorphic, I noticed that f(g) = g^29 is like raising to prime power but i forgot the idea behind why
chilly ocean
29 is coprike to 35
carmine fossil
Ord (g^k)=ord(g)/gcd(k,ord(g))
fading wadi
oh right I forgot about that
Also very sorry but I also can't see why
"If an element g has order 7,it gets mapped to itself
If g has order 5, it gets mapped to g^{-1}"
carmine fossil
g^29=(g^7)^4 *g
fading wadi
oh
im so dumb
yea sorry im very new to this so thx for explaining
now I understand 5
carmine fossil
Don't think this is a easy problem
fading wadi
Another problem I have is like more conceptual
In general, a group (Z/nZ)x is cyclic, so say (Z/mZ)x and (Z/nZ)x have same order even if n,m are different, then these 2 r isomorphic right?
pallid ember
yes cyclics with the same order are iso
fading wadi
aight ty
Also what makes 2 groups (Z/nZ)x and (Z/mZ)x isomorphic if they are not cyclic? If same no. of element have same order?
chilly ocean
i think (Z/nZ)x is almos always cyclic
something like, as long as n is not a power of 2, maybe?
pallid ember
Also what makes 2 groups (Z/nZ)x and (Z/mZ)x isomorphic if they are not cyclic? If same no. of element have same order?
@fading wadi isomorphism implies this
but its not enough to guarantee one
fading wadi
ah
what else is needed to show isomorphism?
to actually show that each element is interchangeable?
chilly ocean
i'm not sure i see why 19^2=1 implies (Z/24Z)x is not cyclic
fading wadi
also @chilly ocean another example would be like (Z/10Z)x right, where 3 has order 4
cyclic means order <a> = order <G>
chilly ocean
hm
ah, nvm
i got it backwards
(Z/nZ)x is cyclic only if n is a p^k (power of a prime) or 2p^k, or 1 or 2 or 4
my b
next obsidian
Wait, it’s cyclic even if it’s a prime power?
I only knew it was cyclic when it’s a prime
next obsidian
Huh, I knew you can show it’s cyclic by showing you have primitive roots mod p for a prime p
But I guess there’s also existence of primitive roots mod p^k for odd primes p?
chilly ocean
i guess so, i have not done any group theory in a while
next obsidian
This sounds like some elementary number theory result to me moreso
sharp sonnet
this is a result of elementary number theory
next obsidian
I think I just assumed primitive roots mod p exist
sharp sonnet
which numbers have primitive roots
next obsidian
Ah, gotcha
sharp sonnet
and the 2^k cases are gross
next obsidian
Wikipedia seems to suggest they don’t exist for 2^k when k > 2
next obsidian
Ahhh, gotcha.
What about for other sorts of composite numbers? I assume you CRT it?
To show non-existence
Hmm... actually that doesn’t sound too right.
¯_(ツ)_/¯ number theory isn’t my thing
chilly ocean
first row, first column?
just a guess
y'know, ultra's gonna quiz you on your name when he wakes up
be ready.
vestal snow
me?
chilly ocean
ah, right
chilly ocean
damn, i do not remember any functional analysis
chilly ocean
or rather, i had not paid attention when i was learning 
next obsidian
I is the set of the top left entries of matrices in J
vestal snow
Ah gotcha
next obsidian
This sounds like fucking
Semisimple ring shit
Or modules or whatever
I did stuff like this that sounds kind of eerily familiar
vestal snow
How's your progress with AG going?
next obsidian
Kind of blech
I was in a rut and didn’t do much
But I’ve been doing it like all day the last few days, but progress is SLOOOOOW
As usual a new section starts
And it feels like I know nothing again
vestal snow
I found it useful to look at chapters on algebraic curves in non-AG books
next obsidian
Every new section of Hartshorne is just “aaaa what is happening” until I do like half the exercises
I’m not even at a point to think about curves haha
I’m just dying in O_X-module hell
vestal snow
I'm not either, but you can look ahead a bit and see how stuff relates
next obsidian
But I’m getting sort of side tracked and in some convoluted way I’m learning about Kan Extensions and coends LOL
I want to understand how a tensor product of modules is a coequalizer
And sources detailing this are scarce
chilly ocean
all of these modules stuff seems sort of dumb to me, just like random symbol manipulations and "obvious" constructions that make almost no sense
next obsidian
That’s why you do it “universally”
chilly ocean
i guess you need to study algebraic geometry and stuff to understand what it means
vestal snow
I personally found that thinking about algebraic curves algebraically rather than geometrically is more intuitive
next obsidian
I think about everything algebraically so ¯_(ツ)_/¯
vestal snow
I meant algebraically instead of sheaves and stuff
next obsidian
Ohhh
vestal snow
At least for the kind of curves I'm thinking about
next obsidian
I just pretend that’s algebraic 
vestal snow
Though I'm pretty sure that that becomes very messy when you do it for general curves
otherwise AG wouldn't really use all that fancy machinery
next obsidian
I suppose so
I guess your curves are all blah blah stuff that says it’s a variety
Maybe they’re smooth ¯_(ツ)_/¯
For as much AG as I know I know like no Algebraic Geometry
I know scheme theory not algebraic geometry I guess lol
vestal snow
The only curves I'm studying atm are hyperelliptic and elliptic
vestal snow
yeah they're pretty cool
next obsidian
You can use them to make curves of arbitrary genus
vestal snow
Well behaved
next obsidian
Something something projective space I don’t remember very well
vestal snow
Yeah so the book i'm using describes those curves using function fields
which is what I meant by algebraic way of thinking about them
next obsidian
What’s a function field in this context?
vestal snow
If K is a field, a function field over K is a finite (sometimes algebraic) extension of K(x)
next obsidian
I see
This sounds well-behaved
It’s late so my brain isn’t all that good right now but
Finite extension of field? Good to go
vestal snow
if you decide to look at this stuff, I am currently using Stichtenoth
It's pretty good
Though it sometimes pulls something like the "If (a_1,...,a_n) is max'l then (a_1^k,...,a_n^k) is max'l" thing you told me about
in Hartshorne
fading wadi
Can anyone help with this?
if a in G is order 11, show C(a) = C(a^3)
Like I know the general idea where you need to show both are subsets of each other but idk how to
fading wadi
bezout relation? You don't mean bezout's identity right?
like the one about gcd(m,n) = 1 iff exist a,b s.t. ma+nb = 1
pallid ember
Like I know the general idea where you need to show both are subsets of each other but idk how to
one direction should be straightforward
nova plank
Let x be in C(a), then xa = ax. Now xa^3 = ||axa^2 = a^2xa ||= a^3x.
This part should be straightforward
wind steeple
yes I mean bezout identity
fading wadi
oh
think i got the idea
ty
yea solved it
also: if orders of elements of 2 groups cannot match up perfectly, then there cannot be isomorphism, true?
and how to argue for it?
pallid ember
wdym by match up perfectly
fading wadi
like one group has 3 elements of order 2, while the other only has 2 elements of order 2
means not perfectly
fading wadi
is it also preserved in homomorphism?
nova plank
Yes
fading wadi
but is there some divisibility rule? like order of elements f(G) would be like a
divisor of order of G, if f was homomorphism
nova plank
Oh, 1 sec
Yes
@nova plank Obvious counterexample is the homomorphism which takes everything to the identity. Mb, mb
pallid ember
the most you can say in general is that order of f(g) divides |G| @fading wadi
fading wadi
alright, and as an obvious consequence so do the order of any element in f(G) divide |G|
tell me if im wrong but thx
fading wadi
is there a theorem that <a> intersect <b> = identity, if so what is it called?
where a,b are elements of some group G
if a and b have different orders
or corollary of some theorem
fading wadi
@pallid ember so suppose like I have |G| = 35, then what's the argument to show that given a^29 = b^29, the order of a and b must be the same?
i think it's related to 35 and 29 being coprime but idk how to explain
queen vine
Show that the order of a equals the order of a^29
fading wadi
another q i have is if f is a group homomorphism from group A to a non-cyclic group B, does it mean group A is not cyclic?
sharp sonnet
no
fading wadi
because I don't understand how you do this
queen vine
Not sure how to use the hint, but one way would be to prove that there exist multiple elements in G that have order 2
fading wadi
@queen vine could you explain how that method would work?
or if anyone else can explain how to use the hint
queen vine
In a cyclic group there is at most one element of order two, in this case we can look at e.g. 2^6-1 and 2^7-1
fading wadi
ic
sturdy marsh
The map from (Z/128Z)* to (Z/8Z)* is surjective
and (Z/8Z)* is not cyclic
which implies the other group is not cyclic
(as if it was, the image of the generator under that map would generate the image)
chilly ocean
Hi guys ! I'm struggling with lie algebra. I don't know how I should see it and what is its link with the group. One thing I think I got is that we can see the generators like a basis for the lie algebra. There is also the adjoint representation in all that mess haha. If someone have simple explanation to remember all that. (It in the context of SU(N) and su(N) for particle physics)
sharp sonnet
the lie algebra of a lie group is the tangent space at the identity
(if that means something to you)
chilly ocean
yes it is defined like that but it is a little abstract to me
sharp sonnet
it's probably the best intuitive way though as it is geometric
personally i think of a lie algebra as a vector space (easy) that is somehow (magic) deeply connected to a (hard) lie group, so i guess i can't help further 🤷
chilly ocean
it's interesting, it is the kind of sentence I'm looking for, but it is quite vague indeed
sturdy marsh
The lie algebra of a lie group can also be defined as the vector space of all left invariant vector fields on the group
along with the lie bracket
astral galleon
Are cyclic groups in general normal?
chilly ocean
u ok?
@astral galleon i guess you mean subgroups of cyclic groups? it doesn't really make sense to call a group normal unless it's a subgroup of something. in that case, yes, because cyclic groups are abelian and all subgroups of abelian groups are normal
however if you interpret it as "are all cyclic subgroups of other groups normal," it's false (e.g. <(12)> in S_3)
astral galleon
Ahhh okay makes sense
paper flint
My wording and solution are both shaky, can someone verify their correctness?
My primary motivation here is that the order of an element leans on how the operation works, and not what S itself contains.
Hence, even though all elements of S may have finite order, the hypothesis does not guarantee that all of these elements have to be in S(I guess?)
nova plank
That works
nova plank
Unless they meant it has to be binary operation and must be closed
paper flint
I think closure is an attribute of the set, not the operation?
I'll retry, thanks for pointing out.
next obsidian
I think you can force the existence of an inverse
paper flint
I now think S must necessarily be a group. Every element of S has finite order, and S is closed under the operation. Finite order and closure together guarantee the existence of identity in S. They also necessitate the existence of inverses.
next obsidian
Err
I meant the existence of an identity
from which you can show inverses
I feel like I've done this for finite sets
It's also annoying but you have to show left and right inverses agree
but this follows easily
paper flint
Since an element has finite order and we have closure, it will eventually produce identity right?
paper flint
Oh, okay.
next obsidian
Yeah, I've done this for finite sets
you can show the existence of an identity since multiplication by any element is a bijection, in the infinite case it's only an injection
sturdy marsh
Hmm, what about the multiplicative part of the ring Z/4Z
oh wait nvm
i was being stupid
next obsidian
It isn't cancellative.
sturdy marsh
ye
carmine fossil
You also have to show an identity exists
paper flint
I still don't understand how cancellation fits into the picture. 
I got the identity by Luna's method.
carmine fossil
i.e., e.a=a.e'=a implies e=e'
paper flint
Aah, is that what I need cancellation for?
next obsidian
uhhh
carmine fossil
You use cancellation to show e and e' exist
next obsidian
how did you get identity from a^n = a^m for n not equal to m?
paper flint
Frick I didn't. I was assuming the identity exists. 
I'm totally lost on this problem now.
next obsidian
So tbh I think I see how to do it
sturdy marsh
i think i got it
next obsidian
you get identity the same way I did for finite ones
by restricting to some "subgroup" of the for {a^n| n in N}
I think that's the idea
sturdy marsh
huh
next obsidian
then you have to show that's identity globally
I think?
I know for sure I can get an "a"-identity that way
next obsidian
You look at a^{m - n} right?
next obsidian
right
sturdy marsh
we need a^(m-n)b = b
sturdy marsh
use cancellation
next obsidian
There's still more work to be done though, you have a left identity
you need a right one, show they agree
then inverses...
it's a pain but doable
sturdy marsh
same argument
next obsidian
Yeah
sturdy marsh
use the same element
next obsidian
Oh right
carmine fossil
Aren't you indirectly using right and left inverses are same?
sturdy marsh
general trick for proving inverses is to show everything has a left inverse
sturdy marsh
we dont need to worry about right inverses then
because the left inverse to whatever element you're looking at also has a left inverse
so the left inverse of the elt you were initially looking at has a left and right inverse
which will then turn out to be the same by the usual proof
next obsidian
So really you weaken the assumption to having any finite sub semigroup
and a counterexample to when none exists is the set of {a^n| n in Z^+} where a is just a formal symbol with the operation given by concatenation or like... addition in the exponents
I guess this is like the fucking... free semigroup on one element???
maybe?
next obsidian
show a counterexample to the statement that it has a group when you don't assume there's a finite sub semigropu
The statement "is a finite sub semigroup" is a generalization of "the set of {a^n} is finite for all a in S"
And you can still obtain the result only from that
sturdy marsh
yeah aight
if you dont assume the finite sub semigroup thing then the multiplicative part of any domain - {0} works
and not a field
next obsidian
minus {1}?
sturdy marsh
So just Z - {0} under multiplication
no
minus {0}
Z - {0} satifies cancellation, assosciative
but not a group
under mult
next obsidian
Actually hold on
I'm no longer convinced what I said about the generalization is true
I don't htink you can show arbitrary inverses
I think you actually need that condition for all a
you can show an identity exists
but I don't see why inverses have to exist, in the finite case you get this by virtue of left multiplication by any element being a bijection
nova plank
@paper flint did you follow what Chmonkey said for the identity? You don't have to assume an identity exists, you multiply a^(m-n) by an arbitrary b and show you get b
next obsidian
This is what was tripping me up with what you said, in that case you get a like cancellative monoid but not a group
sturdy marsh
If you assume for any finite set of elements, there's a finite semigroup containing them then you get it
Maybe just assuming it for each element also works idk
paper flint
did you follow what Chmonkey said for the identity? You don't have to assume an identity exists, you multiply a^(m-n) by an arbitrary b and show you get b
I'm not sure if I follow that yet. 😓 My mind just goes a^m=a^n with m>n--> a^(m-n)=e, I've never seen left and right inverses before(the inverse I've dealt with so far commutes with the element)/
nova plank
Right, there you are applying cancelation, but you can only do that if there is an identity
carmine fossil
a^(m-n)b=b and b(a^(m-n))=b (by a similar argument)
paper flint
And I don't know what to do if there's no identity.
carmine fossil
For all b
nova plank
What you should do is, let m > n then show a^(m-n) is an identity using the definition
Like Drake did
paper flint
Can't I just use a^0 as the identity instead of introducing b?
nova plank
What is a^0?
It's defined to be the identity of a group
But here you don't know there is an identity
So a^0 doesn't mean anything
paper flint
Ah, alright, I finally understand.
nova plank
$a^m = a^n \implies a^{m-n}a^n = ea^n \implies a^{m-n} = e$, by cancellation. But this requires the existence of e.
paper flint
I just have to show that a^(m-n) does what e was supposed to
nova plank
Yes, exactly
paper flint
For any element b in S
cloud walrusBOT
Lunasong:
next obsidian
Exactly
you know it acts as identity on one element
you can extend that to show it acts as identity for all using cancellation
paper flint
Nice. Will that be enough to conclude that S is indeed a group?
next obsidian
No you need inverses
But like okay
So consider an arbitrary a
you know that for that a, there's some n,m so that a^n = a^m with n not equal to m
WLOG n < m
then a^{m - n} acts as identity
so you have like multiple identities floating around for each a
but you can show they all agree
then, just take a^{m - n - 1}
this is an inverse to a
since it gets you to a^{m - n} which is identity
paper flint
Yes, I follow so far.
next obsidian
so that's how you get inverses
I think you technically need to show left inverses = right inverses
but this is mega ez
since you know the identity is just a^{m - n - 1} which works on both sides
paper flint
Yes, makes sense. I'll write this down formally and get it verified here later. 😄
Thanks Chmonkey, Luna, Drake.
next obsidian
What year's putnam is this?
The putnam I looked at from last year had nothing like this
next obsidian
I felt totally powerless on every single problem
huh, I guess they've gotten harder? Or maybe this just happens to be right up my alley lol
There isn't a lot of group theory type problems in general
paper flint
I guess they have gotten harder.
There's another one from '69 which might be easy for you anyway: Prove that a group G cannot be the union of two proper subgroups.
Does the statement remain true if "two" is replaced by "three"?
(That's part of the question)
carmine fossil
(Z,+) is a union of even and odd integers
paper flint
Bruhhh 
paper flint
Odd integers don't form a group
paper flint
Identity
carmine fossil
Mb
paper flint
Yeah lmao
next obsidian
This seems weird
I feel like this should be false haha
Hmm, what's the issue here
I know how to show nothings the union of conjugates of a proper subgroup
for like finite index subgroups
paper flint
I think it should be true, proper subgroups looks like a strong condition.
next obsidian
Okay so
Step 1: for finite
Literally imp;ossible by lagrange
both have size at most 1/2|G|
they intersect
so they can't union to everything just by size considerations
paper flint
Nice
next obsidian
It seems impossible to reduce to the finite case in general though hmmmmm
paper flint
Btw this problem appears in Gallian before Lagrange so there might be alternative arguments based on subgroups, cyclic groups alone.
paper flint
Yeah
next obsidian
I'm trying lol
sharp sonnet
of A and B
choose a in A\B and b similar
consider their product
its either in A or in B, but then b is in A or a in B
carmine fossil
Genius
sharp sonnet
just do it lmao
paper flint
Uh I'm a bit lost, what are A, B? Proper subgroups of G?
sharp sonnet
yeah
paper flint
Okay
next obsidian
then you just win lol
paper flint
LMAO
paper flint
next obsidian
It can for 3
sharp sonnet
ye
next obsidian
So yeah for any n > 3
sharp sonnet
can it for all n though
next obsidian
I mean yes lol
just take more copies of the same ones
if you want distinct
F you
sharp sonnet
😦
next obsidian
but also still yes
just take more products of Z/2Z
Wait hmm
not true
redacted
sharp sonnet
not like i know the answer to this
carmine fossil
Is it false for n not odd prime?
nova plank
Why would n being prime be relevant?
carmine fossil
Because counting shenanigans
sharp sonnet
hm, might also be able to classify all groups for the n=3 case
paper flint
It didn't hold for n=2 either, that was a prime.
sharp sonnet
2 is a weird prime though
golden pasture
yh
paper flint
Agreed.
golden pasture
2 is fake tbh
next obsidian
i like 4 better
first non-prime prime
2 is fake tbh
@golden pasture this meme brought to you by F_2 gang
carmine fossil
So (F_2)^50?
sharp sonnet
but yeah define some kind of function p(G) on groups that is n iff G is the union of n proper subgroups but not of any smaller number
and see if there are groups such that p(G) = n for every n
then classify them
golden pasture
F_2[x_1,x_2,...,screams,...,x_n]
paper flint
Need help on this one.
chilly ocean
look at a clever choice of cyclic subgroup maybe?
that's what i'd try
no clue if it'll work out though 
nova plank
1 = xm + yn for some x and y. I would use this.
paper flint
My intuition was hazily wandering around this.
Alright, I'll give it a try.
\begin{proof}Since $m$ and $n$ are coprimes, we have $$xm+yn=1$$for some integers $x$ and $y$. Notice that $$a^{xm+yn}=a^1\implies a^{xm+yn}=a,$$and hence $$a^{xm}a^{yn}=a\implies (a^m)^x(a^y)^n=a.$$As $a^m=e$, we finally have $$(a^y)^n=a.$$By closure of the group, $a^y\in G$.\end{proof}
cloud walrusBOT
ted-D:
paper flint
Does this work?
nova plank
Yes
paper flint
Nice, thanks!
nova plank
Np, gj
paper flint
sour plume
If $R$ is a unital ring, it's easy to prove that $\text{Hom}_R(R,M) \cong M$ for every $R$-module, but what if $R$ does not assume a unit? Does someone know of a non-unital example like that where the above identity is not true?
cloud walrusBOT
Lartomato:
oblique river
@sour plume how about Z as a module over the ring 2Z?
sour plume
Hm, why does that break it? Isn't then every morphism uniquely determined by how it maps 2, and every choice of image works?
sour plume
I think so, then you get the morphism which maps every even number 2n to n
oblique river
so that's not a ring morphism, but I guess maybe it is a module morphism
sour plume
Oh, I see what you mean, lemme thonk
Aha, I think you're right! I forgot I wasn't just working with morphisms of abelian groups, but with morphisms over the ring 2Z
oblique river
oh wait haha I was gonna say maybe you were right :P
sour plume
Hm, but yeah, maybe it's not actually inconsistent
oblique river
you need to show that phi(2m*2n) = 2m*phi(2n) which I think is true, since phi is just "divide by 2"
sour plume
$2 nm =\phi(2n 2m) = 2n \phi(2m) = 2nm$
cloud walrusBOT
Lartomato:
sour plume
Yeah exactly, it's still fine
But that example made me a bit more careful at least 😄 Perhaps something with Z/2Z works
oblique river
Z/2Z is a unital ring
sour plume
Ye, as a module over something I thought
oblique river
oohhh sorry
oh actually, can't you do things for nonunital rings like
r * m = 0 for all r in R?
since there's no 1 to force 1 * m = m?
at the very least could you do 2Z acting on Z/2Z trivially? as in (2m) * 0 = (2m) * 1 = 0
sour plume
Hm yeah, just working with a super trivial module structure might work
oblique river
is {0} a nonunital ring?
sour plume
Oh yeah, the zero ring is allowed
oblique river
any abelian group would be a module over {0}
and yet every hom from {0} to A would have to send 0 to 0
sour plume
And then it's clear, 'cus there's not so many morphisms from zero to any module
oblique river
because that's an axiom?
sour plume
Yeah, that does it! Thanks!
oblique river
ok great haha
I never think of nonunital rings
so this was harder than I thought it would be haha
sour plume
Yeah, I'm doing spooky things with $C^\infty_c(\mathbb{R})$ right now, so this was a question which came to mind
cloud walrusBOT
Lartomato:
sour plume
So it's at least not something which holds true for every module, 'cause having a unit is a strong condition. Perfect 😄
oblique river
the other thing which is relevant here is that if R is a nonunital ring then R + Z is its unitization
and "modules over R" are the same as "modules over R + Z"
so you could try to look at modules over R + Z and then work in reverse
carmine fossil
What can you say about a group G,$| G |= p^{2}q^{2}$ ,if it is given p sylow and q sylow subgroups are normal?
cloud walrusBOT
DrunkenDrake:
carmine fossil
As in why sylow groups being normal is useful
thorn delta
all sylow subgroups are conjugate to each other, so they are normal iff they are unique
and then if all sylow subgroups are normal, it follows that the group is the direct product of its sylow subgroups.
solemn rain
solvable
thorn delta
nilpotent too
solemn rain
yes what you said implies nilpotency
south storm
If $H ⊆ G$ is a monoid and $G$ a non-Abelian group in which for all $g\in G$ we have $gHg^{-1}=H$ is H necessarily a group? I don’t think so but just something I wondered about and couldn’t find counterexamples.
cloud walrusBOT
27182818284tropy:
next obsidian
damn
Okay so
G non-abelian do you mean not necessarily abelian or actually not abelian
smoky cypress
The whole point of using latex is to use those symbols, so why not just use \subseteq instead of using ⊆
south storm
Actually non-abelian
I know that if G abelian there are counterexamples for both finite and infinite G‘s
The whole point of using latex is to use those symbols, so why not just use \subseteq instead of using ⊆
@smoky cypress thanks
#groups-rings-fields message yup my problem
next obsidian
I know that if G abelian there are counterexamples for both finite and infinite G‘s
@south storm if this is the case I think you can make this work trivially
Why do I think this
So you don’t need G abelian, you just need H to be a subset of the center of G and the “fixed by conjugation” thing is trivial
So I feel like you can embed G as the center of some non abelian group somehow
Or a subset of the center
south storm
How would that work exactly?
next obsidian
Which part?
Why it suffices or how we can embed it?
I know an answer to the first, the second is a hunch
If you have a counterexample in the finite case t suffices to find a group which has a copy of S_n as a subset of its center
And isn’t abelian
Oh wait nvm I’m dumb
Okay so here we go
south storm
I meant how one can embed it yes
next obsidian
Assume H is a subset of G a monoid...
Find a group with a trivial center, I believe Q8 is one such group?
Or maybe some symmetric group
Consider G x G’ where G’ has trivial center
Then the center of this is Z(G) x Z(G)’
G x G’ is non abelian
And H is a subset as H x {e}
south storm
Ah alright, that makes sense
south storm
Yup, thanks again, had a hard time coming up with a counterexample but really wanted to know if it‘s true, not really experienced with group theory but all cool stuff.
next obsidian
Yeah, this is a weird question haha
It seems not true
Tbh the condition that it’s fixed by conjugation to me seemed sort of irrelevant to it being a group or not
I felt like either it’s always a group, or not necessarily
It just didn’t seem connected to the existence of inverses
But I couldn’t figure out a counterexample haha
I guess you can do Z in Q
south storm
Tbh I just reviewed what normal subgroups are and thought of if the conjugation is enough, then finding a counterexample added more condition each time hoping to arrive at some actual result lol.
next obsidian
Ahhh
cinder bone
tips for this?
chilly ocean
Would it ever make sense to think of sets of Turing machines as an algebraic group, could they be distinguished by their languages?
scarlet estuary
@cinder bone are you familiar with lagrange's theorem?
cinder bone
yes!
scarlet estuary
okay, so according to lagrange's theorem, the order of any subgroup divides the order of the "parent" group
so we know 10 divides |G| and 25 divides |G|
and also |G| < 100
so what must |G| be?
scarlet estuary
@chilly ocean what would the group operation be?
we dont really care "what" the elements of a group (or similar algebraic structure) "are", just how they "behave" under the group operation
(this is where the notion of isomorphism comes from)
chilly ocean
I was thinking some type of string substution for the language, but I'm not sure if you'd even need the turing machine for that
scarlet estuary
well a google search of these keyterms brings up this MO post https://mathoverflow.net/questions/88368/can-a-group-be-a-universal-turing-machine
might be worth a read though it doesnt seem to directly address your question
chilly ocean
Thank you!
cinder bone
so for a) I showed K is a subset
not sure what the easiest way to show closure
for K tho
cinder bone
yeah I did that
abelian and every transposition is it’s own inverse
so every element is its own inverse
right
paper flint
Something seems off about my proof since it does not make any use of the fact that G is finite.
Could someone point out the problem with my proof?
sharp sonnet
you don't
chilly ocean
proof looks good btw
sharp sonnet
like, you can easily make this more general
chilly ocean
just the other day i fell for an unnecessary "let G be finite" on my group theory homework lol
sharp sonnet
to give an explicit description of generated subgroups
chilly ocean
maybe you can do my homework for me ted 😌
paper flint
The next chapter is on cyclic groups so I might be dealing with that haha.
@chilly ocean No academic dishonesty but I'd like to see your group theory PSet just to get a feel for how uni stuff looks like. 
chilly ocean
@paper flint lemme get on my computer and ill dm them
paper flint
It was just handwavily introduced in an example, then some computational problems on groups of this kind and then this problem.
@paper flint lemme get on my computer and ill dm them
@chilly ocean Sure, thanks!
sharp sonnet
$<a, b>$ vs. $\langle a, b\rangle$
cloud walrusBOT
Lochverstärker:
solemn rain
@paper flint lemme get on my computer and ill dm them
@chilly ocean same
chilly ocean
typing 7 characters vs 1
next obsidian
I made a command \gen{stuff} which does $\left\langle stuff \right\rangle$
cloud walrusBOT
Chmonkey:
Compile Error! Click the 
reaction for details. (You may edit your message)
coarse storm
I stick to \left<, \left(, \left[, \left\{. A bit easier to remember for me.
And then I found out my editor has a macro to expand <<, etc to the corresponding \left< \right> and put the cursor in the middle.
carmine fossil
yes
Because coset gH has g.1=g in it
Which means g is in H, according to your condition
sturdy marsh
An isomorphism between those two groups is sorta the same thing as choosing another labelling of the elements of the same group
Choosing a different labelling can be encoded by a bijective map from 1,2,..,n to 1,2,...,n
Conjugating any permutation by that map swaps the old symbols for the new ones
So the old permutation representation is "conjugate" to the new one
More concretely if the iso is $\phi: G \rightarrow G'$, and if the permutation representations are given by $\rho_1: G \rightarrow S_n$ and $\rho_2: G' \rightarrow S_n$, then there should be a $\tau \in S_n$ such that $\rho_2 \phi = c_{\tau} \rho_1$
cloud walrusBOT
Brofibration:
sturdy marsh
where $c_{\tau}: S_n \rightarrow S_n$ is given by conjugation by $\tau$
cloud walrusBOT
Brofibration:
sturdy marsh
Another way to see this is by considering $G$ as a $G-$set and $G'$ as a $G-$set via the isomorphism
cloud walrusBOT
Brofibration:
sturdy marsh
Then the isomorphism of groups becomes a G-equivariant isomorphism of G-sets
so the 'representations' are isomorphic
cloud walrusBOT
Token:
latent anvil
the isomorphism
sturdy marsh
Does conjugate mean in this case that the relabeling of the elements should be conjugate in their image under the respective representations? Like if $G\cong H$ and $\phi$ is an isomorphism, then $\phi(g_1)=h_2$ means that $\sigma_{g_1}=t\sigma'_{h_2} t^{-1}$?
@elder condor Yes, like I said, you have $\rho_2 \phi (g) = c{\tau} \rho_1 (g)$ (you switched notation a bit tho).
cloud walrusBOT
Brofibration:
thorn delta
I want to prove "A finite ring R with more than one element and no zero divisors is a division ring."
Do you have to assume that R has identity, or is that not necessary? I have a proof when R does contain 1, but I'm not sure if I was supposed to assume that
hmm i might have an idea
yea, i think this is similar to the fact that semigroups with solutions to ax = b and xa = b for all a,b are groups
sturdy marsh
Yup, and R will actually be a field
not just a division ring
but that's harder to show
it's actually a pretty neat result
I think the section on cyclotomic polynomials in Dummit and Foote guides you through an elementary argument for it
iirc
there's another argument using group cohomology
and one using Noether-Skolem
thorn delta
ah i found it
its crazy that the proof of something so simple sounding can be that involved
sturdy marsh
yup, that argument is super nice
but the other arguments are a lot shorter
they use a lot more machinery though
In my class Noether-Skolem turned up only after doing quite a bit on CSAs and division algebras
nvm it actually turned up a lot earlier than I remembered
ffs online classes have completely screwed up my perception of time
thorn delta
what does CSA stand for?
sturdy marsh
central simple algebra
thorn delta
ah ok, yea never heard of that lol
sturdy marsh
It's an algebra over a field that is central i.e. the field is precisely the center
and it is simple
paper flint
\newtheorem*{prp}{Proposition}\begin{prp}Let $G$ be a group and let $H$ be a subgroup. For any element $g\in G$, define $gH={gh\mid h\in H}$. If $G$ is Abelian and $g$ has order 2, then the set $K=H\cup gH$ is a subgroup of $G$.\end{prp}\begin{proof}We prove this by the two-step subgroup test. First, notice that $K$ is nonempty since $H$ is a subgroup. Next, consider arbitrary elements $a=g^ph\in K$ and $b=g^qh'\in K$. Now, $p,q\in\bZ$ but $g$ being an element of order 2, it is meaningful to say that $p,q\in{1,2}$. Observe that $ab=(g^ph)(g^qh')$, and $G$ being Abelian, $ab=(g^pg^q)(hh')$ which is an element of $K$ by the closure of $H$ and $gH$ as subgroups. Finally, we can see that $a^{-1}=(g^ph)^{-1}=h^{-1}g^{-p}=g^{-p}h^{-1}\in K$. Hence, $K$ is closed and every element in $K$ has an inverse, which implies $K$ is a subgroup of $G$.\end{proof}
cloud walrusBOT
ted-D:
paper flint
Is this proof okay?
thorn delta
Next, consider arbitrary elements a = g^ph \in K and b = g^qh' \in K. .... p, q \in {1,2}
i disagree. arbitrary elements of K have the form g^k h for some k = 1 (when g^k h is in gH) or k = 0 when g^k h is in H
paper flint
Well 2 gives the identity since g has order 2...
So g^k h would either be g^1h=gh or g^2 h=eh=h
I guess both work 🤷
thorn delta
eh sure okay its legal. a bit strange though
paper flint
Yeah, although I'll admit {0,1} has a nice boolean algebra like structure
Addition of exponents could then be interpreted like xor
Is the proof otherwise okay?
thorn delta
yea
paper flint
Thanks for taking a look. 😁
thorn delta
npnp
paper flint
I've asked this before, but I forgot the reasoning so I'll ask again:
Show that the group of symmetries of a circle has an element of infinite order.
Should I pick a point on the unit circle like (cos(sqrt(2)), sin(sqrt(2))?
south storm
I think you can just use the fact that all circles are proportional and if the radius is 1 the circumference is 2π being irrational taking a rotation which shifts every point on the circumference a rational amount.
paper flint
That's a pretty neat argument, thanks!
So essentially for no positive integer n, n*(some positive rational)=(some positive irrational)
south storm
Actually now I’m wondering is the group of the rotational symmetries of a n dimensional sphere isomorphic to SL_n(ℝ), would make sense but the proof would depend on the way you define the group of rotational symmetries. I‘m not entirely sure about it being injective (taking from SL_n(ℝ) to the symmetry group)
Or I think maybe it would be to the group of rotational matrixes probably, but with reflections to the whole SL_n(ℝ).
paper flint
I haven't worked with higher dimensional geometries before, but seems plausible.
south storm
Wait is the orthogonal group hits the group of rotational matrixes?
I never understood what the orthogonal group was about so if so that’d make a lot of sense.
Because I looked n-sphere up on Wikipedia and found this
paper flint
Aah so such a group already exists
south storm
Wdym?
Also I’m not sure it isn’t written if reflectional symmetries are accounted for
Also now I’m wondering about the symmetry groups of ellipsoids generally speaking, is this covered in representation theory does someone know?
paper flint
Wdym?
@south storm I wasn't aware there was a general notion of symmetries of an n-sphere.
south storm
I mean it‘s pretty simple of a concept to come up with, so I don’t think it‘s surprising someone already came up with it.
paper flint
Makes sense.
carmine fossil
I never understood what the orthogonal group was about so if so that’d make a lot of sense.
Orthogonal matrices is the "largest" group that preserves distance between any two points
Rotational matrices preserve distance,so yes
south storm
Huh nice
paper flint
What are Abelian forcing sets?
carmine fossil
paper flint
Lol I am reading Gallian's paper on this, and Google search takes me to his paper
chilly ocean
he defines it in the paper
More generally, let us call a set of integers T Abelian forcing if whenever G is a group with the property that G is n-Abelian for all n in T, then G is Abelian
paper flint
I'm actually having difficulty processing this statement
But nvm, my brain cells are dead at this time, I'll read it in the morning.
leaden finch
how did you get that
leaden finch
omg i still dont see it but it makes sense though
solemn rain
now prove that every field is an euclidean domain
and then find the other roots
or other factors
using long division
gl
leaden finch
wym euclidean domain?/
solemn rain
take any two elements x and y
there exists q and r |r|<y such that x =qy+r
| | would mean deg when talking about polynomiql fields
leaden finch
can someone help me with irreducible polynomials
thorn delta
i am trying to find all of the composition series of S3 x Z2. I (think) I found one: S3 x Z2 > <((123),1)> > <((123),0)> > {e} but how do i know when ive found all of them?
spiral wolf
I have no idea how to go about this. I am in an asynchronous online course and the professor did not upload a lecture discussing polynomial rings.
chilly ocean
Z8[x] is countable, you could brute force it
programming exercise: write a program to brute force this problem
spiral wolf
ahhhhh
I know i could brute force it but i'd like to know how to actually do it
The programming exercise might be good for me... I think I will do that once the full assignment is done
cinder bone
tips? 💀
spiral wolf
so first of all what is a right coset, in this example?
cinder bone
im guessing use this?
spiral wolf
yeah, that is the general idea
why don't you write out what H+15 and H+27 are directly
cinder bone
would that just be {15, 15+-4...} and {27,27+-4...}?
spiral wolf
yes, so are they the same?
cinder bone
yeah seems weird to me too
since H is like a set?
maybe i could show 4m+15 = 4n+27 for any integers m,n?
wouldnt that be the same thing
or
m=n+3
cinder bone
but like wouldnt that be enough to determine theyre indentical
spiral wolf
yes sir
cinder bone
whats a good way i can qualify that
spiral wolf
H, H+1, H+2, H+3
cinder bone
i guess maybe a good way to qualify is that
0=4 mod 4n for all n in Z, 1=5 mod 4n for all n in Z, 2=6 mod 4n for all n in Z, 3=7 mod 4n for all n in Z?
by the thm i showed
cinder bone
oh yeah tru
spiral wolf
i would just say the cosets are H, H+1, H+2, H+3
cinder bone
maybe i can just call the logic of part a)
spiral wolf
I don't think an explanation is necessary
They just ask for the answer it would seem
cinder bone
nah my teacher a stickler bout this shit tbh
maybe just H+a = H+b for 4 | (b-a)
seems simple enough
not sure how to find index
as theyre both infinite groups
spiral wolf
ye
cinder bone
cool
i did this also
i showed that any subgroups of G must divide 161
so any proper subgroups of 161 must haves order 1,7,23
spiral wolf
true
cinder bone
in the case of 1 its the idenitity, which is generated by the identity element, so its cyclic
spiral wolf
true
cinder bone
7 and 23 are both prime so i used a thm that stated they were cyclic and isomorphic to Z_7 and Z_23
spiral wolf
also true
cinder bone
and that seemed sufficient
cinder bone
oh yeah, my b
spiral wolf
that is sufficient
cinder bone
ok i might ask another q again working on this last problem
yeah not sure what to start on this one
lagrange's lets me know it can have an element of order 5
but not that is must
upbeat juniper
by sylow's theorem it must have a subgroup of order 25
cinder bone
hmmm maybe just check each possible one?
upbeat juniper
and you can use the classification of order p^2 groups from there
cinder bone
not sure if weve covered that theorem in class
upbeat juniper
ok that makes things a bit harder
or you can just use cauchy's theorem
idk why I didn't just think of that 🤦♂️
upbeat juniper
but there are many possibilities
upbeat juniper
or you can just use cauchy's theorem
have you learned this in class
cinder bone
not sure, we dont usually get names for stuff just thm numbers
upbeat juniper
it says that if p is a prime dividing |G|, then there is an element of order p
spiral wolf
basically this problem seems like you need to go thru the proof of cauchy's thm but for the specific case of 75 and 5
cinder bone
ok
oh okay i have this
if |G| = k then a^k=e for all a in G
does that help me with the case of |x| = 3?
upbeat juniper
I don't think so
spiral wolf
i also dont think so
carmine fossil
Just prove cauchy for general groups, using sylow and Cauchy for abelian
sturdy marsh
that sorta seems like a cop-out, I think the problem can be attacked more directly
every non-identity element has order 3, 5, 15, 25 or 75 (as the order must divide 75)
If you have an element of order 5, you're done
if you have an element of order 15, 25, or 75, you're also done as the subgroup generated by that element is cyclic, and therefore has an element of order 5 (choose suitable power of the element)
therefore assume that we dont have any such elements, so every non-identity element has order 3
so we need to show that it is impossible to have a group of order 75 where every non-identity element has order 3
oh nvm I did not see the messages above
Every subgroup has order 3
and none of them can be normal, as then we will have a quotient of order 25
hmm
so we have 37 subgroups of order 3
the group acts on the set of subgroups of order 3 by conjugation
so we have 37 = |Fix| + sum over sizes of orbits
there are no fixed elements as none of those are normal
the size of each orbit is divisible by 5 (as the stabilizer has size 1 or 3)
reduce the equation mod 5 and we get 37 = 0 mod 5, which is a contradiction
ye that should do it
I think
thorn delta
and none of them can be normal, as then we will have a quotient of order 25
can u elaborate on this reasoning here? I can see that they are not normal as 3 is the largest power of 3 dividing 75, so subgroups of order 3 are Sylow 3 subgroups and therefore all conjugate to each other, but im not sure i see why having a quotient of order 25 matters?
sturdy marsh
If you have a group homomorphism $\phi: G \rightarrow H$, then for all $g\in G$, $\text{ord}(\phi(g)) \vert \text{ord}(g)$.
cloud walrusBOT
Brofibration:
sturdy marsh
so if we have a quotient of order 25, then we have a surjective morphism from G onto the quotient, which gives us the contradiction
I am not using the sylow theorems as they are stronger than Cauchy's theorem
and I don't want to use Cauchy's theorem as it feels like a cop-out
thorn delta
ah okay i see
sturdy marsh
does the argument sound okay?
im pretty tired rn so it is very possible that i made some massive blunders
thorn delta
i can't find anything wrong with it. very cool proof!
earnest horizon
sturdy marsh
Try showing that there's a copy of the field of p elements inside Fn
Once you have that it's easy
sturdy marsh
If you have an R-module M, then tensoring with M is right exact
Tor measures the failure of exactness
in some sense
uncut girder
In what sense
uncut girder
Its homology of (a free resolution tensored with M)
sturdy marsh
right, a projective resolution
uncut girder
Wrong channel
chilly ocean
yup sorry
uncut girder
The issue I have is that
You talk about failure of tensor with M to be exact
But Tor takes 2 arguments
So what actually is it
next obsidian
Tor(M,N) is the derived functor of both - (x) N and M (x) -
So if you have an exact sequence
0 -> L -> L' -> L'' -> 0 you get an exact sequence either of the form
L (x) N -> L' (x) N -> L'' (x) N -> 0
or M (x) L -> M (x) L' -> M (x) L'' -> 0
you might want to keep going to the left, and let's just deal with the M (x) - business right now since the other one is sorta the same
Then you can extend this to the left by
-> Tor^2(M, L) -> Tor^2(M, L') -> Tor^2(M, L'') -> Tor^1(M,L) -> Tor^1(M,L') -> Tor^1(M,L'') -> M (x) L -> M(x) L' -> M (x) L'' -> 0
This is useful because sometimes you know that say Tor^1(M,L') is 0 for reasons (like L' is free, projective, flat)
This says that Tor^1(M,L'') is identified with the kernel of M (x) L -> M (x) L'
Which if you tensor the right stuff might be say... the a-torsion of M when M is an A-module and a is an element of A
Then, you can calculate Tor^1(M,L'') via projective resolutions (or even flat ones if you like) to figure out what the a-torsion of M is, or more likely just abstractly knowing Tor(M,L'') is the a-torsion let's you do other shit
It's hard to get a feeling for why these things are useful until you do exercises with them IMO
(Also the common notation might be Tor_i(M,N) I forget tbh, it switches between Ext and Tor and it's about chain complexes vs cochain and I never remember what the accepted standard is)
sturdy marsh
also, tor made a lot more sense to me after reading about derived functors
next obsidian
I learned about it via Aluffi, I honestly really like how he sort of gives you a cheap version at first which isn't fully justified you black-box some of and just play with it, come to see why it's useful, how you can use it
then after that introduces Ext in a similar way, and then finally does all the nitty gritty stuff on why it's balanced, why you can calculate it via resolutions...
sturdy marsh
This was also very helpful:
if you ignore the words global sections and just pretend you have a left exact functor
carmine fossil
If N is not normal, conjugation wouldn't be an automorphism on N
stoic dirge
OH ok
spiral wolf
I'm not sure how best to word this, so let me give the idea of what I'm trying to say and maybe someone can offer a professional brief wording.
Let A be a subset of the cyclic group Zn. and consider the group action of Dn = Zn x Z2 on A. I want to say something along the lines that the action of the Zn bit is referred to as transposition, and the action of the Z2 bit is referred to as inversion.
hot lake
um why do you think there are group actions on random subsets of Zns
spiral wolf
Yeah this was kind of my problem
I don't know how to word what i'm going for.... I'm going for like, the collection of all transformations of A from Dn acting on it
idk... it's like instead of G x X -> X i'm looking for G x A -> X where A is a subset of X, if you get me
hot lake
I don't get you
chilly ocean
Acting with the dihedral group on n verticas on a subset of verticas of an n-gon?
Is this what you are talking about?
carmine fossil
The argument is the number of sylow groups is $\mid G: N_{G}(P) \mid = \mid G:C_{G}(P) \mid$ giving us the result
cloud walrusBOT
DrunkenDrake:
upbeat juniper
idk... it's like instead of G x X -> X i'm looking for G x A -> X where A is a subset of X, if you get me
@spiral wolf is this interpretation correct: the action of each element g on X corresponds to a permutation X -> X, and you want to find the restriction of these permutations to A
but you would need A to form a block under the action of G
spiral wolf
let me make a concrete example -- let's say X = Z12, and A = {0,1,3}. Then I want to say something along the lines of D12 acting on A, so we have an equivalence class of sets of A, namely all the transpositions A+1, A+2, A+3 etc and the inversions of each of those (A*-1)
if that makes sense
so like {1,2,4} or {6,7,9} would be transpositions
and like {0,11,9} would be inversion
hot lake
so you are looking at the action of G on P(X) ?
hot lake
and if you have {1 2 3 4}, is {5 6 7 8} a transposition or an inversion ?
spiral wolf
Could be either depending on what elements of D12 are acting on it
Really what I want to say is that adding/subtracting is transposition and reflection/multiplying by -1 is inversion
hot lake
so you can just give those names to the elements of D12 then
upbeat juniper
ahh I see what you mean now
hot lake
transposition is kinda a weird name for it
upbeat juniper
I thought about this before