idk what else you were expecting
#groups-rings-fields
smoky cypress
leaden finch
thats it? LOL
thorn delta
can someone explain where the equalities and isomorphisms come from in the proof of part b that i have circled?
leaden finch
can someone help me with axiom rings
thorn delta
yea go ahead and ask @leaden finch
or if someone can answer my question, feel free to do so
scarlet estuary
@thorn delta according to the text, "theorem-i.8.9"
thorn delta
i think its mislabeled. i.8.9 tells us that if G is the internal direct sum of subgroups, then each element can be written uniquely as a finite sum from finitely many of the subgroups
so that allows us to identify the internal direct sum Ni with the direct sum Gi, but not the isomorphisms.
the fact that $\sum_{i \in I} G_i / \sum_{i \in I} mG_i \cong \sum_{i \in I} G_i/mG_i$ follows from theorem i.8.11, but i don't see why all the stuff in between $G/mG$ and $\sum_{i \in I} G_i/mG_i$ must hold.
cloud walrusBOT
kxrider:
thorn delta
right, but that doesn't justify the isomorphism between the quotients
for example, 3Z is iso to 2Z but Z/2Z is not iso to Z/3Z
chilly ocean
can someone give me a 30 minute rundown of homotopy/cohomology/homology/babby's first theorems of alg top?
@ocean magnet
ocean magnet
I can't teach cohomology 
wind steeple
"30min rundown", "cohomology". Wtf ?
scarlet estuary
this is a bit of a tough ask
considering this is, like, an entire textbook's worth of material
chilly ocean
i know homotopy is something about transforming loops, cohomology is something about groups, and homology is something
chilly ocean
maybe skip homology, i don't hear about that one a lot
supple marsh
homotopy is wiggly-waggly
scarlet estuary
are you familiar with the fundamental group?
chilly ocean
i've heard the name
scarlet estuary
as its the prototype of homotopy groups
chilly ocean
something about loops?
scarlet estuary
im not sure that says much, but yes
its the group of equivalence classes of loops on a space
(where "equivalence" is homotopy equivalence)
informally, homotopy equivalence is being able to "deform loops into each other"
chilly ocean
btw a single point counts as a loop?
scarlet estuary
uh, yes
i mean formally speaking
a loop is a function, not a point
but yes, a function that maps everything in [0, 1] to a single point is a loop
chilly ocean
ok i'm looking at wikipedia, it says fundamental group of a circle is Z, this makes sense to me, but how would i prove rigorously that the loop that is 2pi around is not homotopic to 4pi around?
oh, maybe an intermediate value theorem type thing?
chilly ocean
i was thinking something along the lines like, remove a point of the circle to get a line segment, then one loop becomes just a function from one endpoint to the other, and the longer loop becomes two copies of that, then something something
scarlet estuary
the formal proof here is slightly involved
chilly ocean
ok skip this then
so i recall "algebraic topology is about classifying topological spaces", so different funamental group means non-homeomorphic, right?
is that the only point of fundamental groups?
scarlet estuary
the fundamental group offers a first hint at the idea of "gluing topological spaces together"
(through seifert-van kampen)
and allows us to actually understand these spaces
classification is certainly its main function, but dont underestimate how impactful that actually is
moreover, there is a certain meaning to "going backwards"; start from a group and find spaces that we can associate with it
this slight-of-hand is often done in lie theory
chilly ocean
which spaces have fundamental group Z?
scarlet estuary
many many
you can construct more from the circle through seifert-van kampen
or through deformation retraction
(e.g. this gives you that C \ {0} has fundamental group Z as well)
but there are more examples; for instance, the lie group of unitary matrices
indeed the isomorphism here is actually the determinant onto the circle
chilly ocean
unitary matrices are 1 dimensional?
chilly ocean
if they are isomorphic to the circle via the determinant
scarlet estuary
(for a square complex matrix)
im not saying they're isomorphic, i'm saying their fundamental groups are.
and this fundamental group isomorphism is induced by det
chilly ocean
ah, so with a loop gamma, a point gamma(t) in unitary matrix space corresponds to |gamma(t)| in circle space?
wait a second..
scarlet estuary
its more like
if we map U(n) to U(1), this gives us a morphism of fundamental groups
and its easy to see we can "go back"
by "splitting" U(n) into SU(n) and U(1)
and as previously established (since U(1) is topologically a circle), pi_1(U(1)) = Z
wind steeple
one can introduce cohomology in 30 min but without any example. You'll understand nothing
wind steeple
yup
scarlet estuary
like theres a reason people write entire textbooks on this lmao
chilly ocean
can we go back to seifert van kampen?
how does fundamental group mean we can glue stuff together?
scarlet estuary
how much category theory do you know?
well its more that the fundamental group lets us understand what our glued-together things will look like
chilly ocean
i remember functor and category, sort of, and some exercises from algebra class on so and so's lemma or what not
oh, i guess diagram chasing is not really category theory
if u say like something something universal property then i would sort of know what ur talking about
ah ok, so you use the fundamental groups of the smaller pieces to compute the fundamental group of the bigger thing
is seifert van kampen an easy theorem or a hard theorem?
signal roost
Does anyone here have access to the downloadable MAGMA? As the web version has a time limit set to 120 seconds which is not long enough for my script, if so can someone please run it and tell me the output? Would appreciate it.
chilly ocean
u dont have access to it?
chilly ocean
i thought i might have downloaded it in the past but no
signal roost
2k is way too much for me to pay for it
signal roost
Ah okay
sullen island
hi, im trying to understand why the determinant yields a monic polynomial with variable z of degree n
that would mean that each c_{i, j} can be expressed in terms of z, which im not seeing why
help would be appreciated! thanks!
golden pasture
i too would like to get magma but no access :/
chilly ocean
why would it mean each c_{i,j} can be expressed in terms of z?
sullen island
because the concluding statement states that det(M_z) is a monic polynomial in variable z of degree n
hot lake
Mz's coefficients are called mij, not cij
sullen island
mij are cij if i \neq j, and cij - z if i = j
chilly ocean
isn't this a standard result btw? (determinantn is monic degree n?)
hot lake
characteristic polynomial*
sullen island
degree n in variable z though?
chilly ocean
you just use that one guy's determinant rule (cramer's rule?)
sullen island
well, im getting that its a linear polynomial in z
since z appears once in the matrix
hot lake
chilly ocean
huh, z appears n times
sullen island
?!
there is only one entry in the matrix where i=j
ohhh wait...
ok...
i get it
the whole time i was thinking z appears once
typing that out made me realise the mistake
signal roost
i too would like to get magma but no access :/
@golden pasture if you're a student at NYU you get it for free, go to cims .nyu.edu then search for magma
golden pasture
someone convince my highschool to give us programs then
oh wait i could try to ask from others
john come here
actly if we had like a bunch of ppl wanting magma here
could split cost and patch program to allow multiple isntalls
signal roost
It comes from my country
Australia :)
Unsw
Sydney
I guess I could ask my colleagues
But email takes forever
Patching it is illegal
Its propirtiary
But it is easy to RE it and patch from what I have heard
chilly ocean
how come i can't find a torrent for this thing
chilly ocean
prolly cuz no one uses magma
signal roost
Academics who don't want to ruin the program
Yep
Because sage
A lot of cryptographers use it though
As they're lazy and don't program much
chilly ocean
can mathematica do everything sage/magma can?
signal roost
No
Pari/GP is another good one
But sage uses that
MAGMA is good because it's fast, and solves problems quickly
Just sucks it's proprtitary
Should be a homebrew of it :p
Is anyone here at NYU?
chilly ocean
yeah i don't really like all these proprietary shit
signal roost
Neither
Open source I prefer
But I like MAGMA
Without it being proprtitary
If it wasnt*
chilly ocean
does anyone want to learn cohomology with me? https://maths.tcd.ie/~pedro/PHDCLUB.pdf
learn cohomology in 10 pages
hot lake
oh me I want to learn cohomology
chilly ocean
ok im on section 2.1 just starting, we can work through exercises together or smth
coarse storm
My experience with MAGMA was mm... Not a fan of the syntax. No GUI. Good documentation, which is rather important. So quite usable if you are okay with the terminal.
No idea how much the university was paying for it though. We all used the same install on a server. Normal laptops don't have enough power to use it properly anyway.
Pretty sure it is not good at being multi-threaded either. So thr raw clock cycle speed matters, like LaTeX, and eats a silly amount of RAM.
gray charm
What does the red underlined part mean in this context? My understanding is that φ^-1(g) is some (h,k), but what does that mean when we put an absolute value on it?
gray charm
Is phi^-1 not a map?
chilly ocean
order
gray charm
Order of the element phi^-1(g)?
chilly ocean
yes
gray charm
Ok, thanks!
leaden finch
anyone good with axiom rings?
snow flint
i think @ Whoever#7942 is very good with axiom rings 🙂
ocean magnet
you forgot to show the whole problem
covert vector
what are the two operations?
ocean magnet
leaden finch
what us 3 relating too?
ocean magnet
the operations on the top of the sheet
try to see if there's an additive or multiplicative identity
like a+b=a for all b or a*b=a for all b
ocean magnet
@leaden finch yea
ocean magnet
um yea I don't quite understand the problem's question
there's additive and multiplicative
next obsidian
what are you confused about?
leaden finch
how do i find the multiplicative identity?
next obsidian
It's the thing when multiplied by anything else gives out the same thign?
just look at your table
ocean magnet
yea just do what you did for circle +
with circle x
it's about the top of the page not the table
next obsidian
one element is acting as the identity and the table is reflecting that
leaden finch
how we know?
next obsidian
or wait maybe there is
maybe it's just 1
it isn't 1
but there is one
pretend b is the identity
and set a (x) b = a
solve algebraically for b
you'll get a number
leaden finch
someone posted their solution in my class right now
im cofnused
i thought we could of find it like this?
mint gulch
Which is your doubt? @leaden finch
I have one, a irreducible element is Product of irreducible elements? Like p = p
signal roost
Does anyone here have MAGMA Computational Computational Algebra System downloaded on their computer? As I have a script I wrote to solve a problem, however I do not have access to MAGMA on my PC as it costs around 2k as it is proprietary software. I have used it in the past. But I do not have access anymore.
And the online MAGMA algebraic software for my problem/script is limited to compute for 120 seconds which is too short for my script.
If anyone has the downloaded version, can you please contact me privately in PM or chat. And I will provide the script I wrote to run.
chilly ocean
why cant u write it for another softawre system like sage
golden pasture
sage kinda
sucks af for anything non NT
try implementing like (Z/2Z)x(Z/3Z)x(Z/10Z) and localize it at (1,2,5)
and will start crying
chilly ocean
something looks weird
paper flint
I'm pretty sure there is something weird, I never trust my "proofs" lol
chilly ocean
in the third line, you have (ab)^n = a(ab)(ab)(ba)...(ba)b, there is the term (ab)(ba), so somehow something weird is going on with matching bs and as
i think you could fix it by, in the next line, having (ab)^n = aa(ba)(ba)...(ba)bb, i think
paper flint
I'm flipping elements inside parentheses because the group is Abelian
chilly ocean
you could also rewrite this using induction
paper flint
Maybe I should've made it clearer, I'll improvise that.
chilly ocean
in the line (ab)^n = aa(ba)(ba)...(ba)b, you have more a's then bs
paper flint
you could also rewrite this using induction
@chilly ocean I didn't think of that at all! Thanks, I'll redo this with induction.
Repeating (n-1) sounds cranky anyway
in the line (ab)^n = aa(ba)(ba)...(ba)b, you have more a's then bs
@chilly ocean Yes, I messed up a lil bit. I'll proceed with the induction idea, it ought to be much clearer.
Thanks for the help! :)
Uh a stupid question, but how do I extend induction to negative integers?
chilly ocean
something like, it holds for n=-1, then assume for n=k, and show for n=k-1
paper flint
Thanks, that makes sense.
paper flint
I'm trying to prove that a group $G$ is Abelian iff for all $a,b\in G$, $(ab)^{-1}=a^{-1}b^{-1}$. The forward implication is easy to prove, but for the backwards implication I'm stuck with the fact that for all $a,b\in G$, $a^{-1}b^{-1}=b^{-1}a^{-1}$. Do I need some additional assumption to declare $G$ as Abelian?
cloud walrusBOT
TedNowKaczynski:
next obsidian
This is enough
As you enumerate over a^-1, b^-1 for all a and b in G you go over everything
This is cuz every element of G is the inverse of something
Namely a = (a^-1)^-1
Now if you aren’t satisfied with that, just say a = g^-1 and b = h^-1 for arbitrary g,h and you get the normal statement (for abelian)
paper flint
Thanks, that makes sense! I think a=(a^-1)^-1 is sufficient to justify every element has an inverse, and hence the statement holds for all a, b in G.
next obsidian
Yes
The reason I made a point to state it that way is the following
In a proof in representation theory you do something like a sum g over all g in G
Call this element like f for now
To complete a proof you get something like sum g^-1 over all g in G, and by this same logic you conclude this is equal to f
So knowing that enumerating over all inverses is the same as over all normal elements is important
paper flint
I see. Wait, there's another proposition where I think this could possibly make sense, mind taking a look?
next obsidian
Sure
paper flint
Let $G$ be a group and let $H={x^{-1}\vert x\in G}$. Show that $G=H$ as sets.
cloud walrusBOT
TedNowKaczynski:
next obsidian
Haha yeah this is the same thing
carmine fossil
What did you try?
paper flint
My argument yesterday was a stupid one where I said that the roles of x and x^(-1) are interchangeable, but I didn't understand the stuff explained to me :3
next obsidian
I mean
It’s the same thing at a high level but
For a low level proof just show both are subsets
H is a subset of G since G is closed under inversion
Then just say for any g in G, g = (g^-1)^-1
And that puts g in H
paper flint
I see. It makes much more sense now, I was suggested the same yesterday.
Is the symmetry argument sensible, by the way?
next obsidian
In a sense you saying x and x^-1 is right but
It’s not really interchangeable
really, it’s more like
I think this proposition
Makes the symmetry arguments rigorous
So I think it’s better to show this one by hand
carmine fossil
What do you mean by interchangeable?
paper flint
I mean, what we describe as the inverse is up to us, since every element of G is invertible
paper flint
Yeah, but by similar reasoning you could call x^-1 the inverse of x, so they are inverses of each other, and that's what I meant by interchangeable.
And they are guaranteed to be unique, I've proven that
paper flint
I see.
next obsidian
But saying they’re interchangeable isn’t quite right
You have to qualify that statement
Because changing x to be x^-1 WILL change its value
And can change things
BUT sometimes due to symmetricity it ends up not mattering
Such as doing the sum over all of them
By this problem summing over all inverses actually sums over the same set
But unless you have more reasoning as to how they’re “interchangeable” I wouldn’t say it’s a proof
In the sense that if u came up to me and I asked “why is this true” and you said that I’d ask you wtf you meant
paper flint
I see. I'll be more careful and try not coming up with crank proofs like this one XD
next obsidian
Then if you said well x = x^-1’s inverse you’re getting towards an actual proof
carmine fossil
Also,Be as precise as possible
paper flint
In the sense that if u came up to me and I asked “why is this true” and you said that I’d ask you wtf you meant
@next obsidian Met with that reaction yesterday XD
next obsidian
Agree with Drake
Your intuition on it was right
Altho idk if you came up with that afterwards haha
paper flint
Hmmm, can't be too hand-wavy, noted.
next obsidian
But saying they’re interchangeable doesn’t communicate that idea
paper flint
This was my gut instinct, I somehow just feel inverses have some weird symmetry, since every element is somehow paired to another element to generate the identity.
next obsidian
You’re right about that
But you have to poke at that harder and turn that into a proof
Between G and G^op no?
For the iso
Send g to g^-1
paper flint
I often don't know how to write what I think :p
next obsidian
Then you have to do multiplication backwards
Practice
With this level of group theory
You should feel comfortable like your proof as is
If you explained to ur brother or sister what a group is
That they’d hear the proof and be convinced
paper flint
Also, I'm currently using Gallian's text, and he has a lot of exercises based on groups under modulo operations. I've never done elementary number theory before, so should I take a look at it?
next obsidian
Just focus on doing every little detail for now until you get more experience
I don’t think you need it tbh
The only thing I think you really need is
Bezout’s lemma
paper flint
I often have difficulty proving closure of these groups
Here's a question I couldn't solve
next obsidian
I think that struggle will do the same thing as just learning number theory
If it’s number theory issues
paper flint
Hahaha, maybe you're right about that, but I think the emphasis here is abstract algebra, and I just want results from modular arithmetic to be freely used without being proven.
paper flint
The question I didn't get: An abstract algebra teacher intended to give a typist a list of nine integers that form a group under multiplication modulo 91. Instead, one of the nine integers was inadvertently left out, so that the list appeared as 1,9,16,22,53,74,79,81. Which integer was left out?
I mean, I've seen the group defined as Z_p, another one defined as U(n), but this just seems bizarre
paper flint
Sure
next obsidian
,w totient(91)
cloud walrusBOT
paper flint
Correct
next obsidian
So there’s 72 things coprime to 91
But in this case they said 9 elements and one was left out
So here you’re supposed to multiply those things until you get one that’s not in there
And that’s the missing one
paper flint
So this is just brute force?
next obsidian
Yes
They’re describing a subgroup of U(72)
Which you have 8 out of 9 elements of
sharp sonnet
you can just compute inverses
next obsidian
Right my bad
sharp sonnet
to find the missing one
paper flint
you can just compute inverses
@sharp sonnet Hmm, that makes sense too.
next obsidian
Yeah
You have 7 non-identity elements
So the 8th has to be the inverse of one of those
Much less work
At most you have 7 things to calculate
paper flint
Yes, I should figure that out. Thanks for the help. Uh there's another one I'm stuck on, mind taking a look at that one as well? :3
next obsidian
Sure
paper flint
This will take a few moments to type
Let $p$ and $q$ be distinct primes. Suppose that $H$ is a proper subset of the integers that is a group under addition that contains exactly three elements of the set ${p,p+q,pq,p^q,q^p}$. Determine which of these three are the elements in $H$.
cloud walrusBOT
TedNowKaczynski:
paper flint
There are options as well, should I type them?
paper flint
I mean, this doesn't even seem to be closed under addition for any choice of p and q
sharp sonnet
it will have more elements
next obsidian
I think it’s p, pq and p^q
Since you c an get the latter two from repeated addition of p
paper flint
Owww, that makes sense. But how about 2p? 3p? Clearly these should've been elements as well then?
next obsidian
It’s a bit hard to show this is the only one that works
Yes
It’s saying there’s H which is a subgroup
And of those 5 elements exactly 3 are in H
So there’ll be other stuff too
paper flint
AH, I see. That makes much more sense.
next obsidian
So to show this is the only option
If it weren’t you’d need either p + q or q^p in there as well
Ugh this is really annoying actually
Oh hmm okay so
You can’t have p
Since then you get pq and p^q by adding p over and over
paper flint
Yes, that makes sense.
next obsidian
So you have 3 elements of p + q, pq, q^p and p^q
paper flint
p+q and q^p seem irrelevant then
latent anvil
do we know that subsets of the integers are cyclic?
latent anvil
err wait we don't even need that I think
next obsidian
Since you can subtract off powers of pq
paper flint
I've only skimmed over the next chapters on subgroups, so that knowledge is not formally accessible at this stage.
next obsidian
And factor or something to get p^q and q^p from that I think
latent anvil
so if p is in there then <p> is contained in H
but <p> is maximal among subgroups
although it might be too hard to show that right now
next obsidian
Okay so I got it
paper flint
Maybe I could come back to this problem once I've studied about subgroups then
next obsidian
So suppose you have p + q, pq, and either p^q or q^p
Just do p^q for now
It will work for q^p as well
paper flint
Okay
next obsidian
wait shit
All I managed to show is q^q is in there
Fudge
Can I get q^p from that?
I was gonna say look at (p+q)^q
Subtract p^q
Then everything else besides the q^q term has a pq in it
sharp sonnet
to show that it can't be proper
next obsidian
Okay yes
sharp sonnet
like, the working case is just (p)
and all other cases will be non proper
due to bezout stuff
it's just casework
next obsidian
So everything in (p + q)^q = sum (n choose k)p^kq^n-k
So you can kill the p^q term using p^q since I assumed it’s in there
Besides q^q everything else has pq in it
So subtract off a multiple of pq to kill all those
Then you’re left with q^q in there which is coprime to p^q
So by Bezout the subgroup is everything
This works the same with q^p so it you’re left with p + q, p^q, q^p which is out by coprimeness
Or pq, p^q, q^p
Which is again out by coprime ess
paper flint
I'll have to first take a look at what Bezout's Lemma is, I just remember the name.
next obsidian
It says this
Given a, b integers
And n,k integers
Any number of the form na + kb is a multiple of gcd(a,b)
And that you can reach it
What it means here tho is
There exists integers n,k such that na + kb = 1 if and only if a and b are coprime
paper flint
Ah yes, I remember this, this was talked about in the prelim chapter of Gallian
sharp sonnet
you don't even need the only if
next obsidian
Yeah yeah
But I’d throw it in there
This is the most useful form usually for group theory purposes so I threw it in
sharp sonnet
and then you need to convince yourself that 1 generates all of Z
next obsidian
But anyway this says that you can use sums of a and b to get 1
And as Loch said, having 1 means the group is Z
sharp sonnet
also, you should remember bezout
paper flint
Oww, I see.
sharp sonnet
it's very important
next obsidian
That’s the one number theory thing I said I think u should know
Besides that I think you’re probably fine
paper flint
Noted, maybe I'll skim through an elementary number theory text and look at their proofs.
Thanks for the immense help. Much appreciated.
sharp sonnet
elementary number theory is a trivial corollary of abstract algebra
next obsidian
Abstract algebra which got built out of elementary number theory
paper flint
I don't really understand the relation between the two, is abstract algebra in a sense generalisation of all algebra?
next obsidian
Sorta
Algebra is like doing stuff over R
Using field axioms
Or I guess R(x)
It’s more like
Using basic group theory
You can prove a lot of basic number theory results
paper flint
Hmm, so abstract algebra has an even wider scope? Since group axioms are stronger ig
next obsidian
Like fermat’s little theorem
Group axioms are weaker
Any field is a group
More things are groups
paper flint
Owwww, but not all groups are fields
next obsidian
Fields require two operations
You need a notion of addition and multiplication among other things
paper flint
But in that event, aren't fields just a subclass of groups with an additional operation aligning with field axioms?
sharp sonnet
fields are way cooler than groups ok
paper flint
The name has strong categorical vibes lol
next obsidian
That’s like saying differentiable functions are a subset of all functions from R -> R
This is true, but also not that important
paper flint
Huh, groups seem to be interesting so far, I don't know much about fields.
chilly ocean
no
next obsidian
Since the extra stuff you impose is so strong
paper flint
Hmm, that makes sense.
next obsidian
The stuff you do are completely different
You’ll use group theory knowledge occasionally
But it feels much different
latent anvil
@next obsidian what do you mean by forgetful functor
next obsidian
Send field to it’s underlying group
latent anvil
additive?
latent anvil
hmm I was going to say a dumb thing
sharp sonnet
lol what
latent anvil
But no more
next obsidian
Bends over seductively to pick up a minion
chilly ocean
It was very inappropriate
next obsidian
I heard you’re not supposed to do that because it messes with logs
ionic drum
hey there, does anyone have a copy of humphrey's representation theory of semisimple lie algebras in the bgg category o?
paper flint
Have you tried libgen?
paper flint
I'm almost certain you can find it on Libgen
paper flint
Is $\bQ-{-1}$ closed under the operation $$ defined by $ab=a+b+ab$?
cloud walrusBOT
TedNowKaczynski:
paper flint
In this particular case, Q did not have an inverse for the element -1, so I removed -1 from Q; however, am I guaranteed closure now that I've changed the set itself?
carmine fossil
You have to show a*b is not -1 for any a and b in your domain
paper flint
Hmm, let me try that.
Seems like -1 was supposed to be its own inverse, but that amounts to cancellation by 0 on both sides and hence the -1 has to be excluded. It doesn't affect any other elements apparently.
Cuz $$a+b+ab=-1\implies a(1+b)=-(1+b)$$which gives $a=-1$ or $b=-1$, which are already excluded from my set now.
cloud walrusBOT
TedNowKaczynski:
carmine fossil
Yes
paper flint
Thanks for the help!
paper flint
Is it okay to assume this exponent-like behaviour of repeated group operations?
cloud walrusBOT
TedNowKaczynski:
paper flint
I see. Well yeah, the result does seem trivial, I just want to know what a proof here should look like.
urban acorn
yeah, it's very much fair grounds to assume that in a proof
it won't even be thought of as one of the assumptions
paper flint
Hmmm, okay, but nevertheless I'll try to demonstrate the a^(kn)=(a^n)^k in a more subtle way. Thanks for the input both of you.
paper flint
Let $G$ be an Abelian group and let $H={g\in G\vert g^2=e}$. Is it true that $H$ is a group?
cloud walrusBOT
TedNowKaczynski:
carmine fossil
Do you know when H will be a subgroup?
paper flint
Working with a concrete example makes me think it is, but I don't know about the general example.
I don't know about subgroups yet tbh
carmine fossil
A subgroup is a subset that is also a group
paper flint
I tried using G={1,-1,i,-i} as the example group here, then H={1,-1} and H is indeed a group.
I get that, but I don't know anything useful besides that.
I don't even know how to describe the closure of H
Hmm....
carmine fossil
What do you think closure is?
paper flint
Can you confirm if it is indeed a subgroup? I think I might have a justification then
paper flint
Okay, so for $a\in H$, it follows that $a^2=(g^2)^2=e$ for $g\in G$ such that $g^2=e$. For any iterations of group operation, I'll either get the identity or $a$ itself, so it is indeed closed.
cloud walrusBOT
TedNowKaczynski:
paper flint
Associativity, commutativity seem to trivially follow since G is Abelian.
carmine fossil
This is not sufficient
carmine fossil
Off track
carmine fossil
Do you understand closure?
paper flint
Yes.
carmine fossil
If a and b are elements in H,ab should be in H
paper flint
For all elements $a,b\in H$, $ab=(g_1)^2(g_2)^2$ for some $g_1,g_2\in G$, and since $g_1^2=g_2^2=e$, it follows that $ab=e$ for all $a,b\in H$.
cloud walrusBOT
TedNowKaczynski:
carmine fossil
Why is ab=(g1)^2(g2)^2?
paper flint
Because that's how H is defined?
carmine fossil
a is an element of H if a^2=e
paper flint
Yes, and I demonstrated that every such element does indeed give ab=e for all a,b in H
carmine fossil
a need not be g^2 for some g
carmine fossil
Yes
paper flint
Hmmm, I might have to think harder then. Let me handle this one on my own; if I'm still stuck I'll reach out for help.
Thanks for the help so far!
carmine fossil
np
paper flint
Okay here's something I got: For elements $a,b\in H$, $a$ and $b$ are their own inverses since $a^2=e$ and $b^2=e$. Thus, $a^{-1},b^{-1}\in H$ for all $a,b\in H$. Further, $$ab=(a^2a^{-1})(b^2b^{-1})$$ $$ab=(ea^{-1})(eb^{-1})$$ $$ab=a^{-1}b^{-1}\in H$$This proves closure of $H$ under the group operation. The existence of inverses has already been established, what remains to be proven is associativity.
cloud walrusBOT
TedNowKaczynski:
paper flint
@carmine fossil Sorry for the ping, could you take a look?
carmine fossil
Associativity is inherited from the group,but proof for the closure is not correct. The existence of inverse is fine
You should show (ab) satisfies the group condition
paper flint
Owww, that is correct.
Okay, let me figure that out.
Uhhh I cannot figure it out @carmine fossil
I can't figure out how a^(-1)b^(-1) belongs to H
carmine fossil
focuse on (ab)(ab)
paper flint
But we haven't established that ab is in H?
carmine fossil
Because abab=aabb=ee=e
carmine fossil
Yes
paper flint
Crap. That clears it. Thanks for the help!
carmine fossil
Because treating a and b as elements of G gives a.b=b.a
paper flint
Makes sense.
carmine fossil
And . On H is same except for the domain and range
paper flint
Yes, understood.
Does there not exist a non-Abelian group with n elements for all positive integers n?
I thought the symmetric group doesn't commute in general
paper flint
Owwww I seee, thanks for pointing it out!
carmine fossil
In general, Any group of order p (if p prime) is always commutative
stone fulcrum
Or even better the group of 1 element haha
paper flint
Yeah I thought about that one lol
hard kestrel
any easy way to find n such that f(x) will divide g(x) in [Z/nZ][x]
i saw a method and it was matching coefficients and solving equations to come to an n
any better direct method?
leaden finch
can someone help me a with latex
paper flint
Proposition: A group $G$ is Abelian iff $axb=cxd\implies ab=cd$ for $a,b,c,d,x\in G$.\Proof: We first prove the forward implication. Since G is Abelian, $$axb=cxd\implies xab=xcd.$$By cancellation, it follows that $$axb=cxd\implies ab=cd.$$We now prove the converse. From our hypothesis, it follows that $x=a^{-1}cxdb^{-1}=a^{-1}(abd^{-1})xdb^{-1}\implies x=bd^{-1}xdb^{-1}\implies (x)(bd^{-1})=(bd^{-1})(x)$, and thus it follows that $G$ is Abelian. This completes the proof.
cloud walrusBOT
TedNowKaczynski:
paper flint
My argument for converse seems shaky, could someone verify it?
chilly ocean
what do you mean by "from our hypothesis, it follows that x = a^-1 cxdb^-1"? the hypothesis is that "axb=cxd implies ab=cd", not "axb=cxd"
paper flint
Hmmm
What should I do then?
For the case where axb≠cxd, the implication is vacuously true, for the case where axb=cxd is true, ab=cd must necessarily be true. I just blended the two conditions together, although I have no idea what else could it be 🤷♂️
chilly ocean
here's another way to see the issue: what are a,b,c,d,x, and why is it the case that axb=cxd?
(i didn't follow what you said about if axb \neq cxd, the implication is vaculuos true)
paper flint
Mmmm a,b,c,d could be some elements such that ab=cd, while x could be any element?
Uh if I assume the implication axb=cxd implies ab=cd, then it could possibly be true under the circumstance that axb≠cxd, since that falsifies the hypothesis? I'm screwing up ig :3
chilly ocean
the flow of the argument should sound like "let a, b be arbitrary elements of G. let x, c, d be such and such. we show that axb = cxd. (insert why axb=cxd here.) then by the hypothesis it follows that ab = cd, which implies that ab=ba"
paper flint
Ohkay, I'll try that. Thanks for the help.
I think I should really drop abs algebra for now and work through an intro to proofs book instead :/
mint gulch
Hmm, I would advice learn a subject where you have to prove a lot instead of a intro to proofs book @paper flint
You could study a Discrete Mathematics book
paper flint
I'm taking a discrete maths course as well atm, but it's boringgg, maybe I'll take a brief look at an intro to proofs book, then continue struggling through abstract algebra and analysis.
Thanks for the suggestion though. Any Discrete Maths book you'd recommend?
next obsidian
I'm inclined to say if you enjoy group theory just keep doing it and getting better at proof writing ¯_(ツ)_/¯
deft plume
Yah the only way to get better at proof writing is to just keep writing proofs and with help and changes for mistakes, you’re sure to improve.
paper flint
Thanks, I've decided to stick with abstract algebra and learn a bit of proof-writing along the sidelines.
deft plume
I remember taking discrete maths was pretty fun and challenging, but I definitely benefited more from an intro to proofs book/course; on a personal standpoint.
Dude, abstract algebra is pretty awesome too tho. I feel like the algebra proofs will def help with proof writing improvement too
paper flint
I see. Which text did you consult for intro to proofs?
Yeah, as far as I've seen so far the proofs in abstract algebra aren't mind-blowing in any sense so far, they seem very coherent and understandable. I just need to know how to make mine rigorous and correct, and as has been suggested, I should keep improving.
deft plume
For intro to proofs, I used a book called “a transition to advanced mathematics” by Douglas smith, Maurice eggen, and Richard Andre.
paper flint
I was using "A Passage to Abstract Mathematics", smh, they sound like rip-offs of each other lol
Maybe any such text would do
I should just focus on the writing style and work through problems
deft plume
Everyone has their preferences ahaha. And for the writing style, everyone has their own stylistic variants
Just as long as the argument suffices you’re justification
paper flint
I like Tao's style in his Analysis text, I've adopted his way for induction based proofs so far.
deft plume
Your*
paper flint
Makes sense.
Thanks for the help, I'll definitely look at some intro to proofs text!
deft plume
Yep! If you’re interested at the text I sent too, you should know they have a chapter on algebra and a chapter on analysis as well
paper flint
Oh, that's nice! Libgen time
deft plume
Lol!! Look into 7th or 8th editions.
It isn’t too much but you should get algebra material from the basics of groups to the basics of rings, and for analysis you have completeness, Bolzano wierstrass, and monotone sequence stuff; which I think is good enough as exposure, at least for someone looking for a good intro proofs book with slightly more rigor.
paper flint
That totally is the book I need at the moment, thanks a lot for recommending it!
deft plume
No problem! Glad to hear a fellow mathematician also workin their way up to learning more maths.
paper flint
chrome hinge
Hey there! I wonder if someone could help me with a little problem i got from group theory?
Im trying to prove that $GL(\mathcal{C}^{n}) \cong \mathcal{C}^{} \times SL(\mathcal{C}^{n})$. The thing is that SL is a normal subgroup and $\mathcal{C}^{} $ is clearly isomorphic to the center of GL (the set of non-zero scalar times de identity). However, This last set doesnt intersect SL just in {e}, so im kinda confused how to show this...
cloud walrusBOT
Maikel:
chrome hinge
Here $C^{*}$ is the set of nonzero complex numbers
cloud walrusBOT
Maikel:
thorn delta
what is C^n supposed to be?
chilly ocean
I suppose he meant GL_n(C)?
thorn delta
@chrome hinge assuming you meant GL_n(C), you have an internal direct product.
G = H x N where H, N are normal, HnN = {e}, and HN = G.
chrome hinge
Hi. There are multiple notations for GL_n(C). [GL(n,C), GL(C^n), etc...]
Ya its not exactly an internal direct product
Since the sets you called dont intersect in {e}
chrome hinge
But in ${\lambda id : \lambda \text{is a root of the unit}}$
cloud walrusBOT
Maikel:
chrome hinge
Thats the interesction
If you take determinant of a matrix in that set, you'll get 1
However, i already solved it
thorn delta
ok yea thats true. what did you do?
chrome hinge
Well, i used several isomorphism theorems
well, just two, but multiple times
So basically, lets call the set i just defined $\Lambda^{}$, and lets define $\Lambda = {\lambda id : \lambda \in \mathcal{C}^{}}$
cloud walrusBOT
Maikel:
chrome hinge
Using the 2nd theorem you can see that $\Lambda / \Lambda^{*} \cong \Lambda$
Oh, what happened
cloud walrusBOT
Maikel:
chrome hinge
Nice
Sorry, you need to use 2nd and then 1st theorems to show that
But thats the idea
Then they intersect in {id} modulo an isomorphism
woven delta
@chrome hinge why does that show what you want to show?
chrome hinge
$\Lambda \cong \mathcal{C^{*}}$
cloud walrusBOT
Maikel:
woven delta
I agree that GL_n/SL_n is isomorphic to C*
But that doesn't give you that SL_nxC* is isomorphic to GL_n
chrome hinge
My point (of which im not 100% sure its correct) is that now i found that C* is isomorphic to Lambda which is also isomorphic to $\Lambda / \Lambda^{*}$
cloud walrusBOT
Maikel:
chrome hinge
What
Well. Then you can consider that the intersection between SL_n and C* is $\Lambda^{*}$
cloud walrusBOT
Maikel:
Compile Error! Click the 
reaction for details. (You may edit your message)
chrome hinge
And therefore the identity?
woven delta
The intersection is the nth roots of unity
cloud walrusBOT
Maikel:
chrome hinge
And that set is isomorphic to C*
I agree its kinda weird
But what basically is happening is like the set of nth roots of unity can be considered as the identity right?
woven delta
I'm not sure what you mean
chrome hinge
neither i
lol
Im confused about this
But
The intersection is the nth roots of unity
I thought i could make thats set the identity you know?
By taking the quotient group
And surprisingly the quotient group was isomorphic to C* !
chrome hinge
Well, it was for me
woven delta
Oh okay
chrome hinge
I didnt know that till today
woven delta
Okay here's another example to think of
SL_2(R)x R* is not isomorphic to GL_2(R)
Why is that?
(hint: count square roots of I in GL_2 and SL_2)
woven delta
I don't believe that the statement you made is true
Or if it is it seems to require some work
But yeah try to think about the example I gave above
chrome hinge
What ive found is that GL_n is isomorphic to SL_nxF* with x a semidirect product
woven delta
Have you?
I don't think I believe that either
Especially if you say F for a general field
chrome hinge
At any rate that statement wasnt mine, is an exercise from my algebra teacher
woven delta
Can you post the exercise?
chrome hinge
Its in spanish but i can translate the literal statement
woven delta
Sure
chrome hinge
Show that GL(C^n) is isomorphic to C*xSL(C^n)
woven delta
Okay maybe it's true for that case but it's not obvious from what you are writing
It's not true for the R case in even dimensions
chrome hinge
But what i actually did indeed prove is that the set of scalar matrices quotient the nth roots of unity times de identity is isomorphic to C*
That follows from the second and first theorems
woven delta
Sure
woven delta
Well you have to specify the intersection of what
woven delta
Actually what you're saying now sounds good
You do need to say more
But this is a good idea
chrome hinge
Oh, thanks :D
Yeah actually my big problem is the intersection thing
I want it to be the identity you know?
Then im done
woven delta
Lol
If you can show that the map from GL_n to GL_n given by mapping an element A to A^n is surjective I think you can formalize things
Essentially existence of nth root of invertible matrices
This is true but it's a bit hard to prove I think
Unless you've already done a decent amount of linear algebra
chrome hinge
What if i take the quotient of all three groups by Lambda*?
I can talk of a trivial intersection between the quotients of SL and the set of scalar matrices then
Would that help to prove the initial statement?
woven delta
Do you know like Jordan form?
chrome hinge
Yes
woven delta
Okay I think you can use Jordan form to prove existence of nth roots
Okay let's assume you have that actually
So we have the map GL_n to GL_n given by A to A^n is surjective
What about SL_n?
chrome hinge
Oof
woven delta
Is the image of SL_n exactly SL_n?
chrome hinge
Yes
chrome hinge
Is the image of SL_n exactly SL_n?
@woven delta I actually know its contained in SL_n
woven delta
Okay maybe this is fine actually
chrome hinge
But i really dont get how this map could help me
chrome hinge
Np man thanks for your help
queen vine
Am I overlooking something or can't you just simply define the isomorphism explicitly as $f:GL(\mathbb{C}^n)\to SL(\mathbb{C}^n)\times\mathbb{C}^*, M\mapsto (\det(M)^{-1}M,\det(M))$
cloud walrusBOT
leoli1:
woven delta
Maybe it is that simple
Can you show the map is surjective?
I guess the reason why this doesn't work for R is because in the dimension 2n case this map is not surjective, but I guess it is surjective here
Okay sorry @chrome hinge I guess it is pretty easy
queen vine
Yeah it is surjective because $\mathbb{C}$ contains $n$-th roots. Given a pair $(N,x)\in SL(\mathbb{C}^n)\times\mathbb{C}^*)$, take $M=\sqrt[n]{x}N$ as a preimage.
cloud walrusBOT
leoli1:
woven delta
Yeah
chrome hinge
So basically we were out of the theorems hypothesis
It was way easier to find directly the isomorphism
Thanks leoli1
scarlet estuary
consider the set of integers Z with operation x.y = x * y + 1
where the * here is conventional multiplication
then (3.4)5 = (34 + 1)*5 + 1 = 13 * 5 + 1 = 66, while
3(4.5) = 3(4*5 + 1) + 1 = 3*21 + 1 = 64
so this isnt associative
but its certainly commutative, as x.y = x*y + 1 = y*x + 1 = y.x
since * of real numbers is commutative.
sure, that works too
stone fulcrum
How does one make sense of x.y.z in this case?
scarlet estuary
one doesnt
stone fulcrum
I mean, you'll still need the parenthesis
scarlet estuary
hence my point
stone fulcrum
Oh haha I've answered my own question ignore me
scarlet estuary
formally a binary operation is a function S times S -> S
so it only REALLY makes sense to write either ((a, b), c) or (a, (b, c))
but we can instead represent these bracket pairs as elements joined by an operation symbol *
but only when that doesnt introduce ambiguity
a * b * c is absolutely meaningless unless you assign a canonical interpretation
you COULD assign the interpretation (a * b) * c or a * (b * c)
but this generally doeswnt make much sense to do unless these coincide
(i.e. unless you have associativity)
we do sometimes write A implies B implies C
and this is usually interpreted as A implies (B implies C)
[since implication is not associative]
but i'd generally discourage this
[if you're wondering why this convention is taken, it's because it's notationally handy when https://en.wikipedia.org/wiki/Currying]
quaint ivy
huh, I never would've thought of that.
chrome hinge
Am I overlooking something or can't you just simply define the isomorphism explicitly as $f:GL(\mathbb{C}^n)\to SL(\mathbb{C}^n)\times\mathbb{C}^*, M\mapsto (\det(M)^{-1}M,\det(M))$
@queen vine hey man just realized this is not well defined...
cloud walrusBOT
Maikel:
chrome hinge
It should be defined as $f:GL(\mathbb{C}^n)\to SL(\mathbb{C}^n)\times\mathbb{C}^, M\mapsto (\det(M)^{-1/n}M,\det(M)^{1/n})$ so that the map of an element of $GL(\mathbb{C}^n)$ arrives indeed to $SL(\mathbb{C}^n)\times\mathbb{C}^$
cloud walrusBOT
Maikel:
chrome hinge
I took the nth root that has the least argument
obsidian path
I need to list all the subgroups, I've completely lost faith in my understanding of subgroups. All I'm sure of is using Lagrange's theorem, help?
chilly ocean
all subgroups, up to isomorphism, or actually all of them?
well, i guess even all of them, it is not that annoying, the only subgroups that there are multiple copies of are Z2 i think
obsidian path
explain pls?
chilly ocean
up to isomorphism: Z1, Z2, Z3, Z4, Z5, Z6, Z10, Z12, Z15, Z20, Z30, Z60, or u could list them out by the actual subgroups
nvm i thought there are multiple copies of a particular order subgroup, but i dont think this actually happens
obsidian path
are those just the divisors of 60?
chilly ocean
i was thinking something like V4, where there are multiple subgroups of order 2
yah
obsidian path
right.. oh my god
chilly ocean
note: all subgroups of a cyclic group are themselves cyclic
obsidian path
true
so.. that's it?
wow
gonna kms i spent all day thinking about this
I haven't been confident in my understanding of groups
speaking of isomorphism, could you tell me what it means exactly?
just to be sure of the definition
chilly ocean
if u havent seen the definition then it's probably not worth worrying about
but an isomorphism is a bijective homomorphism
(in the context of groups)
oh wait
Z60 is the additive group, right, not the multiplicative group?
obsidian path
yes
chilly ocean
ok yeah then that should be right
obsidian path
I think it'd be notated with Z60/{0}
if it was multiplication
ok I have a question though
any element mod 60 will return that element right?
nevermind for some reason I thought that made it the inverse of itself but..
I'm just gonna reread some definitions and take a step back lol
tyvm tho
chilly ocean
mhmm
distant rain
Hi. I come across somewhere on maths stackexchange if a a commutative ring $A$ is such that every element is either a unit or zero divisor, then all prime ideals are maximal. But the proof isn’t given. Was raking my mind for hours to no avail. Any hint? Also i come across a stronger condition on a ring for such conclusion to hold true. In fact this condition implies the condition in my original question namely every element $a$ has an integer $n>1$ such that $a^n=a$. Given this stronger condition, it is easier to prove the desired conclusion. But I wonder if this condition is in fact equivalent to the earlier condition? Thanks in advance.
cloud walrusBOT
τφκ:
solemn rain
any nonzero element in a finite ring is either a unit or a zero divisor
queen vine
@chrome hinge You are right, but now we have the problem that the map is not a homomorphism anymore, for $n=2$ and $M=diag(-1,1,1,1)$ we have $f(MM)\ne f(M)f(M)$.
cloud walrusBOT
leoli1:
queen vine
I am not even sure anymore if the statement is correct. It is clear that we have $GL(\mathbb{C}^n)\cong SL(\mathbb{C}^n)\rtimes\mathbb{C}^*$, however the semi-direct product might not be trivial.
cloud walrusBOT
leoli1:
next obsidian
Hi. I come across somewhere on maths stackexchange if a a commutative ring $A$ is such that every element is either a unit or zero divisor, then all prime ideals are maximal. But the proof isn’t given. Was raking my mind for hours to no avail. Any hint? Also i come across a stronger condition on a ring for such conclusion to hold true. In fact this condition implies the condition in my original question namely every element $a$ has an integer $n>1$ such that $a^n=a$. Given this stronger condition, it is easier to prove the desired conclusion. But I wonder if this condition is in fact equivalent to the earlier condition? Thanks in advance.
@distant rain the two conditions are not equivalent. Consider Q x Q, every element is a unit (if both entries are non-zero) or a zero divisors (if either entry is zero). Note however there is no n > 1 such that (1/2,0)^n = (1/2,0)
For the proof of the other thing, where if every element is either a unit or a zero divisor, I have a proof when you assume the ring has finitely many prime ideals. I’ll think more for how to prove it in general, but here’s the proof in this case.
Note that if every element is a zero divisor or a unit, then prime ideals consist solely of zero divisors. One can show that the set of zero divisors is equal to the union of all minimal primes https://mathoverflow.net/questions/20826/when-is-the-set-of-zero-divisors-equal-to-the-union-of-the-minimal-primes-in-a-r
Now assume p is a prime ideal, then p consisting of zero divisors means it’s a subset of the set of zero divisors which is the union of all minimal primes. Since I assumed there are finitely many minimal primes, one can apply the prime avoidance lemma so that p is a subset of a minimal prime q, but by q’s minimality we see that p = q. Thus p is minimal.
What we have shown is that every prime ideal is minimal, which actually is equivalent to showing that every prime ideal is maximal, so we are done.
The hypothesis that our ring has finitely many prime ideals isn’t a terribly strong one either, for example every Noetherian ring has finitely many minimal prime ideals.
Wait, shit I need to assume the ring is reduced to apply this lmfao
uncut cove
can someone reference me or advice me , i dont get how this is "easy to see"
carmine fossil
Try taking some examples
carmine fossil
I guess sigma 1 is a diagonal matrix and for sigma you just add the appropriate number of zeroes to make it a rectangle
uncut cove
my bad
Sigma is a diagonal matrix with eigenvalues
with all >0
Try taking some examples
@carmine fossil Am i suppose to try out diagonalizing a few examples?
carmine fossil
Try some examples of sigma and see if this is true
chilly ocean
bruh what even is that notation ?
obsidian path
I found the order of A and B, but not sure how to do AB. Trying induction but its difficult, can anyone help?
chilly ocean
did you mutiply em?
next obsidian
If you multiply them I’m pretty sure you get an upper diagonal matrix
At least this is the standard thing
And then the thing in the upper left just keeps getting bigger and never goes to 0
paper flint
Proposition: A group G with |G|<=5 is necessarily Abelian.
I tried thinking about it for a while and started enumerating case-by-case, but found it to be too tedious. Can anyone suggest a neater way of proving this?
Any hints are appreciated. Feel free to ping me.
Maybe I should state the proposition as an implication.
Proposition: If G is a group such that |G|<=5, then G is Abelian.
Contrapositive: If G is not Abelian, then |G|>5.
I can't seem to make much progress with the contrapositive either.
sharp sonnet
just list all groups
paper flint
Going case-by-case is the only way to go here?
sharp sonnet
i don't know, but there aren't too many
paper flint
I did it up to 3
sharp sonnet
the smallest non abelian group is S_6
paper flint
Might as well do 4 and 5 then
Thanks for the input; yeah, the instructor did quote S_6(he refers to it as S_3) as a non-Abelian group with 6 elements.
paper flint
Haven't done Lagrange's Theorem yet, I did take a sneak peak and saw this result though.
sharp sonnet
for order 4 there are only 2 possibilities
sharp sonnet
there is Z_4
cyclic group of order 4
and the other is the klein four group
if you don't know that, you can construct it with a cayley table
paper flint
Haven't seen these come up yet :3 The sections on subgroup will now start.
Sounds good, should be able to do that.
But
sharp sonnet
you should know what a cyclic group is?
paper flint
Constructing a Cayley table constitues a proof?
No, they haven't been introduced yet.
sharp sonnet
eh, not really, but
for small orders you will see that there is only one(in this case two) ways to construct a cayley table
that could be turned into a rigorous proof
but it would be like
paper flint
Hmm, so it can just be used as a side visual and then write the arguments rigorously.
sharp sonnet
"if we assume that the product of those two elements is this element, then we reach some contradiction"
and that like a few times
paper flint
Ye, that's what the instructor did, I just thought he was doing things the easy way for understandability.
sharp sonnet
there are only finitely many possibilities how you can turn 4 elements into a group
and all but 2 end in contradiciton
i think in this case writing out the proof does not help understanding
set up the table
make some small calculations on the side
i.e. convince yourself that there are only two possibilities
paper flint
Lol okay, I'll build the Cayley table for a group with 4 elements.
I did like the proof argument too, though.
chilly ocean
yeah, I think a simple script would do the job.
paper flint
Once he proved that {e,a,b} had to be distinct elements for a group of 4 elements, then proving the fourth element is ab=ba seemed like a reasonable process.
He went bit by bit, explaining why a^2!=e, ab!=a, ab!=b and so on
Wait, he did prove a^2=e Maybe I should build the table
sharp sonnet
that's the klein four group then
it has 3 elements besides the identity which are all self inverse
or geometrically, the symmetry group of a rectangle
non-square
(it's also the smallest non-cyclic group)
paper flint
Okay building a Cayley table for {e,a,b,c}, each element distinct has given me a bit more of data than I expected. For example the second row consists of {a,a^2,ba,ca}
Symmetry group of rectangle-that's enlightening
I had only seen the symmetry group for square before
But I can see the V,H,D,D' can be the elements of this group 
Uh Loch I think I'm more confused by the table
cloud walrusBOT
TedNowKaczynski:
paper flint
Uhkay so here's my ugly looking Cayley table
I have to somehow impose the conditions of closure, associativity and invertibility upon the elements in the bottom-left 3x3 grid.
sharp sonnet
the thing is
all of the elements in that table
have to be either e, a, b or c
and each line/row can have each element at most once
(it's kinda like a sudoku)
so you go to 2nd row, look at a^2
it can't be a
so it's either b or c
sharp sonnet
what is distinct?
sharp sonnet
your group has only 4 elements, a^2 is either a, b, c or e
sharp sonnet
it can't be a, because then a would be the identity
and the identity is unique
so it's either b, c or e
paper flint
Makes sense.
sharp sonnet
and the more you fill out the table, the more choices are 'forced' on you
paper flint
Now ba!=a and ba!=b
sharp sonnet
because it's like a sudoku
paper flint
Yepp, it does seem like one now
sharp sonnet
and at some point you will either end up with a valid group structure
or some contradiction
it's a bit tedious tbh
paper flint
Yes, but it's a good exercise. I should do this. Thank you!
sharp sonnet
it's a very abstract way to think about groups
paper flint
I liked the rectangle analogy a lot
That makes things so clear and visual
Isn't there a way to establish some 1-1 correspondence between the two structures and establish being Abelian as equivalent?
sharp sonnet
i mean yeah, usually we want to think of groups as symmetries of some object
you can show that the klein four group (the symmetry group of the rectangle) is abelian that way
but not that there does not exist a group of order 4 that is not abelian
like, i don't think that there is a good reason for this