why not
#groups-rings-fields
viscid pewter
chilly ocean
well what is it defined as?
viscid pewter
so <a^d> is defined as the subgroup generated by a^d where a is some element in D2n such that it fulfils the presentation of D2n (that D2n = <a, x | a^n = x^2 = e, xax^-1 = a^-1 >) and d|n where n is |D2n|
chilly ocean
what's d?
viscid pewter
oh
d|n means d divides n
d is a factor of n
i don't actually see where a is used differently
viscid pewter
like a is defined as an element with the properties given in the presentation
and then used that way consistently
no problems
chilly ocean
in the "special cases" under it, they talk about how (or at least i think) a is used differently
viscid pewter
no they don't
it's just that a fulfils all the properties of the dihedral group and then some other stuff also
chilly ocean
that's what i figured yeah
viscid pewter
i mean symbols can mean different things based on the context
viscid pewter
exactly what it is
in the presentation
an element such that x^2 = e and xax-1 = a-1
a flip, essentially
chilly ocean
aah gotcha
glad juniper
@stone fulcrum "The polynomial that is 0 for all x is the zero vector in the space" As an aside, evaluating to zero everywhere is not the same thing as being the zero polynomial in a general polynomial ring. See, for example https://math.stackexchange.com/a/64053
stone fulcrum
Okay, neat haha. I'll have to give this a slow read
thorn delta
unless im misunderstanding, Z^2/Z is an abuse of notation. What (im guessing?) you really mean is Z x Z/Z x {0} or Z x Z/{0} x Z. WLOG lets work with the first one. Then elements look like (c_1, c_2) + (Z, 0). In other words, two elements of Z x Z are "the same" in ZxZ/Zx{e} if they differ by an integer in the first coordinate and differ by 0 in the second coordinate.
you can only quotient out by subgroups of a group
it doesn't make any sense otherwise
again, when you use "Z" in the quotient, you should really be using a normal subgroup of Z^2. You could have a subgroup isomorphic to Z but its quotient with Z^2 is not iso to Z
viscid pewter
you can just map (c_1 + a_1, c_2 + 0) to (c_1 + a_1)?
sharp sonnet
the other way around
viscid pewter
i mean it's an isomorphism, is the point
i was saying that's why it's isomorphic to Z
viscid pewter
oh that way round
viscid pewter
yeah
sharp sonnet
(c_1 + a_1, c_2) and (c_1, c_2) are equal for all a_1 in the integers
by equal i mean in the same equivalence class
viscid pewter
same coset, sure
sharp sonnet
the equivalence classes look like {(x_1, x_2) | x_1 in Z}
so yes, you are right
yes, that is kinda confusing
you have to make sure n >= m
and "fill up" Z^m with x {0} sufficiently often
you have to, to make sense of that
otherwise it's not a subgroup of Z^n
(technically, you can place the 0s anywhere but it makes no difference modulo isomorphism)
obsidian path
hints for going about:
Which abelian groups are isomorphic to subgroups of S6?
misty wind
I'm watching a series of lectures on linear algebra and the is lecturer making an "analogy" about semirings and how the boolean algebra of semi-rings is analogous to the algebra of matrix multiplication (as viewed as matrices having their own proper vector space structure).....
I'm looking for a crash course on the boolean algebra of semi-rings.....cuz i wanna keep it pertinent to linear algebra.....any youtube videos this chat might recommend?
^that's what i know
These are what i know of the ring axioms (they're like a group with extra steps)
wikipedia and a cursory youtube search assures me this is what a semi-ring is (aka a semirig)
vital quail
did you... make a table in google docs
misty wind
yes
misty wind
vital quail
the government of france wants to know your location
misty wind
??
vital quail
french math pedagogy is similar in the sense that they are quite thorough and linear in the process of teaching students algebraic structures
misty wind
can they provide me an explanation on semi ring, cuz i can whistle the marseillaise
vital quail
starting with magmas, then semigroups, etc
in any case a semiring is just a rig
a ring without (n)egatives
e.g. N
misty wind
so its a semirig, got it
vital quail
no, it's a rig
golden pasture
thonk
thorn delta
monoid under addition, semigroup under multipilcation
golden pasture
people actually learn about magmas and shit
vital quail
@thorn delta no, multiplicative monoid
misty wind
so, when do the ones and zeroes come in to play? how does this relate to bool?
vital quail
it's literally just a ring except the additive group is now a monoid
thorn delta
their def doesn't have 1 
misty wind
their def doesn't have 1 <:thonk:406575732563902485>
are my charts wrong?
vital quail
@misty wind wdym
thorn delta
1 is overrated
vital quail
well your charts should reflect that a semiring is literally a ring except the additive group is a monoid
that is, you don't require additive inverses
golden pasture
so, when do the ones and zeroes come in to play? how does this relate to bool?
0 and 1 is just like "smallest" and "biggest" element
vital quail
misty wind
i deleted the inverses
golden pasture
ik jacobson gives a decent intro to boolean alg
but idk about why semirings
ig in some sense it is a semiring?
misty wind
ig in some sense it is a semiring?
yes this lecturer assures me it is.
golden pasture
very garbage to think of it as a semiring tbh
thinking of it as some sort of lattice is better
misty wind
boolean algebra and matrices are semirings or somefing
misty wind
idk what monoids or lattices are.....
golden pasture
https://en.m.wikipedia.org/wiki/Boolean_algebra_(structure)
have very nice diagram
im lazy to type so here is a copypasted example
The power set (set of all subsets) of any given nonempty set S forms a Boolean algebra, an algebra of sets, with the two operations ∨ := ∪ (union) and ∧ := ∩ (intersection). The smallest element 0 is the empty set and the largest element 1 is the set S itself.
paper flint
If $S$ is a subset of a group, does $\langle S\rangle$ refer to the union of sets generated by each element in $S$?
cloud walrusBOT
Ted
thorn delta
no, it refers to the intersection of all subgroups containing S
paper flint
Uh, I actually need to prove that.
Here's the problem statement: Let $S$ be a subset of a group, and let $H$ be the intersection of all subgroups of $G$ that contain $S$. Prove that $\langle S\rangle=H$.
cloud walrusBOT
Ted
chilly ocean
I guess I am not sure what you mean "union of sets generated by S"
thorn delta
instead of guessing about what <S> means, can't u just look at how ur prof defines it? Its probably something like <S> = {all products of elements of S and their inverses}
paper flint
Each element in S may or may not have finite order. I was guessing <S> then is just the set containing all elements in S, their multiples and products.
paper flint
instead of guessing about what <S> means, can't u just look at how ur prof defin...
I'm self-studying and only have access to textbooks. This appears before the chapter on cyclic groups, I may get more insights after I go through that chapter.
thorn delta
oh then the book should have defined it somewhere. Anyhow, the definition you gave here
I was guessing <S> then is just the set containing all elements in S, their multiples and products.
is pretty safe
paper flint
Ah, okay.
Why can't the intersection possibly contain more elements than in <S>?
I can show <S> is a subset of H, but how do I go other way round?
thorn delta
if x is in H, x belongs to every subgroup of G containing S....
paper flint
Oh nice lol. Thanks for the help!
thorn delta
npnp
paper flint
Okay, so I think I'm a bit confused for two reasons:
- The proposition <S>=H seems to fail when S is empty, since <S> would be empty as well while H would be the trivial subgroup.
- I think <S> may not necessarily contain identity, especially when the order of elements is infinite. H being the intersection of subgroups does contain the identity.
If <S> does indeed contain the identity, the issues can be resolved.
thorn delta
yes, <S> always contains the identity. In particular, the subgroup generated by the empty set is the trivial subgroup
paper flint
Oh, I didn't know about either. Thank you!
thorn delta
npnp
paper flint
Is <S> a subgroup as well?(or could you provide a reference where I can look up for a precise definition of <S>?)
If I'm getting it right, $$\langle S\rangle=\bigcup\langle a\rangle$$for all $a\in S$?
Naah that doesn't work
thorn delta
<S> is a subgroup, yes. <S> is usually defined to be the intersection of all subgroups containing S, although i guess your book defines it differently. In general, the intersection of subgroups is a subgroup, so <S> is a subgroup
paper flint
Owww nice! Thanks a bunch.
I really appreciate helping me out so many times haha, I wish Gallian did a bit more of teaching.
thorn delta
npnp, happy to help
as a remark, if x is an element of G, then <x> = <{x}>, so <S> coincides with the notation we usually use to denote cyclic subgroups
paper flint
I see. That makes sense.
thorn delta
you just have to show that E is nonempty, closed under addition and taking inverses
thorn delta
right
paper flint
Looks good.
thorn delta
why is n in E?
leaden finch
is E known as even or it just a random letter?
paper flint
Yeah, Even integers may be a sensible interpretation, although I've seen 2Z as more standard notation.
You basically choose two arbitrary elements $p$ and $q$ from the given set $E$, and then show that $pq\in E$. Since $p,q\in E$, it follows that for some $a,b\in\bZ$, $p=2a$ and $q=2b$. Hence, $$pq=2a+2b=2(a+b)=2m$$for some $m\in\bZ$. Since $2m\in E$, this proves closure.
cloud walrusBOT
Ted
thorn delta
Ted's got it. Personally, i would not mix these * and + operations together. I know the definition of group picks a symbol for the binary operation, but when ur proving things, you don't have to say things like "a*b = a+b." If you know your operation is addition, just write things with +.
paper flint
Hehehe, I was just trying to keep up with Sunshine's notation.
thorn delta
also, if you're being nit-picky, on the last part, you would say "2m for m = a+b \in Z"
but everyone gets the point
thorn delta
just by putting math in dollar signs. You don't need tick marks or whatever you added to make the code block. There is nothing really to prove for associativity. You don't have a subgroup test or anything? In general, you just have to show 3 things for a subset of a group to be a subgroup:
- nonempty
- closed under the operations (addition in this case)
- closed under taking inverses
leaden finch
is closure the same as nonemty?
thorn delta
nope. a set S is nonempty if there is x such that x is in S.
i.e. S contains a positive (nonzero) number of elements
for example, a proof that the set {1, 2, 3} is nonempty could be "since 3 is in {1,2,3}, the set {1,2,3} is nonempty."
paper flint
Does it make sense to define $\langle S\rangle$ as the smallest subgroup containing $S$? Because in that case proving $\langle S\rangle=H$ is trivial.
cloud walrusBOT
Ted
thorn delta
yes
carmine fossil
How was <S> defined in your text book?
paper flint
It isn't defined at all, smh at Gallian.
carmine fossil
The smallest subgroup definition is what I am familiar with
paper flint
The smallest subgroup definition is what I am familiar with
Oww I kinda made it up on the basis of what kx has told me so far.
thorn delta
in previous convos, we used <S> to mean intersection of all subgroups containing S and set of all products of elements of S and their inverses
err, H was taken to be the intersection of all subgroups containing S, while <S> was the subgroup corresponding to the latter def
paper flint
Correct.
Uh oh.
Okay but this should work. Once I prove <S> is a subset of H, <S> being one of the constituents of the intersection would force H=<S> I guess?
thorn delta
yea pretty much
paper flint
Nice. There's a part (b) to this proposition. I need to prove that $$\langle S\rangle={s_{1}^{n_1}s_{2}^{n_2}\cdots s_m^{n_m}\mid m\geq 1, s_i\in S, n_i\in\bZ}$$
cloud walrusBOT
Ted
paper flint
This again seems to be synonymous with the definition itself.
I don't know what there is to prove.
thorn delta
yea, that was the definition for <S> we agreed on 
paper flint
thorn delta
maybe Gallian uses <S> to mean "smallest subgroup containing S?"
paper flint
Proof: The proposition is true by definition
paper flint
maybe Gallian uses <S> to mean "smallest subgroup containing S?"
I'll look into the subsequent chapters if he defines it at all.
thorn delta
anyway, no need to dwell on it. to show that the intersection of all subgroups containing S equals the set of all possible products of elements of S is probably the most difficult thing you can prove about <S>, and since we took that to be the def earlier, you've already done this
sharp sonnet
It isn't defined at all, smh at Gallian.
lmao what, is this a real book
<S> is just the image of the unique homomorphism from the free group on S into G
paper flint
lmao what, is this a real book
Hehehe, Gallian is actually a good book but I guess he leaves a bit too much to exercises.
thorn delta
i looked up gallian and checked the notation index at the beginning of the book. The page corresponding to <S> is the page with your exercise and there is indeed no definition 
paper flint
thorn delta
wait i actually found it!!! page 66
sharp sonnet
this is still kinda bad
in the sense that it is not a rigorous definition
what does "smallest" mean?
thorn delta
it just means minimial in the set of all subgroups containing S?
sharp sonnet
minimal with respect to what?
golden pasture
you need to show one exists and is unique
sharp sonnet
the answer is set inclusion
but that's not clear from the definition given (for a beginner)
thorn delta
uniqueness should be pretty clear, but ig existence is not really clear under this def.
golden pasture
the direct construction makes it clear
not too sure how to see existence and uniqueness without constructing it ngl
sharp sonnet
you intersect all the groups that have S as a subset
thorn delta
well, you use that the intersection of all subgroups containing S is a lower bound
sharp sonnet
show it's a group still
golden pasture
ah right
sharp sonnet
it's unique by def
thorn delta
and then by that point you might as well have taken it to be the def 
golden pasture
"intersect all subgroups containing S" sounds cuter
sharp sonnet
indeed
paper flint
minimal with respect to what?
Can I say <S> is a subgroup and it is minimal in the sense that of all the subgroups containing S, <S> has the smallest cardinality(I just realised it wouldn't make sense if the subgroups or S itself were to be infinite).
sharp sonnet
yes, that's what i meant
a beginner would maybe think it's smallest with respect to cardinality, but that breaks once you have infinitely many elements in it
hence it's minimal with respect to $\subseteq$
cloud walrusBOT
Lochverstärker
paper flint
True.
So the sensible definition is to think of it as the intersection of subgroups containing S?
sharp sonnet
imo yes
or the explicit one that you did above
with the explicit one, you will need a lemma for uniqueness
with the intersection one you will need to show that the explicit one is actually that
paper flint
Ohh, I see.
sharp sonnet
<S> is just the image of the unique homomorphism from the free group on S into G
thinking about it i actually like this definition now
sharp sonnet
simple idea
take a set
its a natural question to want to turn this into a group
thats the free group
paper flint
How exactly do we turn it into a group in the absence of an operation?
sharp sonnet
we define one
well, the general idea is
for every element in x in the set
x^(-1) also has to be in the set
and all the finite product of all the elements
golden pasture
wait isnt that the construction
sharp sonnet
and some of them can be made shorter
ye
you can also define it categorically
and dont worry about its existence
golden pasture
sharp sonnet
but the construction gives a good idea
golden pasture
ye
sharp sonnet
anyways, you can shorten some formal strings, like xx^-1
and in the end you get a set of "formal strings"
and thats your group
tbh thats one of the good things in aluffi
that he does free group pretty early
free constructions in general are nice
free modules are very important
paper flint
Hmmm, so if I have some finite set, say of cardinality n, then can I just create a free group from that set using a group of order n by relabelling the elements?
@sharp sonnet
sharp sonnet
nah, you will get infinitely many elements in the group
if you care, you should first think about a free group on a single element
lets say {x}
how do you turn this into a group?
well, x^(-1) has to be in it
and arbitrary finite products
so x^n
and (x^(-1))^n
for arbitrary n
and ofc arbitrary products that include both of those
so things like x^42(x^(-1))^3
but even if you know nothing about those elements, you can simplify those
and always arrive at an element of the form x^n or x^(-n)
so the free group on a single element is the cyclic group on 1 element
or in other words, its (Z, +)
other free groups will only be bigger
(except the free group on the empty set, which is the trivial group)
for more details, i think https://en.wikipedia.org/wiki/Free_group#Construction is quite understandable
paper flint
Owww nice, will take a look.
sharp sonnet
tbh if you care about free constructions
paper flint
Also, can I turn any given set into a group with a suitable choice of operation, i.e., can I create groups of arbitrary orders?
Hmmm I actually can, nvm
sharp sonnet
there are groups of arbitrary orders
sharp sonnet
so you can map them any set to a group of the same order
but the thing here is
this assumes relations between the elements
i.e., sure you can map the set {x} to the trivial group
but this assumes x is the identity
with the free construction you don't assume anything
ye, there is weird stuff
the free group on 3 elements is subgroup of the free group on 2 elements
paper flint
with the free construction you don't assume anything
Owwwwww
sharp sonnet
like if i give you an element of a group, say x
and you don't know what group its from
you only know that you can compose it finitely many times with itself
and do the same with the inverse
(and any other group element)
but you don't know about any "simplifications" inherent to the group
that's what a free construction is
paper flint
Makes sense.
It'll take me a while to get to this free group stuff, Gallian covers it under special topics under "Generators and Relations".
sharp sonnet
it's kinda important in algebraic topology
paper flint
That stuff is scary. 
light tusk
they're like... almost almost vector spaces. super familiar, almost almost
golden pasture
this is why jacobson gud he intro free group early
chilly ocean
dummit and foote: "free groups? what are those?"
sharp sonnet
just read aluffi
paper flint
Wait, is the free group generated by a single element isomorphic to (Z,+)?
thorn delta
ya
paper flint
Interesting.
next obsidian
It has one generator with no relations!
But the free group on two is not just Z (+) Z, since you have relations
Or is it...
Wait I need to think now Jesus
carmine fossil
Is a free abelian group not free?(because ab=ba is a relation)
next obsidian
But that exists in the free group on 1 generator
thorn delta
its free in the sense that its free in in AbGrp
paper flint
What's the difference between a free abelian group of rank≥2 and the free group of rank 1? Aren't they isomorphic?
sharp sonnet
free groups on more than 2 generators arent abelian
next obsidian
^
sharp sonnet
or on 2
next obsidian
take two symbols x,y then xy ≠ yx
The free abelian group is the abelianization of the free group
Or you can define it directly
sharp sonnet
just quotient by the commutator
next obsidian
That’s what the abelianization is...
sharp sonnet
my point was "define it directly" is probably easier to understand
paper flint
I'm certainly lacking the background to understand Abelianisation, I did understand the idea of free groups though.
next obsidian
Abelianization is just G/G’
You quotient by the commutator and by the universal property of quotients it has the property that every map G -> A for A abelian factors through G/G’ uniquely
It’s just the “most general” way to get an approximation of G which is abelian
And you make it by saying gh = hg, aka ghg^-1h^-1 = e for all g,h
Idk how important this is for a first pass all things considered
paper flint
And you make it by saying gh = hg, aka ghg^-1h^-1 = e for all g,h
This is the only bit I actually understood here, other stuff is Monka for me.
next obsidian
If you like thinking via universal properties it can be nice, else it’s something you can just come back to
Do you know what a quotient is?
Of groups
paper flint
Nope
paper flint
Still swimming in subgroups and cyclicity before Lagrange lmao, I'm too slow.
next obsidian
Learn about them and their universal property and you shall be able to understand haha
F you loch
paper flint
Okay, so if a free group is perceived as concatenation of letters, than Abelianisation is to do away with permutations, i.e., only the order of elements in the expression matters?
next obsidian
sharp sonnet
fair
paper flint
Lmao I couldn't read XD
next obsidian
No abelianization says that the order doesn’t matter
All that matters is the number of each element present
paper flint
I meant order as in the number of occurences. Confusing times haha.
next obsidian
Oh lol
paper flint
Basically all strings are unique upto rearrangement?(I still don't understand if I'm using "unique up to" correctly)
next obsidian
Yeah the idea is you use the same symbols but you’ve added in relations to say that things commute
And somehow this works by quotient magic
I don’t really understand your sentence
Neither what you mean, nor the usage of unique up to
paper flint
Mmmmm
paper flint
Once I say a free group is Abelianised, $$abca=bca^2=acab$$or whatever if I'm getting it right?
cloud walrusBOT
Ted
next obsidian
Sure, but I don’t see how that has to do with uniqueness
I guess
Actually
What you’re saying does make sense
paper flint
My choice of words is poor, forget it lol.
thorn delta
you could think of it as two strings x,y are the same if xy^-1 is a commutator ghg^-1h^-1 for some strings g, h
next obsidian
But your word choice doesn’t reflect it
wind steeple
next obsidian
Ted where have you seen unique up to... being used before?
I suspect maybe you’ve seen it in the context of free groups
Well, I actually have to do stuff, but I’ll just explain what it means in this case lol
The free group in your head might be this specific group which you made via taking symbols and words and yadda yadda
And maybe you’ve seen the phrase that “free groups are unique up to unique isomorphisms” or something
paper flint
Haha, no worries Chmonkey. I'm just fiddling with these ideas anyway, I'll learn them on a more serious scale when I reach the unit on generators and relations in Gallian.
next obsidian
We’ll ignore the unique isomorphism bit
paper flint
And maybe you’ve seen the phrase that “free groups are unique up to unique isomo...
Likely, yes.
next obsidian
So you want to change what you view the free group as
You’ve seen that free groups have this property that given a set function into a group, there’s a unique map from the free group so some diagram commutes yeah?
paper flint
My entirety of knowledge about free groups currently stems from its Wikipedia article lmao.
next obsidian
So a free group is something you want to view as ANYTHING which has that property
So a free group for a set S is a group F with a map S -> F
sharp sonnet
i should've never started memeing about free groups
next obsidian
Such that for any group G and a map S -> G there’s a unique map F -> G which makes the triangle commute
golden pasture
free groups is image of adjoint to forgetful functor from grp->set
next obsidian
So these things “unique up to isomorphism” are just things which satisfy that property
paper flint
You’ve seen that free groups have this property that given a set function into a...
Uhhh...not really, I'm clinging on to the string of letters analogy used by Wikipedia. It's alright though, I'll learn this stuff on a serious level when I can appreciate what you're explaining. 😛
paper flint
sharp sonnet
😦
paper flint
I still didn't read what Loch said.
sharp sonnet
it's ok
golden pasture
it's ok just need to read jacobson section on free monoids and groups
sharp sonnet
it wasn't relevant to anything mathematics
golden pasture
will learn one
sharp sonnet
i don't know if chmonkey is telling me to f myself because he read it or because he did not read it lol
golden pasture
referring to book 1 here
golden pasture
w
paper flint
monka intensifies
next obsidian
Hurb durb Seifert Van Kampen
sharp sonnet
i learned what free groups are from a meme
"you can't add apples and oranges, sure you can in the free vector space on apples and oranges"
just let the poor guy read his algebra book
let him experience the millions of sylow exercises before doing anything interesting
golden pasture
i literally skipped sylow until rep theory haha
then like
ah trivial
time to skip applications
chilly ocean
sylow 🤢
chilly ocean
all my homies hate sylov
next obsidian
Where will you be when you need Sylow to solve your problems
chilly ocean
You’ll never make it in math
I know.
next obsidian
And when you come begging at your knees for my help I will scoff
sharp sonnet
its the second most important theorem in finite group theory
well, its 3 theorems
wtv
sharp sonnet
did they release sylow 4
next obsidian
I mean...
What are you thinking of?
Existence, conjugation, the thing about being 1 mod p, and dividing the order of the group
sharp sonnet
existence, conjugation and the restrictions on number
next obsidian
Hurb
Okay what about the following
Any p-group is contained in some Sylow
That’s non-trivial
😎
golden pasture
wait what is most important then
next obsidian
Existence
sharp sonnet
lagrange
next obsidian
I think
golden pasture
ah
sharp sonnet
i mean importance of theorems in finite group theory
next obsidian
1st Iso
golden pasture
classification of finite groups
sharp sonnet
lagrange, sylow, something
next obsidian
1st iso > anything else
golden pasture
3rd iso
sharp sonnet
how is that specific to finite groups
golden pasture
same as lagrange theorem isnt specific to finite tbh
sharp sonnet
but it's not really useful
sharp sonnet
i was talking about lagrange
sharp sonnet
my point is lagrange is not really useful for infinite groups
so it is the most important theorem for specifically finite groups
golden pasture
true
sharp sonnet
there is a sylow for infinite groups as well
but i dont think that is useful either
golden pasture
haha yea
sharp sonnet
which makes sylow #2 for finite groups specifically
carmine fossil
there is a sylow for infinite groups as well
How does that work?
thorn delta
i think i remember my professor saying there is a basically complete sylow theory for locally finite groups
sharp sonnet
what's your favorite locally finite group that is not finite
thorn delta
pfff, the only infinite locally finite group i can think of off the top of my head is something like an infinite direct sum of copies of Z/nZ
thorn delta
my prof is a geometric group theorist, so i got to hear all about this stuff xd
obsidian path
Hey guys, for the question "Which abelian groups are isomorphic to subgroups of S6?"
I could use the fundamental theorem of finite abelian groups to find all abelian groups up to order 720 (|S| = 6!)
But then, find the subgroups of S6 that are possibly isomorphic to that.. is there a good way to do this?
next obsidian
My intuition says not many subgroups of S6 are abelian
at least among those which are decently large
Maybe try to make an upper bound on the size of an abelian subgroup of S6, it's very easy to have non commuting permutations so I think this is maybe tractable
I think cycles commute iff they're disjoint?
err, that can't be right, maybe disjoint or the same?
Idk
obsidian path
upper bound on the size?
thorn delta
as a hint, <(1 2)(3 4)> x <(5 6)> is klein four
viscid pewter
disjoint cycles commute
next obsidian
But I don't think its iff
anyway, what I mean is maybe you can say if the subgroup of S6 has size > n
it can't be abelian
or maybe you can do cases or something
I'm just suggesting you might be able to rule out large abelian groups being a subgroup
obsidian path
true
viscid pewter
hmm
next obsidian
so you the only need to look at small ones which makes it easier
And my intuition is that S6 is VERY not abelian
so it should be hard to have a subgroup which is abelian
viscid pewter
yeah wait if you compose a permutation with itself that'll definitely commute lol
viscid pewter
have i made a stupid statement?
golden pasture
are S_n the groups with least abelian subgroups for their cardinality and lower in the sense of #{abelian subgroups}/#{subgroups}
i would write a sage script now
but im too tired to
golden pasture
intuit
next obsidian
and I think their commutator subgroup is equal to them
golden pasture
hiw about A_n🤔
next obsidian
Probably
golden pasture
maybe A_n beats S_n
next obsidian
These are two ways to sort of measure abelianness
Also I think their inner automorphism group is big
I mean
it's S_n hurb
golden pasture
heub
obsidian path
thanks for the hints guys!
golden pasture
S6?
next obsidian
No
Only for a specific one
Inner automorphisms are always gonna be S_n for n >= 4 or so I think
golden pasture
thinking
next obsidian
but only for S_6 I believe you get nontrivial outer automorphisms
golden pasture
something like that
viscid pewter
yeah
next obsidian
You act on some fucked up weird S_5
golden pasture
viscid pewter
wikipedia says it has one particular outer automorphism
next obsidian
yes
and you construct it by doing some non-standard embedding of S_5 into S_6 then doing more bullshit magic
It's annoying lol
thorn delta
S_3 isn't abelian either tho
golden pasture
thono
viscid pewter
hells, it's abelian for s_3?
golden pasture
thonk
viscid pewter
yeah
carmine fossil
S_3 is D_6 and D_6 is def not abelian
next obsidian
(12) and (23) do not commute
thorn delta
Basically, you will want to use the fact that abelian subgroups are the direct products of cyclic subgroups.
obsidian path
oh
thorn delta
and using cycle decomposition stuff, you can count all of these
obsidian path
man I wish we covered that better
vital quail
Also I think their inner automorphism group is big
smh Z(G) bad measure of abelianness
thorn delta
@obsidian path as an example, lets try finding a subgroup iso to Z_3 x Z_2.
viscid pewter
oh yeah that's the way to do it kx, yeah
thorn delta
you need subgroups in S6 that "looks like" Z_3 and Z_2. Can u think of anything like that?
next obsidian
I guess if you embed the direct product then you have the factors as well
viscid pewter
i just have not done fundamental theorem of abelian groups yet
obsidian path
you need subgroups in S6 that "looks like" Z_3 and Z_2. Can u think of anything ...
I'm really bad with isomorphisms so its a bit difficult for me, been thinking about it this whole time lol
I know Z6 is isomorphic to Z3xZ2, and its abelian obviously
its not iso S3
thorn delta
oh yea, oops kind of a dumb example
obsidian path
nono ur good
chilly ocean
Can someone tell me how this is read?
If N is a proper normal subgroup of D2n then D2n/N is a dihedral group.
Because surely it cant be "then D2n divided by N"
chilly ocean
❤️
thorn delta
but this is an opportunity for an analogy, (123) and (45) are disjoint cycles; the first has order 3, and the second has order 2. In this case, <(123)> x <(45)> = <(123)(45)>, the same way Z2 x Z3 is iso to Z6 (the takeaway is that subgroups iso to Z6 don't have to be 6 cycles). So, try something like Z3 x Z3 w/ the same technique
obsidian path
like.. <(045)> x <(123)> ?
thorn delta
um the 0 in (045)? Do u take S6 to be bijections of {0,1,2,3,4,5} or something?
other than that, yes
viscid pewter
i think it's bc they were considering Zn
obsidian path
yeah he's right that's why
thorn delta
aight fair enough. Now i guess you just have to repeat the process for any abelian subgroup you can possibly embed in S6
obsidian path
I'm not sure I understand, does this mean Z3xZ3 is iso to S6 because of that?
obsidian path
oh ok
chilly ocean
hey, another reading question. Anyone know how to read this?
"since A has order n, n | |G|."
comes from this
next obsidian
This is Lagrange’s theorem
viscid pewter
n divides the order of G
next obsidian
The | means divides
chilly ocean
| G | = "order of G"
thnx
last question, and im done with this subject forever (i hope) what does "index" mean in this?
viscid pewter
okay so d | n so n = ad
a is the index of a subgroup with order d in a group with order n?
wait no
more general
it's the number of distinct cosets of H in G where H is a subgroup of G
chilly ocean
ok i think i get that
vestal snow
does every element of every galois group have finite order?
delicate bloom
is your galois group a finite group?
next obsidian
I mean aren't Galois Groups necessarily finite? unless you go to absolute galois groups for infinite extension BS
which I don't know anything about other than it's the automorphism thingy of the like separable closure or something
vestal snow
Yeah I was including absolute galois groups
For finite groups its trivial
I was curious about infinite galois groups
delicate bloom
then no, I can give an example if you like
$\bQ(\zeta_p, \zeta_{p^2}, \zeta_{p^3}, ...)$
cloud walrusBOT
merowo (◡‿◡✿)
delicate bloom
I should explain a little what the automorphisms look like now
where you send the roots will necessarily affect other roots, so let's say you have
$\sigma(\zeta_p)=\zeta_p^2$
cloud walrusBOT
merowo (◡‿◡✿)
delicate bloom
then because $\zeta_p^2 = \sigma(\zeta_p)=\sigma(\zeta_{p^2}^p)= \sigma(\zeta_{p^2})^p = (\zeta_{p^2}^a)^p$
cloud walrusBOT
merowo (◡‿◡✿)
delicate bloom
we have a condition on the higher roots here
a = 2 mod p
and this travels up the chain
and so we end up with the galois group being the units of the ring of p-adic integers $\bZ_p^\times$
cloud walrusBOT
merowo (◡‿◡✿)
delicate bloom
this has elements that are not of finite order so there you go
sturdy marsh
another example is the absolute galois group over a finite field
sturdy marsh
the galois group ends up being the profinite completion of Z iirc
so no element has finite order
sturdy marsh
inverse limit of Z/nZ
vestal snow
Aren't those the same?
sturdy marsh
no
sturdy marsh
inverse limit of Z/nZ ends up being the product of all the Zp s
white nymph
free to ask a question in here?
for part c.), im not seeing how to show the forward direction of \varphi being an isomorphism implying (ad-bc)^2 = 1. i feel like im missing something obvious
delicate bloom
Like the inverse limit of Z/p^nZ
mine's more like the inverse limit of Z/p^nZ haha
next obsidian
As the hint says
it's a matrix
so for it to be an iso the matrix must be invertible
aka the determinant is an invertible element of Z aka is -1 or 1
white nymph
thats the obvious thing... so used to working elsewhere where det just has to not be zero :p
thx much
next obsidian
celest nebula
can someone help me where to start for this
the group operation should be (a,b)*(c,d)= (a+c,b+d)
hot lake
yeah it should be that
leaden finch
can someone help me with this one
Z2 x Z4 determine if it had mult. identity, units, any zero divisors, idempotents
ik our cartesian products would be (0,0) , (0,1), (0,2), (0,3) , (1,0) , (1,1) , (1,2), (1,3)
the multip. identity would be (1,1) ?
sturdy marsh
yes
obsidian path
I is a maximal ideal, implies I is a prime ideal?
vital quail
@obsidian path can you prove it?
leaden finch
can someone help me
delicate bloom
if you want to be sure, try to prove it, begin by writing out the definitions maybe
obsidian path
post ur q
vital quail
@obsidian path so, again, just look at definitionss
another way that you can think about this in a more 'high level' way in some sense is, do you know what R/I is for I maximal, prime respectively?
why dont i tell you, and you prove it
if I is prime, R/I is an integral domain
try proving this now
(and this is actually iff)
leaden finch
this my question : Z2 x Z4 determine if it had mult. identity, units, any zero divisors, idempotents
ik our cartesian products would be (0,0) , (0,1), (0,2), (0,3) , (1,0) , (1,1) , (1,2), (1,3)
the multip. identity would be (1,1) ?
units would be (1,1)
hmm zero divisors would be (0,0) , (0,1) , (0,2) (0,3) (1,0) , (1,1) (1,2) (1,3)?
vital quail
why do you think (1, 1) is the mult id?
also is (1, 3) invertible?
@leaden finch remember that the mult id is characterized by acting as a neutral element in multiplication with all elements
leaden finch
oh thats right, would it be wrong if i wrote (1,1)?
vital quail
that's what im asking you
leaden finch
can i create a table for this?
vital quail
is (1, 1) * (a, b) = (a, b) for every a, b
latent anvil
Is there a theorem which says that any intermediate extension F ≠ C in a tower C(x)/F/C is isomorphic to C(x)?
I think I saw a talk about this but I can't remember the name...
gritty adder
I think this is Luroth's theorem
latent anvil
That's it! Thank you
sturdy marsh
it has a pretty neat proof using the hurwitz formula
I think it was something like
yeah
look at a curve with function field F
the inclusion F into C(x) induces a finite map P1 --> X
where X is the curve
which implies that X is iso to P1
as the hurwitz formula implies genus(X) = 0
latent anvil
I think that's the proof in the talk I saw
Except they called it a riemann surface instead of a curve :P
sturdy marsh
I remember seeing a pretty low-tech proof of it in a galois theory course
don't remember the argument tho 😦
vestal snow
I was thinking about how galois theory uses groups to study field automorphisms and I was wondering if there was something analogous for other algebraic structures
For example, studying automorphisms of a group that act as the identity on a given subgroup
Does anyone know of anything of this sort?
latent anvil
It shows up in topology
If you have some covering map E -> B, the group of all homeomorphism E -> E which permute the fibers of this map is called the deck transform group
There's a lot of analogies between galois theory and covering space theory
sturdy marsh
For example, studying automorphisms of a group that act as the identity on a giv...
this is the same thing as fixed points of some group action on another (abelian) group, group cohomology studies such things to some extent
a lot of things in galois theory/NT are expressed in terms of galois cohomology groups, group cohomology generalizes the mechanism in a particular direction
but galois cohomology can also be thought of as a special case of etale cohomology
so you have a generalization in a different direction too
sturdy marsh
Do you see one of the inclusions?
paper flint
Mmmmm I think I do see <a^k> being a subset of <a^(gcd(n,k))>. Not the other way round.
sturdy marsh
right, so now you want to show the other inclusion
and to do that it is sufficient to prove that a^gcd is in the other set
paper flint
For that bit, I guess I could just say that for $a^m\in\langle a^k\rangle$, I have $a^m=a^{pk}$ for some $p\in\bZ$, and consequently $a^{pk}=a^{pq\gcd(n,k)}$ for some other integer $q$, hence implying $a^m\in\langle a^{\gcd(n,k)}\rangle$?
cloud walrusBOT
Ted
paper flint
and to do that it is sufficient to prove that a^gcd is in the other set
Will try that 👍
Does the first bit of inclusion look okay?
sturdy marsh
yup
paper flint
For $a^m\in\langle a^{\gcd(n,k)}\rangle$, $a^m=a^{p\gcd(n,k)}$ for some integer $p$. I do not know how this would translate into something of the form $a^{kq}$.
cloud walrusBOT
Ted
paper flint
Aahhhh Bezout...
Thank you! 
So $a^{p\gcd(n,k)}=a^{p(nx+ky)}$ for some integers $x,y$ by Bezout's Lemma, and further $a^{pnx+pky}=(a^n)^{px}(a^{pky})=ea^{ck}=a^{ck}$ for some integer $c$. Hence, a^{p\gcd(n,k)}\in\langle a^k\rangle$.
cloud walrusBOT
Ted
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
sturdy marsh
yup
paper flint
Nice. Thank you.
paper flint
Is every finite cyclic group isomorphic to Z_n?(Also, is every infinite cyclic group isomorphic to Z?)
scarlet estuary
yes
(this is, in fact, an alternate way to define cyclic groups used by some authors)
paper flint
Nice, thanks.
paper flint
Can a group of infinite order have a subgroup of finite order?
paper flint
Is this from Dn'F?
carmine fossil
Yes
paper flint
I see. Will try it.
Okay, I found some interesting examples of non-trivial, finite subgroups of infinite groups.
Can't believe I missed out {1,-1} under multiplication lmao.
chilly ocean
yo DD
carmine fossil
Yo godel
paper flint
Need help. :3
Since all elements in H are of the form a^(some integer), can I simply assert it is generated by the smallest positive integer?
fading wagon
hmm...
carmine fossil
Assume a^k is the smallest element in the subgroup and consider the case when there is a element a^r such that r is not a multiple of k
fading wagon
Yeah, then |r|=qk+m and m<k. contradiction
gleaming quarry
How does one show the irreducibility of higher degree polynomials in a finite field?
e.g.
hot lake
well you could always enumerate all smaller degree polynomials and check that they don't divide it
gleaming quarry
hmm
This doesn't seem like a good solution for larger polynomials tho
originally I wanted to check for roots but that only works for degree 2 and 3
hot lake
or you can compute the gcd of the polynomial with the product of all the smaller degree polynomials
in this case you want to check for degree 1 factors and for degree 2 factors
you could try to compute its gcd with x^16-x
gleaming quarry
well you could always enumerate all smaller degree polynomials and check that th...
must the polynomials be elements of the field?
gleaming quarry
wouldn't that be the same thing?
hot lake
a polynomial isn't an element of the field
quaint ivy
How does one show the irreducibility of higher degree polynomials in a finite fi...
you could use what's proved here (an irreducible poly. over an integral domain has a root iff it's linear) https://math.stackexchange.com/a/222251 and then simply check whether any of the 4 elements are a root
thorn delta
are you stuck on part a?
cinder bone
yeah
thorn delta
okay, you have H < H1 < HK and H n K \subset H, so its at least clear that H n K is a subset of HK.
the intersection of subgroups is always a subgroup, so i guess you don't have that result? Basically, if x,y in H n K, x,y is in H (so apply subgroup stuff) and x,y is in K (apply more subgroup stuff) and conclude x+y is in H n K. And similarly for the other subgroup conditions.
cinder bone
H < H1 < HK ?
thorn delta
right, so 1 is in K and h = h*1 is in HK for all h in H
languid meteor
hey all, im kind of confused about how this phi(x) ring hom is working here
the natural projection brings r --> r + I
so phi(x) = x + (x^3)
so phi^2(x) = x + (x^3) + (x^3)?
so everything in the ideal (x^3) is added twice right?
you get like {r+2i | i is an element of (x^3)}
then I dont get how phi^3(x) = 0
thorn delta
you need to compute (x + (x^3))^3
languid meteor
ohhh of course, im not applying the map 3 times that wouldnt make sense since x + (x^3) isnt in Z[x] 🥴
thorn delta
ye, slightly confusing
languid meteor
also not sure how to go about (x+(x^3))^3 because isnt this a set cubed?
so am I cubing the elements of the set? or can I treat this as a binomial
thorn delta
remember the definition of the ring operations in a quotient ring: if R is a ring and I is an ideal, then addition and multiplication in R/I look like
(r+I)(s+I) = rs + I and (r+I)+(s+I) = (r+s) + I
actually, you don't really have to think of it this way either
chilly ocean
hey all, im kind of confused about how this phi(x) ring hom is working here
don't they just mean eta^3 = phi(x)^3 = [x]^3 = [x^3] = [0] in Z[x]/(x^3) ?
languid meteor
oh god, yeah no that makes perfect sense just like that
thorn delta
$\phi$ is a homomorphism so $\eta^3 = (\phi(x))^3 = \cdots $ applying homomorphism properties....
cloud walrusBOT
kxrider
cinder bone
what about b now?
thorn delta
think of it like this: you have H n K < H and H is abelian, so what can you say about H n K?
thorn delta
you cay something stronger than that. H n K is abelian
gleaming quarry
Would appreciate someone checking this now
Check if this polynomial is irreducible
My Solution:
(X^2 + X + alpha) * (X^2 + X + alpha + 1) = X^4 + X +1
polynomial is irreducible
I'm unsure if I'm allowed to use alpha as a constant
or If I'm just allowed to use the elements of F2 as coefficients here
thorn delta
Since H n K is abelian joshua, the elements of H n K commute with the elements of G, and therefore commute with the elements of HK.
thorn delta
ye. Its also just true in general that abelian groups are normal, and therefore normal in every subgroup that contains them.
cinder bone
HK is abelian?
silk wren
does any1 kno
what a triangle pointing to the left means
.
i have to prove that ker ϕ / ker ψϕ
ok i just put it in here & its that
so is that just a fancy way of showing the elements in ker 1st bla bla bla
idk why the triangle was put in the question xd
thorn delta
you mean H < G but with a line at the end of <?
silk wren
yes
thorn delta
it means H is a a normal subgroup of G.
thorn delta
yea, ker phi is a normal subgroup of ker psi phi
thorn delta
npnp
leaden finch
has anyone written any papers in Latex?
chilly ocean
@leaden finch just use some templates in overleaf its super easy usually
Ive written a bit if u have questions ask, but not sure if I will be able to answer since I just google most of that stuff
misty wind
when one speaks of a "canonical isomorphism" is that the same meaning as "naturally isomorphic"?
^just a small language and usage question (soft question):
next obsidian
So like... usually
it also can mean "the obvious map" but like... that's probably also natural
I think it's a bit loosely used but it's probably some sort of categorical universal map
thorn delta
i forgot exactly how, but ik that the canonical isomorphism from a vector space to its double dual can be thought of as a categorical natural isomorphism
misty wind
so imma interpret that as an answer: "Yes, for all purposes where one could reasonably care."
next obsidian
It is
it's literally a natural isomorphism kxrider
from the identity functor to the hom(hom(-,k),k) functor
aka the double dual functor
misty wind
i forgot exactly how, but ik that the canonical isomorphism from a vector space ...
this is actually the context i was asking about, but i didn't know if the usage would differ from one discipline to the next.
TYVM FOLKS!
leaden finch
i googled one but im having trouble with the font format
words are too big @chilly ocean
floral creek
Anyone know of a structure the real projective line would fall under , when R U {infinity} and infinity is 0^-1
with + and multiplication
is it just a wheel if I keep giving things definitions
fervent tulip
im trying to find which abelian groups are isomorphic to the subgroups of S6
could someone help
vital quail
@chilly ocean does
old lava
just write out all the subgroups of S6 <:opencry:586078614865444865>
only 1455, very doable
thorn delta
@fervent tulip did u ask this yesterday?
fervent tulip
no
thorn delta
ah, someone asked this exact question yesterday haha
use classification of abelian groups. i.e. if you have something like Z3 x Z3, you want to find some internal direct product of subgroups of S6 isomorphic to Z3 x Z3.
old lava
a good starting point could probably be cyclic subgroups
chilly ocean
old lava
chilly ocean
i had something in mind
but my monkey brain does not convert thoughts into words well
ok maybe what i said was fine but i'm not gonna repost it because i don't feel like being obligated to follow up on the things i say
chilly ocean
Hello! What would be a good resource to learn groups (abelian groups, cyclic groups, subgroups), rings, and vectorial spaces (and subspaces and bases)?
With solved problems as well, if possible:D
carmine fossil
A algebra book
chilly ocean
haha, sure. Do you have any that you would recommend?
carmine fossil
Check the pins in #book-recommendations
I use dummit and foote,but you may not like it
paper flint
Just go for Gallian.
fading wagon
@chilly ocean if you do olympiad math you can try Infinitely Large Napkin
chilly ocean
thank you! I will look into it
chilly ocean
<@456226577798135808> if you do olympiad math you can try Infinitely Large Napki...
I'm a undergraduate student and I mainly need it for my abstract algebra class, but I might also participate in math student competitions, so I'm going to check out Large Napkin
thank you!
golden pasture
napkin sucks tbh
read like jacobson or check out the pinned message in #book-recommendations
fading wagon
idk it has a wide range of topics, it's not too bad I guess
latent anvil
napkin has bad vibes
that is all I will contribute to this discussion
DD what did I just say
carmine fossil
mb
chilly ocean
read like jacobson or check out the pinned message in <#716264872018706443>
thanks! I've looked at Jacobson's book and it also has lattices, so that's nice
golden pasture
jacobson's part 1&2 combined has like
a lot of info
really once you read till galois
you can start jumping around
chilly ocean
okay, thank you. Do you know if there is also a book of solutions to the exercises?
golden pasture
no
you dont need it lol
solution books are fake news
conspiracy by high school teachers
h a v e
f a i t h
and 6 months later realize your solution is fake
latent anvil
Matsumura's comm alg book has solutions in the back
I was taken aback when I found out
It's so weird lol
It's been useful though!
golden pasture
oo
if you're paranoid use some proof checking program or something
imagine writing all the details
cant be me
latent anvil
I mean
I do both of those things lol
main issue with proof checking programs for me is that their libraries don't have enough math
So I need to build up a bunch of background
golden pasture
i havent actly used a proof checking program before lol
golden pasture
main issue with proof checking programs for me is that their libraries don't hav...
mainly cuz this
scarlet estuary
yes, if you define a function Z_6 -> Z_2 x Z_3 and show it is a bijective homomorphism, that suffices
the reason for this is that "bijective" and "invertible" are the same thing
so if you can define an isomorphism f: Z_6 -> Z_2 x Z_3
you can also define an inverse f^-1: Z_2 x Z_3 -> Z_6
this is why we can just say "these are isomorphic"
rather than "isomorphic in one direction" or whatever
woven obsidian
You need to show that the map is well defined too
sharp sonnet
they wrote down the map, ofc it is well defined
woven obsidian
Because you define it in terms of representative of the class [x]_6
woven obsidian
Ah right
sharp sonnet
(although you could treat this more generally)
leaden finch
do i need well define?
golden pasture
this essentially shows it is well defined but only works cuz the domain is small
could try showing more generally Z_m x Z_n is isomorphic to Z_{mn} when m,n is coprime if you're interested
leaden finch
oh okay i see
chilly ocean
@leaden finch yo nice handwriting
actually I found a mistake in your pics
you wrote 'Homorphism' instead of 'Homomorphism'
old lava
Just go for Gallian. <:hmmm:773801269532819466>
Gallian is too ultra slow and long innit
golden pasture
you wrote 'Homorphism' instead of 'Homomorphism'
proof english is bad
paper flint
Gallian is too ultra slow and long innit
I find it good for self-study, although I think Pinter is even lighter and concise.
old lava
I see, I found gallian to not cover enough content, and the stuff it did cover, it did so not that well (personally)
paper flint
Makes sense, I think Gallian tries to touch the tips of several icebergs, i.e., just give a light introduction to a lot of things within AA.
golden pasture
jacobson does a good job in that
and doesn’t just touch very lightly and go
also very suitable for self study
i think
i mean everything i read have been self studied modulo differing amount of suffering
languid meteor
hey guys ive been trying to figure out how part (b) of this works for some time now. I know that a + b(eta) + c(eta)^2 should be a + b(x + (x^3)) + c(x^2 + (x^3))
can I combine everything to be a + bx + cx^2 + (x^3)
like it makes sense for it to be that way but I dont know if I need to go into more detail as to why
cloud walrusBOT
slimvesus
oblique river
@languid meteor maybe start with the fact that phi is surjective, so every element of R is the image of something in Z[x]. everything in Z[x] is of the form a + bx + cx^2 + dx^3 + ... -- what happens under the quotient map?
@chilly ocean no, S_3 is a counterexample. the element of order p need not generate a normal subgroup. (12) is order p = 2 and (123) is order m = 3 and they both generate S_3 which has order 6, but (12) doesn't generate a normal subgroup
it will be if p > m
and yes it's true but I can't remember the name of the theorem
turns out you can just do it directly, toohttps://math.stackexchange.com/questions/600271/showing-an-exact-sequence-of-groups-is-split