👍
#groups-rings-fields
thorny slate
fallen bluff
Ummm can anybody help me with simplicial complex bois
chilly ocean
fallen bluff
How do we for example construct simplicial model of sphere S^1 ? I start with triangle v0,v1,v2 and then how do I choose orientation of its sides?
I thought the orientation is given by ordering of the vertices (i.e. v0 < v1 for example) but this is apparently wrong
fickle brook
v0 -> v1 -> v2 -> v0?
fallen bluff
Yes that's what my notes say too but it doesn't make sense to me why is that
fickle brook
so that you can go around it following the orientation
fallen bluff
Well then this definition of boundary operator wouldn't make much sense, would it?
Oh wait it would actually, it's just this formulation that's confusing
OK I can understand the definition, but can't understand why triangle with counter-clockwise orientation represents S^1
stone fulcrum
@fallen bluff
Because they're homeomorphic
That is, ultimately, the reason why you want to do any of this, we want to dig out groups that are preserved under homeomorphisms
fallen bluff
@stone fulcrum What do you mean by homeomorphism between S^1 and directed graph?
Yes S^1 is homeomorphic to triangle, but how does its edge orientation come into play in this?
thorny slate
the edge orientation is extra information
it gives a particular, uniform way of taking boundaries
as in representing the boundary as a union of lower dimensional simplices
S1 by itself does not have this
but you can impose it in many ways
but it does not change the underlying theory
it's just a computational tool
just had a chat with tormeson about it in numbers & co
if you wanna check
anyway a homeomorphism between S1 and Delta1 induces this structure on S1
as in, it will induce a simplicial structure on S1
by just copying it from Delta1
fallen bluff
But what orientation do I give to the edges of the triangle S1 is homeomorphic to?
If it's triangle v0,v1,v2, why is the orientation of edges given by v0 -> v1 -> v2 -> v0, as Ann said?
thorny slate
what
this is just a choice
that you make
like
that's the standard simplex for a reason
it's a fixed choice
and you express all other choices in terms of it
it's a choice that's taken because it's easy to describe combinatorially
fallen bluff
well but I can also see S^1 as quotient of disjoint union of three 1-simplices (v0,v1), (v2,v3), (v4,v5) where I identify the vertices in such a way that I obtain a triangle - and this one has any edge orientation I want
fallen bluff
OK so from my understanding, if I choose just ANY 2 delta complexes that are homeomorphic to S1 and then if I calculate their homology groups, they will be isomorphic?
fallen bluff
@thorny slate Forgive me for pinging, this is in case you forgot about me. And it's important for me 
thorny slate
Huh
I dont get the question
Yes theres a bunch of simplicial complex structures on things
Yes, homology doesnt depend on simplicial complex structure
Just on topology
Singular homology is in some sense the limit of every possible simplicial structure
fallen bluff
That is actually pretty fucking cool
Another question about algebraic topology: would we get any benefits whatsoever if we used general R-modules instead of abelian groups in homology?
bleak finch
plain sequoia
I am trying to prove that if $G$ is a group, then $G \text{ is a p-group } \iff |G|=p^n$
cloud walrusBOT
markus:
plain sequoia
The left implication is easy, but I am somehow not able to prove the right implication. I mean, the problem is I dont see how we can deduce that $|G|\neq p^nm$. Any hints are greatly appreciated!
cloud walrusBOT
markus:
bleak abyss
Have you seen Cauchy's theorem? @plain sequoia
plain sequoia
Yes
That states that if a prime p divides the order of the group, then the group has a subgroup of order p?
bleak abyss
Yup, in particular that's an element of order p
plain sequoia
Hm the right implication is given G a p-group then $|G|=p^n$
cloud walrusBOT
markus:
plain sequoia
So we already know the order of all the elements, kind of at least, they are all $p^k$ for some $k$
cloud walrusBOT
markus:
plain sequoia
The left implication is just Lagrange
I am probably just stupid right now; but I don’t see how to use Cauchy to prove that there is no m such that $|G|=p^nm$
cloud walrusBOT
markus:
plain sequoia
Ah nevermind I am stupid
I think I figured it out
Assume that $|G|=p^nq$ for a prime $q$. Then by Cauchy there is a group of order $q$, but that is a contradiction because $G$ is a p-group
cloud walrusBOT
markus:
chilly ocean
Hey
Hey, "4 is a primitive root mod 11".
It is not. It doesn't generate 2.
Why not?
Like, from the wikipedia, there's no known general formula for finding the primitive roots of a number.
But then, what do we know?
And why don't we know that?
Is this somehow related to the distribution of the primes? or...???
chilly ocean
too l8, lmfao
We all make mistakes, lol.
No worries.
You won't hit 2.
Idk if you can program, but you can use a for loop or some kind of recursive structure to figure it out quickly and more generally.
Supposedly 12 has no primitive roots.
solar vessel
essentially
if a number is larger than 12
it can be rewritten 12n+m
where 0<=m<12
so when you multiply by it the n doesn't make a difference in mod 12
chilly ocean
I don't understand what's happening. Idk if my question is clear or not.
I'm saying that I don't understand the algebraic structures happening.
i.e. I should probably study more groups, rings, and fields because I'm v rusty.
stone fulcrum
From Wikipedia, g is a primitive root mod n if it is a generator of the multiplicative group mod n
Duh, now that I think about it.
No general formula is known. Hmm
For small groups it's not too hard to reason out the generators, but really? No way of doing it?
chilly ocean
My friend thinks it's because 4 is a perfect square.
but it's true that 4 isn't a primitive root mod 11
the only primitive roots mod 11 are 2,6,7 and 8
Yes.
So, supposedly, perfect squares can never be primitive roots mod N, where N is an odd prime.
And also
If we have an even integer k > 2, then the multiplicative group mod k has NO primitive roots.
And this is because by the Chinese Remainder Theorem ( CRT ), it's isomorphic to the multiplicative group (Z_k/2 , 2 ) - which has no generators as the 2nd component is "fixed".
It's "fixed" in the sense that if I have an element in that group, it has one of two forms:
(a, 0)
or
(b,1)
but (a^2,0^2) === (a^2 % N, 0)
&
(b^2, 1^2 ) === (b^2 % N, 1 )
So there are no generators for the group.
plain sequoia
I’ve never heard of primitive roots in the context of abstract algebra - only number theory. However, is it fundamentally the same thing ish?
plain sequoia
Ah okay, I see! Thanks 😊
chilly ocean
My math skills have become sooooo shoddy smh.
Would anyone be interested, potentially, in starting a reading group that works through a book in group theory or something?
stone fulcrum
That's not at all a bad idea! But, feel free to ask about any theorems or ideas people here know it all
chilly ocean
Thank you.
turbid owl
@chilly ocean @plain sequoia - How is it stated in the context of number theory?
plain sequoia
Given $a \in \mathbb{Z}$ and $\gcd(a,n)=1$, then $a$ is a primitive root of $n$ if the order of $a$ modulo $n$ is $\varphi(n)$
cloud walrusBOT
markus:
plain sequoia
And order is, just like in group theory, the first time you get back to the identity. So, the order of $a$ modulo $n$ is the first number $k \geq 1$ such that $a^k \equiv 1 \pmod{n}$. Ofc still we must have $\gcd(a,n)=1$ for this to make sense
cloud walrusBOT
markus:
chilly ocean
cool & thnx
chilly ocean
Don't help this guy he's posting everywhere
torn flame
Hadn't read the rules, have now. Sorry for multi-posting
chilly canyon
Hi ! Given a finite (field-)extension L/K, using the primitive element thm, one has L=K(a) for a in L a primitive element.
Let H be a subgroup of Gal(L/K). My goal is to prove that Gal(L/L^H)=H (Artin's lemma).
S. C. Newman considers the polynomial F(X)=\prod_{s \in H} (X - s(a)), and basically says that this polynomial is in L^H[X] "by a familiar argument". Which argument ?
People seem to be proving that t(F)=F for any t in H, by saying t(F(X))=\prod t(X-s(a))=\prod (X-t(s(a)))
Why do they not have \prod (t(X)-t(s(a))) ?
Lang also does the same proof with the same argument, but I cannot understand what's wrong with me
thorny slate
all the coefficients are symmetric functions
on the elements of H
for example
say H = {u,v,w}
(x - u(a))(x - v(a))(x - w(a)) = x^3 - (u+v+w)x^2 + (uv + vw + wu)x + uvw
applying an element of H to any coefficient permutes the summands
for the other question
t doesn't act on X
an automorphism of K extends to K[X] by acting on coefficients
chilly canyon
Mmh... About the second question, I can see the thing, but it looks like cheating tho
The associated polynomial function wouldn't care whether X is the abstract variable of K[X], if one composes with an automorphism, wouldn't it take it into account ? I think it's somewhat messy in my head now ^^"
About the elementary symmetric functions of roots, I remember that ^^
thorny slate
saying that the polynomial is in L^H[X] means that the coefficients are in L^H
X has nothing to do with this
you should try to convince yourself that the action that acts on coefficients and fixes X is the right one to prove this
chilly canyon
Basically, if s is an automorphism, one has s(a_nX^n+...+a_0)=s(a_n)X^n+...+s(a_0) ? That's just about it ?
I mean... Okay, thanks ^^" I feel kinda stupid but this still bothers me ^^" I'll give that a try tomorrow after a good sleep
uncut girder
Let K=<a1,a2,b,c> be the free Z-module generated by those elements.
Let I=<a1+b,a2+b,a1+c,a2+c> be the free Z-submodule generated by those elements.
Is it true that H=K/I is isomorphic to Z/3Z?
thorny slate
in the quotient you have
a1 = -b
a2 = -b
a1 = -c
a2 = -c
the last relation is redundant
and the quotient module is <x>
the free module on one generator
uncut girder
Ty
chilly canyon
Letting Γ∪{π} a Hamel basis (ie a Q-basis of R), is it true that Γ spans R/πZ over Z (ie the Z-module R/πZ is generated by finite Z-combinations of elements of Γ) ?
Trivially, R/πZ is spanned by Γ over Q, but I need it to be the case over Z 😦
thorny slate
Consider the one dimensional Q-submodule piQ/piZ = Q/Z.
this is spanned by a single vector as Q module, but is not even finitely generated as a Z-module
so no
chilly canyon
Oh, sad
Then, is {q*γ, (q,γ)∈QxΓ} spanning R/πZ over Z ? xD
I forgot the elements of πQ
thorny slate
a Z basis will be given by γ/p, p prime numbers, γ in Γ
whoops
okay
that seems better
no
chilly canyon
Holy, that's clever
thorny slate
this isn't a basis
or is it
it's a minimal spanning set for sure
I can't tell if it's a basis
chilly canyon
I'll check this ! Thanks for the tip !
thorny slate
oh hang on
you need the other ones I think
hmmmmmmmm
might be γ/n
for all n
you can check it yourself
by studying the Z-module structure of Q/Z
i believe you need all the 1/n
should be, since the one generated by the 1/p is free
and this thing isnt free
chilly canyon
π/n does not vanish in R/πZ for n≥2
I just added them to the set, that's fine for me ! Thanks, I appreciate the help !
thorny slate
sure
chilly canyon
Wait
Actually, R/πZ has torsion, so it's not a free abelian group, hence there's no Z-basis 😦
thorny slate
Yeah as I said before this minimal Z generating set is not independent
chilly canyon
The family only spans R/πZ
thorny slate
The g/p for g in Gamma, p prime is independent
But not spanning
And the g/n for g in Gamma, n natural is spanning but not independent
chilly ocean
alright so is the Klein group an example where H∩K is in G but HUK is not in G. Like lets say we made H={e,v} and K={e,h} woudl this work?
quiet cave
Yes rhapsody
hv wouldn't be in the union
Whereas the intersection is just the trivial group
fickle brook
@south oxide you have these backwards
south oxide
Ou
chilly canyon
Isn't it the intermediate Z_p^k for k between m and n ?
winter wind
If I translate a graph y=f(x) to y=-f(x-3)-2
X is moved +3
Y coordinates move -2
And the whole graph is only reflected over x axis?
chilly ocean
this isn't abstract algebra. go to #prealg-and-algebra
uncut girder
Actually
It's has something to do with group representations
We talked about this is our algebra class
crimson mango
When you dont know anything and you have an exam coming up thats 40% of your grade 😩 😫 😩
thorny slate
@uncut girder the other obvious restrictions are that it has a subgroup of order p^m, and a subgroup of order at least p^n
they might intersect
I mean you know it's a product of p-groups
i think it has to be only Zp^{n+m} and Zp^n x Zp^m?
uncut girder
False
thorny slate
yeah im not sure about that last line
crimson mango
What are you trying to prove?
uncut girder
It has to have a subgroup Z_p^m and it's quotient by that A/Z_p^m has to be Z_p^n
thorny slate
so the condition I gave was eough
it has a subgroup of order p^m, and a subgroup of order p^n
aka it has a subgroup of order p^max(m,n)
a similar construction as in the Z8 x Z2 case tells you how to do it
chilly ocean
So let G map to H be an isomorphism. If g is in G has order n, prove that Φ(g) also has order n.
stone fulcrum
@chilly ocean
gⁿ = e
φ(gⁿ) = φ(e)
φ(g)ⁿ = e
And thus φ(g) has an order than divides n.
There is no k < n such that
g^k = e
By following the above, there is no k < n such that
φ(g)^k = e
And thus the order of φ(g) isn't lower than the order of g
chilly ocean
thanks
@stone fulcrum how'd you get so good at this?
like did you get this knowledge after the first class?
or did it take time
stone fulcrum
A little bit of both
Later classes will build on homomorphisms so having a very strong understanding of how they work and why we use them will come in time
Mappings between groups are essentially the whole of group theory. You'll get a lot of practice here
chilly ocean
idk im kinda worried cause we've been using some weird IBL learning methods like we solve all the proofs by ourselves without answers and correct each other in class and its been hard to learn without a textbook.
Like I understand isomorphisms fairly well, but Imma probably by a textbook and review over the summer.
stone fulcrum
I'm a fan of that. As much as you're learning group theory, you're also learning how to prove stuff.
Feel free to ask anything if you'd like
chilly ocean
Alright, thanks that helps cause everyone has kinda clicked up in my class and im not an outgoing dude so I was kinda left alone
@stone fulcrum O so also the reason that we show k cant be greater than n is because if k divided then in then it would be n/k which would mean that if k was greater than n then it wouldn't be an integer so it wouldn't fit the definition of the order of an element?
stone fulcrum
@chilly ocean
For example, let's say I showed that φ(g)⁴ = e. Does this prove that the order of φ(g) is 4?
Not if φ(g)² = e as well
chilly ocean
Yeah, that makes sense, but when you say divide does that imply that n must be greater than k, cause if k was greater you'd get a fraction which isn't a positive integer.
stone fulcrum
Yes.
Let's say k is the order of φ(g). Since I showed that φ(g)ⁿ = e, then n/k is an integer
Next is to show that k = n, which is easy if I show that k ≥ n
chilly ocean
ight just making sure
Let a,b in G. If the order of the lemeent ab is n, then prove that the order of ba is also n.
-we'd probably use the same trick as last time, but in a different way right
stone fulcrum
"if the order of the element ab is n"
What does that tell you?
chilly ocean
well then we know from the last theorem that the order of φ(ab) is n right?
@stone fulcrum
stone fulcrum
There is no φ here. No homomorphism exists in this q
chilly ocean
hmm what do you mean by q?
random crag
q=question
chilly ocean
like by itself
random crag
ill give you a start b(ab)a
random crag
yes
chilly ocean
o and since (ab)^n is equal to e by a theorem in my book then you can say b(e)a=ba
random crag
(ba)^n+1=ba
chilly ocean
how'd you get n+1?
random crag
we need n+1 copies of ba to get n copies of ab
b(ab)a one copy of ab two copies of ba
b(ab)(ab)a two copies of ab three copies of ba
chilly ocean
o I see
hmm, so we have |ba|= n+1 but we need to show that it is equal to n. hm, I know about minimality theorems, but not sure about how to reduce it to n.
random crag
consider wat would happen if some k<n is the actual order
umm (ba)^n+1=ba therefore (ba)^n=1
anyway (ba)^k=b((ab)^k)a=1
so $(ab)^k=b^{-1}a^{-1}$
cloud walrusBOT
rockpaperscissors:
chilly ocean
O I get it, cause its like saying that (ba)^n+1=ba^1 which would be the same as ba^n+1-1=1.
random crag
what does that tell you about the order of ab
chilly ocean
O the socks and shoes theorem for inverses
random crag
the what what
chilly ocean
they flip and negate
cause you have to take off your shoes before your socks
idk, thats how I remember it
random crag
new to me
chilly ocean
cause taking off a shock is like inverting the sock cause it goes inside out, idk
welp anyway
chilly ocean
o I see, cause there is a theorem that states that for all g in G |g|=n the |g^-1|=n
b^-1a^-1 must have the same cardinality as ba.
or is there a way to do it without the theorem, just cause im probably gonna forget that in like a year.
random crag
thats what we’re doing now
stone fulcrum
@chilly ocean
Hey, did you get it?
chilly ocean
so how do I turn b^-1a^-1 into ba
Like I got it with using a theorem, but that theorem is kinda trivial and I think I prove it easily so I dont have to memorize it. Idk I just have that feeling
random crag
so $(ab)^k=b^{-1}a^{-1}$
cloud walrusBOT
rockpaperscissors:
random crag
which implies $(ab)^{k+1}=1$
cloud walrusBOT
rockpaperscissors:
chilly ocean
o so I just have to use the theorem then
that states the cardinality of any g is equal to its inverse.
O nevermind
I see it
ab on both sides
ight noice, thanks dude
chilly ocean
- Let φG to H be an isomorphism. Show that φ(Z(g))= Z(H).
- Prove that a finite group G is a union of proper subgroups if and only if G is not cyclic.
thorny slate
1 is trivial
Literally just write it
2 as well
Essentially for 2 think of maximal subgroups
fickle brook
can i just remark how weird i find it that people struggle on questions where unfolding all the definitions makes the question solve itself
there's got to be a name for this kind of question
somber bramble
idk, but it's definitely a good reminder that a good place to start with proving a statement is to write out all the definitions involvee
thorny slate
I mean
it probably means it's their first math course
and they haven't realized that all they have to do is write it out
simple valley
Proof. unfold in *.
timber bay
So I'm looking at the splitting field of x^4-x^2-2 over Q. I found that i x sqrt(2) is a generating element of the splitting field, since +/-sqrt(2) and +/- i are rootsI'm trying to draw the lattice of subgroups. do they expect me to use Q(i, sqrt(2)) as my group? if I used Q(i*sqrt(2)), the lattice would be pretty simple
fickle brook
@simple valley you mean Proof. unfold in *. Qed.
simple valley
nah, if it was immediately Qed after that you wouldn't need unfold
unfold only does delta-transformation of the term to expose it to e.g rewrites
thorny slate
@timber bay im quite sure i sqrt(2) does not generate the splitting field
i + sqrt(2) might
the group is the galois group
of the splitting field, which has Q(i, sqrt(2)) as one presentation
it's an order 4 group
timber bay
@thorny slate yeah i realized my mistake. itll be automorphs that send +/- i and +/- sqrt(2) to each other
thorny slate
exactly
timber bay
there should be 4 automorphs here? because its the size of the min poly?
thorny slate
to be fully rigorous you have to show that they are actual field automorphisms
and that those are the ones
and that they commute
or, you can notice that there are only 2 groups of order 4
timber bay
yeah
thorny slate
which is Z4 and Z2 x Z2
and check that it isn't Z4
because there's two elements of order 2
timber bay
since there are 4 roots wont the gal group be of size 4!?
or do the isomorphism properties impact this
timber bay
is that not the degree of the poly?
thorny slate
in fact this poly is (x^2+1)(x^2-2)
no
it's not
it can be anywhere from d to d!
somber bramble
the galois group doesn’t necessarily have to be all of S4 I don’t think
thorny slate
remember?
somber bramble
it’s some subgroup of it tho
thorny slate
in this case it's a product of two degree 2 polys
so it's between 2 x 2 and 2! x 2!
timber bay
oh yeah
somber bramble
what an amazingly large range
timber bay
true
hmm
so the gal group will just be all the permutations of these four elements? as long as they obey isom properties?
somber bramble
if it was all permutations, that would be 24 of them
thorny slate
yeah it's not the permutations
it's a group of order 4
you described the generators
somber bramble
it is a subgroup of all the permutations, always
thorny slate
i -> -i and sqrt(2) -> -sqrt(2)
yues
and in this case
bexause its a product of 2 polynomials
its a subgroup of S2 x S2
timber bay
that makes sense
okay so these are the automorphs, 1) i -> -i 2) sqrt 2 -> -sqrt 2 3) both 1 & 2, 4) ID
thorny slate
yeah
timber bay
so if i were doing this same procedure for the polynomial x^7-1, it wouldnt be correct to say that the only subgroup of that whole galois group will be the identity, since the power is prime?
i would have to look at between 7 and 7! to find the number of elements in the gal group and see if there are any subgroups?
thorny slate
yeah
well, no
you need to calculate the order of the splitting field
first split the polynomial into irreducibles
it splits as
timber bay
yeah
thorny slate
(x-1)(x6 + x5 + x4 + x3 + x2 + x + 1)
timber bay
yes
thorny slate
and you can check that
is irreducible
then add a root, and see if it splits completely, etc
to calculate the order
it's between 6 and 6!
once you know the order
you can find the generators of the group
timber bay
add a root of the right poly?
thorny slate
yeah
timber bay
and see if that is enough to make all the roots?
thorny slate
adding a root will be an extension of order 6
then see how it splits
once that root is added
in this particular case it's easy because the roots are the seventh roots of unity
and you can describe them very explicitly
timber bay
yeah so (-1)^(1/7) is enough to generate all the other roots of the right poly
thorny slate
yeah
so it's degree 6
now you can find the galois group
by looking for a degree 6 group
timber bay
so these automorphs will send (-1)^(1/7) to its conjugate?
im not making the right connection here
thorny slate
well that's one of them
you need the rest
here's a hint
automorphisms are determined precisely by where each root goes
your field is of the form Q(a)
for a single a
timber bay
oh will it send the powers of (-1)^(1/7) to each other?
thorny slate
so an automorphism is determined precisely by where a goes
yeah, it has to be a subgroup of S6 of order 6
and it's determined by where (-1)^(1/7) goes
there's only 6 possibilities for that
timber bay
so theyll be elements that look like (123456) or (124536)?
is that what the order implies?
thorny slate
well
thorny slate
in principle it can be a group generated by (12) and (345)
or something
but you also know that the galois group has to be transitive
timber bay
i dont think subgroups can have elements both like (12) and (123) in them
thorny slate
aka each root has to go to each of the other roots by some element
in general they can
you can get Sn as a galois group
but the only transitive subgroup of order n of Sn is Zn
generated by an n-cycle
timber bay
hmm
so its a subgroup of s6 that has 6 elements? and thats the whole group for the splitting field of that poly?
thorny slate
it has to have 6 elements
because the extension is order 6
and it has to be a subgroup of S6
timber bay
i could try guessing elements like (14) or (27) and see what they generate but i dont see how doing this process wouldnt generated the whole group
thorny slate
because it's a degree 6 irreducible poly
thorny slate
since the field is principal, Q(a), the automorphism is determined by where a goes
so that's 6 elements as well
another way to see it
it's a bunch of casework usually
but you'll have enough information
to work it out
things can be complicated fast
if the polynomials are complicated enough
timber bay
yeah i could see that
chilly ocean
- Let φG to H be an isomorphism. Show that φ(Z(g))= Z(H).
So for my proof I could say that Z(G) is a subgroup of G let z in Z(G) therefore gz=zg for all g in G. by def of φ: φ(g)= h for all h in H and thus Z(H) is equal to zh=hz
thorny slate
you need to write things out better
it's unclear what you're trying to say in each step
but it looks like you have the right idea
chilly ocean
yeah sorry my "textbook" (its actually a small packet) is basically a blogpost a prof wrote from another university and we have no lectures so, its been a rough year
Im just trying to put everything together, but Im given no direction on how to approach stuff.
thorny slate
your approach is fine
but you need to be more explicit about what you mean
for example you use g two different times, the second one you don't say what it means
or where it comes from
and you use z as an element of Z(G) and as an element of H
chilly ocean
so should I prove this by rewriting the element Z(G) in the form of g then using what φ is defined as. Or should I just establish what Z(G) is and then proof it using the 3 properties of isomorphisms.
thorny slate
your proof is fine if you write it correctly
you wanna show that φ(Z(G)) = Z(H)
so you have to show two things
φ(Z(G)) c Z(H)
and φ(Z(G)) contains Z(H)
chilly ocean
the definition of center?
somber bramble
in what world is “set equality” and “center” the same thing?
what ann means is that two sets A and B are equal if A ⊆ B and B ⊆ A
and usually that’s the easiest way to approach a proof of equality of two sets
take x ∈ A arbitrary, show it‘s also in B, thus A ⊆ B
then take y ∈ B arbitrary, show it‘s also in A, thus B ⊆ A
then A = B
thorny slate
you wanna show that φ(Z(G)) = Z(H)
so you have to show two things
φ(Z(G)) c Z(H)
and φ(Z(G)) contains Z(H)
chilly ocean
O so that second part deals with maximality
chilly ocean
so in this case the A is Z(G) and the B is Z(H)
fickle brook
O so that second part deals with maximality
what
somber bramble
We just covered the fundamental theorem of galois theory and it's pretty nice
I like it
quiet cave
When I saw that for the first time I thought it was very satisfying
fickle brook
given a commutative ring R is it always possible to find a noncommutative ring S that has R as its center
thorny slate
R<x,y>?
chilly ocean
O alright, thanks
remote rampart
Sorry
somber bramble
is actually set theory or sth
chilly ocean
lmfao
fringe nexus
Let D be an integral domain, and let a in D such that a^2 = 1. Prove that a=1 or a = -1
i feel like this is obvious but my brain is not working rn
fickle brook
yup
fringe nexus
ok
also for the factor group $(Z_4 \cross Z) / (2Z_4 \cross 2Z)$ has order 6 right? because $z_4 / 2z_4$ is {(0,2)(1,3)(2,4)} and $Z / 2Z s just Z_2$
cloud walrusBOT
Victoria:
fickle brook
(2,4)?
fickle brook
yes it does
fringe nexus
yea since 2z_4 has order 2 and z_4 has order 4
we know the factor group has order 4/2
oops
fickle brook
and it's iso to Z_2 x Z_2
plain sequoia
Anyone has any hints?
Show that P has no more than $3h^2$ elements is the problem.
cloud walrusBOT
markus:
fringe nexus
Pretty sure you just use the fact that (xa)^3 is the identity to get that a^3 is identity
Because x is in H, and identity exists in h, so then (xa)^3 becomes
A^3 when you let x be identity
chilly ocean
but that may not be necessarily satisfied for all a in G
chilly ocean
oh wait
fringe nexus
Doesnt that mean a^3 is identity then
chilly ocean
lol i read it incorrectly
fringe nexus
Yea and then i think you rearrange the product somehow to use that?
plain sequoia
Hm okay I see
thanks
If a^3 is the identity one may just go ham and multiply it in everywhere and probably get somewhere
fringe nexus
Yea maybe
Find all abelian groups, up to ismorphism, of order40=2^3*5
Why isn't this just z_8 X z_5 byfundamental theorem of finitely generated abelian groups?
is it because of z_2 x z_2
plain sequoia
Remember that $\mathbb{Z}_{nm} \simeq \mathbb{Z}_m\times \mathbb{Z}_n$ if $\gcd(m,n)=1$.
cloud walrusBOT
markus:
fringe nexus
but z_2 x z_2 x z_2 x z_5 is not isomorphic to z_2 x z_4 x z_5 right?
cloud walrusBOT
markus:
plain sequoia
By FTFGAG: Starting from $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_5$, and then "gathering groups" in terms of those who are not isomorphic: $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_5$, $\mathbb{Z}_8 \times \mathbb{Z}_5$
cloud walrusBOT
markus:
plain sequoia
So yes 3
On the last you could just write $\mathbb{Z}_{40}$ if you want
they are isomorphic anyway
cloud walrusBOT
markus:
fringe nexus
oh ok that way is so much simpler
just starting with all the individual cyclic groups
plain sequoia
Yeah thats the smartest way
or I mean its the way that it hardest to fuck up and missing some combinations
chilly ocean
Hey, so some of us are wanting to get a reading group or two going for some introductory algebra.
We are going to be working with either Judson - Abstract Algebra: Theory and Applications
http://abstract.ups.edu/download/aata-20180801.pdf
Though we are considering Aluffi's Ch.0 as well.
marble wagon
aluffi is somewhat special in that it introduces categories early
this might be good or bad depending on what you want
solar wyvern
also not really an intro algebra text tbh
those are two completely different directions btw
marble wagon
it's an intro algebra text if you're really into category theory for some reason
but idk
kinda odd
chilly ocean
Well, if you have interest in joining the Aluffi group or the Judson group, there's a server.
Mniip & Kaynex are in it among others.
marble wagon
not really but hmu if you need help whenever
plain sequoia
I am soon done with my intro course in algebra but I really want to discuss with you anyway, if that is possible?
chilly ocean
with whom?
dim trench
@chilly ocean where's the aluffi server?
chilly ocean
dm
marsh badge
🐷
plucky cradle
so guys im reading a thing online about abstract algebra and im trying to understand this, but it really confuses me, especially the 3 first lines
can someone explain pls
first 2 lines*
fringe nexus
What do you not understand, these follow from the definition of union and intersection, as well as set difference
The union of A with B is the set of elements which are in A or in B
plucky cradle
oh oops, i forgot to write context
but what do they mean by the union of A and A=A
for example
plucky cradle
so everything in A
fringe nexus
yes
plucky cradle
is A
fringe nexus
huh
A U B = {x | x in A or X in B}
this is what union means
so then A U A = {x | x in A or x in A } = A
stone fulcrum
If an element is in A or B, then it is in A ∪ B.
A ∪ B is a combination of A and B, so to speak
plucky cradle
ok
but it is obvious that something x in A or A will be in A so I was wondering if I was missing something
but thx
wind cypress
Sometimes obvious stuff must be stated
plucky cradle
i have learnt that today
wind cypress
And obvious stuff is often very hard to prove.
For example today I tried to prove something very obvious and failed, last week too 😃
I mean, I assumed too much was known.
plucky cradle
indeed
chilly ocean
@plucky cradle
How are you with propositional or aristolean logic?
like truth tables and stuff
plucky cradle
i have not a single idea what the terms you used meant
i pretty much just started reading about abstract algebra
wind cypress
Me too, I guess
chilly ocean
truth tables?
daring wolf
does anyone know how does modulos affect the rank of a matrix?
if I have a matrix that has modulos 3, and I need to construct a 5x5 matrix with rank that is 3, how would it work
let's say one of the entry was 2
and if I multiply it by 1/2 would we be multiplying the 2^-1 of mod 3
or just regular multplication of 1/2
sorry not rank
i meant spark*
chilly trellis
I hope that I am asking the right place, as it is linear algebra (with a twist). I need a book about Finite fields as a vector space over a sub field. A general book about this stuff would be really helpful.
fickle brook
a book about vector spaces specifically over finite fields?
chilly trellis
Yeah - I dont even know if they exists
fickle brook
i'm not sure they do
chilly trellis
The whole "F_{q^n} over F_q has dimension "n" because of stuff seems wierd - im trying to catch up
fickle brook
well i mean idk
that alone isn't terribly counterintuitive, is it?
for any field K, any n-dimensional vector space over K is isomorphic to K^n
chilly trellis
Not at all, but i saw some Dim(F_{q^n}/F_{q^m})=n/m on the internet, and that is infact what I need - i just wanted a proper source for it 😮
marble wagon
Prove it
chilly trellis
Ah sounds simple, i'll just do that 😐
marble wagon
Show that if n|m then Fp^n is a subfield of Fp^m
Show that this makes it into a Fp^m vector space ( and in fact an algebra)
with the field multiplication operation
Using the fact that a k dim VS over a field of size q has size q^k, conclude
The nontrivial claim is the first one, you can find a characterization of finite fields in any galois theory book
chilly trellis
I found a proof of the first one. It however depends on a "fundamental theorem of Galois theory" which is a side track i'd rather not take.. I'll look abit more around - thanks for the help though !
marble wagon
you don't need that much
you need some results about finite fields
the main result is that they are all of the form Fp^n
satisfying the equation x^(p^n)-x = 0
using this equation you can find the subfield of Fp^m that is Fp^n
what you found is probably going the other way
showing that the field of fractions of x^(p^m)-x in Fp^n is Fp^m
so it is an extension
uncut girder
0->Z{p^n} -> A -> Z{p^m} ->0
Classify all abelian groups A that make this exact
We know that A in generared by the image of 1 under the the injection and any element of the premiage of 1 under the surjection
oblique river
If I had to guess, it's just going to be A = Z{p^n} x Z{p^m} and A = Z{p^(n+m)}
woven delta
That's not the case
oblique river
interesting, maybe i'm interpreting the ext group incorrectly
can you give an example?
oblique river
are you sure? the only elements of order 4 in the middle group are of the form (2a, b) for a odd, b = 0 or 1. But it seems like all of those quotients are going to have an element of order 2
woven delta
It would have been a lot nicer
woven delta
Z_4 has an element of order 2
oblique river
so the latter map is (a,b) --> a + 2b?
woven delta
Yeah
oblique river
so I do know that Ext(Z{p^m}, Z{p^n}) = Z{p^(min(m,n))}
but I guess interpreting those extensions is kinda tricky
my guess is something like
the generator of that ext group corresponds to Z{p^(n+m)} (well that's not a guess, I know that's true)
then "p" in the ext group corresponds to Z{p^(n+m-1)} x Z{p}
and then "p^2" corresponds to Z{p^(n+m-2)} x Z{p^2}
and all the way down to the 0 element in the ext group which is Z{p^n} x Z{p^m}
(which, once again, I know for sure that that's the 0 element)
I'm pretty sure that the isomorphism class of the group in the middle (which is weaker than the isomorphism class of the extension) only depends in this case on the order of the element in Ext
i.e. all the generators give teh same group, all the p*gens give the same group, etc
so then you just have to classify what those all are; if you can verify then that the extensions I wrote above of the form Z{p^(n+m-k)} x Z{p^k} are in fact extensions
then some counting argument should show that those are all of them
@woven delta
marble wagon
it's because he didn't post part (a)
this is part (b)
part (a) makes it clear how to construct them all
woven delta
@uncut girder
uncut girder
Ext is too advanced
marble wagon
yeah
uncut girder
Im just a noob
oblique river
yeah but that doesn't change whether or not it's correct lol
really I just wanna know if I "still got it" concerning abelian group extensions hahahaha
marble wagon
there's two conditions which you can get fast
- it has two factors
- one of them is order at least p^max(m,n)
and these are all because you can build all of them
with a generalization of part (a) which is a small case
covert vector
just tell us what part (a) is already 
marble wagon
dont remember but I can look it up
oblique river
do you mean p^min?
woven delta
p^max
oblique river
and yeah I see now
marble wagon
yeah p^max
woven delta
Since you need the bigger group to imbed inside of it
oblique river
yes
marble wagon
hatcher btw
woven delta
If A= Z_p^k \oplus Z_p^i does the map (a, b ) \mappedto a+bp^i work?
And then the kernel is every element of the form (-bp^i, b)
marble wagon
something like that
woven delta
Okay sure
uneven forum
can someone tell me how many elements of order 2 are in the external direct product of the integers modulo 2 with itself thrice?
latex $Z_2\bigoplus Z_2 \bigoplus Z_2$
cloud walrusBOT
vlazer4:
uneven forum
ah you dont need the keyword but whatever
how many elements order 2
oh wait one of those is a 4
i guess it dont matter tho
marble wagon
all of them are order 2
uneven forum
(0,0,0)
uneven forum
order is the \textit{least} integer n such that $a^n=e$\
cloud walrusBOT
vlazer4:
marble wagon
yes I know what order is
every element in Z2 x Z2 x Z2 except the identity is order 2
so it's trivial to count them
in general the elements of order at most k in a product are the products of elements of order at most k
uneven forum
So it's 7
marble wagon
yeah
uneven forum
what if its z4 x z2 x z2
marble wagon
still 7
because of what I just said
there's 2 x 2 x 2 elements of order at most 2
uneven forum
but z4 has elements of at most 4
marble wagon
it has 2 elements of order at most 2
scenic harness
Why does Maschke's theoren proof fail for Field=Rational Numbers?
Where in the proof does it fail?
covert vector
t!wiki maschke's theorem
fossil mangoBOT
covert vector
oof can't use Chrome on data -_-
I'll check later..
scenic harness
It's ok, thanks
oblique river
@scenic harness what do you mean? Maschke's theorem is true for the rationals.
Maschke's theorem just says that if the characteristic of K does not divide the order of the finite group G, then every rep of G over K is semisimple, i.e. a product of irreducible reps
scenic harness
@oblique river
Oh... So im not crazy then. Thanks bananas!
oblique river
haha np
marble wagon
some books will state C right away to avoid conflict
but the two conditions you really want is that char k does not divide |G|, and that the field is algebraically closed
at least for most results in the sharpest case
scenic harness
They created conflict in my mind
scenic harness
Oh my message got eaten. Bad connection
I'm using liebecks
Algebraically closed will come later right? You don't mean that it's too relevant to Maschke's right?
But i can see how its deeply related to submodules
marble wagon
yeah it's not relevant for maschke's theorem
but it makes weddenburn's theorem nicer
for example if you decompose R[G] into irreducibles you'll get stuff with complex numbers and quaternions in principle
while C[G] you'll only get complex stuff
marble wagon
yeah division rings over non algebraically closed fields are annoying
division rings over algebraically closed fields are just itself
scenic harness
I don't understand how R[G] gets you quaternions. R is in C so it should only get you stuff in C[G]
But im not to familiar with this yet i suppose
marble wagon
this is the relevant statement
hmm that's not exactly how it works
but yeah I'm being vague
lyric grove
what is R[G]?
lyric grove
formal sums in G with coefficients in R?
marble wagon
weddenburn's theorem says that k[G] splits as a product of matrices over division rings over k
so you could get matrices over Q
concievably
scenic harness
F algebraically closed didn't have non trivial finite extension because it being a finite extension says the powers of alpha are linearly dependent such that the minimal polynomial of alpha isn't linear right?
Or how do you justify it?
marble wagon
finite extensions are algebraic
so yeah what you're saying is that you can find a polynomial without roots
you can do that
scenic harness
Cool, thanks
marble wagon
👍
chilly ocean
so why or why not can there be a cyclic group with 4 distinct generators?
fringe nexus
Z 5
woven delta
The number of generators is the Euler phi function of the order of the group
That too
fringe nexus
Finite cyclic groups can have more than two generators, but infinite ones can not
woven delta
^
fringe nexus
All finite cyclic groups are isomorphic to zn
woven delta
I think Z_5, Z_8 and Z_10 are the only cyclic groups with 4 generators actually
solar wyvern
can you prove this
scenic harness
Using euler pji?
woven delta
Yeah
woven delta
Oh, also Z_12
marble wagon
ok yeah
woven delta
What's a good lower bound on the phi function?
marble wagon
apparetly ~sqrt(n) https://math.stackexchange.com/questions/301837/is-the-euler-phi-function-bounded-below
fringe nexus
given any two arbitrary rings z_m and z_n, how do you find how many ring homormophisms there are from z_m to z_n ?
marble wagon
are they arbitrary rings or are they Z/kZ
fringe nexus
Z/kz
marble wagon
the additive groups are cyclic
fringe nexus
Sorry worded that kinda bad
marble wagon
all you need to know is where 1 goes
and that determines a unique map of the additive groups
fringe nexus
I get that phi(1) = a implies a = a^2
marble wagon
so it remains to check which of those induce ring maps
fringe nexus
so do i just map the generator to any value that satisfies that
uh say if i had z_4 - > z_ 6
Then i can map phi(1) = 3 because 3^2 = 3 mod 6 right?
and then theres always phi(1) = 0 and phi(1) = 1
fringe nexus
ok
marble wagon
the image of a map Z/kZ -> Z/jZ just needs to be a quotient of Z/kZ
so you can check how to fit those into the image
I dunno
there's a few equivalent ways of checking which ones work
fringe nexus
those are group homomorphisms arent they
I think my example was a ring homomorphism
marble wagon
ring homs are in particular group homs
of the additive and mutiplicative groups
here's the full picture https://math.stackexchange.com/questions/263063/the-number-of-ring-homomorphisms-from-mathbbz-m-to-mathbbz-n
basically two conditions
the additive group condition that I mentioned and the multiplicative condition that you mentioned
fringe nexus
O ok
chilly ocean
O sorry so why can infinite cyclic groups only have 2 generators?
fringe nexus
Because all infinite cyclic groups are isomorphic to Z
and Z only has two generators, 1 and -1
woven delta
@chilly ocean what do you mean?
fringe nexus
isomorphic to <Z, + > yes
plain sequoia
@chilly ocean Z in any binary op wouldn’t even form a group in some cases
Take $\langle \mathbb{Z}, \cdot \rangle $ where $\cdot$ is normal multiplication. Then there are no inverse for numbers other than $1$ and $-1$.
cloud walrusBOT
markus:
scenic harness
Maschke's again
In
https://en.m.wikipedia.org/wiki/Maschke's_theorem
It says the case for arbitrary K is a corollary to the specific case of K=C. Why does the 2nd follow from the 1st?
Is there some sort of homomorphism C->K or sth?
(I understand the theorems fine)
scenic harness
What type of morphism?
scenic harness
In the category of KAlgebras for any K or in the category ofKAlgebras for fixed K?
The first i assume?
scenic harness
So the image is both a KAlg and a CAlg?
marble wagon
yeah
scenic harness
Do you mind saying what orphim of specifically for K={0,1,2}?
marble wagon
what
is that a finite field
needs to be a subfield of C
because then you have an inclusion K -> C
so C is a K-algebra
and you can take the tensor product of K algebras
of K[G] and C
scenic harness
I must be misunderstanding something but im on my phone atm. It helps tough thanks
crisp totem
??
covert vector
(g o f) (x) = g ( (f)(x) ) = g(f(x))
crisp totem
i didnt realy undrestand the part when they said a is an element of a
fickle brook
well
fickle brook
presumably you have f: A -> B, g: B -> C, and h: C -> D
crisp totem
yes
crisp totem
yesh
fickle brook
h . (g . f) and (h . g) . f both have signatures A -> D
fickle brook
no
crisp totem
wait wat
fickle brook
the signature specifies only the domain and codomain.
fickle brook
that's what you're proving.
crisp totem
bruhv
fickle brook
you are now to prove that the values of h . (g . f) and (h . g) . f are the same at all points of A
crisp totem
ok
fickle brook
it's obvious if you give it a little thought
but even the obvious must at some point be spelled out
crisp totem
sorry i slept at 11 for the past few days now i feel drunk
fickle brook
then you should make sure you are well-rested before continuing.
crisp totem
ok
fickle brook
math and lack of sleep don't mix well
crisp totem
:C
wind cypress
This isn't really algebra, I mean, functions like that?
Gotta read some more about groups on my own, otherwise the course I'm on is in trouble
open wraith
It's set theory
that's not algebra
purely set theoretic results
Algebra is about sets endowed with extra structure (groups, fields, vector spaces, etc)
somber bramble
ay we're proving Abel-Ruffini today it seems
brisk granite
How would i show that a symmetric group meets all the criteria to be a group
fickle brook
by a symmetric group, do you mean the full permutation group, i.e. the set of all bijections from {1, ..., n} to itself, with composition as the operation?
if you actually write out all the group axioms they become kinda obvious
brisk granite
I get how an inverse exists
fickle brook
well
brisk granite
An identity permutation clearly exists
fickle brook
yes
brisk granite
But how do I show associativity
fickle brook
associativity follows from the fact that function composition is always associative
fickle brook
for any f: A -> B, g: B -> C, h: C -> D, (h . g) . f = h . (g . f)
it's not that hard to show if you haven't already
((h . g) . f)(x) = h(g(f(x))) = (h . (g . f))(x)
have you not proved that yet
brisk granite
((h . g) . f)(x) --> I don't understand this notation. What does it mean?
fickle brook
the function (h . g) . f, evaluated at x
the dot stands for composition
bc i'm too lazy to look for the small circle symbol
brisk granite
Let n be a positive integer, and let µ_n = {z ∈ C : z^n = 1}. Show that µ_n is a group under multiplication. How many elements does it have?
How many elements does it have?
marble wagon
the elements are the roots of z^n - 1 = 0
brisk granite
so, there are an infinite amount?
somber bramble
for a fixed n
it’s the n-th roots of unity
if you’re familiar with complex numbers, there’s a very easy way to write them all down
also, a polynomial of degree n has at most n roots
brisk granite
I'm not aaaaaaat alll familiar... do you have any good sources for learning about complex numbers
marble wagon
somber bramble
depends a lot on the level you wanna learn about them
marble wagon
read this and whenever you don't know a concept google for the concept
somber bramble
https://acko.net/blog/how-to-fold-a-julia-fractal/ this is the best intro to them I’ve seen so far
I just got a security warning when opening it but the site’s fine
has lots of animations tho so open on pc, not phone
brisk granite
alright, I'll check it out
somber bramble
basically the most important thing about complex numbers is understanding how complex multiplication is just rotating and stretching
(and addition is just vector addition because ℂ is just ℝ² with multiplication)
if you want a much deeper understanding, needham’s Visual Complex Analysis is very good and the first chapter should be understandable without much prerequisite knowledge, but you’d probably profit more of it when you’ve already seen some complex analysis… and despite being easier it’s usually only taught after real analysis
(in particular for complex analysis you wanna be fluent in like… “functions”, and in particular also know some stuff about e.g. power series)
somber bramble
but the basics about complex numbers are really not difficult and if you’re having problems ask away in uh… idk which channel tbh
fickle brook
any of the ten
somber bramble
eh idk, conceputal stuff doesn’t go there in my eyes
marble wagon
oh yeah I guess the first chapter of some complex analysis books usually has good stuff
fickle brook
or you can pm me. i accept pms as long as you don't act like a dick
somber bramble
like, understanding sth
I don’t accept pms from strangers in general but you can ping me
that is, you in particular
:P
fringe nexus
If G is a group, and N is a normal subgroup of G such that G/N is abelian, show that $aba^{-1}b^{-1} \in N \forall a,b \in G$
cloud walrusBOT
Victoria:
fringe nexus
Let $\phi$ be a homomorphism from G-> G/N Such that$ ker(\phi) $= N. Then, since G/N is isomorphic to $\phi(G)$, we have $\phi(aba^{-1}b^{-1}) = \phi (a) \phi(b) \ phi(a^{-1}) \phi(b^{-1}) = \phi(a a^{-1}) \phi(b b^{-1}) $by abelian property of G/N , so $aba^{-1}b^{-1} \in Ker(\phi)$
cloud walrusBOT
Victoria:
uncut girder
I thinks that's good