<@&286206848099549185>
#groups-rings-fields
woeful sand
visual lion
@woeful sand the definite integral is a linear transformation for its real valued interval and the indefinite integral is for integrable functions to differentiable functions, both real valued as before (fixed point)
woeful sand
thanks, some guy was saying to me that derivative isnt the inverse from integration because derivative its a linear transformation but integration its not ..
spark plank
@woeful sand
0) wrong channel, this is for #groups-rings-fields,
- don't post the same question in multiple channels
visual lion
@hot folio it should mean that the function/operation uniquely defines values a,b through the operation given
woeful sand
but its here in #groups-rings-fields
visual lion
Calculus
spark plank
Already answered this time, so don't post again tho
visual lion
Where I linked
spark plank
Already posted in #calculus and #precalculus but I deleted them since it has already been answered
tardy ember
quiet cave
If you can pack an object in a box of any volume you wish, then that means the object had no volume to start with
solar wyvern
@BunnyHops #multivariable-calculus
chilly ocean
hey dumb question but its simple
on finding the orbits of a permutation, if an orbit goes to itself, and is just goes to itself, is it considered an orbit? hope my question makes sense
covert vector
an orbit is a subset of the set that the group is acting on
visual lion
Yeah I'm kind of confused on what you mean by the orbit goes to itself
chilly ocean
let me write something
covert vector
when you say orbits of a permutation, are you saying G is acting on a group?
chilly ocean
say you got it in two-row form
covert vector
Ok
chilly ocean
(1234567)
(7632135) so that untimately = (175) (3) (264). now since three goes to itself does it count as an orbit? Im trying to count the amount of orbits I have, and if three does count its 3 orbits, but if it doesnt its 2 orbits. Hope I cleared myself up
chilly ocean
yes exactly
covert vector
it's confusing because orbits are another group theory word lol
this is a product of 2 disjoint cycles
chilly ocean
ok so it is 2 got it!
covert vector
like, if you had the identity permutation on n elements
you wouldn't call that a product of n cycles
more like 0 cause it doesn't do anything
chilly ocean
How do you go about proving that a subgroup is closed under an operation?
chilly ocean
For instance, showing that $\mathbb{Z_4}$ is closed under addition. I assume you just observe that every power (with respect to the operation) of each element stays within the group itself, and then you just move on.
I dont know how tex works on this server, sorry
stone fulcrum
You can just list them all, yeah
chilly ocean
How about very large finite groups or infinite groups?
stone fulcrum
Since you're dividing the result by 4, the remainder will always be less than 4. So, the operation is closed
chilly ocean
Okay, so invoke the fact that the group is generated by taking mod 4 and show that each element in the group satisfies the remainders of every integer
Which, it does by construction really
stone fulcrum
That result should work for Z146, if you so choose
chilly ocean
Okay, that works for my current problem. I guess in the future if I deal with showing closure under an operation for large groups that I can't list out all the elements of, I can look at how the group is constructed and show that it has to be closed because the construction is injective. Meaning there is no element that can break free of the group using the operation because of how the group was constructed
Anyways, thanks for the help.
stone fulcrum
Np. Feel free to ask if you have anything else!
restive kiln
Does anyone know what it might mean notationally to write out the product of a prime ideal with the localization of a ring with respect to said ideal?
$$\mathfrak{P}R_{\mathfrak{P}}$$ where $$R_{\mathfrak{P}}=S^{-1}R,; S=R-\mathfrak{P}$$
mellow vaporBOT
covert vector
@restive kiln
it is the product of elements in P and elements in R_P
though I guess that isn't very informative
turns out PR_P is the maximal ideal of R_P
wait, notationally
(deleted unnecessary explanations as I misinterpreted)
R_P is a collection of fractions a/b
full blaze
my post doc teacher specializes in Universal Algebra
covert vector
R is taken as a subset of R_P, where you identify r\in R with r/1
full blaze
covert vector
$$R$$ is taken as a subset of
$$R_ \mathfrak{P}$$ notationally as $$r \mapsto \frac{r}{1}$$
mellow vaporBOT
timber bay
dont really know how to begin either of these
i was hearing something about commutativity for 4, but didn't really see it. i saw that distributivity had to be obeyed, but I didn't think there were examples of groups that didn't obey distributivity
taking a modern algebra course
stone fulcrum
By definition, it's a homomorphism if
φ((m, n)+(a, b)) = φ((m, n))φ((a, b))
φ((a+m, b+n)) = φ((m, n))φ((a, b))
g^(a+m)×h^(b+n) = (g^m×h^n)(g^a×h^b)
g^(a+m)×h^(b+n) = g^m×h^n×g^a×h^b
Note that because groups are non-commutative, this statement is generally not true. But if g and h commute, then it works
timber bay
oh okay. i didn't expect you to be adding (m,n) and (a,b). I always see homomorphisms described as f(ab) = f(a)f(b)
where f is like phi ofc
stone fulcrum
I assume the group Z×Z is using addition as its operation, since the math works out well due to this
timber bay
thats been confusing to me, and i should go to office hours to work on that
looking at and understanding the right operations
stone fulcrum
So like you said:
f(ab) = f(a)f(b)
ab is done in the first group, and uses that operation
f(a)f(b) is done in the second group, and uses THAT operation.
timber bay
so I use the domain op for f(ab) and the codomain op for f(a)f(b)?
stone fulcrum
a and b are each in the first group. Multiplying them uses the first group op.
f(a) and f(b) are each in the second group. Multiplying them uses the second group op
ocean marsh
The crux of this is that you need e=phi(0)=phi(1+n-1)=phi(1)phi(n-1)=phi(1)^n=g^n
So g^n = e is necessary. Can you prove that it's sufficient?
timber bay
im not exactly sure what sufficient means in this case
ocean marsh
Sufficient would be proved by showing that if g^n = e then phi is a homomorphism
So for arbitrary a and b in Z/nZ, you want to prove that phi(a+b)=phi(a)phi(b)
timber bay
im not sure really how you come to that conclusion
i see that the domain is finite/cyclic
ocean marsh
Pretty much by using the definition of phi, which is the only thing you have
phi(a)phi(b)=(g^a)(g^b)=....
can you see what to do next?
timber bay
well i think im almost there
i see that one of them, with the power that's the order, and itll cancel out
but im kind of just following a pattern
ocean marsh
admittedly this is a bit tricky to notate because there are multiple types of addition running around
but if a+b>n-1 (in terms of regular integer addition)
I'll use a * for addition in Z/nZ.... then a*b=a+b-n
So phi(a)phi(b)=g^a g^b=g^(a+b) = g^(a+b-n) g^n=g^(a+b-n)=phi(a*b)
And then if a+b < n, you just have phi(a)phi(b)=g^a g^b=g^(a+b) =phi(a*b)
restive kiln
@covert vector I managed to figure it out on my own, but thank you!
timber bay
that makes sense @ocean marsh thanks a lot!
timber bay
yah mane
sonic current
hello
i'm trying to find a group that has got four idempotent elements other than the identity (so, two pairs of idempotent elements)
therefore getting x^2 = id = y^2 (if x and y were such elements)
getting to x = y
is this correct?
i'm not really confident in my thinking here, this seems too easy and I'm wondering if there is a quick group to think of that might have two pairs of idempotents except for the identity
whoa this is not right at all I cant cancel xx = yy to x = y
i'll try to build a cayley table
and see if i can figure out a group with two pairs of idempotent
stone fulcrum
Z2×Z2×Z2 comes to mind.
[1, 0, 0] + [1, 0, 0] = [0, 0, 0]
ect
That's got 7 idempotents though
sonic current
haha nice one
stone fulcrum
Do you want exactly 4?
sonic current
no
i wanted to check if it was possible to have a group of finite order
and more than finite idempotents > 2
basically this
i'm at e)
but is Z2 x Z2 x Z2 a group? i suppose so
stone fulcrum
Stringing more Z2s, you can have as many idempotents as you want
sonic current
cool, thanks
stone fulcrum
Np. Feel free to ask if you have anything else!
sonic current
i'm at the beginning of the course, but I'm doing some simple exercises so I can tutor a friend of mine
stone fulcrum
Very cool! Abstract algebra is a great course imo.
sonic current
well, it's cool but pretty hard to grasp I suppose
if you're new
a lot of concepts and weird techniques
huh
weird notation
is A always a matrix?
jagged gate
here yeah
M_n(R) represents the set of all square matrices of order n with real coefficients
sonic current
yeah but a) and b) are different to c) and d)
worthy kindle
Indeed
sonic current
and the first two are weird. a little unconfortable
makes me think I'm dealing with individual squares instead of the whole matrix
jagged gate
you mean those two are different between a/b and c/d ?
sonic current
yeah
specially without info on the i and j
i realise they're indices but still
seems a bit sloppy
jagged gate
they're essentially the same thing (or correct me @worthy kindle )
worthy kindle
They added the bracket thing becausd they needed names for the matrix coefficients
That's all there is to it
jagged gate
ah c'est pas con ça en effet
sonic current
quick thinking @worthy kindle lol
worthy kindle
Haha
sonic current
this is why I cannot 100% agree algebra is a fun course
do you guys have a strategy for this sort of exercise other than ramming away with random combinations until you arrive at something?
i found the solution online but seriously. question still stands
stone fulcrum
Looking for a group with elements a, b, e, such that a⁵ = e, aba¯¹ = b²? What's the 31 bit?
worthy kindle
Like this I guess
The exercise has a^5 = the neutral element, so you're probably meant to use it in some way, which means you had to iterate a somewhere in your steps
More generally, try to use everything the exercise gives you
sonic current
hmm -- thanks @worthy kindle and @stone fulcrum
worthy kindle
you're welcome
covert vector
I guess the first trick is to recognize that 2^5=32 = 31+1
then try manipulating things to go in that direction
oak perch
OH. nvm
errant drum
If G is a subgroup of S(n) [Symmetric group of size n] how do I show G is either a subset of A(n) [the group of even permutations in S(n)]. Or half of G is in A(n).
I'm stuck on the first bit
hot folio
When checking if a group is abelian, specificly in checking if e is an indentity or checking if x^-1 is an inverse , can i just cut my way through since i already proved that the operation is commutative, and say "since the operation is commutative then x * e = e * x = x " ?
hot folio
yeah but you have to check both cases if it isn't abelian
errant drum
The identity is always abelian
Since a * e = a * (a^-1 * a) = (a * a^-1) * a = e * a
hot folio
okay
pulsar jolt
@errant drum Let H be the intersection of G with A(n). This is still a subgroup, now of both A(n) and of G. If H=G, you're in the first case. If the index of H in G is 2, you're in the second case. Otherwise, let x be in S(n) and not in A(n). Since A(n) has index 2, there are only two cosets. We have that G is the disjoint union of H and xH, and if you're not in the first case both must have the same size, so you're in the second case.
errant drum
Ahh thanks for your help.
chilly ocean
errant drum
I'm only considering the the set polygons that have an even number of sides. So n=2k for some k. You can rotate the shape by an angle equal to it's external angles (e.g. a hexagon has a 60° external angle and rotation of an angle by 60° a group operation)
There are 6 rotations (0° which is the identity, 60° ,120° etc).
Is my proof here sound?
Pages uploaded in reverse order :/
errant drum
<@&286206848099549185>
chilly ocean
What are you trying to prove?
uncut girder
^
errant drum
So for a polygon one of the group operations is rotation right
More specifically rotation by the external angle of the polygon
It's even sided regular polygons
Basically the rotation operations of the dihedral group
For example a hexagon 's exterior angle is 60°
Rotating it 60° is an odd permutation
errant drum
I guess I should have said rotating it by 360/n
I want to show it's an odd permutation
The question said external angle that's why
chilly ocean
I don't think it is an odd permutation
This permutation is a composition of even number of 2 cycles
gentle pendant
(123456)=(12)(13)(14)(15)(16)
chilly ocean
1 ->2; 2->3; ... 2n->1
gentle pendant
odd
chilly ocean
oh, maybe i am wrong
gentle pendant
in general a k-cycle will be even iff k is odd
errant drum
I tried this for a square and a hexagon
Every one rotation it came out odd every 2 rotations it came out even
gentle pendant
👍
chilly ocean
Wait where am I wrong then?
gentle pendant
write (123456) as a product of an even number of transpositions and I will tell you
or (1234)
or (12)
chilly ocean
(12)(23)(34)(45)(56)(16)
gentle pendant
what is your order of multiplication, are we going left to right?
errant drum
Now my idea was that if I can prove it for a 'one' rotation I can compose two of them to make a 2 rotation and since odd composed with odd is even I can show that it alternates
Right to left like functions
gentle pendant
was talking to b4tya
chilly ocean
left to right
gentle pendant
cool, so if you follow where 1 gets sent by your product, you will see it gets sent to itself
not to 2
chilly ocean
got it
errant drum
To determine whether it was even or odd I used this definition..
Basically just counted
uncut girder
Oh that's nice
gentle pendant
Whilst it is more clear that this is well-defined, it is often more convenient to work with the definition:
odd: can be written as a product of an odd number of transpositions.
even: can be written as a product of an even number of transpositions.
After which (123456)=(12)(13)(14)(15)(16) is basically a proof.
(Switch the order of brackets if you prefer right to left)
errant drum
Yeah so for (12....2k)=(12)(13)...(12k) and there are 2k-1 of those
gentle pendant
yep
errant drum
Yeah that's a lot easier can't believe I took the long route 😅
uncut girder
It's important to realize what your saying essentially amounts to the same. Each of those inversions that you counted is represented here as a transposition
errant drum
Yeah
gentle pendant
thats not quite true
if
1->6
2->5
etc
6->1
then every pair of numbers between 1 and 6 is an "inversion".
theres no nice correspondence with the transpositions in general, becase they will be composed with each other.
uncut girder
But in the case of a cyclic permutation there is this correspondence and that's what made it easy to count the number of inversions
In a general permutation you wouldn't count the number of inversions to get the parity. That's too hard
What you would do is break it up into disjoint cycles and then use this idea that it's easy to count inversions in cyclic permutations .
Basically I was just commenting about (1 2 .... 2k)=(1 2)(1 3)...(1 2k)
covert vector
idk what V is
but if it's a reflection
note that reflection + all rotations = all reflections
and (R90)ⁿ = all rotations
so you have all rotations and all reflections, but that's the whole group @vestal needle
also: Today we're covering group group actions in group theory class. the prof defined it so that G has to act on a nonempty set X
and I asked why we forbid X to be empty, and he said there's some category thing you can do but we're not interested in it
so now I'm curious as to what that is
ocean magnet
V is the rectangle group https://www.youtube.com/watch?v=CCThtUg2zBc
timber bay
i have a question about my algebra homework. im almost there, i just wondered if what im doing isnt a logical misstep
im looking at this, and im trying to show that the subgroup has inverses. I'm saying that since H and K are subgroups, they have inverses, so our inverse of HK is h^-1 k^-1. when i operate under this, i get hkh^-1k^-1, but im not sure why i an know that this is the same as h*h^-1 kk^-1 = e
apparently astrices messed up
sry
stone fulcrum
Let h and k be elements in H×K. Then hk is a general element of H×K.
Let's look at its inverse:
(hk)¯¹ = k¯¹h¯¹
H is a normal subgroup, so it commutes.
= h¯¹k¯¹
h¯¹ ∈ H and k¯¹ ∈ Z. So h¯¹k¯¹ = (hk)¯¹ ∈ H×K
@timber bay
timber bay
k^-1 is an integer? Also, K is the normal subgroup, and H isn't necessarily commutable, so does this change anything?
stone fulcrum
Wtf is an integer here?
Sorry, I meant K, that's the one that commutes. Same outcome
@timber bay
chilly ocean
Can someone answer my question on quantum mechanics?
ocean magnet
sure bb
stone fulcrum
Oh shoot I meant K derp
chilly ocean
@ocean magnet fair enough and I don't believe it's standard
Well I'm New because I just finished Soundwave physics
timber bay
Okay no problem. I thought that's what you meant just wanted to make sure.
Thanks! That makes sense.
vestal needle
@covert vector yeah V is a reflection
covert vector
@stone fulcrum @timber bay
H is a normal subgroup, so it commutes.
k¯¹h¯¹ = h¯¹k¯¹
this is not true.
(hopefully I am not too late, turned away from discord to get some work done)
however, it is fixable
timber bay
H isn't necessarily normal?
timber bay
Did my image I sent get deleted?
covert vector
but anyway if H is normal, then for all g in G, ghg¯¹ = h', for some h' in H
so in particular, you can take g in K
no wait 🤔
timber bay
Yeah, so isn't that the same as gHg^-1 = H?
covert vector
ya, but that doesn't mean that ghg¯¹=h
(if that makes sense )
it just maps to something else in H
not necessarily what you had to begin with
I think what you can do is:
k(k¯¹h¯¹)k¯¹ = h¯¹k¯¹
some conjugate of k¯¹h¯¹
hmmmm
idk I remember doing a similar problem before
timber bay
I thought what I said implies it works for every h in H
covert vector
ok hold on
timber bay
So just another h?
covert vector
Ya
covert vector
ok I messed up my conjugate
h¯¹k¯¹ = k¯¹[kh¯¹k¯¹] = k¯¹(h')¯¹
this thing
@timber bay
stone fulcrum
gHg¯¹ = H
Implies
gH = Hg
A normal subgroup is one that commutes with everything in its group
covert vector
wot
normal subgroups are not necessarily abelian
you will have for all g in G, for all h in H, there exists h' such that gh=h'g
@stone fulcrum
but not that h=h'
stone fulcrum
Oh good point, I'm messing up the definition
Ahh, and I see the way you fixed it. Thx!
covert vector
np!
ocean marsh
I think this is a stupid question, but is there generally an injection NH/NK to H/K?
Where they are all subgroups of some group G, with N normal in G and K normal in H.
covert vector
@dense hull
Δ is known as the vandermonde polynomial.
it has the property that if you swap any two variables, the sign flips.
observe that Sn also has the notion of sign of a permutation, which is positive or negative, and a single 2-cycle flips the sign.
furthermore every element of Sn can be decomposed into a composition of 2-cycles.
that should point you in the right direction.
dense hull
so p(x_1, ..., x_n) is p evaluated at x_1, ..., x_n?
dense hull
ok, so it's like
covert vector
for example if n=2, a polynomial may look like
covert vector
p(x,y) = x²+5xy-1
covert vector
np
dense hull
as soon as I read polynomial I assumed single variable
so was so confused
the entire time
covert vector
lol
dense hull
a) also makes so much more sense now
covert vector
👌
timber bay
anybody able to point me in the right direction here?
fallen lion
I have another proof for this statement:
If h in H, then (ah)H = aH, for proof of well-defined we need that aHbH = (ah)HbH,
that is abH = ahbH, so this gives that bH = hbH;
and this implies that H = (-b)hbH, hence -bhb in H => H is normal for any b in G.
fallen lion
Actually, If we have gH ⊆ Hg, then gH(-g) ⊆ H
this implies H = (-g)gH(-g)g ⊆ (-g)Hg = (-g)H(-(-g)) ⊆ H, so Hg = gH
royal cargo
can anyone help me with the steps to reach this?
its too hard for the kids in algebra
royal cargo
fite me 
chilly ocean
If I am wrong, your teacher m
ust
be teaching some other topic and hasn't come to pythagoras theorem
I guess that you are just trolling me now.
royal cargo
no
but you are wrong
im not mad
thts why I light the mood
but you are
incorrect
worthy kindle
Why is this in an algebra channel ;_;
chilly ocean
¯_(ツ)_/¯
sick hearth
is the determinant function linear in multiple rows at the same time?
Is this correct?
I think not
But idk how to prove it
bright fox
=pup is det({a,b,c},{d,e,f},{l,m,n})+det({u,v,w},{x,y,z},{l,m,n})=det({a+u,b+v,c+w},{d+x,e+y,f+z},{l,m,n})
stone fulcrum
@sick hearth
The determinant of an upper triangular matrix is the product of its diagonal entries. This helps create a lot of counter examples
..... [2 0 0] [1 0 0] [1 0 0]
det[0 2 0] ≠ det[0 1 0] + det[0 1 0]
[0 0 1] [0 0 1] [0 0 1]
Lel, amazing formatting
Because 4 ≠ 1 + 1
sick hearth
alriiight, i see it, thank you
sick hearth
For a square matrix A with dimensions nxn
Is this true?
I believe not but I again don't know how to prove it
stone fulcrum
Let A = -cI
LS = det(2cI)
= 2ⁿcⁿ
RS = cⁿ - det(-cI)
= cⁿ - (-c)ⁿ
peak holly
@Raskolnikov#0001 I know this shouldnt be here, but since A = πr^2, that means that 48π = πr^2, right?
so then 48π/π = r^2, and the Pi's would cancel off, leaving you with 48 = r^2.
finally, r = √48, which simplifies to 4√3
and since the diagonal of the square is the diameter of the circle, you would multiply the radius by 2, which would make the diagonal of the square = 8√3
spark plank
@royal cargo has been banned after they left the server. See also: #geometry-and-trigonometry
peak holly
alright
sick hearth
@stone fulcrum thank you
errant drum
If H is a subgroup of G, suppose the elements a,b (which are also in H) are conjugates in G does that mean they're conjugates in H?
covert vector
only if H is a normal subgroup @errant drum
errant drum
I'm struggling to find a counterexample
mellow vaporBOT
covert vector
where $$D_4 = \langle R,T | R^4 = T^2 = (TR)^2 = 1 \rangle$$
mellow vaporBOT
covert vector
then $$T, R^2T$$ are conjugate in $$D_4$$, as $$RTR^{-1} = R^2T$$
mellow vaporBOT
covert vector
but the subgroup $${ 1, R^2, T, R^2T}$$ is abelian, so every element is conjugate to only itself
mellow vaporBOT
errant drum
By T do you mean a reflection?
covert vector
np!
errant drum
How did @covert vector you get to your answer so quickly?
covert vector
by being good at guessing
errant drum
I mean I knew I that I wanted to look for non abelian groups with abelian subgroups
But there are a billion (pun fully inteded) of those
Were there any other criteria for your guessing?
covert vector
@errant drum D4 and Q8 are small groups that generate lots of counterexamples.
errant drum
Ah right I'll keep that in mind
covert vector
it's good to have a collection of easy counterexamples
if you ever have some conjecture to quickly test
chilly ocean
guys I have an abstract algebra question
valid elbow
then ask it
fervent marsh
: (
Was stuck on a proof on my hw all week and didn't bother to ask it here
Turns out it was on the exam -_-
If anyone is interested:
(a in Zn -> a^k = 0 mod n for some k or a inverse exists) iff (n is power of a prime)
ocean marsh
Right to left is mostly just an application of Euler's theorem , and for left to right (by contrapositive) just choose a prime divisor of n and check that it has neither of those properties
fervent marsh
ty det
I don't recall my prof talking about euler's thm
We covered much of Chinese remainder thm and fermat's was mentioned at least
ocean marsh
I suppose it would be a bit harder without Euler's thm (which is a generalization of Fermat's)
near pawn
what is the meaning of rank of a matrix?
worthy kindle
It's the dimension of the vector space generated by its columns (or rows)
worthy kindle
it can be 0 and also can be equal to the number of columns
near pawn
so in a 3x5 matrix the nulity can be 0,1,2,3,4,5?
worthy kindle
what's the nullity?
vestal needle
im having trouble understanding what to do for part d
worthy kindle
0 can divide 0 🤔
vestal needle
i think ive done that part already
covert vector
verify the definition of "divides" @worthy kindle
worthy kindle
o yeah my bad xd
covert vector
so we have that there exists k such that ka=a+b
subtract a from both sides
(k-1)a = b
k-1 is an integer
vestal needle
:o
covert vector
don't divide when you don't need to tbh
well this is kinda divisibility stuff
so, I guess put off actually dividing :P
vestal needle
i see thanks :)
raw moth
... do people not see the number theory channels :^)
spark plank
near pawn
matrix transformation basically means multiply the matrixes right?
near pawn
how do i do this?
mellow vaporBOT
jagged gate
and compute $$\begin{bmatrix} a&b\c&d\e&f \end{bmatrix} \begin{bmatrix} x\y \end{bmatrix}$$
mellow vaporBOT
jagged gate
then identify the different coefficients a,b,c,d,e,f @near pawn
near pawn
how do you determine the dimensions of the matrix?
its it based on [x,y] and [-y,x+6y,3x-9y]?
near pawn
but when we compute it then we get a matrix that equals -y x+6y 3x-9y
i understood
thank you @jagged gate
jagged gate
👌
near pawn
please help
worthy kindle
what exactly is the nullity of a matrix?
near pawn
i dont really understand it
stone fulcrum
The nullity of a matrix is the dimension of its nullspace
This can be as high as the dimension of the matrix itself afaik
stone fulcrum
A matrix like
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
Has a 5d nullspace
gentle pendant
No it doesn't.
stone fulcrum
Or wait, am I flipping my matrices the wrong way?
gentle pendant
The domain is 3 dimensional
stone fulcrum
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
gentle pendant
The rank of an mxn matrix can be as large as the smaller of m and n, the nullspace can be n-r for any possible rank r.
worthy kindle
@near pawn you can just calculate S(T((x1, x2)))
near pawn
do i have to include x1 , x2?
worthy kindle
o wait not sure I even understood the question
I don't think you have to include x1 and x2
near pawn
yeah i have done that but it doesnt work
worthy kindle
have you tried putting the same answer as the one below?
near pawn
i put in the b but it doesnt work
worthy kindle
Do you know if b is correct?
oo I think you did [T]×[S] instead of [S]×[T] @near pawn
worthy kindle
You're welcome
restive kiln
So, I'm trying to use the fact that $$M \oplus N$$ is free in order to prove that $$M$$ is projective. The characterization of projective we're using has to do with surjective maps between modules inducing a surjective map of the morphisms. I.e. If $$M$$ is projective and $$f: P \rightarrow Q$$ is surjective, then there is a surjective map $$i_{f}: Hom(M,P) \rightarrow Hom(M,Q)$$. I feel like I should be able to use the fact that $$M \oplus N$$ is free, and therefore projective, to induce similar behavior on $$M$$. I've been messing with this approach for a few hours and haven't been able to fin a way forward. Is this the correct approach? Or am I doing the wrong thing entirely?
mellow vaporBOT
worthy kindle
Sorry I'm not a native English speaker, can you please explain what does it mean for a vector space to be projective?
restive kiln
Oh, I wasn't clear in my question. M and N are modules
restive kiln
I'm not sure you'll be able to help me, but a left R-module is an abelian group equipped with left multiplication by elements of a ring R
worthy kindle
indeed I won't be able to help you, sorry x')
restive kiln
Anyone else?
I'm still struggling
Looking around online, most discussion of this problem seems to use split sequences. I'm not interested in plagiarizing this strategy, especially since it doesn't appear to make use of the characterization of projective modules that I have on hand.
covert vector
@restive kiln still working on it?
covert vector
(misread)
ya that's a good way to do it
hint that you can take a map f:M →Q, then consider a particular map g:M \oplus N →Q, g(m,n)=f(m)
then write f(m)=g(m,0)
delicate chasm
Oooh fun math!
restive kiln
I did something a bit different where if f: A to B is surjectove, and there’s a map from M to a module B, then said map is only a projection away from being a map from M+N. So there’s a map h from M+N to A such that diagram commutes, but then you end up with something spanned by f{m_i,n_i} being sent to something spanned by the image of {m_i}, meaning that the reduced domain f{m_i,0} should also commute.
@covert vector Argument seems super sketchy though
Idk if it’s even correct
covert vector
that's basically it
I drew a diagram of it last night
then deleted it lol
but if u wanna see
its pretty much what you just said
restive kiln
I kept trying to draw a diagram earlier yesterday, but I wasn't sure how to show that the image of the map from M+N to A didn't include any elements from N
covert vector
except I think I used α and π for the maps
covert vector
we have it in @delicate chasm's mlg pro bot
activation command: ~tex
@restive kiln
restive kiln
Ah okay
If you wouldn't mind, I'd like to see the diagram, just because I couldn't see the answer in my own diagram chasing.
covert vector
@restive kiln
delicate chasm
~tex
\begin{tikzcd}
& A \ar[d, two heads] \
M \oplus N \ar[ur, dashed] \ar[r, "\bar{\alpha}"] \ar[rd, "\pi"] & B\
& M \ar[u, "\alpha"] \ar[uu, dashed, bend right]
\end{tikzcd}
odd forumBOT
Puerøsola:
\begin{tikzcd}
& A \ar[d, two heads] \\
M \oplus N \ar[ur, dashed] \ar[r, "\bar{\alpha}"] \ar[rd, "\pi"] & B\\
& M \ar[u, "\alpha"] \ar[uu, dashed, bend right]
\end{tikzcd}
@covert vector I wish my hand drawn commutative diagrams looked as neat as yours
covert vector
😄
earnest valley
No match.
tacit rune
steep quest
this might be a long shot but does anyone have any recommendations on resources on krohn rhodes theory or even algebraic automata theory in general
sick acorn
I've already shown that the intersection of <a> and <b> cannot be {0}. Any advice for the second half?
i.e., showing that m is lcm(a, b).
(as an aside, I've already shown that lcm(a,b) is a member of the intersection of <a> and <b>.)
sick acorn
@stone fulcrum: <a> is the cyclic subgroup generated by a nonzero integer a
so <a> = {n*a | n is an integer}
rocky spruce
What does it mean for an element to be in the intersection of the cyclic groups?
The goal is to show that any element that is in the intersection is an integer multiple of lcm
sick acorn
Cleaned it up a bit. Thanks again!
Let m be a positive generator of <a> ^ <b>, i.e., <a> ^ <b> = <m>.
WTS: m = lcm(a, b). So WTS: a|m, b|m, and if n is an integer such that a|n and b|n, then m|n.
Clearly, m in <m> = <a> ^ <b>. Then m in <a> and m in <b>. Then m = xa and m = yb, such that x and y in Z. Then a|m and b|m. Let n be an integer such that a|n and b|n. So k1*a = n and k2*b = n, such that k1 and k2 in Z. By definition, n in <a> and n in <b>. Then n in <a> ^ <b> = <m>. So n in <m>. By definition, n = k3*m, such that k3 in Z. So m|n.
We have shown that a|m, b|m, and if n is an integer such that a|n and b|n, then m|n. So m = lcm(a, b).
rocky spruce
Nice!
sick hearth
I don't understand how they made this jump
spark plank
What does union mean?
sick hearth
why does x + y have to belong in either of the subspaces
the union of all the vectors in the subspaces?
sick hearth
but isn't the union a basic mathematical concept?
delicate chasm
We can still define it
sick hearth
i thought you were wondering about what it meant here specifically
delicate chasm
What is the definition of union, in terms of elements
delicate chasm
That is the hint, I guess
First write down the definition of union, then the answer might become clear(er)
(sorry for riding on you Yesus)
sick hearth
but isn't the union able to describe space which wasn't possible in the subspaces by themselves
i mean why can't the sum of two vectors only belong in the union
rather than have to belong in one of the original subspaces
like if we unite a line and a plane we might then be able to describe 3d space
and many of those vectors cannot be described in any of the original subspaces by themselves
delicate chasm
Indeed, the union of two vector spaces isn't usually a vector space
and isn't closed under sums
spark plank
Just answer the question
sick hearth
me?
spark plank
Yes u
sick hearth
idk how to answer tbh
spark plank
Definition of union
delicate chasm
Instead, we can take the smallest vector space containing the union, in the case of a line and a plane (in general position) the thus generated vector space is 3d space
And yeah
definition of union
spark plank
You don't know the definition of the union?
sick hearth
wait
it is all the vectors than can be described as a linear combination of the vectors from the united subspaces?
delicate chasm
Nope, that's the sum of the vector spaces
and it's the smallest vector space containing the union
spark plank
t!wiki union set theory
fossil mangoBOT
sick hearth
i get it
because x+y is like one element and it has to belong to either of them or both
thank you very much for your patience
delicate chasm
👍
spark plank
uncut girder
How to prove fundamental theorem of finite abelian groups easily? Prof is taking way too long to present a proof. It's ridiculous.
earnest valley
Slow down a bit (module, ...).
delicate chasm
Anyone know spectral sequences?
errant drum
I still can't wrap my head around quotient groups any resources I can access for a good explanation?
inner acorn
Hmm, I used Dr Bob's playlist https://youtu.be/hgbnua35tE4
alongside his problem sheets http://mathdoctorbob.org/UR - PS4.pdf
but... he goes very quickly through the content n.n;; almost like bullet revision
http://mathdoctorbob.org/UReddit.html if you're interested
errant drum
@inner acorn I'll check him out TY
snow dew
@delicate chasm spectral sequences?
delicate chasm
Yes
Is the filtration for the anti-diagonal of a homological spectral sequence the reverse of the one for cohomological?
I used that for my last assignment
Hoping it is true
Got an exam which might need it on Tuesday too
snow dew
No idea
Never seen this stiff
From a quick google it looks like arguments I've seen when studying nilpotent linear endomorphisms, in a way more general way
dapper hawk
How do I do 4c?
cerulean rune
Very rough sketch: Fix an element $$x$$ of $$S$$. Use the fact that $$S$$ is finite to find distinct positive integers $$k, l$$, with $$k \geq 2l$$, such that $$x^k = x^l$$. Multiply by a suitable power of $$x$$ to find your element.
mellow vaporBOT
civic linden
Any ideas? I got that <A>= (1,0,n,1) where n is in Z) not sure if thats right
errant drum
Shouldn't it be -n? @civic linden
civic linden
@errant drum ahh yes! my mistake
chilly ocean
@civic linden That's correct
btw even (1,0,n,1) is correct
because as n is in Z, doesn't matter if you forget the - sign
it's false to write A^n=(1,0,n,1)
here it's A^n=(1,0,-n,1)
but to say that <A>={(1,0,n,1) | n in Z}, that's correct
civic linden
yes i was thinking the same, but thank you:)
civic linden
Im have a little problem with a question on order of groups maybe you could help? 😄
thats if you dont mind, and have the time
I'm not really sure what is meant by x^3 etc
civic linden
kind of understand where you're coming from, but still not sure how to approach the question
errant drum
The only hint I'd have is that the laws of exponents apply to an group operation
civic linden
whats the group operation here?
errant drum
E.g (x^3) * (x^2 ) = x^5
I'm taking '*' to be my group operation
It doesn't matter what the group operation is the exponent laws apply
So (x^-4) = (x^4)^-1
civic linden
right
errant drum
And (x^3)^2 = x^6
civic linden
so how does the element x in G with order 16 come into play with this?
errant drum
What does order 16 mean?
civic linden
well the order of an element in a group is the smallest positive integer n
errant drum
That's the first half of the definition
civic linden
such that x^n=1 if any such n exists, if there is no such n then the order of g is defined to be infinite,
civic linden
ah okay
@errant drum yup im still confused xd, i understand how to do this with mod
oh wait a minute
so is the order of x^3 just x^2?
civic linden
anyone have any idea how to do this question?
rocky spruce
So there exists some element that isn't the identity in the group. What is its order?
And, once you've done that, what other elements exist in the group? Then, what happens if there is another element beyond those that you've discovered?
Remember that the dihedral group is generated by 2 elements, one of order 7 and one of order 2
civic linden
ok thanks, ill see what i can come up with 😃
civic linden
mhm what I've come up with is that by lagranges theorem, any subgroup of D14 must have order 1,2,7 or 14. Orders 1 and 14 are ruled out and there the remaing possible order are prime, so any subgroup of these orders must be cyclic
stone fulcrum
Good answer, I like it
civic linden
do you think that is good enough of an answer?
snow dew
Seems good
State the theorem more explicitely in your redaction tho (a group of prime order is cyclic)
civic linden
mhm okay, thanks! 😃
snow dew
(Maybe even prove the above)
civic linden
Let G be a group whose order is a prime p. Since p > 1, there is an element a ∈ G such that
a =/= e. The group <a> generated by a is a subgroup of G. By Lagrange’s theorem, the order of <a>
divides |G|. But the only divisors of |G| = p are 1 and p. Since a =/= e we have |<a>| > 1, so |<a>| = p.
Hence <a>= G and G is cyclic
that any good?
civic linden
Quite confused on this question I understand that D12 consists of the elements {1,r,r^2,r^3,r^4,r^5,s,sr,sr^2,sr^3,sr^4,sr^5} but im unsure on how to determine which element is of which order and hence which are non-cyclic (and the meaning behind non-cyclic) in this context
ocean marsh
Elements can't be non-cyclic. They're just asking you to find an order 4 subgroup of D12 that is isomorphic to the Klein four-group
ocean marsh
You just have to find four elements that satisfy the group axioms
Some intuition about what the group looks like might help, instead of just thinking of it like a list of symbols
civic linden
i understand that the order of a group is the cardinality but in this sense is it still the same?
ocean marsh
Yes it's the same
civic linden
so a subgroup of D12 of order 2 would only contain 2 of those elements?
ocean marsh
correct
civic linden
go lets say G=D8
ok*
the subgroups of G of order 2 are r^2, s, rs, r^2s, r^3s
which is 5 elements?
ocean marsh
Well the subgroups are each of those elements paired with the identity
But yes, there should be 5 subgroups of order 2
civic linden
ah right ok
so what would be the difference of say an order 3 subgroup and an order 4 subgroup?
ocean marsh
One has 3 things in it and one has 4?
civic linden
yeah but what are we finding that makes them different?
ocean marsh
Well they would have totally different element. Nonidentity elements in order 3 groups have order 3, so you would only need to look at the order 3 elements
Nonidentity elements in order 4 groups have order 2 or 4
civic linden
mhm ok
ocean marsh
Have you had any luck?
civic linden
trying to wrap my head around it
so by lagranges theorem we have possible orders being 1,2,3,4,6,12?
civic linden
every subgroup contains the identity
ocean marsh
Yeah that looks right!
civic linden
how do i determine if they're non-cyclic?
ocean marsh
Well a group of order 4 is cyclic if and only if it has an element of order 4
ocean marsh
no worries
civic linden
anybody could help me on how to approach this question would be great: "Let G be a group of order 36, let H be a subgroup of G and let x and y be elements of H such that x has order 12 and y has order 9. Prove H=G".
gentle pendant
order of elements divides order of subgp, so |H| is divisible by both 12 and 9, and hence by their LCM which is 36.
gentle pendant
lol
chilly ocean
I wish this could be structured further
So proving things can feel like playing with lego.
gentle pendant
what?
civic linden
group theory is my worst module :x
chilly ocean
Now these all subgroup and sign mess are meh
gentle pendant
what are you talking about abastro?
yeah I was always a bit of an algebra noob @civic linden
civic linden
not as nooby as me i bet! @gentle pendant XD
gentle pendant
it kind of is, just its a bit harder to put pieces together / there are more kinds of connectors
but I guess the universal connector thing is kind of the defining feature of lego
gentle pendant
examples?
snow dew
That's just for convenience you can use ln only if you x'want
gentle pendant
what does out of place even mean here?
and I can't remember the last time I used a log function with base not e.
snow dew
Yup noone does that in pure math
snow dew
?
steep swallow
that kinda has a reason thou right?
snow dew
What order things?
chilly ocean
Absolute value/norm
Order
steep swallow
cuz it relates to size
chilly ocean
How?
Moreover
|G| and |x| are similar but different
Why do they use the same syntax
snow dew
X >= y iif norm(x)-norm(y) >=0
chilly ocean
Well these are syntactic things so doesn't matter that much
But there is too much of implications in semantics imo.
What is X >= y
steep swallow
i've hated inverse symbol for arcsin and stuff
chilly ocean
I'm going to blame symantics
steep swallow
although it's obvious once u learn it
chilly ocean
sin^2(x) and sin^(-1)(x) lol
chilly ocean
someone with 200IQ designed it
Anyway. One of the two should be wrong, was what I meant
The power is abused
I prefer haskell
civic linden
the area in which u work in
snow dew
"Let (C(R,R),+,*) be the space of continuous function from R to R with term by term multiplication"
snow dew
Most of the choices made for standard notation are well tought tho
chilly ocean
Is it?
steep swallow
i also think that symbols are ingrained in our head after using it for a long time so overusing it even when defined still might confuse readers
chilly ocean
#Doubt
chilly ocean
?
snow dew
2+2 is actually ambiguious cause it can be defined in any ring
full blaze
snow dew
And the fact it can be make these abuse of notation make sense: we generalize stuff to be able to work like we did before
chilly ocean
Can you specify
full blaze
my universal algebra prof is like what is 2+2
snow dew
2x can be defined by x+x as long as you have a plus, and you just need a "1' to say 2 = 2*1
chilly ocean
So?
snow dew
So if we suddenly had another place where 2*something makes sense but differently we wouldn't change our global def
chilly ocean
But you mentioned 2+2
steep swallow
i don't think it's fair to talk about professionals only, i feel like it should be accessible to anyone as much as possible
chilly ocean
Not 2*(something)
snow dew
Cause 2 could be a dog
steep swallow
gg
snow dew
For all i care
chilly ocean
snow dew
If i have a ring of dogs
steep swallow
lol
steep swallow
stuff like mod 4?
chilly ocean
So where is the ambiguity
snow dew
Yeah or mod 2
chilly ocean
Ambiguity wheere
snow dew
In what objects are written
chilly ocean
2+2=4=0 on mod 4 btw
That is abstraction
Not ambiguity
You remove the unnecessary contexts like 'dog'
snow dew
Yeah but not canonically
chilly ocean
Well
I think you are confusing ambiguity and abstraction
Abstraction is that you can remove the details you don't need when you want. Not necessarily confusing.
snow dew
My point was maybe in another space there is a better notion of 2
chilly ocean
While ambiguity means the things which are different in important details are using the same space
snow dew
Likebin funcs
chilly ocean
Wdym by 'better notion of 2
snow dew
For ^
chilly ocean
❔
snow dew
Well it's a dumb example, i'll be clearer: you have a power on functions, it can't be inversed cause it makes a disgusting space of inversible. You have another one which makes a group on bijections and can be inversed
snow dew
Do you create a new notation for cos^-1 when it almost never would make sense as term by term mul?
snow dew
Well where is the abiguity if cos^-1 makes no sense in 99 cases out of 100
It's actually about consistensy with inversion of composition
chilly ocean
Is cosine really inversible?
Also what is consistency on Inversion of composition
snow dew
Yeah on intervals that are used when you use that notation
chilly ocean
You need to give it some interval
Which gets out of nowhere
So.. is it really inversible
Or is it just made it so it could be inversed
snow dew
Nope but -1 still makes sense (i think it's consistent with analytic continuation of that inverse but i'm not 100% sure)
And its its right or left inverse
I think?
restive kiln
I redact my question because I figured out what I need to do.
delicate chasm
It's the latter right, you're defining it on components? My understanding of infinite direct sums and products is shaky as well
What answer did you get @restive kiln?
restive kiln
Oh, I just realized that the question wasn't relevant to the problem I was trying to solve.
delicate chasm
Ah right, fair enough
restive kiln
I was trying to find a natural way to combine R-module homomorphisms on each indexed module into something on their direct sum, but I realized that the morphism I was looking for was something like $$\alpha : (p_i) \mapsto \sum\limits_{i\in I}\alpha_i(p_i)$$
mellow vaporBOT
delicate chasm
~texlisten
odd forumBOT
I am now listening to your tex.
delicate chasm
Right. That is just the induced map for coproducts though right? If you have maps $\func{\alpha_i}{P_i}{A}$ and the coproduct exists then there is a map $\func{\alpha}{\bigoplus P_i}{A}$ which agrees on each component? And that should be exactly the same as what you wrote?
odd forumBOT
Puerøsola:
restive kiln
I'm honestly not sure. It wouldn't surprise me if the correct map turned out to be a natural, category theory thing.
delicate chasm
It should be induced from universal property of the coproduct
But if your indexing set is infinite I'm a lot more unsure
cerulean rune
I am not sure I can add much, but taking the unsureity at face value, can confirm that this is all correct. For me this is a concrete way of thinking of the "reason for" the difference between product and direct sum for R-modules over infinite indexing sets (i.e., direct sum elements have only finitely many nonzero components) -- to satisfy the universal property for direct sum (which is as y'all say the thing you are both describing), you want to write down exactly that sum over a_i(p_i), and for this to make sense in an R-module it had better be a finite sum.
(sorry if butting in if in fact this was all clear.)
delicate chasm
Thanks! :) That does make it clearer!
vestal needle
Can anyone help me with the first part of this question
What are the possible orders of elements in S6 and A6?
i think the method that my teacher showed me and the method in the book are different
vestal needle
<@&286206848099549185>
cerulean rune
I would approach this question by writing down explicitly all the different forms of the cycle decompositions that elements of S_6 can take. (E.g. (12)(345) is a cycle representation; its "form" is "a 3-cycle, a 2-cycle, and a 1-cycle". If you know Young diagrams, it's just writing down all of those, but even if you don't it's straightforward.) Then you can read off the orders of S_6 elements from these. For A_6, just determine which of these are decompositions for elements of A_6 and which are not.
vestal needle
hmm ok thanks
uncut girder
Consider an abelian group of prime power G, with |G| = p^n
Choose an element a of maximum order, ie |a| >= |b| for all b in G.
Then is it true that x^|a| = e (identity) for all x in G?
Please help me understand why this last statement is true.
covert vector
cuz the order of an element divides order of group
so all orders of elements will be prime powers
uncut girder
So?
covert vector
if |a| = p^m, for m≤n, then any other element will have order p^z for z≤m
z an integer
covert vector
p^m = (p^z)^(m-z)
uncut girder
How?
covert vector
o wait
meant
p^m = (p^z)(p^{m-z})
so now if u take b^ both sides
b^|a| = (b^|b|)^(p^{m-z})
b^|b| is identity, and the rightmost thing is a positive integer
reduces to identity to some integer exponent
uncut girder
Oh right
uncut girder
For some reason b^{xy} = (b^x)^y is a little difficult to grasp rn for me
But I got it, thanks @covert vector
covert vector
np
uncut girder
If b^(p^i) = e then b^(p^{i+1}) = (b^(p^i))^p = e^p = e.
Then by induction we get the result.
covert vector
sure
uncut girder
I'm reading this proof for the FToFAbG in Gallian's book and it's so hard to have a global understanding of the proof of some lemmas. It seems the only way I can understand it is on a microscopic level, one step at a time.
I think one common theme is to use induction on the order of G by considering a factor group, invoking the induction hypotheses on this smaller factor group, then recovering the original statement on the full group G
This is basically the proof but the proof of lemma 2 in particular is long-winded
I think I'd rather learn the elegant category theory way of proving this 😆
covert vector
elegant cat theory way?
uncut girder
Idk
little fox
Um hi all, I'm new to this discord (as well as abstract algbera) and notation is something I apparently suck at remembering. Mind if I post a thing and ask what it means?
stone fulcrum
Yeah go for it. I'm the future, don't ask to ask, but simply ask.
stone fulcrum
Q[x] / <x² - m>
is Q[x] mod by the principal ideal generated by x² - m
little fox
hm ok
snow dew
Intuition of this is to make the polynomial you quotient by have roots in the new object. This theorem means that that quotient by that polynomial is the smallest ring extension of Q containing that root (those we have in R)
little fox
That seems to be how they approached it, I'm trying to work through it now, thank you
cerulean rune
You've written it as the product of two transpositions -- (12) and (45) -- so it's even.
vestal needle
sorry what do you mean by transpositions
stone forum
permutation that swaps 2 elements
stone forum
@vestal needle note that any cycle (x1..xn) = (x(n-1) xn)..(x2xn)(x1xn)
so if you have n elements in (x1..xn) then it's a product of n-1 transpositions
(1 6 5 3 2) is a product of 4 transpositions
snow dew
Signature is a morphism, right?
earnest valley
Yes.
cerulean rune
What does it mean to say that "signature is a morphism"? I am not sure what's being asserted