Take an element of G, ANY element of G. Say... 7. What's the operation of this group? It's addition, since Z (our group G) is a group under addition. So grab that 7, and write down all possible sums of 7 with elements of H. So you have 19 (which is 12 + 7), you have 4 (which is -3 + 7), you have 7 itself (which is 0 + 7), and so on. As you can see, you can build a whole subset of Z this way: {..., -5, -2, 1, 4, 7, ...}.
#groups-rings-fields
faint mica
THIS is a coset. There are some elements of Z that, when you use them this way I just described, give rise to the same coset (try it with 4, you'll see that you obtain the very same elements of that coset from my previous message when you add 4 to every number of H), and there are other elements of Z that produce an entirely different coset (for example, 2 will produce the coset {..., -4, -1, 2, 5, 8, ...}.
And then of course you have the coset {..., -6, -3, 0, 3, 6, ...}, which you obtain when the element you pick is inside H: because when you add every element of H to an element of H, what you obtain is... H itself, of course. Since it's a subgroup, so it's closed under addition.
Your aH = H was written in multiplicative notation, but Z is a group under addition, so a better way to write that would be a + H = H. If H is the subgroup of all multiples of 3, what elements a are such that, when they're added to every element of H, produce a subset that's equal to H itself? It's of course... the set of all elements of H, because, as I said earlier, this is a direct consequence of the fact that subgroups are closed under their own operation.
chilly ocean
OHHHHHHHHHHH
So the only way to get a coset of Z that is equal to H is to choose elements of H.
faint mica
Yes, exactly. But after all, it's very easy to prove this:
chilly ocean
So thats why aH = H iff a is in H
faint mica
Suppose there is an element a that is NOT in H, such that aH = H. But then this means that there exist h_1, h_2 in H such that a * h_1 = h_2. But then you can multiply this equality by (h_1)^(-1) on the right to obtain a = h_2 * (h_1)^(-1), which is an element of H because it's the product of elements of H and H is closed under its operation; and this goes against the hypothesis that a was not in H. So a must be in H, if you want that to be true.
chilly ocean
So a is an element of H. And the only way to get H back is to add an element of H to all the other elements of H, and you stay in H. I think I was thinking that a was an entire coset and not just an element of the subgroup.
faint mica
Yeah, it's just an element. 😛
Remember that a coset is a... set, fundamentally, and sets are typically written with an uppercase letter; so if you read 'a', then that has to be just an element.
chilly ocean
That makes sense. So real quick, if there was an element b of G that was not in H, bH would still be an element of a coset of H?
Thats what you were saying up there (like b = 7)
faint mica
Not exactly an 'element': remember that the notation bH stands for a set, the set formed by all possible products of b by elements of H (which is indeed one of the cosets of H anyway, yeah). Thus, in this case, bH wouldn't stand for an element of a coset, but rather for the whole coset that you obtain by multiplying b by all the elements of H.
So... b (lowercase) is just an element; H (uppercase) is a subgroup; bH (lowercase and uppercase) is a coset. 🙂
chilly ocean
Okay, its kind of like cyclic groups where you can have elements that generate cyclic subgroups. So here b would generate bH
faint mica
This is why you can write stuff like aH = H: because both the left-hand side and the right-hand side are sets. It wouldn't make sense to say that an element is equal to a set, if aH were meant to represent an element.
Weeeell... yeah, I guess you can say that. But remember that the cosets of H that are different from H are not subgroups (they all lack the identity element which is only in H because it's a subgroup, and the other ones can't have it since they form a partition of G, and this means that they can't share common elements because this is how partitions work); therefore I wouldn't talk about 'generating' bH, since I mostly use that terminology only for actual groups... but yeah, that's the idea.
chilly ocean
Well yeah. By using b not in H you can create a partition of H known as bH by adding b to every element of H.
if the operation is addition
faint mica
A partition of G. 😛
chilly ocean
Right! so close
This is perfect. Thank you so much @faint mica @muted shoal @stone fulcrum
muted shoal
@faint mica thank you for the explanation : )
upper dome
Okay, so say E is an extension field of F of finite dimension n, with [ E : F ] = n being the degree of E over F....(just definition of degree of a fin. ext.)
I've been under the assumption that I can quickly find the degree of E over F by just looking at the degree of the irreducible polynomial in F with zeros in E.
should I not be making that assumption?
thorny slate
it depends what you mean by "the" polynomial
upper dome
good point
thorny slate
if you mean E = F[x]/p(x), then yes, [E:F] = deg(p)
however, on the other extreme, if E is the field of fractions of a polynomial p(x) in F[x], then n | [E:F] | n!, where n = deg(p)
and it may take all values in between
upper dome
oh for sure
ok that clears some stuff up lemme scribble away for a few. thanks
okay that helps, but also makes what I was trying to do totally wrong
thorny slate
yeah gotta keep track of that stuff
upper dome
I'm just tryin to prove to myself this statement
An extension E of degree 2 over a field F is always a normal extension of F if the characteristic of F is not 2
(nobody spurt the answer pls)
thorny slate
well 2 = 2!
upper dome
oh fuckin hell
thorny slate
lol
upper dome
lol
yea but i gotta show for char 0 and char p, right?
the seperable part for normal extension is chill for that, but im stuck on the splitting field part
thorny slate
the splitting field of a polynomial of degree 2 is an extension of degree at most 2! = 2
upper dome
aight thats an awesome result
but i've never seen it!
im guessing it comes from the fact every polynomial will split to linear factors
thorny slate
it's not hard to prove
take your base field F
and take the polynomial
the polynomial has at most n different roots
so once you add one
you have an extension of degree n
and then a polynomial of degree at most n-1
upper dome
nice
thorny slate
induct, that's n!
upper dome
that is a clean little proof. Im woderin tho, you just made the jump from an extension E of degree 2 of F, and landed on the splitting field of a polynomial of degree 2. Is it natural to switch like that?
cerulean rune
can you elaborate on what is confusing? i'll comment that an order-6 group is isomorphic to Z_6 iff it has an element of order 6 (if a is such an element, then the map sending a to 1 in Z_6 is an isomorphism).
vestal needle
for part c. U(9). the subgroup generated by 2 does not form a subgroup
cause 6 is not a part of U(9)
cerulean rune
U(9)'s operation is multiplication, not addition
cerulean rune
👍
cerulean rune
it has 6
vestal needle
oh shit
cerulean rune
(if you think of it as symmetries of a triangle, there are three orientation-preserving ones and three orientation-reversing ones.)
vestal needle
This is how we were taught to show elements of S3
Sorry sideways lol
The elements have to add up to 3 right
cerulean rune
i am not quite sure what the notation there means, but it looks like you've notated the three different conjugacy classes of S_3? What do you mean by the elements have to add up to 3?
cerulean rune
so let me try to see if we are on the same page, because i am maybe confused
so this looks like cycle notation-ish, to me -- you'd write, for instance, (123) for the permutation that sends 1 -> 2 and 2 -> 3 and 3 -> 1
is that right?
cerulean rune
and then your (3) with a bar over it is trying to denote all the permutations writeable as a 3-cycle
vestal needle
yeah
cerulean rune
which is not a single element of S_3 -- it contains both (123) and (213)
vestal needle
oh i see
cerulean rune
if you're thinking of elements of S_3 as permutations that is just fine, though. actually that is better, the thing i said about triangles is stupid.
in any case it should be clear there have to be 6 elements -- there's 3 places you can send the element 1 and then 2 remaining places you can send the element 2. (Or in general, S_n has order n!.)
and you can just write them all down and figure out their orders
vestal needle
so I’ve written down all the elements of s3 now
cerulean rune
great
cerulean rune
yep
cerulean rune
cool :D
vestal needle
no u
cerulean rune
:p
vestal needle
@cerulean rune followup question. so if each pair of groups has the same number of elements, and they have the same amount of subgroups with equal orders, is that enough to prove they are isomorphic?
for example U(9) and Z(6) both have 6 elements. And they both have 1 subgroup with order 1, 2 subgroups with order 3, 1 subgroup with order 2, and 2 generators. this means they are isomorphic?
cerulean rune
i don't think that in general that's true, although my brain did not immediately come up with a counterexample
vestal needle
so i would have to map a generator and make a cayley table to be sure
cerulean rune
well, any two cyclic groups of the same order are definitely isomorphic -- if you map a generator of one to a generator of the other, it's more or less obvious this map preserves the group operation
vestal needle
cyclic means it has at least one generator?
sorry i forgot the definition of cyclic groups not sure if thats right
cerulean rune
depends on what you mean by generator. there is at least one element whose order is the order of the group; alternatively, there is at least one element a such that the set (subgroup) of all powers of a is the entire group
cerulean rune
sometimes people use "generator" more broadly, like, S_3 "is generated by" the elements (12) and (123), in the sense that the smallest subgroup of S_3 containing both is just S_3 itself
so people might say "a cyclic group is a group that can be generated by a single element"
vestal needle
This lol
cerulean rune
all right
since the operation on H is addition, (h_1)^3 doesn't really "make sense" (although it's clear what you have to mean)
chilly ocean
This is the syntactic weakpoint of group notation
It can easily be confused with power
cerulean rune
(i'm not quite sure whether my complaint is clear. the entire issue is notational -- when H is a group "under addition", the group operation on H is normally written h_1 + h_2)
cerulean rune
what does (h_1)^3 even mean?
vestal needle
is it h_1 + h_1 + h_1
cerulean rune
yes, that's the only coherent thing that it could mean (no one said anything about H having a "multiplication" operation, so all we're left with is that)
people usually write that 3h_1 (for obvious reasons)
cerulean rune
yes
vestal needle
how would i deal with a negative exponent for this?
cerulean rune
well, if H is a group under addition, the inverse of an element h is usually written -h, again for reasons that should be clear
cerulean rune
yes, that is exactly how i would write it :D
vestal needle
awesome!
yeah this question kinda threw me off i didnt think the answer would be that simple as what i wrote before lol
cerulean rune
haha yeah, i think the point of the question is just to try to get familiar translating between the multiplicative and additive notations
cerulean rune
it's useful to have both around (the additive notation for groups is essentially only used when the group is abelian)
chilly ocean
So much for using multiplicative notation
I'd like separate notation since it is not always multiplicative
But that is my opinion
chilly ocean
Something which is neither additive nor multiplicative.
uncut girder
Hi guys. I'm learning ring theory and I came up with this result
|z|^2 = |Z[i]/<z>| for all Gaussian integers z.
uncut girder
I found that this has already been thought about at discussed on math.stackexchange
^this is exactly how I approached the situation but I relied on Pick's Theorem to arrive at the final conclusion
upper dome
number 17
My idea is to hit the entire norm with another element of the Galois group, and hopefully show that the product is unchanged just in a different order
but if the product is full of alpha and it's conjugates, how the heck is that an element of F?
is "element of F" just saying that it is left fixed?
thorny slate
if it's fixed by the galois group then it's in F
you can also show these as the determinant and trace of a certain matrix over F....
which is where the names come from
upper dome
yea i remember em from linear algebra. But that really means its in F? is it just me or is that kinda misleading
upper dome
interesting. I guess it was that alpha is in K, but i guess, yea, the products of all the algebraic elements would be in the base field F
thorny slate
try to make that correspondence explicit
write alpha in a basis of E over F
and show that det/trace are the same as Norm/Trace as defined there
upper dome
right on, thanks again @thorny slate !
thorny slate
👌
cerulean rune
i am kind of very bad at galois theory but i am really confused about a thing here. if K is just a normal extension -- that is, if we don't know that it's separable -- is it really sufficient to check that an element of K is fixed by Aut(K/F), if we want to know it's in F?
i was kind of picturing like F = F_p(t) and K is the splitting field of (x^p - 1), and it wasn't clear to me that K had any nontrivial automorphisms at all.
what am i missing? can anyone describe a nonidentity element of G(K/F) for that example, for instance? or am i just getting mixed up on definitions?
upper dome
normal extension is a seperable extension by defintion, so that might be where you got confused
cerulean rune
like, the phrase "normal extension" means a thing, right? every polynomial that has a root splits into linear factors?
and that's just not the same as separable in general
i should say, it doesn't imply separable in general
upper dome
maybe youre just thinking of a splitting field
cerulean rune
is there a restriction on the kinds of fields F you study? like, characteristic zero or perfect?
upper dome
mostly characteristc 0 and perfect right now for galois theory, but we mess with finite fields here and there too
cerulean rune
finite fields are perfect, right?
cerulean rune
ye ok, if everything is perfect then there's no issue, that may be the explanation
thanks i am like, extremely bad at galois theory and so every time i am epsilon confused i feel completely lost
but i think this helped
upper dome
lol for sure, this stuff is dense, i dig how much it makes me work for results lol
versed gull
what math classes are needed as prerequisites to abstract algebra?
blissful yarrow
At my school, linear algebra 1 and 2 are prerequisites for group theory and ring theory
versed gull
damn. is the linear algebra you do like heavy proof and theoretical stuff so to at least prep you for abstract? because i've heard abstract is pretty much just proof and theoretical stuff
thorny slate
yeah you should probably take a serious linear algebra course first
versed gull
Yeah. I found that out as soon as I opened one of those abstract alg. books
buoyant fox
I get confused when he tries to demonstrate the formula for 10^(k+1)+1
(that is what he's doing right?)
Why does
$$10^{(k+1)+1} + 3 \cdot 10^{k+1} + 5 = 10^{k+2} + 3 \cdot 10^{k+1} + 50 - 45$$
cloud walrusBOT
æshthetic:
buoyant fox
Specifically where do 50 and -45 come from?
@celest wren did you understand this one 👀
uncut girder
You don't NEED linear algebra to start abstract algebra
thorny slate
you don't NEED two cheese pizzas and a shotgun to join chicken gang
buoyant fox
@trail vessel but why did he choose to use 50 - 45? So he could factor out 10?
trail vessel
Possibly. 50 = 10^(1+log(5))
But 5=10^log(5)
So not sure what the point was
Ohh. I scrolled up
Yeah, he wanted to factor out a 10 so he could cancel the other side of the equation
thorny slate
the solution is much nicer if you use some basic properties of divisibility, taking things "mod 9"
celest wren
@buoyant fox what quartetly said
adding 5 is the same as adding 50 and subtracting 45
celest wren
the 50 lets him factor out 10 from a lot of the terms, which leaves the formula for k
in parens
it is pretty neat
try exercise 1
for this chapter
i did it in class
buoyant fox
did you finish reading it
celest wren
i don't think it is
buoyant fox
whats next?
celest wren
idk, don't have it with me rn
celest wren
😎
solar wyvern
is there a way to determine roots of a quadratic over F_{2^n}
polar finch
what theorem is this from? (linear algebra stuff)
thorny slate
that the dimension of R^n is n
buoyant fox
I'm so confused 💀
I don't even understand what's being proved
What's the significance of $$a = bq + r$$ with $$b > 0$$ and $$0 ≤ r < b$$
👀
Not ideal
delicate chasm
It's readable enough :P
cloud walrusBOT
æshthetic:
buoyant fox
Lol
delicate chasm
So
the idea is when you divide $a$ by $b$ you can do it $q$ times and get a remainder of $r$
cloud walrusBOT
Puerøsola:
delicate chasm
Have you worked with any modular arithmetic?
buoyant fox
Hmm okay
delicate chasm
An example might make it clearer
Take a = 13 and b = 5
Then b goes into a twice with a remainder of 3
So $$13 = 5 \cdot 2 + 3$$
cloud walrusBOT
Puerøsola:
delicate chasm
The condition $$0\leq r < b$$ is there so that we don't have remainders larger than what we are dividing by
cloud walrusBOT
Puerøsola:
delicate chasm
That is, it's true that $13 = 5\cdot 1 + 8$, but that isn't all that helpful.
cloud walrusBOT
Puerøsola:
delicate chasm
Similarly with $$13 = 5 \cdot (-1) + 18 = 5 \cdot 3 - 2$$
cloud walrusBOT
Puerøsola:
delicate chasm
For any pair of numbers you can uniquely extract a div (q) and a mod (r) satisfying those conditions
Does that make any sense?
buoyant fox
@delicate chasm hello sorry for the interruption in attention gimme a sec to read and process
delicate chasm
Sure 👍
jagged gate
It's just that you suppose there are two different quotients and remainders, (q,r) and (q',r') for a couple (a,b)
And show they're in fact equal
So (q,r) is unique : very rough draft of the proof
delicate chasm
Good luck!
errant drum
Can anyone recommend any book/website that has a lot of practice problems in group and ring theory?
delicate chasm
Topics in algebra has quite a few problems, it's at a fairly elementary level.
oak perch
hey guys. I have a cool thought about group elements, Z and matrices.
Let G be a group (lets keep examples uncomplicated but non-commutative like D3).
Easily we could have matrices of elements in G that can add as long as the matrix dimensions are the same.
I R2 + F I = F R2
F I I R F R
BUT you can also do a few things with integers.
- g times z could give zg (or for multiplicative groups g^z)
- do this with a matrix of elements of G and a matrix of elements of Z
I R2 + 1 2 = I R
F I 2 3 I I
- you could do something similar when multiplying matrices of g's and z's ...
I R2 X 1 2 = I+R I+I
F I 2 3 F+I I+I
Then you can have all sorts of fun (which I think I've bought up before so I'll keep that part short)
vectors with entries of G could be "dotted" with vectors with integer entries ... possibly allowing a definition of perpendicular between g-vectors and z-vectors. But z-vectors already have a def of perpendicularity. So perpendicularity between g-vectors and other g-vectors could be defined ... g1 perpendicular to z1 to z2 to g2 anyway.
quasi-linear algebra with groups.
Can group matrices take on a geometric form so that our usual idea of "linear transformation" (multiplying by integer matrix) works out?
oak perch
hmm what if it isn't though 😃
thorny slate
But you dont exactly want that I guess
What you are basically doing is condering matrices over the group ring Z[G]
Read on that
oak perch
Hmm Z[G] simplifies to G, though, right? just making sure.
oak perch
okay I'll try that. Thanks.
errant drum
@solar wyvern @covert vector
Thanks for your help
covert vector
np
sonic current
hello
i'm having a little trouble in an abstract algebra exercise
may I ask for help here or should I use one of the other channels?
delicate chasm
You can ask here :)
sonic current
nice! so here it goes
Let G denote a group and let g € G. Show G = <g>
a) if |G| = 12, g^4 ≠ 1 and g^6 ≠ 1
so here's what I did
since g^4 ≠ 1, then g^2 ≠ 1 because g^4 is g^2 squared
same process for g^6 to conclude g^3 ≠ 1
so my reasoning is the only possible order for <g> is 12 since it can't be 1,2,3,4, or 6
what am I missing to prove G is generated by g?
the fact that both have the same order is not enough is it?
sonic current
anyone? 😦
cerulean rune
If the order of <g> is 12, that means it's a subgroup of G with 12 elements. If G only has 12 elements, well, then <g> must be all of G. So the fact that g and G have the same order is indeed enough to prove that <g> = G.
cerulean rune
sure
sonic current
I have the solution to that one
I can guarantee H and K and disjoint (except for the identity) because they are prime?
cerulean rune
what do you mean by "they are prime"?
sonic current
their orders are prime
cerulean rune
give me an argument that that forces them to intersect in just {e}
sonic current
oh I know! I think
since the subgroup order is prime, they are cyclic?
hmm but that should not be enough
H could be {1, g, gˆ2} and K could be {1, g, g^2, g^3, g^4}
ok wait
not so simple
cerulean rune
well, not for the same g, but yes (like, H could be {1, g^2, g^4} and K could be {1, g, g^2, g^3, g^4, g^5})
sonic current
g and g^2 must be each others inverses
your K has order 6 btw
i'm still missing something here I suppose
cerulean rune
so it is true that if p and q are distinct, then H and K only intersect in {e}
sonic current
but nothing guarantees they are
I guess?
by the way, we know there is only subgroup of order 3 and 5
cerulean rune
imagine you have an element g that's inside both H and K -- what's it's order? |g| has to divide both |H| and |K|, but the only number that does that (when p and q are distinct) is 1
sonic current
my question is: does that guarantee it must be the subgroup generated by g?
ok, I see your point
but p and q distinct is not trivial is it?
we must guarantee they are distinct
cerulean rune
it's part of the assumption here, as the claim isn't true for p = q
sonic current
or that they could be the same but then H and K are cyclic or something, in different exponents
hmm let me think a bit
cerulean rune
ok
snow sky
Where is multiplication and division
sonic current
then they might have an element in common. so I think it is necessary to have prime in the assumption
so taking your argument, @cerulean rune, "imagine you have an element g that's inside both H and K -- what's it's order? |g| has to divide both |H| and |K|, but the only number that does that (when p and q are distinct) is 1" i'd say p and q are distinct and prime
I think it's safer that way
cerulean rune
yes, that's right
sonic current
ok, let me just write that down so I can move forward in the exercise
cerulean rune
(the generalization of that particular statement is that it still holds whenever gcd(p,q) = 1 -- that's essentially the definition of gcd -- which you can also think of as saying that no primes divide both p and q)
cerulean rune
👍
stone fulcrum
Euclidean algorithm? Good stuff
celest wren
having trouble understanding why
− r ≤ r' < b.
This is possible only if r' − r = 0.```here's the full proof
just to be clear, I understand how they got to the inequality, I just don't understand why it implies r' - r is 0
stone fulcrum
Let's say a < b, and b divides a. Then a = 0.
If a ≠ 0, then b ≤ a as b has to divide a. But that contradicts a < b.
Note that all numbers divide zero, in the sense that 0 = bk, for some k. Namely, k = 0.
stone fulcrum
Np, lel. Feel free to ask if you have anything else!
Actually, I have a question. Consider the ring of integers. Consider a function that interchanges 2 with 3 and 3 with 2. I believe this is a ring automorphism.
But, consider
2 + 2 + 2² = 2³
That's true. Apply the automorphism on it,
3 + 3 + 3² ≠ 3³
That's wrong. What happened here? I expected that to work.
delicate chasm
I don't think it is a ring automorphism, if it only does that
$15 = 3\cdot 5 = \phi(2)\cdot \phi(5) =\phi(2\cdot 5) =\phi(10)=10$
cloud walrusBOT
Puerøsola:
stone fulcrum
Sorry, the function where
φ(2) = 3
φ(3) = 2
φ(p) = p for any other prime p.
Then something like φ(10) ≠ 10
delicate chasm
My line still holds?
Oh I see
You define it on generators
So it would be defined to be 15
Not sure it works with addition though
I think it is defined by where you send one
Since $\phi(1+1) = \phi(1)+\phi(1)$
cloud walrusBOT
Puerøsola:
delicate chasm
The generator of addition
stoic rose
It's an homomorphism for multiplication but not addition (7=2+5=φ(3)+φ(5)=φ(3+5)=φ(8)=27)
chilly ocean
20
delicate chasm
In general a ring map to Z is defined by where it sends one
chilly ocean
someone help me with my math
delicate chasm
Even a group map for addition
chilly ocean
how do I do a division
delicate chasm
In what ring?
chilly ocean
like a diamond ring
delicate chasm
It's a set with addition and multiplication operators
Satisfying some fairly natural conditions
You can't always divide tough
Though
Like in Z
chilly ocean
ok ok I have a better question
chilly ocean
lets say I am taking a test
and there are 0 questions but I answer the bonus question
what is my percentage
delicate chasm
That depends on a lot of unstated factors
chilly ocean
nope
delicate chasm
This doesn't belong in this channel though
chilly ocean
those are all the factors
stoic rose
That's a stupid test
chilly ocean
stfu
delicate chasm
Does that make sense Kaynex?
stone fulcrum
Oh, welp that's not an automorphism. I'm going to have to find out why I thought it was
chilly ocean
so the end would be 1 out of 0
delicate chasm
It's a nice group map on multiplication though
chilly ocean
which is impossible yet possible because it’s possible to answer a bonus question on a test with no questions
so wtf math
stone fulcrum
I remember reading something about a permutation on irreducible elements forming an automorphism of some kind, but I may be remembering it wrong
snow dew
Well what you said proves it can't be a morphism cause if it was then ...
What seems to surprise you is that being a morphism is a very global property
Permutations (bijection) are often called automorphisms, cause they are automorphisms of blah if you add additional structurure of blah to the thing
@stone fulcrum
But here it's ring automorphism and that's way stronger
Than being a bijection
stone fulcrum
Okay
So this particular construction of a ring automorphism failed, however it doesn't always fail. I know there's something behind it but I don't remember where I read it. Read anything about the permutations of irreducible elements forming an automorphism?
snow dew
Oh you might have failed cause 1 = 2-3 yeah. Maybe if you lean more towards modules than rings i think you can do stuff like that
If you have a base you can build a morphism by switching elements if the base
At least it seems okay, modules can be a pain and i didn't do them in class so i could be wrong
(Vector spaces but the field is no longer a field, just a ring)
(2,3) is a minimal generating part of Z as a Z-module but not free and therefore not a base
So that might have been your confusion
Automorphisms of Z are identity and -identity i think
Well no just identity
sonic current
hello!
I need some assistance in proving normality in the dihedral_4 subgroups
I'm doing this exercise:
Find all the invariant subgroups in D4.
I don't need much help, I suppose, it's more that I want to do it in the most quick way
with some google-fu I've found some helpful links which helped me understand how to find all the subgroups
delicate chasm
Invariant meaning normal?
sonic current
i'm currently exhausting all the order 2 subgroups
yeah
sorry about that
so I currently have the set of order 2 elements, all of which generate subgroups, {b,ba^2,a^2}
my question is, how do I prove normality in a quick way?
or better, what is the standard way to prove normality?
if H is the subgroup, then gHg^-1 € H, for all h in H, g in G ?
I wanted to avoid manual coset calculation
thorny slate
its not too bad here
just try a and b in each case
since they are generators
that is, its enough to check gHg^-1 € H, for all h in some generating set of H, g in some generating set of G
sonic current
I have a corollary: if G = <X>, a subgroup H is normal in G if and only if xHx^-1 is contained in H, for all x in X
is this what you mean?
I suppose so. let me check
sonic current
let me see if I can get the hang of this
so G, in this case, is generated by a and b right?
ok, if H = <b> then it is not normal;
aba^1 = aba^3 = ba^2 which does not belong in H
when H = <a^2>,
aa^2a-1=a^2 which is in H
ba^2b^-1 = ba^2b
let me solve this last one...
wait: baab^-1 = bb^-1 = 1 ?
because aa = 1?
thorny slate
aa isnt 1
sonic current
oh yeah, nevermind
just a sec
hmm
baab^-1 = baab = aaaabaab = aabb = aa
is this correct?
thorny slate
looks ok
sonic current
thank you very much, @thorny slate
i've exhausted the order 2 subgroups, now I'm off to study the order 4 subgroups
sonic current
quick question:
i now want to determine which subgroups are isomorphic to z2*z2
so I must look for two elements of order 2
I know which elements have order 2, those are {b, a^2,ba^2}
so should I pair them up and see which are subgroups? and then, test for normality?
inner acorn
I don't know if this is relevant :P
But there're only 2 groups (up to isomorphism) that have order 4
- cyclic group of order 4
- klein 4-group
Soo if it's easier, if you have a subgroup of order 4, to show it's isomorphic to Z2 x Z2, you could show it's not cyclic ... which upon reading what you wrote I think you already are xD
But yeah, if you have two distinct elements a & b with order 2, then the subgroup <a, b> generated by those two elements should be isomorphic to Z2 x Z2
i.e. <a, b> = {e, a, b, ab}
dapper hawk
How do you find the primes in Z mod n?
snow dew
What's a "prime in Z/nZ ?"
I'ld say coprime with n but i'm not sure that's compactible with Z/nZ irreductible elements as a ring
Well coprimes with n are the invertibles iirc not the irreductibles afaik
thorny slate
they are just the prime divisors of n
unless n is prime
in which case there are no primes
the elements coprime to n are units
so they arent prime
snow dew
Makes sense
deft rapids
I need help
tepid fractal
post the question
wintry eagle
Wrong channel
thorny slate
wrong neighborhood
snow dew
Wrong planet
raw moth
wrong egg
wide zodiac
has high school math basically gone to the point where you select an option on your mobile phone app?
dusky finch
I have a simple question when we write GF(q^m). GF(q^m) this is a field that is made by modding a polynomial, correct? The elements are polynomials. so when a polynomial has a splitting field, it means that the roots of that polynomial are POLYNOMIALS?
thorny slate
F(q^m) is the field with q^m elements
you can see it as a field extension of F(q) if you want
as you say
dusky finch
yes but when a polynomial has roots in it's splitting field what roots do you actually plug in
aren't they equivalence classes?
dusky finch
I see, thank you
@thorny slate If I have GF(2^5) generated by f(x)=1+x^2+x^5 with f(a)=0 and g(x)=1+x^3+x^5+x^6+x^8+x^9+x^10. They say it can easily be verified that a=x^3 is a root for g(x).
So we plug in x^3 in g(x) and we mod by f(x)=1+x^2+x^5 and we should get 0?
thorny slate
yeah
dusky finch
I see, thank you so much
uncut girder
What's a subring of Z+Z that's not an ideal?
uncut girder
@thorny slate what grad classes are you taking this semester
thorny slate
analysis, lie theory, homotopy memes, finite group memes, homology memes
uncut girder
Memes lol
thorny slate
memes means like a topics course
thorny slate
kinda
uncut girder
So which ones do you have left
errant drum
Is group theory still a topic that's researched?
thorny slate
yes
bleak abyss
Geometric group theory is a thing for sure, also representation theory arguably counts
thorny slate
I mean even the theory of finite groups is active still
bleak abyss
I had been under a kind of impression that raw finite group theory was kinda "finished" after CFSG
bleak abyss
Shit maybe I can still live my dream and become a finite group theorist
thorny slate
haha
errant drum
@thorny slate I thought that finite groups were done after Cayley's theorem
thorny slate
lmfao
errant drum
Shows how undergrad I am I guess
thorny slate
yeah probably
there's a lot more to say about them
like you'll see sylow's theorem and stuff
bleak abyss
Sylow is great
thorny slate
I dunno how much it's research daminark but I'm taking a topics class on finite groups right now
with questions on their generation
like what the generator sets looks like
and their orbits under automorphisms
etc
it looks kinda old and not very mainstream
but it has relations to lie groups apparently?
and algebraic k theory????
so I wouldn't know
errant drum
Okay so I just looked up what CSFG was. How exactly does creating a group out of two simple groups even work?
thorny slate
by an exact sequence
basically if you have a group G and a normal subgroup H
then you have
0 -> H -> G -> G/H -> 0
if you don't know what that is, suffice to say that G is somehow built from H and G/H
so that to study G you can study those smaller pieces
the simple groups are the smallest ones, since they no longer have normal subgroups
errant drum
Oh I see
onyx mirage
Is this heaven?
thorny slate
it's hell
onyx mirage
Nice
Was looking so long for a maths channel
What are some of the applications of abstract algebra?
errant drum
Also is there a reason why infinite groups don't seem to appear much. Are infinitely groups just uninteresting?
onyx mirage
I think you just haven't looked at the right places
there's a family of simple groups of infinite order, lemme look the name up again
thorny slate
infinite groups are everywhere
errant drum
Admittedly it was based on what I saw so it was biased
thorny slate
lie groups are infinite
onyx mirage
errant drum
By finitely generated does it mean making an infinite group with finite generators?
snow dew
We have classifications and good theorems for finite groups. But you have a looot of infinite ones everywhere, (R,+) , the unit circle, C, free groups,...
thorny slate
yeah
onyx mirage
You mean the unit circle in the complex plane, right
errant drum
That's interesting I never thought there would be any other than Z
onyx mirage
No one can seems to know the answer as to what the applications are of abstract algebra I see 😂
errant drum
Number theory
thorny slate
Z is the free group on 1 generator
snow dew
^
onyx mirage
I know, from what I know
the entire study got created, because researchers found a lot of the same structures and patterns everywhere, and figured it'd be better to study them generally rather than treating everything as an isolated instance
from what I know
errant drum
That's what I heard as well
onyx mirage
but for something that seems to be so... ubiquitous, it's hard to actually find applications
snow dew
Cristals, pseudocrystals, rubiks cube stuff, rotations, automorphisms of most structures, symetries of some problem in whatever applied science, building invariants of topologial objects.... They are everywhere now
Cause we see them everywhere
stone fulcrum
What DOESN'T involve abstract algebra?
Anything could be made better if you can find a way to put an algebra on it
onyx mirage
Mhm
snow dew
Well a structure on it, an algebra is very nice to have but kinda rare
onyx mirage
I'm in the camp that'd love to make everything mathematical
snow dew
Rarer
onyx mirage
An algebra being something like boolean algebra or linear algebra right
stone fulcrum
Sure, those count
onyx mirage
I'd guess an abelian field with some extra restrictions here and there?
onyx mirage
Hmm neat
stone fulcrum
A vector space has two types of elements.
Scalars from a field
Vectors which are an abelian group.
- scalars distribute over vectors
Vectors distribute over scalars
They associate within eachother.
onyx mirage
Honestly kind of disappointed with how amathematical my compsci courses have been so far xd hope to get my fix from selfstudying a bit of abstract algebra
stone fulcrum
I think that's it. That's from memory but I could be forgetting something
errant drum
Yeah that's it
snow dew
Seems right
errant drum
It just felt odd not seeing the billion vector space axioms
stone fulcrum
Lel I know. I dislike how most sources put the axioms
stone fulcrum
That's a vector space + a bilinear product. So it's even more
snow dew
It's a nice place for shorts. things you call products should almost always be bilinear
When it makes sense to be so ofc
onyx mirage
Hmm cool
Think I'm gonna do a couple of more chapters tomorrow, this quarter seems like it's going to be cake anyways
chilly ocean
Hi big boys
I'm doing exercise II
This is the solution
However I started doing it like this
Can this also result in a good answer
Or am I totally wrong?
Pls help
Thanks in advance
thorny slate
assume it's not maximal
reach contadiction
note that prime ideals in a PID are those generated by a single prime/irreducible element
errant drum
@uncut girder it's pretty easy to show in Z
uncut girder
Hmm lemme think about Z
errant drum
That's all I got
uncut girder
In Z it works because of gcd is a linear combination
Say <a> is nontrivial prime ideal. Then a is prime. Say B is an ideal properly containing <a>. Then there exists b in B not in <a>. So gcd (a,b) = 1. So 1 = Lin combo(a,b) is in B. So B =Z
thorny slate
you are kinda using what you want to prove
it happens because Z is too simple
try doing it in general with the same idea
assume <b> properly contains <a>
and see what happens between b and a
bleak abyss
@thorny slate Z isn't simple though, every subgroup is normal
uncut girder
Stumbled into a proof which uses PIDs are Integral Domains.
<b> properly contains <a>
So a = kb for some k in R
Note b is not in <a> (otherwise <b> is contained in <a>, contradiction)
So k is in <a> and thus a = alb for some l in R.
So 0 = a - alb = a(1-lb) and a is not 0
Since PIDs are IDs 1-lb = 0 so 1=lb is in <b> and <b> =R.
Hence <a> is maximal
vestal needle
can someone please explain this in detail to me
vestal needle
sorry what do you mean by surjective and injective?
thorny slate
onto and one to one
vestal needle
so yeah how would i prove that its onto
stone fulcrum
onto → every element in the codomain can be represented as some f(x)
Also known as surjective
vestal needle
ok yeah thanks
uncut girder
I finally finished my algebra hw
So much procrastination
Took entire day
And only 2 pages long lol
stone fulcrum
@vestal needle
Homomorphism because
√[a × b] = √[a] × √[b]
Isomorphism because it's an invertible homomorphism
Automorphism because it's an isomorphism from a set to itself.
onyx mirage
I don't really know how one would go about formally proving bijectiveness of particular mappings
uncut girder
Use the definition
sullen flint
prove that it's surjective and injective
toxic plinth
if you want to show f:A->B is bijective, prove that
(1) If f(x)=f(y), then x=y.
(2) if b in B, then there is some a in A such that f(a)=b.
sullen flint
all n by n upper triangular matrices with det= 1 supposedly form a group, under matrix multiplication
proof is supposedly:
closure: product of two upper triangular matrices is again upper triangular, and det(AB)=det(A)det(B) =1
associativity: comes from associativity of matrix multiplication
identity: I with 1 on the diagonal and 0 else. det I =1
inverse: if Det A = 1 then det ( A^-1) =1 too
ok. does this suffice?
shouldn't it be proven that the inverse of a triangular matrix is also triangular?
chilly ocean
yeah I think you're right
I thought maybe it sorta followed from closure, but I don't really think that's enough
onyx mirage
How would one go about about finding the period length more easily
Than brute forcing
spark plank
thorny slate
ok
onyx mirage
I figured you could faff a bit with the order of elements
f(5) = f(2) + f(2) + f(2) + f(3) + f(3) + f(1)
onyx mirage
If |f(2)|= 3, |f(3)|=2 and |f(1)|=1 (not possible in this case) we'd know the period would have a length of 6
To find a way of calcing the period length efficiently
Explicitely if possible
thorny slate
give a recurrence relation in terms of matrices
have you seen that one?
let A =
[1 1]
[1 0]
then the powers of A generate F
in fact A^n =
[Fn+2 Fn+1]
[Fn+1 Fn]
therefore the period is exactly the order of A in GL(2,m)
onyx mirage
Mind blown
thorny slate
the order of GL(2,m) is (m^2-1)(m^2-m)
so you have to check maximal divisors
to find the order of A
onyx mirage
I've seen that formula in GL's before
But only for when m is prime
I think
Nvm im mistaken
Formula only works for primes if we're talking about multiplication, that's where I remember it from
@thorny slate Just iterate through the possibilities now?
thorny slate
go through the lattice of divisors of (m^2-1)m(m-1)
to find the order of A
you can do it pretty fast
in polylog
stone fulcrum
@sullen flint
You have the right idea to prove closure, associativity, identity, inverse. But you're just stating the premise for each one
Product of two upper triangular matrices is again triangular is not a proof, it's what you need to prove.
thorny slate
the proof is clear though
stone fulcrum
Is it? I actually don't know how to prove it lel
thorny slate
you just calculate
chilly ocean
the way I can think to do it is to imagine multiplying two matrices AB, the columns of the matrix B are triangular and so at some point they have 0s below, so the dot product of that column vector in B with any row vector in A will result in exactly the same 0s at the bottom of the column vector
I'm sure that can be cleaned up a bit more, maybe even wrapped up into a cute induction argument
I think I said that backwards but whatever not too serious, idea makes sense
thorny slate
(AB)ij is the dot product of a vector with (i-1) zeroes at the start with one with (n-j+1) zeroes at the end.
snow dew
Triangular means that the subspaces you get by taking the k frst vectors of the canonical base are stable
chilly ocean
here I figured out a cuter way just now
snow dew
That translates well with composition
chilly ocean
$ C_{ik} = \sum_{j=1}^n A_{ij}B_{jk}$
mellow vaporBOT
cloud walrusBOT
Empty Axes:
chilly ocean
since A is triangular i>j means A_{ij}=0 and so on
so i>j and j>k means i>k
so C is triangular
sullen flint
I know just stating that AB is triangular for A and B triangular isn't a proof,
I just was wondering whether that, plus the rest of what I listed sufficed to prove it to be a group
because that's what I found in a textbook but it seems to me that it isn't enough
sullen flint
I didn't bother to prove it lol, I just looked at two triangular matrices and supposed its product was going to be triangular lol
also one can intuitively see that the inverse has to be triangular too, if it exists
but I didn't trust my intuition much in that case
thorny slate
you can quickly construct it with row reduction
sullen flint
yup, saw your message before, thanks
also saw a proof that decomposed it into A =D(I+N) where D is the diagonal of A, and N is a triangular matrix with diagonal terms equal to zero.
and as N^n =0 you can construct an inverse easily
chilly ocean
that's cool
sullen flint
namely I-N+N²-N³....
chilly ocean
the geometric series sorta trick
sullen flint
yup
chilly ocean
I'll have to remember that, I swear the geometric series comes up everywhere
I'm still satisfied with myself for coming up with that proof that triangular matrices are closed I think I probably just assumed it up to this point too cause I didn't see any trick lol
toxic plinth
if a^3 = a for all a in some ring R, then R is commutative.
this problem shouldn't be allowed to exist
3hard5me
sullen flint
I have a silly proof for that
please tell me it's wrong
assume xxx=x
for all x
and that ab!=ba for some a and b
then aabb!=abab
(left multiply a and right multiply b)
then right multiply ab and get
aabbab!=ababab=ab
(aabb)ab!=ab
thus aabb!=1
left multiply by a and right multiply by b
ab!=ab
a contradiction, therefore ab=ba
now, (aabb)ab!=ab => aabb!=1
I'm not sure if that can be justified
can you have ab=b with a!=1 ?
I think it could work for b=0
but that would mean that a0!=0a
anyway I think my proof works regardless
cerulean rune
You can't in general conclude from x != y that ax != ay for all a
cerulean rune
yeah, if a has a (left) inverse in the ring R, then it's clearly fine; it's also fine just if a is not a zero-divisor (i.e. there is no z such that az = 0) -- which it is pretty common in working with rings for there to either be no zero-divisors, or for you to have a lot of control over which elements are zero-divisors
so like, in a lot of "typical/real-world" examples there isn't much problem going from ax = ay to x = y. just not in this generality :p
uncut girder
Abba is a pop music group
chilly ocean
fun problem I don't see how to prove it, but I think I have a decent first step,
$ab=ab \ a^3b^3 = (ab)^3 \ aaabbb=ababab$
cloud walrusBOT
Empty Axes:
uncut girder
Mod-p irreducible theorem is powerful
chilly ocean
what's that one, sounds cool
stone fulcrum
A polynomial is irreducible if it is irreducible mod p @chilly ocean
stoic rose
Because (1+1+1)^3 = (1+1+1) we know 1 + 1 + 1 + 1 + 1 + 1 = 0 so the ring have characteristic 2,3 or 6. Maybe this can help
chilly ocean
oh ok
didn't know this had a name
I think that follows from the chinese remainder theorem
chilly ocean
given the functions f : A->B and g: B->C.Their composition is h(x)=g(f(x)) is injection .Prove that if f s not injection
then g is not total ```total function is function that exists
or a bijective function
I am not sure which of these 2 they mean
tough nacelle
total as opposed to a partial function?
chilly ocean
i think total means just a function
uncut acorn
This is not abstract algebra smh
snow dew
Well f must be an injection and i think they want you to prove this by contradiction but its weirdly said
covert vector
suppose a=a³ for all a \in R
then (x+y)=(x+y)³=
x³+y³+(xxy+xyx+yxx+yyx+yxy+xyy) =
x+y+(xxy+xyx+yxx+yyx+yxy+xyy)
then xxy+xyx+yxx+yyx+yxy+xyy = 0 (*)
also substituting y=-y' yields the other equality
yyx+yxy+xyy -xxy-xyx-yxx = 0 (**)
hmm
substitute y=1 in (*) to get
3x²+3x=0
x=1 yields 6=0 so 6z=0 for all z in R
(*)-(**) yields
2(xxy+xyx+yxx)=0
2(xxxy+xxyx+xyxx) = 2(xy+xxyx+xyxx)=0
2(xxyx+xyxx+yxxx) = 2(yx+xxyx+xyxx)=0
subtracting these yields 2(xy-yx)=0
so 2xy=2yx
ok that's pretty damn close to commuting
(I didn't fully come up with this on the spot, I was gonna type some progress here last night but it woulda said "woog is typing..." for like 20 minutes so I did it in my phone notepad instead lol)
thorny slate
gross!
covert vector
3(x²+x) = 0 implies
3((x+y)²+(x+y)) = 3(x²+y²+xy+yx+x+y)
= 3(x²+x)+3(y²+y)+3(xy+yx)
= 3(xy+yx) = 0
3xy+3yx=0
2xy-2yx =0
subtracting these we get
xy+5yx=0
since 6z=0 for all z \in R, setting yx=z we get
xy+5yx = xy+5yx - 6yx = xy-yx = 0
therefore xy=yx for all x,y \in R
@chilly ocean @yaya#4734 got it
oh yaya left
chilly ocean
hot ok I'll read this when I get back from eating dinner
covert vector
yeah
toxic plinth
oof
covert vector
hii
ok the messy xxy stuff not shortening to x²y is cuz I just foiled the whole thing out pretty much
like in the noncommutative case
(x+y)ⁿ is just the sum of all strings of X's and Y's
of length n
toxic plinth
how did you come up with that wtf
covert vector
well when u have equations that are kinda self referential a common technique is to use the axioms of whatever space ur in for hints
and fix some variables, and plug in ez numbers for the others
just see what happens
get a bunch of relations
modify them slightly
test some other stuff, get more relations
and eventually it's enough to break the whole thing down
thorny slate
iis that really a technique
toxic plinth
thing is, this was in the first chapter of an intro AA book lolol
covert vector
that's how u do functional equations
toxic plinth
author trolled the readers big time
covert vector
in competition math
I guess if you have different taste for techniques it may not count for u jaco :p
toxic plinth
ah it must be my severe lack in competition math
oh, it apparently holds for any n not just 3
lol
full blaze
🔫 
toxic plinth
👀 generalization time ?
covert vector
is the exercise to do it for all n?
that is
if aⁿ=a for all n
then R is commutative
I don't mind trying that
toxic plinth
no the exercise is for n=3, but a friend told me it holds for any n
covert vector
I'm somewhat skeptical about how true that is
I doubt it's true tbh but I can doodle some stuff :p
toxic plinth
it's called jacobson's theorem
toxic plinth
i haven't done abstract algebra in a long time, stopped at rings.
im gonna start artin soon
do AA and LA properly
covert vector
nice !!
toxic plinth
now im trying to prove the chain rule
i used the common wrong proof at first. im trying to figure out the correct one
@covert vector did u participate in the putnam
if so, how was it
chilly ocean
I might have found a proof but it seems sheisty
(x-1)x(x+1) multiplied out makes 0
but so does every permutation of (x-1), x and (x+1)
since x commutes with its neighbors
and its neighbors commute with its neighbors
everything commutes
:^D
toxic plinth
proof by neighbors
chilly ocean
lol does that legitimately work
I think there are some issues with the difference of every pair of elements not being an integer
but in that circumstance, they were isolated to begin with, so they probably commute anyways... right?
lmao
toxic plinth
how do you even know 1 exists
chilly ocean
yikes no multiplicative identity, I don't even know of any examples of rings like that
I thought those were called like rngs not rings
toxic plinth
indeed's'td've
stone fulcrum
@chilly ocean
It depends on your study. Depending on the course, a ring can be defined with or without the identity.
snow dew
Polynomials in an element always commute with the element, right?
And with other polynomials of the same element
thorny slate
yes
errant drum
I'm trying to get a list of the irreducibles in the Gaussian integers. So far every prime that's 3mod4 and all complex numbers whose norm is a prime. Is there anything else I've missed?
Like what complex numbers should I consider
supple moss
Hi I don't know the correct place to ask this...
A friend of mine is preparing for an exam. They are going through an old paper + its memo, and don't really agree with one if its answers. It's a linear progamming question.
It's question b and d in the following pdfs
(imho those are not even questions, but nvm)
I also can't really tell why they would put that as the "answer" because my opinion is that, for example in b, they are not even mentioning the $2000 in the memo, and also my opinion is that they are just re-stating the obvious; they are giving the "max amount of a product that can be made if only making that one". We already know that from the given table, don't we? So how is this applicable to "question" b and d? They are briefly mentioning the boolean variables Z_i, but my opinion is that the memo's answer is kind-of missing the point?
chilly ocean
,$ 400(2)^{x+3} = 800^{x+3}?
cloud walrusBOT
joebobbob:
chilly ocean
No.
800^(x+3) = 400^(x+3) * 2^(x+3)
can you do 2^x * 2 ?
I'm not sure I follow?
so i cant multiply things with a variable in the exponent with another thing that doesnt have the same variable?
The exponent rule here is: (ab)^c = a^c * b^c if that's what you're asking?
hmm ok
i see
thorny slate
wtf
chilly ocean
?
thorny slate
wrong channel mate
errant drum
I'm trying to get a list of the irreducibles in the Gaussian integers. So far every prime that's 3mod4 and all complex numbers whose norm is a prime. Is there anything else I've missed?
thorny slate
errant drum
@thorny slate I don't want to look at the answer :/. Just want to know if the above things I mentioned are all of them. Also server lore?
thorny slate
they are all the cases
you should now discard the possibility of an irreducible having norm product of primes which are 1 mod 4
errant drum
Yeah that won't seem difficult.
uneven arch
hey hey, can someone help me with this, I feel stupid just looking at it.
Perry can tar a roof in 12 hours. Nochi can tar the same roof in 10 hours. how long would it take them if they worked together
it sounds so simple yet i can't find the solution
I solved it by multiplying and rounding my sum
I got 100
im not sure if it's correct tho
stone fulcrum
Perry can tar 1/12 of a roof every hour
Nochi can tar 1/10 of a roof every hour.
Together, they can tar (1/12 + 1/10) roofs per hour