okaay
#groups-rings-fields
solemn rain
latent anvil
We can assume that n2 isn't 1, right?
latent anvil
Right, so if n2 is 1 we're done
solemn rain
for contradiction
latent anvil
Right
So we know n2 is either 5 or 25
What about n5?
What are the possibilities?
latent anvil
no it can't
solemn rain
why
latent anvil
Well, what do the sylow theorems tell us about n5?
latent anvil
Why does it divide 5^2?
latent anvil
no
latent anvil
What else do we know about n5?
latent anvil
Yes
solemn rain
well none of those is congrunt to 1 in Z/5Z right?
solemn rain
has to be 1 ig
latent anvil
Yeah exactly
solemn rain
damn
wow im so bad
thanks @latent anvil
groups of order 105 have normal sylow-5 subgroup
n_5 is congrunt to 1 mod 5
and n_5 must divide 21
hence n_5 must be 1
latent anvil
no
solemn rain
y 😦
latent anvil
what are the divisors of 21
solemn rain
fuck my life lmao
latent anvil
yes
solemn rain
yea 21 is congrunt mod 5
so it cant be normal?
ig something happens if n_5 is 21
oh i got it i think
it has njormal sylow 5 and norrmal sylow 7
( for some reason prob cauchy idk tbh ) elements of order 5 are in sylow 5 subgroups
the same for 7
if i count i think this si more than 105 so it cant happen
right argument?
latent anvil
I think you're on the right track
solemn rain
yea
ok but i gotta ask
why do elements of order 5 have to be in sylow-5 subgroupxs
cuz of cauchy?>
latent anvil
No
Because of lagrange
Cauchy tells you that any 5 sylow contains an element of order 5
But if you have an element of order 5
Actually wait, Lagrange isn't good enough
You need to know that any p-subgroup is contained in a sylow p-subgroup
oblique river
I mean in this case, sylow 5 subgroups are order 5
latent anvil
Oh yeah that's true
oblique river
so order-5 subgroups are the sylow subgroups
latent anvil
I was thinking of the general case
solemn rain
You need to know that any p-subgroup is contained in a sylow p-subgroup
okay
latent anvil
so if I remember the proof
solemn rain
i proveed that i think
latent anvil
why what?
An element of order 5 generates a subgroup of order 5
5 is prime
That subgroup is contained in a 5 sylow
solemn rain
cool
tysm i think i got stuff abit
now
@latent anvil i gotta tag you for the thanks bro
olol
toxic zephyr
@latent anvil So through computation I was able to show (not like a formal proof) that for GL2 the center much be a diagonal matrix with equal diagonal elements by choosing two easy elementary matrices. I'm not entirely sure how I would go about formally prove it for GLn (well induction of course). Is it not possible to show algebraically, or do you have to start writing entries?
latent anvil
As I said, think of it in terms of linear transformations
Think about what the two elementary matrices are representating
Because you can probably define functions which do the same thing but for more indices
arctic bear
hi uh I have a question:
can someone help me out with the following problem: if y varies directly as z and inversely as x and y=10 and z=5 when x=12.5, find z when y=37.5 and x=2?
how would i solve this?
scarlet estuary
this is not abstract algebra
ripe crest
is there a name for an algebraic object where a ring R acts on a ring S?
it's more than a R-module
latent anvil
An algebra?
latent anvil
So?
latent anvil
No, it's used more generally than that
latent anvil
ripe crest
ah nice
latent anvil
Really an algebra over a commutative ring R is just a ring A with a morphism R -> A
sorry, a commutative ring A
If you want noncommutative algebras then R should map into the center
Usually
ripe crest
thanks~
ripe crest
Am I making mistake or are there 3 group homomorphisms $\mathbb{Z}/3\mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}$?
cloud walrusBOT
silent flower:
latent anvil
you're making a mistake
latent anvil
lol
ripe crest
hmm
I know there is only 1 homomorphism as Z-modules
but I thought there were 2 as just abelian groups
$1 \mapsto 0$, $1 \mapsto 1$
cloud walrusBOT
silent flower:
latent anvil
Z module homomorphisms and group homomorphisms are the same thing
Z modules and groups are the same thing
If you know what a category is, there's an isomorphism of categories between Ab and Z-Mod
The second one isn't homomorphism because 1 + 1 goes to 0
Err, sorry
ripe crest
I get why abelian groups are z-modules, but why are all z-modules abelian groups?
couldn't the action of Z be different?
latent anvil
So say you have a homomorphism φ from Z/3Z -> Z/2Z such that φ(1) = 1
You're saying such a homomorphism exists, right?
latent anvil
Sure
Also try showing that if A is an abelian group, there is only one Z-module structure on A
ripe crest
1 = φ(4) = φ(2 + 2) = φ(2) + φ(2) = 0 + 0 = 0
latent anvil
\cong
cloud walrusBOT
silent flower:
ripe crest
I know that I have at most min(m, n) homomorphisms
because each homomorphism is completely determined by where 1 is sent
latent anvil
What constraints can you think of on the image of 1?
ripe crest
"The order of φ(x) divides the order of x
Right?"
Is this a group theory thing?
latent anvil
Yeah it's a group theory thing
ripe crest
Drat - took that 2 years ago, remember little lol
latent anvil
It's very easy to prove
n φ(x) = φ(nx)
So if nx = 0, also n φ(x) = 0
In particular take n = |x|
ripe crest
But why is n the smallest value for which that occurs?
latent anvil
it's not
I said the order of φ(x) divides the order of x
Not that it's equal to it
If mx = 0, then m is a multiple of |x|
ripe crest
ah
latent anvil
Use division with remainder
ripe crest
so, |φ(x)| is a multiple of |x| in our case
latent anvil
Yeah
latent anvil
No, it's the other way around
ripe crest
so |φ(x)| divides |x|?
latent anvil
Yes
ripe crest
then |x| is a multiple of |φ(x)|?
latent anvil
Yup
ripe crest
hmm
latent anvil
Let n be the order of x. Then n φ(x) = φ(nx) = φ(0) = 0
So |φ(x)| divides n
That's what I was saying above
latent anvil
Np
ripe crest
been a while
So, |φ(x)| divides |x|. So in our case, |φ(1)| needs to divide |1|
But, |1| = m
So, we need |φ(1)| to divide m
latent anvil
Yup
ripe crest
Since, $\mathbb{Z}/m\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}$
cloud walrusBOT
silent flower:
latent anvil
So that gives an upper bound
ripe crest
Question though
latent anvil
What do you mean?
ripe crest
like, |φ(1)| <= m
ripe crest
so for $Z/3Z \rightarrow Z/9Z$, we cant map 1 to anything greater than 3 in Z/9Z
cloud walrusBOT
silent flower:
ripe crest
$1 \mapsto 0$, $1 \mapsto 1$, $1 \mapsto 3$ would be our 3 homomorphisms
cloud walrusBOT
silent flower:
ripe crest
but 6 > 3
latent anvil
But |6| = 3
ripe crest
ahhh
the order
right
So I have necessary conditions on what the image of 1 can be, but I don't know if it's sufficient yet tho
solemn rain
Prove that if |G| = 132 then G is not simple
|G| = 132 = 2^2(3)(11)
G has sylow-2subgruops of order 4 , sylow-3 subgruops and sylow-11 subgroups
n_3 must divide 44
n_3 = 1 mod 3
this limits n_3 to be either 1 or 4
if n_3 = 1 we are done
now suppose n_3 = 4
now we look at sylow 11-subgorups
n_11 must divide 12 and also n_11 = 1 mod 11
( assume that i assumed G is simple xd )
n_11 must be 12
elements of order 3 are in sylow 3 subgroups and elements of order 11 are in sylow 11 subgroups
im stuck
vestal snow
Let $R$ be a PID and $M$ be a module annihilated by the ideal $(a)$. Let $a={p_1}^{\alpha_1} \cdots {p_k}^{\alpha_k}$ be the factorization of $a$ into unique primes. Let $M_i$ denote the annihilator of ${p_i}^{\alpha_i}$ in M i.e. the set of elements $m$ such that ${p_i}^{\alpha_i} = 0$. Prove $M$ is isomorphic to $M_1 \times \cdots \times M_k$
cloud walrusBOT
Have a Banana Bitch:
vestal snow
The isomorphism I came up with was $f(m) = ({p_2}^{\alpha_2} \cdots {p_k}^{\alpha_k}m, {p_1}^{\alpha_1} {p_3}^{\alpha_3}\cdots {p_k}^{\alpha_k}m, \cdots , {p_1}^{\alpha_1} \cdots {p_{k-1}}^{\alpha_{k-1}}m$
cloud walrusBOT
Have a Banana Bitch:
vestal snow
It is clearly a module homorphism
Injectivity follows from the fact that the gcd of ${p_2}^{\alpha_2} \cdots {p_k}^{\alpha_k}m, {p_1}^{\alpha_1} {p_3}^{\alpha_3}\cdots {p_k}^{\alpha_k}m, \cdots , {p_1}^{\alpha1} \cdots {p{k-1}}^{\alpha_{k-1}}$ is just one, therefore $ \exists c_1, \cdots , c_k$ such that $c_1{p_2}^{\alpha_2} \cdots {p_k}^{\alpha_k}m + c_2{p_1}^{\alpha_1} {p_3}^{\alpha_3}\cdots {p_k}^{\alpha_k}m + \cdots + c_k {p_1}^{\alpha1} \cdots {p{k-1}}^{\alpha_{k-1}}=1$
cloud walrusBOT
Have a Banana Bitch:
vestal snow
Now suppose $f(m) = (0,0, \cdots , 0)$ and use the above condition to get $m=1m=0$
cloud walrusBOT
Have a Banana Bitch:
vestal snow
How would one go about proving $f$ is surjective?
cloud walrusBOT
Have a Banana Bitch:
chilly ocean
Hey everyone, have a quick question on the following problem:$\\$
Let $G$ be a nonabelian group of order $p^3$ where $p$ is prime.$\$
(a) Prove that $G/Z(G) \simeq \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}\$
(b) Prove that $g^p \in Z(G)$ for all $g \in G.\\$
I was able to prove (a), but for some reason cannot prove part (b). If anyone has any insights that would be greatly appreciated!
cloud walrusBOT
wstayman:
chilly ocean
well what is g^p like in Zp × Zp?
Its $([0]_p,[0]_p)$, and that's the angle I was trying at first, but was unable to relate it back to $G$ itself. I can post what I have so far if that'd be helpful
cloud walrusBOT
wstayman:
chilly ocean
have you done something on homomorphisms?
Yes we've covered homomorphisms quite in depth
I guess my question boils down to how can I use the fact $G/Z(G) \simeq \mathbb{Z}_p \times \mathbb{Z}_p$ to prove part (b)?
cloud walrusBOT
wstayman:
chilly ocean
if you proved that 
then you're… done?
so you know how to get a quotient with homomorphisms?
Yes, I guess i haven't proved that $g^p$ is $([0],[0]) \in \mathbb{Z}_p \times \mathbb{Z}_p,$ but that the coset $g^pZ(G)$ is $([0],[0]).$
Since the isomorphism is $G/Z(G) \simeq \mathbb{Z}_p \times \mathbb{Z}_p.$
chilly ocean
so you have this surjective homomorphism $\func{\varphi}{G}{\frac{G}{Z(G)}}$ whose kernel is Z(G)
cloud walrusBOT
tet:
chilly ocean
Okay I can do that. How does that imply $g^p \in Z(G)$ for all $g \in G$? Sorry, a lot of this is fairly new to me
cloud walrusBOT
wstayman:
chilly ocean
first, $\frac{G}{Z(G)}$ is the same as $Z_p \times Z_p$ \
now take any $g \in G$ \
what is $\varphi(g^p)$?
cloud walrusBOT
tet:
chilly ocean
Ahhhh, I think I see now. $\varphi(g^p) = \varphi(g)^p = ([0],[0]) \in Z_p \times Z_p$ and $([0],[0]) \in Z(Z_p \times Z_p) \Rightarrow g^p \in Z(G)$? (Or something roughly along those lines)?
cloud walrusBOT
wstayman:
chilly ocean
yup, remember the kernel is the center
The kernel of $\varphi$ is the center of $G?$
cloud walrusBOT
wstayman:
arctic bear
this is not abstract algebra
@scarlet estuary then where should I ask this question?
scarlet estuary
a generic questions channel, or a channel about high school algebra (ie #prealg-and-algebra )
if you don't know what any of the words here mean, your question doesn't belong here.
ripe crest
isn't the characteristic of a ring and the order of the element in the group the same?
upper pivot
characteristic of a ring is not exactly the order of an element, it is the smallest n such that na = 0 for all a, while order of an element corrosponds only to that element
so you could have for example an element of order 4 in the additive group, but the char of the ring is 12
(3 in Z/12Z for example)
oblique river
I mean, the characteristic of a ring is the order of the element 1
unless the order is infinite in which case we say the characteristic is 0
yeah I was just responding to johndoe
that you can just conside the element 1
instead of all elements
solemn rain
what does 'characteristic' mean
upper pivot
yeah i was just putting a more general anwswer forward
oblique river
oblique river
several people have given definitions in the last 10 minutes lol
because english has a limited number of words?
solemn rain
a subgroup H is characteristic iff f(H) = H for all Aut in Aut(G
oblique river
so some o them get reused?
yes but that's not related to teh characteristic of a ring
oblique river
different people defined those terms at different times
and happened to use the same word
latent anvil
@solemn rain are you still doing the sylow theory problem?
Count the number of elements of order 3 and 11
solemn rain
no i did it
ty
for asking
i wanna try this as my final problem for this section
Determine the Number of Elements of Order 3 in a Non-Cyclic Group of Order 57
wish me luck
ripe crest
gotcha, thanks
so the order of an element is always leq to the characteristic
for finite rings
and more than that, it would divide the characteristic
oblique river
the additive order
and yes
it's also true for infinite rings in positive characteristic
solemn rain
omg i still dont know
how the numebr of elements of ac ertain order is related to sylow
how can i count those?
please help this is frustrating as fuck
ripe crest
@latent anvil , btw I was able to solve that problem except for 1 part...im gonna come back to that later
latent anvil
@ripe crest which problem?
@solemn rain an element of order p is contained in a p-sylow. If your p-sylows have order p (so the order of your group is divisible by p only once) then distinct p-sylows intersect trivially
solemn rain
an element of order p is contained in a p-sylow' why
so elements of order 5 is in a sylow-5subgroup?
elements of order p are in sylow-p subgroups?
how many?
@latent anvil
latent anvil
we talked about this yesterday
A p-subgroup is contained in a p-sylow subgroup
You said you proved this as an exercise
solemn rain
ys
yes
now why are elements of order p
have to be in p-subgroups
ik the converse is obv ig
latent anvil
what's the order of the subgroup generated by g
latent anvil
yes
solemn rain
okay?
i strill dont get it
i am stupid and frustrated
ookay nvm
elements of order 5 are in the groups generated by the elements of order 5 ig XD
what why
latent anvil
why what?
solemn rain
okay so if i generate a group with an element of order p
its a p subgroup
yea obviously im such a fucking moron
and p ssubgroups are in sylow
so order p elements rae in sylow eh?
right?
how cna i count them?
like how can i know the number of elements of order x in each sylow p subgroup for example
am i supposed to argue for this on my own or are the solutions already made
so stupid tbh fuck
latent anvil
yes
You can't do it in general
You have very little control over the number of elements of order p in a p-group
But if your p-sylow has order p or p^2, then you know the group
So you can count them directly
solemn rain
wdym i know the group
latent anvil
What are the groups of order p?
latent anvil
Yes
solemn rain
okay what about p^2
solemn rain
is this classification stuff?
latent anvil
Yeah but it's easy classification stuff
latent anvil
The p^2 case is too hard anyways
You need to reason about when an element is in two p-sylows, but if they have order p then they have to intersect trivially
If they have order p^2 or higher you can't say very much about the intersection
latent anvil
Yeah so suppose you have a group G of order pm where p does not divide m
solemn rain
okay
latent anvil
Let n be the number of p-sylows subgroups
How many elements of order p are there in G?
solemn rain
n_p cuz each element would generate a p subgroup?
solemn rain
n_p is 1 always
latent anvil
You need to be careful about whether elements generate the same cyclic subgroup
latent anvil
Yeah that's true, but there are 4 elements of order p
solemn rain
okay
elements of order p in G are contained in sylow-p subgroups
fuck idk
oh
i forgot this fucking buffoonlagrange
all elements in sylow - ssubgroups must be of order p or 1
cuz they divide the order lmao
so the number of elements of order p in G p(n_p)?
no not p
each p group has p elements
each element must divide p (order)
so either p or 1
so i have (p-1) elements of order p in each sylow p subgroup
okay so its (p-1)(n_p)
am i right?
and i thinnk i get what you mean now , if the sylow p subgroups have order p^2 and ur not sure about their intersection being trivial u lose count cuz u may count an element twice
cool
ripe crest
i forgot this fucking buffoonlagrange
pretty sure he was quite smart
solemn rain
fuck this guy
now he died and nobody gives a shit sad ok boomer
@ripe crest do u know wether im right or wrong?
whether
ripe crest
no
solemn rain
cool ty anyways
ripe crest
and I find it distasteful when you talk about mathematicians like that, or just people in general
solemn rain
i was joking obv
obsidian zealot
where can i get help with a math problem
ripe crest
isomorphic as rings? as groups? as sets?
scarlet estuary
as a ring?
obsidian zealot
I know that Z are countable sets and C are uncountable sets that's why it is not isomorphic since there exists no bijection, but i guess i need a more better proof with numbers
groups
scarlet estuary
i mean, thats your proof
ripe crest
an isomorphism of groups => isomorphism of sets
scarlet estuary
"Z is countable but C is uncountable, so there cannot exist a surjection from Z into all of C"
scarlet estuary
it should be, if you're allowed to assume that Z is countable while Z is uncountable
if you want to be hyper formal you can say
something like ``an isomorphism $f\colon \bZ \to \bC$ cannot exist. this is because, since such an isomorphism would have to be surjective, it must satisfy $\text{range}(f) = \bC$ and therefore $\abs{\text{range}(f)} = \abs{\bC}$; but we know that $\abs{\bZ} < \abs{\bC}$ since $\bZ$ is countabe while $\bC$ is not, so $\abs{\text{range}(f)} < \abs{\bC}$"
ripe crest
isomorphic groups share the same group properties
cloud walrusBOT
Namington:
obsidian zealot
oh I understand
scarlet estuary
if you dont want to use a cardinality argument
oblique river
C is divisible and Z is not
scarlet estuary
e.g. youre not allowed to assume that Z and C have different cardinality
or dont want to assume it fsr
you could just look at like
oblique river
meaning every element of C can be divided by 2 but that's not true for Z
scarlet estuary
literally anything
about the groups
i dont think they share any properties
lmao
[exaggerating but yeah]
obsidian zealot
the definition of isomorphic means that it has to be one-to-one and surjective right
or just surjective
ripe crest
an isomorphism always has a context
an isomorphism of groups is something different from an isomorphism of sets
scarlet estuary
an isomorphism of groups is a homomorphism that is one-to-one (injective) and onto (surjective)
obsidian zealot
got it
scarlet estuary
so in particular, it must satisfy $\varphi(ab) = \varphi(a)\varphi(b)$ for all $a, b$ in its domain
cloud walrusBOT
Namington:
scarlet estuary
as well as being injective and surjective
ripe crest
nand gates ftw
obsidian zealot
thanks for the help guys
gritty cradle
i want to show $K= \mathbb{Q}[x]/(p(x))$ is a field. Since $(p(x))$ is an ideal, $K$ is a quotient ring. So all I need to show is that each element of $K$ has a multiplicative inverse, right?
cloud walrusBOT
h!:
gritty cradle
oh, and p(x) here is a given polynomial
ripe crest
well
bleak abyss
It suffices to show $(p(x))$ is a maximal ideal in $\mathbb{Q}[x]$
cloud walrusBOT
Daminark:
ripe crest
what do u know if p(x) is irreducible?
bleak abyss
Meaning p is an irreducible polynomial
ripe crest
Usually with those types of questions, it's easier to show that P is irreducible or that the quotient is isomorphic to some well-known integral domain like Z, Q, R, or C
gritty cradle
p(x) irreducible means that no element of K would be a zero divisor?
ripe crest
P irreducible <=> Q[X]/(P) a field <=> Q[X]/(P) an integral domain
irreducible means, if P = AB, then either A in Q or B in Q
gritty cradle
sorry yes, i meant to say p(x) irreducible would imply that q(x)+(P) is not a zero divisor right?
bleak abyss
Yes but don't think too directly about it
Have you heard of prime and maximal ideals?
gritty cradle
yeah, it's just been a year since the last time i was taking abstract algebra so i'm refreshing my memories at the same time
maximal ideal means that no other proper ideal contains the ideal
gritty cradle
prime ideal i don't remember
bleak abyss
Prime just means that xy\in I => x\in I or y\in I
gritty cradle
OK
bleak abyss
So yeah if you give me a field, let's take an ideal in it
If that ideal is 0 it's 0
Assume it contains any non-zero element x
(Call the field k)
Well, for any y\in k, we have y = yx^{-1} x, so that's also in the ideal
So all ideals are just 0 or k
If something isn't a field, then give me a non-zero element x which isn't a unit
Then the ideal generated by x can't contain 1
gritty cradle
ahhhh
bleak abyss
So a ring is a field if and only if the only two ideals are (0) and (1)
Meaning an ideal in a general ring is maximal if and only if the quotient is a field
(General meaning commutative with identity lol)
The nice thing about Q[x], or really k[x] for any field k, is that they're PIDs (principal ideal domains)
So prime ideals are maximal. In fact any integral domain with unique factorization has the property that for principal ideals, prime = maximal
So in those cases k[x]/(p) is a field if and only if p is irreducible
gritty cradle
i see
gritty cradle
https://i.imgur.com/BZq4OrD.png Second question, want to check how i'm doing
a) is degree 4, eg. basis ${1, i, \sqrt{2}, i\sqrt{2}}$
b) is degree p-1 eg. basis ${e^(2i\pi n/p) | n \in {1, \ldots, p-1}}$
c) is degree 6?
d) is of infinite degree, basis ${1, u, u^2, \ldots}$
e) I think it is 2? i know $[k(u) : k(u^2)] = [k(u^2) : k] / [k(u) : k]$ by Tower Lemma, but i don't really understand how i would find [k(u^2) : k]
ripe crest
So a ring is a field if and only if the only two ideals are (0) and (1)
@bleak abyss That's not true
need commutativity
and we need 1 != 0
bleak abyss
I'm ignoring obvious stupid assumptions lol
cloud walrusBOT
h!:
Compile Error! Click the 
reaction for details. (You may edit your message)
ripe crest
ok
ripe crest
no
bleak abyss
Yeah actually if it has exactly two ideals
latent anvil
Lol
bleak abyss
Then I'm right
ripe crest
yeah you're right
ripe crest
my bad
bleak abyss
So get fucked
latent anvil
Lmao
bleak abyss
That's what people who try to correct me over nonsense get
ripe crest
errr
bleak abyss
I was accidentally careful
latent anvil
ADMIN ABUSE
ripe crest
I don't think specifying commutativity is nonsense
bleak abyss
Kind of is lol
latent anvil
I usually say ring and mean commutative ring with 1
bleak abyss
Like non-commutative rings are of course a thing but like if a statement holds when a common but unstated assumption is made then it's kind of implicit
ripe crest
i think it depends on the audience
ripe crest
if it's someone new, then not mentioning it is a mistake imo
if it's just relearning, then it's fine because they can easily understand the unstated stuff
shrewd halo
yes sham
bleak abyss
And that happens to be this context
gritty cradle
it would still be infinite degree though right?
bleak abyss
But like in general we're talking about stuff like PIDs
So it's fair to say "rings are implicitly commutative when needed"
latent anvil
@gritty cradle right
ripe crest
fair enough
latent anvil
It would be infinite degree, yeah
And I think your idea with the tower lemma is right
gritty cradle
basis being something like u^n, n in Z then?
latent anvil
Yeah that's exactly right
If that's linearly dependent, you get a polynomial which u satisfies
And it spans it basically by definition
gritty cradle
yeah okay
latent anvil
So we know [k(u) : k] = [k(u) : k(u^2)][k(u^2) : k]
And that the left hand side is odd
Can we say anything about [k(u) : k(u^2)]? Like a rough first approximation
gritty cradle
has to be odd
latent anvil
Yup
gritty cradle
hmmm
latent anvil
I claim we can also get a bound on the degree of the extension
tribal pasture
Why do we define representations of finite groups using the complex field? Is that just an ad-hoc choice or can we consider any field?
gritty cradle
if it's equal that would mean [k(u^2):k] would be 1, i.e. k(u^2) = k
shrewd halo
@tribal pasture having roots of unity makes things very tidy
mild laurel
@tribal pasture There are certain things that are nicer when the field is algebraically closed, but its not true that we only look at representations over C
modular representations are when the representation is over a finite field
tribal pasture
Oh okay thanks thanks, I was just worried because the book just started with saying take the field to be C without any reason
latent anvil
Like, all the big theorems work over an algebraically closed field of char 0
But then you might as well do C
gritty cradle
but then u^2 is an element of k, so u is an element of k, so in that case [k(u):k] would be 1 as well?
tribal pasture
Also char 0 to make maschke work
@latent anvil Is this in response to me? If so is maschke something I will encounter later?
latent anvil
@gritty cradle why would it be 1 if they're equal?
@tribal pasture it was in response to you. I guess so? I would've expected it to be shown before you switch to C
Oh sorry lol
I'm being dumb
tribal pasture
Oh okay. We just started with C
gritty cradle
If [k(u):k] = [k(u):k(u^2)]
latent anvil
Hard to keep track of 3 things at once
tribal pasture
Sorry
I will await
You can first deal with the other person since they were here first
gritty cradle
Uh
latent anvil
Anyways, I don't think this is super helpful
So you get k(u) by adjoining one element to k(u^2), right?
latent anvil
Sure. Think about what [E(α) : E] is in general, and how we can maybe bound it
@tribal pasture so what book are you using?
I think of this stuff in terms of modules
Which may or may not be helpful to you
tribal pasture
This one
tribal pasture
Oh okay
latent anvil
The reason I mention it is that it only works if the characteristic of the field does not divide the order of the group
So if you want to think about e.g. representations of p-groups over F_p, or even F_p bar, you'll have difficulties
tribal pasture
And one more thing, silly question but: Why exactly are we studying the representation, more generally, what does it mean to have a representation? Like I am aware of the formal definitions but am not motivated neither the book develops any motivation
latent anvil
So what I've heard is that it can tell you a lot about the group
Matrices are fairly concrete
But I personally don't get why people care
After a first course in representation theory
It can be used to prove some cool theorems like Burnside
tribal pasture
and what idea is supposed to be captured by "having a representation"
latent anvil
What do you mean?
Every group has a representation
Do you mean like what does a specific representation tell us?
bleak abyss
d i d s o m e o n e s a y r e p r e s e n t a t i o n
tribal pasture
No like what idea is the representation generalising
latent anvil
lol
bleak abyss
tribal pasture
ree- presentation
bleak abyss
tribal pasture
Like for example we have theorems regarding writing the abelian groups as Z-modules and stuff, so that's what my mind goes to when I hear represenation, as in I can present the group in another way. So is that it? Is it like try to find different contexts in which we can find some object which is isomorphic to the group
idk if it makes sense
bleak abyss
You're thinking of the group as a group of automorphisms of some vector space
tribal pasture
Yes
bleak abyss
In principle the representation isn't injective so a priori you're really going through the quotient but yeah
Interestingly enough representation theory of finite groups was developed to solve a very particular problem
tribal pasture
I just dont grasp the word "representation" here like I guess it is just a case of unfortunate terminology and representaion has nothing to do with presenting different contexts isomorphic to the group we are studying
bleak abyss
Namely, take the multiplication table of a group. Replace each g with the variable x_g
You have a square matrix, take the determinant, you get a polynomial
This polynomial factors into irreducibles, and each one comes up with multiplicity equal to its degree
Sound familiar?
And yeah turns out that was what rep theory was developed for and all of a sudden boom it explodes
tribal pasture
Oh I seee I see, thanks for the answer!
ripe crest
why would we want to do that? replace g with x_g
latent anvil
This is probably embarrassing since I've taken a rep theory course but I don't know what that polynomial is
bleak abyss
I don't know if it's a nice polynomial tbh, I said "Sound familiar?" because this to me sounds a whole lot like the regular rep of a group
latent anvil
Oh lol
bleak abyss
Irreps come up with multiplicity equal to degree sorta thing
I'm like 99% sure that's what's gonna prove it lol
leaden schooner
Im trying to do a problem from Pinters abstract algebra where Im asked to find the root field of $x^4 - 2$. I feel like Im missing something because I seem to be doing more work than is required for the first excerise of the chapter. Right away we know the rootfield is $\mathbb{Q}( 2^{1/4}, -2^{1/4}, i2^{1/4}, -i2^{1/4})$ but I think the point is you're supposed to simplify it. So first I simplified $\mathbb{Q}( 2^{1/4}, -2^{1/4}) to \mathbb{Q}( -2^{1/4} + 2(2^{1/4)} = 2^{1/4} ) $. Then I extended that to $\mathbb{Q}( 2^{1/4}, i2^{1/4})$ which is $\mathbb{Q}( 2^{1/4} + i2^{5/4})$. The problem though is I need to then find the minimal polynomial of this for the final extension, but this polynomial is of degree 8 ($z^8 + 30z^4 + 2529 $) and eisenstein, gausses lemma, and crohns don't help. So even if i could check if this is the minimal polynomial, Id still need to check all 8 roots to make sure $2^{1/4} + i2^{5/4} + t(-2^{1/4}) $ isnt equal to a sum of one of the other roots for some value of t, making me check $8 x 4 = 32$ different combinations over all. Is there an easier way to simplify the root field without all this mess?
cloud walrusBOT
deadpan2297:
cold flint
Assuming root field = splitting field, you can see that Q[2^1/4, i2^1/4] is just Q[2^1/4, i]
blissful ice
Hey I'm wondering if this question even makes sense
like, what exactly is S^{-1} I
We've only defined localizations for rings, not ideals
and here it doesn't mention the ambient space
so I'm assuming it's all fractions of the form x/s where x is from I and s is from S
and the sqrt of that is then ( \exists n., (x/s)^n \in S^{-1}I)
cloud walrusBOT
zavzav:
blissful ice
like if that definition even makes sense. I've never seen anyone talk about a localization of an ideal. My prof doesn't define it, neither do exercises
and I can't find it online I think. So if anyone's heard of the precise definition, please help?
or if haven't but done localizations, tell me I guess
leaden schooner
Assuming root field = splitting field, you can see that Q[2^1/4, i2^1/4] is just Q[2^1/4, i]
@cold flint I dont think they are the same, but I just read about splitting fields now on wikipedia so I might be wrong. The root field of a polynomial a(x) in F[x] is F(c_1, c_2, ... ,c_n) where c_i are roots of the polynomial a(x). Is that the same as the splitting field?
blissful ice
yes
mild laurel
One way to think about it is just that
The ideal I is an R module
So you can think of the localization that way
toxic zephyr
What does it mean for two subgroups to "commute with each other"?
toxic zephyr
So if $hk \neq kh$, then there is some problem with $p(h,k)=hk$?
cloud walrusBOT
nix:
mild laurel
It's not a group homomorphism
latent anvil
Can someone think of a group G and a non-normal subgroup H of G such that H is isomorphic to its normal closure?
I was thinking about doing stuff with free groups on countably many generators, but I couldn't figure it out
toxic zephyr
@mild laurel Would it be because of something like this?
$p(h,k)p(h',k')=hkh'k'$. But $p(h,k)p(h',k')=p(hh',kk')=hh'kk'$. So if $p$ is a group homomorphism, then $hkh'k'=hh'kk'$ (i.e. $H$ and $K$ must commute)
cloud walrusBOT
nix:
latent anvil
yes
It might be a little cleaner if you take h = k' = 1
k h = p(1,k)p(h,1) = p(h, k) = hk
latent anvil
Also if anyone was curious a friend came up with an example for my thing
Take G to be the set of permutations of the naturals which fix all but finitely many elements
mild laurel
What's the subgroup?
latent anvil
the same but for even naturals
Conjugation = relabeling
And you only need to relabel finitely many things
vestal snow
Does 22 c) imply that the prime factorization of a in 18) contains all the primes of R i.e. R is finitely generated?
more precisely, R (as a ring) is generated by {p_1,...,p_k}?
oblique river
I'm not sure what these problems have to do with each other
in 18 you're assuming that M has an annihilator
but in 22 it explicitly says that M need not have an annihilator
vestal snow
Yeah, but you can apply 22 to 18
oblique river
22d does imply 18 if that's what you're asking
but it doesn't imply that R is genreated by the p_i
for example this theorem would apply to the ring Z
vestal snow
But if we apply 22 c) to the ring in 18), we get that M is isomorphic to the direct sum of p-primary components as p ranges over all the primes of R
vestal snow
However, in 18, we also proved that M is isomorphic to M_1, M_2,...,M_k where M_i is the p_i primary component
Oh okay that makes sense
So all other primes must have the trivial p-primary component right?
oblique river
yes
tha'ts part of the definition
if p didn't have trivial p-primary component
then it would be a part of the annihilator
oblique river
I need to try to remember the exact statement, but if p divides the annihilator then p will annihilate some elements of M
well I guess the "exact statement" is this problem
if p annihilates some of the elements of M
then the annihilator will be divisible by p
oblique river
is it not true that in a PID, nonzero prime ideals are maximal?
vestal snow
yes
oblique river
like, a nonzero element of M can't be annihilated by two prime ideals simultaneously
because if so then it would be annihilated by their sum which is the whole ring
so if an element is annihilated by some prime ideal p
then p had better show up in the annihilator of M
vestal snow
oh okay
oblique river
or else the annihilator of M wouldn't annihilate that particular element
sorry I hope you'll forgive me being handwavy here I am a little tired
vestal snow
it's all good
oblique river
👍
scarlet estuary
some weird name synergy going on here
vestal snow
I agree
chilly ocean
I understand they can't be equal because Q(3^0.25) can not have complex numbers, but how are they isomorphic?
I don't think they even have same dimensions...
latent anvil
they do have the same dimension
Do you have any techniques or lemmas for showing two fields are isomorphic?
There's one especially useful one I'm thinking of for isomorphisms of the form Q(α) ≈ Q(β)
@chilly ocean
chilly ocean
that when a number is transcendal, it if isomorphic to field of fractions of the polynomials?
$Q(x) \cong Q(\alpha)$ is $\alpha$ if transcendental
cloud walrusBOT
mega:
chilly ocean
@latent anvil
latent anvil
that's true, but not very helpful here
steep hull
What’s the kernel of the (surjective) map Q[x]->Q(a) given by x->a in the algebraic case?
chilly ocean
i think the kernel might be <p(x)> where p(x) is the least polynomial such that p(a) = 0
@steep hull
blissful ice
@mild laurel thanks for the clarification yday btw, I got it (I think)
tribal pasture
Is this true for a finite group |gG| = |G|? I am very sure this is true but cant recall how to prove that any element x of G can be written as ga for some a in G
hot lake
so you don't know how to solve for a in the equation x = ga ?
tribal pasture
): fnx
chilly ocean
Hey everyone, have a quick question on the following problem:
Let G be a group with order 105 = 3*5*7. Prove that if a Sylow 3-subgroup is normal in G, then G is cyclic.
Any insights are appreciated!
fossil vapor
chilly ocean
I'm coming up on a solution, would you mind taking a look?
solemn rain
yea sure
chilly ocean
Thanks one sec
In a previous part of the question, I was asked to come up with a cyclic subgroup of G of order 35. I was able to do that. Let's call it PQ. Then my thought was to take [G: PQ] = 105/35 = 3. And since 3 is the smallest prime dividing 105, this tells us PQ is normal in G. Then with the assumption that G has a normal Sylow 3-subgroup (which we'll call K), we can write $|KPQ| = |K||PQ| = 105$ (since the intersection of $PQ$ and $K$ is trivial since their gcd is 1). I then found a lemma stating that if a group G has 2 normal subgroups A,B with AB = G and $A\cap B = e$, then $G \simeq A \times B.$ Applying this to our situation, we'd have $G \simeq PQ \times K.$ And since $PQ$ is cyclical of order 35, it is isomorphic to $\mathbb{Z}_{35}$ and $K \simeq \mathbb{Z}3.$ And since gcd(3,35) = 1, we can write $G\simeq \mathbb{Z}{35} \times \mathbb{Z}_3$ which implies $G$ is cyclical. Hopefully that's straightforward, I can explain any part more in depth if you need.
cloud walrusBOT
wstayman:
chilly ocean
It's a lemma we covered a while back that I'd forgotten about. Any mistakes as far as you see? Most places I looked online were doing much more heavy lifting which worries me there might be something I'm overlooking
chilly ocean
PQ is normal because [G:PQ] = 105/35 = 3 and 3 is the smallest prime dividing 105. It's another lemma
solemn rain
oh i didnt know that
you gotta prove 3 lemmas here lmao
existence of PQ where supposedly P and Q are subgroups ig
the second lemma
and
PQ is normal because [G:PQ] = 105/35 = 3 and 3 is the smallest prime dividing 105. It's another lemma
@chilly ocean this
if those are all right the nyea i woudl say this is correct
imo
i may be wrong tho
for me i would have done htis problem by looking at G acting on H ( the normal sylow 3 ) by conjugation
but yea ur right
ig
chilly ocean
Yeah we don't have to prove any lemmas we've proved before, just have to cite the assignment they were proved on. Yes, that's how a lot of people online I saw were doing it. I'm not entirely comfortable with those methods yet. This is my first time looking at Sylow groups and it's quite a bit to take in.
chilly ocean
yeah we started Sylow's theorems today. is this your first or second course in abstract?
solemn rain
none i mjust reading dummit and foote
first time looking at the subject ig but i skimmed through other books b4
namely gaillan
which is far easier
any other problemss?
its practice for me aswell so i would appreciate that
chilly ocean
yeah this question came out of dummit and foote, but my proff put some spin on it so we can't google it as easily lol. we learn from beachy and blaire though usually. that's a really nice book
chilly ocean
Not at all a grad text. This is my second course in Abstract. We'll barely touch on Galois theory by the end of the year, but that I'm excited for. Check out Beachy and Blaire though, its awesome. I can send you a pdf of the latest version if you want
solemn rain
yea i was looking for that
just out of curisoity
anyways if any problems please post here i will greatly benefit too
chilly ocean
will do, thanks for the help
solemn rain
send the text
chilly ocean
its a pdf file.. i'm brand new to discord too so idk how to do that
solemn rain
jaboi ur a chemistry major right
latent anvil
@chilly ocean that's exactly right
So you get an isomorphism Q[x]/<p(x)> ≈ Q(α)
Can we use this for your original problem?
steep hull
Yep, it makes the original problem a lot easier
solemn rain
i have the cute problem as a final problem for my sectiopn in direct products
findding a gropu G such that G is isomorphic to G x G
any hints?
all functions i came up with are ill defined lmao
latent anvil
so you can rule some stuff out pretty easily on, right?
G can't be finite
I think it's useful to consider the corresponding question in linear algebra
Can you think of a vector space V such that V and V × V are isomorphic?
solemn rain
no idk any linear algebra
latent anvil
Oh lol nvm then
ripe crest
Hmm, the VS can't be finite dimensional right
latent anvil
yeah
solemn rain
ugh i gave up on this problem
can u guys check my solution for another one?
Let G be a group with at least two elements which has no subgroups other
than itself and the trivial subgroup. Show that G is a cyclic group whose
order is a prime number ?
latent anvil
How long have you been working on it?
solemn rain
10 mins i just gave it up cuz i tried all the basic groups ik about
so its probably some monotonous group
latent anvil
That's stupid
solemn rain
fuck
latent anvil
like, don't just give up on a problem when you try the obvious things and they don't work
Did you try any infinite groups other than Z?
latent anvil
@ripe crest keep in mind that vector spaces are classified by their dimension
V iso W iff dim V = dim W
Actually I think R and R × R are iso
But you'd need to prove it
latent anvil
I don't think you'll be able to
latent anvil
The standard proof uses linear algebra
solemn rain
that f was obv trash
latent anvil
but yeah, just finding one map which isn't an iso isn't good enough
solemn rain
okay can you check my solutions on a problem
then iw il lgt back
u motivated me ig
Let G be a group with at least two elements which has no subgroups other
than itself and the trivial subgroup. Show that G is a cyclic group whose
order is a prime number ?
proof: suppose for the sake of contradiction G is not a cyclic group whose order is prime
ripe crest
"Actually I think R and R × R are iso" - not as vector spaces right, because dim(R) = 1, dim(R x R) = 2
solemn rain
take any non identity element x
latent anvil
@ripe crest not as R-vector spaces
solemn rain
<x> is a subgroup of order |x| which is a contradiction since all subgroups should be trivial
hence G is a cyclic group whose order is prime?
latent anvil
You don't get a contradiction
latent anvil
You get that either <x> is trivial or G = <x>
solemn rain
<x> cant be trivial x is a non identity
solemn rain
i asssumed G isnt cyclic so |x| cant be |G|
solemn rain
okay wait
okay i do this 2 parts
i htin ki can do the later if i have cyclic
okay forget about the prime
G isnt cyclic yes?
by this
G must be cyclic now
latent anvil
What?
solemn rain
i will do this on two parrs
latent anvil
You just showed G is cyclic if it has only two subgroups
solemn rain
show that G is cyclic
latent anvil
Ah yeah okay
solemn rain
and then show G must be of prime order
latent anvil
Agreed
latent anvil
Yup
solemn rain
cool
now suppose G is cyclic
G = <y> for some y in G
for every divisor of n there exists a unique subgroup of that order (fundamental theorem) where n is the order of y
upper pivot
dont forget infinite order
solemn rain
since there exists only the trivial subgroups the divisors are only 1 and itself hence pirme
FUCK
how infinite order?
latent anvil
what?
upper pivot
its just one case lol, Z
latent anvil
Yeah, if you also note that Z has more than two subgroups
latent anvil
Yes
solemn rain
cool
latent anvil
@ripe crest what's the dimensions of R and R^2 over Q?
solemn rain
okay so back to our cool probl;em
@latent anvil wanna teach me some quick linear algebra in context of algebra
to prove R^2 is iso to R
ripe crest
Both should be infinite
latent anvil
Well they are infinite but they have the same dimension
ripe crest
But not sure exactly what the cardinality is
latent anvil
Not all infinite dimensional spaces are iso
ripe crest
yeah
latent anvil
Yeah so the lemma we need here is the following
solemn rain
u talkuing to me?
latent anvil
Let k be a field and V an infinite dimensional k-vector space
Then V ≈ V (+) V
no
I will in a sec
solemn rain
cool tyt
latent anvil
Does that make sense flower?
actually wait aren't you doing module stuff? If so, prove this for V a free module over a ring R
latent anvil
Oh then nvm
that's weird, projective is sort of a generalization of free
P is projective iff P is a direct summand of a free module
latent anvil
free means you have a basis
ripe crest
ah
latent anvil
Or is iso to a direct sum of copies of R
Possibly infinitely many
@solemn rain so do you know what a field and a vector space are?
solemn rain
no
i think a field
is a set F equiped with 2 operations addition and multiplication such that F is an abelian group with addition and a group with multiplication
not sure
latent anvil
if you remove 0 it's a group under multiplication
solemn rain
yea without 0 sorry
latent anvil
And you also need a(b+c) = ab + ac
solemn rain
a vector pace is just something closed under vector addition
latent anvil
But yeah that's it
latent anvil
Yes
So it's an abelian group V under addition
The vector addition is this group operation
solemn rain
okay
latent anvil
Along with a scalar multiplication F×V -> V
So it doesn't make sense to talk about a vector space on its own
You talk about vector spaces over a field F
latent anvil
Yeah exactly
solemn rain
so the scalar multiplication (.) FxV ---> V
latent anvil
So what I defined is an F-vector space