Jeez, and here I was trying to actually work with homogeneous coordinates and shit
#groups-rings-fields
slate forum
sturdy marsh
that might also work
you can move the lines to something like x0 = x1 = 0 and x2 = x3 = 0
using some matrix in PGln
slate forum
Yeah... that's what @chilly ocean was suggesting
I don't actually know how to do a change of coordinates to do that though
sturdy marsh
another way might be to use the fact that L1 and L2 cant be coplanar
slate forum
Yeah, I was wondering about visualizing this in three d space with a bubble around it, that's the "plane" at infinity or whatever
the lines would have to be skew
sturdy marsh
yeah and there's a unique plane containing L1 and P
same for L2 and P
and their intersection would be 1d
you can probably show that using Krull PID thm
the planes cant be the same plane as L1 and L2 would intersect
which can be shown either using Bezout's theorem or directly
slate forum
Don't know Bezout's theorem yet
sturdy marsh
you can do it directly
P2 - L1 is affine
so L2 is affine and projective
which means it is spec k
so just a point
contradiction
the same argument shows that any two hypersurfaces in projective space must intersect
slate forum
uhhh
sturdy marsh
first you can prove that the complement of a hypersurface is affine (reduce to the hyperplane case by taking a veronese embedding), and then use the same projective + affine argument
sturdy marsh
hmm ok, I don't know of another way then lol
slate forum
Your linear algebra way works
sturdy marsh
there's a pretty nice way to attack these sort of incidence problems using grassmanians
slate forum
No idea what those are, like either of them
sturdy marsh
projective space is lines in some affine space
grassmanians are varieties whose points corrrespond to higher dim linear subspaces of affine space
slate forum
oh, so like planes?
sturdy marsh
ye
sturdy marsh
k dim subspaces of F^n
slate forum
Is there some sort of duality when the dimension and codimension match?
Like lines in C3 and planes in C3?
sturdy marsh
sure, the dimension of the grassmanian of k-planes in n space is k(n-k)
i think
so the dimension of lines in C3 and planes in C3 would be the same
slate forum
I meant it kind of like how 1 forms and 2 forms can both be thought of as the same kind of in R3
sturdy marsh
oh, i do not know
sturdy marsh
I think there's a typo after the second equality
but the idea is correct
it's pretty cool that you're TeXing up solutions lol
I've tried multiple times
and I give up after 2 problems lol
leaden finch
anyone know subrings?
vestal snow
Let K be a field. Given an algebraic extension F of K(x), show that K[x_1,...,x_n] is not a field for all x_i in F\K
The proof says that there is a place $P$ such that $v_P(x_i) \geq 0$
cloud walrusBOT
The formerly edible banana:
vestal snow
I don't see why that must be true though
Is it true that every proper subring is contained in some valuation ring?
vestal snow
Figured this one out
There are infinite places and v_P(x_i) is non-zero at only finitely many places
gleaming belfry
Is this the channel in which I'd ask about affine geometry?
It is not true that two affine subspaces that are not parallel must intersect. Does it become true if we consider affine subspaces of the same dimension? Can we say anything about the dimension of the intersection?
Intuitively I'd say that there must be a non-empty intersection and that the dimension will be smaller by 1, but I have no idea how to justify this
next obsidian
this is definitely not the channel for affine geometry
civic linden
The Lie group of rotations in 3-dimensional space R3
is represented by 3 × 3 matrices
R obeying the relations R^TR = RR^T = 1(identity of 3x3 matrix), where T denotes transpose.
The orthogonal
group in n-dimensions is denoted O(n). Determine what type of matrices are the generators
of the corresponding Lie algebra, denoted o(n).
can someone give me a hint on how to answer this pls?
sharp sonnet
well, you have to determine the matrices A, such that exp(A) is orthogonal
i.e. such that exp(tA)exp(tA)^T = exp(tA)^Texp(tA) = 1 for all t
hint: ||compute the derivative with respect to t||
civic linden
mhm okay. I forgot, but why do we use exp in this case again?
sharp sonnet
i assumed this is your definition of Lie algebra of a (matrix) Lie group
civic linden
Yes it is
sharp sonnet
i.e the Lie algebra of a Lie group G is the set of all matrices, such that exp(tA) is in G for all t
sharp sonnet
there are different (more general) definitions, but i assumed you use this
civic linden
yah we do, i just want some intuition behind it :x
sharp sonnet
well, it just turns out that the exponential function is the 'correct' way to translate between a lie algebra and its lie group
my (bad) intuition is that
a lie algebra is a vectors space, so it is 'linear'
a lie group is a smooth manifold, it is 'curved'
and the exponential function turns linear space into curved space
civic linden
ah ok
civic linden
like exp(R) = S^1
@sharp sonnet O
ok ok that makes ense
sense
exp(tA)exp(tA)^T = exp(tA)^Texp(tA) = 1 for all t. for this here, computing the derivative we end up using some product rule right?
sharp sonnet
yeah, you need the product rule
hint 2: ||the lie algebra will be the set of all skew-symmetric matrices||
||checking that all skew-symmetric matrices are in o(n) is easy, but the other implication requires the derivative||
civic linden
ty sir, ok ill be back if im stuck 
Ok so i got $A^T exp{(tA)^T (1+(tA)^T)} + A exp{(tA)^T (1+(tA)^T)}$
cloud walrusBOT
!R. ¬:
civic linden
which shows i guess its a skew symmetric matrix A^T=-A?
sharp sonnet
those are the skew symmetric matrices
civic linden
right, so how does one find the generators of so(3)?
sharp sonnet
it's the set of all skew symmetric matrices ?
i mean so(3) = o(3), because SO(3) is the identity component of O(3)
maybe there is a smaller set of generators but i dunno
civic linden
ive read in a textbook
R = 1_3×3 + iθ^iJ_i + O(θ^2)
for the rotations
no clue whats going on here
some infinitesimal transformations near the identity which is in the neighbourhood of 1_3x3?
and we are just looking at the metric so who cares about O(theta^2)?
civic linden
figured it out! ty
tacit pebble
what is |R?
rugged lance
the real set
tacit pebble
ohhh hahah
smoky cypress
Anything can be a vector, since vector is just an element of a vector space and you can declare anything to be in your vector space
rugged lance
yes but am I allowed to build vector space with sets
tacit pebble
sure why not
smoky cypress
Of course
tacit pebble
as long as it satisfies the 10 VS axioms or whatever
rugged lance
Thanks guys I'll sleep happier
quaint ivy
for instance, any Boolean algebra is also a F_2-vector space: https://math.stackexchange.com/a/1884825/596135
bleak abyss
Yeah, you've got an invariant subspace of a larger space
bleak abyss
Once you are fully explicit about what that means, yes
So W is a subrep of V
Where the map is phi:G->GL(V)
Define phi':G->GL(W)
Which is just phi'(g) = phi(g)\mid_W
Are you thinking of it as a real or complex rep?
As a complex rep 1d is just kinda meh
Some groups are manifolds, not all
So you can think of representations of the circle
bleak abyss
And both are subreps but that's it
The problem is "A subspace of C"
It's just 0 and everything
So there's really nothing going on here
Okay let's consider the dihedral group then
No it's not a subset of the matrices, every single matrix restricted to a subspace
Or better yet I'll give the action of the symmetric group
So $S_n$ acts on $\mathbb{C}^n$ by permuting the coordinates
Yes exactly
cloud walrusBOT
Daminark:
obsidian path
I can't get this to work using induction, are there any combinatorics involved?
next obsidian
This isn't the right channel for this
fossil marsh
hi, i was working on this problem and tried using $\mathbb{F}_3[x]/(x^2+1)$ as an example, but the monics in the field are $x,x+1,x+2$ whose product is $x$ instead of $\pm 1$, so i am wondering if this is an error of some sort
cloud walrusBOT
Rixia Mao:
fossil marsh
oh, A in the problem is a polynomial ring over a finite field
if anyone can help that would be much appreciated
stoic coral
quick question, if an element of a cyclic group of size n has an order of n, does it imply that the element is a generator for that cyclic group?
sturdy marsh
yup
stoic coral
hmmm, i got a question in the textbook that asks for the order of 5 in Z_12, and the order of 5 is 12, but 5 isnt a generator for this group
this is it in particular
how tho, 5^i for any i >=2 is 1mod12
oh lmaooo
ok that clears alot of things up, thanks!
sturdy marsh
What if we modify the problem a bit, and take the product over all nonzero elts of A/P
A/P is a field
so the multiplicative group is everything except 0
and it is cyclic
so pick a primitive, and we are computing g^(1 + 2 + ... + n)
where n+1 is the order of A/P
so they did not demand f to be monic?
yeah it probably doesnt work with monic condition
fossil marsh
yeah it was from rozen's number theory in function fields
that's kind of weird
well, anyways, thank you for confirming
i was going down a different route
trying to partition the ring with the leading coefficient
sturdy marsh
is there a char neq 2 assumption?
fossil marsh
dont think so
sturdy marsh
oh I was trying to prove the corollary
fossil marsh
yep
sturdy marsh
in general if you have a finite abelian group then the product of all elts. is the product of all elts of order 2
right?
fossil marsh
oh hey that's even the same counterexample
bleak abyss
Number theory in function fields interesting
outer estuary
Anyone have any ways to show $(xy-zw)$ is prime in $k[x,y,w,z]$
cloud walrusBOT
freeridebiker:
outer estuary
there is a handy CA theorem I suppose for rings of multiple variables over fields
i guess cause it's a UFD, and (xy-zw) is irreducible that would work right
pallid lagoon
Anybody here interested in a bit of combinatorial group theory?? This problem of embedding a free group of rank 3 in a free group of rank 2 was given by my instructor in class (as a food for thought)...i wrote up a proof based on the fact that reduced words of even length in $F_2$ is isomorphic to $F_3$... now i am curious about whether its possible to use the elementary version of ping pong lemma to cook up three disjoint sets in $F_2$...
cloud walrusBOT
Arpan:
pallid lagoon
For reference ...If $X$ is a nonempty set and $X_1,X_2,X_3$ are three disjoint subsets of $X$ and $a_1,a_2,a_3$ are bijections from $X\to X$ with $a_j^k(X_i)\subset X_j$ for all $i\neq j$ and $\forall k\in \mathbb{Z}\setminus {0}$, then $a_1,a_2,a_3$ generate a free group of rank 3 inside the group $Bij(X,X)$...this is the elementary version of ping pong lemma i am hoping to utilise...
cloud walrusBOT
Arpan:
golden pasture
@outer estuary yup cuz xy-zw is irreducible
alternatively consider k[x,zw/x,z,w]
wind steeple
Is it true for $M$ an $A$-modules that $S^2M \simeq \langle{a\otimes b + b \otimes a|(a,b)\in M^2}\rangle$ ?
cloud walrusBOT
Zak:
wind steeple
It is true for linear spaces bc I can get a basis but I wonder if it is generally true
slate forum
@sturdy marsh Can you help me again? I'm super bad at this projective geometry stuff
stark sigil
@wind steeple If $A$ has characteristic 2, then you will miss all elements of the form $a\otimes a$
cloud walrusBOT
Icy001:
wind steeple
oh yes thx
stark sigil
One should note that $S^2M$ is the quotient of $M\otimes M$ by the relation $x\otimes y=y\otimes x$, and the symmetric tensors (elements of $M\otimes M$ such that the involution swapping the two factors fixes them) map to $S^2M$ by projection. In this case, even with $M=R=\bZ$ we have $1\cdot 3\in S^2M$ but is not the image of any $x\otimes y+y\otimes x$, since the image of any $x\otimes y+y\otimes x$ is $2(x\cdot y)$ and $1\cdot 3$ is odd.
cloud walrusBOT
Icy001:
stark sigil
(nor is it an integer linear combination of such images)
wind steeple
Oh yes I forgot that A is not necessarly a field lol
toxic zephyr
What's the best notation to state that a defined matrix has entries in the prime field F_p?
$(A)_{ij} \in \mathbb{F}_p$?
cloud walrusBOT
nix:
chilly ocean
or why not something like $A \in M_{n,m}(\mathbb{F}_p)$?
cloud walrusBOT
88ddda:
chilly ocean
or whatever the notation is for that (n,m, or nxm?)
toxic zephyr
oh thats sexy
quaint ivy
yeah was going to type something along those lines lol
toxic zephyr
$A\in\mathbb{M}_{m\times n}(\mathbb{F}_p)$
$A\in\bR^{m\times n}(\mathbb{F}_p)$
cloud walrusBOT
nix:
quaint ivy
$\bR^{m\times n}$ is often used for $m\times n$ real matrices. Here you'd have to type $(\mathbb F_p)^{m\times n}$
cloud walrusBOT
derivada.schwarziana:
wind steeple
$A \in \mathcal{L}(\mathbb{F}_p^n)$
cloud walrusBOT
Zak:
wind steeple
big brain time
toxic zephyr
ah right thank you
tidal elbow
I need to proof: "Is the kernel of a lineair variety a lineair variety?" How would I approach this problem?
It is one of the questions of previous exams and it isn't proven in my textbook.
tidal elbow
If $V$ is a vectorspace. A linear variety $L$ is a subset of $V$ in the shape of $$L=\vec{a}+W$$ with $\vec{a} \in V$ and $W \leq V$
cloud walrusBOT
OsamaBinNaughty:
solemn rain
what does the kernel of a subspace mean
tidal elbow
It is a subset, I made a translating mistake.
tidal elbow
Indeed. It doesnt make sense
stoic coral
relating to this question, is it the case that if every element of a cyclic group is a generator for the group, then no other subgroups for the group can exist outside the non-trivial subgroups?
also is there any significant difference between a cyclic group and a regular group? as far as I've seen the only 'difference' is that cyclic groups will often have a generator element that is given
chilly ocean
https://en.wikipedia.org/wiki/Subgroups_of_cyclic_groups might be worth checking out
Moreover, for a finite cyclic group of order n, every subgroup's order is a divisor of n, and there is exactly one subgroup for each divisor.
is already a pretty strong condition
compare with something like S_3 (definitely not cyclic) that has more than one subgroup of order 2
i'm not sure that answers your question about cyclic groups vs other groups (and tbh i'm not sure if there's a right way to answer it)
stoic coral
oh that makes sense
chilly ocean
relating to this question, is it the case that if every element of a cyclic group is a generator for the group, then no other subgroups for the group can exist outside the non-trivial subgroups?
you mean outside the trivial subgroup? (and the whole group ofc)
just wanna make sure i understand your terminology
because that sounds right
(you can also one-line this problem with lagrange, although if you're thinking of what i'm thinking then this might be reproving lagrange in a special case lol)
stoic coral
I just one lined it with Lagrange lol
fossil marsh
are vieta's formulas still valid in finite fields?
for context im working on this problem
and the elements of $\mathbb{F}^*$ are exactly the solutions to $x^{p^k}-1=0$
if vieta's formulas work, i can just claim that the product of all the roots is exactly -1
cloud walrusBOT
Rixia Mao:
quaint ivy
then you're probably being asked to prove that specific case of the formula
which is true in finite fields btw https://en.wikipedia.org/wiki/Vieta's_formulas
paper flint
Is this test simply not equivalent to the definition of a group?(Besides the fact that we're checking for a subset of a given group?) What would constitute a reasonable proof for this theorem?
carmine fossil
That is the definition of subgroup?
woven obsidian
You need to show associativity, existence of inverses, existence of identity and closure
paper flint
Yes, that's what I think, I don't understand why Gallian formulates this as a theorem.
Well associativity follows from the group itself.
woven obsidian
yeah
paper flint
Inverses and closure are covered by the statement of this theorem, I guess it only remains to be shown that the identity exists?
carmine fossil
a (a^-1) is in H
carmine fossil
Yes
woven obsidian
You also need uniqueness of inverse if you want to prove the converse statement
And without proving the converse the statement isn't really useful, because then if the criterion fails it could still be a subgroup
carmine fossil
(I mean The definition of group doesn't involve uniqueness of inverse)
woven obsidian
Yeah uniqueness is a theorem
carmine fossil
Or of identity
paper flint
Yeah, that follows as a theorem. I have proven it previously.
woven obsidian
But if you want to prove the converse, then you need to show that a subgroup has the same inverses as the big group
Which is pretty easy, but has to be done
paper flint
Since H satisfies group axioms and H is a subset of G, I suppose I can say it is a subgroup of G?
Oh, I see.
woven obsidian
Because the a^-1 mentioned in the theorem is the inverse in G
One might think a subgroup could have a different inverse
But yeah, sufficiency is quite straightforward as you said
paper flint
Makes sense, I will do that.
woven obsidian
Just have to show identity is there really
woven obsidian
No problems
This was stated in the answer to one of my problems, I don't completely believe the second to last equality
(I is an ideal of A of course)
Obviously the inclusion < holds, but I'm not so sure about the other direction
If I have some element in the intersection, then I can write it as a sum of element in I multiplied by product of n-1 elements in I for all n geq 2. But how do I know that each product of n-1 elements is also in M?
If it helps, I know that A is Noetherian.
next obsidian
If it helps, I know that A is Noetherian.
@woven obsidian Do you know the ring is local?
If so I think you apply Artin-Rees lemma
woven obsidian
Nope, that's the follow up question, which says that if it is local then the whole thing is 0
next obsidian
Oh actually
You only need Noetherian for Artin-Rees
And then for Krull’s intersection theorem which says it’s 0 as a corollary you need local
Look up the Artin-Rees lemma
woven obsidian
Hmm I'll check it out, we haven't covered it
next obsidian
If you have Lang it’s on page 429
woven obsidian
it seems that it should be a triviality that IM=M
next obsidian
I think it’s in Atiyah-MacDonald
woven obsidian
Once I know that you can just use Nakayama
next obsidian
But I think this is realy important, I think you legitimately need to know this theorem in order to show it
I don’t know an example it fails off the top of my head without Noetherian
But I think it does
woven obsidian
I'll check it out, thanks for the help.
next obsidian
For reference I was looking at the same thing once and remembered I thought you need Artin-Rees
Aka I think your proof with Naka is how you do Krull’s intersection theorem
Which does need the Noethe hypothesis and uses Artin-Rees
woven obsidian
Well this was from an old exam, it would be very uncharacteristic for this lecturer to give a problem requiring a theorem we haven't covered
It doesn't seem like the kind of lemma you come up with during 30 minutes
next obsidian
Yeah
You could try looking at generators of each ideal
There’s only finitely many
woven obsidian
Yeah I thought about that, but I couldn't see how it helped
woven obsidian
Yeah
next obsidian
Hmmm
It might be possible they covered the lemma in a previous time
When it was a bit faster
woven obsidian
I might send him a mail, he has done mistakes in his exams in the past
No they've used the same book always, Reid's book
next obsidian
Is the lemma not in Reid’s book?
I’ve only looked at the beginning parts of Reid’s book
woven obsidian
No, he doesn't say much about Noetherian modules
next obsidian
Huh
woven obsidian
Most of the things he says are corollaries of that if 0->L->M->N->0 is a s.e.s, then M is Noetherian iff L,N is
sturdy marsh
@sturdy marsh Can you help me again? I'm super bad at this projective geometry stuff
@slate forum I don't know much either but I don't mind trying
chrome hinge
Hi there, im struggling a bit with the symmetry group of the tetrahedron
At the moment just want to know the elements in that group to get the cardinality
I know its 24 but im just getting 18... I dont know why
chilly ocean
well, it has to be a subset subgroup of S_4, so it divides 24
chrome hinge
Indeed. But im just at the beggining of the study of this group. Im not supposed to know its isomorphic to S4
Just want to know the elements of the group 😂
Im counting eight vertex-face rotations (four vertex, order three each one), six plane symmetries and three edge-edge rotations
That makes a total of 8 + 6 + 3 = 17 plus the identity i get just 18
chilly ocean
what are plane symmetrices and edge edge rotations?
chrome hinge
well, it has to be a
subsetsubgroup of S_4, so it divides 24
@chilly ocean This work nice to get the order actually, thanks
But not quite to know each elements
Hmm
You know the tetrahedron has pane symmetries?
chilly ocean
i'm not famliar with the terminology, but if you explain what a pane symmetry is i'd probably see it
(actually, i'm not even sure which transformations of the tetrahedron you are even allowed to do in this problem you are talking about. just rigid movements?)
chrome hinge
choose an edge, cut the tetrahedron in two pieces going from that edge all through to the middle of the opposite face
Only symmetries allowed
chilly ocean
then i would think the group is order 12, not 24
chrome hinge
I got an order three rotation for each vertex right? That makes 8 elements
plus the identity 9 actually
And then there are six ways to cut the tetrahedron into two equal pieces. Those "Cutting planes" are reflection symmetric planes
So the order is at least 15 right?
chilly ocean
yeah i think i agree with 8 elements from that, +1 for identity
i guess i don't really follow this stuff about cutting the tetrahedon
ah, so you are allowed reflections?
i didn't know this
(rigid movements would imply 12 i think)
chrome hinge
chilly ocean
yeah i see it now
chrome hinge
The right side
chrome hinge
Of course, reflections are symmetries too
chilly ocean
i guess so, i will admit that i'm not super familiar with the technical terminology
chrome hinge
Lmao its the first time hanging out with the tetrahedron too
Like, im trying to figure out its symmetry group
chilly ocean
ok, so you know that it is order 24, you can probably find some elements that its' generated by, then show that they satisfy some multiplication that is satisfied in S4, then argue that it must be S4 (you're not allowed to use S4? just define it!)
chrome hinge
I'll try to make the analogy with S4 thanks :)
maiden ocean
what exactly is the power rule for D(f^n) supposed to be here lol
D(f^n) = nf^{n-1} D(f)?
yea it was that pretty sure
nvm
smoky cypress
Hmmm does the chain rule hold?
maiden ocean
yea i got it whoever it wasnt hard lol
its the chain rule p much
anyway am i just dumb or is theta^2 + theta like
not a root
of x^3 + x + 1
i get theta^6 + theta^5 + theta^4 + theta^3 + theta^2 + theta + 1 and when i reduce that gets me theta which is definitely not zero in F8
chilly ocean
there should be no +1 right?
maiden ocean
wdym
chilly ocean
er, nvm, im a dumbass
chilly ocean
ok here, is a try, $\theta^3=\theta+1$, so $x^3=(\theta^2+\theta)^3=\theta^3 (\theta+1)^3 =\theta^3 (\theta^3+\theta^2+\theta+1) = \theta^5 = \theta^2(\theta+1) = \theta^3+\theta^2 = \theta^2+\theta+1$, which is the same as $x+1$
cloud walrusBOT
88ddda:
maiden ocean
ty think i got it
thorn delta
Let $G$ be a finite $p$-group and let $H \neq {e}$ be a normal subgroup of $G$. I want to show that $H \cap Z(G) \neq {e}$.
cloud walrusBOT
kxrider:
thorn delta
This is what I've done so far:
scarlet estuary
Z(G) being the centre of G?
thorn delta
yes
scarlet estuary
are you familiar with the class equation? (it's not actually necessary here but it can be used)
thorn delta
yes
scarlet estuary
alright, what have you tried?
thorn delta
If $G = H$, then the conclusion is clear. Otherwise we have $|G| = p^n$ for some $n > 0$ and $|H| = p^k$ for some $0 < k \leq n$. It follows that $G/H$ is a $p$-group, so it has nontrivial center. Let $g \in Z(G)$ s.t. $g \neq e$. Then $aga^{-1}H = gH$. We have $|gH| = p^i$ for some i \leq n-k$ so $g^{p^i} \in H$ and therefore $ag^{p^i}a^{-1} = g^{p^i}$ implies $g^{p^i} \in H \cap Z(G)$. But this doesn't work because $g^{p^i}$ could be $e$, and i haven't found a way to work around that.
cloud walrusBOT
kxrider:
Compile Error! Click the 
reaction for details. (You may edit your message)
scarlet estuary
theres a really slick approach you can use here
hint: how many classes with one element does H have?
thorn delta
conjugacy classes? |Z(H)| many
scarlet estuary
yes
but in particular, H - being normal - is just a union of conjugacy classes
i claim that H contains at least two conjugacy classes with one element
(one of which is the identity)
and furthermore that these must be classes of central elements of G
can you see why those two claims are true? if so, that's the proof
(hint: what's the index of Z(G)?)
an alternate, less slick, proof is based off the class equation
its a similar idea
but a bit more computational
(consider a group action of H on G by conjugation...)
thorn delta
hint: what's the index of Z(G)?
probably not on the right track at all, but [G : Z(G)] is the order of the inner automorphism group of G lol
oh wait, i think im understanding it though. If H only contained a single conjugacy class with one element, then p would not divide the order of H.
steep hull
@scarlet estuary Nice, though it’s essentially the same as the class equation approach.
scarlet estuary
If H only contained a single conjugacy class with one element, then p would not divide the order of H.
yep
this is the core idea
and regardless of whether you use the class equation method or my "slicker" one
this is the key fact of the proof
@steep hull yeah you basically just skip the computations
which is nice since writing out class equations is tedious
(though probably good practice)
steep hull
Yeah. Anyway, the H=G case is good motivation for trying to use the class equation somehow.
That’s what gave it away for me
thorn delta
an alternate, less slick, proof is based off the class equation
its a similar idea
but a bit more computational
(consider a group action of H on G by conjugation...)
what's the idea here? When $H$ acts on $G$ by conjugation, you get a weird class equation. Something like $$|H| = |Z(H)| + \sum_i [H : C_H(x_i)] $$ with each $[H : C_H(x_i)] \geq p$.
cloud walrusBOT
kxrider:
scarlet estuary
oops sorry
i wrote that backwards
that should be G acting on H by conjugation
consider it, uh, extra practice writing the class equation
yeah
thorn delta
ah yea i see. Its reduces to basically the exact same proof that the center of a p group is nontrivial
glossy yoke
``look at the class eqn mod p.'' sounds like good advice when messing with p-groups.
scarlet estuary
yeah haha
hard to go wrong
anyway yeah, as mentioned its essentially the same proof
youre just a little more explicit about it
and get the opportunity to show off your sick sigma-drawing skills, an essential trait for any mathematician.
vestal snow
Let G be a finite abelian group and H a subgroup of G. Prove that G/H is isomorphic to some subgroup of G
I found this proof online
I have a few questions about this
Why does it suffice to establish the result when G is an abelian p-group?
Not all subgroups of A x B have to be of the form A' x B' where A' and B' are subgroups of A and B respectively
solemn rain
Why does it suffice to establish the result when G is an abelian p-group?
@vestal snow a finite abelian group is direct product of p groups
fundamental theorem of finite abelian groups
cyclic groups
vestal snow
That's true, but how does that imply the result?
Say we knew that the result holds for p groups
solemn rain
G = Z_p^a x Z_p^b ....
where a+b+... = p^x =|G|
iirc cuz im kidna rusty with fundamentla theorem
now show that if H is a subgroup then G/H is isomorrphic to some subgroup of G
given that this works for p groups
carmine fossil
Let G be a finite abelian group and H a subgroup of G. Prove that G/H is isomorphic to some subgroup of G
Do you know there is a bijection between subgroups of G containing H and subgroups of G/H?
vestal snow
Yes
carmine fossil
Show that bijection is a homomorphism
vestal snow
What?
carmine fossil
Ok,nvm
solemn rain
hey i think u already said it lmfao
Not all subgroups of A x B have to be of the form A' x B' where A' and B' are subgroups of A and B respectively
@vestal snow give me a counterexample
where everything is abliean th
( as you know AxB abelian <--> A and B abelian )
solemn rain
fuck
yea
it works the other way around tho
if A and B is a subgroup of X and Y then AxB is a subgroup of XxY
vestal snow
Yes
solemn rain
so suppose the proposition holds
for finite baliena p groups
then for each of these 'factors'
there exists a subgroup bla bla bla
therefore they exist for the direct product
aka the group
right?>
vestal snow
I don't think the logic here is correct
We need to show that given an arbitrary subgroup, we can...
But you are only considering the subgroups which can be written as products of subgroups of the factors of G
solemn rain
what we have is that if G is a finite abelian p group and H isa subgroup then there exits a subgroup of G such that G/H is isomoprhic to that subrou
right?
vestal snow
Yes
solemn rain
now G is isomorphic to direct product of cyclic p groups
any G now
any finite abelian G
vestal snow
Yes, say G = prod Z_p_i^k_i
solemn rain
okay now each factor has a subgroup H such that (factor)/H is isomorphic to some subgroup of (factor)
right?
vestal snow
Yes
solemn rain
what happens if you product all these subgroups that are isomorphic?
vestal snow
Wait hold on
solemn rain
is it isomoprphic to G ( the original group ) quotient the direct products of all H's
H's being the subgroups of the factors?
vestal snow
I disagree with your previous claim
solemn rain
which 1
vestal snow
Or rather I don't get what you're saying
vestal snow
yes
solemn rain
and assume the prop holds for anyy p abelian group
now by fundamental theorem G is isomorphic to a direct product
of p abelian groups
say 3
AxBxC
now the prop applies to those 3
as they are finite abelian p groups
apply the proposition --> find the subgroup H such that A/H is isomorphic to some subgroup of A
keep doing this to B and C
and direct product them all up
vestal snow
Wait
" find the subgroup H such that A/H is isomorphic to some subgroup of A"
What does this mean?
solemn rain
what does your theorem say
for a finite abelian p group there exists a subgroup H such that G/H is isomorphic to some subgroup of G
right?
vestal snow
No
solemn rain
the other way around
yea
mb
for any subgroup H there exists a subgroup of G ,A, such that G/H is iso to A
right/
Let G be a finite abelian group and H a subgroup of G. Prove that G/H is isomorphic to some subgroup of G
[12:17 PM]
should be right?
vestal snow
Yes that's the thing I want to prove
solemn rain
now you are having trouble seeing
that if this works for p abelian groups
they work for anay abelian groups
( finite )
right
?
so take any finite abelian group call it G
G is iso to direct product of p abelian groups
for any subgroup H of each of these abelian groups there exists a subgroup such that this mod that is isomorphic to the subgroup
now can you like
direct product them all up
should get you isomorphic to G mod any subgroup
right?
vestal snow
We don't know if H can be written as a direct product of subgroups of the factors of G
solemn rain
really?
vestal snow
Yeah
Z/nZ x Z/nZ
<(k,k)> where k ranges over all elemnts of Z/nZ
this is a counterexample
solemn rain
can you explain to me how
this is ac ounterexample
show me there exists A and B such that A is not a subgroup of X and B is not a subgroup of Y but AxB is a subgroup of XxY
vestal snow
Okay
Z/nZ x Z/nZ
The subgroup {(k,k)| k in Z/nZ} is a subgroup
But you cannot express it as product of subgroups
vital quail
@vestal snow suggestions for what?
next obsidian
Not all subgroups of A x B have to be of the form A' x B' where A' and B' are subgroups of A and B respectively
@vestal snow Pretty sure this is the case when the orders of A and B are coprime which is what's being done here
Take an element (a,b) in H which isn't equal to (e,e)
then |b| is coprime to |a| implying that you can find an n such that n is a multiple of |b| but is equal to 1 mod |a|, then (a,b)^n = (a,e)
and likewise you can find an m such that m is a multiple of |a| and equal to 1 mod |b|, so that (a,b)^m = (e,b)
so you actually get that H is equal to the direct sum of its A and B-parts
carmine fossil
Let's say A and B are groups with presentations for
A=<a,b (ab)^2=a^4=b^3=e>
B=<a,b (ab)^2=a^4=b^3=e,ab=ba>
Can I conclude B is isomorphic to a subgroup of A?
next obsidian
I wanna say no
I gotta think about it a bit more
In this specific case it looks like you're saying B would be the subgroup of A defined by like "the largest thing where everything commutes with everything"
and I feel like that doesn't work well
But I think this is a more general question about group presentations
carmine fossil
Well,What I meant to ask was given a presentation,if you add more relations,will you always get a subgroup
next obsidian
and like "if every relation in one is a relation in the other, is it a subgruop"
where you have to think more abstractly but aaaa
My free group intuition isn't very good
So like
Okay the ideal way to do this is in your specific case
define a map from the free group on 2 generators by just a maps to a, b maps to b
uhh
Say A = F/N
and B = G/M
F and G are just the free group on 2 generators
so you wanted to define this by defining G -> F by a maps to a and b maps to b
then you compose with F -> F/N
then you want everything in M to die when you go G -> F -> F/N
then by universal property stuff you get a map G/M -> F/N
and for this to be injective you need the kernel of G -> F -> F/N to be exactly M
but it isn't even clear that everything in M dies when you go G -> F -> F/N
like... in your specific example with the A,B you chose
the relation ab = ba really is the element aba^-1b^-1
but there's no reason that's 0 in F/N
at least that's what I think is going on, so it seems almost as if the maps wants to go the OTHER way
but then this is clearly wrong as A isn't abelian (probably) while B definitely is
So maybe I've screwed something up (in fact you don't make A by F/N you have to take the normal closure of N and shit ughh)
But this makes me want to say no
carmine fossil
A concrete example would be nice
carmine fossil
Ok
next obsidian
I have an idea in my brain but this is the best way I can communicate it
also
concrete example
and free groups don't really go that well together I think 😅
You could try like
looking on group props
and looking at presentations of specific groups
and see if you can find a counterexample
carmine fossil
but then this is clearly wrong as A isn't abelian (probably) while B definitely is
For B,The generators could both end up being identity because of some internal cancellations
next obsidian
What I'm saying is
Given this setup I suspect if you could show something is a subgroup, it would actually be A which is a subgroup of B
which is not good since B is abelian and A probably isn't
glossy yoke
adding relations doesn't make a subgroup in general. it makes a quotient group. consider, say, <a> and <a | a^2>
snow cliff
So I've read the definition of a quotient group and I don't think I can quite make sense of it. From my understanding, it is the group whose elements are from the fibers of the homorphism. So the elements of a domain in a mapping that map to the same element in the codomain? This feels uncomfortable as a concept.
Does anyone have a better way of explaining it? Or perhaps an example that makes it easier to understand?
chilly ocean
i feel like "set of cosets with a group operation when the subgroup is normal" is pretty clear, but from your wording i feel like you might be thinking of something else (fibers of which homomorphism?)
snow cliff
that feels better
chilly ocean
You can also just think of it as the original group, but eg a quotient G/N, you can just let everything in N be the identity in computations eg g1 * n * g2 just equals g1*g2
snow cliff
@chilly ocean the way my book explains it, it seems to be more about homomorphisms
chilly ocean
what book? im curious
chilly ocean
snow cliff
so that definition was a bit confusing to me
snow cliff
top of 76
chilly ocean
ty
vestal snow
@next obsidian Thanks for the help!
chilly ocean
im in an algebra class that's supposed to follow this book and i didn't see that definition in class, so it'll be good for me to read i guess
vestal snow
One more question: why does Q need to have less than k summands?
I don't get why having less than k generators necessarily implies Q can be written as a product of at most k cyclic groups
chilly ocean
@snow cliff in that definition you can think of the quotient group as being the set of all cosets of the kernel K (for instance, the kernel is the identity, it's the fiber over e_H)
can't go too much into detail rn since im on my phone
any normal subgroup of G the kernel of some group homomorphism defined on G, so this is the same as what i described
(i think, haven't had time to mull on the details yet)
the fiber of h corresponds to (is) the coset gK, where phi(g) = h, to be explicit
maybe that will clear the definition up a little
@vestal snow sorry if i buried your question, feel free to repost
vestal snow
It's all good. I'm sure someone will see it eventually
next obsidian
I don't get why having less than k generators necessarily implies Q can be written as a product of at most k cyclic groups
@vestal snow you need the “unit vector” for each summan
Basically like
Ugh how can I explain this easier lmfao
vestal snow
I think I get what you're saying
next obsidian
It’s not quite that easy but
vestal snow
But I think that the proof uses some stuff from modules?
next obsidian
Yeah lmfao
That’s what I was gonna say
But I think you can translate it to group theory
vestal snow
And we haven't done them yet
next obsidian
I think here’s what you want
vestal snow
Oh really
next obsidian
So like umm
I think if you have k - generators for the product
Blech
This is really gross
My brain is just yelling
Not enough
And telling me that’s rigorous lmfao
vestal snow
Yeah this problem is kinda annoying imo
next obsidian
Okay so observation 1
Requires the specific structure here
I think
Wait no
Still not true
Yeah I’m not gonna lie
The only thing I can think of is module theoretic bullshit type stuff where since Z is a PID it’s almost like a vector space
I guess you need to look really hard at alike
(0,...,1,...,0)
Since that actually make sense since we re dealing with cyclic groups
And figure that if you can get k of those you need k-generators
oblique river
I think you can just say like
snow cliff
@chilly ocean ty I think I need to mull it over some more, may need to clarify later tho
oblique river
there's a surjective map Z^k --> Q, look at the kernel which is a subgroup of Z^k so must be Z^m for some m
(in fact m = k because Q is finite but I don't think we'll need to use that fact explicitly)
the inclusion Z^m --> Z^k is given by a k x m matrix
put that matrix in smith normal form
vestal snow
I don't think we can do that
oblique river
why not?
oblique river
where have I said the word module :P
oblique river
...oh
but wait
weren't you like asking class field theory questions or something at some point
am I just confusing you with someone else?
vestal snow
Unfortunately no
oblique river
seems odd to me
also like, what math grad students don't know group theory? I thought that was like, standard required undergrad math course
chilly ocean
@snow cliff yeah feel free to post it here, if i can take a look at it i will (it's good review for me)
oblique river
sorry I'm just really confused about this whole endeavor haha
vestal snow
It is, but he wants us to use group theory only
or rather, the tools developed in class only
So here's something I tried
oblique river
so... no linear algebra...
vestal snow
Let G be a finite abelian p group and let H be another group. Let G = prod Z_p^k_i and H = Z_p^m_i
There is a lemma that G contains a subgroup isomorphic to H iff k_i>=m_i for all i
k_i, m_i are ordered in decreasing order
So say H was a subgroup of G
them m_i =< k_i
So G/H = prod Z_p^(k_i - m_i)
Since k_i - m_i =< k_i, G/H is isomorphic to some subgroup of G
However, the problem with this is that G iso to K and H iso to L does not imply G/K iso K/H
So we cannot write G/H as prod Z_p^(k_i - m_i)
Maybe someone can fix this?
toxic zephyr
quick question:
when dealing with elements of a prime field, my text doesnt seem too bothered by writing numbers that are less than zero or greater than p-1. As long as it's clear the number we're dealing with is in the prime field F_p, is it understood that when I write a number 'a' outside of that range that it is exactly equivalent to a mod p?
For example, is it bad/improper to write 70 in F_13 or state that over that field 70^-1=8?
golden pasture
it's usually understood you're working in mod 13 so it's ok if somehow you need it
toxic zephyr
Yeah I'm dealing with a matrix equation over that field so numbers outside that range come up frequently
golden pasture
ic
toxic zephyr
Would it still be fine if I said 70^-1=-5?
toxic zephyr
Haha alright thank you
@vestal snow's question doesn't appear to have been answered btw if anyone can help them
#groups-rings-fields message
golden pasture
@vestal snow cant you just quotient away the generators for the other Z/pZ s?
kinda easier to think of it as modules ngl
sturdy marsh
So G/H = prod Z_p^(k_i - m_i)
@vestal snow I'm a bit confused, why is this true?
vestal snow
It isn't
The proof would have worked if it were
So i was hoping someone might somehow fix this
sturdy marsh
What are you trying to prove?
vestal snow
that G/H is isomorphic to a subgroup of G
vestal snow
G is abelian
sturdy marsh
finite?
vestal snow
yup
sturdy marsh
If you're allowed to use the lemma that you mentioned you're pretty much done I think
Assuming you've reduced to the p-group case
Let G/H be a quotient
then G/H is a p-group (not necessarily of the form that you've written above)
So it is of the form $G/H = \prod^{m}_{i = 1} Z/ p^{k_i}Z$.
Assume $G = \prod_{j = 1}^{l} Z/p^{r_j}Z$
sturdy marsh
ffs i cant latex today
okay
assume all in descending order
if you can argue that k_1 > r_1, k_2 > r_2, ..., k_m > r_m then you're done by the lemma
all of those should be > =
alrighty, so we must have r_1 < = k_1 as the order of every element in the quotient divides the order of the element of highest order in G
so the first summand of G has subgroup iso to first summand of G/H
hmm, im trying to see if there's any way to induct from this point
sturdy marsh
oh wait hold on there's no need to induct
assume towards a contradiction that r_2 > k_2
oh wait I got the notation the wrong way around, we need to argue k_1 < r_1, k_2 < r_2, ..., k_m < r_m
all < =
note m < = L is obvious
argument for k_1 < r_1 same as before (with inequalities flipped)
assume k_2 > r_2
then r_2G is cyclic, but r_2G/H is not (as the second summand has elt of greater order)
which means we have a surjection from cyclic to noncyclic
which is a contradiction
that should do it
sturdy marsh
image of generators of G generate G/H
vestal snow
Yeah but we can't use things from module theory
vestal snow
Why does image of generators of G generate G/H imply that the number of factors of G/H is less than or equal to G?
sturdy marsh
if a group has n factors, it cannot be generated by fewer than n generators
i think
p-groups btw
vestal snow
Yeah but the proof of that uses that these are Z-modules and Z is a PID
sturdy marsh
gimme a sec ill try to come up with an argument that doesnt use that
vestal snow
I don't think you really can without implicitly developing the theory for modules over a PID
sturdy marsh
there should be a way w/o modules over PID because this excercise was in my textbook before doing the classification
i ignored the problem thinking it was obvious lol
ooo i think I got it
assume we have decomp Z/p^k_1 \oplus Z/p^k_n
descending order
vestal snow
of G?
sturdy marsh
sure, just some group (im trying to prove fact on generators)
assume we have generating set of size 'r', r < n
now, each generator has order at most p^k_1
ah ffs nvm doesnt work
okay first we'll do a special case
assume we have Z/p \oplus Z/p \oplus ... \oplus Z/p (n -times)
assume we have k generators, k < n
each generator has order at most p
so I think the size of the subgroup they generate has order at most p^k < p^n
yeah now if you have that, then consider Z/p^k_1 \oplus Z/p^k_n
assume we have 'r' generators, r < n
we have a subgroup Z/p^(k_1 -1) \oplus Z/p^(k_2 -1) \oplus ... \oplus Z/p^(k_n -1)
the quotient is iso to Z/p \oplus ... \oplus Z/p (n -times)
the image of the generators generate below
by below i mean the quotient
which is a contradiction
this probably works
it shouldnt need the classification as it was in the groups chapter in Lang
vestal snow
Hold on
Do we even need to show this to begin with?
$G = \prod_{j = 1}^{l} Z/p^{r_j}Z$
cloud walrusBOT
Have a Banana Bitch:
vestal snow
$G/H = \prod_{i = 1}^{m} Z/p^{k_i}Z$
cloud walrusBOT
Have a Banana Bitch:
vestal snow
Then do we need to show m =< l?
Say we just fill the remaining slots, if they exist, with Z/p^0 Z
Since they are arranged in descending order
sturdy marsh
but those are just zeros
vestal snow
Yes
sturdy marsh
they dont contain a subgroup iso to Z/p^k_i
vestal snow
Oh yeah you're right
sturdy marsh
we will need to add zeros to G/H to use the lemma (but that's a triviality, not important)
vestal snow
Why can you not generate Z/pZ x ... x Z/pZ (n times) with less than n elements?
sturdy marsh
I wrote the arguement above
sturdy marsh
another argumentfor the same thing, if you're feeling fancy is that it is a Z/p vector space of dimension n
and generating as a group os the same as span in a vector space in this case
vestal snow
Ah I think you missed something important
"we have a subgroup Z/p^(k_1 -1) \oplus Z/p^(k_2 -1) \oplus ... \oplus Z/p^(k_n -1)
the quotient is iso to Z/p \oplus ... \oplus Z/p (n -times)"
I do not think the quotient is that
We know that there is a subgp isomorphic to $Z/p^(k_1 -1) \oplus Z/p^(k_2 -1) \oplus ... \oplus Z/p^(k_n -1)$ in $Z/(p^k_1) \oplus Z/(p^k_2) \oplus ... \oplus Z/(p^k_n)$
cloud walrusBOT
Have a Banana Bitch:
sturdy marsh
no I mean litterally take the unique subgroup of index p in each summand
and quotient
vestal snow
Oh okay
sturdy marsh
In general the problem of getting weird quitients arises only when there is a nontrivial embedding of the subgroup in the group
Like for example Z has a subgroup iso to Z such that the quotient is not 0
this is because we can embed Z ---> Z given by n -> 2n
which results in the quotient Z/2Z
sturdy marsh
I wrote down the entire argument above
vestal snow
I didn't really understand it
sturdy marsh
which part?
vestal snow
what do r_2G and r_2G/H mean
sturdy marsh
oh those should be p^(r_2)G
and p^(r_2)G/H
take an element
add it to itself p^r_2 times
woven obsidian
In my course book it is mentioned that for modules M-->S^-1M has the universal property for homomorphisms of M to an S^-1A module
What does this mean exactly?
I understand that if I have a map M to an S^-1A-module N, then we should have a unique map from S^-1M ---> N that makes the diagram commute
But what kind of maps are these? Is the one from M to N an A-linear map, and the one from S^-1M--->N S^-1A-linear?
next obsidian
Just means that the map M to the S^-1A module factors uniquely through S^-1M
That’s not even quite what the universal property is, in full generality given M -> N, if every element of S acts as an automorphism of N, aka if mult. by s in S is an automorphism of N then you get a unique map S^-1M -> N through which M -> N factors
woven obsidian
Yeah I know that, my question wad what kinds of "maps" these was
For rings they're just ring homomorphisms
So my guess was that the maps from M to N was an A-linear map and the one from S^-1M to N was an S^-1A-linear map
snow cliff
if I have a group of order $p^\alpha$ where $p$ is some prime and $\alpha$ is some positive integer, $p^\alpha$ is finite and so my group is finite? Just want to double check this. I think this should be true since $p$ is also a positive int and they should be closed under multiplication.
Edit nvm, no inverses, so I can't say closed. Does this mean an int to some power can become infinity?
cloud walrusBOT
camolot457656:
smoky cypress
If it has an integer order then it is finite
snow cliff
I guess what I'm confused on is does $p^\alpha$ have to be finite?
cloud walrusBOT
camolot457656:
smoky cypress
Um
What do you think
It is an integer
So it is finite
But we don't say finite integer because by definition infinity is not an integer
snow cliff
So ints can never reach infinity?
smoky cypress
again that's not really rigorous
the integers can be arbitrarily large
but infinity is not an integer
snow cliff
okay ty
snow cliff
I successfully proved the first part of this problem. What worries me is the way I approached the deduction.
This is what I did:
Assume H is a subgroup of G (G is our group of order p squared) s.t. the index of H is p.
This means that the order of H is equal to the order of G divided by p. The order of G is p squared so the order of H must be p. Furthermore, we proved in the first part that a subgroup with an index of p is a normal group, thus H is a normal subgroup of G of order p.
My concern:
My textbook says that a subgroup of order p does not necessarily have to exist. Therefore I am worried that my initial assumption is invalid and instead I have to prove that the group of size p squared must have a subgroup with an index of p. And if that is the case I would say that a hint is appreciated, as I am not sure how to approach that. I think I should examine cosets of H in G and find a contradiction if there does not exist a subgroup of index p, but I'm not seeing much light there.
woven obsidian
A group of order p exists, that is Cauchy's theorem
smoky cypress
A subgroup of order p exists
tacit saffron
I'm trying to prove that U(8) is isomorphic to Z/2Z x Z/2Z, with operation component wise addition. U is given by U(n) = {0 < a < n | gcd(a,n) = 1} (the relative primes of n), so U(8), the set would be {1,3,5,7}, and if Z/2Z is cosets 2Z and 2Z+1, i don't really know where to go from there
next obsidian
uh
So...
hmm
how long have you done group theory?
based on the answer and some other factors I have 2 different suggestions
tacit saffron
I'm first confused as to what Z/2Z x Z/2Z would be
next obsidian
One of them is really general and you get this as a special case
tacit saffron
not long
next obsidian
oh okay so maybe not
Z/2Z is just pairs (x,y)
okay uh
G x H for groups G, H
is just pairs (x,y) with x in G, y in Y
and (x,y)(x',y') = (xx',yy')
that make sense?
next obsidian
oh yeah
lmfao sorry
yeah I mean
it doesn't matter, I just mean in each component you do the operation
tacit saffron
ok
tacit saffron
ok yes
next obsidian
so Z/2Z x Z/2Z has 4 elements
(0,0)
(1,0)
(0,1)
and (1,1)
and the multiplication table is pretty easy
tacit saffron
oH
next obsidian
yeh
Sounds like something clicked haha
Do you wanna try to prove this again yourself and come back if you still can't get it?
tacit saffron
more like im just stupid and didn't understand Z/2Z x Z/2Z
next obsidian
It's fine lol
tacit saffron
my second level of not understanding is proving the isomorphism
next obsidian
The isomorphism is pretty not that bad
tacit saffron
how would {1,3,5,7} be isomorphic to {(0,0),(1,0),(0,1),(1,1)}
next obsidian
because 1,3,5,7 is under multiplication
so like 1 has to go to (0,0)
since you gotta send the identity to the identity
tacit saffron
hm yes
tacit saffron
(1,0)+(0,1)=(1,1)
tacit saffron
😮
next obsidian
so maybe send 3 to (1,0)
and 5 to (0,1)
and see if that works?
(clearly 7 goes to (1,1))
tacit saffron
wdym see if that works
next obsidian
like is that a group homomorphism?
You've defined a map one way
You took the 4 elements of U(8)
and said where you want to send them
so show that's actually a group homomorphism
and boom, it's an isomorphism since it's bijective
tacit saffron
i mean didn't i just just do that
next obsidian
I mean there's a few other combinations
call the map phi
is phi(3*3) what it should be?
Spoiler
this map works haha
tacit saffron
because thats just same as phi(1)
next obsidian
yeah
tacit saffron
(2,0) is (0,0)
next obsidian
so maybe you can figure out a way to reduce how much you need to check
but that's up to you
I'll leave you with one thing tho okay
This is higher level but
if you tink about it maybe it makes it more obvious what's going on
Z/2Z x Z/2Z can be written the following way
{e,a,b,c}
where a^2 = b^2 = c^2 = 1
and ab = c
bc = a
ca = b
and it's abelian
just let a = (1,0), b = (0,1), c = (1,1)
but actually, (you should do this)
ANY choice of the three non-identity elements for a,b,c work
the same is true of U(8)
the elements 3,5,7 can all be a b or c
so like if you view it more abstractly like that these groups behave exaclty the same
maybe that's a bit confusing rn, but if you mull about it for a while and maybe remember it in a few weeks it'll make a lot of sense
tacit saffron
so in a proof i should specify that my assignments are arbitrary and could be in any combination
no it make pseudo sense
i understand what you saying
rn im just getting caught up definitions and my monke brain go plop when i deal with real numbers
cinder bone
what function would work here
at first i was thinking that obviously i^2 = -1 and i^4 = 1 so i can use that for b and -b
but then i realized (i+1)^2 still isnt an element of the real numbers
next obsidian
I think you send a + bi to exactly that element
IIRC from my complex analysis course this is a way to phrase complex mult
And it’s useful for defining what complex analytic actually means in terms of what it means for matrixes or some shit
Or conversely given a matrix of that form send it to a + bi
next obsidian
Take a matrix of the form in the photo you sent
send that to the complex number a + bi
That’s the map
cinder bone
but it's C-> M right
next obsidian
This is M -> C