groups-rings-fields

[chilly ocean]

no I dont think so

I mean these are kinda different

[tardy hedge]

Its different but its both technically still algebra?

theyre in the same like realm of stuff tho right?

[chilly ocean]

I really dont know what proof based linear algebra is

and what not proof based one is

[tardy hedge]

oh ok so basically the one ive done is linear algebra stuff but the focus is not of proofs

[chilly ocean]

But they are not that much related in terms of like what you have to do

[tardy hedge]

like most questions were just matrix stuff (row reduction alll the damn time) and some conceptual true/false questions

[chilly ocean]

LA and abs alg

[tardy hedge]

oh ok

[stark sigil]

proof based example exam problem: Prove that $I-uu^t$ is whatever if $u$ has such and such properties
non-proof based example exam problem: Calculate the determinant of this matrix

[cloud walrus]

Icy001:

[tardy hedge]

yeah pretty much icy

https://www.math.ubc.ca/Ugrad/pastExams/Files/221_2016WT2.pdf

here it is

[chilly ocean]

yeah I know what it means but why would you have two different calsses for non proof and proof LA

[tardy hedge]

literally the class i took

[chilly ocean]

goddamn US of A

[tardy hedge]

https://www.math.ubc.ca/Ugrad/pastExams/Files/310_2013WT2.pdf

class im taking now

see how the exams are different

im in canada btw lol

[chilly ocean]

goddamn NA

[tardy hedge]

true

[late sedge]

irreducibility, prime, and associated trouble keeping them straight

irreducibile elements are non-unit elements which can be written as a product of two units right?

associated elements have a unit as a divisor

but I dont get this definition of prime? p|ab => p|a or p|b

[golden pasture]

its similar to how you can define prime numbers on the natural numbers

its identical to $p\nmid a$ and $p\nmid b$ implies $p\nmid ab$

[cloud walrus]

Ariana:

[late sedge]

oh ok

lets take 23 in Z which is a prime

23|46 and 46 can be written as 23x2 as well as 23|23 so 23 is a prime

[cloud walrus]

Mortex:

[golden pasture]

wut

[late sedge]

?

[sharp sonnet]

your argument is weird, you could just as well argue that 8 | 16 and 16 = 2 * 8 and 8 | 8, so 8 is prime

which is wrong

because the definition of prime talks about all a, b

the reason why you have to define primes in general rings differently than in the integers, is essentially that there is no notion of "positive" numbers

but you still want things like "unique"(up to something) prime factorization

so you need another notion of primeness

and this works

[chilly ocean]

Hey, can anyone help me understand the proof that number of even permutations in a permutation group $S_n$ is equal to number of odd permutations

[cloud walrus]

mega:

[chilly ocean]

basically, what do they mean by

fix a transposition sigma in S_n

?

[fickle brook]

a transposition is a swap

ie a permutation that just swaps two points

and fixes everything else

[chilly ocean]

So, here they are choosing any random transposition from S_n?

[fickle brook]

an arbitrary one

doesn't matter which

if you wish, you can let it be the transposition that swaps 1 and 2

[chilly ocean]

okay, and the lamda is a function just multiplying an element from A_n with sigma to make it odd?

[fickle brook]

er

well

yeah that's how λ_σ is defined

it's just left-multiplication by σ

[chilly ocean]

oh okay, I got it

Thanks a lot!!

[tribal pasture]

If a group of order 8 has all elements of order 2, is it isomorphic to C2 x C2 x C2?

non-trivial elements*

[mild laurel]

Yes

[cloud walrus]

6fadeymlady:

[mild laurel]

I have never heard the phrase standard k cycle

[mild laurel]

I have no clue what you're saying still

What's the difference between a k-cycle and (1,2,3,...,k)

[mystic raft]

A k cycle goes from element to element in a chain until the chain fepeata

[mild laurel]

Yes and that's what (1,2,3,...,k) denotes

[fickle brook]

do you mean a cycle on k points that aren't necessarily 1 through k in that order

[bleak abyss]

@mystic raft

[mystic raft]

Thanks! So I believe k cycles are having a list of numbers 1,2...,k and that each index in the list is assigned in such a way that

You first discover where 1 is assigned and then search out the number position it has been assigned to, and then continue down the line until a cycle is reached and then that completes that k-cycle however many times that works

[bleak abyss]

Uh, that description is a bit vague, in fact it's vague enough such that I'm not really sure I'm able to decipher it

[mystic raft]

So I’m p sure that’s accurate. Now I have a result I’m trying to understand which talks about using the standard k-cycle, which my guess is that is means 1 assigned to 1, 2->2 and so on to k. Which technically is k 1-cycles.

Anyhow the question is about how to get the commutator between the standard k-cycle and the actual k-cycle?

[bleak abyss]

But basically here's the idea

[mystic raft]

OK

[bleak abyss]

So you give me k numbers between 1 and some n

(a_1,...,a_k)

The a_i have to be different

That gives me a function from the set {1,...,n} to itself

[mystic raft]

3,1,5,2,4 is a 5-cycle fwi gather

[bleak abyss]

How? (a_1,...,a_k)(a_j) = a_{j+1}

[mystic raft]

we could be refering to two different defintions. Apparently there are k cycles & k-cycles

[bleak abyss]

And (a_1,....,a_k)(x) = x if x isn't equal to any of the a_i

[mystic raft]

Ok i think u know more than i do

But 3,1,5,2,4 is 1->2->4->5->3 should be a 5-cycle idk im confused

1 day

[smoky briar]

what notation is this even

[sharp sonnet]

you could just read https://en.wikipedia.org/wiki/Cyclic_permutation

a cycle is just a nice notation for elements of the group S_n

[smoky briar]

ok your notation is like, inverse bijection notation

[sharp sonnet]

and a k-cycle is a cycle of length k

[smoky briar]

which is very messed up

[mystic raft]

I think I understand the basics but idk what a standard k-cycle is + how to take commutator of stnd k-cycle w/ k-cycle

[smoky briar]

commutator is ghg-1h-1

just multiply

[sharp sonnet]

you should find the definition of standard k-cycle

[mystic raft]

Well you’d take a list of 1->k and rearrange the numbers and then observe the cycles

[sharp sonnet]

what is "a list of 1->k"

[mystic raft]

So idk how using the commutator definition can give the same format

A numbered list

[smoky briar]

wtf man

it's just a multiplication

of elements in the group

just do it

[mystic raft]

I don’t believe it’s just multiplying bc what I am looking at is that by taking the commutator of the stndard k cycle with the k-cycle it returns another k-cycle in speicifx cases

[smoky briar]

how can you not believe it?

do you know what a commutator is?

it always returns a k-cycle

conjugation preserves the cycle structure

[mystic raft]

Ok this is asking for a commutator of standard k-cycle with a k-cycle and in whcih instances does that itself create a k-cycle

[sharp sonnet]

it's no multiplication

it's composition

[smoky briar]

multiplication means composition

in this group

[sharp sonnet]

you are looking at function S_n -> S_n after all

[smoky briar]

what is this place

[sharp sonnet]

i think this guy is confused enough

[smoky briar]

im more confused

im going home

[mystic raft]

i am 99% sure im not wrong rn lol idk what the fuss is

[fringe nexus]

[mystic raft]

how would multiplication give the commutator between two different k-cycles, the standard and an actual k-cycle

I could be way wrong i suppose

[sharp sonnet]

the commutator is defined via multiplication

[mystic raft]

Ok so let’s work with the two 3-cycles... 2 3 1 :3 1 2

(2,3,1) & (3,1,2) how would you take the commutator of either of those with the standard k-cycle

[sharp sonnet]

those are the same cycles btw

[mystic raft]

Feel like I’m in the truman show haha the example I saw talked about nothing I’m hearing

and the result I’m trying to understand is speaking about a lotta minor variations in a small area, so it’s confusing how multiplying could reproduce another k-cycle in instances.... and idk how 2,3,1 & 3,1,2 are the same cycles

[sharp sonnet]

because each takes 1 to 2, 2 to 3, 3 to 1 and keeps all other elements fixed

[fringe nexus]

2->3->1->2 for first, 3->1->2->3 for second

[mystic raft]

^

[fringe nexus]

they're the same

[mystic raft]

Oh I see

[fringe nexus]

[sharp sonnet]

in cycles only the order is important

the starting element in notation is arbitrary

[mystic raft]

And there are cases where they may not be the same?

[sharp sonnet]

(1 2 3), (2 3 1) and (3 1 2) are the same, (1 3 2) is different

[mystic raft]

1 3 2 isn’t a k-cycle at all

[sharp sonnet]

it's a 3-cycle

[mystic raft]

Nor is 1 2 3

[sharp sonnet]

it's a 3-cycle as well

[mystic raft]

Must be the standard then ahh ok

I guess? I mean how is 1,2,3 the same as 1 3 2

[fringe nexus]

they aren't

[sharp sonnet]

did you read what i wrote

[mystic raft]

But 1 3 2 isnt a 3-cycle

[smoky briar]

holy fuckj

[fringe nexus]

2->3 in the first one, 2->1 in the second one

[sharp sonnet]

or are you just rearranging the word as you see fit

[fringe nexus]

what is your definition of a 3 cycle

[mystic raft]

1 3 2 is a 1 cycle and a 2 cycle

[sharp sonnet]

wtf is a 1-cycle even

there are no cycles of length 1

unless you want to call the identity that

[fringe nexus]

Im pretty sure the usual definition of a k-cycle is a cycle of length k

[mystic raft]

https://math.stackexchange.com/questions/139721/what-are-k-cycles/139727#139727

[sharp sonnet]

what this person says

and means

[smoky briar]

in case you guys didn't notice, he's using weird as fuck notation that nobody uses

[sharp sonnet]

is that there is difference between "3 cycles" and "3-cycles"

3 cycles means, there is 3 -in words three- cycle

a 3-cycle is a single cycle

of length 1 (one)

we call cycles of length k "k-cycles" to distinguish

[mystic raft]

So a cycle of 9,4,1,3,7,2,8,6,5 is a 1-cycle not 9?

[sharp sonnet]

i think he is reading standard notation and is very, very confused about it

[smoky briar]

his notation is like

if you have (2 4 3 5 1)

then this means

2 comes from second position i.e. 4, 4 comes from 5, 5 comes from 1, 1 comes from 2

so this is actually (2 1 5 4)

which is beyond fucked

I think he's intentionally trying to confuse people who talk to him

so I wouldn't bother

[mystic raft]

1->5->4->2->1 and 3->3 which I believe are a 4-cycle and a 1-cycle

[smoky briar]

that's (2154)

[mystic raft]

ok

And that qualifies as a 4-cycle and we ignore the other?

[sharp sonnet]

you don't write down the elements that stay fixed in cyclic notation

[mystic raft]

So then that’s a 4 cycle^ (2154)?

A k-cycle where k is 4?

[sharp sonnet]

(2154), is a 4-cycle

[mystic raft]

Ok thx. I’ll read more now. Maybe the confusion is because the grammar of what I’m reading about could be read in two different ways

Phew

thx for patience

Idk something still seems amiss lmao

Like wtf is a standard k-cycle. It is being implied to be 1,2,3,...k

Which is either tricky word play or questionable

[sharp sonnet]

i have no idea tbh

i never heard the term

i would assume it is (1 2 3 ... k)

[mystic raft]

This implies that (1) is a 1-cycle

And the commutator of (1) and (1) is also likely (1)

That it remains a 1-cycle. Which assumedly would just be (1)

[sharp sonnet]

well, you usually only define k-cycles for k > 1

because (1) makes no sense

it would be the identity

[mystic raft]

Before multiplying does it matter if I have (2,3) or (3,2)? And when taking commutator with 1,2,...,k does that imply multiplying 2,3/3,2 by standard k or standard k by 23/32?

Thanks u were right

[late sedge]

if an element a is associate with b then a=u*b where u is a part of the unit group

so are both a and b associated

[stone fulcrum]

That is a fact

Associates form a partition of the ring

[fringe nexus]

if we have some polynomial of the form x^n - a in Q[x]

where a isn't any nth power in Q, the galois group of this is generated by automorphisms that map $\phi(\zeta_n) = \zeta_n^{c}$ and $\phi(a^{\frac{1}{n}}) = \zeta_n^{b}a^{\frac{1}{n}}$ where c,b are coprime to n right?

[cloud walrus]

aaaaaaaaaaaaaaaa:

[fringe nexus]

just making sure

[fringe nexus]

actually i think im wrong

With x^6 + 3 over Q the galois group is of order 6

what assumption am i missing in my previous statement?

[smoky briar]

you don't have nth roots

you can't get sixth roots of unity from x^6 + 3

[fringe nexus]

sorry can you elaborate?

I understand the argument that x^6+3 has splitting field Q(a) where a is the 6th root of (-3)

but I'm not sure why we don't have the 6th root of unity in there

[smoky briar]

yes

cuz

you don't lol

[fringe nexus]

ok

[smoky briar]

namely x^6 + 3 and x^6 - 1 are coprime

so you definitely aren't getting roots of the latter

welllllll

that's not a sound argument

but it's something like that

like picture it like this

Q[x]/(x^6 + 3)

is an extension of degree 6

right?

and you have noticed that you have all six roots of the equation

so the splitting field is just degree 6

and not the bigger thing you thought

which would be having the roots of unity

[fringe nexus]

Ok but if we had x^6 - 3

would the galois group be of order 12 then?

[smoky briar]

no

I just told you

[fringe nexus]

[smoky briar]

like

the same thing happens

wait

wait a second

im being stupid

so let the 6th root of -3 be w and the 6th root of 1 be z
then the solutions to x^6 + 3 are w,wz, wz^2, wz^3,wz^4, wz^5 right?

having one doesn't give you the rest

my bad

the splitting field is Q[w,z]

[fringe nexus]

uhh

having the 6th root of -3 should give you everything else

i think

[smoky briar]

no

if you have two of them you have the others

[fringe nexus]

oh

ok

[smoky briar]

wait

jesus christ you're right

[fringe nexus]

[smoky briar]

yeah ok sometimes you can get z from w

[fringe nexus]

because if you adjoin 6th root of -3 you have degree 6 extension already

[smoky briar]

sure

[fringe nexus]

But i guess my general question is how do i even know this

like if i have x^n + a

a is positive

and i don't have that hint

[smoky briar]

hmm

[fringe nexus]

I have 0 trouble computing galois groups for x^n -a but the other one is so much harder for me idk why

[smoky briar]

why is x^n - a easier?

[fringe nexus]

because nth root of a isn't related to i i guess?

so you can't get roots of unity from that

when a is positive

[smoky briar]

yeah ok that works

[fringe nexus]

its straightforward where you adjoin the nth root of unity and nth root of a as long as a is "nice"

[smoky briar]

yeah so your issue is the case x^2k + a

you would have to check whether x^2k + a splits in Q[z]

where z is a 2kth root of unity

but idk if that's any easier

[fringe nexus]

hmm, so is the only time this happens when x^2k +a ?

[smoky briar]

you can say even more

because the only way it splits is that

the degree of Q[z] divides 2k

so phi(2k) | 2k

so this can only happen for x^n + a with n even and phi(n) | n

what else

I don't know if I see something else

https://math.stackexchange.com/questions/204552/computing-the-galois-group-of-polynomials-xn-a-in-mathbbqx

the second answer here is what you want

it's an interesting question

and you won't be expected to know this

[fringe nexus]

oh i see

hmm yea We didn't do anything like this

Would be interesting if he asks this on final

[latent anvil]

This is not abstract algebra

[fringe nexus]

uhhhhh this isn't supposed to be here

[latent anvil]

and you're supposed to wait before calling helpers

Go to #prealg-and-algebra

[chilly ocean]

Well I'm new so where should I take this?

Oh #prealg-and-algebra okay.

[latent anvil]

Also read #info

[late sedge]

wait do quotient rings partition the ring analogous to how quotient groups partition groups?

[fringe nexus]

yes, R/I is the set of cosets { r + I, r_2 +I,...}

[tribal pasture]

Are there any algebraic structures of interests which are non-associative?

[stark sigil]

Lie algebras

[late sedge]

god irreducability and prime still confuse me idky

i get the definitions

but applying them give me a headache

[fringe nexus]

same

For this one, the first part I have the counter example Z[x]/(x^2+5) which is not a UFD

I assume its true for the PID case, but I'm not sure how to prove it or find a counterexample

[tender mist]

I think if R is a PID then f irreducible implies (f) is a maximal ideal

[fringe nexus]

it does, but do we have any way of determining if R/(f) is a UFD based on (f)?

actually wait

[tender mist]

If (f) is maximal then R/(f) is a field

[fringe nexus]

yea i forgot

field implies ufd

[late sedge]

good picture

[stark sigil]

Field implies ufd because every nonzero number is a unit

All units and no primes make for unique factorization up to units

[late sedge]

all PID's are UFD's though

all primes are irreducibles

[fringe nexus]

hate ufds

hate pids

[tender mist]

However your previous exercise was really interesting @fringe nexus, I've been given a very similar one exactly today lol, but it's the case x^n-a
I don't know how to approach it though

[fringe nexus]

x^n - a is rather straightforward

depending on your base field, if its over Q

[tender mist]

Initially I thought the same. But after making some examples the degree of the splitting field isn't always phi(n)n

[fringe nexus]

can you show me those exampls?

maybe im wrong

[tender mist]

Like, x^8-2 has a splitting field of degree 16 = phi(8)8/2

If I'm not wrong

[fringe nexus]

yea its degree 16

aaaaaaaaaaaaaa

[tender mist]

It's a case when sqrt(2) is in Q[ζ_n]

[fringe nexus]

but whats your question

that you were given?

[tender mist]

It's find the degree of the splitting field of x^n-p over Q for p prime

Sometimes the intersecion isn't trivial so I'm stuck

[fringe nexus]

i was reading through the stackexchange the other guy sent

and there was something about this

[tender mist]

Yeah I read that post too, from that answer I could determine some cases

According to them the degree will be phi(n)n/2^s with 2^s a common divisor of phi(n) and n (this last thing because in my problem p is prime)

[fringe nexus]

[stark sigil]

😱

[tender mist]

So if n is odd it's phi(n)n

The problem is if n is even

For p=2 anyway the solution is phi(n)n if 8 doesn't divide n and phi(n)n/2 if 8 divides n

So we can consider p>2 and n even

From what they say
-if p is congruent to 1 mod 4 then sqrt(p) is in Q[ζ_p]
-if p doesn't divide n then Q[ ζ_p] and Q[ζ_n] have intersection just Q.
-if n is even and sqrt(p) isn't in Q[ζ_n], then the degree is phi(n)n

So if p is congruent to 1 mod 4 and p doesn't divide n sqrt(p) isn't in Q[ζ_n] and the degree is phi(n)n

I could't deduce more than this

[stark sigil]

Made the preceding table in SageMath

Only see 2 exceptions here, namely $x^8-2$ and $x^{10}-5$

[cloud walrus]

Icy001:

[tender mist]

Thanks! Interesting
Could you also try x^20-5 @stark sigil ?

[stark sigil]

My table for degree 15 is still computing as of 10 minutes ago

I am just going to interrupt it at this rate

[upbeat burrow]

@stark sigil what are you doing?

[stark sigil]

The $(n,p)$ entry in the table is the splitting field degree of the polynomial $x^n-p$ over $\bQ$

computing it with SageMath

[cloud walrus]

Icy001:

[upbeat burrow]

Ah

Oh wait, I see, mat is the one with the question?

[stark sigil]

yea

[cloud walrus]

Anyone_Someone2018:

Anyone_Someone2018:

Anyone_Someone2018:

Anyone_Someone2018:

[chilly ocean]

What do you folks think about my interpretation?

Am i correct?

[smoky briar]

yeah but the terminology is strange

the analogies are fine but don't call them divisors or prime

also + is used to imply the group is abelian

so try to stick to multiplicative notation

[chilly ocean]

right

[steep hull]

@tender mist I solved it for p=2.

Did you already have a solution

[tender mist]

I had
8 doesn't divide n -> phi(n)n
8|n -> phi(n)n/2

Do you get the same?

[steep hull]

Yes

Did you have a proof

[tender mist]

I'd like to read yours because I can't justify some steps of the one I have

[steep hull]

Okay, I’ll write it out in a little while (too late for me rn). Feel free to share yours (probably in Galois/NT channel).

[tender mist]

Sure, thanks!

[chilly ocean]

how do I show that if p is prime and R is a commutative ring with characteristic p then $\left(a+b\right)^p = a^p + b^p, \forall a,b \in R$?

[cloud walrus]

Godel:

[chilly ocean]

What I tried to do is that p divides pCk for all k={2,3,4...,p-1}, but I couldn't quite do it

maybe theres a better way

[stone fulcrum]

Pretty sure it's all binomial theorem

But let me take a look ieh

Oh yeah your approach works

[chilly ocean]

Does binomial theorem tell us anything about divisiblity of the coefficients though? Because that's what I tried

[stone fulcrum]

pCk = p! / k!(p - k)!

[chilly ocean]

yep

[stone fulcrum]

So divisible by p unless k = p or k = 0

[chilly ocean]

why

I don't see it

[stone fulcrum]

https://en.m.wikipedia.org/wiki/Freshman's_dream

This can explain better than I lol

[sharp sonnet]

p! is divisible by p, k!(p - k)! is divisible by p iff k=p or (p-k)=p, i.e. k=0

so p! / k!(p - k)! is divisible by p unless k=0 or k=0

[chilly ocean]

okay I see

thx

[fringe nexus]

If I have a field extension of K, say K(a), of degree 37, how do i show that the field extension k(a^3) is also of degree 37?

Would it be because of something to do with divisibility

[stark sigil]

It has degree at most 37

[fringe nexus]

Hmm

I was looking at the min poly

of a

[stark sigil]

it's irreducible of degree 37

[wind steeple]

37 is prime

[fringe nexus]

aaaaaaa this should be obvious

im just missing something obvious

does it not have any subfields?

[wind steeple]

it does not have non trivial subfields indeed

[stone fulcrum]

Could think in terms of Galois groups. The only subgroups of a 37 order group is either the group itself or the trivial one, so you'd only need to show that this isn't trivial

[fringe nexus]

wait im so confused

wait im alreayd done

[wind steeple]

you don't have galois correspondance here

[fringe nexus]

if the subfields are trivial

and the min poly works for a^3

it has to be the same field

[wind steeple]

yes

it is

[stone fulcrum]

Oh mb

Forget I say things

Why does it not? I'm trying to understand but I may have gaps

[fringe nexus]

its not a galois

extension

[wind steeple]

bc it's not a galois extension

it's not necessarly normal

[chilly ocean]

Why are all the prime ideals of Z in form of pZ where p is prime? Maybe im not udnerstanding the definition, but why wouldn't 4Z be a prime ideal?

[sharp sonnet]

4 is in 4Z but 2 is not

[chilly ocean]

lol ok

[sharp sonnet]

also notice that 0Z = {0} is a prime ideal, even though 0 is not prime

[chilly ocean]

yeah

[fringe nexus]

is the first one R/(2) isomorphic to Z_2, and R/(2i) isomorphic to Z_4?

so they aren't isomorphic to each other

and R isn't a pid because its not a ufd because 2i*2i =-2*2 = -4

wait no the first one is z_2xz_2?

[chilly ocean]

why is R/(2) iso to Z2?

I dont think so

(2) is all even integers right?

[mild laurel]

no, it's the ideal generated by 2

[chilly ocean]

which is all even integers?

[mild laurel]

No, 4i is in the ideal generated by 2

[chilly ocean]

ohh, yeah forgot you multiply by elements of R, sorry

ok then yeah should be isomoprhic to Z_2

[mild laurel]

No, the first Z_2 x Z_2, 4i is the smallest multiple of i

[fossil vapor]

https://youtu.be/5zLr7hAnrZY

for some history

[delicate bloom]

cool, watching now

[chilly ocean]

why is$\mathbb Z _2 \left[X\right] / \left(X^2+X+1\right)$ a four element field?

[cloud walrus]

Godel:

[chilly ocean]

maybe im not understanding what it means

[delicate bloom]

list 5 of the elements

[chilly ocean]

I cant list 2

[delicate bloom]

oh haha

thought youw ere going in the opposite direction

[chilly ocean]

0 is all the polynomials that have x^2 + x +1 in somewhere in their factorization right?

[delicate bloom]

Z_2[X] is all the polynomials with coefficients in Z_2

[chilly ocean]

yeah

[delicate bloom]

yeah

[chilly ocean]

but I still dont get it

I know what it means, but over (x^2+x+1) Im not sure what the elements of this are

[delicate bloom]

well what's the highest degree polynomial you can have left for starters?

[chilly ocean]

that's what I dont understand, I think its 2, but why cant I have x^4

[delicate bloom]

well if you have x^4 it's really just (x^2)^2

and since x^2+x+1=0 you can rewrite it as

x^2 = -x-1

and reduce this way

of course that really reduces to x^2 = x+1

and so x^4 = (x^2)^2 = (x+1)^2 = x^2+2x+1 = x^2+1= -x = x

or we could have done it as x^2 * x^2 it'd work just as well

try for yourself to see

anything x^2 and higher will get reduced down

[chilly ocean]

ok, and is this some kind of euclid algorithm?

[delicate bloom]

I think you're over thinking it if you're trying to introduce jargon

do you understand what I'm doing

x^2+x+1=0

[chilly ocean]

yes

I think I am overthinking it

Cause I was trying it in a way different way, finding what this ideal is made of

[delicate bloom]

I'm just not convinced yet you know what I've done so far

[chilly ocean]

yes I get it

and now I get why I have 4 elements

[mild laurel]

Eh, I think it is important to note in the long run that this is also euclidean algorithm. If you divide x^4 by x^2 + x + 1, you get x as the remainder

[chilly ocean]

Ok, so the four elements would be 0,1, x^2 +1 and x+1?

[delicate bloom]

why is there an x^2

thought we just talked about that

[chilly ocean]

ok so just x instead of x^2+1

[delicate bloom]

yeah, ok to be fair it doesn't really matter and yeah like zoph said definitely euclidean algo is way to go

[chilly ocean]

hmm so if I have z[x]/f where deg f is n would all the elements of that be at mostof degree n-1?

[delicate bloom]

but it should be kind of obvious in a sense, there are only 2 elements in Z_2 and since you're modding out a degree 2 polynomial it's basically just like 2^2 possibilities from this perspective

[mild laurel]

You need that polynomial to be irreuducible

Otherwise weird things happen

[topaz solar]

if your poly isn't irreducible you better be c a re f u l

[chilly ocean]

And how would you define z[x]/f where z is a field

because I think ive been misunderstanding it the whole time

[topaz solar]

You define it for the field case exactly as for the ring case

because a field is just a ring but with stricter conditions

[chilly ocean]

I guess the equivalence classes are the possibl;e remainders of dividing by f then?

[delicate bloom]

yes

that's why x^2+1 and x didn't matter

they're in the same equivalence class

but it's kind of awkward in the sense of, if you're doing modular arithmetic but not all your representatives are reduced

[chilly ocean]

yeah it seems kind of unintuitive

[delicate bloom]

like working in Z_5 with 0,1,2,3, and 9

it's clearer to use 4 instead but really they're the same equivalence class

[chilly ocean]

Btw, they also state that that thing I posted is also a field

Why is that? Do I have to check like all elements since there arent many or does that follow from something

[mild laurel]

both

[chilly ocean]

wait so f irreducible implies it?

actually its iff

interesting

if f isnt a unit

I actually find rings very interesting, but its starting to get harder since there are so many properties you need to remember to solve problems (even though they are in the most cases not hard to prove, I just forget about a lot of them)

[mild laurel]

True, I have a commutative algebra final on Tuesday

There are a lot of properties for rings, but you learn them and you learn how they fit together and stuff

It becomes a lot easier

[chilly ocean]

This is my biggest problem in this class. I bombed my group theory exam, although I knew thoery pretty well. There were so many theorems and conclusions from them, it was almost impossible for me to learn it well. With rings, I feel like its easier so far since we are dealing with pretty much commutative ones exclusively, which are more intuitive.

[magic owl]

All rings are commutative

[chilly ocean]

Agreed

[waxen iron]

Just a quick question on the Orbit-Stabilizer Theorem:
The index of Gx in G is equal to |G/Gx|, but I'm wondering, what is this set really?
The Stabilizer of x is the set of elements of g that leave x unchanged, but what is gGx? Is there some intuition behind it or it's just a useful tool to analyse stuff?

[waxen iron]

I think there's a problem of notation here 😅
My teacher defined the stabilizer like this:

Hence me using Gx for it, sorry for not being clear

Right, but does that set gGx have any particular meaning or it's just a subset of G?

[waxen iron]

It was more of an intuition thing, trying to piece it all in my head.
My teacher used to G/Gx to prove the OST and I wondered "wtf does it mean to partition G into subsets of gGx"
Many thanks, sorry again for the confusing notation 😅

[chilly ocean]

Hey, I had a doubt: Does the number of transpositions in decomposition of a cycle remains same always?

I'm not asking about whether they remain odd or even, but whether the number of transpositions remain same?

If yes, how do I go about proving it?

There is a nice proof about the odd and even decompositions remaining same in Dummit and Foote, but I wanted to know about the number of transpositions

[sharp sonnet]

what exactly do you mean by decomposition?

i mean if you write any cycle as a product of transposition, just take any transposition (a b) and multiply the cycle by (a b)(a b)

[chilly ocean]

I mean decompostion into transpostions

[sharp sonnet]

(1 2 3) = (1 2)(2 3) = (1 2)(1 3)(1 2)(1 3)

[chilly ocean]

So, the number of transpositions will not change.

Oh okay,

[sharp sonnet]

but it did?

the first decomposition is a product of 2 the second of 4 transpositions

[chilly ocean]

but the multiplication of second gives identity?

[sharp sonnet]

what

[chilly ocean]

if we pass 1 though the second one (from right to left), it gives back 1, and so do 2 and 3

[sharp sonnet]

(1 2)(1 3)(1 2)(1 3) (1) = 2

1 goes to 3, then to 1, then to 2

[chilly ocean]

oh okay sorry, i stopped calculating after i got back 1 thinking the cycle ends

So the number of transpositions isn't fixed

got it

thanks a lot!!

[waxen iron]

So this is probably pretty basic but I'm asked to determine the center of Sn for all n

It feels like for all Sn>=3 only the identity is in the center, but I'm not seeing how to prove this

[sharp sonnet]

take an element from S_n that is not the identity

call it f, it moves at least one letter i, so f(i)=j, then it also moves j, let's say f(j)=k

now try to contstruct a permutation this doesn't commute with

[fading wagon]

hmm

basically conjugates

a^-1ba has the same cycle structure as b

the proof is that all you are doing is replacing numbers

[waxen iron]

Okay I think I got it
let h be another element and h(i)=a
hf(i)=h(j)=a
fh(i)=f(a)
f(a) can't be a because Sn is injective
So, by contradiction we prove Z(Sn) is trivial

Is this correct?

And, @warped bay is it the reason that (123) and (456) are disjointed cycles?
I didn't really understood the question😅

[golden pasture]

well a trivial example is with itself or with its inverse

[waxen iron]

In general no, since the group is not abelian

[golden pasture]

otherwise we have $\alpha\left(a_1a_2\dots a_n\right)\alpha^{-1}=\left(\alpha\left(a_1\right)\alpha\left(a_2\right)\dots\alpha\left(a_n\right)\right)$

[cloud walrus]

Ariana:

[golden pasture]

eg for some not completely trivial commuting permutations
(13)(24)(1234)(13)(24)=(1234)

[chilly ocean]

@waxen iron are you learning abstract algebra from Dummit and Foote, because I stumbled on same question?

[waxen iron]

Nop, just using class notes, and the teacher doesn't follow anyone book @chilly ocean
This particular exercise is from a work sheet he gave out

[chilly ocean]

Oh okay cool 👍

[waxen iron]

I'd send it to you, since he doesn't really mind, but I'm afraid it wouldn't be of much use cuz it isn't in english

[chilly ocean]

oh thanks, but yeah I only understand english 😅

[cloud walrus]

fields:

fields:

fields:

[waxen iron]

Beta(j) is j
Alpha(a) is a
Oooooooh, your way is very clean

[chilly ocean]

can someone help me understand the meaning of this statement?

[waxen iron]

Many thanks @warped bay , you're an angel

[cloud walrus]

fields:

[chilly ocean]

It meant there is no common element between those sets

It is from the proof of Lagrange theorem in group theory

[cloud walrus]

fields:

[chilly ocean]

yeah

[waxen iron]

Socratica as a great video on that proof Riyango

[chilly ocean]

I can't understand why there is i is from 1 ro n in the summation sign but j in $H_{\alpha _j}$

[cloud walrus]

fields:

Riyango:

[golden pasture]

$$\bigcup_{i=1}^n A_i=A_1\cup A_2\cup\dots A_n$$

[cloud walrus]

Ariana:

[chilly ocean]

Oh, I thought it meant something else

[golden pasture]

j was prob typo

[chilly ocean]

but j is in the third line too

[golden pasture]

copy paste

[chilly ocean]

it changed in 4th line

[golden pasture]

exactly

cuz you cant copy paste directly

so need retype

also whats that font it looks ugly af

[chilly ocean]

Thanks for the clearance

[waxen iron]

"Calculate explicitly the parcels of the conjugacy classes of the following subgroups:"

I assume it was supposed to be "the following groups", either way, I got no idea how to do this

[wind steeple]

take an element

conjugate it

see what's the set of its conjugates

[waxen iron]

conjugate it by what?

[mild laurel]

everything

[waxen iron]

Okay that's what I assumed

But I ain't doing that for 16 elements

[mild laurel]

Use some facts you know about conjugacy classes to not

[waxen iron]

My teacher when over this pretty fast, so I probably don't know as much as I should tbh

[mild laurel]

There are some easy things you can figure out

By playing around with it

Like, your first group is abelian

Look at the definition of conjugation, and see what happens if the group is abelian

[waxen iron]

Correction:
All conjugacy classes are singletons and constitute the center

[mild laurel]

Yes

[waxen iron]

Many Thanks!

[fringe nexus]

@mild laurel Are you sure this is z2xz2

https://cdn.discordapp.com/attachments/496784958430380033/655525339862401055/unknown.png

[cloud walrus]

aaaaaaaaaaaaaaaa:

[fringe nexus]

I can't construct a bijection from ${a+2bi|a,b \in Z_2}$ to klein four grou

[cloud walrus]

aaaaaaaaaaaaaaaa:

[fringe nexus]

wait am i understanding wrong

so if i have (1+2i) mapping to (1,1) and (0+2i) mapping to (0,1) then if i multiply (1,1)and(0,1) in the klein four group i get (0,1) ?

wait I can't seem to make work either

1+0i has to be (1,1)

0 has to be (0,0)

[fringe nexus]

in Z[2i] its 2*(a+2bi)

but i can also mod by two first no?

I thought it was z2 x z2 but someone asked me to find the explicit isomorphism and I can't

ok

Huh?

im saying that R/(2)

not R

a+2bi , a,b in Z_2

uhhh

im not 100% sure what that means

but sure

yea

the cosets

I know how to calculate the quotient I'm just not sure which order 4 ring is it isomorphic to

[mild laurel]

Well one argument for why it has to be Z_2 x Z_2 is that every element of R/(2) has additive order 2

So it cannot be Z_4

[fringe nexus]

yes

but there are like

4 rings

of order 4

up to isomorphism

[mild laurel]

Oh

[fringe nexus]

I cannot find the explicit bijection i don't think its z2 x z2

i did

a+2bi , a,b in Z_2

1+2i

2i

1

0

[mild laurel]

Wait I'm not sure what's wrong with what you said

You said that "so if i have (1+2i) mapping to (1,1) and (0+2i) mapping to (0,1) then if i multiply (1,1)and(0,1) in the klein four group i get (0,1) ?"

[fringe nexus]

that isn't wrong

[mild laurel]

But (1+2i)(2i) = 2i

[fringe nexus]

but then i realised

1 has to map to (1,1)

[mild laurel]

Is there something wrong with this

[fringe nexus]

because consider if 1 isn't at (1,1)

[mild laurel]

Ah yeah

What are the other two rings of order 4?

Other than Z_2 x Z_2 and Z_4?

[fringe nexus]

https://math.stackexchange.com/questions/2213201/non-isomorphic-rings-of-order-4

F_4 and some group of matrices

i'm not sure the difference between z_4 and f_4

oh wait

nevermind 4 isn't prime duh

[mild laurel]

Oh hm

Is this ideal maximal

[fringe nexus]

uhhhh

no

wait

maybe?

[mild laurel]

Nah its not

We can also see this cause neither 2i nor 1 + 2i have inverses

in R/(2) I think

So it can't be F_4

[fringe nexus]

wait can you explain why not having inverses imply its not maximal?

oh wait

you meant in the quotient ring

[mild laurel]

yeah

[fringe nexus]

ok nevermind I'm just dumb

ok so its the group of matrices i guess

[mild laurel]

So since R/(2) isn't a field, we know that (2) isn't maximal, but its not really relevant

Yeah I think the matrix group should work

Hm

Yeah I think this does work

[fringe nexus]

i did not know there were more than 2 rings of order 4

until today

[mild laurel]

I also just learned this

Super surprising

I guess there's more than one way to put multiplicative structures on additive groups

[chilly ocean]

Is there a proof that there are only 4 rings of order 4?

[fringe nexus]

i mean yes

its right there

in the stackexchange

[chilly ocean]

Ok I see

[cloud walrus]

fields:

[stark sigil]

Rip messages

[late sedge]

your messages got torn apart like a prolapsed anus

[steep hull]

@bleak abyss :/

[bleak abyss]

@late sedge let's keep things closer to appropriate

[late sedge]

lol

sorry

can i ask you a question @bleak abyss

did you at least chuckle?

[steep hull]

@warped bay multiplication is really a linear transformation on a Q-module (which can be thought of as a vector space).

[chilly ocean]

Hey! I have a doubt

Is it true that a cycle of length k can be written as a product of only k-1 unique transpositions?

If yes, then how do I prove it

because in someplaces this is being used to show that a non-cyclic permutation can be written at most with n-2 transpositions, but there is no direct proof pointing at this

$$(a_1 a_2 ... a_k) = (a_k a_1) (a_{k-1}) ... (a_2 a_1)

[cloud walrus]

mega:
Compile Error! Click the reaction for details. (You may edit your message)

[wind steeple]

that's it

[chilly ocean]

I know about this way of writing, but how does it prove that this is the only way?

[wind steeple]

uh ?

it's clearly the proof

[golden pasture]

(123)=(13)(12)=(21)(23)=(32)(31)

[chilly ocean]

proof that the number of transpostions (unique) are always n-1

uh...Sorry but I didnt understood😅

[golden pasture]

For 2 transpositions to form a cycle they cant be disjoint

So each transposition must have 1 element that is the same as another

So you need minimum n-1

But you can always decompose it to n-1 trivially

[chilly ocean]

So minimum is n-1, and maximum is n-1, so it is n-1 ?

[golden pasture]

yea

[chilly ocean]

Oh okay, thanks!

I was searcing online, and there is a proof about less then n transpositiosn required

The second answer uses induction

https://math.stackexchange.com/questions/1953960/permutation-written-as-product-of-transpositions

I think I might be able to do this with similiar induction

[chilly ocean]

Okay, though it might not be relevant anymore, I found out that we can first prove that any non-cyclic permutation can be written at most with n-2 permutations and then use it to prove that any cycle of k lenght can be written in exactly k-1 permutations, for anyone who is interested

[chilly ocean]

yup, and then we can use this by saying that any non cyclic permutation is product of atleast 2 cyclic ones

[cloud walrus]

All functions are continuous:

[wind steeple]

yes I think

bc you can say, if b is a bilinear form, b(x,y) = b(x,1) + ... + b(x,1) y times = b(x,1)y = (b(1,1) + ... + b(1,1) x times)y = b(1,1)xy

[covert vector]

owo

[upper pivot]

Uwu

[mild laurel]

Given some G that is torsion (so every element has finite order) and some subgroup H such that |G:H| = 5, show there exists an element of G with order 5

[covert vector]

index

number of cosets

of H

in G

[upper pivot]

Wait this is easy?

Bc 5| |G|

And then cauchy?

[mild laurel]

G isn't finite

Or at least, doesn't have to be

[upper pivot]

Oh

[mild laurel]

But you're right

[upper pivot]

Oof nvm im dum on that

[mild laurel]

Idk

[covert vector]

lol

[mild laurel]

This was on my final like four hours ago and I couldn't solve it

[upper pivot]

Hmm ill be back in a few. Will try it then

[covert vector]

hahaha John I thought u were gonna big brain it

with insta solve

:P

[mild laurel]

I didn't think about it

Have more finals tomorrow

Something I wrote down on the test is

[covert vector]

yeah i got it

[mild laurel]

You just want to show that for some g in G\H, that g^5 is the smallest power of g such that g^n in H

[covert vector]

i think I can do this problem

[mild laurel]

I mean, it was on a test, it shouldn't be too hard

[neon coral]

hello is there any good pdf about jordan canonical form calculation algothm?

[mild laurel]

It was 20% of the test

An hour and 45

There were 5 questions

I only needed a 75% to get an A+ so it's okay

[tribal pasture]

What course?

[mild laurel]

I got the other four

[covert vector]

[mild laurel]

http://www.bsmath.hu/CURRENT/AALsyllabus.html

@tribal pasture

Woogo think about my question

[covert vector]

i will

no assumptions on abelianness correct?

[tribal pasture]

Is this like a uni program or what?

[mild laurel]

Nope

It's a study abroad program

[tribal pasture]

Ahh okay

[mild laurel]

There's probably some clever group action here but the obvious ones don't work

Wdym

[upper pivot]

i assume the usual ones

like G acting on G/H

[mild laurel]

Just right multiplication and conjugation

Given some G that is torsion (so every element has finite order) and some subgroup H such that |G:H| = 5, show there exists an element of G with order 5

[solar vessel]

w8 right?

I would've thought you tried left

[mild laurel]

Idk, same thing

[solar vessel]

maybe something dumb like make an element that cycles the cosets

[mild laurel]

I'm not sure how though

Seems possible every element has order 2 with respect to the cosets or something

[rich patrol]

@chilly ocean I didn't see your ping, cause helper kinda masks it

[covert vector]

@mild laurel the action of G on the cosets induces a homomorphism(?) to S_5

either the image or the kernel should have an element of order 5

[chilly ocean]

Ues

Woog

It induces a homomorphism

To S_5

Its transitive

so it has order divisible by 5

[covert vector]

that should do it

[chilly ocean]

so by cauchy it has an element divisible by 5

[upper pivot]

yeah thats clever woog

[chilly ocean]

the order in the image of the homo has to divide the actual order

I just found a way to say homo

And not be banned

[upper pivot]

wait so formalizing the argument, theres a homomorphic map from G to S_5, wait so does it need to be surjective tho

[covert vector]

no

[chilly ocean]

A homeomorphism induced by a transitive action does not imply surjective

[covert vector]

homeo ≠ homo

despite e being identity

[chilly ocean]

HAHAHAHAHA

[upper pivot]

oh yeah i see your argument

cause both ker phi and im phi are groups

ok yeah thats very nice

and either must have factor of 5

[mild laurel]

I'm not sure why either the image or the kernel should have an element of order 5

[upper pivot]

counting formula

wait nvm i dont see how ker phi divides |S_5|

[solar vessel]

this is some black magic I have not seen

woog what did u use

[mild laurel]

@covert vector

[chilly ocean]

By Cauchy???

[solar vessel]

but

cauchy on what

[upper pivot]

if G is infinite, so is ker phi

so cauchy aint gonna cut it on that lol

[chilly ocean]

http://mathworld.wolfram.com/FirstGroupIsomorphismTheorem.html

[mild laurel]

Rudy I really don't know why you're just stating things we all know

[chilly ocean]

You dont know this

You just asked

“I'm not sure why either the image or the kernel should have an element of order 5”

[solar vessel]

w8 a second

[chilly ocean]

Which is

A direct implication

From FIT

[upper pivot]

no its not lmao

so im(phi) | S_5

but ker phi only divides G

so your thing doesnt work

cause im(phi) can literally just not have 5 as a factor lmao

(orders btw ofc)

[solar vessel]

(ofc)

[chilly ocean]

We

wait

What

We arent

Talking about

The kernel

[solar vessel]

rudy

what exactly do you mean then

[upper pivot]

yeah i dont think you know what you are talking about rudy

stop derailing us

[chilly ocean]

Did u read

What i said

Above

[solar vessel]

yes

lmao

[ripe basalt]

Has anyone written out the proof to it yet

[solar vessel]

no

[ripe basalt]

I feel like this would clarify

[upper pivot]

we havent solved it yet

lol

[ripe basalt]

It sounds like what rudy is trying to say is right.

[upper pivot]

it doesnt

[ripe basalt]

What?

[solar vessel]

woog might be writing it now

[chilly ocean]

Its transitive, order is 5. An element is then divisible by 5. That element correspond in G to some element whose order is div by 5

So some power

Of that element

Has order 5

[covert vector]

sorry just got to school

[chilly ocean]

Thats what I had sent???

[covert vector]

was walking from the bus station lol

yeah it doesn't work

[upper pivot]

the image of phi doesnt need to contain the element of order 5 in S_5

[covert vector]

was a nice try tho

[upper pivot]

yeah no it was clever, unfort didnt work

[solar vessel]

[covert vector]

I'm still gonna try again tho, it's a cool problem

well I thought the ker infinite was fine

but seems like it wasn't

[upper pivot]

do the B3 from this year if you want nice LA

[chilly ocean]

I dont understand

What ur not

[covert vector]

there was a cute A5 or A6 a few years ago

[chilly ocean]

Understanding from

The argument

John

[covert vector]

use more space bar pls

[upper pivot]

... i am understandint it fine, i pointed out whats wrong with it

[chilly ocean]

Whats wrong with it?

[covert vector]

@chilly ocean so the problems is

1. G can be infinite
2. in that case the kernel is necessarily infinite, so divisibility arguments don't work on it

[upper pivot]

the argument is that theres a phi such that phi is a homomorphic to S_5

G is*

[covert vector]

the kernel order isnt with respect to s5

[upper pivot]

now if a homomorphism maps to an elemnt of order 5, G must have order 5

however

[chilly ocean]

John

[upper pivot]

the image of phi can simply not contain the element of order 5

[chilly ocean]

Stop

Wait

Hold on

It does have to contain all of them

[upper pivot]

no, as you literally said, it doesnt need to be surjective

[ripe basalt]

Those two things aren't the same.

[placid pond]

@upper pivot every transitive subgroup of S_5 contains all elements in S_5 of order 5

amazing, isn't it

and the left action of a group on the left cosets of a subgroup is always transitive

[covert vector]

aw lit

we're good to go then

nice save 👍

[mild laurel]

Is that easy to show?

[placid pond]

it's called orbit stabilizer + cauchy's theorem

[upper pivot]

oh right ok that makes sense

yeah nvm my bad @chilly ocean

[solar vessel]

I mean if he meant that he could've like said

[chilly ocean]

Thats

Exaclty

What I did say

I explicitly said

By Cauchy above

[covert vector]

lol ok ok enough pointless arguing

[mild laurel]

Well for this, you only need to show that transitive subgroups contain one element of order 5?

[placid pond]

yes

they have order divisible by 5 because of orbit stabilizer

and they contain an element of order 5 because of cauchy

you contain one, you contain them all, since there's only 4 anyhow

[mild laurel]

Okay well while we're at it

Here's the other problem from my final I couldn't fully solve but this requires a little more knowledge

Consider G acting on a set S faithfully and primitively and consider a normal subgroup N of G such that |N| = 39. What is the size of S?

[chilly ocean]

@mild laurel is G solvable?

[mild laurel]

No?

[fringe nexus]

How do i show the sets of all elements of order 7 in A_7 is not a conjugacy class in A_7?

Is this something to do with odd and even permutations and how they relate to conjugacy classes in S_n

[upbeat burrow]

Couldn't you just find a counter example?

[fringe nexus]

are all elements in a_7 of order 7 7-cycles?

i guess? it seems rather cumbersome to look for one, i think theres a better way of doing this

if all elements in A_7 of order 7 are 7-cycles

I know the order of conjugacy class has to divide order of group

we have 6! 7 cycles in A_7

A_7 is of size 7!/2 which is not a multiple of 6! so by lagrange we are done

but idk if my first claim is correct

oh wait they are

because 7 is prime

i think

[mild laurel]

It's true that the only elements in S_7 of order 7 are seven cycles

The order of an element in just the lcm of the length of it's disjoint cycles

[fringe nexus]

yea and since 7 is prime it has to be just 1 7 cycle

[mild laurel]

Yeah

[fringe nexus]

so my proof is right i guess

[mild laurel]

Are there 6! 7 cycles in S_7?

[fringe nexus]

fix 1 element right

permute 6

[bleak abyss]

Fix first element to be 1

[fringe nexus]

like in your cycle notation

[bleak abyss]

So yeah that proof works actually, G acts transitively on a conjugacy class so this violates orbit-stabilizer

[chilly ocean]

@bleak abyss try Zophs

Current problem

Consider G acting on a set S faithfully and primitively and consider a normal subgroup N of G such that |N| = 39. What is the size of S?

[bleak abyss]

Hmm

So I thinking out loud a bit

N acts on S as well, now this action might not be transitive

If it's not we break S into orbits. The question is whether that would violate primitivity

Well, let's say nx = y

Well what about gx and gy?

gy = gnx = n'gx

So yeah

And the point is this action is faithful so this partition isn't into singletons

So actually N is a group of order 39 acting transitively on S

Well, the order possibilities are 1, 3, 13, or 39. Obviously 1 and 3 are out since we're faithful

[chilly ocean]

How did you get that?

[bleak abyss]

Group of order 39 doesn't embed into either S_1 or S_3

Ah well N also has a subgroup of order 13