groups-rings-fields

[void cosmos]

suppose gH = g'H

then phi(gH) = Hg^-1 = Hg'^-1 (why tho?) =phi(g'H)

so phi is well defined

[delicate orchid]

that "why tho?" is like 90% of the work

[void cosmos]

😡

[white oxide]

wtf

ohh ok i see

[delicate orchid]

map is natural as well :hesrightyouknow:

[white oxide]

what is the technical definition of well defined agian

i should know this jesus christ

[void cosmos]

its okay

[delicate orchid]

x = y => f(x) = f(y)

it's like backwards injectivity (which is f(x) = f(y) => x = y)

[white oxide]

$\cong G^{op}$

[cloud walrus]

okeyokay

[void cosmos]

i dont htink

u should know this

its just the dual category

[delicate orchid]

I tried to signal I was memeing by the 3 so trues

[void cosmos]

like take some structure then this is the structure of all functions from the structure to itself

think vector spaces

[delicate orchid]

and it's not it's the opposite group

and that's the dual space not the opposite category

[void cosmos]

ohh

yea makes sense lmfao

[delicate orchid]

there's no reason to explain it anyway I said it as a meme

[white oxide]

so why do you need to show that it's well defined?

[delicate orchid]

because it's not a function if it's not well defined

[void cosmos]

i didnt know why would you say this for this context but turns out ur exactly right lmao

[void cosmos]

u want to prove its a bijective function

[white oxide]

i thought it just suffices to prove injectivity and surjectivity

for a bijection

[delicate orchid]

if it's not a function it can't be a bijection

all bijections are functions, and all functions are well defined

[white oxide]

ah i see

so whenever you're constructing a function you want to show it's well defined first right

[void cosmos]

ye

[delicate orchid]

yeah, it's usually automatic but not when you're dealing with cosets - have to be a bit more careful

[white oxide]

huh ok i never did that ig i should just start doing that lol

okay cool

thanks a lot guys!

i have class rn but i'll be working on it later today but you guys helped a shit ton lol

really appreciate it

[delicate orchid]

just to give you an example of a map that isn't well defined, before you go

[white oxide]

okay sure

[delicate orchid]

if you consider the map f(gH) = g

and two cosets gH = g'H but with g not equal to g'

then f(gH) isn't equal to f(g'H) but gH = g'H

[white oxide]

ohhh okay i see

[delicate orchid]

so it's not well defined

[white oxide]

yeah because in my head like i was thinking x = y means that they are the same element

well yeah ig that's the case

but like physically

like as in 5 = 5

or something

but it's different for groups

[delicate orchid]

yeah which is why it's usually automatic

[white oxide]

didn't think about that

yeah

[void cosmos]

yea man most of the time no 1 does it

cuz most of the time its easy to not think of ill-defined shit

[delicate orchid]

if you're mapping objects that have a choice of representative you need to be careful (g, g' are the representatives here)

[void cosmos]

yea ^

[delicate orchid]

but details details

[white oxide]

cool thanks that really helps my understanding lol

[void cosmos]

its u have two things are the same but under some equivalence class

and the "function" would undo the equivalence class so they wont bee the same

thats lads example

[long geyser]

this is page 2 of the introduction, how far did you even look?

[round sandal]

I was trying to search with copying these symbols, but it found all sorts of other stuff also...

[chilly ocean]

lems is brutal

[void cosmos]

he might have skipped it

happens

[long geyser]

never claimed otherwise, but you still have to take the L

[white oxide]

lems is a cruel man

[void cosmos]

has a quite misleading pfp

haha

[coral shale]

quack

[toxic zephyr]

could we rephrase this last paragraph as "two permutations have the same cycle type if and only if they share an orbit through the conjugation action of the permutation group on itself"?

[void cosmos]

ye

[toxic zephyr]

that's neato

[void cosmos]

yea it is

u can now count shit using orbit stab theorem

once u think of conjugation as the action

cool

[toxic zephyr]

oooh. yeah so we could actually count how many permutations of a given cycle type there are using the orbit stab theorem huh? that's super neato

[void cosmos]

yea prove the original claim

using orbit-stab

no tricks u can d it

do*

two permutations have the same cycle type if and only if they share an orbit through the conjugation action of the permutation group on itself"?

[void cosmos]

oh didnt read whole msg

u would need to count the elemnts in a conjugacy class

for that u need to actually count them

using the good part of math

combinatorics

[sonic coral]

Can I get some help finding the conjugacy classes of D8? Im not sure I understand the definition, [a] = {xax' | x in G}, so a is always in its conjugacy class, but how do i determine what other elements are also in the class?

I have the center of D8 in their respective conjugacy classes since they are in their own class by themselves

would it be kinda like, (a^3)(a)(a^3)' = (a^3)(a)(a) = a^5 = a

so then a^3 and a are in the same class

[delicate orchid]

Yes that’s right

Wait no, sorry I misread

If you did xax^{-1} and got a^3 then a^3 would be in [a]

For some x

[sonic coral]

okay, so whatever i get after doing xax' for some x then goes into [a]?

[delicate orchid]

Yeah you’re just seeing what’s in the set

[sonic coral]

does x have to be different than a

[delicate orchid]

You can do a lot of the conjugation at once btw, for example a^n will commute with a^m for all n, m

[delicate orchid]

No but think about what happens when x = a

[sonic coral]

youre just gonna get a back

since the right two terms collapse

[delicate orchid]

Exactly

[sonic coral]

so a^2 being in the center, the claim is that x(a^2)x' will always give a^2

which is obvious since a^2 commutes with everyone

how do i know when to stop, like if i check ba(b)' = bab = a^3b^2 = a^3, then i know that [a] = {a, a^3}, is it necessary to check all the other 6 non identity elements in D8 to see what else goes in this set?

or is that an experience thing

or maybe just checking one other element and getting something already in the set is sufficient?

[delicate orchid]

well you know that if ax = xa then you have xax^-1 = a

so all powers of a you can rule out pretty quickly (for conjugating a, you'd still have to check for b)

[sonic coral]

how would this apply to D8 then, I would not need to check a^3 after i check a^2 to see if its in a's class?

[white oxide]

for injectivity would you just suppose phi(gH) = phi(g'H), so Hg^-1 = H'g^-1 so by the left cancellation law we see that H = H'?

and how woudl you denote this bijection

$\phi: gH \mapsto Hg$ for any $g \in G$?

[cloud walrus]

okeyokay

[white oxide]

like what is the symbol for the set of all left coset groups ig

[white oxide]

im not entirely sure why other than using the hypothesis to cancel out H from both sides sto obtain g = g' but idk if tha tworks

[delicate orchid]

gH = g’H implies that g = g’h for some h in H

[white oxide]

oh wait can't you just take the identity in H

so like in particular ge = g'e

so g = g'

would that work

[delicate orchid]

No because that’s absolutely not true and isn’t what I said

[white oxide]

are you being sarcastic

[delicate orchid]

No

I didn’t say you could pick any h you want, I’m telling you there is some h such that g = g’h

[white oxide]

how did you get to g = g'h

[delicate orchid]

Sorry you’re right I should be clearer

gH = g’H implies that g’^-1gH = H if and only if g’^-1g is in H, say g’^-1g = h in H, then g = g’h

[white oxide]

ah gotchu

oh wait

so then if gH = g'H

then g'hH = g'H

and since h in H

my b

im being stupid

i have no fucking clue how to do this

[delicate orchid]

Hg’^-1 = H(gh)^-1 = Hh^-1g^-1 = Hg^-1

I’m guessing that’s what you were stuck on

[tawdry crystal]

I’m having trouble visualizing what this Lie algebra looks like

For p+q=n

[white oxide]

ah ok makes sense

thank you

[tawdry crystal]

What properties does an element of this algebra have?

[white oxide]

how did you get to g'^-1 = (gh)^-1?

[delicate orchid]

Boss

Oh ok, just move the apostrophe over w/e same deal

[white oxide]

but g' = h^-1g so Hg'^-1 = H(h^-1g) which is not equal to (gh)^-1

[delicate orchid]

Dude you literally just do the exact same thing except you do gH = g’H => H = g^-1g’H instead of gH = g’H => g’^-1gH = H
I’m very concerned that you’re not able to see the symmetry in the proof, we’re literally just swapping g’ and g

Don’t mean to be insulting, I’m just saying maybe go back and revise certain things with cosets?

[main needle]

If $M$ is the $R-$module over a PID defined as $M = R/(p^2) \oplus R/(p)$ and $N$ is a submodule of $M$ such that $T(N) \subseteq N$ for any $T$ module automorphism. Is there a classification of $N$ ?

[cloud walrus]

ru0xffian

[south patrol]

(If you want an example of a non-abelian group of order p^3, consider the upper triangular matrices over Z/pZ)

[formal ermine]

heisenberg group my beloved

[agile burrow]

You can rewrite this as trying to show that conjugation by a commutator is trivial

[agile burrow]

More specifically, you have a map G -> Inn(G) sending an element to conjugation by that element. You want to show that the image of a commutator is the identity automorphism

But Inn(G) is isomorphic to G/Z(G). Now you can use something about the structure of p groups to reach your conclusion

[delicate orchid]

xyzyx = yxzxy <=> x^-1y^-1xyzyxy^-1x^-1 = z <=> [x,y]z[x,y]^-1 = z

typo central today jesus

[delicate orchid]

yeah nvm somethings not adding up here, walter could u pretty please write out the transformations explicitly

nvm I'm just double skill issuing, it works

[agile burrow]

I'm making sandwich

[delicate orchid]

my excuse is I don't have my glasses on and discord is blurry

~~of course you could always cheat and just classify the p-groups of order p^3 and then check this holds in all 5 of them ~~

[coral shale]

Can someone link me to a decent reference that describes how to compute determinant using elementary row ops; I can't find any. In particular, can't quite remember which row op might change the det

[delicate bloom]

you can write the elementary row ops as elementary matrices, so matrix A becomes EA let's say. Then you can see if it changes the determinant by seeing det(A)=det(EA)=det(E)det(A) forces det(E)=1

[coral shale]

guh lazy

half dead, why can't i find any online resources which describe this decently

[chilly ocean]

any linear algebra book?

[sage lodge]

https://industri.fatek.unpatti.ac.id/wp-content/uploads/2019/03/037-Elementary-Linear-Algebra-Applications-Version-Howard-Anton-Chris-Rorres-Edisi-1-2013.pdf

2.2

[coral shale]

oof thats hefty. ty

[chilly ocean]

i think mero did a decent job

[delicate bloom]

elementary matrices have dead easy determinants to compute tbh lol

[coral shale]

not as dead as me rn

[coral shale]

it came up cus i was helping and almost gave wrong info

[sage lodge]

best introductory linear algebra book imo

[delicate bloom]

when you're alive I expect you to describe what the elementary matrices look like for a,b,c @coral shale

[coral shale]

glub glub glub

[vagrant zinc]

[coral shale]

https://i.imgur.com/OZglohA.png

catching me in a moment of weakness

can't say it's obvious to me why the last det is 1 other than it's a 'shear'

[agile burrow]

p-groups have non-trivial center

Right

If G has non-trivial center, then it has order p, p^2, or p^3, right?

Then G/Z(G) has order 1, p, or p^2, hence Inn(G) does as well

Groups of such order are necessarily abelian

Yes

And it follows that the map G -> Inn(G) sends commutators to the identity map

hence conjugation by a commutator is trivial

[delicate bloom]

seems legit enough to me 😎

[agile burrow]

The map G -> Inn(G) sends g to \gamma_g, where \gamma_g(x) = gxg^-1

it doesn't really matter

because you can always just consider x^-1

that's why some people write commutators as xyx^-1y^-1 and others write x^-1y^-1xy

But hopefully that proof makes sense to you, it's a fun exercise

sure, I might make another sandwich but i'll respond when I get a chance

[agile burrow]

you said |G| = p^3

What did you think a p-group meant?

[rotund aurora]

walter yellow, nice

[agile burrow]

I see

Yeah, for a finite group, an equivalent condition is that the order of G is a prime power

Tbh I forgot infinite p-groups are a thing lol

Happy to help

Yeah, my first intuition when I read the problem was to rewrite the equality as a conjugation thing

Probably just because I've been working with conjugation stuff a lot 😵‍💫

I'll let you know if I think of another solution, but this is the most straightforward way I can see

mhm

The point is that the map G -> Inn(G) is a group homomorphism, right?

And Inn(G) is abelian

Let's give our map a name, say f

Then for a commutator in G, we have f([x, y]) = f(xyx^-1y^-1) = f(x) f(y) f(x)^-1 f(y)^-1

Since Inn(G) is abelian, we can rearrange the terms and it all cancels out to the identity

The identity element in Inn(G) is the identity automorphism

By definition, f([x, y]) is an automorphism of G which sends z -> [x, y] z [x, y]^-1

But since f([x, y]) is equal to the identity map, this means [x, y] z [x, y]^-1 = id(z) = z for all z in G

Right, so conjugation by a commutator is trivial. In particular this means that [x^-1, y^-1] z [x^-1, y^-1]^-1 = z

[x^-1, y^-1]^-1 = [y^-1, x^-1]

oh wait

did i mess this up lol

[delicate orchid]

is this what you're after walter?

[agile burrow]

ah ok

yeah it's just tracing back through that

is that the inverse of the commutator? i thought you just switch whatever you're commutating

[delicate orchid]

well no don't guess - I think there is a problem

[agile burrow]

yeah I messed it up kek

[delicate orchid]

[x, y]^-1 = (xyx^-1y^-1)^-1 = yxy^-1x^-1, which is the thing on the right hand side of the z

so no there isn't a problem

jesus why am I struggling so much with this

[agile burrow]

but on the left side you wrote [x, y] = x^-1 y^-1 x y

[delicate orchid]

ah

uh oh!!

[agile burrow]

i'm devastated

[delicate orchid]

so what we have is like
[x^-1, y^-1]z[x, y]^-1

[agile burrow]

that's right

[delicate orchid]

very interesting

[agile burrow]

ok we can remedy this

can we bound the size of the commutator subgroup/say it's contained in the center?

[molten viper]

J/I is just the set {j + I | j \in J} Yeh? J and I ideals

My memory is fuzzy and I don't have my notebook on hand atm

[chilly ocean]

i can't think of anything else it would be

[delicate orchid]

it's definitely a proper subgroup, as p-groups are solvable

[agile burrow]

right, so the commutator subgroup is abelian

[delicate orchid]

not too useful here I don't think

[agile burrow]

mmm

maybe time for another sandwich

[delicate orchid]

we also have the abelianization is abelian

[agile burrow]

Ok I got it. Recall that Z(G) is non-trivial. If |Z(G)| > p, then G is abelian and the result holds trivially. Otherwise, |Z(G)| = p, so G/Z(G) is abelian. In particular, [G, G] is a subgroup of Z(G), so every commutator is the unique element in its conjugacy class. But [x, y] is conjugate to [x^-1, y^-1] so they must be equal and our earlier solution holds

[delicate orchid]

missed the chance to say "abelian by the 5/8ths theorem" walter, blunder

[chilly radish]

Prove that a group of order p^2 is abelian

[agile burrow]

If |Z(G)| = p^2 then G/Z(G) is cyclic. This implies that G is abelian (you should prove this if you haven't before)

[delicate orchid]

actually wait nvm the centre doesn't have the needed density

[agile burrow]

It's not difficult but it's a good exercise in understanding quotient

Wait, which part are you confused about?

[chilly radish]

Personally it makes more sense to me if we rearrange it as
x'y'zyx=y'x'zxy
Then it's exactly a statement about conjugation by any 2.elements commuting

'=^-1

Than to state it in terms of commutators

[agile burrow]

fair, i just wanted to adapt the earlier proof to work for this situation

Oh wait that makes it so much easier kek

Because we already showed Inn(G) is abelian

[chilly radish]

Exactly lol

[agile burrow]

more sandwiches

[molten viper]

So, I want to show that there is some n where x^5 - x + 1 is irreducible in Zn[x]

how might I go about that

It's not enough to show that it doesn't have any roots, as the degree is greater than 3

I need to somehow account for the case where we factor it as a degree 2 and degree 3 polynomial

[rotund aurora]

mod 3

[molten viper]

Hm.

That seems like a hard thing to prove

[rotund aurora]

to prove that x^5-x+1 is irreducible mod 3?

Not really

There are finitely many irreducible polynomials of a given degree in Zn[x], so you can just check all the possibilities

[molten viper]

That's a fact I'm not really familiar with

[rotund aurora]

there are finitely many polynomials of a given degree in Z/nZ[x]

[molten viper]

I also don't know the irreducible polynomials of degree 3 or 2 in Z3[x] off the top of my head

[rotund aurora]

what I said is not the the best way to do that tho

So you know a quintic, if reducible, it either has roots or factors as a quadratic and a cubic

So first, show it has no roots

[molten viper]

Done on the first count

[rotund aurora]

Then, to show it doesn't factor as a quadratic and a cubic, pretend it factors as such, you obtain a system of equations, and hopefully you can derive some relations

[molten viper]

ok that's what I figured I ought to do

I'm gonna go do that

[rotund aurora]

If you wanted to do what I said, you just have to check divisibility by quadratics right

There are 9 monic quadratics (including reducible ones) in Z/3Z[x]

And you can easily derive a criterion for irreducibility of quadratics in general

so there are not that many

so you would have to do euclid division a few times if you decided to do that, but at least you see that the problem is 100% doable

[molten viper]

this is made a lot easier by the fact that Z3 is a field innit

Because every non-zero element is a unit we can assume it factors as monics

[rotund aurora]

your original polynomial is a monic

You can always assume the factors of a monic polynomial are monic, in a field or not

[molten viper]

mmm, I'm not convinced in the case of a non-field

[rotund aurora]

(assume an integral domain at least)

[molten viper]

I'd have to play around with it

[delicate bloom]

there's a trick you can do for Z/5Z since if r is a root so is r+k for k in {0,1,2,3,4}

then you can look at factoring it as a product of those roots and look at the coefficients, particularly the sum of roots

[molten viper]

This is a way to discover I have no idea how to handle modular systems of equations

[rotund aurora]

in cases like this, just do what you would do with a system of equations over Z or Q, or something like that

but when you see a 5 or something bigger than 5, reduce it

sorry should be 3 if you are doing modulo 3

but you see what Im saying

[molten viper]

yeah

I wanna find some impossible equation like 1 = 0

what a friccen mess

[rotund aurora]

its not

Suppose $x^5-x+1=(x^2+ax+b)(x^3+cx^2+dx+e)$, multiply everything out. If you have two polynomials that are equal, then the coefficients must be equal

[cloud walrus]

Croqueta

[rotund aurora]

You get relations immediately

[formal ermine]

I did this in my exam out of desperation then realized there is an easier way. can't recommend 3/10

[delicate bloom]

to elaborate on this, once you have $$x^5-x+1=\prod_{k=0}^4 (x-r-k)$$ then if you were to factor it, then you'd get some polynomial with coefficients on the second highest term as the sum of roots, with coefficient $ar+b$ with a in {1,2,3,4} but since r is not in Z/5Z, you have a contradiction

[cloud walrus]

Merosity

[molten viper]

Yeah 5 non-linear equations in 5 variables is not fun

[delicate bloom]

that proves it's irreducible

start with f(x)=x^5-x+1 and show that if r is a root of f, then r+k is a root of f in Z/5Z

[molten viper]

I’ll probably set this one aside and work on another problem

My other problem is figuring out the elements of Z2[x]/(x^2 + x + 1)

[formal ermine]

you remove everything that is degree >= 2

[molten viper]

I’m not so sure I can write x^3 = g(x) (x2 + x + 1)

Just for example

[formal ermine]

-x^2 = x + 1
x^3 = x^2x = -(x + 1)x = -x^2 - x = -(x + 1) - x = -x - 1 - x = -1

wait

I messed up the -

there

[molten viper]

Huh???

[formal ermine]

you set x^2 + x + 1 = 0

so -x^2 = x + 1

[molten viper]

I guess that makes sense, but how do I prove that for all polynomials

[formal ermine]

for any monomial x^n with n >= 2 you have x^n = x^(n - 2)*x^2 = -x^(n - 2)(x + 1)

which is degree < n

then by induction everything with degree >= 2 turns into degree < 2

[molten viper]

Fair enough

[delicate bloom]

you can actually solve both problems the same way 😛 write them as f(x)=x^p-x+a for a != 0 mod p

[molten viper]

So my intuition was right, just not my method

[sonic coral]

,rotate

[formal ermine]

lol who tex'd that

[sonic coral]

in the example, doesn’t sylows first theorem also give the existence of atleast one subgroup of order 3 and 3^2?

[rotund aurora]

the devil

[sonic coral]

you should see the handouts my professor gives me lol

[rotund aurora]

some of my profs are latex noobs too, so annoying

[formal ermine]

it depends on your definition of sylow's first lol

some versions of it only say maximal p subgroup

some say p subgroup of any multiplicity

[sonic coral]

i think this is any multiplicity

which means there would be atleast one of order 3 and 9

[formal ermine]

no

it says gcd(whatever) = 1

meaning that the multiplicity is maximal

[sonic coral]

oh okay i see that, thanks.

[formal ermine]

though there also exists a stronger statement of this

[sonic coral]

yeah i can infer that from what you alluded to

[formal ermine]

if p^k is a divisor of |G| then G has a subgroup of order p^k

yeah

[sonic coral]

something tells me that it does

[formal ermine]

k >= 0 and p prime before anyone shouts at me

[sonic coral]

k=0.1

thank you though, just trying to read between the lines a little bit

[formal ermine]

[rotund aurora]

I'm having trouble with this

You can easily show that the right side divides the left, and that both sides have the same degree in each of the coordinates

So res(f,g) is a scalar multiple of the thing in the right

But how to show that that scalar must actually be 1?

[chilly radish]

What's res(f,g)

What is it deginrf as

[rotund aurora]

Classic resultant of two polynomials

[rotund aurora]

I dont understand what you mean here

[chilly radish]

Defined*

[rotund aurora]

ah ok, so that

and the fundamental result is res(f,g)=0 if and only if f and g have common factors of positive degree (at least over a UFD)

[molten viper]

So

One more problem

I have I = (p, x^2 + 1) as my idea

Ideal

I wanna show that Z[x]/ I is isomorphic to Zp[x]/ (x^2 + [1]p)

I think this is just about choosing the right J and using an isomorphism theorem

Specifically J = pZ[x]?

[rotund aurora]

by [1]p you mean 1 mod p ?

This is just one of the isomorphism theorems, no need to do anything fancy

Z[x]/p is Zp[x], and now you quotient (x^2+1) in Zp[x]

[molten viper]

Yep

[rotund aurora]

so you obtain Zp[x]/(x^2+1)

if p = 1 mod 4, then x^2+1 has roots in Z/pZ, and so Zp[x]/(x^2+1)=Zp

[molten viper]

What do you mean by Z[x]/p?

Like the ideal generated by p?

[rotund aurora]

Z[x]/(p), (p) is the principal ideal generated by p in Z[x]

yes

[molten viper]

Not the exact idea I had in mind

I’m using the one that goes

R/I ≈ R/J/I/J

Sorry for the notation

[rotund aurora]

Are you saying this? Because its what I said

https://en.wikipedia.org/wiki/Isomorphism_theorems

[molten viper]

Ok good

Yeah that’s what I’m saying

[rotund aurora]

Essentially, the point is that the order in which you kill stuff doesn't matter

[rotund aurora]

Ok so I think the point is that the monomial ((-1)^nx1...xn)^m is the product of the diagonal, and that's the only place it appears. Since the coefficient is one, and it also appears in the right hand side, the coefficient must also be one

[rustic crown]

https://tenor.com/view/totodile-pokemon-gif-25022266

that's Croqueta

[rotund aurora]

[cloud walrus]

c squared

c squared

[kind temple]

i think i' should be the map taking x to (i(x),0) but i cannot figure out a map pi' which makes this SES work

i dont think projecting onto N works for pi' since we need the kernel of pi' to be im(i) x 0

[oblique river]

(I think that does work)

[rustic crown]

yea

[kind temple]

ig im not seeing why?

it feels like the kernel of pi' would be larger than the image of i'

[oblique river]

Well the trick is to prove that that’s not the case :P

[rustic crown]

(btw are you cat minded? )

[kind temple]

hmm. ill try that

i think i prefer cats to dogs?

[rustic crown]

this is true a little more generally

[oblique river]

Ok i gotta go, det you can take it from here

[rustic crown]

kernels in a pull-back square are isomorphic :3

and since you're pulling-back an epimorphism, it stays an epimorphism

which is how most people think about your question :p

but yee, we'll do it with elements now

[rustic crown]

can you write down your thoughts so we see where it goes weird?

[kind temple]

yea, so if i take y = (i(x),0), then projecting gets me to 0 so we're in the kernel of pi'

[rustic crown]

yee

[kind temple]

but if i take (x,y) in the kernel of pi'

then all ik is that y = 0

and x in M_2

[rustic crown]

nope

you know more because of the definition of M2bar

:3

[kind temple]

oh

pi(x)=0

[rustic crown]

and now?

[kind temple]

x is in the image

: )

[rustic crown]

:3

^w^

[kind temple]

thank you det : )

[rustic crown]

(btw you also need to verify a few more things)

like why pi' is surjective

but i'm sure you can do that

and also that i' in injective

[warm shoal]

Question

P2 = {empty set, {x}, {y}, {x, y}}

are Z4 and P2 isomorphic? I think not, but I'm not sure how to show it.

I showed an additive counterexample:

f(2) = f(1+1) = f(1) + f(1) = {x} + {x}, which isn't even in P2

Therefore f isn't bijective, so Z4 and P2 cannot be isomorphic

Right?

[rustic crown]

how do you define adding two sets?

[warm shoal]

I'm not sure ={

[rustic crown]

only reasonable operation is the symmetric difference

but then it's iso to Z/2Z⊕Z/2Z

see the exercise C on how you define the operation

[warm shoal]

A + B = (A-B) U (B-A)

[rustic crown]

yee that's called symmetric difference

usually denoted with a upper case delta

[warm shoal]

ahah

now we get this

[rustic crown]

try proving this

[warm shoal]

f(2) = f(1+1)=f(1)+f(1)={x}+{x} = empty set

= f(0)

therefore f is not bijective

so Z4 is not isomorphic to P2

😄

?

[rustic crown]

yep

[warm shoal]

awesome 🙂

now we proving things are NOT isomorphic

is additive operation

always a good one to try?

[rustic crown]

just one thing

[warm shoal]

yes

[rustic crown]

you don't know f(1) = {x}

[warm shoal]

i don't?

[rustic crown]

if you wanna do this argument, you'll have to say this generally

but since A Delta A = empty

it follows

[warm shoal]

how do i say it generallty

generally

[rustic crown]

f(2) = f(1+1) = (f(1) \ f(1)) u (f(1) \ f(1)) = empty

\ is the set minus

f(1) = A can be anything

it has 4 options right

why did you directly assume it was {x}

[warm shoal]

well now im confused because

f(2) could be empty right?

so what does that show

@rustic crown

[rustic crown]

okie so how much group theory do you know?

if you just know the definitions, then the way you're is the nicest way

but you should elaborate why you know stuff

[warm shoal]

we know the definitions

and don't need to elaborate much from here

but okay, we showed (f2) = empty

[rustic crown]

yep

[warm shoal]

but you're saying that could be valid?

[rustic crown]

and teh same logic shows that f(0) is also empty

[warm shoal]

like what if f(2) actually maps to empty

[rustic crown]

f(2) = f(1 + 1) = f(1) Δ f(1) = {}

f(0) = f(0 + 0) = f(0) Δ f(0) = {}

so such a homomorphism f can never be injective

[warm shoal]

ah yes brilliant 😄

now @rustic crown

for Z4 and V

they seem pretty isomorphic upon first inspection lol

HOWEVER

proving isomorphism seems hard

any tips for this?

cause thus far we've just been finding counterexamples

[rustic crown]

define an explicit map

Z/4Z --> V

and use properties of exponents to argue this is a homomorphism

this would be nice and bijective

so will be an iso :3

[warm shoal]

can u elaborate on this?

[rustic crown]

try thinking about the function

[warm shoal]

we don't let about homomorphims until chapter 14

we're on chapter 9 my good brother ❤️

[rustic crown]

oh

weird

so you directly define isomorhpism?

but not homomorphism?

[warm shoal]

yes

lemme show ya

[rustic crown]

yee don't worry

just argue it's an iso

[warm shoal]

here we go!

[rustic crown]

but first you need to come up with a nice function

if f satisfies the identity 3, it's called a homomorphism

[warm shoal]

ahhh gotcha

okay

alright

time to first make an educated guess

let's use the additive operation

does that sound good?

@rustic crown

[rustic crown]

oh oops

det playing a game >.<

yee find a nice function

[warm shoal]

hey it's all good

@rustic crown also can i just say that

any two groups

of a fixed order

than are also cylic

are isomorphic?

[rustic crown]

yep

[warm shoal]

yay ")

🙂

is that what u normally use

[tender wharf]

i mean that alr works

if G,H are cyclic

and have the same order

they are isomorphic lol

[warm shoal]

V denotes the group of the four complex numbers {i, −i, 1, −1} with respect to multiplication.

Z4 = {1, 2, 3, 4}

Z4 is obviously cyclic

Though I must show that V is too

Any ideas?

[stone hamlet]

what does it mean for a group to be cyclic? start from the definition and see if any aspects of your set satisfy that definition

[summer path]

Considering it's a small set, you could kind of get away by just computing the order of each element if you can't just see which one generates the group

[warm shoal]

computing the order of each element u say

[summer path]

Each element of V

[rustic crown]

gomen

det is back

[warm shoal]

okay guys

this is all new to me

so im taking everything your saying into account

3. every group of order four is isomorphic to Z4 or Z2 × Z2 = V4.

OOPS

[rustic crown]

yee

[warm shoal]

so we have this right

so i just need to prove that

V is cyclic

and that the order of some element of G

is some finite number

correct?

[summer path]

Note that g here is specifically the generator of G, so it's not just some arbitrary element of G

[warm shoal]

gotcha okay

well V = {i, -i, 1, -1} w/ respect to multiplication

and I want to show that it's cyclic, aka generated by a single element g

but im not sure how to show that

[summer path]

So V has 4 elements and we want to show that one of them actually generates the entire group

[warm shoal]

yes

i know i x i = -1

i x 1 = i

i x -1 = -i

i x -i = 1

=[

[summer path]

Ok so recall that $\langle g \rangle = {g^n : n \in \mathbb{Z}}$

[cloud walrus]

Tubular Cat

[warm shoal]

yes

and g in G right?

[summer path]

Yes

So try to compute <i>

[warm shoal]

TRUE!

wooooo

V is cyclic because V = <i> and i in V

[charred crescent]

is it true that if H is a subgroup of a finite group G, and the cardinality of H equals the cardinality of G, then H = G?

i don't see how it couldn't be the case, but i can't prove it

[delicate orchid]

Forget groups, that’s true for subsets of finite sets too

[tacit hemlock]

can someone please explain why $$g^{3} = g^{11}$$?

[cloud walrus]

normalAtmosphericPa=101,325

[delicate orchid]

I’m guessing g has order 8

[charred crescent]

^

[summer path]

So you have V is cyclic and a generator of order 4, so you can apply your theorem to conclude

[delicate orchid]

Or 4, or 2!! But probably 8

[tacit hemlock]

What is an intuitive understanding of the "order" of an element of a group?

srry, this is all very new to me

[summer path]

g in G has order n if n is the minimal positive integer such that g^n = 1

[tacit hemlock]

very intuitive

[charred crescent]

the number of times you need to operate the element with itself to get the identity

[warm shoal]

now, how does the generator have order 4

[delicate orchid]

Has order n if g^n = 1 and there isn’t a m < n where g^m = 1

[warm shoal]

here's our definition of generator my good brother

[delicate orchid]

Alternatively it’s the size of <g>

[warm shoal]

here g is i, no?

[summer path]

You should have been able to see that i^2= -1, i^3=-i, and i^4=1

[tacit hemlock]

I just finished single var calc

I'm used to having 2d and 3d graphs for intuition

: )

[warm shoal]

but i^0 = 1

[summer path]

Yes hence minimum positive integer

[warm shoal]

yes

so where does the 4 come from

[delicate orchid]

i^4=1…

[warm shoal]

yeah but so does i^8

so is this of order 8?

im confused af =[] sorry!

[delicate orchid]

Quite literally in the definition that it’s the smallest number

“Least positive integer”

[warm shoal]

0 is the smallest number

i^0 = 1

right?

[summer path]

0 is not positive

[warm shoal]

oh

im clown

sorry!

[delicate orchid]

Rip -1

[warm shoal]

lmao

im sorry fellas!

[summer path]

xd

[warm shoal]

im not the sharpest tool here

but u guys are so helpful

[delicate orchid]

And rip 1 in the wrong way ordering

[warm shoal]

i have lots of hours left of this homework

hope u stay tuned!

[delicate orchid]

No you’re just new it’s fine

We were all in your position at some point

[warm shoal]

bless ❤️

now i gotta check if Z2 x Z2 is isomorphic to P2

P2 = {empty set, {x}, {y}, {x, y}}

[delicate orchid]

With what operation lol

Cause that looks more like a topology

[warm shoal]

true

well

to show that Z4 and P2 are NOT isomorphic, we used the symmetric difference operation as an example

[summer path]

I guess the most intuitive way you can think about this (which only really works for cyclic groups) is say ord(g) = n, then each time you multiply g to itself you (start at some point 1 and) move some angle the way around a circle and n is the smallest number of iterations to return to where you began

[warm shoal]

so perhaps i should go the same route?

though that is all the instruction that the problem has

[warm shoal]

what should i do about that?

[delicate orchid]

See chapter 3 exercise C

[warm shoal]

all it says is the symmetric difference

like with set addition

A+B = (A-B) U (B-A)

[delicate orchid]

Ah ok yes that makes sense

[warm shoal]

i take it these 2 are not isomorphic

in which i can hit a little counter example

[delicate orchid]

Lemme think

yup, see if you can figure out why they’re not isomorphic though

There’s a property (relating to the order of elements!) that Z4 does have but P2 doesn’t

[warm shoal]

something to do with being abelian or cyclic?

[summer path]

*note that isomorphisms preserve the order of elements

[warm shoal]

what does the order of elements even mean

like

for g in G

|g|?

[summer path]

This

[warm shoal]

okay

let me try to find the orders

of each of them

but that's only if

the groups are cyclic

so i first gotta show that is what you're alluding to?

[summer path]

You can talk about order of elements even if the group is not cyclic

[summer path]

This might also be a good exercise

[warm shoal]

bet okay

[warm shoal]

we're talking

Z2 x Z2

and P2 rn

@summer path and @delicate orchid

[delicate orchid]

Yes it’s Z2xZ2

Because every element in P2 (that isn’t the identity) is order 2, but there’s an element in Z4 with order 4, so they can’t be isomorphic!

[warm shoal]

by Z4 do u mean Z2 x Z2?

[summer path]

no

[delicate orchid]

No I mean Z4

[warm shoal]

oh

[delicate orchid]

I was referring back to this

[tender wharf]

these are not isomorphic

[warm shoal]

correct i agree

[delicate orchid]

I thought you were trying to show it wasnt isomorphic to Z4 as opposed to showing it was isomorphic to Z2xZ2, apologises

[tender wharf]

oh I got confused too my bad

[warm shoal]

oooo well wait

[summer path]

(in fact any group of order 4 is isomorphic to Z2 x Z2 or Z/4Z)

[warm shoal]

p2 iso Z4

Z4 not iso z2 x z2

ah nvm

i am trying to show that

z2 x z2 and P2

are not isomorphic

[delicate orchid]

But they are isomorphic

[warm shoal]

oh shoot

okay im sorry

so what's the best way of showing that they are indeed isomorphic

there just seem like so many options

this is crazy

[delicate orchid]

This is a good exercise, also G_1 iso to G_2 <=> G_2 iso to G_1

[summer path]

it seems to me that you are having a hard tiem keeping track of which ones are/aren't isomorphic: it might help to write down the 4 groups in considering in a kind of rectangle shape and draw arrows between them to indicate whether you have shown they are/aren't isomorphic, just as way to keep track of what you have done/need to do

[delicate orchid]

Construct an isomorphism between them, at your stage that’s basically the only way

So find a bijection f from P2 to Z2xZ2 (or the other way around if you prefer) such that f(ab) = f(a)f(b)

It might be useful to write out the Cayley tables of both the groups (if you know what a Cayley table is , don’t worry if you don’t)

[warm shoal]

damn this is tough

[summer path]

i remember writing out an 8x8 Cayley table before, it was not a fun time

[delicate orchid]

I’m guessing D_8?

If it’s Q_8 I will cry

[summer path]

i don't remember tbh

now that you mention it, maybe i've done it twice

xd

[warm shoal]

okay

with z2 x z2 and P2

any tips?

which set should I make the domain?

which operation should I try?

just seems like a black box to the boy rn

[delicate orchid]

What do the elements of Z2xZ2 look like

[warm shoal]

(a, b)

(0, 0), (1, 0), (0, 1), (1, 1)

[delicate orchid]

You may see the connection now?

P2 = {{}, {x}, {y}, {x,y}}

[warm shoal]

wow!

they look so similar ahahah

look at that!

[delicate orchid]

So what’s your idea for the map between them?

[warm shoal]

i want the simplest little thing like

last time with P2 we used symmetric difference

which works well with the sets

and since these look so similar

maybe try that

and also since they're so similar

[delicate orchid]

The actions on the sets are fixed we can’t change that

[warm shoal]

doesn't matter which is domain and which is codomain

ooo

[delicate orchid]

If we change the action we might get a different group

[warm shoal]

gotcha

well

maybe some simple addition?

[delicate orchid]

Like I’m literally just talking about a function, f(a,b) = something in P2

[warm shoal]

well they're so similar like

f(a, b) = something in P2 +- 0

yo 💀

[delicate orchid]

I don’t think you’re really understanding

so I’ll help you out a little

f(0, 1) = {x}

[warm shoal]

yes

that's what i ideally want correct?

f(0, 1) = {x} and only {x}

[delicate orchid]

Functions can only have one output

[warm shoal]

right

[delicate orchid]

that’s automatic

[warm shoal]

well idk how to map f(0, 1) = {x}

like

[delicate orchid]

I literally just did

[warm shoal]

if this was coding i'd base it on index

ahahah but what's the operation

[delicate orchid]

f is a function that takes (0,1) to x

sorry, {x}

[warm shoal]

so

but u said find a bijection

like do i literally just write

[delicate orchid]

try and figure out what f(1,0) should be

[warm shoal]

f(0, 0) = empty set, f(1, 0) = {x}, etc.

[delicate orchid]

Yes! f(0,0) has to be the empty set as the identity has to be mapped to the identity by a group homomorphism

so that’s 2 out of 4

[warm shoal]

and f(0, 1) = {y}, f(1, 1) = {x, y}

boom

bijection?

[delicate orchid]

It should be easy to see this is surjective and injective

[warm shoal]

yes

then f(0, 0 * 1, 0) = f(0, 0) * f(1, 0) = empty set * {x}

[delicate orchid]

So now we just need to check that f(ab) = f(a)f(b) for all a, b in Z2xZ2

Ah you’ve already started

[warm shoal]

lol

do we need to check each case?

that seems a little redundant right

also we didn't define an operation

i just said *

isn't that too ambiguous

cause a function needs an operation, no?

[delicate orchid]

We did define an operation on both groups, standard multiplication in Z2xZ2 and symmetric difference in P2

[warm shoal]

so now i j check all the cases?

[warm shoal]

righttttttttttt

[delicate orchid]

Yeah so for example
f(0, 1)+f(1, 0) = {x}+{y} = {x, y} = f(1, 1) = f((0,1)+(1, 0))

[warm shoal]

gotcha okay 🙂

[delicate orchid]

Good luck

[warm shoal]

yay did it

thank you 🙂

now Z2 x Z2 and V

i should just go about it same way yeah?

tho

these dont seem isomorphic!

or i could do

P2 and V

im tryna get to where i have enough isomorphisms to start making the connections u know

[delicate orchid]

I thought you’d already shown Z4 was iso to V

[warm shoal]

yes

[delicate orchid]

And Z2xZ2 isn’t iso to Z4… so

[warm shoal]

oh shoot

so if

G1 iso G2 and G2 not iso G3, then G1 not iso G3?

[delicate orchid]

Yup

cause if Z2xZ2 was iso to V, since V is iso to Z4, that would imply that Z2xZ2 is iso to Z4 - a contradiction!

[warm shoal]

yay

now we're done

[sonic coral]

i’ll wait

[warm shoal]

thank you @delicate orchid @summer path

u guys are the best

[delicate orchid]

No worries, I’m going to bed now cause it’s 2:30am

[warm shoal]

❤️

sweetest of dreams.

[sonic coral]

oh trust me you wanna do more math until 3

anyways, if i’m applying sylows theorem on a group of order 100, we have 2^2 * 5^2 for the prime factorization, and so there exist atleast one subgroup of order 2^2, and 5^2.

but then the p-subgroups are denoted n_p, and so we would denote them by n_2 and n_5

even though the existence of atleast one subgroups are of order 4 and 25? i guess because those aren’t prime so we denote it by the prime that gives us these orders?

[delicate orchid]

We call groups of order p^k p-groups, hence the convention

Ok bed for real now

[sonic coral]

gotcha, thanks

dream about some algebra for me

also, is the only way to show a group of order n is not simple by sylows theorem, to produce exactly one p-subgroup?

[summer path]

You can also find some nontrivial normal subgroup

[sonic coral]

how would i go about that?

[vivid tiger]

"term induction"? i have not heard of that and can't find a source on google. if your statement is about general sets and elements and functions, it looks false, but it sounds like you do mean something else

ahh, thank you

[white oxide]

say that i'm trying to prove that a finite cyclic group of order n has exactly one subgroup of each order d dividing n. would this be sufficient?

Let X and Y be subgroups of a finite cyclic group G with the same order d that divides n. Since X and Y are cyclic, they each have generators, say x in X and y in Y. Then x^d = e = y^d, so x = y and hence X and Y are the same subgroup

[thorn delta]

well, the meat of the statement is not uniqueness. its existence

[white oxide]

ah ur right

but say we knew it existed

would that proof of uniqueness suffice

[thorn delta]

no

[white oxide]

why not

[thorn delta]

Then x^d = e = y^d, so x = y
this just doesn't follow

[white oxide]

and for existence can't we just take the trivial subgroup {e} which has order 1 which clearly divides n

[white oxide]

why not

[white oxide]

wait no that's not right

[thorn delta]

for example, in the klein four group (Z/2 x Z/2), there is x != y such that x^2 = y^2 = e

[white oxide]

oh ok

[thorn delta]

yeah, that only proves the d=1 case

as a hint, you'll want to make sure you are actually using cyclic-ness

[white oxide]

so like for example, would i have to prove in the case of n = 10 that a finite cyclic group of order 10 has exactly one subgroup of each order d dividing 10

so like there's only one subgroup of order 1, one subgroup of order 2, and the same for 5 and 10

[thorn delta]

yes

[white oxide]

ok

would induction be useful here or nah

[thorn delta]

nah

[white oxide]

aighty

wait how would you even prove existence other than subgroups of order 1 and n lmfao

because n has to be positive

oh wait

maybe it has to do something with n being prime

and maybe i could make use of the fact that every group of prime order is cyclic or smt

idfk

[thorn delta]

like for example, {0, 2, 4, 6, 8} \subset Z/10 is a subgroup of order 5

{0, 5} \subset Z/10 is a subgroup of order 2

[white oxide]

so since i proved d = 1 and d = n

could i just let 1 < d < n

and then try to prove it for that

[white oxide]

ye i see that

i'm too stupid to see the pattern though let me think on it

oh could we just say that it's isomorphic to Zn since it's a finite cyclic group

and then use properties that we know about Zn

[thorn delta]

if that helps you then sure

[white oxide]

more precisely that if G is a cyclic group with n elements generated by a, and b in G and b = a^s, then b generates a cyclic subgroup of H containing n/d elements

from which existence immediately follows

[thorn delta]

what is s?

[white oxide]

sorry d is gcd of n and s

and s is some integer

[thorn delta]

huh. i don't see where s is coming from though. you are given d dividing n. you want to find an element of G that generates a subgroup of order d

[white oxide]

idk i just wrote down the theorem

isn't it just saying that b can be expressed as some integer power of the generator

which makes sense since a generates the group

[white oxide]

wait i thought we had to prove the existence of a subgroup of order d dividing n

[thorn delta]

yeah, that's what you need to do. to do that it suffices to find an element which generates a subgroup of order d dividing n

[chilly ocean]

it's universal algebra stuff

[vivid tiger]

(i see that now)

(and is why i confirmed that i didn't know what i was talking about before not answering)

[white oxide]

well by the theorem we know that there is a cyclic subgroup H that has order n/d

and clearly n/d divides n

here let me copy and paste the theorem

[warm shoal]

@thorn delta it's me, Toucan!

ur the goat

[thorn delta]

hi toucan

[thorn delta]

yea, so this is telling you the order of a subgroup generated by a^s where G = <a>

so you can think of your problem as finding the right s

[white oxide]

ok i got it!

thank you so much bro ur the goat

[thorn delta]

nice, and np

[warm shoal]

@thorn delta check ya DM my boy

[past temple]

if x is an element in a field of characteristic p such that x^{p^k} = 1 for some k >=1, is x = 1?

[next obsidian]

(x-1)^{p^k} = x^{p^k} - 1

Use this to prove ur fact

[past temple]

hm well that fact makes it obvious

[next obsidian]

Yeah

[past temple]

but how does one prove that fact

[next obsidian]

Prove it just for p

[past temple]

say, knowing that (x-1)^p = x^p - 1

[next obsidian]

Follows by induction

Show that (x+y)^p = x^p + y^p

[past temple]

ah

i see

the base case is easy

[warm shoal]

hey

Why is the group S3 non-abelian?

[past temple]

inductively, assume that x^{p^k} - 1 = (x - 1)^{p^k}

[warm shoal]

S3 is just a symmetric group of 6 elements right?

[next obsidian]

Write down noncommutjng elements

[delicate bloom]

does that proof work outside of Z/pZ?

[lethal dune]

works on F_p²

[next obsidian]

Are u asking me or pdk

[delicate bloom]

anybody

[next obsidian]

Yes it works

[delicate bloom]

maybe I'm misunderstanding the idea in mind

[past temple]

i just verified it yeah

[next obsidian]

I mean in char p

Frobenius is additive

Then if (x-1)^{p^k} = 0 cuz field

This says x -1 = 0

So x = 1

[delicate bloom]

ok I see, I misunderstood what you meant by induction earlier

[lethal dune]

for other char, the poly becomes separable

[next obsidian]

I meant induct to show that you can do p^k

For showing this breaks across sums

Like u want

[tawdry crystal]

I have this matrix which is an element of SO(4)

So when I take the logarithm, I should expect to get a traceless skew-symmetric matrix (it's Lie algebra) right?

But when I compute it I get this

Which does not seem to be correct because the diagonal should be all zeros

[tribal moss]

Those numbers are all pi/2, right?

[tawdry crystal]

Yes i*pi/2

[tribal moss]

Oh, didn't see the i's for all the decimals.

How did you compute that?

[tawdry crystal]

I diagonalize the first matrix (call it U). So U = PDP^-1

Then I take the logarithm of D

Since U is unitary, the diagonal is just e^itheta

So taking the diagonal gives diag{i * theta_1, ..., i * theta_n}

And then log(U) is just P diag{i * theta_1, ..., i * theta_n} P^-1

[tribal moss]

Hmm, guessing wildly here, but if you use pi·i as the logarithm of one of the -1 eigenvalues and -pi·i for the other, it feels like you ought to at least get "traceless" back after the dust settles.

[tawdry crystal]

I think traceless is implied because you expect your matrix to have zero along the diagonal

[tribal moss]

It would be skew-Hermitian and traceless though.

[tawdry crystal]

Well U is actually SO(n) so you can get a smaller subgroup than skew Hermitian

[tribal moss]

Which puts it in $\mathfrak{su}(4)$ rather than $\mathfrak{so}(4)$.

[cloud walrus]

Troposphere

[tawdry crystal]

Why can’t it be in so(4) again?

[tribal moss]

I'll admit I'm not exactly sure.
My guess: You have a 2-dimensional eigenspace with eigenvalue -1. You can change the basis for that space without ruining the diagonalization before you take the two different logarithms of the -1s, and I suspect that if you choose that basis just right, you might be lucky enough to get a purely real logarithm out of it -- which would then be in so(4).

[tawdry crystal]

Hmm I’ll see what I can do

[warm shoal]

im so confused

how do I show that

Z6 is of order 6

[tribal moss]

Something like using ½(i,i,-i,-i) and ½(i,-i,i,-i) as the eigenvectors for -1 might work. (No, that won't do.)

[tribal moss]

What is your definition of "Z6"? And of "order"?

Yeah, but this is very much a "follow the definitions" problem, so Circle has to actually look at the definitions in his book.

[tawdry crystal]

Ah I see you’re trying to motivate the answer

[warm shoal]

im so confused rn

and i have so much math to do

im sorry

[past temple]

https://math.stackexchange.com/questions/4430019/the-irreducible-representations-of-a-p-group

im a bit confused by the conclusion of the comment here

why can we conclude that the representation has a fixed point?

[delicate bloom]

I'm guessing since an eigenvector with eigenvalue 1 is unchanged

[white oxide]

can somebody help me understand why if there was a generator of Z_3 x Z_3 that generator added to itself successively could only give the identity after nine summands?

im a bit confused because i suppose i'm trying to supply a generator of Z_3 x Z_3 to see this but if there's no generator idk how i can see it in my head

[agile burrow]

By definition, if a group G is cyclic then there exists some element g in G such that if we repeatedly add g to itself, we eventually get every element in the group, right?