#groups-rings-fields

406252 messages · Page 449 of 407

latent anvil
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Sorry it's 1:38 am here and I'm also doing homework

potent lynx
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im really sorry for beinf stupid

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for disturbing u

latent anvil
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You really aren't

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This is a normal level of competency

potent lynx
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i losg all hope from this problem xd

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we been doing it for like w5 mins

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and im replying stupid dtuff

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u dont have to be nice xd

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tysm

latent anvil
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Okay so h(1) = 0

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Right?

potent lynx
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yea

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didnt u say this

latent anvil
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What if 1 = f(x) for some x?

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Then h(f(x)) = 0 ≠ 1 = g(f(x))

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I didn't

potent lynx
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So two functions are the same iff they're equal on all inputs

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this

latent anvil
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This is true

potent lynx
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g is not ewual to f on all inputs

latent anvil
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I'm saying h°f and g°f are not equal either

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So it's not a counterexample

potent lynx
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oh fuck

latent anvil
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Let's think about how we could patch h to fix this issue

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Don't give up! A good problem fights back

potent lynx
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ok we want h =1 for x =0 then ig

latent anvil
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Yeah

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Wait

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No

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Because the issue h(t) for t = 1

potent lynx
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we want h(1) = 1

latent anvil
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Okay, can you prove that for me?

potent lynx
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whats definiton of h again

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oj

latent anvil
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You just gave it

potent lynx
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nvm

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i got it

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h cant ne 0

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its like by exhaustion

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be*

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i think

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lao

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lmao

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h(f(x))

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f(x) cant be 0

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thats what we assumed

latent anvil
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Yup

potent lynx
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so by definition h(t) = t for t not 0

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so h(1) = 1

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ohh

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that swhy

latent anvil
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Okay, so can we prove this when Y = {0, 1, 2}?

potent lynx
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f not being surjec

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is a problem

latent anvil
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Yeah, you see it now?

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Or even for any Y

potent lynx
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yea adlong as

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there exists a y not =

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f(x)

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ig

latent anvil
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Right, we're assuming f isn't surjective

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So we want functions g, h which are a counterexample to f being epic

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Like we did for Y = {0,1}

potent lynx
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g(t) = t

latent anvil
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What was the property that made the h we chose work before?

potent lynx
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that h =! g for some t

latent anvil
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I'm gonna

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Sleep

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Now

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I think if you think about this you can prove the general case

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And also to repeat: you're not stupid or bad at math

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This stuff is hard

potent lynx
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tysm for helping

latent anvil
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The transition into advanced math was harder than the actual hard math for me

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Once I got it things became doable again

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Not impossible but not incredibly like futile and frustrating

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I would suggest focusing more on fundamentals right now

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Not trying to read Rudin or aluffi

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But find a book which covers basic set theory and logic

potent lynx
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i tried how to prove it

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i finished it

latent anvil
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This isn't because you're bad, just inexperienced

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Did you look up the problems?

potent lynx
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no i jist did similar problems for other ppl on discord

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i just*

latent anvil
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Okay, then just keep in mind that math is hard and everybody struggles

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You kept saying you were being stupid

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But you weren't

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This is a nontrivial problem

potent lynx
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i thought it was tbh

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anyways what text should i do then

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for algebra

latent anvil
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Maybe try Artin or Pinter?

potent lynx
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this tkme i will completly

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have a diff approach for probmems

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problems

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pinter is a bit low lvl ig

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like i can actually do the problems

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habaha

latent anvil
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/shrug

potent lynx
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i did like untill chap 7 or 8

latent anvil
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Whatever works

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Ah okay

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Then yeah, give aluffi a shot

potent lynx
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prob will use your problem set

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tysm

latent anvil
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👍

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They're good

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And challenging

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And I can hopefully solve all of them

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Since I wrote it lol

potent lynx
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oh they are yours

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lmao

latent anvil
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Mine and a friend

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The website is the friend's

potent lynx
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yea

latent anvil
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The other problem I was going to give

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Was to show that the inclusion ι : Z -> Q

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Is an epimorphism in the category of rings

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Despite being super not surjective

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It's a ring homomorphism because addition and multiplication works the same

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In Z and Q

potent lynx
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yea whats the diff

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between lkke

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category of sets and category of rings

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just the obj?

latent anvil
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No, not just the objects

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Categories have maps too

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And the maps we care about between rings are...?

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Any guesses?

potent lynx
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homomorphs?

latent anvil
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Yup

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So to be epic in the category of rings

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You need to satisfy the right cancelation property for ring homomorphisms g and h

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So to show ι is epi, you need to show that for any ring R and homomorphisms g, h : Q -> R, if g°ι = h°Ι, then g = h

potent lynx
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oj

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cool

brisk granite
mild laurel
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First show that H must be abelian I think

latent anvil
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Ufhhh I forget how to do this but have done so before

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Lemme check

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Okay so

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You want to show it's abelian

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That's equivalent to [N, N] = 1

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Why's that true?

brisk granite
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yeah, but even if G is Abelian, how do I know about it's isomorphic to (Z/pZ)^n

latent anvil
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By doing more things after that lol

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Let's step back

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So you know how normality isn't transitive right?

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You can have A normal in B and B normal in C, but A not normal in C

brisk granite
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yep

latent anvil
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What extra condition on A tells us that A is normal on C?

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Assuming it's contained in B and B is normal in C

brisk granite
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umm, I'm not sure

latent anvil
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So a subgroup H of G is called characteristic if every automorphism of G fixes H

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Does that make sense?

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As in φ(A) = A

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Not φ(a) = a for all a in A

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Exercise: prove that if A is characteristic in B and B is normal in C then A is normal in C

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Exercise: prove [G, G] is characteristic in G

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Exercise: prove that if G is solvable, then [G, G] is proper subgroup of G

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@brisk granite

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Use this to conclude [N, N] = 1 by minimality and solvability of N

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Now suppose p is a prime dividing the order of N. Because N is abelian, the function which sends x to x^p is a homomorphism.
Exercise: prove that the kernel and image of this map is characteristic in N

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Use this to show g^p = 1 for all p in N

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Conclude the result by linear algebra (G is a vector space over Fp)

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Does that proof sketch seem possible?

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@mild laurel does that make sense to you? I want to make sure this works

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But don't want to actually go through it

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In full depth

mild laurel
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Yeah, this is vaguely what I was thinking

latent anvil
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I guess it's really a result about groups which have no nontrivial characteristic subgroups and are not perfect

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I.e. the only such groups are these products

mortal stream
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Why?

hot lake
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no

sharp sonnet
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it's wrong and not really algebra, so?

golden pasture
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isnt even mildly related to alg

brisk granite
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what does the arrow mean?

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Also, I saw a numberphil video that said this but they assumed a bunch of divergent series converged

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twas weird

fading wagon
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@brisk granite

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@mortal stream

mortal stream
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I just can’t understand how the sum of integers and positive numbers could give such a result

fading wagon
sturdy charm
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@mortal stream i think that it has to deal with our understanding of the positive and negative values...
let's consider a car moving at 60km/h, and then it accelerates a little bit, and so it moves at 70km/h. 70-60=10km/h, so we can say that the car accelerated by 10km/h.
now let's consider a car moving at 60km/h, and then it slows down a little, and so it moves at 50km/h. 50-60=-10km/h, so we can say that the car accelerated by -10km/h.
Actually i don't know how to explain it, but the concept of accelerating is the same as slowing down, it just depends on the value (whether it's negative or positive). in physics, we don't even say slow down, we always say accelerating.
it's actually weird when you think about it. let's say you're in a car with a friend. the car starts slowing down, and you tell your friend : "the car accelerated". he's not going to understand it easily! but we can compare this situation to your reaction to the fact that the sum of all integers is -1/12. we give the word "addition" a bigger "sense of understanding", relating it somehow to some kind of soustraction.

fading wagon
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I don't think that has anything to do with this

mild laurel
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Yeah

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That has literally nothing to do with this

sturdy charm
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i think it has based on the fact that this idea comes from a mathematician, and sorry for my bad english tho

mild laurel
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What

wind steeple
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She says that his idea was based on a mathematician thoughts

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French english ♿

sturdy charm
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frenglish* 😩

wind steeple
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Pas de chance

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T'as vu la vidéo de micmaths c'est ça ? XD

jagged gate
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franglais powa

wind steeple
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Fr >>>>> all

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😎

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:kappa:

chilly ocean
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Can we talk about rings now?

chilly ocean
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Guess not

chilly ocean
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sure

cloud walrusBOT
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Anyone_Someone2018:

Is $(\mathbb R, +)$ cyclic? I mean if there is some infinitesimal element $g$, we can write all other elements as an integer factor of it.
📎 393131698277056519.png
sharp sonnet
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no

chilly ocean
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cyclic groups have countable orders

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the reals are uncountable in number

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hence that doesn't work

wind steeple
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or simply, g<2g<...<ng

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there's a real betwean 2g and g

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3g/2

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so it doesn't work

simple agate
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is (A/I)/I = A/I for a ring A with ideal I?

chilly ocean
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If $\mathbb{R}$ (additive) were cyclic, that is, if $\mathbb{R} = \left<g\right>$, we could represent every real number $x$ as $kg$ where $k$ is an integer (hence a bijection, specifically an isomorphism would exist $\phi : \mathbb{R}\to\mathbb{Z}, kg \mapsto k$ but that implies $\left|\mathbb{R}\right|=\left|\mathbb{Z}\right|$). From this assumption you can see you'll run into more problems than just the order not being countable, but that's definitely the biggest glaring issue

cloud walrusBOT
chilly ocean
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@simple agate the ideals of A/I are of the form J/I where J is an ideal of A containing I

simple agate
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hmm I'm trying to prove that if P is a prime ideal of A containing I then P/I is prime

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I've done it all apart from proving that P/I =/= A

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so far my approach is using proof by contradiction - consider P/I = A then A = {p + I : p in P} which implies P = A (and that A is a quotient ring), however P is prime in A thus that cannot be the case

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is that valid @chilly ocean ?

chilly ocean
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first of all, you would need to have P/I = A/I

simple agate
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yeah I kinda put that in parenthesis, that A would need to equal A/I

chilly ocean
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no

simple agate
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which relates to my original question

chilly ocean
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there is literally no reason why A/I would be A in general

simple agate
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can I accept that without proving it?

chilly ocean
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but the easiest way would probably be to pick any x in A\P

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then assume x+I is in P/I

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well if A/I = A, then I is the trivial ideal

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if x+I is in P/I then hmm

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well then x is in P

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and since that is a contradiction, there is at least one element in A/I not in P/I

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or idk maybe you could also use noether and the result that an ideal P is prime iff A/P is an integral domain

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assuming that A is commutative ofc

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then you could do it very simply

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A/P = (A/I)/(P/I)

simple agate
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the thing is that stuff is in the next questions so I don't want to use it

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I think I'm just gonna string together something and wait for my professor's feedback & official solution. thanks though

chilly ocean
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and conclude that P/I is a prime ideal of A/I iff P is a prime ideal of A

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do you have the result saying that there is a bijection between the ideals of A containing I and the ideals of A/I?

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an order preserving bijection

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ordering given by inclusion

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cause then you could use that

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P is not A, so it is contained in some maximal ideal M of A

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so P/I is contained in M/I

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and hence not A/I

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or just simply do what i recommended first and pick any x not in P and show that x+I isn't in P/I

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but yeah, do whatever you wish

simple agate
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@chilly ocean thanks for the help but my professor just emailed us to let us know he wrote the question wrong 🤦 lol

chilly ocean
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what course is this for, abstract II?

cloud walrusBOT
upbeat burrow
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I'm not understanding the 1/2 thing.

latent anvil
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Suppose that we had a Q module structure on Z

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Then we could multiply 1 (in Z) by 1/2 (in Q)

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Call this element x

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Then x + x = 1/2.1 + 1/2.1 = (1/2 + 1/2).1 = 1.1 = 1

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@upbeat burrow make sense?

upbeat burrow
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Yes

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I see now

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Thanks!

latent anvil
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Np

chilly ocean
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If you do -x/2 you get 1/2 right?

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(ik it's a dumb question im dumb 🤷🏿 )

latent anvil
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Is this a high school algebra question?

chilly ocean
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tfw my abstract exam today involved a question requiring some good knowledge of conjugacy classes to prove, but the prof mentioned the concept like once during the previous lecture and didn't even define it properly

chilly ocean
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Tfw you only realize why conjugacy was important in abstract algebra and thought it was useless in linear algebra

latent anvil
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Conjugacy good

magic owl
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Conjugacy is just free homotopy

latent anvil
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that sounds very fake

magic owl
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It’s true if by conjugacy you mean the bs fundamental group version

latent anvil
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Oh then sure

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Conjugacy in pi is like change of basepoint in my head

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Kind of

magic owl
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The free homotopy perspective is probably better imo

latent anvil
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Could you elaborate on it?

magic owl
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But this is pretty similar to a change of baseline

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Baseline

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Basepoint holy shot

latent anvil
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Basepoint but we're in a moduli space of lines

magic owl
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Anyway any free homotopy (not preserving endpoints) induces a conjugacy of loops

latent anvil
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What is free homotopy?

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I don't know that term

magic owl
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A conjugacy that doesn’t preserve base points

latent anvil
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Oh sure, a homotopy that moves around the basepoint

magic owl
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Yeah

latent anvil
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Yeah I am aware of this

magic owl
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You can write down this proof explicitly

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It’s a cute exercise

latent anvil
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I just hadn't heard the term free homotopy

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I needed it at one point for something

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But can't remember what

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Oh

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Nullhomotopic as a function iff path homotopy to the identity

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I think

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Although I don't remember why that came up

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Oh yeah Brown constructed a homotopy which didn't preserve the basepoint and didn't say anything

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And I got really mad after I figured out what was going on

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Because he hadn't mentioned anything about this

chilly ocean
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it was fun learning about homotopy as an undergrad in an independent study course

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(am still undergrad obv cuz I sound like a noob)

latent anvil
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Homotopy is pretty cool

chilly ocean
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@simple agate and the mistake was having P/I be an ideal of A instead of A/I?

simple agate
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Yep

chilly ocean
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weSmart🚬

simple agate
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lol

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it suddenly became a lot easier

chilly ocean
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well it should

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because you would be proving a false claim

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which is impossible

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;)

simple agate
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my favourite type of proof

chilly ocean
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yeah

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absurd algebra > abstract algebra

potent lynx
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Prove that G cannot have a subgroup H with IHI = n-1, where n = IGI > 2.

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since orders are integers

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can i do induction?

final gulch
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I don't see how induction would help

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presumably you know that the order of a subgroup must divide the order of the group?

potent lynx
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yes

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^

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this is what i wanted to dow ith induction

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assume if |G| = k and |H| = k-1

final gulch
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doesn't that give you what you want right away?

potent lynx
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well

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wtf

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XD

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k-1 never divides k

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for all k right?

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for all k >2 atleast

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yea

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lmao

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sorry

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ty

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i have a problem with proving

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2 groups arent isomorphic

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do you have any advice

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for how to like

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think to solve these problems

final gulch
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er

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what are the groups?

potent lynx
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R and C

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nonzero ofc

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but i mean like

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in general

final gulch
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I guess you can use the fact that the multiplicative group of nonzero complex numbers has four elements of order 4 but R-{0} only has two? edit: should have said solutions to x^4 = 1 rather than order 4

potent lynx
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i mean

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that seems not easy to see lmao

final gulch
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really? it's pretty clear you should try to make use of i somehow

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i and -i are the 4th roots of unity that aren't in R

potent lynx
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what the fuck

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xd

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yea its acatually clear

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sorry

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but in general what type of propertis

final gulch
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as for your other question

potent lynx
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are enoguh to show that 2 groups arent isomorphic

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enough*I

final gulch
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I mean there's not going to be one answer

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for example if you google it you can get this big list

potent lynx
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mmm

final gulch
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just gotta find some property that one satisfies but not the other

potent lynx
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i think the easiest ones are like

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being abelian

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or like being cyclic

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ig

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right?

final gulch
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right that would work

potent lynx
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yea

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ty

magic owl
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The things I look at normally are like

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Centers, element orders, cardinality if applicable

fossil vapor
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Just came across this, might be good

chilly ocean
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subgroup orders as well

latent anvil
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Sylow data too

chilly ocean
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and what subsets normalize each other can help, especially using generators to make the calculations faster if you have explicit elements in front of you

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*+centralize

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if it was a problem involving quotient groups you could use the fourth iso thm

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sorry that was a huge tangent i just realized lol

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After I’m done with Abstract I this semester I’m thinking of reading thru the ring theory unit of Dummit & Foote since I’ll have finished the earlier units

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is there one anyone likes better or has any recommendations about that

magic owl
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Jacobson

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Is a common rec here

chilly ocean
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imma stick with d&f for now lol just peeped Jacobson and d&f is still daddy tbh

bleak abyss
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D&F takes years to say anything lmao

simple agate
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I found gallian better for my first course on group theory

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but it's a more visually appealing book and I'm a sucker for illustrations and pretty presentation

chilly ocean
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if d&f is slow then what is pinter 😂

merry island
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a casual read before bed

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jokes aside i like that Pinter puts a lot of material in the exercises but makes them easy/gradual

magic owl
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DF is so boring

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i dont understand how anyone can use it as anything but a reference

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but maybe this is because i think the subject is try

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without topology algebra is useless

mild laurel
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?????????

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That's the hottest take I've ever heard

stone fulcrum
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I also find DF can't keep my focus

wind steeple
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Df ?

latent anvil
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Dummit and foote

magic owl
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what? df being boring

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is a super common take

bleak abyss
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Lmao Max

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Expert avoidance

chilly ocean
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Nah I love d&f

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It prepares you to the point you kinda need to start proving shit in your head real fast while reading and that’s good and important, the exercises are all adequate to help you get to that point

bleak abyss
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Do you need to prove things in your head fast while reading?

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Just read for another 30 pages and eventually he'll get to the proof

magic owl
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lmfao

urban acorn
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"so here we will prove that even numbers times any integer are also even"
"nah I'll just wait till he proves it, i don't feel like breaking my head right now"

potent lynx
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whats wrong with df gt?

faint elm
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Let $\lambda = (\lambda_1, ..., \lambda_k)$ be a partition of $n$, so $\sum \lambda_i = \lambda$, let $T$ be a young tableaux associated to $\lambda$ and $P_\lambda$ be the subgroup that permutes the rows of $T$. Do you know why $Ind_{S_{\lambda_1} \cross ... \cross S_{\lambda_k}}^{S_n}(\text{trivial}) = \mathbb{C}S_n a_{\lambda}$ where $a_{\lambda} = \sum_{w \in P_T} w$?

cloud walrusBOT
faint elm
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so they have the same dimension as vector space over $\mathbb{C}$, but I can't see an explicit representation isomorphism between them.

cloud walrusBOT
faint elm
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Anyone?

faint elm
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Ok I figured that out, let $H = S_{\lambda_1} \cross ... \cross S_{\lambda_k}$ and $\sigma in S_n/H $ then, up to representation isomorphism, $sigma = gh$ for a fixed $g$. Then I just send $\sigma\mathbb{C} \mapsto g\cdot a_{\lambda}$. It's $S_n$-linear and a subjection, hence the thesis.

cloud walrusBOT
faint elm
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What about the general case? We have a group $G$, a subgroup $H$ and a $1$-dimensional $H$ representation $\rho : H \to \mathbb{C}$. What is $Ind_{H}^{G}(\rho)$?

cloud walrusBOT
keen moat
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Is induction abstract algebra

mild laurel
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no

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Not even close

chilly ocean
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yes

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the induced modules

faint elm
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well there is no section for representation theory, so I posted here

mild laurel
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He's not talking about you

faint elm
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k

clear obsidian
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I'm guessing that C bar here is the extendend complex plane, but what is SO(2) (I'm guessing this is a definition of CSO(2), right?)?

📎 unknown.png
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but how would you take the cartesian cross product of the positive reals and 2x2 matrices?

woven delta
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In the obvious way

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Scaling

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@clear obsidian

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Think of what cauchy riemann tells you in terms of the real Jacobian

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And the matrix representative of a complex number

fossil vapor
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Might be $\otimes$

cloud walrusBOT
woven delta
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No

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They literally just mean the complex plane minus a point

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Ie S^1 times R^+

cloud walrusBOT
upbeat burrow
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Can someone explain how this is true to me?

blissful root
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what you have to prove is that (qa,b) = (a,qb) for any q

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write q = c/d

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now note

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(qa,b) = (ca/d, b) = (a/d,cb)

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first because the tensor is over Z

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now

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(a/d, cb) = (a/d, cbd/d) = (ad/d, cb/d) = (a, cb/d)

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essentially if you multiply one side by d/d then you can move the upper d

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so this gives a net result of moving a 1/d

upbeat burrow
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@blissful root so you get to move the c to the right because it's over Z? Over Z means I can only move integers right?

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Q tensor Q over Q is just isomorphic to Q right?

latent anvil
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Yes to the first question

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For the second question, yes R (×)_R M is always just M

chilly ocean
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hi, i'm not sure whether this is the correct place to ask questions about functions and equivalence classes / partitions, but could someone point me towards a bijection between real numbers and the set of their equivalence classes?

potent lynx
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there isnt one in general

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there are many

urban acorn
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@chilly ocean Equivalence classes under what equivalence relation?

chilly ocean
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under some arbitrary equivalence relation given f(x) = f(y) where f is surjective from R -> R

potent lynx
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show f is injective then?

urban acorn
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The equivalence classes given by an equivalence relation that looks like f(x) = f(y) are the fibers of the elements in the image of f.

#

The image of f is R in this case since that's its codomain and it's surjective

#

Q.E.D.

chilly ocean
#

well i'm just not really grasping the concept here, how can there be a bijective mapping from R to the factor set of R over A (the equivalence relation)

#

wouldn't that only ever be the case if the factor set only consisted of sets with one element each?

urban acorn
#

Infinite sets act somewhat unintuitively when it comes to things related to cardinality

#

For example, consider the set of even numbers.

#

It's in bijection with all of the integers (consider x -> 2x)

#

despite being contained in all of the integers, and there being infinitely many integers that aren't contained in it

#

Also, consider the equivalence relation that partitions the integer into {..., {-6, -5}, {-4, -3}, {-2, -1}, {0}, {1, 2}, {3, 4}, {5, 6}, ...}.

#

Clearly, these partitions of the integers are in bijection with the integers, despite almost all of them containing more than one element.

chilly ocean
#

yea

#

okay thanks, I might have found something

urban acorn
#

The bijection you're looking for is $x \in \bR \rightarrow {y \in \bR | f(y) = x}$.

cloud walrusBOT
chilly ocean
#

also something rather stupid, is the set of archetypes of a surjective mapping always a partition on it's domain?

#

or am i wrong to assume that

urban acorn
#

what's an archetype real quick

#

i can probably figure out the answer once i know that

chilly ocean
#

uh the set of elements that map to a certain element in the codomain

urban acorn
#

Oh, I know that as fibers.

chilly ocean
#

yea it's probably not a perfect translation

urban acorn
#

Yes, it's always a partition on its domain.

chilly ocean
#

okay thanks

urban acorn
#

Intuitively, a collection of subsets is a partition on their superset if and only if every element of their superset is contained in exactly one subset.

#

Every element in the domain gets mapped to precisely one value in the image, and thus the fibers of the elements in the image must be a partition.

chilly ocean
#

yep

fringe nexus
#

Why is every Z module an abelian group?

#

I get the other way round

#

do i just not understand modules properly

bleak abyss
#

Uh, I think you're asking about the other way around

#

Every anything module is an abelian group under addition

fringe nexus
#

o

#

my

#

god

#

i forgot the definition requires the group to be abelian

#

nevermind

bleak abyss
#

Lmao it happens 😛

barren delta
#

I'm trying to remember how to figure if two direct products are isomorphic

#

$\mathbb{Z}2 \times \mathbb{Z}{12} \simeq \mathbb{Z}_4 \times \mathbb{Z}_6$

cloud walrusBOT
barren delta
#

can I just willy nilly break non-prime boyes into their prime factorizations?

#

I'd have Z2 x Z4 x Z3 and Z4 x Z2 x Z3 which would then definitely be isomorphic, yeah?

#

ignoring that the Z4 can also be broken down

mild laurel
#

The thing is that power of primes cannot be broken down

#

Z_4 is not isomorphic to Z_2 x Z_2 as you can check

#

But otherwise, yes, the chinese remainder theorem says you can break it down into powers of primes

barren delta
#

ah right derp

fossil vapor
latent anvil
#

How many groups of order less than 187,500 are there?

#

I'm debating trying to do exhaustive search to show there's no counterexample

#

Oh I can also assume the group has order divisible by 60

blissful root
#

about 38 metric shittons

latent anvil
#

Hmmm

#

I've eliminated all the p groups

#

That's lots of them

latent anvil
#

I only need to consider 3125 orders

#

That's not very many

#

If I start right now I can probably prove it by midnight Wednesday

blissful root
#

good luck with order 8192x15

visual turret
#

I'm trying to prove that if a group G has order p^2, then it must have a subgroup of order p (without use of Cauchy's Theorem). I imagine that this follows from Lagrange's Theorem in some way, but am having trouble seeing it.

mild laurel
#

Lagrange's theorem doesn't really say anything about this

visual turret
#

From lagrange's theorem we know that the only possible orders for subgroups are 1, p, and p^2, no?

#

This seems like it would be useful in showing that there must be a subgroup of order p

mild laurel
#

Sure that's true

wind steeple
#

Take an arbitrary non neutral element of your group and study its order

#

You may have your answer

mild laurel
#

But lagrange's theorem can't tell you that anything that there exists an element of order p

wind steeple
#

Why taking an element suffices here ? Because a group of order p is cyclic

#

So we are reduced to study orders of elements of G x)

visual turret
#

That makes a lot of sense, I will spend some time thinking about why there must be an element of order p.

stone fulcrum
#

I would personally start by assuming there's a group that has no element of order p

stoic plaza
#

Hello, I am doing a number of exercises but this question bothers me. It states: Suppose that x is an element of G with order 16. What is the order of x^3? What is the order of x^(−4)? What is the order of x^6?

solar vessel
#

What have you tried?

stoic plaza
#

I've Tried modulo. I've tried assuming x^2 = e

#

But they are still wrong

solar vessel
#

e being the identity?

stoic plaza
#

Correct.

solar vessel
#

ok so x^16=e

mild laurel
#

why did you assume that x^2 = e?

solar vessel
#

also that can't be true lol

#

if x²=e then x has order at most 2

mild laurel
#

What does the order of an element mean

thorn delta
#

x has order 16. assume x has order 2 :p

mild laurel
#

Some people do mean order to just mean that x^16 = e, not necessarily minimal

stoic plaza
#

I believe it's the least positive integer such that x^n=e

mild laurel
#

But beginning students usually don't use it in taht sense

solar vessel
#

ok so you knew 16 was the least positive integer n such that x^n=e

stoic plaza
#

"why did you assume that x^2 = e?" Because this is part D of question 1. It's a follow up question

solar vessel
#

and you assumed 2 (a positive integer less than 16) was such that x²=e

stoic plaza
#

Yes correct.

solar vessel
#

Do you see the problem with this?

stoic plaza
#

Oh shit................................

#

xD

#

How can I be so blind...

#

Gotcha. So the answers are 16, 4 and 8 then

solar vessel
#

yes

stoic plaza
#

Cheers.

solar vessel
#

np

junior edge
#

Hi all, I'm a bit stuck on a polynomial division question if anyone is available to help. I have a polynomial which I am trying to prove has no divisors of degree 1, and I understand that I need to use the polynomial division algorithm to prove that when I divide my p(x) by x-a, I am left with p(a), which leaves me to prove that there is no a such that p(a) = 0 so that I may conclude that there are no degree 1 divisors of p(x). I'm stuck here as I'm not sure how to prove that in a general sense. Any hints?

mild laurel
#

It's probably easier if you just post the question

#

This is kind of hard to answer without knowing the full context as what you would do really depends on teh question

junior edge
#

Fair enough.
The question as a whole is: Prove that f = x^4-2x^3+2x^2-4x+1 does not have any degree 1 divisors

mild laurel
#

Ah yeah

#

This is where you use the rational root theorem

junior edge
#

Ahh, so it would suffice to show that there are no divisors of the constant term such that the polynomial evaluated at that term would equal zero? So, f(1) = -2 == 0, which means that my polynomial is irreducible

mild laurel
#

Uh

#

You've shown it has no rational polynomial divisors of degree 1

#

That doesn't mean a polynomial is irreducible

#

The polynomial (x^2 + 1)(x^2 + 1) has no rational polynomial divisors of degree 1, but it is definitely reducible

junior edge
#

Ok ok, thank you for the correction lol. That actually helps a lot as a later part of the question asks me to prove that f is irreducible in Q[x]

visual turret
#

I'm having trouble showing that if a group has order 77, then it must have an element of order 7 and an order of element 11. I have been trying to show this by contradiction, but ran into a wall. It is pretty simple to find a contradiction in the case of there not existing an element of order 7 nor an element of order 11, but I can't seem to get the other cases.

latent anvil
#

@visual turret do you know Lagrange's theorem?

visual turret
#

Yeah

latent anvil
#

Okay, so it has to have an element of order 7 or order 11

#

Otherwise all the elements have order 1 or 77

#

Right?

visual turret
#

Yeah, I get that bit

latent anvil
#

And they can't all have order 1

#

Okay

#

I'm thinking about it, sorry

visual turret
#

So my problem from there is why they can't all have order 77

#

No worries

latent anvil
#

Oh

#

Well I can answer that

#

Suppose your group had an element x of order 77

#

Then it's cyclic, and so it has elements of order 7 and 11

#

Or even quicker, |x^7| = 11 and |x^11| = 7

#

So you know there's an element of order 7 or an element of order 11

visual turret
#

That makes sense, thanks. Do you see how we get to order of 11 and order of 7 from there?

latent anvil
#

No, I'm thinking about it

#

In like a week or two you'll learn the sylow theorems

#

Which will give you this automatically

#

So I'm not used to thinking about this kind of thing without those

#

Do you know the orbit stabilizer theorem?

visual turret
#

I don't

latent anvil
#

Well no worries, that's not actually what I meant to aks lol

#

*ask

#

But you probably won't know the other theorem either

#

Okay so like

#

Let's think about what happens when |x| = 7 for all x

#

What subgroups can G have?

#

@visual turret

#

Wait that's not what I mean to ask exactly

#

What I'm trying to say is that all cyclic subgroups are C7

#

And they have to intersect trivially

#

And there's no subgroups of order 11

#

Since their elements orders won't divide the subgroup order

#

Also, the cyclic subgroups are maximal

#

Since they have prime index

#

So all the subgroups are just cyclic of order 7

#

(and G and 1)

#

Suppose G has a normal subgroup N

#

Then |N| = 7

#

Given any element x outside of N, we know N<x> is a subgroup by 2nd iso, and so it has to be G

#

Since it properly contains N

#

By lagrange stuff

visual turret
#

We don't have isomorphisms yet unfortunately

latent anvil
#

Oh hm

#

Do you know |AB| = |A|*|B|/|A intersect B|?

visual turret
#

Nope

latent anvil
#

Hmmm

visual turret
#

It should also be doable without normal subgroups, but we can use them, since we just covered them in class

latent anvil
#

I'm sure there's a nice simple solution

#

But I know too many techniques to find it

#

Does that make sense?

visual turret
#

Yeah, I getcha

latent anvil
#

I'll keep thinking though

#

I just finished my group theory take home midterm

#

So what I said above is still true

#

The nontrivial proper subgroups are all of the form <x> where |x| = 7

latent anvil
#

Oh yeah I think you can just like

#

Count

#

Pick an element x in G\{1}

#

Take away <x>

#

This new set will size 77 - 1 - 7

#

Do this again

#

You'll get 77 - 1 - 6 - 6

#

And so on

#

But 77 isn't 1 mod 6

#

@visual turret does that make sense?

#

Every element generates a subgroup of order 7

#

And they intersect trivially

#

So earlier removals won't overlap with later ones

#

Think about G as like a flower

#

Since 77 is also not 1 mod 10, this works for the other case too

#

Each of the "petals" in that diagram is a subgroup of order 7 btw

#

I assume that's clear because it's a beautiful diagram

cobalt pilot
#

A question about the Jacobsen Radical. How can we say that if x \in J(A) and y\in A then 1-xy cannot be in any maximal ideal?

latent anvil
#

@cobalt pilot suppose it was in a maximal ideal m. Then x is in m too, and since ideals are absorbtive also xy in m. Thus 1 = (1-xy) + xy is in m. Since m is maximal, it is proper, and so this is a contradiction

#

Does that make sense?

cobalt pilot
#

Maybe, I'm thinking 🙂

latent anvil
#

Np, sorry

#

x is in m since the Jacobson radical is contained in all maximal ideals

cobalt pilot
#

Still unsure how you can set it equal to 1

#

I get the x in m part

latent anvil
#

(1-xy)+xy?

cobalt pilot
#

yeah

latent anvil
#

Add them

cobalt pilot
#

oh, lol

#

yeah

latent anvil
#

Math is fucked lmao

junior edge
#

Can someone help me real quick with a proof? I think I'm mostly done, I would just like someone to verify I didn't mess up somewhere lol. It's probably too wordy, but I just wanna make sure I'm on the right path

📎 20191106_0340371.jpg
cobalt pilot
#

@latent anvil yeah, I get it now. Thanks a lot!

#

Sometimes you understand the hardest crap, and obvious implications fly over your head

latent anvil
#

Yeah, I definitely empathize

#

@junior edge I'm confused by your proof. Why can't we take e.g. d = 1, r = 0, and q = a?

junior edge
latent anvil
#

I haven't finished reading it

#

But you seem to be doing Euclidean division without dividing by anything

#

Oh sorry

#

You're assuming d is a common factor of a and b

#

I misread

#

I think the start is pretty unclear

#

Why not just say "assume d is a common divisor of a and b....then d divides 1"?

junior edge
#

Yea, I suck at proofs lol. I tend to try to 'over-prove' things

latent anvil
#

That plus the fact that 1 is a divisor of a and of b is actually enough to tell you it's the gcd

#

Proofs are hard, don't be so down on yourself!

#

I'm really bad at writing too much

#

I tend to over elaborate too

#

The logic looks right though

junior edge
#

Your explanation makes sense and is what my initial gut feeling was, but then I started to bog myself down with trying to explain everything. I'll rework it to be a bit more concise

latent anvil
#

No worries. Feel free to repost when you do so!

steep hull
#

Yeah, it looks fine (exclude the last part as uniqueness is implied already).

cobalt pilot
#

Another question about the converse part of the prof

#

So we assume that 1-xy is a unit and we want to prove that x has to be in J(A)

#

we do this by assuing x not in m for any maximal ideal and reach a condridiciton

latent anvil
#

For all y

cobalt pilot
#

then I can skip some because it does not realte to my question

#

yeah

#

my question though

#

hang on

latent anvil
#

Sorry

cobalt pilot
#

Is we prove that our assumption that x is not in any maximal ideal is wrong. How does this prove that x is in ALL maximal ideals? Won't this just prove that it's in SOME maximal ideal?

latent anvil
#

Yeah I was also going to correct that lol

cobalt pilot
#

So the proof is just wrong?

latent anvil
#

Post a screenshot?

cobalt pilot
#

(I have not typed out the prof though)

latent anvil
#

You might be misreading the quantifiers or something

cobalt pilot
#

I did not write the last concluding sentence since I had this question

latent anvil
#

So what you prove therr is actually the correct statement

#

You assume you have some m such that x not in m and reach a contradiction

#

You just wrote the wrong thing at the start

cobalt pilot
#

wait, what? What did I write wrong at the start?

#

oh, when I typed and not posted?

#

(Sorry, English is not my first language)

latent anvil
#

If you write "suppose x not in m for some maximal ideal m"

#

(no worries, your English is fine)

cobalt pilot
#

ah, ic

latent anvil
#

Then the proof is valid

#

As is it doesn't actually prove the claim

#

But the claim is the wrong one anyways

cobalt pilot
#

Difference between "some" and "all"

latent anvil
#

Yup

cobalt pilot
#

but hang on

#

I assume x is not in ANY maximal ideal. That is contridiced. Still, how does it prove it's in ALL?

latent anvil
#

Right, I'm saying you made the wrong assumption

#

And your proof doesn't actually derive a contradiction

#

You work with a specific maximal ideal m

#

But you start by assuming it works for all of them

#

Like you essentially prove "for any m, if x isn't in m then we have a contradiction"

#

Which is equivalent to saying "for any m, x is in m"

cobalt pilot
#

Let me look over it

#

You sure that's what I'm saying?

#

oh

#

hang on

#

"Which is equivalent to saying "for any m, x is in m"" - so if I can wrap my head around that. I've done the proof right

latent anvil
#

Your proof is wrong

#

The problem is your first sentence

cobalt pilot
#

Assume 1-xy is a unit??

latent anvil
#

Change "any" to "some"

#

Sorry, next sentence

#

If you change any to some it becomes correct

cobalt pilot
#

ok, I think I got it

#

thanks

latent anvil
#

Np

cloud walrusBOT
cloud walrusBOT
mild laurel
#

@latent anvil

#

Question for you that showed up on my algebra test

#

Show that a group G with order 3 * 5^2 * 7^2 *11

#

Has a subgroup of order 33

faint elm
#

Is a group of such order always not simple?

simple agate
#

I have the question "if A[x] is a PID then briefly explain why A must be an integral domain". I can show it by expanding elements of A[x] but I doubt that's what is meant by brief. can anyone give any pointers?

woven delta
#

A ring is an integral domain iff it imbeds in a field

#

A imbeds in A[x] and A[x] imbeds in a field

#

So you're good

#

@simple agate

simple agate
#

I'm not quite sure what imbeds means but I can look it up lol

woven delta
#

There is an injection

#

Into a field

simple agate
#

oh right

woven delta
#

(the field of fractions)

#

This proof actually shows that A[x] integral domain implies A integral domain

#

Which is stronger than what you need

#

Assuming you know that PID implies Integral domain

#

Which I'm assuming you do

simple agate
#

yeah I was just confused by the statement "explain briefly" rather than proving it

woven delta
#

🤷

#

It doesn't need to be formally proven

#

It's kind of trivial

simple agate
#

but thanks a lot, I can definitely use that reasoning

junior edge
#

Given that I've already proven that a polynomial f (of degree 4) in Z[x] has no degree 1 divisors and that the homomorphism of the polynomial from Z[x]->Z_5[x] factors as the product of gh where g is irreducible of degree 3 and h is irreducible of degree 1, how can I use that to prove that f is irreducible in Q[x]? I'm thinking I need to use part of the proof of Eisenstein's criteria, but I'm unsure. Any clues?

queen vine
#

If your polynomial f was reducible, then it would have to factor as gh where g and h are irreducible polynomials in Z[x] of degree 2 ( I assume that f is primitive and that 5 doesn't divide its leading coefficient). What would happen if you reduce f=gh modulo 5?

tribal pasture
#

@mild laurel By the prime factorization of the G, we have that there exists a 3-Sylow H and a 11-Sylow K. Since H is a normal group inside G thus HK is a subgroup of G. Since HK has order 33 thus we conclude there eixsts a subgroup of order 33.

cloud walrusBOT
fringe nexus
#

How would you show that the only group of order 35 is Z_35 ?

#

up to isomorphism

#

is the only way sylow

blissful root
#

yeah

#

it's the practical way anyway

#

since 35 is small you can do it "by hand"

#

but the argument holds for any group of order pq

fringe nexus
#

isnt there another restriction

#

like p cant divide q-1

#

wait nvm

#

Im probably thinking of something else

#

also how would i prove that any group of order 30 has normal subgroup of order 15?

#

I know any subgroup of order 15 would be normal since the index is 2

#

is it sylow again

blissful root
#

yeah

#

that's groups of order pqr in general

#

but index being 2 makes life easier

#

you just take the product of the groups of order 3 and 5

#

and it's normal

fringe nexus
#

pandaOhNo gonna fail qual

#

even though its super easy

blissful root
#

qual?

fringe nexus
#

compared to other

blissful root
#

you in gradschool?

fringe nexus
#

like this

#

well

#

basically for our grad class final

#

we take it

#

and he grades us on it

blissful root
#

oh

fringe nexus
#

for final grade

blissful root
#

I see

fringe nexus
#

its like theres a sylow problem on every exam

blissful root
#

yeah it's kinda obligatory

ripe basalt
#

If this is on a qualifying exam

#

I would laugh

mild laurel
#

@tribal pasture uh

#

7 is one mod 3 lmao

#

So is 55

#

So is 49

#

So is 55 times 49

#

So is 25

#

And 25 times 49

#

Did you even check any of the options lmao

tribal pasture
#

Oh shit........ Nvm....

#

Just maybe redo the whole argument for Sylow K

mild laurel
#

Nope

#

3 * 5^2 * 7^2 is 1 mod 11

#

But since this is the other other option, maybe we can do something with Burnside's theorem

bleak abyss
#

Did someone say Sylow?

mild laurel
#

The question on my algebra midterm was

#

Show that a group G with order 3 * 5^2 * 7^2 * 11 has a subgroup of order 33

whole basalt
#

So, quick question (I hope). An algebra over a field F is basically a vector space where there's also a bilinear product right... Is there any special name given to the easy example of F^n with componentwise multiplication?

bleak abyss
#

Oh so are you trying to say it either has a single subgroup of order 3 or 11

mild laurel
#

It's not true

bleak abyss
#

F

mild laurel
#

For Syl_11(G), either 1 or 3 * 5^2 * 7^2 works

#

and there are a ton of options for Syl_3(G)

#

But since the only other option for 11 was that

#

It's true that there must exist a normal p complement for 11

#

Since Burnside's theorem, but I don't think that helps

#

Other idea is if you can show a normal p complement exists for 5 and 7, you can take the intersection of those two to get your subgroup

whole basalt
#

My guess was "free algebra" but that seems to be different

latent anvil
#

I'll work on that

#

It looks fun

#

Sorry, I'm pretty much just getting up

mild laurel
#

This is like what you've been doing for the past couple weeks so

latent anvil
#

This is true

#

I am free of group theory as of last night

#

Oh yeah @bleak abyss i realized my nick is wrong. Can you change it to "all diagrams are commutative"?

bleak abyss
#

So the set of elements not of order 11 is a subgroup of order 3675=3*25*49

#

If there's more than one subgroup of order 11

faint elm
#

This one should be more easy: let G be a finite group, p a prime such that p|o(G) and for all a, b in G we have (ab)^p = a^p b^p. Show that the p-sylow is normal

wind steeple
#

the elements of order p^n H form a subgroup of G, and is normal in G because if a is order p^n, gag^-1 is order p^n for all g in G. But p doesn't divide G/H since if it does, there an element a of order p in G/H by Cauchy and a^p in H so a is order p^n for some n and a is in H. H is order p^km where p^k||G| and p^(k+1) doesn't divide |G|. m=1 because H is compouned of only order p elements. Then H is a p-Sylow and G. Since every p-Sylow in G are conjugate, there's only one p-Sylow H which is normal. @faint elm

faint elm
#

Yes

latent anvil
#

lol, the Jacobson radical problem I helped someone with last night is on my algebra homework for this week

mild laurel
#

Let me repost this because I still have zero clue how to do it:
Show that a group G with order 3 * 5^2 * 7^2 * 11 has a subgroup of order 33

latent anvil
#

Can you post the sylow bashing?

#

I'm too lazy to do it

mild laurel
#

for 3, there are a ton of options

#

for 11 only 1 and 3 * 5^2 * 7^2 works

faint elm
#

Ok assume the second one

mild laurel
#

Okay I'll rehash some of the ideas I thought about

#

Well if Syl_11(G) is 1, then we're done since its normal and you can just multiply it with any sylow 3-subgroup

#

Otherwise, by Burnside's theorem, we know that a normal p-complement for 11 must exist

#

Which shows the existence of a normal subgroup of size 3 * 5^2 * 7^2

latent anvil
#

If the normal subgroup has a normal 3 sylow, it'll actually be characteristic, so you get a normal subgroup of order 3 in your big group and thus one of order 11

#

There could be 1,7,49, or 175 sylow 3s in the normal subgroup

fringe nexus
#

Can someone give me a hint for the part about finding an example?

📎 unknown.png
bleak abyss
#

Oh this is interesting

fringe nexus
#

I thought it would be rotation but rotation is normal

woven delta
#

Rotation on C^2?

fringe nexus
#

o

#

wait

woven delta
#

You have a nice theorem about normal operators

#

They can be diagonalized

fringe nexus
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yup

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spectal theorem?

woven delta
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So you can use that fact to say that something is not normal lol

fringe nexus
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wait im thinking about wrong question

woven delta
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Just as part of the problem

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Oh, you're looking at the first one?

fringe nexus
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no im looking at part 2 to find an example

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I was just thinking about some other problem dw about it

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does knowing it cant be diagonalized help at all?

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I'm not sure how to prove that there is no such operator that can satisfy that equality

bleak abyss
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Is there a, say quadratic, polynomial p such that p(S) is diagonalizable for any S?

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Actually does every operator have a square root? I feel like the answer is no

fringe nexus
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has to be positive

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right?

bleak abyss
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Over C, a diagonalizable operator should have a square root

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Positive is only relevant if you're worrying about the real case

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So if T^2 = 0

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And there's a matrix S such that S^2 = T

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Then S^4 = 0, so the minimal polynomial of S divides x^4 but has degree at most 2

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So that does it

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@fringe nexus and @woven delta

uncut girder
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I have shown i => ii => iii

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I need help showing iii => i

uncut girder
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<@&286206848099549185>

latent anvil
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Suppose there was a nonconstant common factor h(x)

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So f(x) = h(x) p(x) and g(x) = h(x) q(x)

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Then q(x) f(x) - p(x) g(x) = 0

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Since h is nonconstant, p and q have degree smaller than m and n

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@uncut girder does this help at all?

uncut girder
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Yes!

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Thanks

latent anvil
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Np

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Also, did I get pinged by helpers?

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Cool

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@bleak abyss did helpers get changed or something?

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I usually don't get notifications except for pings

bleak abyss
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Maybe you got ghost pinged

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:0

steep hull
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Lol he already helped

latent anvil
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/shrug

mild laurel
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Depends in which set

cobalt pilot
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Since all maximal ideal are prime, would it be correct to say that J(A) \subset N (N being the nilradical)?

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(in com alg, not neccearnly in general alg)

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If so, all elements in J(A) should also be nilpotent?

mild laurel
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Yes

cobalt pilot
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So the xy-1 is a unit is in addition to x being nilpotent

cobalt pilot
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I ment 1-xy

chilly ocean
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both of them are units

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if a(1-xy)=1

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then -a(xy-1)=1

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how would you use the nilpotence of x to construct an inverse for 1-xy?

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it's pretty simple

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and involves a symbol you would see on a first aid kit

chilly ocean
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How would I find all normal subgroups of D5? (Dihedral group with 10 elements) Is there a 'fast way'? Or do I jsut find all subgroups and just multiply by elements from both sides?

mild laurel
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That would work

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Alternatively, you can look for homomorphisms from D5 to other groups

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Since you know that kernels of homomorphisms are normal subgroups

chilly ocean
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hmmm

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Cause I also will ahve to find all homomorphisms of D5 -> Z26 x Z26, so how would finding homomorphisms help here?

stone fulcrum
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The kernel is a normal subgroup of D5

mild laurel
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Yep

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Subgroup generated by x,y which implicitly means that its generated by x,y x^{-1} and y^{-1}

strong grail
steep hull
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Wrong channel. Go to precalc.

strong grail
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Wtf

mild laurel
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This is not abstract algebra material

strong grail
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Fine

chilly ocean
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@bleak abyss No, in general operators do not have square roots. Here is a counterexample that works over any field: Fix a basis (e_1, \dots, e_n) (n > 1) and let T be the endomorphism that maps e_i to e_{i-1} (and e_1 to 0).
Then this map cannot have a square root. Indeed, if S is a square root of T, then we have S(e_1) = 0 and S(e_2) \in Span(e_1), hence ker(T) = ker(S^2) contains e_1 and e_2, which is a contradiction since T(e_2) = e_1 <> 0.

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I think there is probably a general criterion in terms of Jordan forms

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At least over C, any invertible matrix has a square root, then you have to look at nilpotents

bleak abyss
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Yeah I pretty much had the special case of that in mind when I was talking about choosing a matrix such that T^2 = 0

lone niche
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how would you do c)

chilly ocean
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It follows from b) that Ker((D-a)^n) = E_a(Ker(D^n))

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But Ker(D^n) is fairly easy to understand

lone niche
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so (1e^ax,...,x^n-1 e^ax)

chilly ocean
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yep

lone niche
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how do i show it spans

chilly ocean
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E_a is an iso

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and (1, x, ... x^{n-1}) is a basis of Ker(D^n)

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so this is a basis of Ker((D-a)^n)

lone niche
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ah good point

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thank you

maiden ocean
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im a bit confused about the proof of this lemma, that if two elements of an abelian group have relative prime orders than the order of their product = the product of their orders

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Let g and h be elements onf an abelian group G having finite relatively prime orders m and n. suppose $(gh)^r = 1$. Then $k = g^r = h^{-r} \in <g> \cap <h>$

cloud walrusBOT
maiden ocean
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i dont understand why we can say that g^r = h^-r

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wait nvm im a brainlet

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disregard this

solemn hollow
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on a similar note, let's call (gh)^2=g^2 h^2 The Group Theory Freshman's Dream :)

maiden ocean
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im kinda confused as to why we can say that $o(k) \mid m$ tho

cloud walrusBOT
maiden ocean
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i get that if a is the order of k we have $k^a = g^{r^a} = g^{ar}$

cloud walrusBOT
delicate bloom
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not sure what your notation or original problem is, but the answer is probably cause the order of an element is equal to the size of the cyclic subgroup it generates and the size of the subgroup divides the size of the group

maiden ocean
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im trying to prove that if g and h are elements of an abelian group G, with relative prime orders m and n, o(gh) = mn

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also @delicate bloom that second part was proved but not the first part yet

delicate bloom
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having trouble figuring out which part you're having trouble with

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you can show that (gh)^{mn}=e and then you're trying to show there's not something less than that which works?

maiden ocean
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im trying to just follow the proof of this lemma : p

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given in the book

delicate bloom
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what's the original lemma as it's written, I don't see where all the stuff is you're talking about came from

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like "k" came outta nowhere it seems but maybe I missed it earlier

maiden ocean
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Let g and h be elements of an abelian group G having finite relatively prime orders m and n respectively. Then o(gh) = mn.

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Suppose $(gh)^r = 1$. Then $k = g^r = h^{-r} \in <g> \cap <h>$. Then $o(k) \mid m$ and $o(k) \mid n$ and thus $o(k) = 1$

cloud walrusBOT
maiden ocean
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im stuck on the o(k) | m part

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wait i think i get it

delicate bloom
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I see, I thought k was referring to something earlier on, but it's just a definition here

maiden ocean
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if we say $o(k) = a$ then we have $g^{r^a} = 1$ and thus $g^{ar} = 1$ and since $ar \mid m$ we have $a \mid m$

cloud walrusBOT
delicate bloom
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that doesn't look right, if o(k)=a then that means you have k^a=1

maiden ocean
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wait no

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we have m | ar

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yah but k = g^r so k^a = 1 implies (g^r)^a = 1

delicate bloom
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that's better looking

maiden ocean
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hmm

delicate bloom
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before you had $g^{r^a} = 1$ not $(g^r)^a=1$

cloud walrusBOT
maiden ocean
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sry implies parenthesis

delicate bloom
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feel like maybe that wasn't clear why I was saying that earlier

maiden ocean
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o