#groups-rings-fields

406252 messages Ā· Page 434 of 407

bleak abyss
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So yeah the only hole here is H_3(K(Z/p,1))

marble wagon
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surprising

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well I haven't done much topology

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but it makes sense that group cohomology would be useful for this stuff

bleak abyss
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Yeah if I remember right there was some theorem about how... okay don't quote me on this but I think odd-dim group homology of Z/p worked nicely?

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We just Googled hard until we found something

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Anyway I have an exam in about an hour and a half so I should probably stop procrastinating and get back to studying, so see you

marble wagon
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yeah you can just write the stuff and it works out quickly

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for cyclic groups

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see ya

misty aspen
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Hello fellows. I'm studying from Dummit and Foote here and I got a quick question on the necessity of an assumption in an exercise. The question is " R is a commutative ring with 1. Show that an R-module is irreducible if and only if M is isomorphic (as an R-module) to a quotient R/I for some maximal ideal I of R". Is commutativity of R really needed here or is it just added to avoid choosing between left or right modules?

marble wagon
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the latter

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since left/right ideals could be different

misty aspen
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makes sense, just wanted to confirm. Thanks!

bleak finch
full blaze
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so listen, this question asked Show that * is a binary operation on S (ok good so far no problem), then the second part was prove that (S, * ) is a abelian group

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do i need to first prove its a group or just show commutativity of the binary operation

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just curious

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Also the set hasn't been defined as a group or anything yet

bleak abyss
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You need to show it's a group

full blaze
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alright

full blaze
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i proved it lets go baby

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Question: Can two groups be isomorphic if the operation under one group is different than the operation under the other group

mild laurel
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The positive real numbers under multiplication and the real numbers under addition are isomorphic groups.

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I don't know if this really answers what you're asking but

topaz solar
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@full blaze yes

full blaze
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i thought so just checking and nice example

topaz solar
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plus just changing the symbol works too

full blaze
woven delta
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@full blaze learn model theoryopencry

full blaze
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i learned some in discrete math tinktonk

woven delta
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really?

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thats surprising

full blaze
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since my teach was a universal algebraist

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ye hes retarded

woven delta
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oh thats pretty cool

topaz solar
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model theory in discrete math
prof was universal algebraist
bruh what

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10/10

full blaze
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dood hes a boss

oak perch
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šŸ˜›

full blaze
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Dr. William Demeo Cu Boulder

topaz solar
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I've been meaning to look into model theory but i feel like i should finish my anal stuff

woven delta
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they're totally unrelated

full blaze
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i completely forgot anything he taught me

woven delta
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although model theory and topology are deeply related

topaz solar
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They are unrelated, hence why I should finish up my whole adventure into intro analysis first

full blaze
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lul

woven delta
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heres a cute baby model theory problem

full blaze
woven delta
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prove $(\mathbb{Q}, +)$ and $(\mathbb{Q}^+, *)$ are not isomorphic

cloud walrusBOT
full blaze
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proof by induction

woven delta
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oof

full blaze
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i see set of rational numbers under addition and set of positive rational numbers under multiplication

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ok continue

topaz solar
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inb4 some dumb shit about UFD

full blaze
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i know they are both groups

woven delta
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no

full blaze
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oh

woven delta
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@topaz solar

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not you gage

topaz solar
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oof, sounds fun

woven delta
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hint: there is a first order sentence which one satisfies and the other doesn't

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this is not a hard problem btw

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don't overthink

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@topaz solar any thoughts?

topaz solar
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no i am dumdum

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first order hurts me

woven delta
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lol

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what is the corresponding operation to division by 2

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if you convert the symbol + to *

topaz solar
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h m m m

woven delta
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well

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you're on the right track

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gage hbu?

topaz solar
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Would it be like the ^1/2 thing? seems wrong since you said "right track" to just ^ but idk anything else

woven delta
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yeah

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that's what it is

topaz solar
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smh here I was thinking it was wrong since "right track"

woven delta
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think about the definition of division by 2

topaz solar
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yeah, just apply some homomorphism boyes

woven delta
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yeah

topaz solar
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smh

woven delta
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but just replace the + in the definition with *

topaz solar
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well yeah since it has to convert over

woven delta
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and you get the definition of square root

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now how do you use that?

topaz solar
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the whole square root thing implies that division by 2 from the additive group is taking ^1/2 in the multiplicative group, which clearly doesn't work due to sqrt(2) not being nicely rational

woven delta
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yeah

topaz solar
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or some other arbitrary not-so-square rational

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clearly it misses at least one

woven delta
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yes

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so they are different

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you can also express this difference in FOL

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$\forall x \exists y\ y+y=x$

cloud walrusBOT
woven delta
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$\forall x \exists y\ y*y=x$

cloud walrusBOT
woven delta
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an isomorphism would make these statements equivalant

topaz solar
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ooh, a sort of isomorphism between models or some shit?

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fancy

woven delta
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well group homomorphisms lend themselves well to model theory

topaz solar
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Interesting

solar vessel
woven delta
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you can define a homomorphism in logic (a bit) more generally as a function that preserves some constants and some particular first order formulas

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this is equivalant to the standard definition for groups

topaz solar
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So it preserves some truth stuff and conjunctions or something?

woven delta
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yeah

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it preserves something called atomic formulas

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from that you can derive that it preserves a lot of other formulas as well

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for more details see enderton's a mathematical introduction to logic

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or mendelson's introduction to mathematical logic

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or hodges' model theory

solar vessel
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time to lawfully purchase those books

uncut girder
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Help

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R (Ɨ)_Z Z^n

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How to prove R^n satisfies the universal property of tensor product

oblique river
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What else do you know about tensor products? Do you know that they distribute over direct sum?

bleak oxide
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Hey, can someone tell me why (N,+) is a group? Mustn't a group have an element e such that for every a in the set we have a b that a + b = e ?

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Or maybe the book... is wrong. For (Z,+) I understand since for every a, there exists b=-a for which a+b=0=e.

somber bramble
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(ā„•, +) is not a group. depending on your definition of ā„• there wonā€™t be a neutral element (0) and thereā€™s never an inverse

bleak oxide
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There is a neutral element for the monoid group, i.e., a + 0 = a. But for the inverse nope @somber bramble ty

marble wagon
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what is this book?

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seems awful

fickle brook
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(N, +)
group

mild laurel
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N* isn't a group either?

fickle brook
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what's N*

mild laurel
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Positive integers I was thinking but

fickle brook
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what operation

mild laurel
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Multiplication cause of the cdot

bleak oxide
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(N^*,x) isn't a group either, I think :/

marble wagon
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that's because ^* means taking group of units, not just taking 0 out

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the book doesn't say this cuz it sucks

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N* is {1}, Z* is {-1, 1}

fickle brook
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the book is crap

bleak oxide
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šŸ˜¦

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Oh ok thanks @marble wagon

uncut girder
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@oblique river I got it by showing it satisfies the universal property, but distribulivity of the tensor product is something I didnt think of thanks

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Now for this problem, the last line in the image above. I need a counter example

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A is a nonzero commutative ring

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Obviously A has to be infinite

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And not finitely generated

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So take A = (+)_{n in N} Z

marble wagon
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you're almost done

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i mean, if you want unit you want the product instead of the sum

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but that's a good one to use

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but in this case A^2 = A so uh

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I don't know how you're proving the first part

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I imagine you're not showing the full problem

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oh I guess it works as A-modules

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you didn't mean as rings

uncut girder
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@marble wagon apparently its true...

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Also if A^2 = A that would give a counterexample to the 2nd to last line?

marble wagon
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actually you can have R^2 = R as R-modules

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so uh

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I don't know what's going on here I'll just walk away

marble wagon
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im being dumb cuz that can't happen in commutative rings

full blaze
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what makes a coset distinct, because im calculating the left cosets of a subgroup {p_0, u_2} of the group D_4

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using the table and calculated all the left cosets i get that

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dont get how only 4 of them are distinct when they all look distinct

fringe nexus
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if the elements are the same

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they're not distinct

full blaze
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i get that

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but

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dealing with permutations as elements

fringe nexus
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what is p?

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rotation?

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what is u?

full blaze
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ye

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ye

fringe nexus
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flip?

full blaze
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uh

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p is for rotations u is for mirror images

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d is for diagonal flips

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and D_4 has 8 permutations so ye

fringe nexus
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did you write out all the cosets?

wicked mason
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uh doesnt that set generate the whole group

full blaze
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i did

fringe nexus
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you can send a pic and i can show you which ones are the same

full blaze
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asked for all left cosets i wrote 8 left cosets

fringe nexus
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@wicked mason reflections and rotations generate d_n

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wait one sec

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yes n is even in this case

full blaze
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ye i think if i read this section in the book it shall prevail the answer

fringe nexus
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i dont think thats a normal subgroup if its just a rotation and a reflection

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for even dihedral the normal subgroups are {r^2,f} {R^2,rF} {e} and {r^d} if d | n

full blaze
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havent learned def of normal i know its number of right and left cosets being equal

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but i think we learn that tomorrow

fringe nexus
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ok but is that a subgroup though?

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its the whole group

full blaze
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yes

fringe nexus
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because as @wicked mason said reflection and rotations completely generate the dihedral group

full blaze
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the little set of 2 permutations is a subgroup

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of D_4

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of course by lagranges theorem

fringe nexus
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uhhhhhhhh

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a subgroup has to be closed right?

full blaze
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yes

fringe nexus
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under whatever operation the supergroup has

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ok so we have reflection and rotation

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then rotation ^2 should be in the group as well

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and rotation ^3 .. and so on

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it generates the whole group

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maybe i just dont know what the symbols mean

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if u_2 means r^2 then you're fine

full blaze
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those are permutations

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1 2 3

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2 1 3

fringe nexus
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ok so what does the permutation u_2 do

full blaze
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permutation table right there but thats just an example that doesnt really matter

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what doe you mean do

fringe nexus
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pandaRee like in cycle notation is it (1 3 5 7) (2 4 6 8)

full blaze
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i take an element from the group and multiply it to the left of the given subgroup

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p_0 is

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(1)(2)(3)(4) identity permutation

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i mean i got it i think

fringe nexus
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ok

full blaze
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the book shades different sections

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of the group table of D_4

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i think that helps determine if they are distinct

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i figured it out the first 4 left cosets consist of all the elements of D_4

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actually im retarded

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the other 4 cosets are the same as the other 4 just in a different arrangement

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and the arrangement of elements within a set doesnt matter in a fuckin set lulul so of course those first 4 are the distinct ones

full blaze
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question or i find it peculiar that the definition of both right and left coset make it seem like it the operation on the group does not matter, makes it seem like its just multiplication of an element from the group multiplied by the right or left of the given subgroup set

mild laurel
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What do you mean?

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That's kind of the point of defining things for all groups. We define it so it doesn't matter what group and what operation we're talking about

full blaze
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definition is like gH = { gh l h element of H }

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is the definition of a left coset

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just looks like multiplication

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but when finding the cosets of 4Z of Z

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use addition not multiplication i know these groups have the binary operation of addition but not obvious you would apply addition to the coset definition

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ya know

wicked mason
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It's always the group operation

full blaze
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i get that now

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but it never said anything about that

mild laurel
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Just like normal multiplication

full blaze
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also my professor didnt tell us that either

mild laurel
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People get lazy at writing the group operation and just put things next to each other

full blaze
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true

mild laurel
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And so it's assumed to be the group operation

full blaze
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ye i need to get in that mindset

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and he probably expects me to know that actually

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fuck

wicked mason
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the lazy mindset šŸ¤£

full blaze
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it is lazy

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your totally right when dealing with permutation groups the operation is composition of functions

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but it just looks like multiplication of two functions

wary sphinx
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Is there any reason a ring homomorphism has to map unity to unity? I remember hearing that but it seems like it's not a true thing.

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I feel like I'm maybe remembering something that only holds for isomorphisms.

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But I'm also running on no sleep during finals so I want to make sure I'm not just crazy.

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Ohhhh, I'm remembering f(1)=f(1*1), which is true and maybe does imply the thing I was saying.

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Well no, because f(1)=0 can be a thing.

topaz solar
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@wary sphinx What properties do ring homomorphisms sustain between operations like a+b and ab?

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or preserve I guess is better

brittle basin
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Hello

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is there a way to calculate the point of 2 intersecting rays using only the coordinates of the rays

chilly ocean
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If the rings have unity then ring homomorphisms builds the unity to the unity

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But rings don't have to have unities

somber bramble
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maybe yours don't

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@wary sphinx let f be a ring hom, r an element in the ring. then f(r) = f(1r) = f(1)f(r), so f(1) is the multiplicative identity of the target ring

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hm that's not 100% a solid proof yet actually. if the target is a domain, then you can cancel f(r) to get id = f(1)

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but if it's not?

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oh, you can do $f(1) = f(1)f(1)$, and then multiply on both sides by $f(1^{-1}) = f(1)^{-1}$ to get the desired result

cloud walrusBOT
chilly ocean
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multiplicative inverses dont need to exist in rings

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theres a reason that in the definition of a homomorphism, f(e_1)=e_2

somber bramble
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certainly, but if 1 exists then its inverse does too

chilly ocean
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probably because it doesnt follow from the othe properties

somber bramble
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as far as I can tell you can either demand it preserved the identity, or inverses (where they exist) and either will follow from the other

chilly ocean
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thing is, if the rings do have unities, it makes no sense to have homomorphisms not build unities to unities

somber bramble
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well yea. you just don't necessarily have to put it as an axiom

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all semantics, as far as I see it

wary sphinx
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Yeah. I think the thing I was remembering is that unity has to map to an idempotent element

somber bramble
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in the end, the important thing is that f(1) is 1

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(the other one, ofc)

marble wagon
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The reason is you want to preserve a module structure as well

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Where rings are modules over themselves with 1 as the identity

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And you could ask again why

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And well, it seems to be the right construction

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It works well with what we know about rings

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And gives us powerful results

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Every ring can be taken to have a unit, rings without unity can be included in rings with unity

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So in some sense this is a completely general notion

restive ocean
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Can someone give me an example for a non-abelian group?

marble wagon
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S3

restive ocean
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?

marble wagon
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the set of bijections (permutations) on 3 elements {1,2,3}

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under composition

restive ocean
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Ohkay,thankss

magic owl
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Matrix multiplication is also a good one

somber bramble
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my favourite group is nonabelian, the rubikā€™s cube group

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(itā€™s just all the rotations of a rubikā€™s cube that leave the centers fixed in place. itā€™s highly nonabelian, and if it was abelian then the rubikā€™s cube would be trivial)

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(itā€™s a subgroup of S48)

tulip barn
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Rotations in 3 space are non-abelian.

somber bramble
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and relatedly, the quaternion group is a nonabelian gruop with 8 elements

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{Ā±1, Ā±i, Ā±j, Ā±k} with multiplication, obeying the laws iĀ²=jĀ²=kĀ²=ijk=-1

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that one is actually ā€œas abelian as possible without being abelianā€ in some sense

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(look up the 5/8th theorem if you wanna know what I meant with that)

raw moth
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that 5/8ths theorem is cool

bleak abyss
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Oh yeah it was on my rep theory final

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Didn't end up figuring it out sadly

full blaze
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that theorem is cool actually wtf

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take to elements from a group if they commute or something with a 62.5% chance then the group is abelian

marble wagon
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that formulation kinda obscures what's going on

bleak abyss
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Yup

marble wagon
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the point is that the center is of index at least 4

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because of an easy result that it can't have prime index

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center index 4 is the same as the 5/8 bound

full blaze
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ok i see a little coset talk going on

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take the order of the group divide it by the order of the subgroup and get the index value

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of course if they are divisible by each other

marble wagon
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they are

mild laurel
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is there a name for the ring of rationals such that some prime p doesn't divide the denominator?

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I've seen this ring come up a lot

marble wagon
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localization

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Z_(p)

mild laurel
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ah

marble wagon
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you basically want to zoom in on the ideal (p)

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so you invert (add inverses of) all the elements which are outside

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and now you have a local ring with maximal ideal (p)

uncut girder
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@mild laurel where do you see it?

mild laurel
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As an example of a local ring

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Also in neukirch

uncut girder
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Funny, I just started learning this stuff in atiyah-Macdonald

mild laurel
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Think it's mentioned in there

mild laurel
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Feel like this is a dumb question but

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Is it true that all endomorphisms of an abelian group is a power map, i.e. x^d for some d

fickle brook
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no

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you

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edited your previous message?

tame bear
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i saw that

mild laurel
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Sorry meant to send that as a new message

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But edited accidentally instead

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Nothing changes if we specify that the group is the multiplicative group of a field right

solar wyvern
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@mild laurel have you thought of a counterexample yet?

mild laurel
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Because Silverman makes this statement

solar wyvern
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oh, the endomorphism ring

mild laurel
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I mean I'm sure that it's a ring homomorphism, but I can't see why it's surjective

hearty oyster
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What is the cardinality of the permutation subgroup <(1,2,6,5), (1,5,8,4), (1,2,3,4)>?

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(it is a subset of the symmetric group of order 8 isomorphic to a quotient group)

All it says is the it's isomorphic to a quotient group of a variation of a different question but doesn't tell me what

oblique river
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@mild laurel are you thinking of G_m as a scheme or just as an abelian group?

mild laurel
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I'm pretty sure it's just an abelian group

oblique river
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uhhh no it is a scheme

mild laurel
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I might be wrong here though

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Oh then that's why

oblique river
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It's Spec(K[t, t^(-1)])

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(also a group scheme for that matter)

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but it's certainly not true if you only consider G_m as an abelian group

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for example, take k = C, the complex numbesr

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then C* is isomorphic to S^1 x R* which is isomorphic to S^1 x R

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as abelian groups

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and R has many many endomorphisms as a Q-vector space

mild laurel
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Yeah okay, I also was thinking it wouldn't be injective for finite fields

oblique river
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correct; for x in F_q, we always have x^{q-1} = 1

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so M_{q-1} would just be the identity map

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so that's why we need to consider it as a scheme

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so the question, translated, is asking you to find all the k-endomorphisms of k[t, t^{-1}]

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which is equivalent to finding all of the units of k[t, t^{-1}]

mild laurel
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fk okay I have a lot to learn

oblique river
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you'll get there

mild laurel
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thanks

wicked mason
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@hearty oyster if you're still wondering that subgroup has size 5040

hearty oyster
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@wicked mason tysm

thorn delta
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Any hints on showing that AB does not have finite order? I thought about diagonalizing AB to make it more "apparent" that this is the case, but surely there is a better way....

šŸ“Ž unknown.png
bleak abyss
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You can compute the characteristic polynomial of AB, if it's not of the form t^n - 1 then AB doesn't have finite order

tame bear
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the powers of AB should have nice structure

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induction would be quick

bleak abyss
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Yup

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"There are two types of people"

tame bear
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matricies in latex scare me

thorn delta
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yeah that makes sense sigma. @bleak abyss I'm not sure I understand your explanation though

bleak abyss
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Oh I guess that wasn't exactly right

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Okay so

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Let's write AB = M

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If M^n = I, then the minimal polynomial of M must divide t^n - 1

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Not char poly, sorry

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So in small cases you can factor the polynomial, e.g. in the 2x2 case t^2 - 1 = (t+1)(t-1)

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So if a 2x2 matrix has finite order, it must have minimal polynomial t+1, t-1, or t^2 - 1

thorn delta
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i don't know what minimal polynomial is tbh

bleak abyss
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Now in this particular scenario yeah either you're \pm I, or the min poly is quadratic so the char poly is as well, so yeah

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Oh it's not bad, do you know what an ideal is?

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Or nah I won't bother with ideals

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Okay so

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Do you know the Cayley-Hamilton theorem?

thorn delta
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rip no :c

bleak abyss
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Okay so

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Let's say A is a matrix with characteristic polynomial f

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In general if you give me a polynomial I can stick in a matrix. If you give me the polynomial p(t) = t^2 + t + 1 and you give me the matrix A, then I write p(A) = A^2 + A + I (identity matrix)

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Theorem: f(A) = 0

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This takes some work to prove, the naive thing is to say "Well if f(t) = det(tI - A) then f(A) = det(AI - A) = det(A-A) = det(0) = 0" but this doesn't type check

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The outline of the proof I know: diagonal matrices are ez, you can extend to diagonalizable matrices, and those are dense in M_n(C)

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Anyway take this as a given for now

thorn delta
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all right

bleak abyss
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This is one way to see that for any matrix A, there's some polynomial p such that p(A) = 0

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Exercise: think of an easier way

thorn delta
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"think of an easier way" to do what?

tame bear
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the 0 polynomial

bleak abyss
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To prove that if A is a matrix, then there's some polynomial p such that p(A) = 0

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Okay non-zero polynomial if you wanna be a smartass lmao

tame bear
topaz solar
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dangit sigma

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smfh

tame bear
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how are you meant to prove that

fickle brook
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@bleak abyss ||{x, Ax, A^2x, ..., A^nx} can't be LI||

thorn delta
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if we assume that l and k are relatively prime, then 240 is the answer because 15 and 16 must both divide m.

But we don't know anything about k and l other than that x = g^k = h^l. I can't think of any implications of this that lead to a definite answer... thonkzoom

wicked mason
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I'd go about this differently. Elements of <g> have orders dividing |g| = 15 and similar thing with h. So if something is in the intersection...

thorn delta
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something in the intersection would have order dividing 15 and 16

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or multiples of 15 and 16....

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240 would be the smallest number that does this. yea makes sense

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and then in general if |g| = m and |h| = n or something, then the intersection of the cyclic subgroups generated by g and h would have order mn/gcd(m, n), right?

wicked mason
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in the first line, yes the order has to divide both 15 and 16

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and we know they're relatively prime

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you're thinking of lcms with 240

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plus an intersection should be smaller than the original things

thorn delta
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oh yea, i got mixed up. it would be order 1 then.

magic owl
#

Does anyone have an intuition as to why ā€œsplitā€ is used to describe split functions, split exact sequences, etc?

#

Is there an image Iā€™m not seeing? (I know the formal defs)

#

Split morphisms might be a better term

topaz solar
#

Well, for morphisms in particular, you have splits where it's the identity on one side, but not necessarily the other

#

so it's "split" between identity and not so identity

magic owl
#

By other you mean depending on composition order?

topaz solar
#

well yeah

magic owl
#

Just making sure

topaz solar
#

yeah fair

marble wagon
#

cuz it literally splits the middle thing into a direct sum

#

in abelian cats

topaz solar
#

ooh

magic owl
#

So thereā€™s not really a unifying picture? Those make sense individually

marble wagon
#

0->A->B->C->0, if either map is split, then B splits as B = A+C

#

(so the sequence splits as 0->A->A->0 and 0->C->C->0)

magic owl
#

Huh. Never made that full connection

topaz solar
#

interesting

full blaze
#

question lads

#

why would a group lets say has order 120 not have a subgroup of order 60

#

what would prevent a group to not have a subgroup of a certain order

#

of course a subgroups can have an order that divides the order of the group

marble wagon
#

consider this: Sn is generated by an order 2 and an order n element

#

few elements can generate huge groups

#

if the group is far for being abelian

fickle brook
#

ok but S_5 contains A_5

marble wagon
#

yes this isn't a (120,60) example

#

although it is a (120, 15) example as S5 doesn't have an order 15 subgroup

#

no order 30 subgroup either

topaz solar
#

The Lagrange (? idk i probably butchered that spelling hardcore) thing only says the subgroups divide the order, not that all factors have at least one corresponding subgroup

marble wagon
#

if you want an (n, n/2) example, consider any simple group of even order

#

they don't have order n/2 subgroups as index 2 subgroups are normal

fickle brook
#

lagrange

topaz solar
#

Thank

tame bear
#

monster group is simple of even order
So it works as a (808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,000,
808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,000/2) example

marble wagon
#

every nonabelian simple group is actually of even order

#

by a deep theorem

bleak abyss
#

@marble wagon Atiyah proved that in just 12 pages šŸ™ƒ

marble wagon
#

lmfao

#

legit? or after he wasn't so good

#

I know the answer šŸ˜¢

full blaze
#

lads

#

yall went in a completely opposite direction

#

let me restate why can a group not have a subgroup of an order that divides the order of the group

#

e.g a group of order 120, prove that it cannot have a subgroup of order 60

marble wagon
#

your question was answered above

full blaze
#

no

marble wagon
#

wow

#

way to waste my time

bleak abyss
#

It was, depends on the group

marble wagon
#

fucking brainlet

full blaze
#

hm

topaz solar
#

I mean, you obviously can have a group with order 120 that has a subgroup of order 60, like the classic C_60 x C_2 which i'm p sure fits that bill

full blaze
#

it does not have a subgroup of order 60

#

thats what im trying to prove

topaz solar
#

Proving that a group doesn't have a subgroup of that order would require additional information about the specific group at hand

full blaze
#

ok

#

lemme post it

topaz solar
#

since there certainly exist groups with order 60 subgroups

bleak abyss
#

Why did you not post the whole question from the start?

full blaze
#

just wanted a general way of going about it so i could solve it

bleak abyss
#

There's no reason for there to be a general way, it depends on the specifics

topaz solar
#

^

full blaze
#

got it

bleak abyss
#

Okay that's easy

#

A subgroup of order n/2 is normal

#

But the normal subgroups have size 1, 2 (the center), and 120

#

gg

topaz solar
#

g g

full blaze
#

explain further im not grasping it

bleak abyss
#

If G is a group of order n and H is a subgroup of order n/2, then H is a normal subgroup of G

marble wagon
#

index 2 subgroups are normal as was said above

full blaze
#

i get that yes

bleak abyss
#

But we already know the orders of the normal subgroups of SL(2,5)

full blaze
#

what reflexive said how does that help tho

#

ok

#

true

#

ah

bleak abyss
#

The only normal subgroups are the trivial subgroup of order 1, Z(G) which has order 2, and the whole group which has order 120

full blaze
#

yes

bleak abyss
#

So we're screwed

full blaze
#

ah i see

topaz solar
#

That also shows that a simple group can't have a normal subgroup of order n/2, since the whole "simple" part

#

not so nice for the whole "no subgroup at all" part but it kicks a full case out

#

(and since it's always normal you get a big oof)

full blaze
#

so what i would say is

#

since there are 3 normal subgroups and there isnt a normal subgroup of order 60 there cant be a subgroup of order 60

#

or no

topaz solar
#

You say there's no subgroup of order 60 since any subgroup of index 2 is normal

#

(with some fill in the blank about normality)

full blaze
#

oh shit

#

brother

#

thanks

bleak abyss
#

A subgroup of order 60 would be a normal subgroup of order 60, but there is no normal subgroup of order 60

full blaze
#

appreciate it i got it now

#

well done lads

#

šŸ‘

#

just remembered he said that there would be a question about normal subgroups and index of 2

#

but couldnt piece it together thanks

thorn delta
#

so, some $a^{km}$ is in the intersection if $n$ divides $km$. Then wouldn't the generator of this subgroup be $a^c$ where $c$ is the least common multiple of $m$ and $n$?

cloud walrusBOT
mild laurel
#

Yes

chilly ocean
#

"ys" here is meant to mean s(y)

#

Then you would go onto prove that all "integral systems" are isomorphic

#

and hence define natural number sequence as an integral system

oblique river
#

that's basically the peano axioms

#

(not for arithmetic, but for the naturals)

#

which is the "standard" way to define the naturals

chilly ocean
#

Peano axioms huh

topaz solar
#

That's literally just peano (sans the whole induction and arithmetic part) but generalized so it's not about the particular model but rather the univeral structure as a thing with a unary and nullary operator

oblique river
#

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

Peano axioms

In mathematical logic, the Peano axioms, also known as the Dedekindā€“Peano axioms or the Peano postulates, are axioms for the natural numbers presented by the 19th century Italian mathematician Giuseppe Peano. These axioms have been used nearly unchanged in a number of metam...

chilly ocean
#

that will come up later in this book I think as well

#

explicitly

topaz solar
#

But yeah it's literally just peano

oblique river
#

waht do you mean "sans induction"

topaz solar
#

there's not the whole induction axiom schema thing

#

though I guess you could make that an integral system actually

oblique river
#

the statement I_2 is induction

topaz solar
#

Yeah it is i'm rarted

#

f u c k

oblique river
#

I don't think induction is an axiom schema

#

it's just an axiom

#

wait nvm

topaz solar
#

there's a second order induction axiom, but iirc the first order one is a schema

oblique river
#

yeah you're right

#

I do my best to never do any math that requires me to distinguish between first- and second-order logic haha

topaz solar
#

Though technically I2 is just induction but saying theres an isomorphism with integral systems or some dumb wording of that

#

Like uh I2 is the whole "yeah induction works"

#

then you just say "lmao integral system"

#

$\forall p(p_1 \land (p_n \implies p_{n^\prime})\implies \forall n p_n)$ something right?

cloud walrusBOT
oblique river
chilly ocean
#

oh yeah , and for the above screen shot I posted with the <N,',0> integral system , ' is a function with it being '={<n,n'>| $ n \in \mathbb{N}$}

cloud walrusBOT
chilly ocean
#

oh ffs

#

ahhhhhh

oblique river
#

lmao

chilly ocean
#

this was my first time using uh

topaz solar
#

yeah it probably goes in #proofs-and-logic or some such, but the way it's formulated in his text seems to take a slightly univeral algebra/model approach

chilly ocean
#

latex in a convo

#

lOL

#

LOL

#

this shti is slow

#

yeah I have no idea where to ask these questions

#

its just uh

#

set theory to me so idk shit

topaz solar
#

so it just says an integral system is an algebra with a signature (1, 0) where the unary operation is bijective to X\{nullary boy}

#

smh

#

there's probably a nicer way to formalize that statement but forget it

#

Anyhow, that formulation is literally just peano

#

so that's something for you to look up if you need more stuff on it

chilly ocean
#

I'm into this book, its pretty old and I dont even understand these terms like "bijective" lmao

#

when other people use them

#

even though its covered a fair bit so far

topaz solar
#

Bijective: the function is injective and surjective

chilly ocean
#

yeah

#

its like when you ask a doctor for a definition and they give you more medical terms, but I can google it ofc

#

I think this book is weird

#

but its good so

#

ill keep going

topaz solar
#

nice

#

Oh and bijective is the way you handle "same number of elements"

chilly ocean
#

alright

#

I know what the problem is

#

its just a tiny bit abstract

#

all of this

#

just a little bit

thorn delta
#

Prove that $\bbZ_n$ has an even number of generators for $n>2$.

cloud walrusBOT
thorn delta
#

tiny hint maybe?

oblique river
#

a standard way to prove that some set is even is to find a way to pair up elements of the set

#

so what you should do is try to find a way to pair up the generators of Z_n

tame bear
#

it would be better phrased as "has an even number of possible generators" right?

oblique river
#

I see what you mean but I think the current phrasing is fine

thorn delta
#

thonkingrealhard

topaz solar
#

The current phrasing made me take a thonk for about a second till i'm like "oh yeah, things that can generate"

thorn delta
#

ok after a lot of staring i managed to figure this out.

For any m relatively prime to n, m is a generator of Z_n. Since m is relatively prime to n, n-m is relatively prime to n as well.

I can't think of a good proof for the last part other than that gcd(m,n)=1 implies that there is nothing that can be factored out of n-m that can also be factored out of n. Just kinda obvious šŸ¤·

mild laurel
#

well what is n - m (mod n)

thorn delta
#

-m?

tame bear
#

yeah, so its also relatively prime

#

only off by a -1

thorn delta
#

yep i see

brisk granite
#

what is a transposition?

#

Like, in the context of symmetric groups

fickle brook
#

a permutation which swaps two points

brisk granite
#

could you give me an example

#

?

chilly ocean
#

123 to 213

#

1 and 2 swapped and 3 didn't move

fickle brook
#

let me clarify

#

a transposition is a permutation which swaps two points and does nothing to all the others

brisk granite
#

k thx

chilly ocean
#

Who can help me with figural reasonning?

woven delta
#

This is probably not the right channel for you

full blaze
#

the symmetric group S_n is generated by transpositions

mild laurel
#

A_n is generated by 3 cycles

full blaze
#

tru

mild laurel
#

Prove it

fringe nexus
#

just write out the two cases and you're done pandaThink

woven delta
#

Hey Victoria

fringe nexus
#

hey liquid

woven delta
#

Haven't seen you in a while

fringe nexus
#

pandaOhNo I havent done math in a while

#

i've been taking a break

woven delta
fringe nexus
#

It is summer after all :(

full blaze
#

A_n is the set of all even permutations

fringe nexus
#

yes

#

so take two transpositions

#

and theres two cases here

#

and you're done

full blaze
#

and any k - cycle where k is odd produces even number of transpositions

fringe nexus
#

o

full blaze
brisk granite
#

Consider the finite abelian group G consisting of the set of elements ${a_1, a_2 , a_3 , \dots, a_m, b_1, b_2, b_3, \dots, b_n}$ - $m$ of which are of order two and $n$ of which aren't. The elements of order two are their own inverse where as those of a greater order aren't. I.e. for all elements $b_k$, there exists a unique inverse $b^{\prime}{k}$ such that $b_k \neq b^{\prime}{k}$. \

Hence, $$a_1 a_2 a_3 \dots a_m b_1 b_2 b_3 \dots b_n = a_1 a_2 a_3 \dots a_m e = a_1 a_2 a_3 \dots a_m$$

cloud walrusBOT
brisk granite
#

does this look ok?

#

Does it make sense?

marble wagon
#

that's pretty botched up

#

it isn't clear what you're saying

#

and the little things that are clear are redundant

brisk granite
#

ok, I'll try to redo it

#

I'm trying to show that the in a finite abelian group, the product of all elements = the product of elements of order two

marble wagon
#

oh okay

#

the idea is fine, you just need to write it properly

#

writing it as b' k is misleading, you mean b k'

#

since it's just a different index that depends on k

brisk granite
#

oh ok

marble wagon
#

and be more explicit about the idea that you're pairing off each element with its inverse

brisk granite
#

ok

marble wagon
#

and that you're taking the product of all the elements

#

and showing it's the product of order 2 elements

brisk granite
#

ok, will do

#

thankyou

brisk granite
#

\newcommand{\mysubparagraph}[1]{\subparagraph{#1}\mbox{}\}
\mysubparagraph{Product of the Elements of an Abelian Group}
Consider the finite abelian group G consisting of the set of elements ${a_1, a_2 , a_3 , \dots, a_m, b_1, b_2, b_3, \dots, b_n}$ - $m$ of which are of order two and $n$ of which aren't. The elements of order two are their own inverse where as those of a greater order aren't. I.e. for all elements $b_k$, there exists a unique inverse $b_{k^{\prime}}$ such that $b_k \neq b_{k^{\prime}}$. \

Hence, we can effectively pair up the elements $b_k$ with their inverses to show that
$$a_1 a_2 a_3 \dots a_m b_1 b_2 b_3 \dots b_n = a_1 a_2 a_3 \dots a_m e = a_1 a_2 a_3 \dots a_m$$

\mysubparagraph{Proof of Wilson's Theorem }
Wilson's Theorem:
$$ \text(p-1)! \equiv -1 \mod p $$

Consider the group $\mathbb{Z}/p\mathbb{Z}$ (integers modulo $p$) under multiplication (Note that this is a group because $p$ is prime). Any element of this group may be defined as $p-k$ such that $k \in [1, p-1]$, and it holds that $(p-k)^2 \equiv k^2 \mod p$. Hence, the only element of order two is $p-1$ and $1$. From the previous sub-paragraph, one may then conclude that the product of the elements of $\mathbb{Z}/p\mathbb{Z}$ is $p-1$. I.e. $(p-1)! \equiv p-1 \mod p \implies (p-1)! \equiv -1 \mod p$

cloud walrusBOT
brisk granite
#

does this look ok?

marble wagon
#

hence, the only element of order two is p-1 and 1
can't see how you concluded this

#

instead of saying "previous subparagraph", call that result a lemma

brisk granite
#

oh ok

marble wagon
#

the rest looks fine

woven delta
#

Is that even true?

marble wagon
#

yeah cuz it's a field

brisk granite
#

Is Wilson's theorem true?

marble wagon
#

x^2 - 1 has 2 roots

woven delta
#

Oh right

#

Lol

#

I was thinking of using lagranges theorem to show that the number of elements of order 2 is odd lol

marble wagon
#

monkas

woven delta
#

But I guess it's much more trivial

#

I feel like that argument should work for a generalization of this statement

marble wagon
#

the product of elements in k* is -1?

#

sure

woven delta
#

Hmm yeah I guess

#

Eh whatever

#

Not so interesting

marble wagon
#

you could generalize in principle to finite UFDs

#

but those are fields so

brisk granite
#

what is a character table?

lethal gazelle
#

You can break a character table up into five parts

#

Character tables contain information about how functions transform in response to the operations of the group

hot lake
#

for a finite group it is a table where rows are indexed by irreducible characters (up to isomorphism) and columns are indexed by conjugacy classes of the group, and the entries are the trace of the representation of any element of the conjugacy class

lethal gazelle
#

@brisk granite Is this clear to you?

marble wagon
#

lol wtf

lethal gazelle
#

sh xD

lone shard
#

sh: xD: No such file or directory

lethal gazelle
#

xD

#

;)

#

Christopher C. Cummins is evil.

#

Math?

full blaze
#

ring theory squad

#

whats more challenging or important ring or group theory

marble wagon
#

these people

stone fulcrum
#

Commutative algebra (study on rings) is pretty difficult, but groups are where its at

full blaze
#

i see kayn

prisma ibex
#

from my biased opinion, commutative algebra is one of the most important topics in algebra

#

commutative and representation theory are the two most important topics

brisk granite
#

I would like to find certain subgroups of $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$ where p is any arbitrary prime. Specifically, those subgroups of order $p$. Since p is prime, I know that the order of every element is $p$, and, hence, I can generate subgroups of order $p$ by considering any element in $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$ and creating a cyclic group from said element. Each cyclic group will have $p-1$ elements (excluding the identity) and there are exactly $p^2 -1$ elements other than the identity in the group. Thus, there must be at least $\frac{p^2 -1}{p-1} = p+1$ subgroups, but I have no idea how to conclusively say that this the total amount of subgroups of order $p$.

cloud walrusBOT
brisk granite
#

how to?

marble wagon
#

just be more careful in your reasoning

#

it contains all the ingredients already

brisk granite
#

wdym?

marble wagon
#

your paragraph already has the answer

#

you just need to stare at it more carefully

brisk granite
#

how do I know that all subgroups of order $p$ are cyclic though?

cloud walrusBOT
brisk granite
#

that's the part I'm stuck on

marble wagon
#

oh

#

prove that a group of order p is cyclic

brisk granite
#

ok, I'll try that

#

Is that true for any arbitrary group?

marble wagon
#

ye

brisk granite
#

k

fickle brook
#

no, it is not true that any arbitrary group is cyclic @brisk granite

smoky cypress
#

Any arbitrary group with prime order is cyclic

full blaze
#

by theorem

chilly ocean
#

ya the only groups of prime order are simple cyclic groups

brisk granite
#

@fickle brook I meant with a prime order. I was trying to clarify whether he meant it was specifically for the subgroups I mentioned above

#

Is there any way to prove it without Lagrange?

#

I haven't learned it yet; so, I'm not sure if it would be fair to use it

mild laurel
#

Prove what with Lagranges? @brisk granite

tame bear
#

prime order groups are cyclic

mild laurel
#

Yeah you can

#

Take any element of your group and look at the subgroup it generates

#

E.g. take some element a \in G and look at {a, a^2, a^3,\dots , a^n}. We know that this must be a finite set since our group is finite

#

Furthermore, you can pretty easily show that this subset of your group forms a subgroup as well

#

But Lagrange's theorem says that the order of any subgroup must divide the order of the group, which in this case, is p

#

Thus, we either have that the subgroup is of order 1 or order p

#

The identity will always be in this subgroup, so as long as we take the element a not to be the identity, this subgroup has more than one element

tame bear
#

without lagranges

mild laurel
#

So thus the subgroup has to have order p and the subgroup is cyclic

woven delta
#

Lol

#

He knows how to prove it with lagranges

#

But apparently his class hasn't done lagrange yet

#

@mild laurel

mild laurel
#

o

#

im illiterate

woven delta
#

Just prove cauchys theorem lmao

#

Ez

hollow peak
#

Or prove Lagrange weSmart

dense patio
#

Hi guys I'm good at problem solving but not quite there yet...
I was hoping that there might be some epic maths boss here that could teach me more techniques over discord...
please DM me

full blaze
brisk granite
#

seems a little vague

#

I would consider myself an epic math boss though

jagged gate
brisk granite
#

is this the right place?

#

are there rules against this kinda stuff?

#

def cool pics though

jagged gate
#

yah :/

brisk granite
#

Note the all groups consisting of only elements (barring the identity) of the same order as the group are cyclic. To illustrate this, consider a group $G$ - defined such that $\forall x \in G, ; |G| = |x| = n$ - and a single element $x$ in $G$. It holds that $|\langle x \rangle| = n$, and, thus, $G = \langle x \rangle$.\

Now consider the group $\underbrace{\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}...\times \mathbb{Z}/p\mathbb{Z}}_{n \text{ times}} = G$ for any arbitrary prime $p$. Since $p$ is prime, it holds that the order of any element (other than the identity) must be $p$ and that all subgroups of order $p$ are cyclic.\

Then, notice that there are exactly $p^{n}-1$ elements in $G$ (excluding the identity) and $p-1$ elements (excluding the identity) in any subgroup of order $p$ of $G$. Hence, there must be exactly $\frac{p^{n} -1 }{p-1}$ possible subgroups of order $p$ in $G$

cloud walrusBOT
brisk granite
#

So, I extended the answer I presented before to a more generic case. does this look complete and correct?

hollow peak
#

No

#

It's not clear when you say "Hence" in the last paragraph

marble wagon
#

yes

#

it's clear

hollow peak
#

oh

#

i didnt see "cyclic" . Yes it is mb

marble wagon
#

the part where you prove an order p group is cyclic is very confusing

wind steeple
#

e has order 1 and can't be order n...

chilly ocean
#

ya

#

you need for all x in G\e

cloud walrusBOT
clear heath
chilly ocean
#

That's equivalent to showing that K is normal in G, right?

clear heath
#

i have no idea what that means sorry

chilly ocean
#

ahh gotcha

#

well we know that K is a subgroup of G and that for all k in K, f(k) = eH, right?

clear heath
#

yeah

chilly ocean
#

and we know f is a homomorphism, so for all g in G, f(k ā€¢Ā g) = f(k) ā€¢ f(g)

clear heath
#

yeah

chilly ocean
#

see if that helps you out

#

er one more step for yaā€” we then know from the above that f(k ā€¢ g) = f(k) ā€¢Ā f(g) = eH ā€¢ f(g) = f(g)

clear heath
#

f(k g) = f(k) f(g) = eH f(g) = f(g)

#

i was just writing that lmao

chilly ocean
#

right

clear heath
#

f(g) = f(g) eH = f(g) f(k) = f(g k )

#

?

chilly ocean
#

yes

#

and this is true for all k in K

clear heath
#

yes so the left and right cosets are equal

chilly ocean
#

tada

clear heath
#

thats a much simpler solution than what they gave

chilly ocean
#

right

clear heath
#

i dont understand what they do

chilly ocean
#

if you're taking a course on abstract algebra right now you'll learn what a normal subgroup is pretty soon

#

and what (i) is showing is that K is a normal subgroup of G

#

which implies that its left cosets are the same as its right cosets

#

as well as some other nice properties

clear heath
#

well this is in high school

chilly ocean
#

damn that's impressive

clear heath
#

in IB, the set theory option

chilly ocean
#

gotcha

clear heath
#

so the understanding they provide us with is very shallow in all the options

chilly ocean
#

I see I see

clear heath
#

i dont really have an idea of a lot of bigger concepts, and the same is seen with most topics

chilly ocean
#

well it's a good way to dip your toes into the things you'd get in an introductory group theory course

clear heath
#

how exactly would you get from knowing that K is a normal subgroup to the left and right cosets being equal

chilly ocean
#

the lazy answer is by copying and pasting the proof for 10.1 from my abstract algebra textbook

clear heath
#

sure lmao

#

i just wanna see something tangible

sharp sonnet
#

it depends how you define normal subgroup

clear heath
#

šŸ˜¦

sharp sonnet
#

sometimes normal subgroup is defined by right and left cosets being equal

chilly ocean
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Sorry, 11.1

sharp sonnet
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but any book on algebra has some part "the following conditions are equivalent"

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and it will mention a few things that are equivalent to being a normal subgroup

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each of those could be used as the definition as well

clear heath
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Ahhh i was thinking of something like that but I lacked the capability to actually do it

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Oh interesting

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Thanks!

chilly ocean
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Book in question is Dan Saracino Abstract Algebra ā€” A First Course 2nd Ed.

clear heath
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When is an abstract Algebra course normally taken?

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Second year?

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I looked at my first year stuff and it's pretty much all calculus and linear algebra

chilly ocean
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I took it at the end of my sophomore year, but I go to a liberal arts college which doesn't emphasize math really at all (and since we're right next to CalTech, the zealous mathy students go over there to take tough classes)

clear heath
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Sophomore = ?

chilly ocean
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I think most students at CalTech take it their first year but that's also bc they usually come in having had all of multivariable calculus and linear algebra

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Sophomore = 2nd year

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4th semester

clear heath
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Oh okay

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Wait why would they already know those things?

chilly ocean
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because if you get into caltech and plan on doing math it probably means you're a real whiz and have had lots of crazy classes already

clear heath
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Oh that's a scam for international students isn't it

chilly ocean
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there's some math prep school in the bay that teaches introductory real analysis the first year of high school

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prooobably

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idk so much about how these things affect international students

sharp sonnet
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in europe you dont take calculus at all

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math students start with real analysis and linear algebra

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in your first year

clear heath
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I'll be done with AP calc content and linear algebra, and basic set theory, and first year stats by the end of my high school

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But defs not multivariable

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At least not formally

chilly ocean
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might be better convo for #math-discussion or another channel at this point, but that's better than I had at the end of high school

clear heath
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Sure

full blaze
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indeed multiple ways to define if a subgroup is normal

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such as if you have a kernal of a known homomorphism, that is a normal subgroup of a group

brisk granite
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what is a normal subgroup?

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Like, what's the difference between a normal subgroup and a subgroup, I mean

chilly ocean
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that's a formal definition

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a more intuitive definition is: a normal subgroup is a subgroup which has some nice properties and let you do nice things. One of the nice things is that it's the foundation of the quotient group, where you basically are dividing your group by a normal subgroup

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if the subgroup's not normal, the quotient group doesn't form a group, but if it is normal, then it does form a group, so it has lots of nice properties that are useful for people

marble wagon
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you wanna put a group structure on the cosets gH of a group G

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for this it should be true that g1H g2H = g1 g2 H

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in other words, that H g2 = g2 H

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this is the defining property of a normal subgroup

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left and right cosets agree

brisk granite
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Is there any way to show two groups are isomorphic without actually defining an isomorphism and showing it?

mild laurel
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Find an inverse homomorphism

full blaze
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many ways to show 2 groups are isomorphic

tame bear
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assume they arent and find a contradiction

full blaze
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look at abelian vs non abelian, also subgroup orders/# of subgroups have to match with isomorphic groups that is a structural property, also countable/order, also if you know a large infinite group such as C (set of complex numbers) you know the reals, and integers, and rationals are subsets of that big set which is also a group

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@brisk granite

brisk granite
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I can't find an isomorphism that works.

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Actually, I only tried $f((m,n)) = \tau^{n}\rho^{m}$

cloud walrusBOT
brisk granite
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I think that only works $\phi$ is the trivial homomorphism though

cloud walrusBOT
marble wagon
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no, that works

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well

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the order 2 element is rho here?

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your map is the natural choice, and it works

brisk granite
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The order 2 element is tau

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Tau is the reflection

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Rho is the rotation

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@marble wagon are you sure it works?

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cuz, I think I showed that f(ab) not equal to f(a)f(b). so, that would suggest it isn't even a homomorphism

marble wagon
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do it again

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it works fine

brisk granite
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ok

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$f((a,b))f((c,d)) = \tau^{b} \rho^{a} \tau^{d} \rho^{c} = \tau^{b+d}\rho^{c - a}$\
$f((a,b)(c,d)) = f((a \phi(b)(c), bd)) = \tau^{bd}\rho^{a \phi(b)(c)}$

cloud walrusBOT
marble wagon
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the first line only holds if d = 1

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do it case by case, with d = 1 and d = 0

brisk granite
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oh shit

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thx

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also, in the second row, that should say b+d and a + phi(b)(c)

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when d = 1, I don't see how it works

marble wagon
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try an example

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fix n = 3

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for example

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and look at both groups

brisk granite
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when d = 1 and b = 0, if the top and bottom rows were equal, then $$\tau\rho^{c - a} = \tau\rho^{a + c}$$

cloud walrusBOT
brisk granite
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How does that make any sense?

marble wagon
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oh you're doing it the other way around

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try rho^a tau^b instead

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for the morphism

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you know, it's one of the two

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didn't realize this one didn't work but it makes sense

uncut girder
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What are the radical ideals in the multivariate polynomial ring over an algebraically closed field?

marble wagon
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finitely generated by irreducibles

misty aspen
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possibly dumb question: if b is not a root in a field F and a+sqrt(b)= c+sqrt(d) for some a,c, and d in F. Does sqrt(b) = sqrt(d)? all I deduce is that Sqrt(b) -sqrt(d) is in F

hot lake
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what does it mean to not be a root in a field ?

misty aspen
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sorry i meant not a squard

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square

hot lake
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yeah it should imply b=d

misty aspen
#

just verified this, thanks for replying

brisk granite
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@marble wagon I tried it the other way around and it worked!

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Like, where I have rho tau instead of tau rho

marble wagon
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yeah

uncut girder
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Can someone explain why the last sentence works?

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Why is A/q^c isomorphic to a subring of B/q

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Consider the quotient map pi: B->B/q.
Now consider g=piā€¢f: A->B/q. It is a ring homomorphism so by an isomorphism theorem, A/ker g is isomorphic to g(A) which is a subring of B/q. Notice also that ker g = f^-1 (q)=q^c.
So A/q^c is isomorphic to a subring of B/q.

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šŸ‘ šŸ‘

full blaze
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is this a theorem or corollary : any field is a field of fractions of itself

marble wagon
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a remark

full blaze
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really

marble wagon
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yes

full blaze
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ok]

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ill just put stated by prof

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quite self evident

marble wagon
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remark is a simple fact that's easily verified

full blaze
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indeed

marble wagon
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proposition is a heavier result, which takes some work and is of value of itself. if it's used as a key ingredient for a bigger result it's called a lemma. and a big result is a theorem

full blaze
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PepoG clean

brisk granite
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So, I'm having trouble proving that an element that can be decomposed into an even number of transpositions can only be decomposed into an even number of transpositions

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I've been stuck for a while tbh

mild laurel
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one way to do it is by looking at the number of inversions

brisk granite
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the inversions?

mild laurel
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pairs (i,j) such that i < j but sigma(i) > sigma(j)

brisk granite
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what is sigma tho?

full blaze
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if you have a odd k-cycle permutation

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that same permutation can be expressed in terms of an even number of transpositions

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aka a 2-cycle

brisk granite
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what is an odd k-cycle?

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the odd part I mean

full blaze
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for example if you have the permutation (1 2 3 4 5 ) = (15)(14)(13)(12)

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you see

brisk granite
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yep

full blaze
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there you go

brisk granite
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what is an odd cycle though?

full blaze
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has an odd number of length

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k is represented as the number of numbers within the permutation

brisk granite
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are you sure you read my thing correctly

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?

full blaze
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(1 2 3 4 5 6 7 8) is an 8 -cycle

brisk granite
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ok

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how does this help me though

full blaze
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i just showed you

brisk granite
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showed me what

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than an odd cycle can be decomposed into the product of an even number of transpositions?

full blaze
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a cycle that has an odd k value can be written as a even product of transpositions

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that is what i showed

brisk granite
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ok

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that is not what I'm tasked with showing

full blaze
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if k is even then you will have an odd product of transpositions

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post the question

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jesus

brisk granite
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"So, I'm having trouble proving that an element that can be decomposed into an even number of transpositions can only be decomposed into an even number of transpositions"

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this what I posted earlier

full blaze
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i meant take a picture

mild laurel
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I just don't think you understand his question

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It's pretty clear already

tame bear
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i dont see how inversions help

mild laurel
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||The parity of inversions is equivalent to the parity of the number of transpositions||

tame bear
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ofcourse i didnt see it, i messed up the example i was thinking of

brisk granite
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so, I don't understand what orbits are for a symmetric group. like, if I had a permutation f in S_n, then what would orbit(f) mean?

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I think I understand what an orbit is, but I have no idea what it means here

bleak abyss
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You don't have orbits of group elements

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You have a group acting on a set, and can ask about the orbit of the element of the set

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So S_n acts on {1,...,n}

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And you can ask what's the orbit of, say, 3?

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In this case, it's the whole set, since if you give me any other element in S_n, say 7, I just take the transposition (3 7)

brisk granite
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the second answer uses stuff like orbit(s)

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what does that mean?

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And, s is an element of the symmetric group...

bleak abyss
#

Oh

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This is shit terminology

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Fraleigh needs to get shot

brisk granite
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lmao

bleak abyss
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I don't care if the rest of the book is good tbh