#groups-rings-fields

406252 messages · Page 417 of 407

solar wyvern
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does SA know what normal subgroups are yet thonker

raw moth
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or quotient groups

covert vector
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no probably not

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well u don't need quotients here 🤔

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cuz it's not normal a priori

raw moth
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o lol I misread

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didn't see the order signs

solar wyvern
covert vector
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or you can write |G:H| = p

raw moth
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ya

covert vector
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this is the index of H in G

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how many slices of H there are in your loaf of G

solar wyvern
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For $H \leq G$ meaning $H$ a subgroup of $G$,
$$aH:= {ah: h \in H \leq G}$$
$$G:H := Card({aH:a \in G})$$

simple valley
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that would be / not : would it

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ah

covert vector
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now how would we denote double loafs for double cosets thonkeyes

cloud walrusBOT
solar wyvern
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what's a loaf thonker

covert vector
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it's like bread

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t!wiki loaf of bread

fossil mangoBOT
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📖 | ** https://en.wikipedia.org/wiki/Bread_Loaf **

Bread Loaf

Bread Loaf may refer to:

Bread Loaf, Vermont
Bread Loaf Mountain in Vermont
Bread Loaf School of English at Middlebury College
Bread Loaf Writers' Conference, held annually at the Bread Loaf Inn, near Bread Loaf Mountain in Vermont
Loaf of Bread Butte, a mountain in Montana

solar wyvern
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🍞

covert vector
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Ya that

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that's how I think of cosets

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where the loaf is the whole group

solar wyvern
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maybe you could think of each loaf as a slice of a bigger loaf

raw moth
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4D bread

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I like it

covert vector
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no they are more like rounded loaf, I don't think I like that higher level of bread

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but anyway if u go to the bakery u can ask for different thickness of slices, and they can slice the bread for you

solar wyvern
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or it could be like french toast sticks where each "slice" is a loaf

covert vector
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so in this way you can get cosets for different subgroups

solar wyvern
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with bred

covert vector
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and youd have to be a master baker to convince me when you slice a big loaf of bread that the slices are closed under conjugation

solar wyvern
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double cosets are like HaK right?

covert vector
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therefore not all subgroups are normal cathonk

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ya

spark plank
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nope, what's normal subgroups and quotient groups cathonk

raw moth
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ignore quotient groups for now (probably a short amount of time)

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a subgroup H is normal in G if and only if gH = Hg for all g in G

spark plank
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so H is...?

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a group?

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an element?

raw moth
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a subgroup of G

spark plank
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what's gH and Hg mean then

raw moth
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o

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if g is an element of G, and H is a subgroup of G

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then Hg = {hg : h in H}

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and gH = {gh : h in H}

spark plank
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kk

raw moth
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these are known as cosets

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Hg is the right coset of H with respect to g, gH is the left coset of H

spark plank
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what does "H is normal in G" mean?

raw moth
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it means gH = Hg for all g in G

spark plank
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oh right

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kek

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I feel like woog's question is above my paygrade

raw moth
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hmm try this

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(separate problem)

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prove that if H is a subgroup of G, then two left cosets xH and yH for x,y in G are either disjoint or identical

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oh I guess also prove that a coset of H has the same (cardinal) number of elements as H as well

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but that one's probably simpler

spark plank
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thonker that sounds too trivial

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the cardinal question

raw moth
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ya it is pretty trivial

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but we'll need the result for something after you do the coset one :^)

spark plank
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$xh_1=yh_2$\$xh_1h_1'h_3=yh_2h_1'h_3$\$xh_3=yh_2h_1'h_3$\Similarly,\$xh_1h_2'h_3=yh_3$

cloud walrusBOT
spark plank
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for any h_3 in H

raw moth
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h_1' is inverse?

spark plank
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yeah

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since H is a group, products of h's are in H

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So if xH shares an element with yH, every element of xH is an element of yH and vice versa

raw moth
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ya since it's any h_3 in H you get the entire coset

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so now it's time for lagrange's theorem

spark plank
raw moth
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if H is a subgroup of a finite group G, then |H| divides |G|

spark plank
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|H| is the cardinality of H?

raw moth
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it's called the order but yes

spark plank
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thonker I feel like notation abuse

covert vector
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:3

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if you want, it's less abusive if you define order of element g as |<g>|

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where <g> is all powers of g

spark plank
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kk

sullen flint
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<g> would be the cyclic group generated by g right?

spark plank
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I thonk so

solar wyvern
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ya

sullen flint
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it's better to define it like that I guess

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that way there isn't order of elements and order of groups

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just order of groups

spark plank
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mmm I dunno how to proof @raw moth

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hint?

raw moth
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use the two things we just proved

spark plank
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thonker or is this a nuke

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hm

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kk

raw moth
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if you want a different hint

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for an arbitrary element g in G, is g necessarily in a left (right) coset of H?

spark plank
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g in gH?

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cuz H has identity element

raw moth
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ye

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so the cosets gH cover the entire group G

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however, we also know that the cosets are either disjoint or identical

spark plank
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So there's a subset G° of G including e and for every g in G°, we have gH = H

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and for every g in G\G°, gH and H are disjoint

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we also know that |gH| = |H|

raw moth
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ya

spark plank
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hm, still dunno

raw moth
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well, pick some g in G/G°

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say, g_1

stone fulcrum
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Oh wow we're doing quotient groups already k then

raw moth
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and consider g_1H

spark plank
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oh okay ic

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wait nvm

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I imagine we want to do something about the amount of g's such that g_1H = g_2H

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and show that there's an equal amount of g_2's for each g_1

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or something like that

raw moth
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if g_1H is in g_2H then since they're disjoint or identical, g_1H = g_2H

spark plank
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you mean subset(?)

raw moth
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sorry

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I meant if g_1 is in g_2H

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because g_1 in g_1H also

spark plank
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Ah

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and vice versa, meaning there are only |H| many g_1 for each g_2 such that g_1H = g_2H

raw moth
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ye

spark plank
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that means you can divide cosets of H into sets with cardinality |H|

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and altogether they equal |G|

raw moth
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ye

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you can write the elements of G as the union of disjoint (left or right) cosets of H

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by just taking a coset under some element of G that isn't in any of the previous cosets

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and then since the cosets all have cardinality |H| and they're disjoint

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|H| divides |G|

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the number of distinct cosets of H is called the index of H in G, and denoted by |G : H| (or sometimes [G: H])

stone fulcrum
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Get Lagrange'd

spark plank
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but yeah I get that

raw moth
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cool theorem

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I guess another way of writing it is |G| = |H| |G : H|

stone fulcrum
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Well SA you're like a month into the course already how do you feel

spark plank
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seems trivial enuf

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but I smell matrices 🤢

stone fulcrum
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Lel, fair enough!
No, no matricies it doesn't need to happen

spark plank
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eyesb great

stone fulcrum
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Representations have matricies but we don't need those

spark plank
stone fulcrum
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Only 62 matrices

spark plank
stone fulcrum
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Or matriarchs

spark plank
solar wyvern
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sa into representations already?

spark plank
cloud walrusBOT
raw moth
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another way is the dihedral groups - groups of symmetries of regular polygons

spark plank
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I haven't gone through an explicit examples of noncommutative groups

raw moth
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it isn't

spark plank
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what's the symmetric group of degree 3 thonker

raw moth
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group of all permutations of 3 objects

spark plank
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k

stone fulcrum
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All bijections between two sets of three elements is a group

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S3

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Where function composition is the operator

covert vector
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Ya so function composition is associative

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and if you just look at bijective functions on a set to itself

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you get invertibility and identity from that

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so bijections on X→X form a group

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in general function composition here is not commutative, unless X has ≤ 2 elements

snow dew
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It's generated by transpositions, and by every connex "graph of transpositions"

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(When X is finite at least)

solar wyvern
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what does convexity mean here thonker

spark plank
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?

stone fulcrum
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Why not say "apple"? It's even shorter

snow dew
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Lel

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Connexity here means that your family ((ai bi)) generates SX iif it is connex as a graph on X.
Which means that for every two points of X you can find a path of transpositions from the first point to the second

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(Weakly connex, you don't see it as a graph with arrows cause if you have a transposition you automatically can get it backwards (inverse of (a b) is (b a))

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Stuff like "(1 2),(1 3),(1 4),(1 5) generates S5"

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Or "what's the minimum number of transpositions needed to generate Sn"

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Are easier to see that way

solar wyvern
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is connex something different from convex? real question

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@snow dew

snow dew
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There's no def of convex for graphs. Convex means the segment between two ppints is still in the thing. For parts of a topo space, Connex means that for two points there's a continuous path from one to the other. Convex => connex but not the other way around

solar wyvern
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yeah...i think that's called path-connected in english

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wiki says there's like a connex relation which means no pair of elements is incomparable but that's all I'm finding

spark plank
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connex sounds like connects

narrow crown
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Is it true there’s an algebra study group for the Milne text? are people up to a specific chapter or should I just jump in, if I want?

covert vector
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study group where :0

narrow crown
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nnutannep#7247 Simple_Art#1200  started up milne's group theory today in #abstract-algebra  if you want to join in on that
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(sorry for the ping!)

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not a study group per se, my bad

bleak abyss
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Oh it's a study ring, i c

narrow crown
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a study groupoid 😧

raw moth
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currently it's a study monoid because it only has one element

sullen flint
prisma ibex
sonic current
spark plank
sonic current
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@chilly ocean how many exams will you be taking this semester?

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what is the other one?

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hmm okay

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just wondering!

bitter mauve
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u say we'll have a exam?

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in here?

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wait

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what u talking about

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whayt is stydy group

sullen flint
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a handful of people that get together to tickle each other

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and study sometimes

bitter mauve
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when does this start

narrow crown
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@bitter mauve i think its not rly like a study group, its just 1 person working on it in this channel. my bad

raw moth
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tbh I'd be down to also work through it when I have more time

narrow crown
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yeah same. ive never even done a basic algebra course tho

raw moth
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I haven't either, idk about SA

snow dew
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Oh yeah connex was path connected pardon my french

raw moth
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yes

cloud walrusBOT
prisma ibex
cloud walrusBOT
prisma ibex
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what on earth are you trying to do anyways?

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you want to construct an isomorphism between which two groups?

thorny slate
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me: prove G and H are isomorphic
student: first, we show that G is isomorphic. [some gibberish]. now, we show that H is also isomorphic. [more gibberish]

bleak abyss
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Lol

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But yeah I mean, there are some methods you can use

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Sometimes there's an obvious map and you just need to verify that it has the properties

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Other times there are clever things you can do (e.g. non-abelian group of order 6)

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Well what are the groups?

cloud walrusBOT
bleak abyss
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Okay life tip

thorny slate
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is that just Z

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yeah

bleak abyss
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Let's say H is a subgroup of G, and K is another group. A map G/H to K is the same thing as a map G to K whose kernel contains H

thorny slate
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I know it's Z cuz it doesn't have torsion and has dimension 1

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:)

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oh wait

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im stupid

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it has torsion

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it's Z x Z/2

bleak abyss
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Lol I was gonna say

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Tfw (1,1)

thorny slate
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lole

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finite groups do

prisma ibex
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No

bleak abyss
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Aside from the identity

thorny slate
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(1,1) isn't the identity

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(0,0) is the identity

bleak abyss
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The point is that you say a group has torsion if it has non-trivial torsion

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Which this guy has

thorny slate
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it's generated by (1,0) and (1,1)

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those are also generators

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but not very nice ones

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(0,1) = (1,1) - (1,0)

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so you have (0,1) there as well

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yeah

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cuz after the change of basis
(1,0) -> (1,0)
(0,1) -> (1,1)

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your group is going to be

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Z x Z / (0,2)

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which is Z x Z/2

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it just means you should show an isomorphism to Z x Z/2

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where (1,0) goes to (1,0) and (1,1) goes to (0,1)

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?

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whatever you wanna do, the nice presentation of this group is Z x Z/2

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either

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but you have to do it through an isomorphism

cloud walrusBOT
thorny slate
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yeah

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you go through (Z x Z) / (0,2)

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with the map
(1,0) -> (1,0)
(1,1) -> (0,1)

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no

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I mean you could

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but the map would be different

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what

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it's just a map from (Z x Z) / (2,2) to (Z x Z) / (0,2)

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I dont get it

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you have two groups

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and you are defining a map between them

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you just have to check that this map is well defined

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and that it's an isomorphism

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so you have a map

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f: (Z x Z) / (2,2) to (Z x Z) / (0,2)

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given by

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(1,0) -> (1,0)
(1,1) -> (0,1)

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it's easier if you consider the map
f: (Z x Z) to (Z x Z) / (0,2)

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given by the same equations

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and prove the kernel is (2,2)

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yeah it's a bit tricky

civic linden
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I know that this corresponds to G=D20 the symmetry group of a regular 10-gon

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So number of necklace configurations is given by 10C3=120

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what now?

sullen flint
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I wanna say 120/20

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but I think that's too few

civic linden
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Not the ans

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but thanks

raw moth
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should just be 10C3 unless I'm misunderstanding what it means by different necklaces

civic linden
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Think its an application of Burnsides Lemma

sullen flint
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necklaces with 10 beads have decagon symmetry

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so many of those 120 configurations would be equivalent

raw moth
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o right I see it's asking for the joined-up necklace

civic linden
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Yas

sullen flint
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8!

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?

civic linden
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nop

sullen flint
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oof

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damn that's huge

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actually

civic linden
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ans is a small value

sullen flint
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8+7+6+5+4+3+2+1

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?

civic linden
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No

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ans < 10

sullen flint
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hmmm

civic linden
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Involves orbiter-stabiliser theorem

civic linden
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<@&286206848099549185>

civic linden
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if we take the reflections in D20

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it contains 5 reflections across axes passing through two opposite vertices right

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How would one calculate Fix(g)?

civic linden
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8?

wicked basin
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How many different necklaces can be made from seven black beads and three white beads?

civic linden
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ye

wicked basin
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I'm guessing

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10!/7!3!

raw moth
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no

wicked basin
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Yes baka

raw moth
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doesn't work because the neckless is joined up

wicked basin
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Jk

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Divide by 10

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Then

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10!/7!3!10

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== 10!/((7!)(3!)(10))

mellow vaporBOT
# 
Runtime error in iterm_1
On line 1 at position 10

10!/((7!)(3!)(10))
         ^
5‚040 is not a function
raw moth
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that's 12

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but I think some of these might be reflections of eachother

civic linden
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wrong ans

wicked basin
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Divide by 2

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6

civic linden
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XD wrong

raw moth
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not all of them have reflections I guess :^)

wicked basin
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do u already know the answer tho

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Burnsides lemma

civic linden
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Yeah its Burnsides Counting Theorem

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But I want to know how to apply it.

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Why does Fix(g) = 8 when it contains 5 reflections across axes passing through two opposite vertices?

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Fix(g)=|{x in X : gx=x}|

wicked basin
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idk I haven't done much with it I only recognized it :/

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Just add back the cases where it was symmetrical

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And u get the answer

civic linden
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mhm

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What about the 5 other reflections across axes passing through the middle of opposite edges

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what would Fix(g) be?

wicked basin
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idk just ask mudkip

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He's like too good for us plebs

raw moth
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lol

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I mean I'll try and figure it out

wicked basin
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15+ problems solved

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I can only do like 6 of them lol

raw moth
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I haven't done the burnside challenge problem tho

wicked basin
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It's ok ik u can do it

raw moth
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well ok lets think

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the identity leaves all 120 in place

civic linden
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Yes

raw moth
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if we rotate by something coprime to 10 then none are kept the same

civic linden
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Yup

raw moth
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in fact I don't think any of the rotations will keep it the same because it's 7 and 3

civic linden
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none are

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rotations do nothing here

raw moth
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ya ok

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for a reflection through two points

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ok got it

civic linden
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ok

raw moth
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so since there are an odd number of each colour

civic linden
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yes

raw moth
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you must have one of each colour on the line of reflection

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leaving 3 black and 1 white on each side of the reflection

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since it's reflected it's entirely determined by one side and the position of the ones on the line of reflection: there are thus 4C1*2C1 = 8 that are kept fixed

civic linden
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reeeeeeeeee

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yeah thats right

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8 is the ans

raw moth
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for the number of different necklaces?

civic linden
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yes

raw moth
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ok I need to try and slightly change this logic then

stone fulcrum
#

We're not there yet

raw moth
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ya

civic linden
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i know, im just saying

raw moth
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o and then there are the other 5 reflections

stone fulcrum
#

Are you sure you need to change the logic? It makes sense so far

raw moth
#

through the midpoints of sides

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nah it's fine I just forgot about the other kind of reflection :^)

civic linden
#

1/20 (120+ (5x8))=8

raw moth
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which don't keep any fixed because both kinds are odd

civic linden
#

mhm

raw moth
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so the average amount fixed is 1/20 (120 + 5*8) = 8 ya

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so there are 8 different necklaces

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nice

civic linden
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wanna try the other xD

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much harder

stone fulcrum
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👏

raw moth
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8 and 4 no ty not rn :^)

civic linden
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yah I like your logic in working it out

raw moth
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well ok I'll give it a go because I have some time to kill until I get up

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12C4 configurations = 12*11*10*9/24

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which is u

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h

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495

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I think

civic linden
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yes

raw moth
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ok

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so identity keeps 495 fixed, rotation coprime to 12 keeps none fixed

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if we rotate by 2 we have groups of order 6 that must be all the same colour

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so none

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if we rotate by 3 we get groups of order 4, and it's entirely determined by any 3 adjacent beads

civic linden
#

yes

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that sounds right

raw moth
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we need 2 of the 3 adjacent to be black and the other one white

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so I want to say 3 kept the same

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yeah I think it would be

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because if we label one of the beads as the first one which is what we do to get the 495

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then it's obviously 3C1

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if I knew the theorem better I could probably know if this were justified

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but it looks like it should be

civic linden
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ans is 3c1?

raw moth
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ya

civic linden
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ans is 29

raw moth
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lol

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I meant the number fixed for this specific case

civic linden
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oh

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I tihnk we missed some rotations

raw moth
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ya I'm not finished yet don't worry

civic linden
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oh baby

raw moth
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if we rotate by 4 we get groups of order 3 so we can't make 8 and 4

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if we rotate by 6 we get groups of order 2 so we need 4 groups of black beads and 2 of white beads

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so there are 6C2 = 15 fixed

civic linden
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yessss

raw moth
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now for reflections

civic linden
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ok i got the same for rotations

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Reflections is where i screw up

raw moth
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we'll first consider the reflections through the midpoints of the sides

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entirely determined by one side, and there must be 4 black and 2 white on each side

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so there are 4C2 fixed for each reflection, so for totalling for the 6 reflections there are 36

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now for the reflections through 2 points

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the points must be the same colour

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if they're both black we need 3 black and 2 white on each side, so 5C2

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if they're both white we need 4 black and 1 white on each side so 5C1

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so there are 5C2 + 5C1 = 15 per reflection and 6 reflections, so 90 fixed

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so overall there are

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lol this isn't right

civic linden
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every reflection fixes 15

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I think

raw moth
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if it does then the rotations are wrong

civic linden
#

I dont know I only have the end ans (just the number)

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not the working out

raw moth
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there are 12 rotations and 12 reflections total right

civic linden
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yes if u count e as a rotation

raw moth
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ya

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ok I'm not going mad

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but I'm also not getting something divisible by 24 as the total fixed

civic linden
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ah

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we forgot r9?

raw moth
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right yes

civic linden
#

270 degrees

raw moth
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and 8

civic linden
#

nu

raw moth
#

and a lot of other stuff

civic linden
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mhm

raw moth
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because I did factors rather than non-coprime

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ok, that's good

civic linden
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I think only r^3, r^6 and r^9 fix configurations

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r^3 fixes 3 configurations

raw moth
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r^6 fixes 15

civic linden
#

yes

raw moth
#

r^9 fixes 3

civic linden
#

yeah

raw moth
#

certainly if every reflection gives 15 would give 29

civic linden
#

mhm im having trouble seeing that tho

raw moth
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yeah I don't think they do

civic linden
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ans mbe wrong?

raw moth
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well what I have if it isn't that still doesn't give an integer

raw moth
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lol I found the mistake

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not 4C2 for the ones through midpoints

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6C2

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which is indeed 15

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so the answer is 1/24 (495 + 3 + 15 + 3 + 6*15 + 6*15) = 29

#

@civic linden

civic linden
#

@raw moth ah yes! great thanks

raw moth
#

np

#

nice to finally learn burnside (ish) after hearing it mentioned so much

stone fulcrum
#

Same, I've never known it before. That was a cool display

chilly ocean
#

More algebra sangry

cloud walrusBOT
sick acorn
#

Z_2 is Z/2Z

#

right?

#

ya

#

i think Z_2 is Z/2Z

#

just different notation

earnest valley
#

Parenthesis are missing.

cloud walrusBOT
sick acorn
#

wait what

#

the right is

#

the integers mod 2

#

right is 0, 1

#

Z/2Z is the integers mod 2

#

so 0, 1

#

yes

#

that is 2Z

#

but Z/2Z is the integers mod 2

#

it's a quotient group

#

right?

bitter mauve
#

Z/nZ means modulo n group dudr

#

i dont think thats a minus sign

#

dummit and foote

sick acorn
#

so we know

#

what quotient groups are, right?

#

we know the additive group Z

bitter mauve
#

teach me too senpai

sick acorn
#

and we know, for instance, 7Z is a subgroup of Z

#

yes where'd you gooo

bitter mauve
#

he lost

sick acorn
#

sonuva

bitter mauve
#

G/N is a quotient group

#

where N is the normal right?

#

...

#

@sick acorn teach me

#

i think so

#

im shaky abt what i know

sick acorn
#

back

#

i forget about N

#

but uh

#

so we know 7Z is a subgroup of Z

#

we know Z / 7Z is a quotient group

#

it contains all the left cosets of Z

#

(for the subgroup 7Z)

#

so they're of the form z + 7Z, where z is an integer

#

right?

#

i did an arbitrary choice

bitter mauve
#

same thing dude

sick acorn
#

we're just discussing notation atm

#

it would be

#

7Z \ Z

#

for right cosets haha

#

btw

#

the N is normal

#

it's a left coset

#

because Z/7Z is abelian

#

but we need to establish what Z/7Z is first

#

haha

#

just so you can get familiar with the notation

#

so Z/7Z is a quotient group

#

it's the set of left cosets of Z

#

the definition of a left coset (for a multiplicative group G and subgroup H) = {gh | h in H}

#

right?

#

while G / H = {gH | g in G}

bitter mauve
#

is it really 7Z/Z for right?

#

what is this

sick acorn
#

i would guess so? because H \ G = {Hg | g in G}

#

for right cosets

bitter mauve
#

fot normal N, every left coset is a right coset

sick acorn
#

yep

#

but yes uses the Z_2 notation

#

i'm just trying to explain what the Z/2Z notation is atm

#

haha

#

yes, we agree that the left coset gH = {gh | h in H}

#

right?

#

and that the set of left cosets G / H = {gH | g in G}

#

(for a group G and subgroup H of G)

bitter mauve
#

i read that G/N means collection of right cosets of N in G.

sick acorn
#

oh the way i learned it was

bitter mauve
#

so Z/2Z means

sick acorn
#

G/N is the collection of left cosets

bitter mauve
#

all elements x in the form 2a+x

#

so that's just 0,1

#

and thats certainly integer modulo n

sick acorn
#

don't assume that

bitter mauve
#

where is H

sick acorn
#

yet

bitter mauve
#

idk, when we define G/N, we assume that N is normal right?

sick acorn
#

for now, it's better to assume that N is a subgroup

#

nah we don't need to assume it's normal

#

normal just means coset multiplication is well-defined

#

you can, yes

#

(also, re: notation, i found some hw assignments that mention both)

bitter mauve
#

i think we need the lemma that "product of two right cosets of a normal subgroup is a right coset in G" when forming the group G/N

sick acorn
#

we don't haha

#

well

#

the GROUP yes

#

you're right

#

we're just talking about the set

#

G/N

#

for now

#

i shouldn't have used the term quotient group yet; my bad

bitter mauve
#

o well, i'm dumb. i only say what i read lol

sick acorn
#

so we start with the set (G/H) and then choose a well-defined operation (coset multiplication) to make it into a group

#

but really fast

bitter mauve
#

dafuq bro

sick acorn
#

G = Z

#

H = 7Z

bitter mauve
#

give me ur ques

#

ill do it

sick acorn
#

Z / 7Z = set of left cosets

#

0 + 7Z = {..., -7, 0, 7, 14, ...}

#

1 + 7Z = {..., -6, 1, 8, 15, ...}

#

all the way to 6

#

Z/7Z is just another notation for Z_7, and different authors use either one

#

but knowing why Z/7Z is written the way it is

bitter mauve
sick acorn
#

is helpful for internalizing what a quotient group is

bitter mauve
#

sorry about this

sick acorn
#

oh man my prof used different notation

#

hahaha

bitter mauve
#

here he says right cosets and uses the same notation

sick acorn
#

yeah

bitter mauve
#

so i dont think it matter

sick acorn
#

hahahha

#

ugh that difference in books bothers me

bitter mauve
#

which was ur college

sick acorn
#

probably just a bad collegeeee

bitter mauve
#

i dont see the right coset notation u showed

sick acorn
#

oh, in example 2

#

he says Z/7Z has right cosets blahblah

bitter mauve
#

yes

#

yea same thing since left cosets are also some right cosets

#

it has to be normal

#

right?

sick acorn
#

it's normal if and only if left cosets are equal to right cosets

bitter mauve
#

he says H is the kernel

sick acorn
#

(in the case of Z/7Z, it is)

bitter mauve
#

so it's fine

#

since kernels are normal

#

speak, dude

sick acorn
#

hahaha

#

i'm confused as to the kernel part

#

like... is there a function / homomorphism

#

oh are we just going with

#

the quotient homom

bitter mauve
#

u talking about the second link right?

sick acorn
#

oh whoops didn't realize you were talking about the second link

#

ya

bitter mauve
#

even in first link

sick acorn
#

we also use addition for Z/7Z elements, right

#

which would be abelian

bitter mauve
#

that notation means H is a normal subgroup of G

sick acorn
#

it does, yeah

bitter mauve
#

what was our topic of discussion

sick acorn
#

what Z/7Z was hahaha

#

yes is used to Z_7

#

(the notation)

bitter mauve
#

what was his question

#

do u know

#

the main question

sick acorn
#

Let $G\ceq \mathbb Z \times \mathbb Z$. Prove that $G/\langle(2,2)\rangle \cong \mathbb Z_2 \times \mathbb Z$

cloud walrusBOT
sick acorn
#

is what he said

bitter mauve
#

oh damn

#

i don't even understand it

sick acorn
#

my initial thoguht is

#

(and i'm not gonna try to solve b/c i'm gonna shower and go to bed soon lol) is

#

first isomorphism theorem.. hm

#

find a homomorphism f with Image f = Z_2 x Z

#

and Kernel f = <(2,2)>

#

where f goes from G to.... G?

bitter mauve
#

what does <(2,2)> mean

sick acorn
#

the cyclic set generated by (2,2)

#

so

#

n(2,2) where n is an integer

#

(0,0), (2,2), (-2,-2), etc

bitter mauve
#

oh

#

go on

#

why u stop

sick acorn
#

showering brb

bitter mauve
#

yes sir

worthy kindle
#

What

#

Z/2Z isn't a set of numbers

#

a set of equivalence classes

#

the equivalence classes do contain numbers tho

#

yea

#

[69]

#

Sorry, keep going

#

[420]

#

in Z/2Z, [0] = [420] = [222] = [-36]

#

Ya

#

Uhh

#

$\bbZ/2\bbZ^\times ={[1]}$

cloud walrusBOT
worthy kindle
#

you meant this?

#

alright then $\bbZ/2\bbZ\setminus{[0]}={[1]}$

cloud walrusBOT
worthy kindle
#

but $\bbZ/2\bbZ \neq {[1]}$

cloud walrusBOT
worthy kindle
#

$\bbZ/2\bbZ = {[0], [1]}$

cloud walrusBOT
worthy kindle
#

$\bbZ = [0]\cup [1]$

cloud walrusBOT
worthy kindle
#

Different things

bitter mauve
#

what's the definition for quotient group

worthy kindle
#

Idfk what you believe what the Z/nZ are

#

$\bbZ/n\bbZ = {{z\in\bbZ\ |\ z\equiv k\mod n}\ |\ k\in {0,...,n-1}}$

cloud walrusBOT
worthy kindle
#

or $\bbZ/n\bbZ ={ n\bbZ+k\ |\ k\in{0,...,n-1}}$

cloud walrusBOT
worthy kindle
#

if you prefer

bitter mauve
#

Z/nZ is a set of sets?

worthy kindle
#

yea

bitter mauve
#

oh

worthy kindle
#

but operations on it make it behave a lot like a set of numbers

bitter mauve
#

ok

#

why u ppl still discussing about notation

sick acorn
#

because ideally it'll help him become more familiar with what the idea of a coset is haha

worthy kindle
#

seemed like there were some misconceptions

sick acorn
#

and yeah, what tuong said

#

yes, you know what Z_2 is, right?

#

[0] and [1]

#

the integers mod 2

#

the equivalence classes* sorry

cloud walrusBOT
worthy kindle
#

doesn't look like very homogenous

#

The elements of Z × Z/2Z × 2Z are triplets
The elements of Z/2Z × Z/2Z are duplets

#

(a, b, c) things

#

That's what I meant by not homogenous

#

gg

cloud walrusBOT
earnest valley
#

$\phi:\mathbb{Z}^2\to(\mathbb{Z}/2\mathbb{Z})\times\mathbb{Z}, (a, b)\mapsto\left(a\pmod{2}, b-2\left\lfloor\frac{a}{2}\right\rfloor\right)$.

cloud walrusBOT
worthy kindle
#

idk, I know how to read / only with equivalence relation

#

like E/~, but (A×B)/(C×D) idk

earnest valley
#

The kernel is <(2, 2)>.

#

Go with the first.

earnest valley
#

That's a morphism.

#

No, a set isn't a morphism. A morphism is an application which preserves the structure.

earnest valley
#

That phi is a morphism.

earnest valley
#

Yes.

cloud walrusBOT
earnest valley
#

Yes.

#

$G=\mathbb{Z}^2/\ker{\phi}$.

cloud walrusBOT
earnest valley
#

The surjective part seems easy, maybe some guessing to find the morphism.

cloud walrusBOT
thorny slate
#

you can do that as well if you want

#

but it's awkward

#

because you have to check that the map is well defined in the quotient

#

you would have to check that the kernel is trivial once you do so

#

it's easier to use the map Z^2 -> .. and check that the kernel is (2,2)

cloud walrusBOT
earnest valley
#

The theorem states that the second morphism phi is injective but if the first morphism f is surjective then phi is bijective.

#

$\begin{tikzcd}
G \arrow[d, two heads] \arrow[r, "f"] & {G'} \
G/\ker{(f)} \arrow[r, "\phi"]& \arrow[u, hook] \im(f)
\end{tikzcd}$.

cloud walrusBOT
earnest valley
#

Where f, phi are group morphisms.

stone fulcrum
#

@chilly ocean
Phi tilda and phi are the same thing, cuz first iso theorem

#

You've started by finding φ, then used it to show isomorphism. Backwards from the way the theorem states it, but valid

earnest valley
#

Yes, injective.

#

Example: $f:\mathbb{Z}\to\mathbb{Z}, n\mapsto n\pmod{p}$, then:
$\begin{tikzcd}
\mathbb{Z} \arrow[d, two heads] \arrow[r, "f"] & \mathbb{Z} \
{\mathbb{Z}/p\mathbb{Z}} \arrow[r, "\phi"]& \arrow[u, hook] {0, ..., p-1}
\end{tikzcd}$.

#

With p > 1, f is a group morphism.

cloud walrusBOT
earnest valley
#

Small mistake, now it's good.

#

That's the inclusion map.

prisma ibex
#

there's no group inclusion from Z/p to Z

earnest valley
#

You are right, it does work maybe with a finite group Z/nZ.

#

Example: $f:\mathbb{Z}/15\mathbb{Z}\to\mathbb{Z}/15\mathbb{Z}, n\mapsto 3n$, then: \
$\begin{tikzcd}
{\mathbb{Z}/15\mathbb{Z}} \arrow[d, two heads] \arrow[r, "f"] & {\mathbb{Z}/15\mathbb{Z}} \
{(\mathbb{Z}/15\mathbb{Z})/(5\mathbb{Z}/15\mathbb{Z})} \arrow[r, "\phi"]& \arrow[u, hook] {0, 3, 6, 9, 12}
\end{tikzcd}$.

cloud walrusBOT
earnest valley
#

This should work.

#

Yes.

stone fulcrum
#

Wat be q @chilly ocean

civic linden
#

I think something to do with 1st iso theorem

solar wyvern
#

@chilly ocean what happens if Im(f) \cong G'

earnest valley
#

If $f:G\to G'$ is surjective then $\im(f)=G'$.

cloud walrusBOT
earnest valley
#

You don't need \textit{three} functions, there are $f$, the projection $\pi$ and $\phi$ which verify $f=\phi\circ\pi$.

cloud walrusBOT
earnest valley
#

You don't need f injective.

simple valley
#

what's the question

#

I haven't been following

#

I've been too intrigued by hyperplane projections and partitions

#

this is about the first isomorphism theorem right?

cloud walrusBOT
simple valley
#

that should be easy enough

#

define a surjection $\mathbb{Z}^2 \to \mathbb{Z}_2 \times \mathbb{Z}$ whose kernel is $\langle (2, 2) \rangle$

cloud walrusBOT
simple valley
#

do you see how $\mathbb{Z}^2$ is generated by $(0, 1)$ and $(1, 0)$?

cloud walrusBOT
prisma ibex
#

yes

#

do you want me to just spoil it for you lmao

#

it's super duper obvious

#

I'll give you a hint

#

k

simple valley
#

@chilly ocean if a group is generated by a set of generators

#

a homomorphism from that group is determined by the values of generators

prisma ibex
#

yeah of course not

#

if you send (a,b) to (a mod 2,b mod 2)

#

the kernel will be <(0,2),(2,0)>

#

besides you won't have the right codomain in this case

#

yes

#

let me be more explicit

#

the map sending (a,b) in Z^2 to (a mod 2, b mod 2) in (Z/2)^2

#

has <(0,2),(2,0)> as kernel

#

Z_2 is bad notation but sure

#

I mean a lot of people use Z_2 to refer to integers mod 2

#

But it's bad because there's also something called the 2-adic integers

#

which is also written Z_2

#

but anyways

#

so the thing I mentioned

#

the map sending (a,b) in Z^2 to (a mod 2, b mod 2) in (Z/2)^2

#

has kernel <(0,2),(2,0)>

#

should I explain this?

#

or can you see it?

#

Hint: what do the elements of the subgroup <(0,2),(2,0)> look like?

#

right and <(0,2),(2,0)> is the subgroup of (a,b) in Z^2 where a and b are both even

#

so clearly the whole thing is in the kernel

#

I mean the whole subgroup <(0,2),(2,0)>

#

yea

#

on the other hand

#

everything outside this subgroup is not in the kernel

#

since the compliment of <(0,2),(2,0)> in Z^2 are those (a,b) in Z^2 with a or b odd

#

so you can see how the kernel of this map f is the subgroup <(0,2),(2,0)>, yeah?

#

okay

#

right I was just giving an example

#

let's think of another map

#

same lmao

bitter mauve
#

ooo wait @prisma ibex

prisma ibex
#

yea

#

nnutannep?

bitter mauve
#

yea i had one ques too

#

basic

#

ques 12 looked stupid then i saw ques 13

#

then it looked hard

#

i know that to prove that G is a group we need to have a.y(a) = y(a).a = e

prisma ibex
#

sure

bitter mauve
#

for the first i tried

$a.y(a) = e \Rightarrow y(a).a.y(a).a = y(a).a$

cloud walrusBOT
prisma ibex
#

12 isn't hard; I won't spoil it for you, it's literally just working out the axioms and it's maybe a couple of lines

#

13 is uhhh

#

I would have to think about it

bitter mauve
#

so by using it with i) i said that y(a).a=e

#

but if i do the same with 13, i get a.y(a)=e

#

so i must be doing sth wrong in 12

#

right?

#

however, i don't see any mistake with what i did in 12

#

it looked valid

prisma ibex
#

hmmm

bitter mauve
#

how would you have done 12

prisma ibex
#

I don't think your step for 12 works

#

I would just do

bitter mauve
#

yea my logic for 12 proves that 13 is a group too

prisma ibex
#

lmao I don't want to write this out

#

shower time

#

😂

bitter mauve
#

ok

simple valley
#

let G be the set of injective functions under composition, with e the identity function

#

hmm, not a good example

#

maybe if we consider injective functions N -> N

bitter mauve
#

if we do that, a . e =e.a =a

simple valley
#

injective => left-invertible

#

but not a group

#

oh nevermind

#

the inverse isn't injective

bitter mauve
#

how would u do 12

#

whats wrong with my logic

#

is it because e might not be unique

#

thats the only thing i can think of

simple valley
#

I think 12 isn't a group either?

#

consider the free noncommutative monoid over N

#

with equations n s(n) = e

#

for all n in N

#

consider the operation y(m n ... k) = s(k) ... s(n) s(m)

#

(s obviously preserves the quotient structure)

#

this monoid, with its identity and y as a right invertor satisfies the axioms of question 12 I think?

#

but isn't a group as 0 has no left inverse

bitter mauve
#

im not sure what u talking about @simple valley

#

:(

simple valley
#

wait crap

#

I think I constructed C_2

bitter mauve
#

o well

#

I interpreted the ques as

i) $a.e=a $ so may be let's say $e = a^(-1).a$
ii) $a.y(a)=e$

so from i and ii we have inverses satisfying both sides.

However i cant find a way to show it while actually proving.

cloud walrusBOT
bitter mauve
#

i tried with the logic i mentioned way above, but is apparently wrong.

#

<@&286206848099549185>

#

do, with that logic, if i'm to do ques 13, I'll need to find a closed set associative under product whose identity works only on the right side, because if it worked both sides it'll make 13 ii) a valid assumption for defining group

#

but i guess my intuition is flawed in this one

#

idk

#

anyone help

bitter mauve
#

nvm, i got it

earnest valley
#

What was it ?

oak perch
#

Let n be rational. {(n, 2n)} under the following multiplication:

(j, 2j) X (c, 2c) = (jc, 2jc)

(1, 2) is the identity on the right.

#

Suppose (b, 2b) is a left identity. For all c, (b, 2b) X (c, 2c) ==> (bc, 2bc) = (b, 2b). So b = cb and c = 1. c = 1 is not a good restriction for a universal left identity.

#

Given a = (c, 2c), y(a) = (1/c, 2/c) ... (1/c, 2/c) x (c, 2c) = (1, 2)

earnest valley
#

With y(a)=(1/c, 2/c).

oak perch
#

y(a) is the inverse of a?

#

oh. typo. thank you

earnest valley
#

It's a group.

analog patio
#

All the proofs i see for linear maps being distributive assume the maps inside the brackets are also linear. Is there a counter example when the maps are non-linear because it seems to hold for me..

#

e.g. $$\int sinx+cosx = \int sinx + \int cosx$$ etc.

cloud walrusBOT
thorny slate
#

The integral is linear

analog patio
#

right but cos and sin arent

#

all the proofs say S(T1+T2) = ST1 + ST2, where S T1,T2 are linear, but i can't find a reason why T1 and T2 cant be non-linear in general

#

and in axler too

thorny slate
#

Because the integral is linear

#

This you linked is something else

#

The integral isnt being composed

#

Its a functional in the space of functions in R

covert vector
#

Ya sin and cos are just elements of a vector space @analog patio

#

perhaps a very large vector space, consisting of continuous functions on R

analog patio
#

Makes sense. Thanks

covert vector
#

here the "non linearity" of sin and cos isn't really something you really care about, these are just vectors

#

np 👍

cyan quarry
#

do dual's span all of R necessarily?
(or more precise, the field over which the vector space is)

thorny slate
#

What

cyan quarry
#

If I have a vector space V, and it's dual space V*, is there necessarily a v* of V* such that v*(v)=x, for some v of V, x is any real number

#

sorry for awful format and perhaps slang, I don't study in English so I'm not sure I'm using the correct terms

thorny slate
#

Yes

#

Take an element v in V

#

Then there is an element w in V* with wv = 1

#

So awv = a

earnest valley
#

w(v)=1, w(a*v)=a*w(v)=a.

cyan quarry
#

thanks, just wanted to be sure

#

and I appreciate the rigorous additivity and homogenous, Quantic

earnest valley
#

You are welcome.

cyan quarry
#

for real, it only hit me 2 days ago while I was watching this https://www.youtube.com/watch?v=kxOpozNkUg4, looking for something that will help me understand dual space better.

I'm like "wtf is the point, this guy is so anal about this, pedantic like my professor. Doesn't help me.." and while it hasn't helped my understanding too much, I did finally get some appreciation for why it is important - or rather, how I didn't understand at all these small distinctions and their meaning.

YouTube
Ben1994
Dual Vector Spaces Part 1

In this video we discuss the algebraic dual of a vector space.

▶ Play video
cyan quarry
#

How do I find a dual basis for a subspace, in which case I can't just find the invertible matrix?

#

for reference, I have 2 vectors that span a subspace of R3

#

I can intuitively look and figure out the matrix equivalent that will give me the kroneker delta, but what is the actual technical way?

civic linden
#

Z(G) the centre of a group

#

it's property is being abelian for all g in G?

bleak abyss
#

For all G you mean? If so then yes

civic linden
#

ty.

bleak abyss
#

Wait what do you mean for all g in G?

civic linden
#

z in G and for all g in G

#

we have the abelian property gz=zg?

bleak abyss
#

Okay wait

#

Z(G) is an abelian group for every group G

civic linden
#

Oh it's something that's always present?

bleak abyss
#

Being abelian is a property of groups, not of elements of a group

civic linden
#

Oh

bleak abyss
#

A group is abelian if any two elements commute, so you'd either ask whether it's true that any two elements of Z(G) commute or whether the group Z(G) is an abelian group

#

Makes sense?

civic linden
#

Right okay, I got it

#

Thank you, G.

#

Okay quick question

#

If I had to prove Z(G) is a normal subgroup of a group G

#

I'd go by proving the axioms of subgroups using the property zg=gz

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i.e. the identity, ex=xe=x for all x in G, hence e in Z(G)? @bleak abyss

bleak abyss
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Yeah if you haven't already verified just that it's a subgroup then do that

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And then normality is another step

civic linden
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yes so after proving all 3 axioms, prove that conjugation holds?

bleak abyss
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Yeah

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One thing I will say

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There's a slick way to verify that a subset of a group is a subgroup

civic linden
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Ooo I'm listening

cloud walrusBOT
stone fulcrum
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I'm listening too I don't think I know this

bleak abyss
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Err whoops I started off the sentence being like eh I'm not gonna TeX it so I left the letters in non-math mode

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Too late to edit it now but yeah always have symbols in math mode, don't be me

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But yeah this is a fairly straightforward verification of the axioms. It's a bit of a gimmick tbh, usually it's no more mental work to just go and be like, okay closed under multiplication, you have inverses, and identity.

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But it may or may not be faster to write up. Turns out for finite groups, any non-empty a closed under multiplication is a subgroup

simple valley
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because each element's inverse is its power?

bleak abyss
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Yup

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There's actual content to this statement so it's epsilon work saved but still technically > 0

solar wyvern
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epsilons in finite groups catThink

raw moth
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let epsilon >= 1

sullen flint
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@chilly ocean

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I tried to prove what you said in the chill channel

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I started by proving the cancellation law holds

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if for each x,y exists some unique g and h such that
xh=gx=y

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that and associativity were your axioms

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let's say we have ab=ac

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let's call ab=ac= y

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for a an y there's an unique h such that ah=y

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it follows that b=c=h

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and I guess the same holds when the common factor is on the right, because why not

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now the axioms provides for some x unique a and b such that ax=xb=x

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then we could have that a ax= ax= x

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by cancellation it follows that a²=a

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now let's say we have some x' such that bx'=xb

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x'exists by assumption

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now left multiply and right multiply by b

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and you have bx'b=bxb

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by cancellation twice you have x'=x

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therefore bx=ax=x and that implies again by cancellation b=a

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the left and right identities of x are Equal

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now pick any (unrelated to the preceding) x and y

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q must exist such that qx=y

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but y = y i

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where i is y's own identity

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and so qxi=yi =y

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qxi=qx=qxe where e is x's own identity

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by cancellation i=e

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so the identity is unique

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only rests to prove the existence of the inverse but I don't want to clutter the chat too much

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hope I didn't make any mistak

oblique river
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what are you trying to prove @sullen flint ?