Group Thoery

[obtuse orchid]

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[serene ruin]

timestamp of what you're referring to?

I dont see how your concern relates to the proof tbh

there's no relation to begin with

nor any ordered pairs

there's no function in where you linked to

he just wants to ensure multiplication in the quotient is well defined

same as any algebraic quotient

you mod out by some special subset of the structure, then try to endow it with structure

yeah, there he's proving the equivalence of modding out by a normal subgroup makes the quotient a group

showing something is well defined just means showing the choice of representation doesnt matter

wdym

a function $f:X\to Y$ is well defined iff $a=b\implies f(a)=f(b)$

[dawn palm]

Omegabet_

[serene ruin]

proving the multiplication on the quotient is well defined amounts to showing if (aH,bH)=(cH,dH), then (ac)H=(bd)H

since multiplication as a function on whatever set $X$ is

$\ast: X\times X\to X$

[dawn palm]

Omegabet_

[serene ruin]

f(a) is whatever element in Y a maps to

likewise f(b)

take a map whose graph is y^2=x, this isnt well defined

but when x=1, f(1)=+-1 since it's 1 to many

so it's not a well defined map since 1=1, but f(1) != f(1) necessarily

since I can easily make it be 1=-1

that's why I didnt originally define it in terms of mappings

since notationally it's weird

any time the domain of a function involves quotients you have to worry about well definedness, since equivalence classes have in theory infinitely many representatives

maybe idk

the proof Michael gives is pretty straightforward and standard for showing well definedness

but your statement doesnt make use of the representative choice so my guess is that doesnt prove it's well defined

your set afaik is just a set with no real meaning

keep in mind, equivalence classes require you to make a choice on what element you use to represent it

as defined, multiplication makes direct use of the specific representative

that's why you need to go through the hurdles with checking well defined ness, the operations as written appear to depend on the representatives of the class

[valid kraken]

@plain carbon has given 1 rep to @serene ruin

[serene ruin]

yes

the map doesnt care what representative you choose

[tight ferry]

@plain carbon

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