Thread: function from quotient group to orbit, well defines, 1-1, onto

it's often customary to write g.s instead of gs, just so we don't get confused and think gs is a product in G.

here's the thing: we want to make sure that f(gG_s) only depends on the coset gG_s, and not on which "representative" g we pick.

(in other words, we might have gG_s = g'G_s, for different g,g' and we need to be sure that we get the same value for f, that gs = g's).

now gG_s = g'G_s DOES NOT MEAN g = g'. what it DOES mean is that: g'^-1g is in G_s.

and (by the definition of G_s) that means:

g'^-1gs = s.

now what we CAN do, is have g' act on both sides (both sides are elements of S). this gives:

g'(g'^-1gs) = g's (*)

now, for ANY action of a group G on a set S, for g,h in G, we have:

g(hs) = (gh)s. in our situation that means (looking at the LHS of (*)):

g'(g'^-1gs) = (g'(g'^-1g))s = ((g'g'^-1)g)s = (eg)s = gs, so NOW we have gs = g's, as desired.

***********

for your second problem, we need to pick two elements f(gG_s) in the orbit of s (Os), and show that if they are equal, they are images of the same coset of G_s.

in short, we need to show gs = g's implies gG_s = g'G_s (as luck/fate would have it, this is the REVERSE implication of showing f is well-defined).

now you can't "divide by s" (S is just a set, it may not have any "multiplication/division" structure).

what you CAN do, is act on both sides by g'^-1.

this gives:

g'^-1(gs) = g'^-1(g's), and BECAUSE WE HAVE AN ACTION:

(g'^-1g)s = (g'^-1g')s = es = s (because with a group action es = s, for all s in S).

but this tells us that g'^-1g stabilizes s (or fixes s, when we act on s by g'^-1g, nothing happens, we "stay" at s).

and THIS means that therefore g'^-1g is in the stabilizer of s, which is what G_s is.

but if g'^-1g is in G_s, then the two cosets g'G_s and gG_s are equal, and we're done.

finally, what we need to show is that the function f is ONTO Os.

that is, given an element in Os, say, t, for example, we need to find "some coset" of G_s, gG_s with f(gG_s) = t.

since t is in the orbit of s, t = gs, for some g in G. so why not try the coset gG_s (for that same g).

well, then we have f(gG_s) = gs (by the definition of f), and gs = t, so, yep, that works.

**************

i suspect that part of your trouble is you are having difficulty seeing how this all works. so let's find a simple G and a simple set S to actually work with.

let G be this subgroup of S₄: G = {e, (1 2 3 4), (1 3)(2 4), (1 4 3 2)}. this is a cyclic group of order 4.

what set shall we have G act on? well an obvious choice is S = {1,2,3,4}.

so for g = (1 2 3 4), and s = 3, for example, gs = 4. note that the only element of G that stabilizes anything is e, so for all s, G_s = {e}.

what is the orbit of 1? well 2 = (1 2 3 4)1, 3 = (1 3)(2 4)1, and 4 = (1 2 3 4)1, so the orbit of 1 is all of S.

so what we did above shows we have a bijection between G/{e} (= G) and O1 = S:

e ---> 1
(1 2 3 4) ---> 2
(1 3)(2 4) ---> 3
(1 4 3 2) ---> 4

let's pick a different G, and a different S. this time we'll pick S = G, and we'll let G = {e,a,a²,b,ab,a²b}, where a³ = b² = e, and ba = a²b. this time, our action will be:

g.x = gxg^-1.

we can ask, what is the orbit of a? let's find out:

eae^-1 = a
aaa^-1 = a
a²aa^-2 = a
bab^-1 = bab = (a²b)b = a²
(ab)a(ab)^-1 = (ab)a(ab) = a(ba)ab = a(a²b)(ab) = a³(bab) = bab = a²
(a²b)a(a²b)^-1 = (a²b)a(a²b) = a²ba³b = a²b² = a².

so the orbit of a is {a,a²}.

looking above, we see that the elements of G that stabilize a, are {e,a,a²}, which is <a>. and what we did above says we have a bijection between:

G/G_a = G/<a> and Oa.

it is easy to see that the (left) cosets of <a> are <a> and b<a>. the bijection we established above is:

<a> <---> {a}
b<a> <---> {a²}

Okay...this is beginning to make some sense. I have some follow up questions.

In the section about well defined, you wrote, "now gGs = g'Gs DOES NOT MEAN g = g'. what it DOES mean is that: g'-1g is in Gs." I don't understand why g'-1g is in Gs if gGs = g'Gs. Is that a definition or a property I just don't know or remember? or did you multiply both sides by g'-1? If so, you get g'-1gGs=Gs. So that means that g'-1g is an element of Gs. I don't understand why.

In the section on 1-1, you wrote but if g'-1g is in Gs, then the two cosets g'Gs and gGs are equal, and we're done. I think my question here is the same as the one above. I think I am missing something fundamental, and I don't know where to find it...

you wrote, "i suspect that part of your trouble is you are having difficulty seeing how this all works. so let's find a simple G and a simple set S to actually work with.

let G be this subgroup of S4: G = {e, (1 2 3 4), (1 3)(2 4), (1 4 3 2)}. this is a cyclic group of order 4.

what set shall we have G act on? well an obvious choice is S = {1,2,3,4}.

so for g = (1 2 3 4), and s = 3, for example, gs = 4. note that the only element of G that stabilizes anything is e, so for all s, Gs = {e}."

I don't understand how gs = 4. Actually, I don't understand how any of this actually works. I feel like I need remedial abstract algebra...

it comes from THIS theorem:

two (left) cosets of a subgroup H of a group G, aH and bH, are equal if and only if b^-1a is in H.

proof: suppose aH = bH. since a = ae is in aH, and aH = bH, we have a = bh, for some element h in H.

thus:

b^-1a = b^-1(bh) = (b^-1b)h = eh = h is in H.

on the other hand, suppose that b^-1a is in H, so b^-1a = h, for some h in H.

then a = ea = (bb^-1)a = b(b^-1a) = bh.

so for any element ah' of aH, we have ah' = (bh)h' = b(hh').

since hh' is in H (because subgroups are closed under multiplication), ah' is in bH. thus aH is contained in bH.

on the other hand, if b^-1a = h, then (b^-1a)^-1 = a^-1b = h^-1 is also in H.

this means b = ah^-1.

so for any element bh" in bH, we have bh" = (ah^-1)h" = a(h^-1h"), where h^-1h" is in H,

so bh" = a(h^-1h"), which is in aH. thus aH is contained in bH. since the two sets contain each other, they are equal.

*****

to sum this all up, saying aH = bH or saying b^-1a is in H are equivalent conditions, one is always true when the other is.

in fact, we can define an equivalence (called "equivalence modulo H") on G by: a~b if b^-1a is in H.

the equivalence class of a under this equivalence is just the left coset of a, aH = {ah: h is in H}.

an equivalence is just a fancy name for a partition of a set: any partition generates an equivalence relation:

a~b if a,b lie in the same subset of a partition.

and any equivalence generates a partition of the set it is on: the equivalence classes.

*****

now to your second question. let me explain what "cycle-notation" of a permutation means. when i write: (1 2 3 4), i mean the following function:

f:{1,2,3,4}-->{1,2,3,4} given by:

f(1) = 2
f(2) = 3
f(3) = 4
f(4) = 1

as you can see, this is a 1-1, onto function on the set {1,2,3,4}. so (1 2 3 4) should be read as:

"1 goes to 2, 2 goes to 3, 3 goes to 4, 4 goes back to 1".

this is called a 4-cycle, and you could picture it like a clock dial, with 1 at the 12 o' clock position, 2 at the 3 o' clock position, 3 at the 6 o'clock position, and 4 at the 9 o'clock position, and arrows between them (going...clock-wise. see what i did there?).

what happens if we perform f twice?

f(f(1)) = f(2) = 3
f(f(2)) = f(3) = 4
f(f(3)) = f(4) = 1
f(f(4)) = f(1) = 2

so f² is the function:

1-->3
2-->4
3-->1
4-->2

in cycle notation, 1-->3, and 3-->1 so we start with the 2-cycle (1 3) (this means 1 and 3 change places). 2 and 4 also change places, so f² = (1 3)(2 4).

ok, what about f performed 3 times in a row?

f(f(f(1))) = f(f(2)) = f(3) = 4
f(f(f(2))) = f(f(3)) = f(4) = 1
f(f(f(3))) = f(f(4)) = f(1) = 2
f(f(f(4))) = f(f(1)) = f(2) = 3, so the function f³ is:

1-->4
2-->1
3-->2
4-->3

in cycle notation this is (1 4 3 2).

finally, let's look at f performed FOUR times in a row:

f(f(f(f(1)))) = f(f(f(2))) = f(f(3)) = f(4) = 1
f(f(f(f(2)))) = f(f(f(3))) = f(f(4)) = f(1) = 2
f(f(f(f(3)))) = f(f(f(4))) = f(f(1)) = f(2) = 3
f(f(f(f(4)))) = f(f(f(1))) = f(f(2)) = f(3) = 4

so we see that f⁴ is the function:

1-->1
2-->2
3-->3
4-->4

which is the identity function on the set {1,2,3,4}.

so when i write (1 2 3 4)3, what i MEAN is f(3), which is 4.

similarly, when i write (1 3)(2 4)1, what i mean is f²(1) which is f(f(1)) = f(2) = 3.

********

permutations of a set S = {a,b,c....,x} are just "rearrangements of that set". we can identify a permutation (re-arrangement) uniquely with a bijective function on S.

for example, if our set is {a,b,c} we can identify the re-arrangement abc --> cba with the function:

a-->c
b-->b
c-->a

so if we want to call our function f:

f(a) = c
f(b) = b
f(c) = a.

the things we're "re-arranging" don't really matter, all that really matters is how MANY of them there are. so it's customary to use the set {1,2,...,n} where:

1 = "first thing"
2 = "second thing"
etc.

now, the reason i'm going on at length about permutations is this:

an ACTION of a group G on a set S, is really this:

for every g in G, we get a PERMUTATION of S, defined by:

s--->gs

the group of permutations on a set S, forms a group, called Sym(S) (the symmetric group of S), and our action is really a homomorphism from G to Sym(S).

that is for every ELEMENT g of G, we get a MAPPING g_ :S-->S with g_(s) = gs.

this let's us "multiply" set-elements by a group element we set g.s = g_(s) = gs. this multiplication takes a pair (g,s) and spits out an element gs in S.

if S is a FINITE set, with n elements, then Sym(S) is isomorphic to S_n.

and yes, elements of S_n (which are bijective functions on the set {1,2,...,n}) have a natural action on this set,

if the permutation g takes a to b, then we set ga = b.

in naive terms, we use a group G to "shuffle" elements of S. the possible places an element s may wander during this shuffling, is called the ORBIT of s.

now some elements of G may not touch s at all during this shuffling process. these elements are called the stabilizer of s, and it turns out they form a subgroup of G.

an example (again).

using S₃, consider the permutation:

1-->3
2-->2
3-->1 this is usually called (1 3).

if we let this group act on {1,2,3} in the natural way, we can ask: what is the stabilizer of 2?

well the identity stabilizes 2 (since it stabilizes EVERYTHING, nothing moves under the identity). and as you can see (1 3) also stabilizes 2 (it only moves 1 & 3, by swapping them).

it turns out that everything else moves 2, so

Stab(2) = {e, (1 3)}.

(Stab(x) is another way that G_x is written).

now the permutation:

1-->2
2-->3
3-->1 ...this is usually written (1 2 3)

takes 2 to 3. so (1 2 3) is not in the stabilizer of 2, and also it shows that 3 is in the orbit (under S₃'s action) of 2.

and the permutation:

1-->1
2-->1
3-->3 ...this is written (1 2)

takes 2 to 1, so it's not in the stabilizer of 2, and it shows 1 is in the orbit of 2.

so the orbit of 2 is {1,2,3} which is our entire set.

so from what we JUST proved (from your original post), there had better be just 3 cosets of Stab(2).

the theorem we are talking about is called "the orbit-stabilizer theorem" and it says:

"the size of an orbit of an element s, is equal to the index of the stabilizer of s"

(the index of a subgroup is the number of cosets there are).

for a finite group G, the index of H in G, [G:H] is equal to |G|/|H|.

so the orbit stabilizer theorem (for a finite group) can be written:

[G:G_s] = |Os|, or, if we like:

|G| = |Os|*|G_s|.

in our example for G = S₃, S = {1,2,3} and s = 2, this becomes:

6 = |S₃| = |{1,2,3}|*|Stab(2)| = 3*2.

********

just for completeness' sake, let's compute the left cosets of H = Stab(2) = {e, (1 3)}. they are:

H = {e, (1 3)}
(1 2 3)H = {(1 2 3)e, (1 2 3)(1 3)} = {(1 2 3), (2 3)}

here is how we compute (1 2 3)(1 3). first we take the function (1 3):

1-->3
2-->2
3-->1

next we apply the function (1 2 3):

1-->3-->1
1-->2-->3
3-->1-->2, we see the result is:

1-->1
2-->3
3-->2, which is (2 3) (2 and 3 swap places).

and our last coset:

(1 3 2)H = {(1 3 2), (1 2)}

what is the bijection between these cosets and the orbit of 2?

H<--->2 (these elements of S₃ send 2-->2)
(1 2 3)H <---> 3 (these elements of S₃ send 2-->3)
(1 3 2)H <---> 1 (these elements of S₃ send 2-->1)

everything works out just as the theory says it should.

This has been very helpful. Thanks so much for taking the time to explain this to me.