Results 1 to 5 of 5

Math Help - function from quotient group to orbit, well defines, 1-1, onto

  1. #1
    Junior Member
    Joined
    Oct 2012
    From
    Austin, Texas
    Posts
    36

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

    f: G/Gs------>Os, defined by f(gGs)=gs.
    Os={s'in S st s'=gs for some g in G}
    Gs={g in G st gs=s}

    I am supposed to show f is well defined, 1-1, and onto.

    I know that to be well defined, if a=b then f(a)=f(b). In this case I think that would mean, if g1Gs=g2Gs, then g1s=g2s. So I assume that
    g1Gs=g2Gs is true. Does that mean that g1=g2? If so, can I then substitute a generic g in for g1 and g2?

    I know 1-1 means that if f(a)=f(b), then a=b. For this case, if
    g1s=g2s, then g1Gs=g2Gs. So I assume g1s=g2s, and show g1Gs=g2Gs. Since g1s=g2s, g1=g2? Am I allowed to divide by s? I know I can't multiply by the reciprocal because S isn't a group, it's a set.

    I know that onto means that for every element in the codomain-y
    , there exists and element in the domain-x with f(x)=y. So in this case, for every element in Os, the orbit of s, there must exist an element in the domain, the quotient group of Gs in G (which is the (left) cosets of Gs in G) with f(g'Gs)=g's? I really don't know what to do here?

    Any guidance would be appreciated.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: 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(gGs) only depends on the coset gGs, and not on which "representative" g we pick.

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

    now gGs = g'Gs DOES NOT MEAN g = g'. what it DOES mean is that: g'-1g is in Gs.

    and (by the definition of Gs) 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(gGs) in the orbit of s (Os), and show that if they are equal, they are images of the same coset of Gs.

    in short, we need to show gs = g's implies gGs = g'Gs (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 Gs is.

    but if g'-1g is in Gs, then the two cosets g'Gs and gGs 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 Gs, gGs with f(gGs) = t.

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

    well, then we have f(gGs) = 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, Gs = {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/Ga = 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}
    Last edited by Deveno; November 23rd 2012 at 08:59 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2012
    From
    Austin, Texas
    Posts
    36

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

    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...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

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

    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 Sn.

    and yes, elements of Sn (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 Gx 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:Gs] = |Os|, or, if we like:

    |G| = |Os|*|Gs|.

    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2012
    From
    Austin, Texas
    Posts
    36

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

    This has been very helpful. Thanks so much for taking the time to explain this to me.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. quotient group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 13th 2010, 03:06 AM
  2. Replies: 1
    Last Post: June 29th 2010, 10:39 PM
  3. Replies: 0
    Last Post: October 30th 2009, 12:36 PM
  4. show this series defines a continuous function on (0,1]
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 3rd 2009, 10:58 PM
  5. Replies: 4
    Last Post: September 14th 2008, 12:57 PM

Search Tags


/mathhelpforum @mathhelpforum