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Math Help - finding equivalence classes

  1. #1
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    Angry finding equivalence classes

    If R is an equivalence relation on S, and a is an element of S, define the equivalence relation of a by [a]={x in S: aRx} for the following relation.

    Let H be a subgroup of G, define R on G by saying aRb iff a-b is an element of H.

    Find the equivalence class of a.

    I have already proved that the above relation is an equivalence relation. I just don't know how to find the equivalence class of a. It can't be that hard, but I'm not sure where to start.

    Thanks.
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  2. #2
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    Re: finding equivalence classes

    Quote Originally Posted by amyw View Post
    If R is an equivalence relation on S, and a is an element of S, define the equivalence relation of a by [a]={x in S: aRx} for the following relation.
    Let H be a subgroup of G, define R on G by saying aRb iff a-b is an element of H.
    Find the equivalence class of a.
    I don't know what notation your text/professor uses.
    Consider a+H=\{a+h: h\in H\}.
    Does that do it?
    Last edited by Plato; October 15th 2012 at 12:26 PM.
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  3. #3
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    Re: finding equivalence classes

    I really don't know. We don't have a textbook and we spent about 6 minutes on equivalence classes since the start of the semester...
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    Re: finding equivalence classes

    Quote Originally Posted by amyw View Post
    I really don't know. We don't have a textbook and we spent about 6 minutes on equivalence classes since the start of the semester...
    You must have done cosets? I am assuming that + is the group operation and you are using -x as the group inverse of x. So when you wrote aRb\text{ iff }a-b\in H you really meant a+(-b)\in H.

    If x\in a+H does a-x\in H~? (note that means that the operation is not necessarily commutative).
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  5. #5
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    Re: finding equivalence classes

    I think that cleared that up for me.
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  6. #6
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    Re: finding equivalence classes

    first of all, some technical quibbles:

    you really can't just say: "a group G", you need an OPERATION to go along with the SET.

    you haven't defined what "a-b" MEANS.

    it is more common to use the notation:

    a*(b-1) unless it is KNOWN that * is commutative, in which case the notation a - b means a + (-b).

    let's prove the relation aRb iff a*(b-1) is in H (a subgroup of G) is actually an equivalence relation.

    (if one is using additive notation this becomes: a - b is in H).

    1)for all a in G: aRa (R is reflexive):

    multiplicative notation:

    aRa means a*(a-1) is in H. but a*(a-1) = e, the identity of G, which always lies in any subgroup H. thus R is reflexive.

    additive notation:

    aRa means a - a is in H. but a - a = 0, which is in H.

    2) for all a,b in G: aRb implies bRa (R is symmetric):

    multiplicative notation:

    aRb mean a*(b-1) is in H. since H is a subgroup, H cointains all inverses, so:

    (a*(b-1))-1 = (b-1)-1*(a-1) = b*(a-1) is in H, so bRa. thus R is symmetric.

    additive notation:

    aRb means a - b is in H, thus -(a - b) = b - a is in H, so bRa.

    3) for all a,b,c in G, aRb and bRc together imply aRc (transitivity of R).

    multiplicative notation:

    from aRb, and bRc, we have:

    a*(b-1) in H, b*(c-1) in H. since subgroups are closed under *:

    (a*(b-1))*(b*(c-1)) = [a*(b*(b-1))]*(c-1) = (a*e*)(c-1) = a*(c-1) is in H, so aRc.

    additive notation:

    aRb and bRc means:

    a - b is in H, b - c is in H, so (a - b) + (b - c) = (a + ((-b) + b)) - c = (a + 0) - c = a - c is in H.

    so R is, in fact, a bona-fide equivalence relation.

    therefore, one can ask, what does [a], the equivalence class of a, look like?

    if b is in [a], so that aRb, then a*(b-1) = h, where h is some element of H. doing some algebraic manipulation, we have:

    b*(a-1) = h-1 = h' (because inverse signs are annoying to write)

    b = h'*a, that is b is an element of the set Ha = {h*a: h is in H}.

    this shows that [a] is contained in Ha.

    suppose that x is in Ha.

    then x = h*a, for some h in H, so x*(a-1) = (h*a)*(a-1) = h*(a*(a-1)) = h*e = h.

    thus xRa, so aRx (since R is symmetric), so x is in [a]. thus Ha is contained in [a], so we have [a] = Ha.

    in additive notation:

    if b is in [a], then a - b is in H, so b - a is in H, so b is in the set a + H = {a + h: h in H}

    (technically, this should be H + a, but when + is used, it usually means the operation + is commutative, so a + h = h + a, and thus a + H = H + a. it is IMPORTANT to know that in general, the cosets Ha and aH NEED NOT BE EQUAL if * is NOT commutative).

    as above this shows that [a] is contained in a + H, and a similar argument as above shows that a + H contains [a]. hence [a] = a + H.

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

    this is a "standard example":

    let G = (Z,+), and let H = (3Z,+), the subgroup of all multiples of 3.

    we define k~m if k - m is a multiple of 3.

    the equivalence classes are:

    [0] = 0 + 3Z = 3Z = {.....-9,-6,-3,0,3,6,9,.....}
    [1] = 1 + 3Z = {.....-8,-5,-2,1,4,7,10,......}
    [2] = 2 + 3Z = {.....-7,-4,-1,2,5,8,11,......}

    note that the "a" in [a] isn't UNIQUE, for example: [1] = [7], that is: 1 + 3Z = 7 + 3Z, even though 1 and 7 are clearly different integers.

    we say that a is a REPRESENTATIVE for [a], and when talking about [a], we want to be sure that what we say is true for ANY b in [a], and not just a itself.

    for example:

    any element of 1 + 3Z has a remainder of 1 upon division by 3 is a TRUE statment.

    any element of 1 + 3Z is odd is a FALSE statement (even though 1 itself happens to be odd).
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