Math Help - finding equivalence classes

1. 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.

2. Re: finding equivalence classes

Originally Posted by amyw
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?

3. 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...

4. Re: finding equivalence classes

Originally Posted by amyw
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).

5. Re: finding equivalence classes

I think that cleared that up for me.

6. 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.

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.

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.

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.

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).