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.
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.
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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).