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Cosets are either equal or disjoint

  1. Jul 17, 2004 #1
    (G is a group, and H is a subgroup of G). I've just read in a book, that all distinct (left or right) cosets of H in G form a partition of G, i.e. that G is equal to the union of all those cosets. Apparently, this follows from the fact that two cosets are either equal or disjoint (I've proved that), but I just can't figure out how the whole partition thing follows. It must be either very hard or very easy to prove, as it is stated without proof in the book... Can anyone shed some light on this?
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  3. Jul 17, 2004 #2
    Its just because cosets are equivalence classes for the congruence defined by:

    s is congruent to r if r = hs for some h in H.

    Here obviosuly H is a subgroup of the group G in mind.
    (so that we can make sH with elements s and h in H)
    Last edited: Jul 17, 2004
  4. Jul 17, 2004 #3
    that it is symmtric is pretty clear:

    if r is congruent to s then we have s = rh
    so that r = sh inv
    but clearly h inv is in H (its a(sub) group) so s is congruent to r

    Reflexivity is easy too:

    r = re (e is identity of H) so r is self congruent

    Transitivity takes a tiny bit more thinking (at least for me) as it usually does:

    if r is cong to s and s is con to p then s = rh1 and p = sh2 for some h1, h2 in H.

    Then we can write p = rh1h2 but h1h2 is in H since H is a closed set under the mult.
    So then r is cong to p.

    Since an equivalence realtion partitions a set (a fact from set theory)then (left) cosets parititon a group.
    Last edited: Jul 17, 2004
  5. Jul 17, 2004 #4


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    Isn't the definition of "partion of a set" simply "a collection of sets such that every member of the original set is in one and only one of them"?

    Let p be any member of the set. Suppose p were in two distinct partions. The two sets are not identical (since they are distinct) nor is they disjoint (they p is in both).

    Of course you would still need to show that p IS in one of the sets (as opposed to none).
  6. Jul 18, 2004 #5
    Thanks, but I can't find a proof of this either (everyone just says "it can be shown"... grr ;)), so this basically reduces the original problem to an identical one ;)

    Yep, which is what I'm having trouble with.
  7. Jul 18, 2004 #6


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    So, the statement you are trying to prove is:

    Let g be an element of G. There is an f such that g is in fH....
  8. Jul 18, 2004 #7
    ...and one might simply choose f = g to prove the existence of such an f... Woah, that was hard. ;) Thanks.
  9. Jul 18, 2004 #8
    I don't understand. Are you taking an abstract alegbra course in which you aren't assumed to know basic set theory?
    Or is this just a case of "you wanted to see it yourself"?
    The reason why there is a partiton is because it makes an equivalence relation.
    An existence is null issue.
  10. Jul 19, 2004 #9
    I'm just reading a book on abstract algebra. It /does/ assume knowledge of set theory, and I do know (some of) it, just not the proof of this particular theorem.

    "Just"? Is it so bad to want to see proofs of non-trivial statements?
  11. Jul 19, 2004 #10
    No its not bad at al...just was a linking word more than a value judgement implying word.

    Butfor the sake of a 'course" of abstract algebra that statement if noit trivial would be considered at least assumed and fundamnetal.

    Anyway the point is a coset is a useful object and therefore worth defining and repeating becaue it partitions the sets as it does.
  12. Jul 19, 2004 #11
    The biggest point the book is trying set up is one of the most important theorems in group theory: Langrange's thm. that the order of a subgroup must divide the order of a group.
  13. Jul 22, 2004 #12


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    Suppose "=" is some equivalence relation on set A. That means:
    1) (Reflexive) For all x in A, x= x.
    2) (Symmetic) For all x, y in A if x= y then y= x.
    3) (Tranistive) For all a, y, z in A if x= y and y= z, then x=z

    A subset, B, of A is an "equivalence class" if all members of the subset are equivalent and there are no members of A, equivalent to something in B, that are not in B.

    First, by the reflexive law, every member of A is in at least one equivalence class.
    Second, suppose a is in two distinct equivalence classes,B and C. The a= b for some b in B and a= c for some c in C. But then, by the transitive property, b= c so b and c must be in the same equivalence class, a contradiction.
  14. Sep 17, 2004 #13


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    think of all the major league baseball teams. If two people name one team each, then they either name the same team or different teams. so the teams they name either have all the same players or no common players at all.; I.e. any two teams named are either equal, or disjoint.

    that just means that all the teams in baseball give us a "partition" of all the players. I.e. the various teams serve to divide up all the players into disjoint sets of players.

    In group theory the cosets are the teams and the elements are the players. so it is exactly because two cosets are either equal or disjoint that they do divide up all the elements into disjoint cosets, i.e that they "partition" the group.

    this is not a deep idea. just two ways of saying the same thing.

    lagranges theorem is based on this and the additional idea that all cosets have the same number of elements. the mconclusion is that the total number of elements is a multiple of the number of cosets.

    i.e. if all baseball teams carried exactly the same number of players, say 50, then the total number of players would equal 50 times the number of teams. i.e. the number of teams would divide the total number of players.

    this is not deep stuff, but it does have beautiful consequences.
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