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Math Help - Coset Representatives

  1. #1
    MHF Contributor Swlabr's Avatar
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    Coset Representatives

    Let G be a group and suppose <a> = H \leq G is a cyclic subgroup which is not normal. I am unsure how this question generalises, but I am specifically interested in the cyclic case so it is posed thataway.

    Now, let S be a set of (right) coset representatives of G/H. My question is this,

    -Does there always exist some such set S where every s \in S does not contain any copies of a?

    My first thought was...well, obvously there must! That is the point of the cosets! However, when I thought about it my opinion changed. For example,

    G = F(a, b), the free group generated by a and b. Let H = <a>. Then what would be the representative for, say, a^bH? Or even abH? Is S essentially just all the words which do not end in an a^{\epsilon} up to free equivalence?

    And so I was wondering if anyone could clear this up for me.

    Also, another reason the answer should be `No' is Burnside's Problem (as then one can just `pull' every generator out of a product and induct to get a positive solution, which is a contradiction).
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  2. #2
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    Quote Originally Posted by Swlabr View Post
    Let G be a group and suppose <a> = H \leq G is a cyclic subgroup which is not normal. I am unsure how this question generalises, but I am specifically interested in the cyclic case so it is posed thataway.

    Now, let S be a set of (right) coset representatives of G/H. My question is this,

    -Does there always exist some such set S where every s \in S does not contain any copies of a?

    My first thought was...well, obvously there must! That is the point of the cosets! However, when I thought about it my opinion changed. For example,

    G = F(a, b), the free group generated by a and b. Let H = <a>. Then what would be the representative for, say, a^bH? Or even abH? Is S essentially just all the words which do not end in an a^{\epsilon} up to free equivalence?

    And so I was wondering if anyone could clear this up for me.

    Also, another reason the answer should be `No' is Burnside's Problem (as then one can just `pull' every generator out of a product and induct to get a positive solution, which is a contradiction).

    What do you exactly mean by "any copies of a"?? If <a>=H\leq G and if [G:H]>1\Longleftrightarrow H\neq G , then any

    coset different from H itself does not contain a or any of its powers, of course, since different

    cosets are disjoint...

    In you example, a representative for a^bH is, of course, a^b ...

    Tonio
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tonio View Post
    What do you exactly mean by "any copies of a"?? If <a>=H\leq G and if [G:H]>1\Longleftrightarrow H\neq G , then any

    coset different from H itself does not contain a or any of its powers, of course, since different

    cosets are disjoint...

    In you example, a representative for a^bH is, of course, a^b ...

    Tonio
    I mean copies of a in the word s. I should probably point out that I also wanted a to be a generator of the group. So, for example, G=F(a, b, c) then a is not a subword of bcb^4cb, or even of bcbabb^{-1}a^{-1}c because we are working up to free reduction, but a is a subword of abc^5, of a^b and of ab.
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