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Thread: Coset Representatives

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

    Let $\displaystyle G$ be a group and suppose $\displaystyle <a> = H \leq G$ 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 $\displaystyle S$ be a set of (right) coset representatives of $\displaystyle G/H$. My question is this,

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

    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,

    $\displaystyle G = F(a, b)$, the free group generated by $\displaystyle a$ and $\displaystyle b$. Let $\displaystyle H = <a>$. Then what would be the representative for, say, $\displaystyle a^bH$? Or even $\displaystyle abH$? Is $\displaystyle S$ essentially just all the words which do not end in an $\displaystyle a^{\epsilon}$ 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 $\displaystyle G$ be a group and suppose $\displaystyle <a> = H \leq G$ 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 $\displaystyle S$ be a set of (right) coset representatives of $\displaystyle G/H$. My question is this,

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

    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,

    $\displaystyle G = F(a, b)$, the free group generated by $\displaystyle a$ and $\displaystyle b$. Let $\displaystyle H = <a>$. Then what would be the representative for, say, $\displaystyle a^bH$? Or even $\displaystyle abH$? Is $\displaystyle S$ essentially just all the words which do not end in an $\displaystyle a^{\epsilon}$ 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 $\displaystyle <a>=H\leq G$ and if $\displaystyle [G:H]>1\Longleftrightarrow H\neq G$ , then any

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

    cosets are disjoint...

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

    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 $\displaystyle <a>=H\leq G$ and if $\displaystyle [G:H]>1\Longleftrightarrow H\neq G$ , then any

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

    cosets are disjoint...

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

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