Originally Posted by

**Swlabr** 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).