Let
be a group and suppose
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
be a set of (right) coset representatives of
. My question is this,
-Does there always exist some such set
where every
does not contain any copies of
?
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,
, the free group generated by
and
. Let
. Then what would be the representative for, say,
? Or even
? Is
essentially just all the words which do not end in an
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).