1. ## 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).

2. 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).

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

3. Originally Posted by tonio
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 $\displaystyle ab$.