Let $G$ be a group, and let $g_1,g_2\in G$ be nontrivial elements that do not commute.
If $g$ and $h$ are not free as group elements, then the only a priori information this provides us is that there exists a sequence of alternating indices $i(1)\neq i(2)\neq\cdots\neq i(s)$ (with $i(1),\ldots,i(s)\in\{1,2\}$) and exponents $m(1),\ldots,m(s)\in\mathbb Z\setminus\{0\}$ such that
- $g_{i(l)}^{m(l)}\neq e$ for all $l\leq s$; and
- $g_{i(1)}^{m(1)}\cdots g_{i(s)}^{m(s)}=e,$
that is, a reduced word in $g_1$ and $g_2$ that is equal to the neutral element $e$.
My questions are the following:
Assuming that $i(1)=i(s)$, is it possible to generate another reduced word $g_{j(1)}^{n(1)}\cdots g_{j(t)}^{n(t)}$ such that $g_{j(1)}^{n(1)}\cdots g_{j(t)}^{n(t)}=e$ and $j(1)\neq j(t)$?
Assuming that $i(1)\neq i(s)$, is it possible to generate another reduced word $g_{j(1)}^{n(1)}\cdots g_{j(t)}^{n(t)}$ such that $g_{j(1)}^{n(1)}\cdots g_{j(t)}^{n(t)}=e$ and $j(1)=j(t)$ for both $j(1)=1$ and $j(t)=2$?
So far, I have been completely unable to answer the first question, and I can only partially answer the second question as follows:
- If the orders of both $g_{i(1)}$ and $g_{i(s)}$ are strictly greater than two (or infinite), then it is possible to find $p,q\in\mathbb Z$ such that $g_{i(1)}^p,g_{i(1)}^{m(1)+p},g_{i(s)}^q,g_{i(s)}^{m(s)+q}\neq e$, yet $$g_{i(1)}^{m(1)+p}\cdots g_{i(s)}^{m(s)}g_{i(1)}^{-p}=e$$ and $$g_{i(s)}^{-q}g_{i(1)}^{m(1)}\cdots g_{i(s)}^{m(s)+q}=e$$
- If at least one of the orders of $g_{i(1)}$ and $g_{i(s)}$ is equal to $2$, I don't know.