Let $S_n$ be the symmetric group, and $e\in S_n$ the identity permutation. Let $\pi_i\in S_n$ be transpositions in the group. Define a transposition series as a composition of transpositions: $$\overset{n!}{\underset{i=1}{\bigcirc}}\pi_i=\pi_{n!}\circ...\circ \pi_1=e$$ And define its partial transposition series $\sigma:\mathbb{N}_{<n!}/\{0\}\rightarrow S_n$ as: $$\sigma(k)=\overset{k}{\underset{i=1}{\bigcirc}}\pi_i=\pi_{k}\circ...\circ\pi_1$$ Where $k\in \mathbb{N}_{<n!}/\{0\}$.
My question is: Is it possible to prove that there exists a transposition series for each nontrivial $S_n$ such that its partial transposition series $\sigma$ is injective?
Considerations
The first three symmetric groups, $S_n$, for $n=1,2,3$ are trivial:
For $n=1$ the group is trivial.
For $n=2$, we define the complete series as: $$\overset{2}{\underset{i=1}{\bigcirc}}\pi_i=\tau\circ \tau=e$$ Where $\tau=\begin{pmatrix}1 &2\\2&1\end{pmatrix}$. Its partial series only has one element, so it must be injective.
For $n=3$ we need to define the following transpositions: $$\rho_1=\begin{pmatrix}2&1&3\\1&2&3\end{pmatrix}, \rho_2=\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix},\rho_3=\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix}$$ It is then straightforward to check that the complete transposition series: $$\overset{6}{\underset{i=1}{\bigcirc}}\pi_i=\overset{6}{\underset{i=1}{\bigcirc}}\rho_i=e$$ Has an injective partial series if we let $\rho_4=\rho_1$. (Alternatively, the index $i$ can be let to run twice mod 3 in $\frac{\mathbb{Z}}{3\mathbb{Z}}$ for $\rho_{i+1}$).
I'm still working on the 4th case, and i suspect induction can be used to construct a proof for the general case, $S_n$. My conjecture is that an injective partial series $\sigma$ exists for every $S_n$, but the complete one might not; i.e., $\nexists \pi \in S_n$: $$\pi\circ\overset{n!-1}{\underset{i=1}{\bigcirc}}\pi_i=e$$ Got any ideas?