TITLE: Long identity-free sequences of permutations
QUESTION [5 upvotes]: Let $N = [n]$ and for any subset $A \subseteq N$, let $S_A$ denote the subgroup of the symmetric group $S_n$ that fixes all objects outside $A$.  Say that a sequence $A_1, \dots, A_k \subseteq N$ is "identity free" if the equivalence
$$s_1 s_2 \dots s_k = \mathop{id} \qquad \text{where } s_i \in S_{A_i} \text{ for all } i$$
has no solutions, except for the trivial one where $s_1 = s_2 = \dots = s_k = \mathop{id}$.
Have these been studied under any name that I can search for?  While I would be interested in any discussion at all, I am particularly interested in the question of the extremal length $L$ of the longest identity-free sequence when all $A_i$ are constrained to have some fixed size $\alpha$.  The only upper bounds on $L$ I have been able to observe so far come from the easy observation that no two $A_i$ may overlap on more than one element (and then apply Cauchy-Schwarz).

REPLY [5 votes]: Construct a bipartite graph $G$ where one part is $[n]$ and the other is $[k]$ such that there is an edge between $i\in[n]$ and $j\in[k]$ iff $i\in A_j$. Then $A_1, \dots, A_k$ are identity free if $G$ is acyclic. The converse may not hold though (as pointed out by Jan Kyncl).
If $|A_j|=\alpha>1$ for all $j\in[k]$, then $G$ has $\alpha k$ edges. In this case, $G$ may be acyclic only if $\alpha k \leq n+k-1$, i.e. $$k \leq \frac{n-1}{\alpha-1}.$$
This inequality does not have to hold if $G$ is not acyclic.