A non-trivial finite group $G$ of order $n$ is said to be sequenceable if its elements can be arranged in a sequence ($g_{1}, g_{2}, \dots ,g_{n}$) in such a way that the partial products ($a_{1}, a_{2}, \dots , a_{n}$), where $a_{i}=g_{1}g_{2} \dots g_{i}$ are distinct. The sequence ($g_{1}, g_{2}, \dots g_{n}$) is called a sequencing for G.
I was trying to prove that the problem of testing whether a group is sequenceable(GS) is self-reducible, or may be random self-reducible. Or can it be proven that this problem is not self-reducible.
One approach that I thought was to consider a variant of this problem PREFIX SEQUENCEABILITY (PS), where we have a given ordering of some of the prefix elements of $G$, say $P=(g_{1}, g_{2}, \dots g_{k}$), and we want to check whether there is a sequence of G which extends P. It can be easily proved that PS is self-reducible.
One way to show GS is self-reducible would be to show a reduction from PS($G,P$) to GS($G′$) . If we could show such a reduction, then, since, PS is self-reducible, so GS is also self-reducible. We want to construct the group $G′$ in such a way that G′ is sequenceable if and only if $G$ contains a sequence that extends P.
Can someone provide some ideas on how to do this construction, or give ideas on proving GS is self-reducible in some other way ?