Given a finite group $G$, write $K(G)$ for the complete digraph on the elements of $G$. Label the edge from $g$ to $h$ by element $g^{-1}h$.


**Question**: For what groups does there exist a Hamiltonian path in $K(G)$ whose edge labels exhaust the elements of $G$, apart from the identity?


Some observations:


1. If $G={\Bbb Z}/12{\Bbb Z}$, mathematical music theory calls such paths "all-interval rows."
2. No such paths exist for cyclic groups of odd order greater than 1 because the sum of all elements in such a group equals the identity. More generally, when $G$ abelian has such a path, I believe that $G$ must have exactly one factor of even order when expressed as a product of cyclic groups, or equivalently, a unique element of order 2. I don't know the status of the converse.
3. Any Hamiltonian path determines a sequence of $|G|-1$ non-identity edge labels. Heuristically, a random such sequence has probability $(|G|-1)!/ {(|G|-1)}^{|G|-1}$ of having no repeated labels. This predicts that the desired paths exist in great profusion, at least absent any global obstruction as in the previous comment. Where I have made exhaustive searches either no desired paths turned up at all, or I saw a total reasonably consistent with the heuristic. Can one prove or disprove anything along these lines? Even-order cyclic groups have paths of the desired sort, but I don't have any interesting bounds on the total counts even in this case.
4. The case of dihedral times ${\Bbb Z}/2{\Bbb Z}$ has mathematical music theory interest.
Paths exist in profusion with the dihedral group factor having order 6, 8, 10, 12 or 24 (the musically most interest case!), but I've yet to see any desired paths searching orders 14, though this might simply reflect the vast size of the search space. (I have now found examples for 24, but only by using an ad hoc hack whose effectiveness I don't understand.)



Paths in groups
Groups with this property are known as *sequenceable*, although the standard
definition of a sequenceable group looks a little different: namely, a finite
group $G$ is called sequenceable if its elements can be arranged in a
sequence $(g\_1,\ldots,g\_n)$, where $n=|G|$, so that all partial products
$a\_1:=g\_1,\ a\_2:=g\_1g\_2,\ \ldots,\ a\_n:=g\_1\dotsb g\_n$ are pairwise distinct.
The link to your property is established by the simple observation that in
this case $(a\_1,\ldots,a\_n)$ is a Hamiltonian path with $a\_i^{-1}a\_{i+1}$
pairwise distinct, and this is reversible: if $(a\_1,\ldots,a\_n)$ is a
Hamiltonian path with $a\_i^{-1}a\_{i+1}$ pairwise distinct, then $g\_1:=e,\
g\_2:=a\_1^{-1}a\_2,\ \ldots,\ g\_n:=a\_{n-1}^{-1}a\_n$ is an arrangement of the
group elements with all partial products pairwise distinct.


Abelian sequenceable groups have been characterized completely. If
$(g\_1,\ldots,g\_n)$ is an arrangement of the elements of a finite abelian
group $G$ of order $n$ with all partial products pairwise distinct, then none
of $g\_2,\ldots,g\_n$ can be equal to the neutral element of the group $e$.
Hence, $g\_1=e$, and therefore $g\_1\dotsb g\_n\ne e$; that is, the product of
all group elements is distinct from the neutral element of the group. It is
not difficult to see that an abelian group has this property if and only if
it is a group with exactly one involution. Conversely, it was shown by B.
Gordon (*Sequences in groups with distinct partial products*, 1961) that any
finite abelian one-involution group is sequenceable. To sum up,



> 
> In the case where $G$ is abelian, the necessary and sufficient condition
>  for a Hamiltonian path with the property in question to exist is that $G$ has
>  exactly one involution.
> 
> 
> 


The non-abelian case is subtler, but there are some partial results in this
direction, too. For a comprehensive and up-to-date account see a recent
[paper by Ollis](http://www.combinatorics.org/ojs/index.php/eljc/article/viewFile/DS10/pdf). Related stuff and more references can be found
[here](http://math.haifa.ac.il/~seva/Papers/Hamilton.dvi).


