Let $G=\{g\_1,g\_2,...,g\_n\}$ be a group with $e=g\_1$ and $n$ is odd,


Set $$a\_1=g\_1$$
$$a\_2=g\_1g\_2$$
$$a\_3=g\_1g\_2g\_3$$
$$a\_n=g\_1g\_2...g\_n$$


I am looking for example that all $a\_i$ are different from each other i.e. $G=\{a\_1,a\_2,...,a\_n\}$. By the way it is clear that $a\_i\neq a\_{i+1}$.


**Note** I had asked this question [there](https://math.stackexchange.com/questions/833474/can-all-of-them-be-different) but I think it is suitable for here. I require that $n$ is odd since there are many example when $n$ is even yet I found no example in the case $n$ is odd.



Enumeration of a finite group
A group $G$ with the property is called sequenceable. For a survey, see [this paper](http://www.combinatorics.org/ojs/index.php/eljc/article/viewFile/DS10/pdf) by M. A. Ollis, which also tells that sequenceable groups are related to constructing row-complete latin squares. It is conjectured by Keedwell that $D\_6,D\_8$ and $Q\_8$ are the only non-abelian non-sequenceable groups (see page 17); in particular, there should be none with odd order. It is known that an abelian group is sequenceable if and only if it has a unique element of order 2 (see page 5 for a proof). The article gives a list of groups that are known to be sequenceable. Apparently the question is not completely solved even in the case where $|G|$ has two prime factors. 


However, some groups of odd order are known to be sequenceable, so an example to your question would be for instance the non-abelian group of order 21 (page 5; there are other examples as well).


