Can all of them be different?

Question

Edit: Cross-posted to MathOverflow here (and resolved).

---

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 $a_i\neq a_{i+1}$.

Answer 1

This question was cross-posted on MathOverflow where it was answered by Joni Teräväinen. I have reproduced the answer below.

A group $G$ with the property is called sequenceable. For a survey, see this paper 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).