# Enumeration of a finite group

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 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.

• Why are you looking for an example of this? – Tobias Kildetoft Jun 14 '14 at 18:01
• I think for odd case, there is no group satisfying this condition but I am not sure and it is related the a cayley graph of the group. – mesel Jun 14 '14 at 18:05
• In the commutative case $a_n=e=a_1$ for $n$ odd. So $G$ is not commutative. – GH from MO Jun 14 '14 at 18:10
• So if one was to prove this cannot happen (with the group non-trivial), it suffices to show it when no initial sequence (except the one of length one) forms a subgroup. It is probably possible to do some further reductions like that. – Tobias Kildetoft Jun 14 '14 at 20:03
• @KevinO'Bryant The OP writes s/he hasn't found any example when $n$ is odd. When $n$ is even, consider $\mathbb{Z}/4\mathbb{Z}$ where $g_1 = 0, g_2 = 1, g_3 = 2, g_4 = 3$. Then $a_1 = 0, a_2 = 1, a_3 = 3, a_4 = 2$. – Benjamin Dickman Jun 14 '14 at 20:34

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.