Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
Why are you looking for an example of this? –  Tobias Kildetoft Jun 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 at 18:05
6  
In the commutative case $a_n=e=a_1$ for $n$ odd. So $G$ is not commutative. –  GH from MO Jun 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 at 20:03
2  
@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 at 20:34

1 Answer 1

up vote 44 down vote accepted

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).

share|improve this answer
    
Thank you joni, it will help a lot. –  mesel Jun 15 at 8:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.