@Tracy Hall .. Please see my comment to Mariano. I am sorry for the sloppy comment regarding connectivity of $i$ and $j$.

@Mariano Regarding your first question: sorry for the confusion, I meant the following: Let's say $lc_i$ and $lc_j$ were to act on a specific graph, $G_a$. Then if there exists an edge, $ E_{ij} $ between the two vertices $ i$ and $j$ of $G_a$ then $m_{ij}=3$. Otherwise it is $2$. Now consider the orbit that is generated by all the $\{lc_i|i=1,\cdots,n\}$ (i.e. All the simple graphs related by local complementation), as you say, then $lc_i$ induces an involution on these graphs. Now for $lc_i lc_j$ some of the graphs will have $i$ and $j$ connected and some don't. Hence the $m_{ij}=6$.

@ Zaimi Actually that is a question we've wondering about too. These are however what we know: $lc$'s are reflections. The above relations are the only ones mentioned in literature and ones we work with. However the right answer is that we don't know! What we do know is the transformation on the adjacency matrix, $A_G$ of a graph, $G$: $lc_k: (A_G)_{ij} \rightarrow (A_G)_{ij}+(A_G)_{ik}(A_G)_{jk}+Diagonal[(A_G)_{ij}+(A_G)_{ik}(A_G)_{jk}]$. Where the addition is $mod(2)$.

@Holt that property is satisfied as I commented above.

@Humphreys 1) Local complementation is as Hahn mentions 2) We have a set of generators of a group, which seem to obey the conditions for a coxeter set of generators. So we want to know what group we're dealing with. 3)I am not sure! For now we have only done some numerical work. And we are getting a pattern; only the product of the alternating group and or symmetric group is observed (see my question on "local complementation group of simple graphs" asked before). So we are wondering if there is some graph/group theory property which would clarify what we have.

@Zaimi they can easily be worked out. The $m_{ij}=2$ is for when $i=j$ or $i$ and $j$ are disconnected, and $m_{ij}=3$ when $i$ and $j$ are disconnected. So for a given $(lc_i lc_j)^{m_{ij}}=1$,$m_{ij}=2,3 or 6$.

Thank you. I am going through these to see if I can find the answer to what I am looking for.