Let $Q_n = \{0,1\}^n$ be the hypercube graph. (Points are adjacent iff they differ by 1 in exactly 1 coordinate.) I am interested in the group of automorphisms of $Q_n$. I see the description of this group like $Q_n = Z_2^n \rtimes S_n$. I would like to think of it slightly differently.
Thinking of $Q_n$ as a subset of $\mathbb R^n$, some of these automorphisms are orientation preserving, and some reversing. The "flip" map $F(t_1,t_2,\dots,t_n) = (1-t_1,t_2,\dots,t_n)$ is reversing. Any "basic rotation" $R_{i,j}(t_1,\dots,t_i,\dots,t_j,\dots,t_n) = (t_1,\dots,1-t_j,\dots,t_i,\dots,t_n)$ is orientation preserving.
Am I correct that the group $Aut(Q_n)$ is generated by $F$ together with the set of $R_{i,j}$ for various $i,j$? In particular I would like to say that any orientation preserving automorphism is a composition of basic rotations, and any orientation reversing automorphism is $R \circ F$, where $R$ is an orientation preserving automorphism (and thus a composition of rotations).
I assume this has all been worked out but I can't find any reference. Everything I see is in terms of symmetric groups and $Z_2^n \rtimes S_n$ (Also my graph theory knowledge is very limited, so I don't really know where to look) Thanks-
Best Answer
Yes, what you say is correct. The subgroup of orientation preserving automorphisms is just the intersection of ${\rm Aut}(Q_n)$ with the alternating group on $Q_n$, and it has index $2$ in the ${\rm Aut}(Q_n)$.
It is not hard to see that this subgroup is generated by the $n-1$ basic rotations of the form $R_{i,i+1}$ for $1 \le i < n$. Since $S_n$ is generated by the transpositions$(i,i+1)$, the subgroup generated by the $R_{i,i+1}$ projects onto $S_n$.
Note also that $R_{i,i+1}^2$ is the double flip $i \mapsto -i$, $i+1 \mapsto -(i+1)$, and these generate a subgroup of index $2$ in the base group $Z_2^n$ of the wreath product.