We had an exercise on the lecture to show that Q = $Cay(Z_{2} ^n; e_1,e_2,...e_n)$ is a normal Cayley graph. And also to calculate |Aut(Q)|. Also before we had a theorem that states that :
If G ia abelian,finite group , S- inverse closed generating set of G, not containing identity. If S satisfies the condition $\forall s_1,s_2,s_3,s_4 \in S$ such that $s_1 + s_2 = s_3 + s_4 $ implies {$s_1$,$s_2$} = {$s_3$,$s_4$}.
I have a question, firstly what mean {$s_1$,$s_2$} = {$s_3$,$s_4$} ? Just that this two sets are equal? And what if there does not exist elements such that $s_1 + s_2 = s_3 + s_4 $ ? Can we conclude that the theorem holds?
Also could someone please help, and give a hint on how to show that Q = $Cay(Z_{2} ^n; e_1,e_2,...e_n)$ is normal Cayley graph?
$Q$ is the $n$-dimensional cube. It's well known that $\mathrm{Aut}(Q)=\mathbb{Z}_2^n\rtimes S_n$, where $\mathbb{Z}_2^n$ is the copy you started with, and $S_n$ permutes the generators (this is also called the hyperoctahedral group). It follows that it's a normal Cayley graph.
It's quite obvious that at least $\mathrm{Aut}(Q)\leq\mathbb{Z}_2^n\rtimes S_n$. I'm not sure what the easiest way to prove the other inequality is, but the way I usually think about/remember it is that it's the Cartesian product of $n$ copies of $K_2$, and there are results giving the automorphism groups of Cartesian products.