# Cayley graph is normal

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.