$i=1$ then $O_1$ is exactly a base of $Z_2^n$, and so is a generating set. If $i=3$, then $O_3$ contains $x_1=(1110\cdots 0)^T$, $x_2=(1101\cdots 0)^T$ and $x_3=(1011\cdots 0)^T$. We have $e_1=(1000\cdots 0)^T=x_1+x_2+x_3$. Since $O_3$ is an orbit under $W$, we have $O_1\subset\langle O_3\rangle$. Thus $O_3$ generates $Z_2^n$. Similarly we can prove that $O_i$ generates $Z_2^n$ for other odd number.

It is well known that a Cayley graph $Cay(G,S)$ is connected if and only if $S$ is a generating set of $G$, that is, $\langle S\rangle=G$. The following result is immediate from Lemma \[lem-z-1\].

\[cor-z-1\] The orbit Cayley graph $Cay(Z_2^n,S)$ ($S\ne O_0$ or $O_n$) is connected if and only if $S$ contains $O_i$ for some odd $i$.

In what follows, we focus on the spectra of connected orbit Cayley graphs. Note that the eigenvalues of Cayley graphs over an abelian group have been given by Babai.

\[lem-z-2\] Let $G$ be an abelian group of order $n$ and $S$ a subset of $G$ such that $1\not\in S$ and $S^{-1}=S$. If $\chi_1,\ldots,\chi_n$ are all irreducible characters of $G$, then the eigenvalues of the Cayley graph $Cay(G,S)$ are $\lambda_i=\sum_{s\in S}\chi_i(s)$ for $1\le i\le n$.

To get the eigenvalues of orbit Cayley graphs over $Z_2^n$, we 