# Is 1 the only generator of $\mathbb{Z}$?

I am trying to prove that $\mathbb{Z}/2 \times \mathbb{Z}$ is not cyclic. But I am not quite sure - is $\mathbb{Z}$ can only be generated by $1$?

Thank you very much!

• -1, and I'm not saying I downvoted... – RghtHndSd Sep 4 '13 at 1:35

No, $-1$ is also a generator of $\mathbb{Z}$.
• Thank you Rebecca. So only $\pm 1$ right? – Tumbleweed Sep 4 '13 at 1:37
• Yes, that's right. $\langle n \rangle$ generates $n\mathbb{Z}$, which will be $\{0\}$ if $n=0$ or the integers divisible by $n$ otherwise (in the case when $|n| \geq 2$, we thus have $\langle n \rangle$ is a proper subgroup). – Rebecca J. Stones Sep 4 '13 at 1:38
• In the additive group $\mathbb{Z}$, the group generated by $n$ is defined as $\langle n \rangle=\{kn:k \in \mathbb{Z}\}$ where $kn$ is shorthand for $n+n+\cdots+n$ ($k$ times), if $k>0$, defined as $0$ if $k=0$, and $-n-n-\cdots-n$ ($|k|$ times), if $k<0$. (Using multiplicative notation $\langle g \rangle=\{g^k : k \in \mathbb{Z}\}=\{\ldots,g^{-2},g^{-1},g^0,g,g^2,\ldots\}$.) – Rebecca J. Stones Sep 4 '13 at 1:42