1
$\begingroup$

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!

$\endgroup$
  • 11
    $\begingroup$ -1, and I'm not saying I downvoted... $\endgroup$ – RghtHndSd Sep 4 '13 at 1:35
10
$\begingroup$

No, $-1$ is also a generator of $\mathbb{Z}$.

$\endgroup$
  • 1
    $\begingroup$ Thank you Rebecca. So only $\pm 1$ right? $\endgroup$ – Tumbleweed Sep 4 '13 at 1:37
  • 1
    $\begingroup$ 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). $\endgroup$ – Rebecca J. Stones Sep 4 '13 at 1:38
  • $\begingroup$ Sorry I got confused - how could 1 generate -1? $\endgroup$ – Tumbleweed Sep 4 '13 at 1:39
  • 1
    $\begingroup$ 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 $\langle g \rangle=\{g^k : k \in \mathbb{Z}\}=\{\ldots,g^{-2},g^{-1},g^0,g,g^2,\ldots\}$.) $\endgroup$ – Rebecca J. Stones Sep 4 '13 at 1:42
  • 1
    $\begingroup$ In general, any cyclic group whose order is not one or two, will always have at least two generators, as the inverse of a generator will itself be a generator. $\endgroup$ – David Ward Sep 4 '13 at 7:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.