0

Let $G$ be a group generated by the finite set $X=\{x_1,\ldots, x_n\}$. Now suppose that $G$ is a finite cyclic group. It is clear that $G$ need not be generated by $x_i$ for any $i$. What additional hypothesis ensures that $G$ can be generated by an $x_i$ for some $i$?

1 Answers1

1

The obvious one: that $\;ord(x_i)=ord(G)=|G|\;$ . Now, if you already have a generator, say

$\;G=\langle x\rangle\;$ , then $\;x^k\;$ is also a generator iff $\;gcd(k,|G|)=1\;$

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • $(1,0), (0,1)$ are generators of $Z_2\times Z_3\cong Z_6$, but neither $(1,0)$ nor $(0,1)$ is a generator of $Z_2\times Z_3$. So my question is when will such a thing not happen in the general case, hence what additional hypothesis will ensure that such a thing does not happen. I understand what you write above. Is there something more. –  Mar 12 '16 at 02:36
  • For example if the group is of prime order, then any nonzero element is a generator, so any of the $x_i$ will generate it. –  Mar 12 '16 at 05:30
  • @user114539 Thank you. What does your example in your first comment has to do with anything? I didn't say one of your $;x_i$'s HAS to be a generator. You asked "what additional hupothesis ensures $;G;$ is generated by an $;x_i;$$?", and I correctly answered what I did: THAT is the additional, sufficient and necessary condition that'll ensure what you want. When none of the several generators of the group is a single generator of it is precisely, as your example shows, when none of them has the right order... – DonAntonio Mar 12 '16 at 09:40