a) Let $G = \langle a \rangle$ be a finite cyclic group. Prove that for each $b\in G$, $\langle b \rangle=G$ if and only if order of $b$ equals order of $G$.

b) The previous part does not hold if $G$ is an infinite cyclic group. Why not?


(a) How many elements can $b$ generate? How many elements are in $G$?

(b) What is the prototypical infinite cyclic group?

In future, please give some indication of what you've tried out, and what you are stuck on.

  • $\begingroup$ Sorry, this is all i was given: The order of G equals n which is less than infinity and the order of b has k elements i.e. {e, b, b^2, ...b^k-1} $\endgroup$ – user77158 May 10 '13 at 10:30
  • $\begingroup$ My answer was a hint, not a question. You've pretty much just written down the answer for (a). $\endgroup$ – not all wrong May 10 '13 at 10:32
  • $\begingroup$ I am confused as to show how G also has k elements, when i know G has n elements $\endgroup$ – user77158 May 10 '13 at 10:36
  • $\begingroup$ If $<b>$ generates at least $n$ distinct elements then obviously it generates the whole group. If not, then what can you say about $b^{n-1}$ and lower powers of $b$? What does this imply about the order of $b$. $\endgroup$ – not all wrong May 10 '13 at 11:06

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