If $g$ has order $n$, then $\langle g\rangle=\langle g^2\rangle=\cdots=\langle g^{n-1}\rangle$.

This should be fairly easy but somehow I just couldn't prove it. I only managed to prove the case when $n$ is prime (which might be wrong, I haven't checked carefully). Say $x=g^k$ for $k<n$ and we want to show that it is in $\langle g^m\rangle$, then we want to write $x$ as $g^{mp}$, but then I don't see how to proceed. Any hint?

Update: so I found a counterexample when $n$ is not prime. Sorry for the confusion. Anyway, if anyone could provide a proof when $n$ is prime, still appreciated.

  • 2
    $\begingroup$ This is only true, when $n$ is prime. $\endgroup$ – MooS May 7 '15 at 7:21
  • $\begingroup$ Are you sure $n$ is not only prime? $\endgroup$ – 9301293 May 7 '15 at 7:21
  • $\begingroup$ To be honest I'm not sure, I'm reading the notes of my friends and in the problem, n is 5. $\endgroup$ – 3x89g2 May 7 '15 at 7:22
  • $\begingroup$ For example, consider $Z_n$ when $n$ isn't prime. Then you can find an element in $Z_n$ for counterexample. $\endgroup$ – hamid kamali May 7 '15 at 7:23
  • $\begingroup$ And $5$ is certainly prime. A cyclic group is generated by any non-trivial element if and only if its order is prime. If you have shown the result for $n$ prime, you could show that it does not hold for $n$ composite. This is very easy. $\endgroup$ – MooS May 7 '15 at 7:23

Hint. If $ n $ is prime, given any $ k < n $, we have ${\rm gcd} (k, n)= 1$. Hence, we can write $1= an + bk $ for some $ a, b \in \mathbf Z $. Then $$ g = (g^n)^a \cdot (g^k)^b. $$

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