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Let $G$ a cyclic group of order $N$, and $g$ an element of $G$ of order $N$. Show that $g^n$ has order $\frac{N}{\gcd(n,N)}$.

Attempt

I see that $g^{\frac{nN}{\gcd(n,N)}}=1$ so the order of $g^{n}$ divide $\frac{N}{\gcd(N,n)}$. Suppose that the order of $g^n$ is $k<\frac{N}{\gcd(n,N)}$.

I indeed have $k\mid \frac{N}{\gcd(n,N)}$, but I can't have a contradiction. I also tried to prove that $\frac{N}{\gcd(n,N)}\mid k$ but with out success.

José Carlos Santos
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user352653
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1 Answers1

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Suppose that $(g^n)^k=1$. This means that $g^{nk}=1$ and therefore $N\mid nk$. But this implies that $\frac N{\gcd(n,N)}\mid nk$. Note that $\frac N{\gcd(n,N)}$ and $n$ are relatively prime. Therefore, $\frac N{\gcd(n,N)}\mid k$. Since you proved $(g^n)^{\frac N{\gcd(n,N)}}=1$, this proves that the order of $g^n$ is indeed $\frac N{\gcd(n,N)}$.

José Carlos Santos
  • 427,504