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Let $\langle g\rangle$ be a cyclic group of order $n$. Suppose $1\leq q \leq n-1$, I want to show that $g^q$ generates $\langle g\rangle$ if and only if $\gcd(n,q)=1$.

Suppose $g^q$ generates $\langle g\rangle$, then

$$ 1,g^q,g^{2q},...,g^{(n-1)q} $$

are all distinct and $g^{nq}=1$, since $n$ is the order of the group. Note that this is the lowest multiple of $q$ for which this is the case. Therefore $\operatorname{lcm}(q,n)=qn$ and it follows that $q$ and $n$ are coprime.

On the other hand, if $\gcd(q,n)=1$, then $\operatorname{lcm}(q,n)=qn$, we have $g^{nq}=1$ and hence all of

$$ 1,g^q,g^{2q},...,g^{(n-1)q} $$

are distinct again and $g^q$ is a generator.

Is my proof correct at all and is there maybe a more elegant argument?

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Your proof seems correct to me.

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